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Unified d Desig gn of Steel S S Structures, 3rd Ed d.

Lo ouis F. Geeschwind dner Forrmer Vice Prresident of Engineering E g and Researrch, Americaan Institute of Steel Connstruction; Prof ofessor Emerritus of Arch hitectural En ngineering, T The Pennsylvvania State U University; aand Senior Con nsultant, Pro ovidence Eng gineering Co orporation

Judy Liu Prof ofessor, Scho ool of Civil and a Construcction Engineeering, Oreggon State University

Ch harles J. Carter C President, Amerrican Institu ute of Steel Construction C

State Co ollege, Pennsyllvania

Dedication This book is dedicated to our mentors, teachers and former students, each of whom taught us something and, in the process, contributed to who we are today.

Copyright © first and second editions 2008, 2012, John Wiley & Sons, Inc., third edition 2017, Louis F. Geschwindner. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without prior written permission of the copyright holder. Anyone making use of the information presented in this publication assumes all liability arising from such use. Cover photo: Rutgers University School of Business, Piscataway, NJ Photo courtesy of WSP Parsons Brinckerhoff

Printed by CreateSpace, an Amazon.com Company Available from Amazon.com, CreateSpace.com, and other retail outlets For additional information see www.SteelStuff.com ISBN-13: 978-1543207521 ISBN-10: 1543207529 Library of Congress Control Number: 2017906318 CreateSpace Independent Publishing Platform, North Charleston, SC

Preface INTENDED AUDIENCE This book presents the basics of design of steel building structures, and is based on the 2016 unified specification, ANSI/AISC 360-16 Specification for Structural Steel Buildings. It is intended primarily as a text for a first course in steel design for civil and architectural engineers. Such a course usually occurs in the third or fourth year of an engineering program. The book can also be used in a second, building-oriented course in steel design, depending on the coverage in the first course. In addition to its use as a textbook, it provides a good review for practicing engineers looking to learn the provisions of the latest specification or to convert their practice from any of the old specifications to the new specification. Users are expected to have a firm knowledge of statics and strength of materials and have easy access to the AISC Steel Construction Manual, 15th Edition.

UNIFIED ASD AND LRFD A preferred approach to the design of steel structures had been elusive over a 20 year period from 1986 to 2005. In 1986, the American Institute of Steel Construction (AISC) issued its first Load and Resistance Factor Design (LRFD) Specification for Structural Steel Buildings. This specification came after almost 50 years of publication of an Allowable Stress Design (ASD) specification. Unfortunately, LRFD was accepted by the academic community but not by the professional engineering community. Although AISC revised the format of the ASD specification in 1989, it had not updated its provisions for over 25 years. This use of two specifications was seen as an undesirable situation by the professions related to the building industry, and in 2001 AISC began the development of a combined ASD and LRFD specification. In 2005, AISC published its first unified specification, combining the provisions of both the LRFD and ASD specifications into a single standard for the design of steel building structures. That specification, ANSI/AISC 360-05 Specification for Structural Steel Buildings, reflected a major change in philosophy by AISC, one that made the use of ASD and LRFD equally acceptable as approaches for the design of steel buildings. That has been the case for the past 11 years. Now, the newest specification, ANSI/AISC 360-16 Specification for Structural Steel Buildings continues that philosophy of equal status for design by ASD and LRFD. The reader familiar with past editions of the separate ASD and LRFD specifications but not with either the 2005 or 2010 editions will undoubtedly question how these two diverse design philosophies could be effectively combined into one specification. This is a reasonable question to ask. The primary answer is that the 2005 specification was not a combination of the old ASD and LRFD provisions. It was a new approach with a new ASD and a new LRFD that used the best of both previous approaches. The first unified specification took a different approach. It was based on the understanding that the strength of an element or structure, called the nominal strength in the specification, can be determined independent of the design philosophy. Once that nominal strength is determined, the available strength for ASD or LRFD is determined as a function of that nominal strength. Thus, the available strength of the element is always based on the same behavior and no inconsistency in behavior results from the use of ASD or LRFD. This important aspect of the unified specification is further explained in Chapter 1. The 2016 specification shows AISC’s commitment to maintaining both design approaches. With only one strength equation for both methods, any updates are applied equally to the ASD and LRFD provisions. The designer continues to be given the opportunity to apply engineering judgment in deciding what approach to use for design, either by ASD or LRFD. iii

iv Preface AISC has also issued ANSI/AISC 341-16 Seismic Provisions for Structural Steel Buildings, the standard that guides the design of steel building structures to resist seismic loads.

CHANGES IN BUILDING LOADS In addition to the provisions for steel design issued by AISC, structural engineering has seen many changes in the area of loads for which buildings must be designed. The American Society of Civil Engineers (ASCE) is continually revising ASCE 7 Minimum Design Loads for Buildings and Other Structures, its standard for building loads. The International Code Council (ICC) continues to issue its International Building Code (IBC) on a three year cycle, and it in turn is adopted by a local jurisdiction on the schedule chosen by that jurisdiction. The IBC adopts ASCE 7, ANSI/AISC 360, ANSI/AISC 341, and other structural standards. The major changes for the structural engineer brought about by these continually changing standards are the inclusion of requirements for consideration of seismic loading, which applies to the entire country, and revised wind and snow load requirements. For the calculation of loads within this text, ASCE 716 provisions are used. For any actual design, the designer must use the loadings established by the governing building code. ANSI/AISC 341 is discussed in Chapter 13.

UNITS ANSI/AISC 360-16 is, as much as possible, a unitless specification. In those rare instances where equations could not be written in a unitless form, two equations are given, one in U.S. customary units and one in SI units. The Manual presents all of its material in U.S. customary units. The construction industry in this country has not adopted SI units in any visible way, and it is not clear that they will in the foreseeable future. Thus, this book uses only U.S. customary units.

TOPICAL ORGANIZATION Chapters 1 through 3 present the general material applicable to all steel structures. This is followed in Chapters 4 through 9 with a presentation of member design. Chapters 10 through 12 discuss connections and Chapter 13 provides an introduction to seismic design. In Chapter 1, the text addresses the principles of limit states design upon which all steel design is based. It shows how these principles are incorporated into both LRFD and ASD approaches and shows how reliability varies with live-to-dead load ratio for both approaches. Chapter 1 also provides a description of ANSI/AISC 360-16, the Steel Construction Manual and the AISC web site. These are the major source documents for all that is presented in this book. Chapter 2 introduces the development of load factors, resistance factors, and safety factors. It discusses load combinations and compares the calculation of required strength for both LRFD and ASD. The calibration of ASD and LRFD and the resulting system reliability is also addressed. Chapter 3 discusses steel as a structural material. It describes the availability of steel in a variety of shapes and the grades of steel available for construction. Once the foundation for steel design is established, the various member types are considered. Tension members are addressed in Chapter 4, compression members in Chapter 5, and bending members in Chapter 6. Chapter 7 covers plate girders, which are simply bending members made from individual plates. Chapter 8 treats members subjected to combined axial load and bending as well as design of bracing. Chapter 9 deals with composite members, that is, members composed of both steel and concrete working together to provide the available strength. Each of these chapters begins with a discussion of that particular member type and how it is used in buildings. This is followed by a discussion of the specification provisions and the behavior from which those provisions have been derived. The LRFD and ASD design philosophies of the 2016 specification are used throughout. Design examples that use the specification provisions

Preface v directly are provided along with examples using the variety of design aids available in the AISC Steel Construction Manual. All examples that have an LRFD and ASD component are provided for both approaches. Throughout this book, ASD examples, or portions of examples that address the ASD approach, are presented with shaded background for ease of identification. The member-oriented chapters are followed by chapters addressing connection design. Chapter 10 introduces the variety of potential connection types and discusses the strength of bolts, welds, and connecting elements. Chapter 11 addresses simple connections. This includes simple beam shear connections, light bracing connections and direct bearing connections. Chapter 12 deals with moment-resisting connections. As with the member-oriented chapters, the basic principles of limit states design are developed first. This is followed by the application of the provisions to simple shear connections and beam-to-column moment connections through extensive examples in both LRFD and ASD. The text concludes in Chapter 13 with an introduction to steel systems for seismic force resistance. It discusses the variety of structural framing systems available and approved for inclusion in the seismic force resisting system.

NEW TO THIS EDITION This third edition is based on ANSI/AISC 360-16 Specification for Structural Steel Buildings, the 2016 edition of the AISC specification. The committee responsible for developing the specification was charged by its chair to develop a new specification that would show minimal changes. Thus, changes in the specification upon which this book is based are limited to those areas where the committee felt important knowledge had been gained since the 2010 edition. As a result, the changes in this book related to the specification provisions are also somewhat limited. Other changes have been implemented here to better explain some topics and to expand others while revising the entire book wherever needed to reflect the latest specification. Throughout, the use of Specification and Manual equation numbers has been implemented to assist the reader in navigating the Specification and Manual. Also, all examples have been revised to provide the actual equations being used prior to entering the numbers for the calculations. Example Problems for ANSI/AISC 360-16: Twenty new examples have been added and all other examples have been updated to the new specification and the new table of member properties, as well as being revised to improve clarity of intent. New Homework Problems: Fifty-five homework problems have been added for a total of 489 and a large percentage of old problems have been revised to expand the opportunity for instructors to provide meaningful activities for their students. Problems continue to be included that carry over from one chapter to another so that an opportunity exists to link concepts of design to one or two specific structures. The following addresses the changes to be found in each chapter: Chapter 1 includes an expanded discussion of structural integrity along with a discussion of the timing of adoption of the new provisions into the International Building Code. The integrated project introduced in this chapter for use throughout the book has been relocated to a new city from the 2nd edition and the framing system modified. This will provide new homework options for those who have implemented this project. A computer model using the RAM Structural System will be available on the book website to support inclusion of the integrated project in courses. Finally, an expanded discussion of reliability and statistics as it applies to structural steel design has been included.

vi Preface Chapter 2 provides an expanded discussion of snow, wind and seismic loads and additional calculations for these environmental loads using ASCE 7. Chapter 3 discusses the new steels approved by the 2016 Specification and the new approach taken by ASTM to the specification of high strength bolts. Chapter 4 addresses tension members. The provisions have not changed, but there has been a revision in standard hole sizes for bolts. These new sizes have been implemented in the examples where appropriate. Chapter 5 looks at compression members and the Specification nomenclature change of KL to Lc has been implemented. A section and an example have been added to address gravity-only columns and their influence on the effective length of columns in lateral load resisting systems. The completely new approach for treatment of columns with slender elements, introduced with the 2016 Specification, is addressed. Single angle compression members and built-up compression members are discussed and examples provided. Chapter 6 on flexural members includes a discussion of the shape factor and its significance. The use of Manual Table 3-10, the beam curves, with Cb not equal to 1.0, is expanded and a new example is included to illustrate the use of Manual Table 3-2, the economy tables, for noncompact beams. The treatment of tees, single angles and double angle beams has been expanded and examples included. Determination of shear strength for wide-flange members when the reduced resistance factor or increased safety factor must be used is now illustrated. Chapter 7 addresses plate girders as doubly symmetric I-shapes built up from plates. It now includes a discussion of these plate girders with compact webs. The completely revised treatment of shear in plate girders included in the 2016 Specification has been incorporated, and the corresponding stiffener design has been expanded. Chapter 8 has an extensively expanded discussion of stability analysis and design including comparisons between the direct analysis method, the effective length method and the first-order analysis method. The discussion of beam column design has been revised to reflect the changes in Manual Part 6, and an example of interaction of tension and bending is included. Chapter 9 now includes a discussion of partial composite action, recognizing the concern expressed in the Specification with low levels of partial composite action and the corresponding ductility demand. Emphasis is given to the plastic neutral axis locations in Manual Table 3-19 and the corresponding percent of composite action. Composite column discussion is expanded to include examples for composite HSS. Chapter 10 has added discussions of the new ASTM bolt specifications and the new high strength Group C bolts. A discussion of bolt installation has been included. The new interpretation of the controlling bolt limit states is discussed, and examples have been revised to reflect this new approach. Slip-critical bolts in combined shear and tension are now addressed. The discussion of weld strength when loaded at an angle and strength at the base metal has been expanded. Chapter 11 has been revised to account for the new interpretation of the controlling bolt limit states. Weld strength in the base metal has been included in all examples where applicable and examples have been expanded to add clarity.

Preface vii Chapter 12 continues to address moment connections and also reflects the changes in controlling bolt limit states as applicable. Base metal weld strength is also treated more directly and overall discussion within the examples has been expanded for clarity. Chapter 13 continues to outline the application of the Seismic Provisions. Throughout the book, new figures are included to better illustrate the corresponding material.

EXAMPLES AND HOMEWORK PROBLEMS IN LRFD AND ASD The LRFD and ASD design philosophies of the 2016 specification are used throughout the book. It is anticipated, however, that instructors and students will concentrate on only one design approach. This will be the most effective use of time and resources except for situations where it will be valuable to illustrate the differences and similarities of the LRFD and ASD philosophies. ASD is presented in this book primarily for those who are looking to compare methods in order to better understand the concepts of the unified specification and designers who have been practicing with ASD and wish to move their practice to LRFD. LRFD should be the primary approach used to teach structural steel design to new students. LRFD and ASD in Examples Design examples that use the specification provisions directly are provided along with examples using the variety of design aids available in the AISC Steel Construction Manual. All examples that have an LRFD and ASD component are provided for both approaches. Throughout this book, ASD examples, or portions of examples that address the ASD approach, are presented with shaded background for ease of identification. EXAMPLE 4.10b Tension Member Design by ASD

SOLUTION

Goal:

Select a double-angle tension member for use as a web member in a truss and determine the maximum area reduction that would be permitted for holes and shear lag.

Given:

The member must carry a dead load of PD = 67.5 kips and a live load of PL = 202.5 kips. For the load combination PD + PL, the ASD required strength is Pa = 270 kips. Use equal leg angles of A36 steel.

Step 1:

Determine the minimum required gross area based on the limit state of yielding where Ω = 1.67: Ag

Step 2:

min

= Pa

( Fy

Ω ) = 270 ( 36 1.67 ) = 12.5 in.2

Based on this minimum gross area, from Manual Table 1-15, select 2L6×6×9/16 with Ag = 12.9 in.2

LRFD and ASD in Homework Problems Each chapter includes homework problems at the end of the chapter. These problems are organized to follow the order of presentation of the material in the chapters. Several problems are provided for each general subject. Problems are provided for both LRFD and ASD solutions. There are also problems designed to show comparisons between ASD and LRFD solutions. These problems show that in some instances one method might give a more economical design, whereas in other instances the reverse is true.

viii Preface WEBSITE Additional resources are available from the book website at www.SteelStuff.com . The following resources are available on the student section of the website. ● Answers: Selected homework problem answers are available on the student section of the website. ● Errata: We have reviewed the text to make sure that it is as error-free as possible. However, if any errors are discovered, they will be listed on the book website as a reference. If you encounter any errors as you are using the book, please send them directly to the authors ([email protected]) so we may include them on the website, and correct these errors in future editions.

RESOURCES FOR INSTRUCTORS All resources for instructors are available through an Instructor link on the website at www.SteelStuff.com . The following resources are available to instructors who adopt the text: • Solutions Manual: Solutions for all homework problems in the text. • Integrated Building Project RAM Structural System: Computer model and example output for one solution of integrated Building project • Textbook Figures: Select figures available in PowerPoint format Visit the Instructor link on the website at www.SteelStuff.com to register and request access to these resources.

ACKNOWLEDGEMENTS The previous two editions of this book were under the sole authorship of the first author. Many individuals contributed to those editions and we want to thank them for their contributions to those works as those contributions clearly are carried over to this new edition. This third edition has been produced through the cooperative efforts of three individuals who have a long standing common academic and professional relationship. For the many individuals who have motivated and guided each of us, we acknowledge your contribution to all that we do. We thank those who have provided input and clarity to our understanding of structural steel design through our involvement in the development of the AISC Specification and Manual. Finally, we acknowledge the contributions of our spouses and families through whose support we are able to accomplish this project. Louis F. Geschwindner Judy Liu Charles J. Carter

Contents 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7

1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 2 2.1 2.2

2.3

2.4 2.5 2.6 2.7 3 3.1 3.2 3.3 3.4

Introduction 1 Scope 1 The Specification 1 The Manual 4 AISC Web Site Resources 5 Principles of Structural Design 5 Parts of the Steel Structure 6 Types of Steel Structures 11 1.7.1 Bearing Wall Construction 11 1.7.2 Beam-and-Column Construction 12 1.7.3 Long-Span Construction 14 1.7.4 High-Rise Construction 15 1.7.5 Gable-Frame Construction 16 Design Philosophies 17 Fundamentals of Allowable Strength Design (ASD) 18 Fundamentals of Load and Resistance Factor Design (LRFD) 19 Inelastic Design 20 Structural Safety and Integrity 20 Limit States 26 Building Codes and Design Specifications 27 Integrated Design Project 27 Problems 29 Loads, Load Factors, and Load Combinations 30 Introduction 30 Building Load Sources 31 2.2.1 Dead Load 31 2.2.2 Live Load 31 2.2.3 Snow Load 32 2.2.4 Wind Load 32 2.2.5 Seismic Load 33 2.2.6 Special Loads 33 Building Load Determination 34 2.3.1 Dead Load 34 2.3.2 Live Load 35 2.3.3 Snow Load 37 2.3.4 Wind Load 38 2.3.5 Seismic Load 40 Load Combinations for ASD and LRFD Load Calculations 43 Calibration 51 Problems 53

3.5 3.6

3.7 3.8 4 4.1 4.2 4.3 4.4

4.5

4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 5 5.1 5.2 5.3

42

Steel Building Materials 56 Introduction 56 Applicability of the AISC Specification 56 Steel for Construction 58 Structural Steel Shapes 62 3.4.1 ASTM A6 Standard Shapes 63 3.4.2 Hollow Shapes 65 3.4.3 Plates and Bars 66

5.4 5.5

5.6 5.7 5.8

3.4.4 Built-up Shapes 68 Chemical Components of Structural Steel 68 Grades of Structural Steel 70 3.6.1 Steel for Shapes 70 3.6.2 Steel for Plates and Bars 74 3.6.3 Steel for Fasteners 75 3.6.4 Steel for Welding 77 3.6.5 Steel for Headed Stud Anchors 78 Availability of Structural Steel 78 Problems 78 Tension Members 80 Introduction 80 Tension Members in Structures 80 Cross-Sectional Shapes for Tension Members 82 Behavior and Strength of Tension Members 84 4.4.1 Yielding 85 4.4.2 Rupture 85 Computation of Areas 87 4.5.1 Gross Area 87 4.5.2 Net Area 87 4.5.3 Influence of Hole Placement 4.5.4 Effective Net Area 96 Design of Tension Members 101 Block Shear 104 Pin-Connected Members 112 Eyebars and Rods 115 Built-Up Tension Members 116 Truss Members 116 Bracing Members 117 Problems 120

91

Compression Members 124 Compression Members in Structures 124 Cross-Sectional Shapes for Compression Members 124 Compression Member Strength 126 5.3.1 Euler Column 126 5.3.2 Other Boundary Conditions 130 5.3.3 Combination of Bracing and End Conditions 131 5.3.4 Real Column 134 5.3.5 AISC Provisions 137 Additional Limit States for Compression 145 Length Effects 146 5.5.1 Effective Length for Inelastic Columns 152 5.5.2 Effective Length when Supporting Gravity Only Columns 154 Slender Elements in Compression 157 Column Design Tables 163 Torsional Buckling and Flexural-Torsional Buckling 169

ix

x Contents 5.9 5.10 5.11 5.12

Single-Angle Compression Members Built-Up Members 177 Column Base Plates 181 Problems 182

6 6.1 6.2 6.3

Bending Members 188 Bending Members in Structures 188 Strength of Beams 189 Design of Compact Laterally Supported WideFlange Beams 194 Design of Compact Laterally Unsupported WideFlange Beams 201 6.4.1 Lateral-Torsional Buckling 201 6.4.2 Moment Gradient 206 Design of Noncompact Beams 216 6.5.1 Local Buckling 216 6.5.2 Flange Local Buckling 218 6.5.3 Web Local Buckling 219 Design of Beams for Weak Axis Bending 222 Design of Beams for Shear 224 Continuous Beams 227 Plastic Analysis and Design of Continuous Beams 229 T-Shaped Members in Bending 232 6.10.1 Yielding of Tees 232 6.10.2 Lateral-Torsional Buckling of Tees 233 6.10.3 Flange Local Buckling of Tees 234 6.10.4 Stem Local Buckling of Tees 234 Single-Angle Bending Members 237 6.11.1 Yielding 238 6.11.2 Leg Local Buckling 238 6.11.3 Lateral-Torsional Buckling 239 Double-Angle Members in Bending 243 6.12.1 Yielding of Double Angles 243 6.12.2 Lateral-Torsional Buckling of Double Angles 243 6.12.3 Leg Local Buckling of Double Angles 244 Members in Biaxial Bending 247 Serviceability Criteria for Beams 247 6.14.1 Deflection 248 6.14.2 Vibration 248 6.14.3 Drift 248 Concentrated Forces on Beams 250 6.15.1 Web Local Yielding 251 6.15.2 Web Local Crippling 252 Open Web Steel Joists and Joist Girders 256 Problems 259

6.4

6.5

6.6 6.7 6.8 6.9 6.10

6.11

6.12

6.13 6.14

6.15 6.16 6.17 7 7.1 7.2

7.3

174

Plate Girders 264 Background 264 Homogeneous Plate Girders in Bending 266 7.2.1 Noncompact Web Plate Girders 267 7.2.2 Slender Web Plate Girders 271 7.2.3 Compact Web Plate Girders 279 Homogeneous Plate Girders in Shear 281 7.3.1 Nontension Field Action 282

7.4

7.5 8 8.1 8.2 8.3 8.4 8.5 8.6 8.7

8.8 8.9 8.10 8.11 8.12

8.13 8.14 9 9.1 9.2 9.3 9.4 9.5

9.6 9.7

9.8 9.9 9.10

7.3.2 Tension Field Action 284 Stiffeners for Plate Girders 286 7.4.1 Intermediate Stiffeners 286 7.4.2 Bearing Stiffeners 288 7.4.3 Bearing Stiffener Design 291 Problems 295 Beam-Columns and Frame Behavior 297 Introduction 297 Second-Order Effects 298 Interaction Principles 300 Interaction Equations 301 Braced Frames 305 Moment Frames 312 Specification Provisions for Stability Analysis and Design 324 8.7.1 Direct Analysis Method 325 8.7.2 Effective Length Method 325 8.7.3 First-Order Analysis Method 326 8.7.4 Geometric Imperfections 326 8.7.5 Comparison of Methods 327 Initial Beam-Column Selection 333 Beam-Column Design Using Manual Part 6 336 Combined Simple and Moment Frames 337 Partially Restrained Frames 349 Stability Bracing Design 358 8.12.1 Column Bracing 359 8.12.2 Beam Bracing 360 8.12.3 Frame Bracing 361 Tension Plus Bending 363 Problems 366 Composite Construction 372 Introduction 372 Advantages and Disadvantages of Composite Beam Construction 375 Shored versus Unshored Construction 375 Effective Flange 376 Strength of Composite Beams and Slab 377 9.5.1 Fully Composite Beams 378 9.5.2 Partially Composite Beams 383 9.5.3 Composite Beam Design Tables 386 9.5.4 Negative Moment Strength 391 Shear Stud Strength 391 9.6.1 Number and Placement of Shear Studs 392 Composite Beams with Formed Metal Deck 394 9.7.1 Deck Ribs Perpendicular to Steel Beam 395 9.7.2 Deck Ribs Parallel to Steel Beam 396 Fully Encased Steel Beams 403 Selecting a Section 403 Serviceability Considerations 408 9.10.1 Deflection During Construction 408 9.10.2 Vibration Under Service Loads 409 9.10.3 Live Load Deflections 409

Contents xi 9.11 9.12 9.13

Composite Columns 412 Composite Beam-Columns Problems 426

10 10.1 10.2 10.3 10.4 10.5

Connection Elements 429 Introduction 429 Basic Connections 429 Beam-to-Column Connections 430 Fully Restrained Connections 432 Simple and Partially Restrained Connections 433 Mechanical Fasteners 434 10.6.1 Common Bolts 434 10.6.2 High-Strength Bolts 435 10.6.3 Bolt Holes 436 Bolt Limit States 438 10.7.1 Bolt Shear 439 10.7.2 Bolt Bearing and Tearout 440 10.7.3 Strength at Bolt Holes 441 10.7.4 Bolt Tension 442 10.7.5 Slip 448 10.7.6 Combined Tension and Shear in Bearing-Type Connections 450 10.7.7 Combined Tension and Shear in SlipCritical Connections 451 Welds 451 10.8.1 Welding Processes 452 10.8.2 Types of Welds 453 10.8.3 Weld Sizes 454 Weld Limit States 454 10.9.1 Fillet Weld Strength 455 10.9.2 Groove Weld Strength 461 Connecting Elements 461 10.10.1 Connecting Elements in Tension 461 10.10.2 Connecting Elements in Compression 462 10.10.3 Connecting Elements in Flexure 462 10.10.4 Connecting Elements in Shear 462 10.10.5 Block Shear Strength 463 10.10.6 Connecting Element Rupture Strength at Welds 466 Problems 467

10.6

10.7

10.8

10.9 10.10

10.11 11 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8

421

Simple Connections 471 Types of Simple Connections 471 Simple Shear Connections 471 Double-Angle Connections: BoltedBolted 473 Double-Angle Connections: WeldedBolted 492 Double-Angle Connections: BoltedWelded 496 Double-Angle Connections: WeldedWelded 498 Single-Angle Connections 498 Single-Plate Shear Connections 509

11.9 11.10 11.11 11.12 12 12.1 12.2 12.3

12.4

12.5 13 13.1 13.2 13.3

Seated Connections 517 Light Bracing Connections 524 Beam Bearing Plates and Column Base Plates 536 Problems 542 Moment Connections 546 Types of Moment Connections 546 Limit States 548 Moment Connection Design 549 12.3.1 Direct-Welded Flange Connection 549 12.3.2 Welded Flange Plate Connection 556 12.3.3 Bolted Flange Plate Connection 565 Column Stiffening 578 12.4.1 Flange Local Bending 578 12.4.2 Web Local Yielding 579 12.4.3 Web Local Crippling 580 12.4.4 Web Compression Buckling 580 12.4.5 Web Panel Zone Shear and Stiffeners 581 Problems 589

13.7 13.8

Steel Systems for Seismic Resistance 591 Introduction 591 Expected Behavior 592 Moment-Frame Systems 594 13.3.1 Special Moment Frames (SMF) 595 13.3.2 Intermediate Moment Frames (IMF) and Ordinary Moment Frames (OMF) 597 Braced-Frame Systems 597 13.4.1 Special Concentrically Braced Frames (SCBF) 598 13.4.2 Ordinary Concentrically Braced Frames (OCBF) 601 13.4.3 Eccentrically Braced Frames (EBF) 601 Other Framing Systems 603 13.5.1 Special Truss Moment Frames (STMF) 604 13.5.2 Buckling-Restrained Braced Frames (BRBF) 604 13.5.3 Special Plate Shear Walls (SPSW) 605 13.5.4 Composite Systems 606 Other General Requirements 606 13.6.1 Bolted and Welded Connections 606 13.6.2 Protected Zones 607 13.6.3 Local Buckling 607 13.6.4 Column Requirements 607 13.6.5 Column Bases 608 Conclusions 608 Problems 608

Index

610

13.4

13.5

13.6

xii Contents This page left intentionally blank.

Chapter

1

Introductio on

One Worlld Trade Centeer, New York Phooto Courtesy M Michael Mahessh, Port Authorrity of NYNJ

1.1

SCOP PE A wide variety v of desiigns can be ch haracterized aas structural ssteel design. This book deeals with the design of steel structu ures for build dings as goveerned by the ANSI/AISC 3360-16 Specif ification for Structura al Steel Build dings, publish hed by the Am merican Instiitute of Steel Constructionn (AISC) in 2016, and referred to as the Speciffication in thiis book. The aareas of appliication given throughout this book k specifically y focus on thee design of stteel building structures. T The treatment of subjects associateed with bridgees and industrrial structuress, if addressedd at all, is keppt relatively brief. The T book ad ddresses the concepts andd design critteria for the two design approaches detailed by b the Specifi fication: Load d and Resistannce Factor Deesign (LRFD)) and Allowabble Strength Design (A ASD). Both methods m are discussed d laterr in this chapter. In I addition to o the Specifica ation, the prim mary referencce for this boook is the 15thh edition of the AISC C Steel Constrruction Manu ual. This referrence handbook contains tables of the bbasic values needed for f structurall steel design n, design tabbles to simpliify actual deesign, and thhe complete Specifica ation. Through hout this book k, this is referrred to as the Manual.

1.2

THE SPECIFIC CATION The ANS SI/AISC 360-1 16 Specificatiion for Structtural Steel Buuildings is thhe latest in a llong line of standard specification ns published by b the Ameriican Institute of Steel Connstruction forr the design and consstruction of structural s steeel buildings. The first ediition was pubblished in 1923. For the reader in nterested in th he historical aspects of thhese specificaations, AISC C has two ressources that provide detailed d guid dance on the historical struuctural steel standards. Thhe first is AIISC Design Guide 15, AISC Reh habilitation and Retrofit Guide: A R Reference foor Historic SShapes and Specifica ations. This Design D Guide provides outlline comparissons of the prrovisions in thhe different editions of the Specifi fication. The second s resourrce is found oon the AISC web site, ww ww.aisc.org, AISC Sp pecifications 1923–2010, 1 which w containns a searchabble compenddium of all of the AISC Specifica ations for Stru uctural Steel Buildings B prooduced from 11923 through 2010.

1

2 Chapter 1

Introduction Current design is carried out under the provisions published in the 2016 edition of the AISC Specification. In addition to the detailed provisions, the Specification contains User Notes and a detailed Commentary that provides insights into the source and application of the provisions. The reader interested in additional background on the provisions discussed in this book is encouraged to investigate the materials cited in the appropriate sections of the Commentary. The Specification contains 14 chapters and 8 appendices. To provide a concise guide to the use of the Specification, a brief description is given here. Chapter A: General Provision. This chapter provides the scope of the Specification and summarizes all referenced specifications, codes, and standards. It also provides the requirements for materials to be used in structural steel design and the design documents necessary to communicate that design. Chapter B: Design Requirements. This chapter gives the general requirements for analysis and design that are applicable throughout the entire Specification. It provides the charging language needed for application of the subsequent chapters. Chapter C: Design for Stability. This chapter, along with Appendix 7, addresses the requirements for the design of structures to ensure stability. It details those factors that must be taken into consideration in any analysis and design. Chapter D: Design of Members for Tension. This chapter applies to the design of members subjected to axial tension resulting from forces acting through the centroidal axis. Chapter E: Design of Members for Compression. This chapter addresses members subjected to axial compression resulting from forces applied at the centroidal axis. Chapter F: Design of Members for Flexure. This chapter applies to members loaded in a plane parallel to a principal axis that passes through the shear center or is restrained against twisting. This is referred to as simple bending about one axis. Chapter G: Design of Members for Shear. This chapter addresses webs of singly or doubly symmetric members subject to shear in the plane of the web. It also addresses other shapes such as single angles and hollow structural sections. Chapter H: Design of Members for Combined Forces and Torsion. This chapter addresses design of members subject to an axial force in combination with flexure about one or both axes, with or without torsion. It also applies to members subjected to torsion only. Chapter I: Design of Composite Members. This chapter addresses the design of members composed of steel shapes and concrete working together as a member. It addresses compression, flexure, and combined forces. Chapter J: Design of Connections. This chapter addresses the design of connections, including the connecting elements, the connectors, and the connected portions of members. Chapter K: Additional Requirements for HSS and Box Section Connections. This chapter addresses requirements in addition to those given in Chapter J for the design of connections to hollow structural sections and built-up box sections of uniform thickness. Its also addresses connections between HSS and box members.

Introduction

Chapter 1 3

Chapter L: Design for Serviceability. This chapter summarizes the performance requirements for the design of a serviceable structure. Chapter M: Fabrication and Erection. This chapter addresses the requirements for shop drawings, fabrication, shop painting, and erection. Chapter N: Quality Control and Quality Assurance. This chapter addresses the requirements for ensuring quality of the constructed project. Appendix 1: Design by Advanced Analysis. The body of the Specification addresses design based on an elastic analysis. This appendix addresses design by alternative methods generally referred to as advanced methods. It includes the classical plastic design method and design by direct modeling of imperfections. Appendix 2: Design for Ponding. This appendix provides methods for determining whether a roof system has sufficient strength and stiffness to resist the influence of water collecting on the surface and forming a pond. Appendix 3: Fatigue. This appendix provides requirements for addressing the influence of high cycle loading on members and connections that could lead to cracking and progressive failure. For most building structures, fatigue is not an issue of concern. Appendix 4: Structural Design for Fire Conditions. This appendix provides the criteria for evaluation of structural steel subjected to fire conditions, including (1) the prescriptive approach provided for in the model building code and most commonly used in current practice and (2) the engineered approach. Appendix 5: Evaluation of Existing Structures. This appendix provides guidance on the determination of the strength and stiffness of existing structures by load tests or a combination of tests and analysis. Appendix 6: Member Stability Bracing. This appendix details the criteria for ensuring that column, beam and beam-column bracing has sufficient strength and stiffness to meet the requirements for member bracing assumed in the provisions of the Specification for design of those members. Appendix 7: Alternative Methods of Design for Stability. This appendix, along with Chapter C, provides methods of designing structures to ensure stability. Two alternative methods are provided here, including the method most commonly used in past practice. Appendix 8: Approximate Second-Order Analysis. This appendix provides a method for obtaining second-order effects by an amplified first-order analysis. The provisions are limited to structures supporting load primarily through vertical columns. Each chapter of this book will identify those chapters of the Specification that are pertinent to that chapter. The reader is encouraged to become familiar with the organization of the Specification.

4 Chapter 1

Introduction

1.3 THE MANUAL The AISC Steel Construction Manual, 15th edition, is the latest in a series of manuals published to assist the building industry in designing safe and economical steel building structures. The first edition was published in 1928 and the ninth edition in 1989. These manuals addressed design by the allowable stress method. In 1986 the first edition of the load and resistance factor design method manual was published, with the third edition published in 1999. The next in this unbroken string of manuals published in support of steel design and construction was the first manual to unify these two design methods and was published in 2005 as the 13th edition. The current edition of the Manual is the 15th. Students who purchase the Manual through the AISC Student Discount Program also have an opportunity to apply for a free AISC Student Membership at the same time. Students are encouraged to become AISC Student Members in order to take full advantage of all free member benefits. As is the case for the Specification, AISC has two resources to assist in addressing the historic aspects of steel design and construction. The first is, again, AISC Design Guide 15, AISC Rehabilitation and Retrofit Guide: A Reference for Historic Shapes and Specifications. This Guide provides properties of beam and column sections as old as the wrought iron shapes produced as early as 1873. The second resource is the electronic AISC Shapes Database. This database is available through the AISC web site www.aisc.org. It is a searchable database with properties for all shapes produced since 1873, consistent with the printed data in Design Guide 15. Access to the electronic shapes database is free to AISC members. The Manual is presented in 17 parts as follows: Part 1: Dimensions and Properties Part 2: General Design Considerations Part 3: Design of Flexural Members Part 4: Design of Compression Members Part 5: Design of Tension Members Part 6: Design of Members Subject to Combined Forces Part 7: Design Considerations for Bolts Part 8: Design Considerations for Welds Part 9: Design of Connecting Elements Part 10: Design of Simple Shear Connections Part 11: Design of Partially Restrained Moment Connections Part 12: Design of Fully Restrained Moment Connections Part 13: Design of Bracing Connections and Truss Connections Part 14: Design of Beam Bearing Plates, Column Base Plates, Anchor Rods, and Column Splices

Introduction

Chapter 1 5

Part 15: Design of Hanger Connections, Bracket Plates, and Crane-Rail Connections Part 16: Specifications and Codes Part 17: Miscellaneous Data and Mathematical Information Each chapter of this book identifies those parts of the Manual that will be used with the material to be addressed. In many instances, the user will need to look in several parts of the Manual to fully understand the topics or solve the problems presented.

1.4

AISC WEB SITE RESOURCES Another primary resource is the AISC web site, where there is information that is free to all visitors and additional electronic resources that are free to members only. Students will find a great deal of useful information on the AISC publications web site, www.aisc.org/epubs. The primary resources include electronic versions of the Specification, the Shapes Database, the Steel Construction Manual References, and the Steel Construction Manual Design Examples. The Specification, as described in Section 1.2 and the historic Shapes Database, as mentioned in Section 1.3, are available free to all through the web site. The 15th edition Steel Construction Manual Shapes Database is also available free to all. The AISC web site also includes an extensive array of journal and proceedings papers. All of the references cited in the Commentary and the Manual, for which AISC owns the copyright, are accessible under Steel Construction Manual Resources; Interactive Reference List. Probably the most valuable aspect of the AISC web site for readers of this book is the complete set of the 15th edition Steel Construction Manual Design Examples. These examples are presented in four sections. Section I: Examples Based on the AISC Specification. This section contains examples demonstrating the use of the specific provisions of the Specification, organized by Specification chapter. Section II: Examples Based on the AISC Steel Construction Manual. This section contains examples of connection design using the Specification and the tables found in the Manual. Section III: System Design Examples. This section contains examples associated with the design of a specific building and the application of the system-wide requirements. Section IV: Additional Resources. This section provides design tables for higher-strength steels than published in the printed Manual. Although the topics covered in this book are supported by calculated example problems, the reader might find the electronic Steel Construction Manual Design Examples helpful for further understanding of some of the specific provisions or design aids described in the book. In addition, some of the Design Examples go beyond the coverage in this book and provide additional useful information regarding typical design or detailing. The reader is encouraged to investigate what the AISC web site has to offer through both free and member only publications.

1.5

PRINCIPLES OF STRUCTURAL DESIGN From the time an owner determines a need to build a building, through the development of conceptual and detailed plans, to completion and occupancy, a building project is a multi-faceted

6 Chapter 1

Introduction task that involves many professionals. The owner and the financial analysis team evaluate the basic economic criteria for the building. The architects and engineers form the design team and prepare the initial proposals for the building, demonstrating how the users’ needs will be met. This teamwork continues through the final planning and design stages, where the design drawings, specifications, and contract documents are readied for the construction phase. During this process, input may also be provided by the individuals who will transform the plans into a real-life structure. The steel detailer, fabricator and erector all have a role in that process, and add their respective expertise to make the design constructible. Thus, those responsible for the construction phase of the project often help improve the design by taking into account the actual on-site requirements for efficient construction. Once a project is completed and turned over to the owner, the work of the design teams is normally over. The operation and maintenance of the building, although major factors in the life of the structure, are not usually within the scope of the designer’s responsibilities, except when significant changes in building use are anticipated. In such cases, a design team should verify that the proposed changes can be accommodated. The basic goals of the design team can be summarized by the words safety, function, and economy. The building must be safe for its occupants and all others who may come in contact with it. It must neither fail locally nor overall, nor exhibit behavioral characteristics that test the confidence of rational human beings. To help achieve that level of safety, building codes and design specifications are published that outline the minimum criteria that any structure must meet. The building must also serve its owner in the best possible way to ensure that the functional criteria are met. Although structural safety and integrity are of paramount importance, a building that does not serve its intended purpose will not have met the goals of the owner. Last, but not least, the design, construction, and long-term use of the building should be economical. The degree of financial success of any structure will depend on a wide range of factors. Some are established prior to the work of the design team, whereas others are determined after the building is in operation. Nevertheless, the final design should, within all reasonable constraints, produce the lowest combined short- and long-term expenditures. The AISC Specification follows the same principles. The mission of the AISC Committee on Specifications is to “develop the practice-oriented specification for structural steel buildings that provide for life safety, economical building systems, predictable behavior and response, and efficient use.” Thus, this book emphasizes the practical orientation of this Specification.

1.6 PARTS OF THE STEEL STRUCTURE All structures incorporate some or all of the following basic types of structural components: 1.

Tension members

2.

Compression members

3.

Bending members

4.

Combined force members

5.

Connections

The first four items represent structural members. The fifth, connections, represents the contact regions between the structural members, which ensure that all components work together as a structure.

Introoduction

C Chapter 1 7

Detailed D evalu uations of thee strength, beehavior, and ddesign criteria for these m members are presented d in the follow wing chapterss: Tension T mem mbers:

Chapteer 4

members: Compression C

Chapteer 5

Bending B mem mbers:

Chapteers 6 and 7

Combined C forrce members:

Chapteer 8

Connections: C

Chapteers 10, 11, andd 12

The T strength and behavio or of structurral frames coomposed of a combinatioon of these elementss are covered in Chapters 8 and 13, and the special considerationss that apply too composite (steel and d concrete wo orking togeth her) constructiion are presennted in Chaptter 9. An intrroduction to the desig gn of steel stru uctures for eaarthquake loaading is presennted in Chapter 13. The properties of structural steel and th he various sh hapes commoonly used are discussed inn Chapter 3, and a brief discussio on of the types of loads and d load combinnations is presented in Chaapter 2. Tension T memb bers are typiccally found ass web and choord members in trusses andd open-web steel joissts; as diagon nal members in structurall bracing sysstems; and ass hangers forr balconies, mezzanin ne floors, and d pedestrian walkways. w Theey are also ussed as sag rodds for purlins and girts in many bu uilding types, as well as to t support plaatforms for m mechanical eqquipment andd pipelines. Figures 1.1 1 and 1.2 illlustrate typicaal applicationns of tension m members in acctual structures. In I the idealizeed case, tensiion members transmit conncentric tensille forces onlyy. In certain structures, reversals of o the overalll load may cchange the tennsion membeer force from m tension to compresssion. Some members m will be b designed ffor this actionn; others will hhave been designed with the assum mption that th hey will carry tension only.. The T idealized d tension mem mber is analyyzed with thee assumption that its end cconnections are pins,, which prev vent any mom ment or sheaar force from m being transsmitted to thhe member. Howeverr, in an actuall structure, th he type of connnection norm mally dictatess that some beending may be introd duced to the tension t memb ber. This is aalso the case when the tennsion memberr is directly exposed to some form m of transversse load. Mom ments will alsso be introducced if the eleement is not perfectly y straight, or if the axial loaad is not appliied along the centroidal axxis of the mem mber. The T primary load l effect in n the tension member is a concentric aaxial force, w with bending and shearr considered as a secondary effects. members aree also known as columns, sstruts, or postts. They are uused as web Compression C and chorrd members in trusses and a joists andd as verticall members inn all types oof building structures. Figure 1.3 shows a typiccal use of struuctural comprression membbers.

Figure 1.1 1

Use of Tension T Memb bers in a Trusss

Photo cou urtesy Ruby + Associates A

8 Chapter 1

Introduction n

Figure 1.2 Use of Tenssion Memberss as Hangers The idealized co ompression member m carriies only a cooncentric, coompressive foorce. Its strength is heavily h influeenced by thee distance bettween the suppports, as weell as by the support conditions. The T basic collumn is thereffore defined aas an axially loaded membber with pinnned ends. Historically,, design ruless for compresssion memberss have been bbased on the bbehavior and strength of this idealiized compression memberr. The basic colum mn is practicaally nonexisttent in real sstructures. Reealistic end ssupports rarely resem mble perfect pins; the ax xial load is nnormally nott concentric, due to the w way the surrounding g structure traansmits its load to the meember; and beeams and sim milar componnents are likely to be connected to o the column n in such a w way that mom ments are intrroduced. All of these conditions produce p bendiing effects in the member,, making it a combined forrce member oor beamcolumn, as distinct d from the t idealized column. The primary loaad effect in the pinned--end column is thereforee a concentrric axial compressivee force accom mpanied by thee secondary eeffects of bending and sheaar. Bending memberrs are known n as beams, girders, joistts, spandrels, purlins, linttels, and girts. Altho ough all of these t are bending membbers, each naame implies a certain sttructural application within w a build ding:

umns in a Buiilding Frame Figure 1.3 Use of Colu

Introoduction

C Chapter 1 9

1. 1 Beams, girders, and jo oists form parrt of commonn floor system ms. The beam ms are most often considered as thee members thhat are directlly supported bby girders, whhich in turn are usuallly supported by b columns. JJoists are beam ms with fairlyy close spacinng. A girder may geneerally be conssidered a highher-order bennding memberr compared w with a beam or joist. However, H variiations to this basic schemee are commonn. 2. 2 The bending members that form thee perimeter off a floor or rooof plan in a bbuilding are known ass spandrels or o spandrel b eams. Their design may be different from other beams and d girders becaause the load comes primaarily from onee side of the m member. 3. 3 Bending members in roof systemss that span bbetween otheer bending m members are usually reeferred to as purlins. p 4. 4 Lintels arre bending members m that span across tthe top of oppenings in waalls, usually carrying the t weight off the wall aboove the openiing as well ass any other looad brought into that area. They ty ypically are sseen spanninng across the openings forr doors and windows. 5. 5 Girts are used in exterrior wall systtems. They trransfer the laateral load froom the wall surface to o the exterior columns. Thhey may also assist in suppporting the weight of the wall. Figure F 1.4 sh hows beams and girders in an actuall structure unnder construcction. The idealized d beam is sho own in the figure fi as a m member with a uniform looad supportedd on simple supports.. The T basic ben nding member carries traansverse loadds that act inn a plane conntaining the longitudiinal centroidaal axis of thee member. Thhe primary looad effects arre bending m moment and shear forrce. Axial forcces and torsio on may occur as secondaryy effects. The T most com mmon combin ned force meember is know wn as a beam m-column, im mplying that this strucctural elemen nt is simultan neously subjeccted to bendiing and axiall compressionn. Although the preseence of both h bending an nd axial tenssion represennts a potentiaal loading caase for the combined d force memb ber, this case is not as criti cal or commoon as the beam m-column loaading case.

Iddealized bendiing member

Figure 1.4 1

Building g Structure Sh howing Bendiing Members

100 Chapter 1

Introductio on

R n of Steel Fraames in Whicch the Verticaal Members A Are Figure 1.5 Schematic Representatio o Axial Loadss and Bending g Moments Subjected to Figu ure 1.5a is a schematic illu ustration of a multi-story steel frame w where the beaams and columns aree joined with rigid connecctions. Becausse of the geo metric configguration, the types of connections, and the loaading pattern n, the verticaal members aare subjectedd to axial looads and bending mom ments. This is a typical ap pplication of ppractical beam m-columns; otther examples are the members off the gable fraame shown in n Figure 1.5bb and the verttical componeents of a singgle-story portal framee shown in Fig gure 1.5c. The beam-colum mn may be reg garded as the general strucctural elemennt, where axiaal forces, shear forces, and bending g moments acct simultaneouusly. Thus, thhe basic colum mn may be thoought of as a special case, represeenting a beam m-column wiith no momennts or transveerse loads. Siimilarly, the basic ben nding membeer may be thought of as a bbeam-columnn with no axiaal load. Therefore, the consideratio ons that must be accounted d for in the deesign of bothh columns andd beams mustt also be applied to beeam-columnss. Because of the generalized g nature n of the combined foorce elementt, all load efffects are considered primary. p How wever, when the ratio of aaxial load to axial strengtth in a beam-column becomes hig gh, column behavior b will overshadow other influennces. Similarlly, when the ratio of applied mom ment to mom ment strength h is high, beaam behavior will outweiggh other effects. The beam-colum mn is an elem ment in which h a variety off different forrce types inteeract. Thus, ppractical design appro oaches are no ormally based d on interactioon equations. Con nnections are the collectio on of elemennts that join the memberss of a steel sstructure together. Wh hether they co onnect the axially loaded m members in a truss or the bbeams and columns

Introdduction

Figure 1.6 1

Chhapter 1 11

Building g Connectionss

Copyrightt © American Institute I of Steeel Constructioon. Reprinted w with permissionn. All rights resserved

of a multti-story framee, connectionss must ensuree that the strucctural membeers function toogether as a unit, con nsistent with th he assumption ns made in thhe design. The T fastenerss used in stru uctural steel connections today are alm most entirelyy limited to bolts and d welds. The load effects that the variious elementss of the connnection must resist are a function of the specifi fic connection n type being cconsidered. T They include aall of the possible forces and mom ments. Figuree 1.6 illustratees a variety of connectionns. The ideallized represenntations for connectio ons are presen nted in Chaptters 10, 11, annd 12.

1.7

TYPE ES OF STEEL STRUC CTURES It is diffi ficult to classiify steel strucctures into neeat categoriess, due to the wide variety of systems availablee to the design ner. The elem ments of the sttructure, as deefined in Secttion 1.6, are ccombined to form thee total structu ure of a build ding, which must safely and economiically carry aall imposed loads. Th his combinatio on of members is usually rreferred to as the framing ssystem. buildings com Steel-framed S me in a widee variety of shapes and sizzes and in coombinations with otheer structural materials. m A few f exampless are given inn the followinng paragraphss, to set the stage for the applicatio on of structurral design preesented in subbsequent chappters.

1.7.1

Bea aring Wall Constructio C n Bearing wall constru uction is prim marily used ffor one- or ttwo-story buiildings, such as storage warehouses, shopping g centers, offiice buildings,, and schoolss. This system m normally usses brick or concrete block mason nry walls, on which are plaaced the endss of the flexuural memberss supporting the floor or roof. The flexural mem mbers are usuaally hot-rolledd structural stteel shapes, allone or in

122 Chapter 1

Introductio on

Figuree 1.7 Bearinng Wall Buildding Photo ccourtesy Dougglas Steel Fabriicating Corporaation

n with open web w steel joistts or cold-form med steel shap apes. An exam mple of a bearring wall combination building is shown s in Figu ure 1.7.

1.77.2

Beam--and-Colum mn Construcction Beam-and-ccolumn constrruction is the most commoonly used systtem for steel structures todday. It is suitable for large-area bu uildings such as schools annd shopping ccenters, whichh often have nno more than two sto ories but may have a large number of sppans. It is alsoo suitable for buildings wiith many stories. Colu umns are plaaced accordin ng to a reguular, repetitiouus grid that supports thee beams, girders, and joists, which h are used forr the floor andd roof system ms. The regulaarity of the floor plan lends itself to t economy in i fabrication n and erectionn, because moost of the meembers will bbe of the same size. An A example of this type of structure is shhown in Figuure 1.8. For multi-story buildings, th he use of com mposite steell and concreete flexural m members affords addiitional saving gs. Further ad dvances can bbe expected ass designers bbecome more familiar with the use of compositee columns and d other elemeents of mixed construction systems.

Figure 1.8 Beam-and-C Column Build ding Photo courtessy Douglas Steeel Fabricating Corporation

Introdduction

Chhapter 1 13

Figure 1.9 1 Braced Beeam-and-Colu umn Buildingg Photo cou urtesy Douglas Steel Fabricatting Corporatioon

Beam-and-col B lumn structurres rely on eitther their connnections or a separate braccing system to resist lateral l loads. A frame in which w all connnections are m moment resistant provides rresistance against th he action of laateral loads, such s as wind and earthquak akes, and overrall structural stability, through the t bending stiffness of thee overall fram me.

Figure 1.10 1 Idealizeed Illustration n of Several T Types of Beam m-and-Colum mn Framed Strructures (a) moment--resistant fram me; (b) truss-b braced frame;; (c) core-bracced frame; (dd) floor plan oof shear wall and core-braced building; b (e) floor f plan of bbuilding withh a combinatioon of braced and unbraced d bents. A frame with hout memberr-end restrainnt needs a seeparate lateraal load resistiing system, which is often afford ded by having g the elemennts along onee or more of the column llines act as

144 Chapter 1

on Introductio braced fram mes, as seen in n Figure 1.9.. One of the most commoon types of bbracing is the vertical truss, which h is designed to take the loads imposeed by wind aand seismic aaction. Other bracing schemes inv volve shear walls and reinfforced concrette cores. The latter type m may also be refferred to as a braced core c system and a can be hig ghly efficientt because of thhe rigidity of the box-shapped cross section of th he core. Thee core serves a dual purppose in this ccase: in addittion to providding the bracing systtem for the bu uilding, it serv ves as the verrtical conduitt in the comppleted structurre for all of the necessary services,, including eleevators, stairccases, electriccity, and otheer utilities. Com mbinations off these types of o constructioon are also coommon. For eexample, fram mes may have been designed d as moment resistaant in one direection of the bbuilding and as truss braceed in the other. Of course, such a choice c recogn nizes the threee-dimensionall nature of thee structure. Figu ure 1.10 shows an idealiized represenntation of sevveral types oof beam-and-column framed strucctures.

1.77.3

Long-S Span Consttruction This type of o constructio on encompassses steel-fram med structurres with longg spans betw ween the vertical load d-carrying eleements, such as covered aarenas. The loong distancess may be spaanned by one-way tru usses, two-way y space trussees, or plate annd box girderrs. A long-spaan structure iis shown in Figure 1.1 11. Arches orr cables could d also be usedd, although theey are not connsidered here. Lon ng-span constrruction is also o used in builldings that reequire large, ccolumn-free iinteriors. In such casees the buildin ng may be a corec or othe rwise bracedd structure, whhere the longg span is the distance from the exteerior wall to the t core. Man ny designers would also characterize c ssingle-story rrigid frames aas examples of longspan constru uction system ms. Dependin ng on the geoometry of thee frame, suchh structures ccan span substantial distances, d ofteen with excelllent economyy.

Figure 1.11 Long-Span Structure Photo courtessy Douglas Steeel Fabricating Corporation

Introdduction

Chhapter 1 15

Figu ure 1.12 Higgh-Rise Buildding Struucture Photto courtesy Doouglas Steel Faabricating Corpporation

1.7.4

gh-Rise Con nstruction Hig High-risee construction n refers to mu ulti-story buiildings of signnificant heighht; an example is shown in Figure 1.12. The large heightts and uniquue problems encountered in the desiggn of such structures warrant treeating them in ndependentlyy from typicaal beam-and-column consstruction. In addition, over the passt 40 years seeveral designners have devveloped a num mber of new concepts in multi-sto ory frame desiign, such as th he super com mposite colum mn and the steeel plate shear wall. Particular P carre must be ex xercised in thhe choice andd design of tthe lateral loaad resisting system in n high-rise co onstruction. Itt is not just a m matter of extrrapolating froom the principples used in the analy ysis of lower-rrise structures, because maany effects pllay a major roole in the desiign of highrise build dings but hav ve significanttly less impacct on frames of smaller hheight. These effects are crucial to o the proper design d of the high-rise h struccture. Some S of thesee effects may y be referred tto as second--order effects,, because theyy cannot be quantifieed through a normal, n linearrly elastic anaalysis of the fframe. Althouugh second-oorder effects are preseent in all strucctures, they may m be more significant inn high-rise sttructures. Foor example, when a structure is displaced d lateerally, additioonal momentt is induced in a column due to the eccentriccity of the column loads. When W added too the momentts and shears produced by gravity and wind loaads, the resu ulting effectss may be siggnificantly laarger than thhose computted without consideriing the secon nd-order effeccts. A designeer who does not incorporaate both typees of effects will be making m a serio ous and perhaaps unconservvative error. Framing F systeems for high--rise buildinggs reflect the increased im mportance of lateral load resistance. Thus, attem mpts at makin ng the perimeeter of a buildding act as a uunit or tube hhave proven quite succcessful. Thiss tube may bee in the form of a truss, ass with the John Hancock Building in Chicago,, Illinois, show wn in Figure 1.13a, or a frrame, as in thhe former Worrld Trade Cennter in New York Citty, shown in Figure F 1.13b; a solid wall ttube with cuttouts for winddows, as usedd in the Aon Center in n Chicago, sh hown in Figurre 1.13c; or seeveral intercoonnected or bbundled tubess, such as in the Williis Tower in Chicago (form merly known aas the Sears Tower), shownn in Figure 1.13d.

166 Chapter 1

Introductio on

(a)

(b)

(c)

(d)

Figure 1.13 High-Rise Buildings (a)) the John Haancock Centerr; (b) the Worrld Trade Cennter; (c) the Aon Center; (d) th he Willis Tow wer. Photo (b) cou urtesy Leslie E. Robertson Asssociates, RLL LP

1.77.5

Gable--Frame Con nstruction Many design ners include the t single-storry frame as ppart of the lonng-span constrruction categoory. The metal buildiing industry has h capitalizeed on the usee of this system through ffine-tuned deesigns of frames for storage s wareh houses, indusstrial buildinggs, temporaryy and permannent office buuildings, and similar types t of strucctures. The gaable frames arre typically buuilt-up, web ttapered membbers that can be optim mized for thee forces preseent. Memberrs are conneccted in the fiield using bolted end plate connecctions. Metall buildings, co ommonly refferred to as prre-engineeredd metal buildiings, are now availab ble as custom engineered for fo specific appplications. A An example off a gable-fram me metal building is seen s in Figuree 1.14.

Introdduction

Chhapter 1 17

Figure 1.14 1 Gable-F Frame Metal Building Photo cou urtesy Metal Bu uilding Manufa facturers Assocciation

1.8

DESIIGN PHILO OSOPHIES A successsful structuraal design resu ults in a struucture that is safe for its ooccupants, caan carry the design lo oads without overstressing g any componnents, does noot deform or vibrate excesssively, and is econom mical to build d and operate for its intendded life span. Although ecoonomy may aappear to be the primary concern of o an owner, safety must be the primaary concern oof the engineeer. Costs of labor an nd materials will vary frrom one geoographic locaation to anoother, makingg it almost impossib ble to design a structure th hat is equally economical iin all locationns. Because thhe foremost task of th he designer is i to produce a safe and s erviceable strructure, desiggn criteria suuch as those published d by the Am merican Institu ute of Steel Constructionn are based oon technical m models and consideraations that prredict structurral behavior aand material rresponse. Thee use of thesee provisions by the designer d will dictate the economy e of a particular ssolution in a particular loocation and business climate. To T perform a structural deesign, it is neecessary to qquantify the ccauses and efffects of the loads thaat will be exeerted on each element throoughout the life of the struucture. This iis generally termed th he load effectt or the requirred strength. It is also neccessary to acccount for the behavior of the mateerial and the shapes that compose theese elements.. This is refeerred to as thhe nominal strength or capacity of o the elementt. In I its simplesst form, strucctural designn is the deterrmination of member sizees and their correspon nding connecctions, so thatt the strength of the structuure is greater than the loadd effect. The degree to o which this is i accomplish hed is often teermed the maargin of safetyy. Numerous approaches for accom mplishing thiss goal have beeen used overr the years. Although A passt experience might seem to indicate thhat the structtural designerr knows the exact maagnitude of th he loads that will w be applieed to the struucture, and thee exact strenggth of all of the structtural elementts, this is usuaally not the caase. Design looads are provvided by manyy codes and standardss and, althou ugh the valuess given are sspecific, signiificant uncerttainty is assoociated with those maagnitudes. Loaads, load factors, and load combinationns are discusseed in Chapterr 2. As A is the casee for loading, significant uuncertainty is associated w with the determ mination of the behav vior and stren ngth of structtural memberrs. The true inndication of lload-carryingg capacity is given by y the magnitu ude of the loaad that causess the failure of a componnent or the strructure as a whole. Failure F may eiither occur ass the physicall collapse of ppart of the buuilding, or be considered to have occurred o if deflections, d fo or instance, eexceed certainn predeterminned values. W Whether the failure iss the result of o a lack of strength s (colllapse) or stifffness (deflecction), these phenomena

18 Chapter 1

Introduction reflect the limits of acceptable behavior of the structure. Based on these criteria, the structure is said to have reached a specific limit state. A strength failure is termed an ultimate limit state, whereas a failure to meet operational requirements, such as deflection, is termed a serviceability limit state. Regardless of the approach to the design problem, the goal of the designer is to ensure that the load on the structure and its resulting load effect, such as bending moment, shear force, and axial force, in all cases are sufficiently below each of the applicable limit states. This ensures that the structure meets the required level of safety or reliability. Three approaches to the design of steel structures are permitted by the AISC Specification: 1.

Allowable strength design (ASD)

2.

Load and resistance factor design (LRFD)

3.

Design by inelastic analysis

The design approaches represent alternative ways of formulating the same problem, and all have the same goal. All three are based on the nominal strength of the element or structure. The nominal strength, most generally expressed as Rn, is determined in exactly the same way, from the exact same equations, whether used in ASD or LRFD. Some formulations of design by inelastic analysis, such as plastic design, also use these same nominal strength equations whereas other approaches to inelastic design model in detail every aspect of the structural behavior and do not rely on the equations provided through the Specification. The use of a single nominal strength expression for both ASD and LRFD permits the unification of these two design approaches. It will become clear throughout this book how this approach has simplified steel design for those who have struggled in the past with comparing the two available philosophies. The following sections describe these three design approaches, any one of which is an acceptable approach to structural steel design according to the AISC Specification.

1.9 FUNDAMENTALS OF ALLOWABLE STRENGTH DESIGN (ASD) Prior to 2005, allowable strength design was referred to as allowable stress design. It is the oldest approach to structural design in use today and has been the foundation of AISC Specifications since the original provisions of 1923. Allowable stress design was based on the assumption that under actual load, stresses in all members and elements would remain elastic. To meet this requirement, a safety factor was established for each potential stress-producing state. Although historically ASD was thought of as a stress-based design approach, the allowable strength was always obtained by using the proper combination of the allowable stress and the corresponding section property, such as area or elastic section modulus. The current allowable strength design approach is based on the concept that the required strength of a component is not to exceed a certain permitted or allowable strength under normal in-service conditions. The required strength is determined on the basis of specific ASD load combinations and an elastic analysis of the structure. The allowable strength incorporates a factor of safety, Ω, and uses the nominal strength of the element under consideration. This strength could be presented in the form of a stress if the appropriate section property is used. As a result of doing this, the resulting stresses will most likely again be within the elastic range, although this is not a preset requirement of the Specification. The magnitude of the factor of safety and the resulting allowable strength depend on the particular governing limit state against which the design must produce a certain margin of safety.

Introduction

Chapter 1 19

Safety factors are obtained from the Specification. This requirement for ASD is provided in Section B3.2 of the Specification as

Ra ≤

Rn Ω

(AISC B3-2)

which can be stated as

Required Strength (ASD) ≤

Nominal Strength = Allowable Strength Safety Factor

The governing strength depends on the type of structural element and the limit states being considered. Any single element can have multiple limit states that must be assessed. The safety factor specified for each limit state is a function of material behavior and the limit state being considered. Thus, it is possible for each limit state to have its own unique safety factor. For example, the limit state of yielding of a tension member is given by Pn = Fy Ag

where Fy is the steel yield strength and Ag is the gross area of the member. The safety factor is Ω = 1.67. Thus, for steel with a yield strength of 50 ksi, the allowable strength is Pn 50.0 Ag = = 30 Ag 1.67 Ω

Design by ASD requires that the allowable stress load combinations of the building code be used. Loads and load combinations are discussed in detail in Chapter 2.

1.10 FUNDAMENTALS OF LOAD AND RESISTANCE FACTOR DESIGN (LRFD) Load and resistance factor design explicitly incorporates the effects of the random variability of both strength and load. Because the method includes the effects of these random variations and formulates the safety criteria on that basis, it is expected that a more uniform level of reliability, and thus safety, for the structure and all of its components will be attained. LRFD is based on the concept that the required strength of a component under LRFD load combinations is not to exceed the design strength. The required strength is obtained by increasing the load magnitude by load factors that account for load variability and load combinations. The design strength is obtained by reducing the nominal strength by a resistance factor that accounts for the many variables that impact the determination of member strength. Load factors for LRFD are obtained from the building codes for strength design and will be discussed in Chapter 2. As for ASD safety factors, the resistance factors are obtained from the Specification. The basic LRFD provision is provided in Section B3.1 of the Specification as

Ru ≤ φRn

(AISC B3-1)

which can be stated as Required Strength (LRFD) ≤ Resistance Factor × Nominal Strength=Design Strength

20 Chapter 1

Introduction Again considering the limit state of yielding of a tension member, Pn = Fy Ag and the resistance factor is φ = 0.90 . For steel with a yield strength of 50 ksi, the design strength is φPn = 0.90(50) Ag = 45 Ag LRFD has been a part of the AISC Specifications since it was first issued in 1986.

1.11 INELASTIC DESIGN The Specification permits a wide variety of formulations for the inelastic analysis of steel structures through the use of Appendix 1. Any inelastic analysis method will require that the structure and its elements be modeled in sufficient detail to account for all types of behavior. An analysis of this type must be able to track the structure’s behavior from the unloaded condition through every load increment to complete structural failure. The only inelastic design approach that will be discussed in this book is plastic design (PD). Plastic design is an approach that has been available as an optional method for steel design since 1961, when it was introduced as Part 2 of the then current Specification. The limiting condition for the structure and its members is attainment of the load that would cause the structure to collapse, usually called the ultimate strength or the plastic collapse load. For an individual structural member this means that its plastic moment capacity has been reached. In most cases, due to the ductility of the material and the member, the ultimate strength of the entire structure will not have been reached at this stage. The less stressed members can take additional load until a sufficient number of members have exhausted their individual capacities so that no further redistribution or load sharing is possible. At the point where the structure can take no additional load, the structure is said to have collapsed. This load magnitude is called the collapse load and is associated with a particular collapse mechanism. The collapse load for plastic design is the service load times a certain load factor. The limit state for a structure that is designed according to the principles of plastic design is therefore the attainment of a mechanism. For this to occur, all of the structural members must be able to develop the yield stress in all fibers at the most highly loaded locations. There is a fine line of distinction between the load factor of PD and the safety factor of ASD. The former is the ratio between the plastic collapse load and the service or specified load for the structure as a whole, whereas the latter is an empirically developed, experience-based term that represents the relationship between the elastic strength of the elements of the structure and the various limiting conditions for those components. Although numerically close, the load factor of plastic design and the factor of safety of allowable stress design are not the same parameter.

1.12 STRUCTURAL SAFETY AND INTEGRITY The preceding discussions of design philosophies indicate that although the basic goal of any design process is to ensure that the end product is a safe and reliable structure, the ways in which this is achieved may vary substantially. In the past, the primary goal for safety was to provide an adequate margin against the consequences of overload. Load factor design and its offshoots were developed to take these considerations into account. In real life, however, many other factors also play a role. These include, but are not limited to the following: 1.

Variations of material strength

Introdduction 2. 2

Variaations of crosss-sectional sizze and shape

3. 3

Accurracy of metho od of analysiss

4. 4

Influeence of workm manship in shhop and field

5. 5

Presence and variaation of residuual stresses

6. 6

Lack of member sttraightness

7. 7

Variaations of locattions of load aapplication pooints

Chhapter 1 21

These factors consideer only some of o the sourcees of variationn of the strenggth of a struccture and its componeents. An even n greater sourrce of variatioon is the loadding, which iss further com mplicated by the fact that t different types t of load have differennt variational characteristiccs. Thus, T a metho od of design that does nott attempt to inncorporate thhe effects of sstrength and load varriability will be burdened d with unacccounted-for ssources of unncertainty. T The realistic solution, therefore, is to deal with safety as a prrobabilistic cooncept. This iis the foundattion of load and resisstance factor design, wheere the probaabilistic charracteristics off load and sstrength are evaluated d and the ressulting safety y margins dettermined statiistically. Eacch load type is given its own speccific factor in n each combin nation, and eaach material liimit state is aalso given its own factor. This metthod recognizzes that theree is always a finite, thouggh very smalll, chance thaat structural failure will w actually occur. o Howev ver, this methhod does not attempt to atttach a speciffic value to this probability. No sp pecific level of o probabilityy of failure is ggiven or impllied by the Sppecification. In I ASD, the variabilities of load andd strength arre not treatedd explicitly as separate parameteers. They are lumped togeether through the use of a single factorr of safety. Thhe factor of safety vaaries with eaach strength limit l state buut does not vvary with loaad source. A ASD can be thought of o as LRFD with a singlee load factor.. LRFD desiggns are generrally expectedd to have a more uniiform level of o reliability than ASD deesigns. That is, the probaability of failuure of each element in an LRFD design d will be the same, reegardless of tthe type of looad or load coombination. Howeverr, a detailed analysis a of reeliability undeer the LRFD provisions shhows that reliability still varies un nder different load combin nations. In AS SD there is noo attempt to aattain uniform m reliability; rather, th he goal is to simply havee a safe struccture, althouggh some elem ments will bee safer than others. For F the development of LR RFD, load efffect (memberr force), Q, aand resistancee (strength), R, are assumed to each have a variability that caan be describbed by the norrmal distributtions shown by the bell-shaped b curves c in Fig gure 1.15. Sttructures can be considerred safe as llong as the resistance is greater th han the load effect, e R > Q. If it were apppropriate to cconcentrate soolely on the mean vallues, Qm and Rm, it would be relativelyy easy to ensuure a structuree’s safety. Hoowever, the full reprresentation off the data shows an area where the ttwo curves ooverlap. Thiss area

Figure 1.15 1

Probability Distributtion, R and Q

222 Chapter 1

Introductio on

Figure 1.16 Probability y Distribution n, (R – Q) represents cases where th he load effectt exceeds the resistance annd therefore identifies occuurrences of failure. Saafety of the sttructure is a function f of th e size of this region of oveerlap. The sm maller the region of ov verlap is, the lower l the prob bability of faiilure. Ano other approacch to presentiing the data iis to look at the differencce between reesistance and load efffect. Figure 1..16 shows thee same data aas Figure 1.155 but presentss it as (R – Q)). For all cases where (R – Q) < 0, the structure is said to havve failed, and for all cases where this diifference is positive, the t structure is i considered d safe. In this presentation of the data, tthe shaded areea to the left of the orrigin represen nts the probab bility of failurre. To limit thhat probabilityy of failure, thhe mean value, (R – Q)m, must bee maintained at an approprriate distancee from the orrigin. This disstance is shown in Fig gure 1.16 as βσ β (R–Q), wheree β is the reliaability index aand σ(R–Q) is thhe standard ddeviation of (R – Q). A th hird represen ntation of thee data is shoown in Figurre 1.17. In thhis case, thee data is presented ass ln(R/Q). Thee logarithmic form of the ddata is a welll-conditioned representatioon and is more useful in the derivaation of the factors fa requireed in LRFD. If we know tthe exact disttribution of the resisttance and loaad effect data, the probabbility of failuure can be directly related to the reliability in ndex β. Unforrtunately, wee know the acctual distribut utions for relaatively few reesistance and load efffect componeents. Thus, we w must relyy on other chharacteristics of the data, such as means and standard s deviaations. The statistical an nalyses requirred to establissh an approprriate level off reliability haave been carried out by b the approp priate specificcation commiittees, and thee resulting looad factors, reesistance factors, and safety factorss have been established. e L Load factors aare presented in the buildinng codes whereas resiistance factorrs and safety factors f for eacch limit state are given in tthe Specificattion.

Figure 1.17 7 Probability y Distribution n, ln(R/Q)

Introduction

Chapter 1 23

Since the load combinations and resistance and safety factors have been established, the reliability can be determined for specific design situations. The reliability index, β, is given in the Specification Commentary as

β=

ln ( Rm Qm )

(AISC C-B3-2)

VR2 + VQ2

where Rm is the mean resistance, Qm is the mean load effect, as discussed earlier, and VR and VQ are the coefficients of variation of resistance and load effect respectively. Design according to LRFD is given by Ru ≤ φRn (AISC B3-1) where the required strength, Ru is another term for the load effect, Q, Rn is the nominal strength and ϕ is the resistance factor. The reliability of design is determined when the required strength is exactly equal to the available strength. Thus, Equation B3-1 can be rewritten as Q = φRn . The load effect will depend on the load combination being considered. Thus, for the LRFD live load plus dead load combination, written in terms of the live-to-dead load ratio, L/D, Q = 1.2 D + 1.6 L = (1.2 + 1.6 ( L D ) ) D = φRn

(1)

From Ravindra and Galambos1 the mean resistance is given by

Rm = Rn M m Fm Pm

(2)

and the coefficient of variation of the resistance is given by

VR = Vm2 + VF2 + VP2

(3)

Mm is the mean of the ratio of the actual yield stress to the specified yield stress and VM is the coefficient of variation; Fm is the mean of the ratio of the actual section property to the Manual value and VF is the coefficient of variation; and Pm is the mean of the ratio of the test specimen strength to the predicted strength using the Specification equations and the actual material and geometric properties and VP is the coefficient of variation. Solving Equation 1 for Rn and substituting into Equation 2 yields Rm =

(1.2 + 1.6 ( L D ) ) D M φ

m

Fm Pm

(4)

Rearranging Equation 2 yields

M m Fm Pm =

Rm Rn

(5)

Thus, combining Equations 4 and 5 gives 1

Ravindra, M.K. and Galambos, T.V. (1978), “Load and Resistance Factor Design for Steel,” Journal of the Structural Division, American Society of Civil Engineers, Vol. 104, No. ST9, September, pp. 1,337–1,353.

24 Chapter 1

Introduction ⎛ R ⎞ Rm = D (1.2 + 1.6 ( L D ) ) ⎜ m ⎟ ⎝ φRn ⎠

(6)

From Ravindra and Galambos the mean load effect for dead load plus live load is

Qm = Dm + Lm

(7)

which can, with some manipulation, be rewritten as Qm = Dm + Lm = ( Dm D + ( Lm L )( L D ) ) D

(8)

They also give the coefficient of variation of the load effect which can be written as a function of the live-to-dead load ratio as

VQ =

(( D

D )VD ) + ( ( Lm L )( L D )VL ) 2

m

2

Qm

(9)

If Equations 6 and 8 are substituted into Equation C-B3-2 the reliability index, β, will be given in terms of the live-to-dead load ratio, L D , and the resistance factor, ϕ. Thus, β=

⎡R ln ⎢ m VR2 + VQ2 ⎢⎣ φRn 1

⎛ ⎞⎤ 1.2 + 1.6 ( L D ) ⎜⎜ ⎟⎟ ⎥ ⎝ ( Dm D ) + ( Lm L )( L D ) ⎠ ⎥⎦

(10)

For LRFD, the live plus dead load combinations are 1.4D and (1.2D + 1.6L). The effective dead load factor as a function of the live-to-dead-load ratio can be taken as

1.4 ⎡ ⎤ γ LRFDi = max ⎢ ⎥ ⎣1.2 + 1.6 ( L D )i ⎦

(11)

and the mean load effect dead load multiplier as

Qmi = Dm D + ( Lm L )( L D )i

(12)

Thus, Equation 10 can be generalized to address other LRFD load combinations as follows: ⎡R ⎛ ⎞⎤ ⎡ R ⎛γ 1.2 + 1.6 ( L D ) ⎞⎤ 1 1 ln ⎢ m ⎜⎜ ln ⎢ m ⎜ LRFDi ⎟ ⎥ β= (13) ⎟⎟ ⎥ = 2 2 2 2 VR + VQ ⎢⎣ φRn ⎝ ( Dm D ) + ( Lm L )( L D ) ⎠ ⎥⎦ VR + VQi ⎣ φRn ⎝ Qmi ⎠ ⎦

where γ LRFDi is the effective LRFD load factor for the load combination under consideration, Qmi is the mean load effect multiplier for that load combination, and VQ is the coefficient of variation of the load effect, all as a function of the varying load ratio as indicated by the subscript i. To convert Equation 13 for use with ASD load combinations, γ LRFDi is replaced by γ ASDi and φRn is replaced by Rn Ω . Thus, Equation 13 becomes i

Introduction

β=

⎡ Rm ⎛ γ ASDi ln ⎢ ⎜ ⎢⎣ ( Rn Ω ) ⎝ Qmi V +V 1

2 R

2 Qi

⎞⎤ ⎟⎥ ⎠ ⎥⎦

Chapter 1 25

(14)

Based on extensive studies for A992 steel (Bartlett et al.2) and the original work for the development of the 1986 AISC Specification by Ravindra and Galambos, the following values can be used: Mm = 1.055; VM = 0.058 Fm = 1.00; VF = 0.05 Pm = 1.02; VP = 0.06 Thus, Rm = Rn (1.055 )(1.00 )(1.02 ) = 1.076 Rn VR =

( 0.058)

2

+ ( 0.05 ) + ( 0.06 ) = 0.097 2

2

Based on Galambos et al.3 (1982), the ratio of mean to code specified dead and live loads can be taken as, for dead load Dm/D = 1.05; VD = 0.10 and for live load Lm/L = 1.00; VL = 0.25 The values for Qmi and VQi will be functions of the live-to-dead load ratio. The reliability index, β, based on Equations 13 and 14, for a live-to-dead load ratio from 1.0 to 5.7 is presented in Figure 1.18 for a compact wide-flange beam under uniform moment for both LRFD and ASD. The figure is based on the load combination of live load plus dead load and statistical variations consistent with those used in the development of the Specification. It is seen that the reliability of design by LRFD is somewhat more uniform for this condition than design by ASD and that at a live to dead load ratio of approximately 3, the two approaches yield the same reliability. The higher the reliability index is, the safer the structure. Regardless of the numerical value of β, any structure that meets the requirements of the Specification will be sufficiently safe. A more detailed discussion of the statistical basis of steel design is available in Load and Resistance Factor Design of Steel Structures.4 Since the introduction of the 2005 AISC Specification, design by ASD and LRFD have essentially been equivalent and differ only by the effect of load combinations.

2

Bartlett, R.M., Dexter, R.J., Graeser, M.D., Jelinek, J.J., Schmidt, B.J. and Galambos, T.V. (2003), “Updating Standard Shape Material Properties Database for Design and Reliability,” Engineering Journal, American Institute of Steel Construction, Vol. 40, No. 1, pp. 2–14. 3 Galambos, T.V., Ellingwood, B., MacGregor, J.G. and Cornell, C.A. (1982), “Probability-Based Load Criteria: Assessment of Current Design Practice,” Journal of the Structural Division, American Society of Civil Engineers, Vol. 108, No. ST5, May, pp. 959–977. 4 Geschwindner, L. F., Disque, R. O., and Bjorhovde, R. Load and Resistance Factor Design of Steel Structures. Englewood Cliffs, NJ: Prentice Hall, 1994.

26 Chapter 1

Introduction 4.0

Reliability Index, β

3.5 3.0 2.5 2.0 1.5 ASD LRFD

1.0 0.5 0.0 0

2 4 6 Live-to-Dead Load Ratio, L/D Figure 1.18 Reliability Index vs Live-to-Dead Load Ratio for Compact Simply Supported Wide-Flange Beams with Uniform Moment

General structural integrity requires a continuous load path to the ground for resisting all gravity and lateral loads that might be applied to the structure. With the introduction of the 2016 AISC Specification, provisions that address structural integrity beyond these general requirements have been introduced. The requirements in Section B3.9 are beyond normal strength requirements and are intended to improve the connectivity of the structure and thus the performance of the structure under undefined extraordinary events. These requirements apply only to a small set of structures where additional structural integrity is mandated.

1.13

LIMIT STATES Regardless of the design approach, ASD or LRFD, or the period in history of the design’s execution, 1923 or 2018, all design is based on the ability of a structure or its elements to resist load. This ability is directly related to how an element carries that load and how it might be expected to fail, which is referred to as the element’s limit state. Each structural element can have multiple limit states, and the designer is required to determine which of these limit states will actually limit the structure’s strength. There are two types of limit states to be considered: strength limit states and serviceability limit states. Strength limit states are those limiting conditions that, if exceeded, will lead to collapse of the structure or a portion of the structure, or to such serious deformations that the structure can no longer be expected to resist the applied load. Strength limit states are identified by the Specification, and guidance is provided for determination of the nominal strength, Rn, the safety factor, Ω, and the resistance factor, φ. Examples of the more common strength limit states found in the Specification are yielding, rupture, and buckling. Serviceability limit states are not as well defined as strength limit states. If a serviceability limit state is exceeded, it usually means that the structure has reached some performance level that someone would find objectionable. The Specification addresses design for serviceability in Chapter L and defines serviceability in Section L1 as “a state in which the function of a building, its appearance, maintainability, durability, and the comfort of its occupants are preserved under typical usage.” Chapter L lists deflections, drift, vibration, wind-induced motion, thermal expansion and contraction, and connection slip as items to be considered, although no specific limitations are given for any of these limit states.

Introduction

Chapter 1 27

Strength and serviceability limit states will be addressed throughout this book as appropriate for the elements or systems being considered.

1.14 BUILDING CODES AND DESIGN SPECIFICATIONS The design of building structures is regulated by a number of official, legal documents that are known commonly as building codes. These cover all aspects of the design, construction, and operation of buildings and are not limited to just the structural design aspects. The model code currently in use in the United States is the ICC International Building Code. Model codes are published by private organizations and have been adopted, in whole or in part, by state and local governments as the legal requirements for buildings within their area of jurisdiction. In addition to the model codes, cities and other governmental entities have written their own local building codes. Unfortunately, since the adoption of a building code is in great part a political activity, the regulations in use across the country are not uniform. A new International Building Code is published every 3 years but not adopted as quickly as issued. Thus, building codes with effective dates from 2003 to 2015 are still in use. In addition, governmental bodies will often adopt a model code with local amendments. Because of the technical nature of the AISC Specification, local amendments normally do not affect those aspects of steel design but they often do modify the loading definitions and thus do ultimately affect steel design. To the structural engineer, the most important sections of a building code deal with the loads that must be used in the design, and the requirements pertaining to the use of specific structural materials. The load magnitudes are normally taken from Minimum Design Loads for Buildings and Other Structures, a national standard published by the American Society of Civil Engineers (Structural Engineering Institute) as ASCE/SEI 7. The loads presented in ASCE/SEI 7 may be altered by the model code authority or the local building authority upon adoption, although this practice adds complexity for designers who may be called upon to design structures in numerous locations under different political entities. Throughout this book, ASCE/SEI 7 will be referred to simply as ASCE 7 as it is most commonly referred to in the profession. The AISC Specification is incorporated into the model building code by reference. The Specification, therefore, becomes part of the code, and thus part of the legal requirements of any locality where the model code is adopted. Locally written building codes also exist and the AISC Specification is normally adopted within those codes by reference also. Through these adoptions the AISC Specification becomes the legally binding standard by which all structural steel buildings must be designed. However, regardless of the Specification rules, it is always the responsibility of the engineer to ensure that their structure can carry the intended loads safely, without endangering the occupants.

1.15

INTEGRATED DESIGN PROJECT This section introduces a building to be used in subsequent chapters of this book as an integrated design project. It is a relatively open-ended design project in that only a limited set of design parameters are set at this point. Several options will be presented in subsequent chapters so that the project can be tailored at the desire of the instructor. The building is a four-story office building with one story below grade. It is located in Downers Grove, Illinois, at approximately 42°N latitude and 88°W longitude. This is a 102,000 ft2 building with approximately 25,500 ft2 per above-grade floor. For the first three floors, the floor-to-floor height is 13 ft 6 in. For the top floor, the floor-to-roof height is 14 ft 6 in. The below-grade floor-to-floor height is 15 ft 6 in. The façade is a lightweight metal curtain wall that extends 2.0 ft above the roof surface, and there is a 6.0 ft screen wall around the middle bay at the roof to conceal mechanical equipment and roof access. All steel will receive spray-applied fireproofing as necessary.

288 Chapter 1

Introductio on Based on prelimiinary discussiions with the architecturall design team m, the design w will start with bay sizzes of 30.0 ftt in the east-w west directionn and 45.0 ft,, 30.0 ft, andd 45.0 ft in thhe northsouth directiion, as shown n in Figure 1.19. Represenntative floor aand roof fram ming plans aree shown in Figures 1.20 1 through 1.22. To acco ommodate a two-story atrrium on the ffirst floor, thee second floor framin ng plan shows an opening bounded by column liness A, C, 4, annd 5. The lateeral load resisting sysstem consists of a pair off three-bay m moment frames in the east--west directioon and a pair of one-b bay chevron braced b framess in the north--south directiion.

Figure 1.19 1 Schemattic Plan for In ntegrated Design Problem P

F Figure 1.20 R Representativve Second-Flooor F Framing Plann

ural analysis and design i s accomplishhed through tthe use of inntegrated Much of today’s structu d design softtware. Figure 1.23 shows an example of a compleete three-dim mensional analysis and model of thee given prelim minary framin ng system deeveloped using RAM Strucctural System m. Figure 1.24 shows the results of o the same computer c moodel with thee gravity-onlyy structural eelements removed.

F Figure 1.21 Representativ R e First-, Third d-, and F Fourth-Floor Framing Plan n

Figu ure 1.22 Reprresentative Rooof Framing Plann

Introdduction

Figure 1.2 23 Three-Dim mensional Com mputer Modell of Compleete Structure from f RAM Sttructural System

11.16 PROBLEMS 11. Where could d one find information about the provisionss oof the 1961 AIS SC Specificatio on? 22. What resourrce would be most likely to o assist in thee ddetermination of o properties of o a steel mem mber found in a bbuilding built in n 1954? 33. Which chaapter of the AISC A Specifica ation providess innformation abo out: a. b. c. d. e.

geeneral requirem ments for analy ysis and design n deesign of memb bers for flexure deesign of connections deesign of memb bers for combin ned forces and d teension reequirements fo or design of structures to o en nsure stability

44. In the AISC Steel Construcction Manual, where can onee ffind: a. th he AISC Speciffication b. deesign consideraations for boltss c. diimensions and d properties for structurall stteel shapes d. deesign of comprression membeers 55. List and deffine the three basic b goals of a design team m ffor the design of o any building g. 66. All structurees are composed of some or all a of five basicc sstructural typ pes. List theese five basic structurall ccomponents and provide an ex xample of each h.

Chhapter 1 29

Figurre 1.24 Threee-Dimensionaal Computer M Model Show wing Only Latteral Load Reesisting System ms from RAM M Structural Syystem mple of each oof the followinng types of 7. Prrovide an exam consttruction. To tthe extent poossible, identiffy specific builddings in your ow wn locale. a. b. c. d. e.

Bearinng wall Beam m-and-column Long--span High--rise Gablee-frame

8. W What type of structural system uses the combined propeerties of two oor more differrent types of m materials to resistt the applied looads? 9. Liist and describbe two types of lateral loaad resisting systeems commoonly used in beam-aand-column consttruction. 10. IIn designing a steel structture, what muust be the primaary concern off the design enggineer? 11. P Provide a simpple definition oof structural dessign. 12. D Describe the ddifference betw ween a strengthh limit state of a sstructure and a serviceability limit state. 13. G Give a descripttion of both thhe LRFD and A ASD design approoaches. What is the fundam mental differencce between the m methods? 14. P Provide a brieff description off plastic designn (PD). 15. IIdentify three ssources of variiation in the strrength of a struccture and its com mponents. 16. P Provide three eexamples of strrength limit staates. Provide three eexamples of seerviceability lim mit states. 17. P

C Chapterr 2

L Loads,, Load d F Factorrs, and d L Load C Combiination ns

Septem mber 11 Memorrial Pavilion, N New York Coopyright Buro--Happold

2.11

INTROD DUCTION Material deesign specificcations, like the AISC SSpecification,, do not norrmally prescrribe the magnitudes of loads thatt are to be ussed as the baasis for designn. These loadds vary basedd on the usage or typ pe of occupan ncy of the bu uilding, and thheir magnituddes are dictatted by the appplicable local, region nal, or state laaws, as prescrribed through the relevant bbuilding codee. Buillding code lo oads are given n as nominal values. Thesse values are to be used inn design, even though h it is well known k that th he actual loaad magnitudee will differ from these sspecified values. Thiss is a commo on usage of th he term nomiinal, the same as will be uused for the nominal depth of a stteel member, to be discusssed later. Theese nominal vvalues are dettermined on tthe basis of material dimensions d and a densities for dead loadd, load surveyys for live loads, weather data for rain, snow and a wind load ds, and geolo ogical data forr earthquake or seismic looads. These looads are further described in Secttion 2.2. To be b reasonablyy certain that these loads aare not exceeded in a given structu ure, code loaad values hav ve generally bbeen higher th than the actuaal loads on a random structure at an arbitrary point p in time. This somew what higher looad level also accounts forr the fact that all strucctural loads will w exhibit som me random v ariations as a function of ttime and loadd type. To properly p addrress this rand dom variationn of load, an aanalysis refleecting time annd space interdependeence should be b used. This is called a stoochastic analy lysis. Many sttudies have deealt with this highly complex c phen nomenon, esp pecially as it ppertains to livve load in buiildings. Howeever, the use of timee-dependent loads l is cum mbersome andd does not aadd significanntly to the safety or economy off the final design. d For most m design ssituations thee building coode will speecify the magnitude of o the loads as a if they werre constant o r unchangingg. Their time and space vaariations are accounteed for throug gh the use off the maximuum load occuurring over a certain referrence or return period. As an exam mple, the Am merican live l oad criteria aare based on a reference pperiod of 50 years, wh hereas the Can nadian criteriia use a 30-yeear interval. The geographicaal location off a structure pplays an impoortant role foor several loaad types, such as tho ose from snow w, wind, or earthquake. These loads also are speecified based upon a reference peeriod appropriiate for the ty ype of load.

300

Chapter 2

2.2

Loads, Load Factors and Load Combinations 31

BUILDING LOAD SOURCES Many types of loads may act on a building structure at one time or another, and detailed data for each are given later. Loads of primary concern to the building designer include: 1. 2. 3. 4. 5. 6.

Dead load Live load Snow load Wind load Seismic load Special loads

These primary load types are characterized as to their magnitude and variability by the building code, and are described in the ensuing paragraphs. 2.2.1

Dead Load Theoretically, the dead load of a structure remains constant throughout its lifespan, unless modifications are made. The dead load includes the self-weight of the structure, as well as the weight of any permanent construction materials, such as stay-in-place formwork, partitions, floor and ceiling materials, machinery, and other equipment. The dead load may vary from the magnitude used in the design, even in cases where actual element weights are accurately calculated. The weight of all dead load elements can be exactly determined only by actually weighing and/or measuring the various pieces that compose the structure. This is almost always an impractical solution and the designer therefore usually relies on published data of building material properties to obtain the nominal dead loads to be used in design. These data can be found in such publications as ASCE 7, the model building codes, and product literature. Some variation will thus likely occur in the real structure. Similarly, differences are bound to occur between the weights of otherwise identical structures, representing another source of dead load variability. However, compared to other structural loads, dead load variations are relatively small and the actual mean values are quite close to the published data.

2.2.2

Live Load Live load is the load on the structure that results from all of the non-permanent installations. It includes the weight of the occupants, the furniture and moveable equipment, plus anything else that the designer could possibly anticipate might occur in the structure. The fluctuations in live load are potentially quite substantial. They vary from being essentially zero immediately before the occupants take possession to a maximum value at some arbitrary point in time during the life of the structure. The magnitude of the live load to be used in a design is obtained from the appropriate building code. The actual live load on the structure at any given time may differ significantly from that specified by the building code. This is one reason why numerous attempts have been made to model live load and its variation and why measurements in actual buildings continue to be made. Although the nominal live loads found in modern building codes have not changed much over the years, the actual use of buildings has, and load surveys continue to show that the specified load levels are still an adequate representation of the loads the structure should be designed to resist. The actual live load on a structure at any given point in time is called the arbitrary pointin-time live load (Lapt). Figure 2.1 shows the variation of the live load on a structure as might be obtained from a live load survey. The load specified by the building code is always higher than

322 Chapter 2

Loads, Loaad Factors and d Load Comb binations

Figure 2.1 V Variation of L Live me Load with Tim the actual load found in n the building survey. In addition, a portion of thhe live load remains constant. Th his load comees from the reelatively perm manent fixturees and furnishhings associaated with a particular occupancy and a can be reeferred to as the sustainedd live load (S SLL). The occcupants who enter an nd leave the space s form an nother part off the live loadd, raising andd lowering thee overall live load maagnitude with time. This vaarying live loaad is called a transient livee load (TLL).. 2.22.3

Snow Load L now load miight be consiidered a form m of live loaad, unique cconditions goovern its Although sn magnitude and a distributio on and it is trreated separattely in the loaad combinatioons used in ddesign. It is the primaary roof load d in many geographical g areas and heeavily depennds on local climate, building exp posure, and bu uilding geometry. Snow load data are a normally based on surrveys that ressult in isoline maps showinng areas of equal deepth of grou und snow. Using U this m method, annuaal extreme ssnowfalls havve been determined over a perio od of many years. Thesee data have bbeen analyzeed through sttatistical models and the expected d lifetime max ximum snow loads estimaated. The refeerence period is again the 50-year anticipated life of the struccture. A major m difficultty is encounttered in transslating the groound snow looad into a roof snow load. This iss accomplisheed through a semi-empiriccal relationshhip whereby thhe ground snnow load is multiplieed by factorss to account for propertties such as roof geomettry and the thermal characteristiics of the roo of. In additio on to the unifform snow looad determinned using thee ground snow load, drifts often lead to areass of roofs w with significanntly higher loads which m must be or. Work con ntinues to be done d to improove the methhod of snow lload computaation and accounted fo to collect sn nowfall data.

2.22.4

Wind Load L By its very nature, wind d is a highly dynamic nattural phenom menon. For thhis reason it iis also a complex pro oblem from a structural peerspective. W Wind forces fl fluctuate signiificantly and are also influenced by b the geomettry of the stru ucture, includiing the heighht, width, deptth, plan and eelevation shape, and the t surroundin ng landscape. The basic b uilding code approach to wind load annalysis is to treat wind d as a static load l problem m, using the B Bernoulli equuation to transslate wind sppeed into wind pressu ure. In an app proach similaar to that use d for snow load, a semi-eempirical equuation is used to givee the wind lo oad at certain n levels as a function of a number off factors representing effects of features such ass wind gusts, topography, and structuraal geometry. The data used for determin ning wind l oads are baased on meaasured wind speeds. Meteorologiical data for 3-second win nd speed gustts have been accumulatedd over the conntiguous United Statees and corrected to a stand dard height oof 33 ft. Thesse data are theen used to m model the long-term characteristicss for a mean recurrence iinterval of 300, 700, 17000 or 3000 yyears, as

Chapter 2

Loads, Load Factors and Load Combinations 33

required. ASCE 7 and the model codes provide maps to be used as the foundation of wind force calculations. Because local site characteristics often dictate wind behavior, there are locations for which special attention must be given to wind load calculation. In addition, some buildings require (or can benefit from) special attention in determining wind load magnitude. In these cases it might be valuable to conduct wind tunnel tests before the structural design for wind is carried out. 2.2.5

Seismic Load The treatment of seismic load effects is extremely complicated because of the high variability of this natural phenomenon and the many factors that influence the impact of an earthquake on any particular structure. In addition, because the force the building is subject to is the result of the reaction of the mass of the building responding to the ground moving, inertia effects must be considered. For most buildings it is sufficient to treat seismic effects through the use of equivalent static loads over the height of the building, provided that the magnitudes of these equivalent static loads properly reflect the dynamic characteristics of the seismic event. Many characteristics of the problem must be quantified in order to establish the correct magnitudes of these static loads. These characteristics include such factors as the ground motion and response spectra for the seismic event and the structural and site characteristics for the specific project. Meanwhile, the extension of seismic design requirements to all areas of the country through the current model building codes is making seismic design compulsory for many more structures. Earthquakes are no longer a design consideration only on the West coast.

2.2.6

Special Loads Several other loads are sometimes important. These include impact, blast, and thermal effects. Impact: Most building loads are static or essentially so, meaning that their rate of application is so slow that the kinetic energy associated with their motion is insignificant. For example, a person entering a room is actually exerting a dynamic load on the structure by virtue of his or her motion. However, because of the relatively small mass and slow movement of the individual, the kinetic energy is essentially zero. When loads are large and/or their rate of application is very high, the influence of the energy brought to bear on the structure as the movement of the load is suddenly restrained must be taken into account. This phenomenon, known as impact, occurs as the kinetic energy of the moving mass is translated into a load on the structure. Depending on the rate of application, the effect of the impact is that the structure experiences a load that may be as large as twice the static value of the same mass. Impact is of particular importance for structures where machinery processes and similar actions occur. Cranes, elevators, and equipment such as repetitive-action industrial machinery could all produce impact loads that would need to be considered in a design. In addition, vibrations may be induced into a structure either by these high-magnitude impact loads or by the normally occurring occupancy loads. Although normal live load occupancy, such as walking, is not likely to produce increased design load magnitudes, the potential for vibration from these activities may require consideration in the design. Additional guidance may be found in AISC Design Guide 11, Vibrations of Steel Framed Structural Systems Due to Human Activity. Blast: Blast effects on buildings have become a more important design consideration during the first years of the twenty-first century. Prior to that time, blast effects were primarily considered to be accidental. These types of blast do not occur as often as impact for normal structures, but should be considered under certain circumstances. Many structures designed for industrial

34 Chapter 2

Loads, Load Factors and Load Combinations installations, where products of a volatile nature are manufactured, are designed with resistance to blast as a design consideration. When the structure is called upon to resist the effects of blast, a great deal of effort must be devoted to determining the magnitude of the blast to be resisted. The threat of terrorism has been increasingly recognized since the attacks on the World Trade Center and Pentagon on September 11, 2001. In order to take that threat into account, owners must determine the level of threat to be designed for and design engineers must establish the extent to which a particular threat will influence the design of a particular structure. Generally speaking, analysis and design data for blast effects are somewhat limited. Researchers continue to work toward establishing design guidelines that help determine blast effects and member strength in response to blast. Guidance may be found in AISC Design Guide 26, Design of Blast Resistant Structures and AISC Facts for Steel Buildings 2, Blast and Progressive Collapse. Thermal Effects: Steel expands or contracts under changing temperatures and in so doing may exert considerable forces on the structure if the members are restrained from moving. For most building structures, the thermal effects are less significant than other loads for structural strength. Because the movement of the structure results from the total temperature change and is directly proportional to the length of the member experiencing the change, the use of expansion joints becomes important in long dimensions of structural framing. When expansion or contraction is not permitted, the resulting forces must be accommodated in the members. Additional guidance may be found in AISC Design Guide 3, Serviceability Design Considerations for Steel Buildings and AISC Design Guide 7, Industrial Buildings: Roofs to Anchor Rods. The AISC Specification includes guidance on the design of steel structures exposed to fire. Appendix 4 provides criteria for the design and evaluation of structural steel components, systems, and frames for fire conditions. In the current building design environment, design for fire is usually accomplished by means of a prescriptive approach defined in the Specification as design by qualification testing. If the actual thermal effects of a fire are to be addressed, the Specification permits design by engineering analysis. Additional guidance may be found in AISC Design Guide 19, Fire Resistance of Structural Steel Framing.

2.3 BUILDING LOAD DETERMINATION Once the appropriate building load sources are identified, their magnitudes must be determined. Methods to determine these magnitudes are set by the applicable building code for each load source. The following sections provide general guidance to determine the building load magnitudes, but for specific details, the applicable building code must be consulted. 2.3.1

Dead Load Building dead load determination can be either quite straightforward or very complex. If the sizes of all elements of the structural system are known before an analysis is conducted, actual material weights may be determined and applied in the structural analysis. Selected unit weights of typical building materials are given in Table 2.1. Manual Table 17-13 and ASCE 7 Table C3-1 provide the weights of building materials, and product catalogs provide weights of things such as building mechanical equipment. If the final sizes are not known, as is normally the case in the early stages of design, assumptions need to be made to estimate the self-weight of the structure. This necessitates an iterative process of refinement as the design and its corresponding weight are brought together.

Chapter 2 Table 2.1

Unit Weights of Typical Building Materials Weight (lb/ft3)

Material Aluminum Brick Concrete Reinforced, with stone aggregate Block, 60 percent void Steel, rolled Wood Fir Plywood 2.3.2

Loads, Load Factors and Load Combinations 35

170 120 150 87 490 32–44 36

Live Load As discussed earlier, live load magnitudes are established by the applicable building code. Table 2.2 provides values for the minimum uniformly distributed live loads for buildings for selected occupancies. In design, how much of the load is supported by a particular element can be determined by multiplying the uniformly distributed load (lb/ft2) by the tributary area, AT (ft2). The tributary area method is a simplified approach for visualizing the load on a structural element without performing the actual equilibrium calculations. It does, however, provide the same result because it is fundamentally based on an equilibrium analysis. Simplified tributary areas for some structural members are given in Figure 2.2. Although the concept of the tributary area can be used to determine the load on a member, an equally important concept is the influence area, AI. The influence area is significant because it reflects the area over which any applied load has an influence on the member of interest. The member under consideration will experience no portion of the load applied outside of the influence area. Table 2.3 provides the relationship between tributary area and influence area, as defined by ASCE 7, for several specific structural elements, where AI = KLLAT. Several of the values in this table are simplified relative to the actual relationship, and it is always permissible to calculate the actual influence area. As the influence area increases for a particular member, the likelihood of the full codespecified nominal live load actually occurring on it decreases. Because the code cannot predict the likelihood of that full area being loaded, the magnitude of the specified load is set without consideration of loaded area. Thus, the tabulated values are referred to as the unreduced nominal Table 2.2

Minimum Uniformly Distributed Live Loads for Building Designa

Occupancy or use Roof, flat Residential dwellings, apartments, hotel rooms, school classrooms Offices Corridors above first floor Auditoriums (fixed seats) Retail stores Bleachers Library stacks Heavy manufacturing and warehouses a Data are taken from ASCE 7.

Load (lb/ft2) 20 40 50 80 60 75–100 100 150 250

366 Chapter 2

Loads, Loaad Factors and d Load Comb binations

Figu ure 2.2

Sim mplified Tribu utary Areas foor Some Strucctural Membeers

live load. To T account fo or the size of the influennce area and thereby provvide a more realistic predictor off the actual live load on th he structure, a live load reeduction facttor is introducced. For 2 influence arreas greater than t 400 ft the live loadd may be redduced accordding to the liive load reduction eq quation: ⎛ 15 ⎞ L = Lo ⎜⎜ 0.25 + (2.1) ⎟ AI ⎟⎠ ⎝ Table 2.3 Live Load Ellement Factorr, KLLa Element Interior colu umns Exterior colu umns withoutt cantilever sllabs Edge columns with cantillever slabs Corner colum mns with can ntilever slabs Edge beamss without canttilever slabs Interior beam ms All other meembers not id dentified abov ve, including Edge beeams with can ntilever slabs Cantilev ver beams One-way slabs Two-waay slabs Memberrs without pro ovisions for continuous c sheear tran nsfer normal to o their span Note: In lieu u of the values above, calcu ulation of KLLL is permittedd. a Data are tak ken from ASC CE 7.

KLL 4 4 3 2 2 2 1

Loads, Load Factors and Load Combinations 37

Chapter 2 where L = reduced live load Lo = code-specified nominal live load AI = influence area = KLLAT

Limitations on the use of this live load reduction are spelled out in ASCE 7. The most important one is that L may not be taken less than 0.50Lo for members supporting one floor nor less than 0.40Lo for members supporting more than one floor. Other limitations and exceptions are given for loads over 100 lb/ft2, passenger vehicle garages and assembly buildings. 2.3.3

Snow Load Roof snow load calculations start with determination of the ground snow load for the building site. Table 2.4 provides typical ground snow load values for selected locations. The complete picture of ground snow load is provided in the appropriate building code. In many locations, however, the snowfall depth is a very localized phenomenon and the variability is such that it is not appropriate to map those values. In these situations, local building officials should be consulted to determine what the local requirements are. The determination of roof snow load is more complex and there are many acceptable approaches. Roof snow load on an unobstructed flat roof, as given in ASCE 7, is pf = 0.7CeCtIspg (2.2) where pg is the ground snow load determined from the appropriate map, as shown in Figure 2.3. Ground snow load by location may also be found using the website http://snowload.atcouncil.org/. The exposure factor, Ce, varies from a low of 0.7 for a windswept site above the tree line to a high of 1.2 for a sheltered site. The thermal factor, Ct, accounts for the melting effect of heat escaping through the roof. It varies from 0.85 for continuously heated greenhouses to 1.3 for freezer buildings. The importance factor for snow, Is, varies from 0.8 for buildings that represent a low risk to human life, risk category I, to a high of 1.2 for buildings designated as essential, risk category IV. Buildings of normal risk fall into risk category II where the snow importance factor is 1.0. Numerous other factors enter into the determination of roof snow load, including roof slope, roof configuration, snowdrift, and additional load due to rain on the snow. The applicable building code or ASCE 7 should be referred to for the complete provisions regarding snow load determination. Table 2.4

Typical Ground Snow Loads, pga

Location Whittier, Alaska Talkeetna, Alaska Portland, Maine Minneapolis, Minnesota Hartford, Connecticut Chicago, Illinois St. Louis, Missouri Raleigh, North Carolina Memphis, Tennessee Atlanta, Georgia a Data are taken from ASCE 7.

Load (lb/ft2) 300 120 60 50 30 25 20 15 10 5

388 Chapter 2

Loads, Loaad Factors and d Load Comb binations

Figurre 2.3 2.33.4

Groun nd Snow Load d Map from A ASCE 7 (Usedd with permisssion from ASC CE)

Wind Load L As with sno ow load and other geograaphically linkked environm mental loads, tthe starting ppoint for wind load calculation is map based. In the case off wind loads, maps are proovided based upon 3second gust wind velocitties in the buiilding code (ssee Figure 2.44). Different wind speed m maps are provided forr the differen nt risk categorries of buildiings. The diffferent maps take the placce of the importance factor used in n the determin nation of som me other desiggn loads such as was illustrrated for snow. Thus there are fou ur basic wind speed maps ffor the four rrisk categories. Table 2.5 pprovides the wind speeed data for several s selectted locations with varyingg wind velociities for risk ccategory II. Wind speeed by locatio on may also be found usingg the website http://windsppeed.atcounciil.org/. These data must be transform med into wind pressures onn a given buildding to determ mine the appropriate design wind d loads. This transformatioon must takee into accounnt, in additionn to the map, such faactors as importance of the buildiing included through seleection of the appropriate m height abov ve the ground d, relative sh heltering of tthe site, topoography, and the directionn of the dominant wiinds. ASCE 7 offers seeveral approacches to the d etermination of design wiind load for tthe main wind force resisting r systeem (MWFRS S). The directtional proceduure found in ASCE 7, Chhapter 27 for building gs of all heigh hts, where thee load must bbe applied too the windwarrd, leeward, aand side walls converrts the mappeed data to velo ocity pressuree at height z: qz = 0.00256 0 K z K zzt K d K eV 2 (2.3) where V is the wind sp peed obtained d from the aappropriate rrisk category map. The eexposure coefficient, Kz, ranges from f 0.57 at grade for thhe least criticcal exposure to 1.89 at a 500 ft elevation forr the most criitical exposure. The topogrraphic factor, Kzt, is a calcuulated factor w which is intended to account for th he wind speed d-up in locatiions of a hill, ridge or escaarpment. For all other locations, Kzt = 1.0. The directionality y coefficient, Kd, for the M MWFRS is taaken as 0.85 for both buildings an nd componentts and claddin ng. For other structural typpes it may vaary up to a m maximum of 1.0. The ground elev vation factor,, Ke, is intennded accountt for air denssity at the siite. It is permitted to o take Ke =1.0 0 for all locattions. For a siite with grounnd elevation at 6,000 ft abbove sea level, Ke =0..80, its lowestt value.

Chapter 2

Loads, Loadd Factors and Load Combinnations 39

Figure 2.4 2 Basic Wind W Speed, 3--Second Gustt at 33 ft Abovve Ground foor Risk Categoory II Exposuree Category C from ASCE 7 (Used with PPermission from m ASCE) Once O the velo ocity pressuree is determinned through E Equation 2.3, it must be converted to the extern nal design wiind pressure. For F the MWF FRS, this is giiven by p = qGCp – qi(GCpi) (2.4) where p = q = qi = G = Cp = GC G pi =

design n wind pressu ure velocity pressure from f Equationn 2.3 as a funnction of wall location and height velocity pressure from f Equationn 2.3 at meann roof height gust-eeffect factor taken t as 0.85 for rigid buildings extern nal pressure coefficient c dettermined as a function of bbuilding geom metry intern nal pressure coefficient takken as ±0.18 for enclosed buildings

The T actual forces applied to the structuure are then ddetermined byy multiplyingg the design wind preessures by thee tributary areeas at each leevel. Wind prressure distribbution on the main wind force resisting system m will vary with height accoording to the variation of tthe exposure coefficient, Kz. The pattern of disstribution is illustrated i in Figure 2.5. T To account foor the fact thaat wind may come fro om any directiion, ASCE 7 requires that the wind loadd be determinned for eight w wind 5 Approxim mate Represen ntative Wind Velocities annd Resulting D Dynamic Presssuresa Table 2.5 Dynam mic pressure (llb/ft2) Wind velo city (mph) Location n V = 0.00256V 2 Miami, Florida F 1770 74.0 Houston,, Texas 1330 43.3 36.9 New Yorrk, New York k 1220 33.9 Chicago,, Illinois 1115 31.0 San Fran ncisco, Califorrnia 1110 a Data aree taken from ASCE A 7 for Risk R Categoryy II.

400 Chapter 2

Loads, Loaad Factors and d Load Comb binations

Figure 22.5 Wind Looad Distributiion over Buillding Height directions at 45° intervaals in addition n to three quuarters of thee load along the principall axis in conjunction with a torsio onal moment. This results in the wind lload being appplied to the sstructure in 12 different direction ns and magn nitudes withoout considerration yet off load combiinations. Because buiilding codes have differen nt requiremennts for wind load determ mination, the ddesigner ASCE 7 must review w the provisions specified in i the governning code. If tthere is no buuilding code, A should be ussed. 2.33.5

Seismicc Load Perhaps thee most rapidlly fluctuating g area of buuilding load ddeterminationn is that for seismic design. Alth hough there have h been maany advances in the use off dynamic annalysis for earrthquake response, co ommon practiice is still to model the phhenomenon uusing a static load. Seismicc design requirementts are based on n the seismic design categgory, A througgh F, with A bbeing the leasst severe and requirin ng no consideeration of seissmic loads. F For a large m majority of buuildings in all seismic risk categories, ASCE 7 permits the equivalent laateral force prrocedure for determinatioon of the building basse shear throu ugh the expresssion V = C sW (2.5) where V = seismic base shear Cs = seismic resp ponse coefficiient W = effective seismic weight of o the buildinng The seismic response coeefficient is dettermined by

Cs =

SDS

( I) R

(2.6a)

e

but for T ≤ TL neeed not be greeater than

Cs =

S D1 T (R / Ie )

(2.6b)

Cs =

S D1TL T (R / Ie )

(2.6c)

and for T > TL 2

Chapter 2

Loads, Loadd Factors and Load Combinnations 41

where SDS = SD1 = T = TL = R = Ie =

design n spectral response accelerration for shoort period design n spectral response accelerration for 1 seec period fundaamental buildiing period long period p transition period response modificattion factor seism mic importancee factor whichh is 1.0 for risk categories I and II

The T U.S. Geo ological Surveey provides a web based toool to determ mine site speciific data for seismic design that precludes p thee need to usee the maps ffound in ASC CE 7. It is aavailable at http://earrthquake.usgss.gov/designm maps/us/appliccation.php. Once O the seism mic base sheaar is determinned it must bee converted to a horizontaal load to be applied to t the seismiic force resissting system. For a buildding with uniiform story hheights and uniform seismic weight distributiion over the height of thhe building, this results in a linear variation n of load from m a maximum at the top to zzero at the boottom as illusttrated in Figuure 2.6. Each E seismic force resistin ng system is aassigned a response modiffication factoor, R, by the building code, accord ding to its ab bility to resisst seismic forrces through a ductile ressponse. The most ducctile steel sysstems, e.g., special s momeent frames, arre assigned R = 8 while less ductile systems are assigned d lower valuees. Higher R values leadd, through Eqquation 2.6, to a lower required base shear and a thus to lo ower lateral lload on the sstructure. Thee resulting loower lateral load, ho owever, leadss to higher demand forr ductility aand corresponnding speciaal detailing requirem ments. In casees where app propriate, thee selection of R = 3 usuually permits a building structure to be design ned according g to the AISC C Specificatioon without use of any speccial seismic provision ns. If a valuee of R greaterr than 3 is uused in designn of a buildinng structure, or in cases where sp pecifically req quired, the design d must pproceed accorrding to the additional prrovisions of ANSI/AIISC 341-16 Seismic Provvisions for Strructural Steeel Buildings. T This is discusssed further in Chapteer 13. As A with the other o environmental loads discussed heere, the details of load determination for seism mic response must be foun nd in the apppropriate buildding code. A ASCE 7 requiires that the seismic load l be applieed to the struccture in such a way as to prroduce the moost critical load effects.

Figurre 2.6 Seism mic Force Disstribution over H Height for Buuilding with U Uniform Story Height and S Story Seismicc Weight

42 Chapter 2

Loads, Load Factors and Load Combinations

2.4 LOAD COMBINATIONS FOR ASD AND LRFD In addition to specifying the load magnitudes for which building structures must be designed, building codes specify how the individually defined loads should be combined to obtain the maximum load effect. Care must be exercised in combining loads to determine the most critical combination because all loads are not likely to be at their maximum magnitude at the same time. For instance, it is unlikely that the maximum snow load and maximum wind load would occur simultaneously. Another unlikely occurrence would be a design earthquake occurring at the same time as the maximum design wind. Thus, building codes specify which loads are to be combined and at what magnitude they should be considered. The designer must exercise judgment when combining loads in situations where the normal expectations of the building code might not be satisfied or where some particular combination would result in a greater demand than previously identified. The two design philosophies addressed in the AISC Specification are the direct result of the two approaches to load combinations presented in current building codes and ASCE 7. ASD uses the load combinations defined in ASCE 7 as being for allowable stress design and LRFD uses the load combinations defined as being for strength design. The provisions in ASCE 7 for allowable stress design combine loads normally at or below their nominal or serviceability levels, the load magnitudes discussed in Section 2.3. These load combinations were historically used to determine the load effect under elastic stress distributions, and those stresses were compared to the allowable stresses established at some arbitrary level below failure, indicated by either the yield stress or the ultimate stress. Including only dead, live, wind, and snow loads, the load combinations presented in ASCE 7 Section 2.4 for ASD can be written as 1. 2. 3. 4. 5. 6. 7.

Dead Dead + Live Dead + Roof Live Dead + 0.75 Live + 0.75 Roof Live Dead + (0.6 Wind) Dead + 0.75 Live + 0.75 (0.6 Wind) + 0.75 Roof Live 0.6 Dead + 0.6 Wind

In load combinations 3, 4, and 6, Roof Live refers to a code-specified roof live load, a snow load, or a rain load, whichever is greater. Used with the current AISC Specification, these load combinations are not restricted to an elastic stress distribution as was the practice in the distant past. Even recent past ASD editions of the AISC Specification had transitioned to use strength past the elastic level in many cases. The current Specification is a strength-based specification, not a stress-based one, and the requirement for elastic stress distribution is no longer applicable. This has no impact on the use of these load combinations but may have some historical significance to those who were educated primarily with this previously used interpretation. The second approach available in ASCE 7 combines loads at an amplified level. These combinations, referred to as strength load combinations, permit one to investigate the ability of the structure to resist loads at its ultimate strength. In this approach, loads are multiplied by a load factor that incorporates both the likelihood of the loads occurring simultaneously at their maximum levels and the margin against which failure of the structure is measured. Again, using only dead, live, wind, and snow loads, the load combinations presented in ASCE 7 Section 2.3, if the live load is not greater than 100 lb/ft2 (psf), for LRFD are

Chapter 2 1. 1 2. 2 3. 3 4. 4 5. 5

Loads, Loadd Factors and Load Combinnations 43

1.4 Dead D 1.2 Dead D + 1.6 Liv ve + 0.5 Rooff Live 1.2 Dead D + 1.6 Roof Live + (0.55 Live or 0.5 Wind) 1.2 Dead D + 1.0 Wiind + 0.5 Livee + 0.5 Roof L Live 0.9 Dead D + 1.0 Wiind

As was the case for ASD load combinations c , Roof Live is intended to be taken as a codespecified d roof live loaad, a snow loaad, or a rain looad, whicheveer is greatest. For F both ASD D and LRFD, load combinnations that innclude wind m must be appliied with the wind loaad acting ind dependently in each of tthe directionns specified in ASCE 7. For those combinattions that include seismic load, that loaad must also be applied inn the directionns specified by ASCE E 7. Load casses with a red duced dead looad in combiination with w wind or seism mic load are used to account a for siituations wheere the appliccation of deadd load might reduce the eeffect of the lateral lo oad. This cou uld be importaant if the actuual dead loadd is less than anticipated. The design method to t be used, an nd thus the ch hoice of load combinationns, is at the diiscretion of thhe designer. All curreent building codes c permit either ASD or LRFD, annd the AISC Specificationn provisions address all a limit statees for each ap pproach. As discussed in Chapter 1, thhe resulting ddesign may differ slig ghtly for each h design philo osophy becauuse the approaach taken to eensure safety is different, but safety y is assured when w the applicable buildiing code and the AISC Specification arre followed, regardlesss of the desig gn approach.

2.5

Load d Calculation ns To underrstand the im mpact of thesee two approacches on analyysis, it is helppful to compuute the load effect forr a variety off structural members m accorrding to bothh ASD and LR RFD load com mbinations. The floorr plan of a tw wo-story officee building is ggiven in Figuure 2.7. Load case 2 for dead plus live load is considered c fo or several beaams and coluumns. The bbuilding is ann office buildding with a nominal live load of 50 5 psf and a calculated deead load of 700 psf. Assum me all beams and girders are simplly supported.

Figure 2.7 2

Floor Plaan of a Two-S Story Office B Building

44 Chapter 2

Loads, Load Factors and Load Combinations

Example 2.1 Girder Load Calculation

Goal:

Determine the required moment strength for a girder from the framing plan shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.

Given:

Girder AB on line 2-2 if the floor deck spans from line 1-1 to 2-2 to 3-3, resulting in a uniformly distributed load on Girder AB:

SOLUTION

Step 1:

Determine the live load reduction from Equation 2.1 Tributary area: AT = (40)(20) = 800 ft2 Influence area: AI = 2AT = 2(800) = 1600 ft2 Live load reduction: 15 15 0.25 + = 0.25 + = 0.625 ≥ 0.50 1600 AI Since this member supports one floor the maximum reduction is 0.50.

For LRFD Step 2:

Determine the required moment strength for the LRFD load combination LRFD loads per foot: 1.2 Dead = 1.2(70 psf)(20 ft) = 1680 lbs/ft 1.6 Live = 1.6(0.625)(50 psf)(20 ft) = 1000 lbs/ft wu = 1.2 Dead + 1.6 Live = 1680 + 1000 = 2680 lbs/ft = 2.68 kips/ft Required moment strength (LRFD), w l 2 2.68(40) 2 Mu = u = = 536 ft-kips 8 8

For ASD Step 2:

Determine the required moment strength for the ASD load combination ASD loads per foot: Dead = (70 psf)(20 ft) = 1400 lbs/ft Live = 0.625(50 psf)(20 ft) = 625 lbs/ft wa = Dead + Live = 1400 + 625 = 2030 lbs/ft = 2.03 kips/ft Required moment strength (ASD), w l 2 2.03(40) 2 Ma = a = = 406 ft-kips 8 8

Example 2.2 Beam Load Calculation

Goal:

Determine the required moment strength for a beam from the framing plan shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.

Given:

Floor beam 2-3 on line D-D if the floor deck spans from line C-C to D-D to E-E, resulting in a uniformly distributed load on floor beam 2-3:

SOLUTION

Step 1:

Determine the live load reduction from Equation 2.1 Tributary area: AT = (20)(30) = 600 ft2

Chapter 2

Loads, Load Factors and Load Combinations 45

Influence area: AI = 2AT = 2(600) = 1200 ft2 Live load reduction: 15 0.25 + = 0.683 ≥ 0.50 1200 Since this member supports one floor the maximum reduction is 0.50. For LRFD Step 2:

Determine the required moment strength for the LRFD load combination LRFD loads per foot: 1.2 Dead = 1.2(70 psf)(30 ft) = 2520 lbs/ft 1.6 Live = 1.6(0.683)(50 psf)(30 ft) = 1640 lbs/ft wu = 1.2 Dead + 1.6 Live = 2520 + 1640 = 4160 lbs/ft = 4.16 kips/ft Required moment strength (LRFD), w l 2 4.16(20) 2 Mu = u = = 208 ft-kips 8 8

For ASD Step 2:

Determine the required moment strength for the ASD load combination ASD loads per foot: Dead = (70 psf)(30 ft) = 2100 lbs/ft Live = 0.683(50 psf)(30 ft) = 1020 lbs/ft wa = Dead + Live = 2100 + 1020 = 3120 lbs/ft = 3.12 kips/ft Required moment strength (ASD), M a =

wa l 2 3.12(20) 2 = = 156 ft-kips 8 8

Example 2.3 Beam Load Calculation

Goal:

Determine the required moment strength for a beam from the framing plan shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.

Given:

Floor beam 2-3 on line C-C if the floor deck spans from line B-B to C-C to D-D, resulting in a uniformly distributed load on floor beam 2-3:

SOLUTION

Step 1:

Determine the live load reduction from Equation 2.1 Tributary area: AT = (20)(30+40)/2 = 700 ft2 Influence area: AI = 2AT = 2(700) = 1400 ft2 Live load reduction: 15 0.25 + = 0.651 ≥ 0.50 1400 Since this member supports one floor the maximum reduction is 0.50.

For LRFD Step 2:

Determine the required moment strength for the LRFD load combination LRFD loads per foot:

46 Chapter 2

Loads, Load Factors and Load Combinations 1.2 Dead = 1.2(70 psf)(30 ft + 40 ft)/2 = 2940 lbs/ft 1.6 Live = 1.6(0.651)(50 psf)(30 ft + 40 ft)/2 = 1820 lbs/ft wu =1.2 Dead + 1.6 Live = 2940 + 1820 = 4760 lbs/ft = 4.76 kips/ft Required moment strength (LRFD), w l 2 4.76(20) 2 Mu = u = = 238 ft-kips 8 8 For ASD Step 2:

Determine the required moment strength for the ASD load combination ASD loads per foot: Dead = (70 psf)(30 ft + 40 ft)/2 = 2450 lbs/ft Live = 0.651(50 psf)(30 ft + 40 ft)/2 = 1140 lbs/ft wa = Dead + Live = 2450 + 1140 = 3590 lbs/ft = 3.59 kips/ft Required moment strength (ASD), w l 2 3.59(20) 2 Ma = a = = 180 ft-kips 8 8

Example 2.4 Column Load Calculation

Goal:

Determine the required axial strength for a column from the framing plan shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.

Given:

Interior column D-2 to support one floor regardless of deck span direction:

SOLUTION

Step 1:

Determine the live load reduction from Equation 2.1 Tributary area: AT = (30)(20) = 600 ft2 Influence area: AI = 4AT = 4(600) = 2400 ft2 Live load reduction: 15 0.25 + = 0.556 ≥ 0.50 2400 Since this member supports one floor the maximum reduction is 0.50.

For LRFD Step 2:

Determine the required axial strength for the LRFD load combination LRFD load entering column at this level: 1.2 Dead = 1.2(70 psf)(600 ft2) = 50,400 lbs 1.6 Live = 1.6(0.556)(50 psf)(600 ft2) = 26,700 lbs 1.2 Dead + 1.6 Live = 50,400 + 26,700 = 77,100 lbs = 77.1 kips

For ASD Step 2:

Required axial strength (LRFD), Pu = 77.1 kips Determine the required axial strength for the ASD load combination ASD load entering column at this level: Dead = (70 psf)(600 ft2) = 42,000 lbs

Chapter 2

Loads, Load Factors and Load Combinations 47

Live = 0.556(50 psf)(600 ft2) = 16,700 lbs Dead + Live = 42,000 + 16,700 = 58,700 lbs = 58.7 kips Required axial strength (ASD), Pa = 58.7 kips

Example 2.5 Column Load Calculation

Goal:

Determine the required axial strength for a column from the framing plan shown in Figure 2.7 due to live load of 50 psf and dead load of 70 psf.

Given:

Exterior column D-4 to support one floor regardless of deck span direction:

SOLUTION

Step 1:

Determine the live load reduction from Equation 2.1 Tributary area: AT = (30)(10) = 300 ft2 Influence area: AI = 4AT = 4(300) = 1200 ft2 Live load reduction: 15 0.25 + = 0.683 ≥ 0.50 1200 Since this member supports one floor the maximum reduction is 0.50.

For LRFD Step 2:

Determine the required axial strength for the LRFD load combination LRFD load entering column at this level: 1.2 Dead = 1.2(70 psf)(300 ft2) = 25,200 lbs 1.6 Live = 1.6(0.683)(50 psf)(300 ft2) = 16,400 lbs 1.2 Dead + 1.6 Live = 25,200 + 16,400 = 41,600 lbs = 41.6 kips

For ASD Step 2:

Required axial strength (LRFD), Pu = 41.6 kips Determine the required axial strength for the ASD load combination ASD load entering column at this level: Dead = (70 psf)(300 ft2) = 21,000 lbs Live = 0.683(50 psf)(300 ft2) = 10,200 lbs Dead + Live = 21,000 + 10,200 = 31,200 lbs = 31.2 kips Required axial strength (ASD), Pa = 31.2 kips

The floor plan of a 10-story office building is given in Figure 2.8. As with the previous examples, load case 2 for dead plus live load is considered for several columns to illustrate the application of live load reduction to members that support loads from two or more floors. The building is to be designed for flexibility so the nominal live load will be 80 psf and the calculated dead load is 100 psf.

488 Chapter 2

Loads, Loaad Factors and d Load Comb binations

Fiigure 2.8 Floor Plan of a 100-Story Office Building

Exxample 2.6 Coolumn Load d Caalculation

Goa al:

Deterrmine the req quired axial sttrength for a column from m the framingg plan shown n in Figure 2..8 due to live load of 80 pssf and dead looad of 100 psff.

Giv ven:

Interior column C--3 supporting one level witth 18K3 steell joists spanniing as shown n in all panelss:

SO OLUTION

Step p 1:

Deterrmine the livee load reductioon from Equaation 2.1 Trributary area pper floor: AT = (30)(24) = 720 ft2 In nfluence area pper floor: AI = 4AT = 4(7200) = 2880 ft2 l reduction n: Live load 15 0..25 + = 0.530 ≥ 0.5 0 2880 Sincee this memberr supports onee floor the maaximum reducction is 0.50.

Forr LRF FD Step p 2:

Deterrmine the requ uired axial strrength for thee LRFD load ccombination LRFD D load enterin ng column at tthis level: 1.2 Dead = 1.2(100 psf) f)(720 ft2) = 86,400 lbs 1.6 Live = 1.6(0.530)(880 psf)(720 ft2) = 48,800 lbbs 1.2 Dead + 1.6 Live = 886,400 + 48,8800 = 135,0000 lbs = 135 kiips Requiired axial streength (LRFD)), Pu = 135 kiips

Forr ASD D Step p 2:

Deterrmine the requ uired axial strrength for thee ASD load coombination ASD load entering g column at thhis level:

Chapter 2

Loads, Load Factors and Load Combinations 49

Dead = (100 psf)(720 ft2) = 72,000 lbs Live = 0.530(80 psf)(720 ft2) = 30,500 lbs Dead + Live = 72,000 + 30,500 = 103,000 lbs = 103 kips Required axial strength (ASD), Pa = 103 kips

Example 2.7 Column Load Calculation

Goal:

Determine the required axial strength for a column from the framing plan shown in Figure 2.8 due to live load of 80 psf and dead load of 100 psf.

Given:

Interior column C-3 supporting five levels with 18K3 steel joists spanning as shown in all panels:

SOLUTION

Step 1:

Determine the live load reduction from Equation 2.1 Tributary area per floor: AT = (30)(24) = 720 ft2 Tributary area for five floors: AT = 5(720) = 3600 ft2 Influence area per floor: AI = 4AT = 4(720) = 2880 ft2 Influence area for five floors: AI = 5(2880) = 14,400 ft2 Live load reduction:

0.25 +

15 = 0.375 < 0.40 14, 400

Therefore the live load reduction is limited to 0.40 since the member supports more than one level. LRFD Step 2:

Determine the required axial strength for the LRFD load combination LRFD load in the column from five levels: 1.2 Dead = 1.2(100 psf)(3600 ft2) = 432,000 lbs 1.6 Live = 1.6(0.40)(80 psf)(3600 ft2) = 184,000 lbs 1.2 Dead + 1.6 Live = 432,000 + 184,000 = 616,000 lbs = 616 kips Required axial strength (LRFD), Pu = 616 kips

ASD Step 2:

Determine the required axial strength for the ASD load combination ASD load in column from five levels: Dead = (100 psf)(3600 ft2) = 360,000 lbs Live = 0.40(80 psf)(3600 ft2) =115,000 lbs Dead + Live = 360,000 + 115,000 = 475,000 lbs = 475 kips Required axial strength (ASD), Pa = 475 kips

Example 2.8 Column Load Calculation

Goal:

Determine the required axial strength for a column from the framing plan shown in Figure 2.8 due to live load of 80 psf and dead load of 100 psf.

Given:

Exterior column D-4 supporting seven levels with 18K3 steel joists spanning as shown in all panels:

50 Chapter 2 SOLUTION

Loads, Load Factors and Load Combinations Step 1:

Determine the live load reduction from Equation 2.1 Note that the exterior column supports an 8 in. (0.67 ft) portion of slab outside the column center line. This must be included in the tributary and influence areas. Tributary area: AT = 7(24)(15) + 7(24)(0.67) = 2630 ft2 Influence area: AI = 4(7)(24)(15) + 2(7)(24)(0.67) = 10,300 ft2 Live load reduction: 15 0.25 + = 0.398 < 0.40 10,300 Therefore, the live load reduction is limited to 0.40 since the member supports more than one level.

For LRFD Step 2:

Determine the required axial strength for the LRFD load combination LRFD load in column from seven levels: 1.2 Dead = 1.2(100 psf)(2630 ft2) = 316,000 lbs 1.6 Live = 1.6(0.40)(80 psf)(2630 ft2) = 135,000 lbs 1.2 Dead + 1.6 Live = 316,000 + 135,000 = 451,000 lbs = 451 kips Required axial strength (LRFD), Pu =451 kips

For ASD Step 2:

Determine the required axial strength for the ASD load combination ASD load in column from seven levels: Dead = (100 psf)(2630 ft2) = 263,000 lbs Live = 0.40(80 psf)(2630 ft2) = 84,200 lbs Dead + Live = 263,000 + 84,200 = 347,000 lbs = 347 kips Required axial strength (ASD), Pa = 347 kips

The 10-story building given in Figure 2.8 and just discussed for gravity load has uniform story heights of 13.5 ft. The wind load is determined to be 40 psf over the upper 40 ft and the lateral load is to be resisted by the perimeter moment frames in each direction. Use the load case that includes dead, live and wind loads.

Example 2.9 Wind Load Calculation

Goal:

Determine the wind force applied to the wind resisting frame for the framing plan shown in Figure 2.8.

Given:

For each 90 ft wide frame the force applied at the roof level of each of two frames:

SOLUTION

Step 1:

Determine the area of loading:

Chapter 2

Loads, Load Factors and Load Combinations 51

Tributary area for upper half of the top story: AT = (145.3/2)(13.5/2)= 490 ft2 For LRFD Step 2:

Determine the lateral load for the LRFD load combination LRFD wind load, load case 4: 1.0(Wind) = 1.0(40 psf) = 40 psf Wu = (40 psf)(490 ft2) = 19,600 lbs

For ASD Step 2:

Determine the lateral load for the ASD load combination ASD wind load, load case 6: 0.75(0.6Wind) = 0.75(0.6(40 psf)) = 18 psf Wa = (18 psf)(490 ft2) = 8,820 lbs

Example 2.10 Column Load Calculation

Goal:

Determine the wind force applied to the wind resisting frame for the framing plan shown in Figure 2.8.

Given:

For each 145 ft wide frame the force applied at the first level below the roof level of each of two frames:

SOLUTION

Step 1:

Determine the area of loading:

For LRFD Step 2:

Tributary area for the first level below the roof: AT = (91.3/2)(13.5)= 616 ft2 Determine the lateral load for the LRFD load combination LRFD wind load, load case 4: 1.0(Wind) = 1.0(40 psf) = 40 psf Wu = (40 psf)(616 ft2) = 24,600 lbs

For ASD Step 2:

Determine the lateral load for the ASD load combination ASD wind load, load case 6: 0.75(0.6Wind) = 0.75(0.6(40 psf)) = 18 psf Wa = (18 psf)(616 ft2) = 11,100 lbs

2.6 CALIBRATION The basic requirements of the ASD and LRFD design philosophies were presented in Sections 1.9 and 1.10 and Equations B3.2 and B3.1. The required load combinations for

52 Chapter 2

Loads, Load Factors and Load Combinations

ASD and LRFD, as found in ASCE 7, were presented earlier in the chapter. This section establishes the relationship between the resistance factor, φ, and the safety factor, Ω, as found in the AISC Specification. Early development of the LRFD approach to design concentrated on the determination of resistance factors and load factors that would result in a level of structural reliability consistent with previous practice but more uniform for different load combinations. Because the design of steel structures before that time had no particular safety-related concerns, the LRFD approach was calibrated to the then-current ASD approach. This calibration was carried out for the live load plus dead load combination at a live-to-dead-load ratio, L/D, of 3.0. This can be seen in Figure 1.18, where for a live to dead load ratio of 3.0 the reliabilities for ASD and LRFD are essentially the same. It was well known that for any other load combination or live-to-dead-load ratio, the two methods could give different answers for the same design situation. The current Specification has been developed with this same calibration, which results in a direct relationship between the resistance factor of LRFD and the safety factor of ASD. For the live load plus dead load combination in ASD, using Equation B3.2, and representing the load effect simply in terms of D and L, ( D + L) ≤

Rn Ω

This same combination in LRFD, using Equation B3.1 is (1.2D + 1.6L) ≤ φRn If it is assumed that the load effect is equal to the available strength and each equation is solved for the nominal strength, the results for ASD are Ω(D + L) = Rn and for LRFD, (1.2 D + 1.6 L ) = Rn φ

With L/D taken as 3, the above equations are set equal. They are then solved for the safety factor, which gives Ω=

1.5 φ

(2.7)

The resistance factors in the Specification were developed through a stochastic analysis to be consistent with the specified load factors and result in the desired reliability for each limit state. More detail on the development of these resistance factors can be found in Section B3.1 of the Commentary to the Specification. Once the resistance factors were established, the corresponding safety factors were determined using Equation 2.7. This relationship has been used throughout the Specification to set the safety factor for each limit state. Although the relationship is simple, there is actually no reason to use it to determine safety factors. This has already been done in the Specification, which explicitly defines resistance factors and safety factors for every limit state.

Chapter 2

22.7

Loads, Loadd Factors and Load Combinnations 53

PROBLEMS

11. Name and deescribe five baasic types/sourcces of building g looads. 22. Categorize th he following loads as dead load, l live load,, ssnow load, win nd load, seismicc load, or speciial load. a. Lo oad on an office o floor due to filing g cabineets, desks, and computers. b. Load on a ro oof from a permanent p airr handliing unit. c. Lo oad on stadiu um bleachers from studentss jumpin ng up and dow wn during a co ollege footballl game. d. Load on a buildiing caused by an explosion. e. Weight W on a steeel beam from a concrete slab b that it is supporting. f. Lo oad experienceed by an officce building in n Califo ornia as it shakees during an eaarthquake. g. Lo oad on a skysscraper in Chiccago on a day y with blustery b condittions causing the t building to o sway back b and forth.

11. Iff a response m modification facctor of 3 is choosen in the desiggn of a steel bbuilding to reesist seismic looads, what desiggn specificationn should be connsulted? 12. W Which designn approach coombines loads that are norm mally at their noominal or serviiceability levells? 13. S Strength load combinations that are incorpporated by the L LRFD method ttake into accouunt what two faactors? 14. U Using ASCE 7, determine the minimum uniformly distriibuted live loadd for a hospitall operating rooom. 15. U Using ASCE 7, determine the minimum uniformly distriibuted live loadd for library staacks. 16. U Using ASCE 7, determine the minimum uniformly distriibuted live loadd for an apartm ment building. 17. D Determine thee nominal uniiformly distribbuted selfweigght of a 6 in. thiick reinforced concrete slab.

33. What is onee source you can c consult to find the snow w looad data for a particular regiion as well as maps showing g w wind gust data to allow you to o calculate win nd loads? 44. Where in th he AISC Manual can you find f a table off sselected unit weights of comm mon building materials? m 55. What analyssis method allo ows the design ner to visualizee thhe load on a particular structural eleement withoutt pperforming an actual a equilibriium calculation n? 66. What is the rationale for th he live load reduction factor,, w when can this reduction facto or be used, and what are thee liimitations on its i use? 77. In determinin ng the snow lo oad on a structu ure, what valuee thhat can be obttained from thee applicable bu uilding code iss m multiplied by a series of factors to obtain the roof snow w looad? 88. Identify and briefly describ be three factorss that affect thee m magnitude of th he snow load on o an unobstruccted flat roof. 99. Name four factors f that mu ust be taken in nto account in n cconverting win nd speed data referenced by y the building g ccode into wind pressure on a given g building g. 110. If the locall building codee specifies a deesign load thatt ddiffers from what is stated in n ASCE 7, wh hich documentt sshould you folllow?

P2.188 18. A building hass a column layyout as shownn in Figure P2.188 with 30 ft baays in each dirrection. It musst support a unifoorm dead load of 90 psf and a uniform live load of 80 psf. Determine thhe required mooment and shear strength for bbeams and girdders, and axial strength for coolumns, as notedd below for design by (a)) LRFD and (b) ASD. Calcuulate with annd without llive load redduction, as approopriate. i. The beeam on columnn line 3 betweeen column lines A annd B if the decck spans from line 2-2 to 3-3 to 4-4.

554 Chapter 2

Loads, Lo oad Factors and a Load Com mbinations

ii. The girder on n column line C between n colum mn lines 3 and 4 if the deck spans from linee B-B to o C-C to D-D. iii. The T column at the corner on lines 4 and A and su upporting one floor. f iv. The T column on n the edge at th he intersection n of linees C and 4 and supporting onee floor. v. Th he interior co olumn at the intersection i off colum mn lines D and 3 and supportin ng one floor. 119. If the fram ming plan show wn in Figure P2.19 P were forr thhe roof of a sttructure that carried a dead load of 55 psff aand a roof live load of 30 psf, determine the required d m moment and shear strength for beams and girders, and d aaxial strength for columns, as required below for (a)) ddesign by LRF FD and (b) dessign by ASD. Do not reducee thhe roof live loaads. i. Th he girder on column line A beetween column n lines 1 and 2 if the deck d spans fro om line A-A to o B-B.

below w for (a) desiggn by LRFD and (b) designn by ASD. Calcuulate with annd without llive load redduction, as approopriate. i. The bbeam between column liness 2 and 3 along line D. ii. The girder on coolumn line 3 between column linne E and midw way between liines D and C. iii. The bbeam on the lline between liines C and D and coluumn lines 3 annd 4. iv. The ccolumn at the corner on linees 1 and E that suppoorts eight levelss. v. The ccolumn on the edge at the inntersection of lines 4 aand A that suppports eight levvels. vi. The interior colum mn at the interrsection of column liine 4 and the point midwaay between lines C annd D that suppoorts three levelss.

ii. The beam on co olumn line 3 beetween column n lines B and C if the deck spans frrom line 2-2 to o 3-3 to 4-4. iii. The T column at the t corner on lines l 1 and E. iv. The T column on n the edge at th he intersection n of linees 1 and B.

P2.200 21. IIntegrated Deesign Project.. For a Downners Grove, Illinoois, office building, determiine the dead, live, snow and w wind loads baased on ASCE E 7 and the coonstruction sugg ested below. A reasonable aassumption forr the constructtion of the rooof is the folloowing:

v. Th he interior co olumn at the intersection i off colum mn lines C and 2. 2 P P2.19 220. The framin ng plan shown in Figure P2.2 20 is for an 18-sstory office buiilding. It must support a dead d load of 80 psff aand a live load of 50 psf. In all a cases, the deecking spans in n a direction fro om line A tow ward line E. Determine D thee rrequired momeent and shear strength for the t beams and d ggirders, and ax xial strength for f the column ns, as required d

Roofing Insulation Deck Beams Joists Misc. Total

5 ppsf 2 ppsf 2 ppsf 3 ppsf 3 ppsf 5 ppsf 20 psf

Dow wners Grove is at an elevationn of approximaately 690 ft so a case-specific snow load callculation is noot required. walls, snow Becaause of the perrimeter parapett and screen w drift loading must bbe considered.

Chapter 2

A reasonable assumption for the floor construction is Slab and deck 60 psf Beams and girders 8 psf Misc. 12 psf Total 80 psf Since no information has been given regarding office layout, it is prudent to use the corridor loading over the entire floor plan. This will permit architectural flexibility. The two-story area bounded by column lines A, C, 4, and 5 should be treated as a lobby. For wind load, this is a rigid enclosed building located on a site where there are no special terrain features to consider nor is air density a concern since the elevation is approximately 700 ft above sea level. Thus, to determine the velocity pressure, Kd = 0.85, Kzt = Ke = 1.0 and Kz at the top of the building is 1.09. To determine the external design wind pressure use G = 0.85 for a rigid building, Cp = 0.8 for the windward wall and GCpi = +0.18 for an enclosed structure. a.

Develop a table showing the gravity loads in all columns based on tributary area and live load reductions. Remember that the columns in the lateral load resisting systems will also carry wind or seismic forces. Wind loads will be determined in part c; seismic loads are in the Chapter 13 Integrated Design Project problem.

b. Develop a table showing beam loading based on tributary area and live load reductions. Remember that beams will see a uniform load and girders will see concentrated loads and some uniform load due to their self-weight. Beam and girder self-weight is included in the construction loads given above. Perimeter beams and girders will also support the lightweight metal curtain wall at each level. A reasonable assumption for the weight of the curtain wall is 18 psf. c. Using a uniform wind pressure as determined at the top of the building and spread over the entire windward wall, determine the concentrated wind force at each level for each main wind force resisting system in each direction.

Loads, Load Factors and Load Combinations 55

C Chapterr 3

S Steel Buildin B ng Ma aterials

wer, Cape Canaveral, FL Exxploration Tow Photo courteesy Thornton T Tomasetti

3.11

INTROD DUCTION Steel has beeen produced in the United States sincee the 1800s. Its first use iin a bridge w was for a railroad brid dge across th he Mississippii River in Stt. Louis, builtt in 1874 by James B. Eaads. The bridge, know wn as the Eaads Bridge an nd shown in Figure 3.1, iis still an insspiring steel sstructure crossing thee river in the shadow s of the St. Louis G Gateway Archh. The first skkyscraper is ggenerally considered to t be the Hom me Insurance Building, dessigned by Willliam LeBaroon Jenney andd erected at 135 South h La Salle Strreet, Chicago o. The buildinng, shown in Figure 3.2, w was started onn May 1, 1884, and completed c in the fall of 1885. 1 It was originally a 10-story buillding but lateer had 2 stories added. The origin nal structural design d calledd for wrought iron beams bbolted througgh angleiron bracketts to cast iron columns. As the framewoork reached thhe sixth floor,, the Carnegiee-Phipps Steel Comp pany of Pittsb burgh, Pennssylvania, indiicated that thhey were now w rolling “B Bessemer Steel” and requested r perrmission to su ubstitute steeel members fo for the wrougght iron beam ms on all remaining floors. f This was w the firstt use of steell beams in a building. T The Home Innsurance Building waas demolished d in 1929. The first all-steell skyscraper was w the Randd-McNally Buuilding at 1655 West Adam ms Street in Chicago, designed by Daniel Burnh ham and Johnn Root. This 110-story buildding, shown inn Figure 3.3, was bu uilt from 1888 8–1890 and was w construccted of built-uup members made from sstandard rolled steel bridge shapees that were riveted togetther. It markeed the beginnning of a conntinuous evolution in n steel buildin ng structures that t is ongoinng today as nnew ideas are brought into play by architects an nd engineers who w build witth steel. Thiss evolution in steel build dings encomppasses the m materials usedd, the applicaations of innovative designers, d and d the specificaations that dirrect their desiigns.

3.22

APPLIC CABILITY OF THE AISC SPECIIFICATION N The specificcation that gu uides the dessign of our m modern steel buildings waas first publiished by AISC in 192 23. Its purpose was to prom mote uniform practice in thhe design of ssteel buildingss. At

566

Chappter 3 Steel Building Matterials

Figure 3.1

57

Eads Bridge, St. Louis, Missouri M

Photo courtessy Luke Gatta

Figure 3.2 Home Insuraance Building g, Chicago, Illlinois. Copyright © American A Instiitute of Steel Construction C Reprinted witth Permission. All Rights Resserved

Figu ure 3.3 Raand McNally B Building, Chiccago, Illinois Coppyright © Amerrican Institute of Steel Connstruction Reprrinted with Perrmission. All R Rights Reseerved

58 Chapter 3

Steel Building Materials that time, numerous approaches were being used across the industry. Steel producers each had their own standard for design, and the larger cities also required that their own standards be used. This multiplicity of standards meant no standard at all. It led to a confusion of approaches in which designers were continually called upon to change the way they designed, depending on where their current building project was to be located. The 1923 Specification defined “the practice adopted by the American Institute of Steel Construction for the design, fabrication, and erection of structural steel buildings.” It went on to provide direction on how to obtain a satisfactory structure. The following requirements were to be fulfilled: 1. 2. 3. 4. 5.

The material used must be suitable, of uniform quality, and without defects affecting the strength or service of the structure. Proper loads and conditions must be assumed in the design. The unit stresses must be suitable for the material used. The workmanship must be good, so that defects or injuries are not produced in the manufacture. The computations and design must be properly performed so that the unit stresses specified shall not be exceeded, and the structure and its details shall possess the requisite strength and rigidity.

The Specification also provided guidance on the material to be used, stating, “Structural steel shall conform to the Standard Specifications of the American Society for Testing Materials for Structural Steel for Buildings, Serial Designation A9-21, as amended to date.” These principles from 1923 are still important to steel construction almost a century later. The 2016 AISC Specification for Structural Steel Buildings supersedes all previous AISC Specifications. Over the years, the Specification has lost the terms fabrication and erection from its title, nonetheless, the Specification continues to address issues pertinent to design, fabrication and erection of steel structures. In addition, the AISC Specification has regularly been used to guide the design of structures other than buildings. In recognition of this practice, and to ensure that the Specification is properly applied, the scope of the Specification states, “The specification sets forth criteria for the design fabrication and erection of structural steel buildings and other structures, where other structures are defined as structures designed, fabricated, and erected in a manner similar to buildings, with building-like vertical and lateral load resisting elements.” Additionally, the Specification indicates that it “shall apply to the design, fabrication and erection of the structural steel system or systems with structural steel acting compositely with reinforced concrete, where the steel elements are defined in Section 2.1 of the AISC Code of Standard Practice for Steel Buildings and Bridges.” In that document, structural steel is defined as “the elements of the structural frame that are shown and sized in the structural design drawings, essential to support the design loads,” and they are given here in Table 3.1. A quick review of the table contents indicates that elements are structural steel if they are part of or attached to the structural frame. Examples of the use of many of these elements are shown in Figure 3.4. All elements discussed in this text will meet the above definition.

3.3 STEEL FOR CONSTRUCTION Since the introduction of the first AISC Specification, a variety of steels have been approved for use in steel construction. The specific steels approved at any given time has changed along with the techniques of manufacture and steel chemistry. Steels available for use in construction have increased in strength as manufacturing has become more refined. One important aspect of all steel is that it generally behaves in a uniform and consistent manner. Thus, although the strength might be different for different grades of steel, the steel can be expected to behave the same, regardless of grade, up to its various strength limits.

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Table 3.1 Definitions of Structural Steela Anchor rods that will receive structural steel. Base plates, if part of the structural steel frame. Beams, including built-up beams, if made from standard structural shapes and/or plates. Bearing plates, if part of the structural steel frame. Bearings of steel for girders, trusses, or bridges. Bracing, if permanent. Canopy framing, if made from standard structural shapes and/or plates. Columns, including built-up columns, if made from standard structural shapes and/or plates. Connection materials for framing structural steel to structural steel. Crane stops, if made from standard structural shapes and/or plates. Door frames, if made from standard structural shapes and/or plates and if part of the structural steel frame. Edge angles and plates, if attached to the structural steel frame or steel (open-web) joists. Embedded structural steel parts, other than bearing plates, that will receive structural steel. Expansion joints, if attached to the structural steel frame. Fasteners for connecting structural steel items: permanent shop bolts, nuts, and washers; shop bolts, nuts, and washers for shipment; field bolts, nuts, and washers for permanent connections; and permanent pins. Floor-opening frames, if made from standard structural shapes and/or plates and attached to the structural steel frame or steel (open-web) joists. Floor plates (checkered or plain), if attached to the structural steel frame. Girders, including built-up girders, if made from standard structural shapes and/or plates. Girts, if made from standard structural shapes. Grillage beams and girders. Hangers, if made from standard structural shapes, plates, and/or rods and framing structural steel to structural steel. Leveling nuts and washers. Leveling plates. Leveling screws. Lintels, if attached to the structural steel frame. Machinery supports, if made from standard structural shapes and/or plates and attached to the structural steel frame. Marquee framing, if made from standard structural shapes and/or plates. Monorail elements, if made from standard structural shapes and/or plates and attached to the structural steel frame. Posts, if part of the structural steel frame. Purlins, if made from standard structural shapes. Relieving angles, if attached to the structural steel frame. Roof-opening frames, if made from standard structural shapes and/or plates and attached to the structural steel frame or steel (open-web) joists. Roof-screen support frames, if made from standard structural shapes. Sag rods, if part of the structural steel frame and connecting structural steel to structural steel. Shear stud connectors, if specified to be shop attached. Shims, if permanent. Struts, if permanent and part of the structural steel frame. Tie rods, if part of the structural steel frame. Trusses, if made from standard structural shapes and/or built-up members. Wall-opening frames, if made from standard structural shapes and/or plates and attached to the structural steel frame. Wedges, if permanent. a From Code of Standard Practice for Steel Buildings and Bridges, AISC 2016.

Steel Elements

Reprinted from CISC Code of Standard Practice for Structural Steel, with permission, Copyright © Canadian Institute of Steel Construction All rights reserved Construction.

Figure 3.4

600 Chapter 3 Steel Build ding Materialss

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61

The T characterristics of steell that are impoortant to the sstructural enggineer can be determined through a simple un niaxial tensio on test. Thi s standard ttest is conduucted accordding to the requirem ments of ASTM M A370, Stan ndard Test M Methods and D Definitions foor Mechanicaal Testing of Steel Products. A sp pecimen of specific s dimeensions is suubjected to a tensile forcce, and the resulting stress and sttrain are plottted for the duuration of thee test. The sttress, f, and strain, ε, are plotted in n Figure 3.5 and a defined ass follows:

f =

P A

aand ε =

ΔL L

where A = L = P = f = ΔL Δ = ε =

cross--sectional areea at start of teest length h of specimen n at start of teest tensile force axial tensile stress chang ge in length of specimen unnder stress axial strain

The curve shown in Fig gure 3.5 is tyypical of mildd carbon steell. Several chaaracteristics of this sttress-strain cu urve are worth h noting. Firsst, the initial portion of the curve, whicch indicates the respo onse that wou uld be expecteed under mosst nominal or service loadiing conditions, follows a straight line l up to a point p called th he proportionnal limit. For structural steeel with yieldd stresses of 65 ksi orr less, this pro oportional lim mit is the poinnt where the ccurve first deviates from liinear and is called the yield point. The ratio off stress to straain in this reggion is constannt and is calleed Young’s modulus or the modu ulus of elasticity, E. All strructural steelss exhibit the same initial sstress-strain behaviorr and thus hav ve the same E value. The vvalue of E obttained throughh a large num mber of tests is consisstently betweeen 29,000 ksi k and 30,0000 ksi. For aall calculatioons based onn the AISC Specifica ation, E = 29,000 ksi has historically bbeen used. W Within the straaight-line porrtion of the curve, th he material is said to behav ve elastically. A load can bbe applied and then removved with the structure returning to its original co onfiguration, showing no ppermanent deformation. After A reaching g the yield strress, the stresss-strain curvee for mild carrbon steel exhhibits a long plateau where w the strress remains essentially coonstant whilee the strain increases. Thiis region is called thee plastic region. Any struccture that is looaded into thhis region exhhibits a permaanent plastic deformattion as show wn by the unlloading line in Figure 3.5. The lengtth of this plaastic region depends on the particu ular type of stteel but typicaally is 15 to 220 times the strain at yield..

3 Figure 3.5

Typical Stress-Strain Plot for Mildd Carbon Steeel

622 Chapter 3

Steel Build ding Materialss

At the t end of the plastic regiion, the curvee again rises with increasing stress and strain. This increasse is called strain s harden ning and conntinues until the specimeen reaches itss tensile strength, or ultimate stress, Fu, at the peak of the sstress-strain ccurve. Once tthe tensile strrength is pidly sheds lo oad and increaases strain unntil complete rrupture occurrs. reached, the specimen rap Yielld stress, tenssile strength, and moduluus of elasticityy are the enggineering dataa values used throug ghout design to fully desscribe the m material and tto determine the strengthh of the structural ellements. Thee ratio of thee tensile strenngth to the yield stress is also an im mportant characteristiic of steel. Itt is used to control c the baasic material behavior so that at varioous limit states, the ex xpected behav vior can be en nsured. Figu ure 3.6 showss the lower strain region oof the stress-strain curves for three steeels with different yieeld stresses: 36 3 ksi, 50 ksi,, and 100 ksi . Elastic behaavior for the hhigher-strenggth steels is the same as for the lo ower-strength steels. As allready noted, E = 29,000 ksi for all stteel. The differences occur o after th he proportionaal limit is reaached. For steeels with a yieeld stress lesss than or equal to 65 ksi, the plateeau defining the plastic reegion can be expected to occur. Howeever, for steels with a yield stress greater than 65 ksi, it is expected thaat no well-deffined yield pooint will exist and no o well-defined d plastic platteau will occuur, as illustraated for the steel having Fy = 100 ksi. For thesse steels it iss necessary to o define yieldd strength by some other means. ASTM M A370 provides forr yield streng gth determinaation by the 0.2 percent offset methood or the 0.5 percent elongation method. m In eitther case, the stress-strain ccurve must bee obtained annd the specifieed offset or elongatio on used to dettermine the ap ppropriate strress value. Thhe results of tthese two appproaches are shown in n Figure 3.6, and the two methods m woulld yield slighttly different yyield strength values.

3.44

STRUCTURAL ST TEEL SHAP PES Structural stteel design seerves to deterrmine the apppropriate shappe and amouunt of steel neeeded to carry a given n applied load. This is norrmally accom mplished by seelecting, from m a predeterm mined list of availablee shapes, th he lightest-weeight membeer. Howeverr, options coould also innclude a combination n of steel elements in some s particuular desired form. The early days of steel construction n witnessed very v little stan ndardization oof available sshapes. Althoough each milll would produce its own o shapes, the t variety off available shaapes was limited and most structural meembers

Figure 3.6 Enlarged Ty ypical Stress-Strain Curvess for Steels w with Different Yield Stressees

Chappter 3 Steel Building Matterials

63

were com mposed of theese available shapes riveteed together. O One of AISC C’s original ggoals was to standardiize the shapees being prod duced. Over tthe years, shhapes becamee standardizedd and more shapes, designed d speccifically for the t needs of building connstruction, beccame availabble. Modern productio on practices now n make a wide w variety of shapes avvailable to thee designer so that design can almo ost always be accomplished d by selectingg one of thesee standard shaapes. In situattions where these staandard shapess do not meett the needs oof a project, m members com mposed of plaate material can be prroduced to caarry the impossed loading. 3.4.1

AST TM A6 Stand dard Shapes The firstt standard shaapes to be disscussed are th those definedd by ASTM A A6: W-shapess, S-shapes, HP-shapees, M-shapess, C-shapes, MC-shapes, M aand L-shapess. Cross-sectiions of these shapes are shown in n Figure 3.7, where w it can be b seen that W W-, M-, S-, aand HP-shapees all take thee form of an I. C- and d MC-shapess are called channels c and take the form m of a C, whhile L-shapess are called angles. Part P 1 of the Manual M contaiins tables of pproperties forr all of the stanndard shapes. W-Shapes a wide-flangee shapes and are the mostt commonly uused shapes W-shapes are usually referred to as in buildin ngs. They haave two flang ges with essenntially paralleel inner and outer faces aand a single web locaated midway along a the flan nges. The oveerall shape of the wide flannge may vary from a

3 Figure 3.7

Structuraal Shapes

64 Chapter 3

Steel Building Materials fairly deep and narrow section, as shown in Figure 3.7a, to an almost square section, as shown in Figure 3.7b. These shapes have two axes of symmetry; the x-axis is the strong axis and the y-axis is the weak axis. Wide-flange shapes can be as deep as 44 in. and as shallow as 4 in. A typical wide-flange shape is designated as a W16×26, where the W indicates it is a W-shape, the 16 indicates it has a nominal depth of 16 in., and the 26 indicates its weight is 26 pounds per foot. The nominal depth indicates an approximate member depth but not its actual depth. The production of wide-flange shapes results in their grouping in families according to the size of the rolls used to produce the shape. All shapes in a family have the same dimension between the inner faces of the flanges. The different weights are achieved by increasing the actual depth of the member and thus the flange thickness. Manual Table 1-1 provides the dimensions and section properties needed for design for all W-shapes. HP-Shapes HP-shapes are similar to wide-flange shapes and normally used as bearing piles. These shapes also have parallel face flanges, but unlike the W-shapes, their webs and flanges are of the same nominal thickness and they are all close to being square, as shown in Figure 3.7c. An HP14×117 is an HP-shape with a nominal depth of 14 in. and a weight of 117 pounds per foot. Manual Table 1-4 provides the dimensions and section properties needed for design for all HP-shapes. S-Shapes S-shapes are American Standard beams and were previously referred to as I-beams. They were the standard shapes used in construction prior to the development of the rolling process that permitted the introduction of the wide-flange shapes. Although these shapes are still available, their use today is infrequent and their availability should be confirmed prior to their specification. These shapes have narrow flanges in relation to their depth, and the flanges have a sloping interior face, as shown in Figure 3.7d. The Manual lists 28 S-shapes and gives their properties in Table 1-3. As with the shapes previously discussed, the numbers in the name refer to the nominal depth and the weight per foot. In all cases except for two S24s and two S20s, the nominal depth and the actual depth are the same. M-Shapes M-shapes are miscellaneous shapes that do not fit the definitions of W-, HP-, and S-shapes. The Manual lists 18 miscellaneous shapes. They are not particularly common and should be used in design only after confirmation that they are economically available. A typical designation is M12×11.8; as with the other shapes, the 12 indicates the nominal depth and the 11.8 indicates the weight per foot. Dimensions and properties for M-shapes are found in Manual Table 1-2. C-Shapes C-shapes are American Standard channels and are produced using essentially the same process as for S-shapes. They have two flanges and a single web located at the end of the flanges, as shown in Figure 3.7e. These shapes have only one axis of symmetry and, like the W-shapes, the x-axis is the strong axis and the y-axis is the weak axis. As with the S-shapes, the flanges have sloping inner faces. One of the 32 C-shapes found in Manual Table 1-5 is a C8×18.75. All C-shapes have an actual depth equal to the nominal depth.

Chapter 3 Steel Building Materials

65

MC-Shapes MC-shapes are miscellaneous channels that cannot be classified as C-shapes. Their designations follow the same rules as the previous shapes, with a typical shape being an MC6×18. Manual Table 1-6 lists 40 MC-shapes, and their sizes fall into the same overall range as the C-shapes. L-Shapes L-shapes are angles that can have equal or unequal legs. The largest angle legs are 12 in. and the smallest are 2 in., with the dimension taken from heel to toe of the angle. A typical angle designation is L6×4×7/8, where the first two numbers are the dimensions of the legs and the third is the leg thickness. Leg dimensions are actual dimensions, and the leg thickness is the same for both legs. For unequal leg angles, the longest leg is given first. Equal leg angles have one axis of symmetry, whereas unequal leg angles have no axis of symmetry. All angles have two sets of two axes of interest to the designer: the geometric and principal axes, which are illustrated in Figure 3-7(f). The geometric axes are parallel to the faces of the legs with the x-axis, parallel to the short leg, and the y-axis, parallel to the long leg. For equal leg angles, the x- and y-axes are similarly oriented even though one cannot distinguish between longer and shorter legs. Alternatively, the principal axes can be used. The minor principal axis, which for equal leg angles is perpendicular to the axis of symmetry, is the z-axis; and the major principal axis, the axis of symmetry for equal leg angles, is the w-axis. The principal axes are similar for unequal leg angles, except there is no axis of symmetry. Manual Table 1-7 provides the dimensions and section properties for the design for all angles. For some special cases, properties that may be needed for the w-axis are not available in Table 1-7. These can be found in the 15th edition electronic shapes database. WT-Shapes WT-shapes are tees that have been cut from W-shapes. They are also called split tees because that is how they are produced – by splitting a W-shape. These shapes are designated as WT5×56, where both numbers are one-half of the corresponding numbers of the parent W-shape they were cut from. Dimensions and properties for WT-shapes are given in Manual Table 1-8. MT-Shapes and ST-Shapes MT-shapes and ST-shapes are tees that have been cut from parent M- and S-shapes. The properties and dimensions for these shapes are found in Manual Tables 1-9 and 1-10. 3.4.2

Hollow Shapes Another group of shapes commonly used in building construction are the hollow shapes commonly referred to as tubes or pipes. These shapes are produced by bending and welding flat plates or by hot rolling to form a seamless section. For all hollow structural sections (HSS), ASTM specifications set the requirements for both the materials and the sizes. Round HSS Round hollow structural sections are manufactured through a process called “formed from round,” which takes a flat strip of steel and gradually bends it around its longitudinal axis and joins the edges by welding. Once the weld has cooled, the round shape is passed through additional shaping and sizing rolls to fix the final diameter. An example of a round HSS is HSS5.563×0.258, where the first number is the outside diameter and the second is the nominal

666 Chapter 3

Steel Build ding Materialss wall thickneess. For round d HSS, the diameter d and nnominal walll thickness arre always shoown as a decimal num mber to three places. p Thesee shapes are fo found in Manuual Table 1-13. Square and d Rectangular HSS Square and rectangular HSS may iniitially be form med from rouund, with thee final sizing used to also change the formed shape into a rectangle, orr formed from m a flat platee through a ““formedsquare weld d-square” proccess, in which h the plate is gradually beent into its neaar final size. Another process starrts with two flat f pieces that are each bbent, with thee two half sections then jooined to form the fin nal shape. A typical t rectan ngular HSS iss the HSS12× ×8×1/2. The ffirst number iindicates the actual heeight of the seection, the seccond the actuual width, andd the third the nominal thickness of the section wall. w For square and rectan ngular HSS, tthe nominal w wall thicknesss is always shhown as a fraction. Manual M Tablees 1-11 and 1-12 1 provide the dimensioons and section propertiess needed for the desig gn of rectangu ular and squarre HSS-shapees, respectivelly. Steel Pipes Steel pipes are a another ho ollow round section used in building construction. T They are prodduced to different maaterial standaards than the round HSS. Pipes are avvailable as staandard weighht (Std.), extra strong (x-Strong), and a double ex xtra strong (xxx-Strong), whhich refer to tthe wall thickkness for a given outsside diameterr. The standaard designatioon for a pipee section is inn the form Pipe 5 xStrong, indiicating that itt has to meett the pipe maaterial standarrds, have a nnominal 5 in.. outside diameter, an nd a thicknesss correspondin ng to the “exttra strong” deesignation. Thhis particular pipe has an actual ou utside diameteer of 5.56 in. and a nominnal wall thicknness of 0.3755 in. Manual T Table 114 provides the propertiees for steel piipes. Note thaat many rounnd HSS cross sections are made to match the steel s pipe crross sections.. Although tthey are dim mensionally innterchangeabble, it is important to o remember that round HSS and steeel pipes aree produced tto different material standards.

3.44.3

Plates and a Bars In addition to the shapees already disscussed, steell is availablee as plates annd bars, as shhown in Figure 3.8. These elemeents are rarely y used alone as shapes buut are combined to form built-up shapes or ussed alone as connecting eleements to joinn other shapess.

Figure 3.8 Plate and Baar Products

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Table 3.2 2 Preferred Dimensions for Plates andd Bars Rangee of Thicknessses/Diameterrs Product t ≤ 3/8 in. 3/8 in. < t ≤ 1 in. 1 in. < t Plates 1/16 1//8 1/4 Square an nd rectangulaar bars 1/8 1//8 1/8 Circular bars 1//8 1/8 1/8 Note: Taable gives incrrements in thiickness or diaameter. Plates Plates arre flat rectang gular elemen nts that are hoot-rolled to a given thickkness and sheeared to the appropriaate width. Att one time, plates were allso available that were rollled to a giveen width as well as a given thicckness. These plates werre called universal mill plates. Becaause of the manufactturing processs, these plaates had patteerns of residdual stresses that differedd from the patterns in sheared plates and ressulted in low wer strength. C Current manuufacturing prractice is to produce all plates as sheared s platess. By industryy definition, pplates are a m minimum of 8 in. in width and may vary in thick kness from 3/1 16 in. up. Thee designation for a typical plate is PL 1//2×10×2 ft– 4 in., wh here the first number n is thee thickness, thhe second thee plate width, and the thirdd the length. Table 3.2 2 reflects the preferred p stan ndard practicee for plate thiickness increm ments. Bars Bars are available in n rectangular, circular, andd hexagonal shapes, with the rectanguular bar the most com mmonly used d shape in bu uilding constrruction. The only differennce between rectangular bars and plates is the width. Any rectangular r soolid element less than 8 inn. in width is technically referred to as a bar. Because the distinction bbetween barss and plates iis not signifiicant to the designer,, the designattion for these narrow elem ments is the saame as for a pplate. Thus, P PL 1/2×6×2 ft–4 in. iss a 6 in. wide bar.

3 Examp ples of Built--up Shapes Figure 3.9

68 Chapter 3 3.4.4

Steel Building Materials

Built-up Shapes Other shapes are available and may be found in the Manual, but they are of limited application in building construction. The Manual also contains tables for combinations of standard shapes that have, over the years, been found to be useful to the designer. Figure 3.9 shows a variety of builtup shapes formed from combining plates and shapes.

3.5 CHEMICAL COMPONENTS OF STRUCTURAL STEEL The basic mechanical properties of structural steel were presented in Section 3.3 with a limited discussion of the types of steel available for use by the building industry. Essentially three types of steel are used for shapes in the construction industry: carbon steel, sometimes referred to as carbon-manganese steel, high-strength low-alloy steel (HSLA), and corrosion-resistant highstrength low-alloy steel. For plates and bars, quenched and tempered steels are also available. The chemical composition of steel significantly influences the properties that are of ultimate importance to the engineer. Steel is primarily made of iron but also contains such other elements as carbon, silicon, nickel, manganese, and copper. The primary element that influences the characteristics of steel, not counting iron, is carbon. The addition of carbon increases steel strength but decreases ductility and weldability. Even though carbon is the most significant component of steel (after iron), it still represents a very small percentage of the final product. Steels generally have a carbon content of not more than 0.3 percent by weight. Steels that contain strengthening elements in addition to carbon and manganese are referred to as HSLA steels although there are no strict rules that apply to that designation. Although the formula for a specific steel might be different from that of any other steel, certain elements are required in order to meet a specific set of criteria. These specifications come from the ASTM standards for each steel type and are discussed in Section 3.6. However, the chemical elements that may be found in the most dominant steel for wide flange shapes are reviewed here. The specific percentage requirements for ASTM A992 steel are given in Table 3.3. Carbon Carbon (C) is the most common element, excluding iron, found in all steel. It is the most economical element used to increase strength. However, it also decreases ductility. Carbon content usually ranges from about 0.15 to 0.30 percent. Anything lower than 0.15 percent would produce steel with too low a strength, and anything higher than 0.30 percent would yield steel with poor characteristics for use in construction. Table 3.3

Chemical Requirements for A992 Steel

Element Carbon, max Manganese Silicon, max Vanadium, max Columbium, max Phosphorus, max Sulfur, max Copper, max Nickel, max Chromium, max Molybdenum, max

Composition, % 0.23 0.50 to 1.60 0.40 0.15 0.05 0.035 0.045 0.60 0.45 0.35 0.15

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Manganese Manganese (Mn) has an effect on strength similar to that of carbon. It is a necessary component because of the way it combines with oxygen and sulfur and its impact on the rolling process. In addition, manganese improves the notch toughness of steel. It is added to steel to offset reductions in notch toughness due to the presence of other elements. It has a negative effect on material weldability. Silicon Silicon (Si) is an important element for removing oxygen from hot steel. Vanadium Vanadium (V) is another strengthening element. It refines the grain size and thus increases strength. Its biggest advantage is that while increasing strength, it does not negatively impact weldability or notch toughness. Columbium Columbium (Cb) is a strengthening element that, in small quantities, can increase the yield point and, to a lesser extent, the tensile strength. However, it has a significant negative impact on notch toughness. Phosphorus Phosphorus (P) increases strength and decreases ductility. It improves resistance to atmospheric corrosion, particularly when used in combination with copper. It has a negative impact on weldability that is more severe than that of manganese. It is generally an undesirable element but is permitted in very limited quantities in all steel. Sulfur Sulfur (S) is also permitted in very limited quantities in all steel. It has a negative impact on weldability comparable to that of phosphorus. Generally, steelmaking practice works to remove as much sulfur as possible. Copper Copper (Cu) in limited quantities is beneficial to steel. It increases strength with only a limited negative impact on ductility. If its content is held relatively low, it will have little effect on weldability. It is the most significant contributing element in the production of corrosion-resistant steel. Nickel Nickel (Ni) can provide a moderate improvement in strength and enhances corrosion resistance. It can also improve resistance to corrosion for steel subjected to seawater when used in combination with copper or phosphorus. It generally leads to a slight improvement in notch toughness.

70 Chapter 3

Steel Building Materials Chromium Chromium (Cr) is typically used in combination with copper to improve corrosion resistance. It also provides some strengthening of steels containing copper and vanadium. Chromium is an integral component of stainless steel. Molybdenum Molybdenum (Mo) increases strength but significantly decreases notch toughness. However, this negative impact can be controlled by appropriate processing or balancing with other elements.

3.6 GRADES OF STRUCTURAL STEEL 3.6.1

Steel for Shapes Many more grades of steel are produced than are approved by AISC for use in structures. A unique ASTM number designates each type of approved steel. The steels approved for structural shapes are grouped as carbon steel (A36, A53, A500, A501, A529, A709, A1043 and A1085), high-strength low-alloy steel (A572, A618, A709, A913, A992, and A1065), and corrosionresistant high-strength low-alloy steel (A588, A847, and A1065). Figure 3.10 lists these approved steels, their minimum yield and tensile stresses, and the shapes to which they are applicable. The table also identifies which combination of shape and material specification are preferred, such as A992 for W-shapes, and which material specifications do not apply for particular shapes. A36 Steel A36 steel was the most commonly available structural steel for many years. It was introduced in the 1961 AISC Specification and until the late 1990s was the steel of choice for most steel shapes except for HSS, pipe, and plates. It is a mild carbon steel, so it is well suited for bolted or welded construction. Even when higher-strength steels were used for members, this steel was the usual choice for connecting elements. It continues to be the preferred steel for M-, S-, C-, MC-, and Lshapes. It has a minimum yield stress, Fy = 36 ksi and a tensile stress, Fu = 58 to 80 ksi, although Fu = 58 ksi is used for calculations throughout the Specification, Manual and AISC Design Examples. A53 Steel A53 steel is the single standard for steel pipes approved for construction. It is available in three types and two grades; however, they are not all approved for structural applications. These pipes are generally intended for mechanical and pressure applications, and the only grade approved for construction is Grade B. This grade is available as Type E, which denotes electric-resistance welding of the seam, or Type S, which is a seamless pipe. A53 Grade B has a minimum yield stress, Fy = 35 ksi and a minimum tensile stress, Fu = 60 ksi. A53 Grade B steel pipe are treated as HSS in the Specification A500 Steel A500 steel is a carbon steel used for structural tubing in round, square, and rectangular shapes, otherwise known as HSS. It comes in two grades approved by AISC for construction: Grade C, which is the preferred grade, and Grade B. The standard permits either welded or seamless manufacture. Round HSS Grade C has a minimum yield stress, Fy = 46 ksi and a minimum tensile

Chappter 3 Steel Building Matterials

Figure 3.10 3

Applicaable ASTM Specifications S s for Various Structural Shhapes

Copyrightt © American Institute I of Steeel Constructioon. Reprinted w with permissionn. All rights resserved.

71

72 Chapter 3

Steel Building Materials stress, Fu = 62 ksi; whereas rectangular HSS Grade C, has a minimum yield stress, Fy = 50 ksi, with a minimum tensile stress, Fu = 62 ksi. A501 Steel A501 steel is a carbon steel similar to A36 but used for round and rectangular HSS. It is approved in Grade A, which has a minimum yield stress, Fy = 36 ksi, and a minimum tensile stress, Fu = 58 ksi, and in Grade B, which has a minimum yield stress, Fy = 50 ksi, and a minimum tensile stress, Fu = 70 ksi. A529 Steel A529 steel is a carbon-manganese steel available in grades 50 and 55. It is approved for the smaller shapes with flange thickness no greater than 1.5 in. A529 Grade 50 has a minimum yield stress, Fy = 50 ksi and a tensile stress, Fu = 65 to 100 ksi whereas Grade 55 has a minimum yield stress, Fy = 55 ksi, and a tensile stress, Fu = 70 to 100 ksi. A709 Steel A709 steel represents two types of steel, carbon steel, and high-strength low-alloy steel. Seven grades in four yield strengths are included. This standard is developed specifically for bridges where requirements necessary for bridge applications are added to the basic requirements for A36, A572, A992, and A588 steels. In buildings, the corresponding grade of A709 steel may be used in place of these four steels. A1043 Steel A1043 steel is a carbon steel available in two grades. Grade 36 has a yield stress, Fy = 36 to 50 ksi and a minimum tensile strength, Fu = 58 ksi. Grade 50 has a yield stress, Fy = 50 to 65 ksi and a minimum tensile strength, Fu = 65 ksi. This steel is available when a maximum yield-to-tensile ratio of 0.80 is required. W-shapes with a flange width of 6 in. or greater and plates up to 5 in. thickness are included. A1085 A1085 steel is a carbon steel for hollow structural sections. For HSS produced according to this standard, the wall thickness is permitted to be no more than 5% under the nominal wall thickness and the mass cannot be more than 3.5% under the nominal mass. This standard was developed so that the design properties for HSS could be developed based on the nominal dimensions of the section. For other HSS material specifications, the tolerance on thickness is so large that design must be carried out using properties determined by using 0.93 times the nominal thickness, rather than the nominal thickness. Grade A is the only grade provided for in the standard. It is available with a yield stress, Fy = 50 to 70 ksi, and a minimum tensile stress, Fu = 65 ksi. A572 Steel A572 is a high-strength low-alloy steel, also referred to as columbium-vanadium structural steel, available in five grades. It is a versatile high-strength steel with good weldability. Availability of shapes and plates is a function of grade, generally depending on element thickness. It is available in all shapes other than HSS and pipe. The full range of minimum yield stress is 42 to 65 ksi, depending on grade, and the minimum tensile stress ranges from 60 to 80 ksi, again depending on grade. A572 Grade 50 is the preferred steel for HP-shapes.

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A618 Steel A618 is a high-strength low-alloy steel used for HSS. Grades I, II, and III are approved for use in structures by AISC. It is one of two high-strength low-alloy steels available in HSS. Grade II has limited atmospheric corrosion resistance, and Grade III can be produced with increased corrosion resistance if required. The minimum yield stress depends on the particular product and may vary from 46 to 50 ksi. The minimum tensile stress varies from 65 to 70 ksi, again depending on grade and product wall thickness. A913 Steel A913 is a high-strength low-alloy steel produced by quenching and self-tempering. It is available in grades 50, 60, 65, and 70. The minimum yield stress ranges from 50 to 70 ksi and the minimum tensile stress ranges from 65 to 90 ksi. A992 Steel A992 steel is a high-strength low-alloy steel that has become the steel of choice for wide-flange shapes. It was first approved for use in 1998 as a replacement for steel that had come to be dualcertified as meeting both A36 and A572 Grade 50. This standard was developed partly as the result of an improved understanding of the impact of material property variations on structural behavior, and partly as the result of the changes in properties caused by the use of scrap as the main resource for steel production. The chemical components for A992 steel were given in Table 3.3 and discussed in Section 3.5. It has a minimum yield stress, Fy = 50 ksi, and a minimum tensile stress, Fu = 65 ksi. An additional requirement is that the yield-to-tensile ratio cannot exceed 0.85. A1065 Steel A1065 steel is a high-strength low-alloy steel specifically for HSS. This steel is produced to the standards for steels previously approved for use as plates. Thus, the tolerances are tighter than those for other common HSS steels such as A500 and A501. HSS produced according to this standard are first formed into two channels on a press brake and then welded together to form the tube. This material is produced in Grade 50 and in Grade 50W, a corrosion resistant grade, with a minimum yield stress, Fy = 50 ksi, and a minimum tensile stress, Fu = 60 ksi. A1065 steel appears in the 2016 Specification for the first time. Before specifying this steel, the designer should confirm its availability. A588 Steel A588 is a high-strength low-alloy corrosion-resistant steel with substantially better corrosion resistance than carbon steel with or without copper. It is available for all shapes, except HSS and pipe, as well as plate. For all shapes, and for plates up to 4 in., it has a minimum yield stress of 50 ksi and a minimum tensile stress of 70 ksi. Plates up to 8 in. are available at reduced stress values. A847 Steel A847 is a high-strength low-alloy corrosion-resistant steel used for HSS. It has the same minimum yield and tensile stresses as A588.

744 Chapter 3

ding Materialss Steel Build

Figure 3.11 Applicablee ASTM Speccifications forr Plates and B Bars Copyright © American Insttitute of Steel Construction. C R Reprinted with Permission. A All rights reservved.

3.66.2

Steel for Plates and Bars Many of thee steels alread dy discussed for f shapes aree also availabble for plates aand bars. Figgure 3.11 shows the ASTM desig gnations, the correspondiing yield andd tensile streesses, and thhe plate

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thickness for which they apply. The only steels available for plates and bars that are not also available for shapes are A283, A242, A514, and A1066. A283 Steel A283 steel is a low to intermediate strength carbon steel that is available in Grade C and Grade D. Grade C has a minimum yield stress, Fy = 30 ksi, and a tensile stress, Fu = 55 to 75 ksi while Grade D has a minimum yield stress, Fy = 33 ksi, and a tensile stress, Fu = 60 to 80 ksi. Although there is no thickness limit specified, plates thicker than 2 in. may be more difficult to obtain and the designer should check availability prior to specifying. A242 Steel A242 is a high-strength low-alloy corrosion-resistant steel also called weathering steel. It was one of the first corrosion-resistant steels and has a corrosion resistance approximately four times that of normal carbon steel. It is available in three grades but is now less common than the newer A588. The minimum yield stress ranges from 42 to 50 ksi and the minimum tensile stress ranges from 63 to 70 ksi. A514 Steel A514 is a high-yield strength-quenched and tempered alloy steel suitable for welding. It is available as plate material up to 6 in. There are 14 different grades, which vary according to the chemical content and maximum thickness. The minimum yield stress is either 90 or 100 ksi, and the ultimate tensile stress ranges from 100 to 130 ksi. This is the highest-yield-stress steel approved for use, except for fasteners, according to the AISC Specification. A1066 Steel A1066 steel is a high-strength low-alloy steel produced in 5 grades, Grade 50, Grade 60, Grade 65, Grade 70, and Grade 80 with a minimum yield stress corresponding to the grade designation. Minimum tensile strength for Grade 50 is 65 ksi, Grade 60 is 75 ksi, Grade 65 is 80 ksi, Grade 70 is 85 ksi and Grade 80 is 90 ksi. Grades 50 and 60 are available up to 4 in. thickness, Grade 65 to 3 in. Grade 70 to 2 in. and Grade 80 to 1 in. 3.6.3

Steel for Fasteners Fasteners for steel construction today include high-strength bolts, common bolts, threaded rods, and anchor rods. In addition, nuts, washers, and direct-tension indicators must be specified. The ASTM steels approved for these elements are listed in Figure 3.12. Many grades of steel are appropriate for the variety of mechanical fasteners used in steel construction, but only the three steels commonly specified for bolts are discussed here. A307 Bolts A307 bolts are also called common bolts or black bolts. Although the ASTM standard specifies three grades, only Grade A is approved for use as bolts in general applications. These bolts have an ultimate tensile strength of 60 ksi and are thus at a strength level similar to A36 steel. Although these bolts continue to be listed by AISC, they are rarely used in steel-to-steel structural connections.

766 Chapter 3

Steel Build ding Materialss

Figgure 3.12 Applicable A AS STM Specificcations for Vaarious Types of Structural Fasteners Coopyright © Am merican Institutee of Steel Consstruction. Reprrinted with Perm mission. All riights reserved.

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F3125 Bolts F3125 is a standard that consolidates what had previously been given in multiple separate standards. It provides for quenched and tempered bolts manufactured in two strength grades, two types, and two styles. Thus, there are 6 grades specified, two of those six being metric equivalents to two other grades. Grade A325 Bolts Grade A325 bolts are heavy hex head bolts that are the predominant high-strength bolts used in construction. Two types are available: type 1, the normal medium carbon bolt; and type 3, which is the same bolt provided in a weathering steel. Bolts are available from 1/2 in. to 1-1/2 in. diameter and the minimum tensile strength is 120 ksi. These bolts are commonly referred to as A325 bolts. Grade A490 Bolts Grade A490 also designates a heavy hex head structural bolt. These fasteners are used when a higher tensile strength is required. As with the Grade A325 bolts, they are available as type 1 or type 3 with the same distinction. Again, the bolts are available from 1/2 in. to 1-1/2 in. diameter and have a minimum tensile strength of 150 ksi. These bolts are commonly referred to as A490 bolts. Grade F1852 Bolts Grade F1852 provides the standard specification for “twist-off” tension control bolt-nut-washer assemblies with a tensile strength of 120 ksi. These structural fasteners are unique in that they do not have a hex head but rather have a splined shank that permits installation through the use of a special torque wrench. Grade F1852 connectors are essentially Grade A325 bolts but are manufactured as a different grade because their geometric characteristics differ from those of heavy hex head bolts. As with Grade A325, the tensile strength of these fasteners is 120 ksi for the diameters available, from 1/2 to 1-1/4 in. These bolts are commonly referred to as A325 TC bolts. Grade F2280 Bolts Grade F2280 provides the standard specification for “twist-off” tension control bolt-nut-washer assemblies, with a tensile strength of 150 ksi. These connectors are essentially Grade A490 bolts but, like Grade F1852 connectors, must be manufactured as a separate grade because their geometric characteristics differ from those of heavy hex head bolts. These fasteners are available in diameters of 1/2 to 1-1/4 in. These bolts are commonly referred to as A490 TC bolts. 3.6.4

Steel for Welding Steel used for welding is called filler metal because it essentially fills the gap between the base metal pieces it is joining. The most critical aspect of selecting filler metal, which actually corresponds to the welding electrode, is matching the electrode with the base metal. In all cases, the weld must not form the weak part of the joint. The American Welding Society provides the specification for appropriate matching of the base metal and electrodes in Table 3.1 of their standard ANSI/AWS D1.1 and the AISC Specification includes some information in Chapter J. The most commonly used weld strength is 70 ksi. A discussion of welding processes and material matching is presented in Chapter 10.

78 Chapter 3

3.6.5

Steel Building Materials

Steel for Headed Stud Anchors Steel headed stud anchors, more commonly known simply as shear studs, are mechanical fasteners welded to structural shapes and embedded in concrete that permit steel and concrete to work together. This is called composite construction. Because these studs are welded to the steel shape, their properties are specified jointly between AWS and ASTM. Shear studs are specified as given in AWS D1.1 Clause 7, with material as required in Clause 7.2.6. Type B is usual and the corresponding mechanical requirements are stated in AWS D1.1 Table 7.1 (Fy = 51 ksi, Fu = 65 ksi).

3.7 AVAILABILITY OF STRUCTURAL STEEL Structural engineers normally use the list of shapes found in the AISC Manual as the basis for design. Unfortunately, all shapes are not equally available in the marketplace and the selection of shapes that are difficult to obtain could negatively impact the overall cost and schedule of a project. Some shapes are available from a wide variety of producers; for instance, a W10×30 could be obtained from six different mills as of April 2016. However, the largest shapes, such as the W44×335, are not produced by all mills. Also, several of the smaller M-shapes are not rolled by any mill. Shape availability data are maintained by the mills on the AISC Web site at www.aisc.org/steelavailability. Another important source – increasingly the primary source – of steel is the steel service center. These organizations are warehouses throughout the country that obtain steel directly from the mills; many stock nearly the full range of shapes. Although the task of obtaining the steel needed for any given project falls to the steel fabricator, it is always beneficial to the engineer to have some knowledge of availability. AISC maintains a list of steel service centers at www.aisc.org which in April 2016 identified 20 AISC Member Steel Service Centers and the states they serve.

3.8 PROBLEMS 1. When was the first AISC Specification published and what was its purpose? 2. In addition to buildings, what other types of structures are included in the scope of the 2010 AISC Specification? 3. Sketch and label a typical stress-strain curve for steel subjected to a simple uniaxial tension test. 4. What is the value of the modulus of elasticity used for calculations according to the AISC Specification, and what does this value represent in relation to the graph of stress versus strain for steel? 5. What happens to a steel element when it is loaded beyond the elastic limit and then unloaded? 6. Describe the difference between the yield stress and ultimate stress of a steel element. 7. Sketch and label 10 different structural shape cross sections whose properties are given in the AISC Manual.

8. What are the nominal and actual depths of a W36×135 wide-flange member? What is the weight of this member per linear foot? (Hint: Use your AISC Manual.) 9. What are the nominal and actual depths of a W16×100 wide-flange member? What is the weight of this member per linear foot? (Hint: Use your AISC Manual.) 10. What are the nominal and actual depths of a W18×130 wide-flange member? What is the weight of this member per linear foot? (Hint: Use your AISC Manual.) 11. What are the nominal and actual depths of a W14×730 wide-flange member? Compare these to the nominal and actual depths of a W14×145. (Hint: Use your AISC Manual.) 12. What are the nominal and actual depths of a W12×336 wide-flange member? Compare these to the nominal and actual depths of a W12×16. (Hint: Use your AISC Manual.)

Chapter 3 Steel Building Materials 13. What are the nominal and actual depths of a W10×112 wide-flange member? Compare these to the nominal and actual depths of a W10×12. (Hint: Use your AISC Manual.) 14. What are the actual depth, flange width, and flange thickness of a W14×132? (Hint: Use your AISC Manual.)

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27. What is the outside diameter of a round HSS10.750×0.375? What are the nominal and design wall thicknesses? (Hint: Use your AISC Manual.) 28. What is the distinction between a round HSS and a pipe?

15. What are the actual depth, flange width, and flange thickness of a W27×84? (Hint: Use your AISC Manual.)

29. What are the outside diameter and nominal and design wall thicknesses of a Pipe 4 xx-Strong? (Hint: Use your AISC Manual.)

16. What are the actual depth and flange width of an HP12×84? What are the thicknesses of the web and flange? (Hint: Use your AISC Manual.)

30. What are the outside diameter and nominal and design wall thicknesses of a Pipe 10 Std.? (Hint: Use your AISC Manual.)

17. What are the actual depth, average flange thickness, and web thickness of a C12×30? (Hint: Use your AISC Manual.)

31. What is the difference between a rectangular bar section and a plate?

18. What are the actual depth, average flange thickness, and web thickness of a C9×15? (Hint: Use your AISC Manual.)

32. What are the three preferred types of steel for shapes in the construction industry? 33. What effects does the addition of carbon have on steel?

19. What are the cross sectional area, leg dimensions and thickness of an L8×8×5/8? (Hint: Use your AISC Manual.)

34. Name three elements that improve the corrosion resistance of steel.

20. What are the cross sectional area, leg dimensions and thickness of an L6×4×1/2? (Hint: Use your AISC Manual.)

35. What grade of steel is most commonly used today in the production of W-shapes, and what are its yield stress and tensile stress?

21. What are the cross sectional area and weight per linear foot of an L5×3×1/2 member? (Hint: Use your AISC Manual.)

36. What grade of steel is most commonly used today in the production of round HSS? Rectangular HSS?

22. What are the actual depth, flange width, flange thickness, and stem thickness of a WT7×45? Compare to the properties for a W14×90. (Hint: Use your AISC Manual.) 23. What are the actual depth, flange width, flange thickness, and stem thickness of a WT10.5×22? Compare to the properties for a W21×44. (Hint: Use your AISC Manual.) 24. What are the outside dimensions of a rectangular HSS12×10×3/8? What are the nominal and design wall thicknesses? (Hint: Use your AISC Manual.) 25. What are the outside dimensions of a rectangular HSS8×6×1/2? What are the nominal and design wall thicknesses? (Hint: Use your AISC Manual.) 26. What is the outside diameter of a round HSS6.000×0.500? What are the nominal and design wall thicknesses? (Hint: Use your AISC Manual.)

37. What are the differences between an A500 Grade C rectangular HSS and an A1085 rectangular HSS? 38. What grade of steel is preferred for the fabrication of most structural shapes other than W-shapes, and what are its yield stress and tensile stress? (Hint: See Figure 3.10.) 39. What grade of steel is typically used for high-strength bolts in construction? 40. What resources can be consulted to determine the availability of a particular steel structural shape?

C Chapterr 4

T Tension Mem mberss

111 South M Main, Salt Lake City, UT Photo courteesy Hassett Enggineering

4.11

INTROD DUCTION The most effficient way to o carry a forcee in steel is thhrough tensioon. Because teension forces result in a fairly unifform stress disstribution in the t member ccross section, all of the maaterial is able to work to its fullest capacity. Th he normal assu umption that tensile forcess are applied to a member through the centroid of the cross section mean ns that other sstructural actiions, such as buckling or bbending, are not norm mally present to reduce thee material’s abbility to carryy load. Thus, tension mem mbers are perhaps the simplest to deesign and a good starting ppoint for studdying structuraal steel designn. Ten nsion memberrs are fairly common c elem ments in buildding structurees, although they are not found in i every stru ucture. The structural meembers considered in thiis chapter arre those subjected to o a concentricc tensile forcce as their prrimary force.. Secondary effects, such as load misalignmen nt and the in nfluence of co onnections, w will be addreessed; howeveer, the interaaction of tension and bending is saaved for treatm ment in Chaptter 8. Table 4.1 lists th he sections off the Specificaation and part rts of the Mannual discussed in this chapter.

4.22

TENSIO ON MEMBE ERS IN STR RUCTURES A wide varriety of tensiion memberss can be fouund in buildiing structurees. Among thhe more important arre members of trusses, braccing memberss, hangers, annd sag rods. Ten nsion memberrs are found in trusses ass chords, diaagonals, and verticals. Figgure 4.1 shows a typ pical simply supported tru uss, with the tension bottoom chord and two of thee tension diagonals in ndicated. Ten nsion membeers used as bbracing for sstructures aree normally loong and slender, as seen s in Figuree 4.2. Becausee these slendeer members arre relatively fflexible, they must be carefully designed and errected, particcularly if therre is any channce of load reeversal, which would call on them m to carry a compression load. Even thee smallest coompressive foorce in a mem mber that has been deesigned as a tension-only y member caan cause signnificant strenngth or servicceability problems in the final stru ucture.

800

Chapter 4 Table 4.1

Tension Mem mbers

81

Sections of o Specification and Parts of Manual Coovered in thiss Chapter Specificattion

B3 B4.2 B4.3 D1 D2 D3 D4 D5 D6 J3.2 J3.5 J3.6 J4.1 J4.3 J7 M2.5

Design Basiis Design Walll Thickness fo for HSS Gross and Net N Area Deteermination Slenderness Limitations Tensile Stren ngth Effective Neet Area Built-up Meembers Pin-Connectted Members Eyebars Size and Usee of Holes Maximum Spacing S and E Edge Distancee Tensile and Shear Strenggth of Bolts annd Threaded P Parts Strength of Elements E in T Tension Block Shearr Strength Bearing Streength Bolted Consstruction Manuall

Part 1 Part 5

Dimensions and Propertiees Design of Teension Membbers

Other O examples of tensio on members are hangers that connect lower floors to some support above, a as seen n in Figure 4.3, and sag roods that suppoort purlins in the roof struucture or the girts in the t walls of a steel-frameed building. IIt is easy to see the importance of m most tension memberss because they normally carry c an obvioous, direct looad. For mem mbers like sagg rods, their failure caan produce unsightly u displacements in the walls, annd could causse stability prroblems for the purlin ns or girts, bu ut they are nott likely to be found carryinng significantt direct loadinng.

Figure 4.1 4

A Simply Supported Truss T with Teension Membbers Indicatedd

Photo cou urtesy Matt Meelrose/LERA Consulting C Struuctural Engineeers

822 Chapter 4

Tension Members

Figure 4.2 Tension Braacing Membeers

ngers Figure 4.3 Tension Han

4.33

CROSS--SECTIONAL SHAPE ES FOR TEN NSION ME EMBERS Tension meembers can be structural steel s shapes, plates, or coombinations oof shapes andd plates; eyebars and d pin-connected plates; rod ds and bars; or wire ropee and steel caables. Wire rrope and steel cables are not coverred by the Speecification noor consideredd here, althouggh they are im mportant elements in the special strructures wherre they occur . Eyebars are not in i common use today but ccan be foundd in older applications, partticularly in bridges, trrusses and sim milar structurres, as seen inn Figure 4.4. A Although rareely used, theyy are still covered in the t Specificattion. The pin--connected pllate shown inn Figure 4.5 iis actually a ppart of a connection. This configu uration is used in a varietyy of applicatiions, such as industrial strructures, anchors for tension t memb bers, and con nnections in brridge girders..

Chapter 4

Figure 4.4 4

Eyebars in an Historicc Building Rooof Structure

Figure 4.5 4

A Pin-Co onnected Mem mber

Tension Mem mbers

83

Photo cou urtesy of LeJeu une Steel Co.

Several S comm mon shapes used u for tensiion memberss are shown iin Figure 4.66, and some typical bu uilt-up shapes are given in n Figure 4.7. T The solid rounnd bar is freqquently used, either as a threaded rod or weldeed to other members. m The threaded endd provides a ssimple connection to the gn must take into account the reductionn in cross-secttional area caaused by the structure, but the desig threads. Rods R with up pset ends are occasionally used instead of the normaal rods; the ennlarged end permits threading t witthout reducin ng the cross-ssectional areaa below the m main portion of the rod. The diffeerences betweeen these two types of rodss can be seen in Figure 4.88. Square, S rectan ngular, and circular c HSS have becomee more comm mon as tensioon members over the past few yeears, largely due to their attractive apppearance annd ease of m maintenance. Howeverr, the end connections may m become complicatedd and expenssive, dependding on the particular application.. HSS are especially usefull for longer teension compoonents, when slenderness and related serviceability consideraations may bee important. Single S angless, as shown in i Figure 4.77a, are used eextensively inn towers, succh as those supportin ng cellular telephone com mmunications and high-volltage power lines. Double angles and double ch hannels, as sh hown in Figu ure 4.7b and dd, are probablly the most ppopular tensioon members for planaar trusses due to the fact thaat gusset plattes can be connveniently plaaced in the space

844 Chapter 4

Tension Members

Figure 4.6 Common Sh hapes for Ten nsion Memberrs between th he individuall shapes. Th he end connnections forr these mem mbers are ttherefore straightforw ward to design n and fabricatee, and allow ffor symmetryy in the verticaal plane. Larg ge tensile forcces may requ uire cross secttions with an area dictatingg that the meember be made from a wide-flangee shape, a tee,, double channnels, or builtt-up shapes, ssuch as those given in Figure 4.7e and f. Built--up cross secctions were m more commonn in the past,, when rolledd shapes were compaaratively smalller in cross section s and thhe cost of labbor was loweer; today largeer rolled shapes are available a and labor costs mean m it is moore economiccal to use them. Current sttructural applications of such elem ments are fou und in long- span roof truusses, bridge trusses, and bracing members in large industriial structures..

4.44

BEHAV VIOR AND STRENGTH H OF TENS SION MEM MBERS Tension meembers are co overed in Ch hapter D of thhe Specificattion. The firsst requiremennt given, Section D1, is not really a requiremen nt at all but reccognition thaat in the past tthere had beenn a limit on the slend derness, the raatio of length h to radius off gyration, L/rr, for tension members. Thhat limit was really a recommend dation since very slenderr members m may be difficcult to handlee during construction n but it had no o impact on th he strength off tension mem mbers. Thus, tthe user note ssuggests that a limit on o L/r of 300 would be reaasonable.

Figure 4.7

Typicaal Built-up Shapes Used as Tension Mem mbers

Chapter 4

Figu ure 4.8

Tension Mem mbers

85

Thrreaded and Up pset Rods

Two T possiblee limit states are defined iin Section D22 for tensionn members: yyielding and rupture. The controllling limit staate depends oon the abilityy of the mem mber to undeergo plastic deformattion. Both of these failure modes repressent limit stattes of strengthh that must bee taken into account in the desig gn of the ten nsion membeer. The desiggn basis for ASD and L LRFD were presented d in Sections 1.9 and 1.10 0, respectivelly. Equations B3-2 and B3-1 are repeaated here to reinforcee the relation nship betweeen the nominnal strength, resistance faactor, and saafety factor presented d throughout the Specificattion. The T requirement for ASD is i

Ra ≤

Rn Ω

(A AISC B3-2)

The T requirement for LRFD D is

Ru ≤ φRn 4.4.1

(A AISC B3-1)

Yield ding Yielding occurs when n the uniform mly distributeed stress throughout the ccross section reaches the yield streess over the length of the member. Althhough the meember will coontinue to ressist the load that caussed yielding to t occur, it wiill undergo exxcessive strettching, and thhis elongationn will make the mem mber unusable. The longer the t member, the greater thhe elongationn. Because thee limit state of yieldiing on the grross section of the membber is accomp mpanied by thhis large defoormation, it readily seerves as a waarning of any impending i faailure. The T yield limiit state is defiined as (A AISC D2-1) Pn = Fy Ag where Pn = nominal tensile yield strength Fy = yield streess Ag = gross areea of the mem mber

4.4.2

The desig gn strength an nd allowable strength are tto be determinned using φt = 0.90 (LRFD) Ωt = 1.667 (ASD) Rupture Holes in a member wiill cause stresss concentrati ons to develoop under the sservice load, aas shown in Figure 4.9. 4 Elastic theory show ws that the stress conccentration ressults in a ppeak stress approxim mately three times t the average stress. As the peakk stress reachhes the yieldd stress, the member will continuee to strain and d load can conntinue to incrrease. With inncreasing loadd, the strain in the reg gion of the ho ole increases into the strainn hardening rregion, and thhe member ruuptures once the stresss in this areaa exceeds thee ultimate streength. Althouugh the mateerial in the reegion of the holes yieelds initially, it does so over o a very shhort length, rresulting in a small total elongation. Thus, the material caan reach its ultimate u strenngth through strain hardeening, withouut excessive on, and failuree occurs throu ugh rupture. T The limit statee of rupture oon the effectivve net area elongatio

866 Chapter 4

Tension Members

Figure 4.9 Stress Conccentration Duee to Hole in M Member of the crosss section is accompanied a by small deeformations ffrom yieldingg, giving littlle or no warning of the impendin ng sudden failure, and off ffering limitedd opportunitiies to take coorrective action beforre the rupture.. The rupture limitt state is defin ned as Pn = Fu Ae (AISC D2-2) where Pn = nominal tensile rupture sttrength Fu = ultimate streess Ae = effective nett area of the member m The design strength s and allowable a streength are to bbe determinedd using ASD) φt = 0.75 (LRFD) ( Ωt = 2.00 (A If the two lim mit states werre to result in n the same avaailable strengtth, using the L LRFD formuulation, 0.75 0 Fu Ae = 0.990 Fy Ag

(4.1)

Ae Ag = 0.90 Fy 0.75 Fu

(4.2)

or

The limit staate of yieldin ng on the grosss section govverns when thhe right-handd side of Equaation 4.1 is less than the t left-hand side. Using Equation E 4.2, yyielding on thhe gross section governs iff (4.3) Ae Ag > 0.90 Fy 0.75 Fu and rupture on the effectiive net section n governs if Ae Ag < 0.90 Fy 0.75 Fu

(4.4)

Steel with a small (Fy /F Fu) value, such h as ASTM A A36, with 0.99Fy/0.75Fu = 0.9(36)/(0.755(58)) = 0.74, will allow more off the cross seection to be removed in tthe form of bbolt holes beefore the ns than will stteel with a hiigher (Fy /Fu) value, such aas ASTM A992, with rupture limitt state govern 0.9Fy/0.75F Fu = 0.9(50)/(0 0.75(65)) = 0.92. The comparisonss discussed ab bove are appllicable only ffor normal boolted connectiions and their corresp ponding areass. The Specifiication equatiions are not inntended to coover tension m members with large cutouts. c Thesee require speccial design coonsiderations,, and are beyyond the scope of this book becausse they are no ot common in most buildinng structures.

Chapter 4

Tension Members

87

Although welded connections do not normally require the removal of material from the cross section, the placement of the welds and the type of cross section may require a reduction from the gross area to determine the effective net area.

4.5 COMPUTATION OF AREAS The design of tension members uses the following cross-sectional area definitions: 1. 2. 3.

gross area, Ag net area, An effective net area, Ae

The criteria governing the computation of the various areas required for tension member analysis and design are given in Sections B4.3 and D3 of the Specification. They are discussed in further detail here. 4.5.1

Gross Area The gross area of a member might also be thought of as the full cross-sectional area. A section is made perpendicular to the longitudinal axis of the element, along which the tensile force is acting, and the gross area, Ag, is the area of that cross section. No holes or other area reductions are considered when calculating the gross area. In the case of plates, bars, and solid circular shapes, the value of Ag is found directly as the value of width times thickness, bt, for plates and bars, and as πd 2 4 for circular shapes, where d is the diameter. For structural steel shapes commonly used in construction, the Manual provides values for gross areas in Part 1. However, in lieu of using the tabulated values, Ag may be approximated as (4.5) Ag = Σwi ti

where wi and ti are the width and thickness, respectively, of the rectangular cross-sectional element, i, of the shape. Equation 4.5 applies only to shapes that are composed of flat plate components, such as wide flanges. The calculation for hollow circular shapes is similarly straightforward. The gross area of HSS meeting the requirements of ASTM A500 is determined using 93 percent of the nominal wall thickness of the shape, as indicated in Section B4.2. Because HSS are consistently manufactured with a thickness at the low end of the tolerance limit, the values provided in the Manual are all based on this reduced thickness. For channels and S-shapes, both with sloping flanges, the inaccuracy of this approach means that the approach is rarely used. For all hot-rolled shapes, the area is easily obtained from the tables in Part 1 of the Manual. The procedure for angles requires a slight modification. The angle may be treated as an equivalent flat plate, where the effective width is taken as the sum of the leg dimensions less the thickness, and the gross area is this effective width times the angle thickness. An angle and its equivalent flat plate are shown in Figure 4.10. 4.5.2

Net Area

The net area is obtained by subtracting the area of any holes occurring at a particular section from the gross area at that section. Thus, holes associated with mechanical fasteners, such as bolts, and with welds, such as plug welds and slot welds, are considered. These fastening elements are normally used only to connect tension members to the adjacent parts of the structure, and the reduced areas therefore normally appear at member ends. However, if holes occur at any point along a tension member, their effect must be considered.

888 Chapter 4

Tension Members

Figu ure 4.10 An ngle and its Equivalent E Flaat Plate Plug g and slot welds are made through holees in a membeer and are relatively uncom mmon in most structu ures in which the framin ng members are made frrom structuraal shapes andd plates. Normal wellded joints do o not involvee making holles in membeers. Thus, weelds do not nnormally reduce the cross-sectionaal area of tension members, so the net arrea is equal too the gross areea. In the t computattion of net area a for a teension membber with boltted end connnections, determining g the size of the holes is important. T The criteria foor standard, oversize, andd slotted holes are cov vered in Speccification Secttion J3.2. H Standard Holes Normal steeel constructio on requires th he specificati on of fasteneer size ratherr than hole siize. The hole is then n sized accord ding to whatt is required to accommodate the fastener. The maanner in which the ho ole is fabricatted is also crittical. In recognizing th he needs for fabrication an and erection ttolerances, staandard bolt hholes for bolts up to 7/8 7 in. in diam meter are mad de 1/16 in. laarger in diameeter than the bolt to be insserted in the hole. Fo or 1 in. diameeter bolts and d larger, the sstandard hole may be 1/8 in. larger in ddiameter than the bolt to be inserrted in the hole. h This inccrease for larrger bolts is an option thhat some fabricators may m deem un nnecessary; ho owever, sincee that may noot be known bby the designeer ahead of time, the larger hole sh hould be used d in calculatioons. Thus, a 3/4 in. bolt rrequires a holle with a (3/4+1/16) = 13/16 in. diameter d and a 1 in. diameeter bolt requuires a 1-1/8 diameter holee. In the case of pun nched holes, the punching g process maay damage soome of the m material imm mediately adjacent to the t hole. Thaat material maay not be connsidered fully effective in ttransmitting lload and must also bee deducted fro om the gross area along wiith the materiial that has acctually been reemoved. Figure 4.11 shows an ang gle leg with a hole along w with the punchhed out piece.. As the punchh is

g with Puncheed Hole and P Plug that was removed by P Punching Figure 4.11 Angle Leg

Chapter 4

Tension Members

89

applied to the material, the edges around the hole are deformed, as seen in the taper of the punched out piece. In discounting this region, the effective hole diameter is increased by another 1/16 in., regardless of bolt diameter, according to Specification Section B4.3b. The deduction is conservatively applied to all holes, since one generally does not know whether holes will be punched or drilled. The most common practice for producing holes is through punching. However, any method that meets the hole surface quality requirements of Specification Section M2.5 is permitted. Because the decision to punch or drill a hole is a function of the steel fabricator’s equipment capacity, for the design of tension members it is standard practice to deduct for holes with a diameter 1/8 in. greater than the specified bolt size up to 7/8 in. and 3/16 for 1 in. bolts and above. The following examples demonstrate gross and net area calculations for several shapes.

EXAMPLE 4.1 Gross and Net Area

Goal:

Determine the gross and net areas of a plate with a single line of holes.

Given:

A single line of standard holes for 3/4 in. bolts is placed in a 6×1/2 plate, as shown in Figure 4.12a.

SOLUTION

Step 1:

Determine the gross area at Section 1-1. Ag = 6 (1 2 ) = 3.0 in.2

Step 2:

Determine the effective hole size for a 3/4 in. diameter bolt.

Step 3:

Determine the net area at section 2-2. An = ( b − d e ) t = ( 6.0 − 7 8 )(1 2 ) = 2.56 in.2

de = ( 3 4 + 1 16 + 1 16) = 7 8 in.

EXAMPLE 4.2 Gross and Net Area

Goal:

Determine the gross and net areas of a plate with a double line of holes.

Given:

A double line of standard holes for 7/8 in. bolts are placed in a 10×3/4 plate, as shown in Figure 4.12b.

SOLUTION

Step 1:

Determine the gross area at section 1-1. Ag = 10 ( 3 4 ) = 7.5 in.2

Step 2:

Determine the effective hole size for a 7/8 in. diameter bolt.

Step 3:

Determine the net area at section 2-2. An = ( b − nd e ) t = (10.0 − 2 (1.0 ) ) ( 3 4 ) = 6.00 in.2

de = ( 7 8 + 1 16 + 1 16) = 1.0 in.

900 Chapter 4

Tension Members

Figure 4.12 Plates and Shapes with Holes for Us e with Exampples 4.1 throuugh 4.4

EX XAMPLE 4.3 4 Grross and Net Arrea

Goa al:

Deterrmine the gross and net areeas of an anggle with a singgle line of hooles in each leg. l

Giv ven:

A sin ngle line of sttandard holess for 7/8 in. bbolts is placed on each legg of a 6×6×¾ ¾ angle as sh hown in Figurre 4.12c.

SO OLUTION

Step p 1:

Deterrmine the grosss area at secttion 1-1 from m Manual Tabble 1-7. Ag = 8.46 in.2

Step p 2:

Deterrmine the effeective hole sizze for a 7/8 inn. diameter boolt.

Step p 3:

Deterrmine the net area at sectioon 2-2. An = Ag − nd e t = 8.46 − 2 (1.0 ))( 3 4 ) = 6.96 in.2

de = ( 7 8 + 1 16 + 1 16) = 1.0 in.

EX XAMPLE 4.4 4 Grross and Net Arrea

Goal:

Determine the gross and net are as of a channnel with muultiple lines oof holes.

Given:

Four lines of standard d holes for 1 in. bolts are placed in an MC12×31, aas shown in n Figure 4.12 2d. Two liness are in the w web and one line is in eacch flange.

SO OLUTION

Step 1:

Determine the gross area a at sectionn 1-1 from M Manual Table 1-6.

Ag = 9.12 in.2 Step 2:

Determine the effective hole size ffor a 1.0 in. ddiameter bolt.

Chapter 4

Tension Mem mbers

91

de = (1.0 + 1 8 + 1 16) = 1.19 in. Step 3:

Deterrmine the nett area at sectioon 2-2. First determine thee web and flaange thicknesses from Mannual Table 1-6. tw = 0.3700 in. and tf = 00.700 in. An = Ag − nw d e t w − n f d e t f = 9.12 − 2 (1.119 )( 0.370 ) − 2 (1.19 )( 0.7000 ) = 6.57 in.2

Oversizeed and Slotteed Holes Section J3.2 J of the Specification S gives the reequired measuurements forr larger-than-standard or oversized d holes, as well w as for sh hort-slotted aand long-slottted holes. Figgure 4.13 illuustrates the criteria that t apply fo or nominal bolt b diameterrs of 5/8 in., 3/4 in. annd 7/8 in.; rrefer to the Specifica ation for data for other bollt sizes. Thesse types of hooles are used to facilitate tthe erection of the strructure and, in n some casess, to permit laarger rotationss or deformattions to take pplace under loading. Short Co onnecting Ellements Tension members witthin connectio ons are usuallly short connnecting elemeents such as liinks, flange plates, orr gusset platees. When the member is sshort, and thee net area andd gross area aare close to equal, th here may not be sufficient length for thhe load to sprread to a uniform distribuution on the entire cro oss section. In I this case, the t area that is the first too yield may rreach rupture at an early stage, an nd the rupturee limit state would w therefoore be reachedd prematurelyy. This is an undesirable mode off failure, prim marily because it is not duuctile and couuld occur sudddenly, with little or no warning. Section J4.1 indicates thaat the effectivve net area in these short cconnecting eleements may be limiteed due to strress distributiion as calcullated by methhods such ass the Whitmoore section, which is discussed in Chapter 11. 4.5.3

Influ uence of Holee Placement The exam mples in Sectiion 4.5.2 reprresent simple cases in which the net areea is found inn the section that prod duces the larg gest reduction n in area, usuually the sectiion with the llargest number of holes. Howeverr, hole placem ment does no ot always folllow simple ppatterns wherre every secttion has the same num mber of holess. It is sometim mes advantaggeous to use a pattern of staaggered holess, such as

Figure 4.13 4

Size Crriteria for 5/8 to 3/4 in. Bollts in Oversizze and Slottedd Holes

922 Chapter 4

Tension Members

Figure 4.14 Staggered Hole Patternss in Plate andd Angle those shown n in Figure 4.14. Figure 4.14a 4 shows aan arrangemeent of staggerred holes for a plate, and Figure 4.14b 4 shows an example for an angle. When there are multiple holes, the ceenter-tocenter distan nce between adjacent holes in the direection paralleel to the prim mary applied force is defined as th he pitch, s. When W there is more than onne line of holes parallel to the line of foorce, the center-to-cen nter distance between adjjacent holes in the directtion perpendiicular to the primary applied force is the gage, g. It iss not clear fro om Figure 4.1 14 what the ggoverning nett section wouuld be for eithher case. For the platee, sections 1--1 and 2-2 giv ve identical A n values, in w which a deduuction for onee hole is taken for eaach line. Ano other possibillity would bee to follow a line that inccorporates tw wo holes, starting alon ng line 1 an nd ending along line 2, as shown byy the diagonnal dashed liine. The Specification n refers to thiis line as a “chain” becaus e it links indiividual holes. In this case, the area of two holess would be deducted d from m the gross crross section. H However, thiis approach w would be no differentt than if both h holes weree along the ssame straightt line, line 1 or line 2. IIt seems reasonable in this situatio on that an app proach that woould deduct bboth holes woould deduct tooo much, because thesse holes are staggered s and d are not alonng the same straight line. The correct solution should be so omewhere between the ressult of deductting for one hhole and that of deductingg for two holes. A siimplified approach to addrress the interaaction of stagggered holes w was adopted llong ago by previouss AISC Speciifications. Allthough numeerous studiess have been conducted siince this original sim mplification was w introducced, none haave proposedd a significaantly more accurate approach thaat is equally easy e to implem ment. The Specification n approach, given g in Sectiion B4.3b, reequires that evvery potentiaal failure line be assesssed with the full area of eaach intersecteed hole deduccted and someething added back for the increaseed strength prrovided by th he diagonal ppath. For every diagonal oon a potentiaal failure path, the quaantity s 2 4 g is added bacck into the nett width to acccount for the overestimatioon of the required ded duction when n a full adjaceent hole has bbeen deductedd. Examples 4.5 and 4.6 sshow the application of the staggeered hole criteerion in the m middle of the member where the tensioon flows through the member. Wh hen holes are at the end off a member aas part of a coonnection thee flow of the tension force f through h the connecto ors should be considered.

Chapter 4

Tension Mem mbers

93

ure 4.15 Holle Pattern forr 18 in. Plate U Used in Exam mple 4.5 Figu

EXAMPLE E 4.5 Net Width and Net Area of Pla ate

Goal: G

Deetermine the net n width of a plate with sttaggered holees. Then deterrmine the nett area.

Given: G

The hole pattern n for an 18×77/8 in. plate with holes foor 3/4 in. bollts that is on as shown in Figure 4.115. A and F rrepresent the edges of loaaded in tensio thee plate and d B, C, D D, and E represent the hole llocations.

SOLUTION N

Step S 1:

Ch hain ABF (a sttraight line thhrough one hoole): Deeduct for one hole h

( 3 4 + 1 8) = −00.875 in.

Step S 2:

Ch hain ABCF: Deeduct for two holes h

2 ( 3 4 + 1 8) = −1.75 in.

Forr BC add

s 4 g = ( 2.0 ) 2

2

Tottal deduction

Step S 3:

= −1.50 in.

Ch hain ACEF Deeduct for two holes h Forr CE add

2 ( 3 4 + 1 8) = −1.75 in. s 2 4 g = ( 2.5 )

( 4 (10.0 ) ) = +0.16 in.

2

= −1.59 in.

Tottal deduction

Step S 4:

Ch hain ABCEF Deeduct for threee holes

3( 3 4 + 1 8) = −2.63 in.

( 4 ( 4.0 ) ) = +0.25 in. 4 g = ( 2.5 ) ( 4 (10.0 ) ) = +0.16 in.

Forr BC add

s 2 4 g = ( 2.0 )

Forr CE add

2

Tottal deduction

( 4 ( 4.0 ) ) = +0.25 in.

s2

2

= −2.22 in.

94 Chapter 4

Tension Members Step 5:

Chain ABCDEF Deduct for four holes

4 ( 3 4 + 1 8) = −3.50 in.

( 4 ( 4.0 ) ) = +0.25 in. 4 g = ( 4.5 ) ( 4 ( 6.0 ) ) = +0.84 in. 4 g = ( 2.0 ) ( 4 ( 4.0 ) ) = +0.25 in.

s 2 4 g = ( 2.0 )

2

For CD add

s2

2

For DF add

s2

For BC add

2

= −2.16 in.

Total deduction Step 6:

Deduct the largest quantity to obtain the least net width. For chain ABCEF, the net width is bn = 18.0 − 2.22 = 15.8 in.

Step 7:

Use the least net width to determine the net area An = bn t = 15.8 ( 7 8 ) = 13.8 in.2

EXAMPLE 4.6 Gross Area and Net Area

Goal:

Determine the gross and net areas of an angle with staggered holes.

Given:

An L6×4×1/2 with holes for 7/8 in. bolts placed in the middle of the member as shown in Figure 4.16.

SOLUTION

Step 1:

Determine the width of the equivalent flat plate representing the width of the angle. we = l1 + l2 − t = 6 + 4 − 1 2 = 9.50 in.

Step 2:

Determine the gross area of the equivalent plate. Ag = we t = 9.50 (1 2 ) = 4.75 in.2 This is the same as found in Manual Table 1-7

Step 3:

Determine the gages for each bolt line. The workable gages for the holes as found in Table 1-7A are shown in Figure 4.16. The gage between the holes closest to the heel of the angle in the two legs must be adjusted to account for the angle thickness. Thus,

( g + g1 − t ) = 2.50 + 2.25 − 0.50 = 4.25 in.

Step 4:

Determine the net area. The governing net section will be section 2-2 or section 2-1-2. There is no need to consider section 1-1 because bn1 will clearly be greater than bn2. For section 2-2, the net width is bn 2 = ( 9.5 − 2 ( 7 8 + 1 8 ) ) = 7.5 in. and the net area is

An 2 = 7.5 ( 0.5 ) = 3.75 in.2

Chapter 4

Tension Mem mbers

95

Fo or section 2-1 1-2, this chainn has two staaggers of boltt holes, and bboth have thee same pitch (s = 2.50 in)). The gages are differentt, with one att 4.25 in. an nd the other att 2.50 in. Th he net width for f this chain is ⎛ 2.50 2 ⎞ ⎛ 2.50 2 ⎞ bn 3 = 9.5 − 3(7 8 + 1 8) + ⎜ ⎟+⎜ ⎟ = 7.49 in. ⎝ 4(2.5 ) ⎠ ⎝ 4(4.25)) ⎠ an nd the net areaa is

An 3 = 7.49(0.5) = 3.75 in. Step S 5:

Seelect the least net area. west net areaa controls. In this case, thee two chains yield the The low same net n area, An = An 2 = An3 = 3.75 in.2

Figure 4.16 4

Hole Paattern for L6× ×4×1/2 Used in Example 44.6

Figure 4.17 4

Concep ptual Basis forr Shear Lag R Reduction Facctor

96 Chapter 4 4.5.4

Tension Members

Effective Net Area

When all elements of a tension member are attached to connecting elements, the entire member participates fully in transferring the load to the connection. However, when not all elements are attached to connecting elements, they cannot all participate fully. Figure 4.17 shows an angle with one leg attached to a connecting element and the other, the outstanding leg, unattached. To account for the inability of this unattached leg to transfer load, the net area used in calculating the rupture strength must be reduced to an effective net area, Ae. This phenomenon occurs because the uniform stresses, occurring near the mid-length of the member at some distance from the connection, must be transferred through the more restricted area where the connection is located. The portion of the member area that is participating effectively in the transfer of force is smaller than the full net area. Thus, the net area is reduced to the effective net area. This general behavior is called shear lag. Since its introduction into the Specification, it has been approximated by the use of the shear lag reduction factor such that Ae = UAn Specification Table D3.1 provides values of the shear lag factor, U, for a wide variety of elements. For all tension members, except plates, HSS and those with longitudinal welds, when the tension load is transmitted to some but not all of the cross-sectional elements, the effective length of the connection is reduced to l′ = l – x , where x is the distance from the attached face to the member centroid and l is the length of the connection, as shown in Figure 4.17. The reduction in net area is then taken in proportion to the reduction in effective length, l′/l. Thus, the reduction becomes l′ l − x x (4.6) U= = =1− l l l Figure 4.18 shows the definition of connection length, l, for both a bolted and a welded connection. For a welded connection, the length of the connection, l, is the length of the weld. For a bolted connection, l is the distance between the bolt holes at each end of the connection. Table D3.1 of the Specification also provides simplified approaches to finding the shear lag factor when certain criteria are met. Case 7 of Table D3.1: for W-, M-, S-, and HP-shapes or for tees cut from these shapes, the following apply: Flange connected with three or more fasteners per line in the direction of loading

bf ≥ 2/3d bf < 2/3d

Web connected with four or more fasteners per line in the direction of loading

U = 0.90 U = 0.85 U = 0.70

Case 8 of Table D3.1: for single and double angles: With four or more fasteners per line in the direction of loading With three fasteners per line in the direction of loading

U = 0.80 U = 0.60

Chapter 4

Figure 4.18 4

mbers Tension Mem

97

Definitiion of Connection Length,, L, for Boltedd and Weldedd Connection

If I U is calculaated using Caase 7 or Casee 8 as well as Case 2, the llarger value iis permitted to be useed. For any other o conditio ons, the tablee values shouuld be used. H However, for open cross sections (such as W-,, M-, S-, C-, HP-, WT-, aand ST-shapees, single anggles, and doubble angles), n need to tak ke U as less th han the ratio of the gross area of the coonnected elem ments to the there is no member gross area. This will also prove useeful when thhere is only one row of connectors mber axis. transversse to the mem When W tension n members arre connected with staggerred bolts andd not all elem ments of the member are connecteed, the effeccts of bolt sttagger, shearr lag and tennsion flow tthrough the on must be co onsidered. connectio

Figure 4.19 4

Single-Angle Tensio on Member foor Example 4.7

(a)

Figure 4.20 4

(b)

A WT Welded W to a Gusset G Plate ffor Examples 4.8 and 4.9

98 Chapter 4

Tension Members

EXAMPLE 4.7 Tensile Strength of an Angle

Goal:

Determine the design strength (LRFD) and the allowable strength (ASD) of an angle.

Given:

Consider an L4×4×1/2 attached through one leg to a gusset plate with 3/4 in. bolts in standard holes as shown in Figure 4.19. Use A36 steel.

SOLUTION

Step 1:

Determine the gross and net areas needed for calculations. Ag = 3.75 in.2 (from Manual Table 1-7) An = 3.75 − ( 3 4 + 1 8 )(1 2 ) = 3.31 in.2

Step 2:

Determine the shear lag factor and the effective net area. First using Case 8 from Table D3.1, there are only three fasteners in a line for this connection, U = 0.60. The minimum U for this open shape connected as shown is Gross area of connected leg 4.0(0.5) U= = = 0.533 Gross area of angle 3.75 Based on connection length, the shear lag factor for a 6 in. connection length and an angle with x = 1.18 in. (from Manual Table 1-7) is 1.18 x U =1− =1− = 0.80 6 L Thus, use U = 0.80, the largest permitted value, and

Ae = 0.80(3.31) = 2.65 in.2 For LRFD Step 3:

For the limit state of yielding, Pn = Fy Ag = 36 ( 3.75 ) = 135 kips φPn = 0.90 (135 ) = 122 kips

Step 4:

For the limit state of rupture, Pn = Fu Ae = 58 ( 2.65 ) = 154 kips φPn = 0.75 (154 ) = 116 kips

Step 5:

For ASD Step 3:

Determine the design strength. The limit state of rupture controls and the design strength is φPn = 116 kips

For the limit state of yielding, Pn = Fy Ag = 36(3.75) = 135 kips

Pn 135 = = 80.8 kips Ωt 1.67 Step 4:

For the limit state of rupture,

Chapter 4

Tension Members

99

Pn = Fu Ae = 58(2.65) = 154 kips Pn 154 = = 77.0 kips Ωt 2.00

Determine the allowable strength. The limit state of rupture controls and the allowable strength is Pn = 77.0 kips Ωt Note that the controlling limit state is the same for ASD and LRFD. Step 5:

EXAMPLE 4.8 Tensile Strength of a Tee

Goal:

Determine the design strength (LRFD) and the allowable strength (ASD) of a WT.

Given:

Consider a WT6×32.5 attached to a gusset plate with equal length longitudinal welds as shown in Figure 4.20a. Use A992 steel.

SOLUTION

Step 1:

Determine the gross and net areas: Ag = 9.54 in.2 (from Manual Table 1-8) Because the force is transferred by welds only, there are no holes and An = Ag

Step 2:

Determine the shear lag factor and the effective net area. The minimum U for this open shape connected as shown is Gross area of connected flange 12.0(0.605) U= = = 0.761 Gross area of WT 9.54 Based on connection length, the shear lag factor for a 12 in. connection length with longitudinal welds only and a tee with bf = 12.0 and x = 0.985 in. (from Manual Table 1-8) is 2 3 (12.0 ) 3l 2 ⎛ x⎞ ⎛ 0.985 ⎞ U= 2 1 − = 1− ⎟ ⎟ = 0.688 2 2 ⎜ 2 ⎜ 3l + w ⎝ l ⎠ 3 (12.0 ) + 12.0 ⎝ 12 ⎠ Thus, use U = 0.761 and

Ae = 0.761(9.54) = 7.26 in.2 For LRFD Step 3

For the limit state of yielding, Pn = Fy Ag = 50 ( 9.54 ) = 477 kips φPn = 0.90 ( 477 ) = 429 kips

Step 4:

For the limit state of rupture, Pn = Fu Ae = 65 ( 7.26 ) = 472 kips φPn = 0.75 ( 472 ) = 354 kips

100 Chapter 4

Tension Members Step 5: For ASD Step 3:

Therefore, the limit state of rupture controls and the design strength is φPn = 354 kips

For the limit state of yielding, Pn = 50(9.54) = 477 kips Pn 477 = = 286 kips Ωt 1.67

Step 4:

For the limit state of rupture, Pn = 65(7.26) = 472 kips Pn 472 = = 236 kips Ωt 2.0

Step 5:

Therefore, the limit state of rupture controls and the allowable strength is Pn = 236 kips Ωt

EXAMPLE 4.9 Tensile Strength of a Tee

Goal:

Determine the design strength (LRFD) and the allowable strength (ASD) of a WT.

Given:

Consider a WT6×32.5 attached to a gusset plate with unequal length longitudinal welds as shown in Figure 4.20b. Use A992 steel.

SOLUTION

Step 1:

Determine the gross and net areas: Ag = 9.54 in.2 (from Manual Table 1-8)

Step 2:

Because the force is transferred by welds only, An = Ag Determine the shear lag factor. The minimum U for this open shape connected as shown is Gross area of connected flange 12.0(0.605) U= = = 0.761 Gross area of WT 9.54 Based on connection length, 6 + 18 l +l l= 1 2 = = 12 in. 2 2 the shear lag factor for a 12 in. connection length with longitudinal welds only and a tee with bf = 12.0 and x = 0.985 in. (from Manual Table 1-8) is 2 3 (12 ) 3l 2 ⎛ x⎞ ⎛ 0.985 ⎞ U= 2 1 − = ⎜ ⎟ ⎜1 − ⎟ = 0.688 3l + w2 ⎝ 12 ⎠ l ⎠ 3 (12 )2 + 12.0 2 ⎝ Thus, use U = 0.761

Chapter 4 Step 3: For LRFD Step 4:

For ASD Step 4:

Tension Members

101

Determine the effective net area. Ae = 0.761(9.54) = 7.26 in.2

The effective net area is the same as it was for Example 4.8, thus tension rupture will again control φPn = 354 kips

The effective net area is the same as it was for Example 4.8, thus tension rupture will again control Pn = 236 kips Ωt

4.6 DESIGN OF TENSION MEMBERS To design a structural steel tension member, the member size must be determined and then the appropriate limit states checked. The only additional issue to address is the slenderness of the member. For tension members, slenderness is defined as the member length divided by the least radius of gyration, L/r. The Specification has, in the past, placed a limit on the slenderness of tension members. However, since slenderness of a tension member is primarily a serviceability limit state, there is currently no specified limitation on tension member slenderness, as indicated in Section D1. The designer should exercise caution when selecting tension members with very high slenderness ratios, that is, those near the former limit of L/r = 300, because these members could easily be damaged during erection and might cause other problems due to their flexibility in the transverse direction. Because the main task in tension member design is to determine the area of the member, the two limit states of yielding and rupture can be used to determine minimum gross and net areas such that P ΩP Ag min = u (LRFD) or Ag min = t a (ASD) Fy φt Fy and P ΩP Ae min = u (LRFD) or Ae min = t a (ASD) Fu φt Fu Because connection details are not normally known in the early stages of member selection, it may not be possible to determine the actual deductions necessary to obtain the exact effective net area of the member being designed. One approach would be to assume a fixed percentage deduction for the effective net area. The designer would decide the magnitude of this deduction and then confirm the adequacy of this deduction at the completion of the design. Part 5 of the Manual provides tables for tension member design that give the strength of tension members based on the limit states of yielding on the gross area and rupture on an effective net area equal to 0.75Ag. This is a reasonable estimate of a value of Ae that is practical to achieve with typical end connections, not a minimum value or a prescriptive way to calculate Ae. If the actual effective net area differs from this assumed value, the designer can simply adjust the strength accordingly.

102 Chapter 4

Tension Members

EXAMPLE 4.10a Tension Member Design by LRFD

SOLUTION

EXAMPLE 4.10b Tension Member Design by ASD

SOLUTION

Goal:

Select a double-angle tension member for use as a web member in a truss, and determine the maximum area reduction that would be permitted for holes and shear lag.

Given:

The member must carry a dead load of PD = 67.5 kips and a live load of PL = 202.5 kips. For the load combination 1.2PD + 1.6PL, the LRFD required strength is Pu = 405 kips. Use equal leg angles of A36 steel.

Step 1:

Determine the minimum required gross area based on the limit state of yielding where ϕ = 0.9: Ag min = Pu φFy = 405 ( 0.9 ( 36 ) ) = 12.5 in.2

Step 2:

Based on this minimum gross area, from Manual Table 1-15, select 2L6×6×9/16 with Ag = 12.9 in.2

Step 3:

Determine the minimum effective net area based on the limit state of rupture where ϕ = 0.75: Ae min = Pu φFu = 405 ( 0.75 ( 58 ) ) = 9.31 in.2

Step 4:

Thus, the combination of holes and shear lag may not reduce the area of this pair of angles by more than Ae Ag = 9.31 12.9 = 0.722

Goal:

Select a double-angle tension member for use as a web member in a truss and determine the maximum area reduction that would be permitted for holes and shear lag.

Given:

The member must carry a dead load of PD = 67.5 kips and a live load of PL = 202.5 kips. For the load combination PD + PL, the ASD required strength is Pa = 270 kips. Use equal leg angles of A36 steel.

Step 1:

Determine the minimum required gross area based on the limit state of yielding where Ω = 1.67: Ag min = Pa ( Fy Ω ) = 270 ( 36 1.67 ) = 12.5 in.2

Step 2:

Based on this minimum gross area, from Manual Table 1-15, select 2L6×6×9/16 with Ag = 12.9 in.2

Step 3:

Determine the minimum effective net area based on the limit state of rupture where Ω = 2.00: Ae min = Pa ( Fu Ω ) = 270 ( 58 2.00 ) = 9.31 in.2

Thus, the combination of holes and shear lag may not reduce the area of this pair of angles by more than Ae Ag = 9.31 12.9 = 0.722 Note that since the ratio PL/PD = 3.0, the required areas for the LRFD and ASD solutions are the same. Step 4:

Chapter 4

Tension Members

103

EXAMPLE 4.11a Tension Member Design by LRFD

Goal:

Select a WT9 for use as a tension member and determine the maximum area reduction that would be permitted for holes and shear lag.

Given:

The member must carry an LRFD required strength of Pu = 818 kips. Use A992 steel.

SOLUTION

Step 1:

Determine the minimum required gross area based on the limit state of yielding where ϕ = 0.9: Ag min = Pu φFy = 818 ( 0.9 ( 50 ) ) = 18.2 in.2

Step 2:

Based on the minimum gross area, from Manual Table 1-8, select WT9×65 with Ag = 19.2 in.2

Step 3:

Determine the minimum effective net area needed to resist the applied force where ϕ = 0.75: Ae min = Pu φFu = 818 ( 0.75 ( 65 ) ) = 16.8 in.2

Step 4:

The combination of holes and shear lag may not reduce the area of this WT by more than Ae Ag = 16.8 19.2 = 0.875

EXAMPLE 4.11b Tension Member Design by ASD

Goal:

Select a WT9 for use as a tension member and determine the maximum area reduction that would be permitted for holes and shear lag.

Given:

The member must carry an ASD required strength of Pa = 545 kips. Use A992 steel.

SOLUTION

Step 1:

Determine the minimum required gross area based on the limit state of yielding where Ω = 1.67: Ag min = Pa ( Fy Ω ) = 545 ( 50 1.67 ) = 18.2 in.2

Step 2:

Based on the minimum gross area, from Manual Table 1-8, select WT9×65 with Ag = 19.2 in.2

Step 3:

Determine the minimum effective net area needed to resist the applied force where Ω = 2.00: Ae min = Pa ( Fu Ω ) = 545 ( 65 2.00 ) = 16.8 in.2

Step 4:

The combination of holes and shear lag may not reduce the area of this WT by more than Ae Ag = 16.8 19.2 = 0.875

1004 Chapter 4

Tension Members M

(a) ( Figure 4.21 Example of o Block Sheaar Failure of a Plate

((b)

Photo courtessy Robert Driv ver

Figgure 4.22 B Block Shear G Geometry forr Exaample 4.10

4.77

BLOCK K SHEAR mber tears ou ut as shown iin Figure 4.221a, a combinnation of tenssion and When a porrtion of a mem shear resultss as shown in n Figure 4.21 1b, and the faailure is know wn as a blockk shear failurre. Even though this failure f mode is primarily the t result of a connection ffailure, it mayy possibly conntrol the overall stren ngth of a tenssion member. The resistannce to tear-ouut is provided by a combinnation of shear on the plane paralleel to the tensio on force and ttension on the plane perpeendicular to itt. Rup pture will alw ways be the co ontrolling mo de on the tennsion face of tthe failure bloock, due to the relativ vely short len ngth of material that will bee available too yield. The coontrolling lim mit states on the shearr face will be either yieldin ng or rupturee, whichever hhas the lowerr strength. Unnlike the situation forr overall mem mber strength h, in which thhe yield and rrupture limit states were ffound to have different resistance and safety faactors, block sshear uses thee same valuess for both lim mit states. mple comparisson of nomin nal strengths is appropriatte for determ mining the conntrolling Thus, a sim limit state. Section S J4.3 of o the Specificcation gives thhe block sheaar strength as Rn = 0.6 Fu Anv + U bs Fu Ant ≤ 00.6 Fy Agv + U bbs Fu Ant (AIS SC J4-5) where i shear Agv = gross area in

Chapter 4

Tension Members

105

Anv = net area in shear Ant = net area in tension Ubs = 1.0 if the tension stress is uniform and 0.5 if the tension stress is not uniform. For tension members, the tensile stress is assumed to be uniform. Thus, Ubs = 1.0 will be used (Ubs = 0.5 is addressed in Chapter 10.) The design and allowable strengths are determined using φ = 0.75 (LRFD) Ω = 2.0 (ASD) The block shear strength given by Equation J4-5 can be simplified by recognizing that it simply requires the combination of the minimum shear strength, yielding or rupture, with the tensile rupture strength. It is also important to note that the calculation of net area for both tension and shear is determined by deducting the area of the holes from the gross areas.

EXAMPLE 4.12 Gusset Plate Tension Strength

Goal:

Determine whether the gusset plate has sufficient strength in block shear.

Given:

The gusset plate shown in Figure 4.22 has a plate thickness of 1/2 in. The required strength for LRFD is Pu = 225 kips and for ASD is Pa = 150 kips. The steel is A36 and the holes are punched for 7/8 in. bolts.

SOLUTION

Step 1:

Determine the areas needed to perform the calculations. Ant = ( 6 − ( 7 8 + 1 8 ) ) (1 2 ) = 2.50 in.2 Agv = 2 (11)(1 2 ) = 11.0 in.2

Anv = 2 (11.0 − 3.5 ( 7 8 + 1 8 ) ) (1 2 ) = 7.50 in.2

Step 2:

Determine the nominal block shear strength: Determine the tension rupture strength Fu Ant = 58 ( 2.50) = 145 kips Consider shear yield and shear rupture and select the least nominal strength; thus, 0.6Fy Agv = 0.6 ( 36)(11.0 ) = 238 kips

0.6Fu Anv = 0.6 ( 58)( 7.50) = 261 kips Selecting the shear yield term and combining it with the tension rupture term gives a block shear nominal strength , using Ubs = 1.0, of Rn = 238 + 1.0 (145 ) = 383 kips For LRFD Step 3:

For LRFD, the design strength is φRn = 0.75 ( 383) = 287 > 225 kips Because the design strength is greater than the required strength of 225 kips, the gusset plate is adequate to resist this force based on block shear.

For ASD Step 3:

For ASD, the allowable strength is

1006 Chapter 4

Tension Members M Rn Ω = 3883 2.00 = 192 > 150 kips

Because the allo owable streng gth is greater than the reqquired strengtth of 150 kipps, the gussset plate is ad dequate to resiist this force bbased on blocck shear.

Figure 4.23 Spliced Teension Membeer for Exampple 4.13

EX XAMPLE 4.13a 4 Teension Stren ngth of Sppliced Memb bers by LR RFD

Go oal:

Deteermine the deesign strengthh of a splice bbetween two W W-shapes.

Giiven:

Two o W14×43 A992 wide flannges are splicced by flangee plates, as shhown in Figu ure 4.23, witth 7/8 in. diaameter bolts arranged as shown in standard holees. The LRFD D available sstrength of a group of six bolts in a fllange is 227 kips. The plaates will be seelected such tthat they do nnot limit the m member ngth. stren

SO OLUTION

Sttep 1:

Deteermine the deesign strengthh for the limit state of yieldding of the W W14×43. m Manual Taable 1-1, Ag = 12.6 in.2 From Thu us, Pn = Fy Ag = 50 (12.6) = 630 kips

φPn = 0.90 ( 630) = 567 kips Sttep 2:

Deteermine the net area of thee W14×43. T The area to bbe deducted ffor each flan nge, with tf = 0.530 0 in., is 2 ( 7 8 + 1 8 )( 0.530 ) = 1.06 in.2 Thu us, with the deeduction for tw two flanges frrom the gross area An = 122.6 − 2 (1.06 ) = 10.5 in.2

Sttep 3:

Deteermine the sh hear lag factoor. The W14× ×43 is treatedd as two tee seections, each h a WT7×21.5. The x foor each WT iis found in M Manual Tablee 1-8 as

Chapter 4

Tension Members

107

1.31 in. With L = 2(3.0) = 6.0 in., U =1−

1.31 = 0.782 6.0

Specification Table D3.1 provides that for this case, if b f ≥ 2d 3 , a value of U = 0.90 may be used and if b f < 2 d 3 U = 0.85. From Manual Table 1-1, bf = 8.0 in. and d = 13.7 in. thus: b f = 8.0 < 2 (13.7 ) 3 = 9.13 in. Therefore use U = 0.85 Step 4:

Determine the design strength for the limit state of rupture. Ae = UAn = 0.85 (10.5 ) = 8.93 in.2

Pn = Fu Ae = 65 ( 8.93) = 580 kips φPn = 0.75 ( 580) = 435 kips Step 5:

Determine the design block shear strength of the flanges. The block shear limit state must be checked for tear-out of the flanges, as shown in Figure 4.24. The calculations will be carried out for one block as shown in the figure and the total obtained by adding the results for all four flange sections. Rupture on the tension plane: Fu Ant = 65 ( 2.0 − (1 2 )( 7 8 + 1 8 ) ) ( 0.530 ) = 51.7 kips Yield on the shear plane: 0.6Fy Agv = 0.6 ( 50)( 8.00)( 0.530) = 127 kips Rupture on the shear plane: 0.6 Fu Anv = 0.6 ( 65 ) ( 8.00 − 2.5 ( 7 8 + 1 8 ) ) ( 0.530 ) = 114 kips Because shear rupture is less than shear yield, the design strength for a single block shear element is, with Ubs = 1.0, Rn = (114 + 1.0 ( 51.7 ) ) = 166 kips

φRn = 0.75 (166) = 125 kips and the block shear strength of the W14×43 is φRn = 4 (125) = 500 kips Step 6:

Compare the design strength for each limit state. Bolt design shear strength (given, 2 flanges ×227 kips/(bolt group)) Yielding of member Rupture of the member Block shear for the member

454 kips 576 kips 435 kips 500 kips

1008 Chapter 4

Tension Members M

Step 7:

Rup pture of the member m contrrols the desiggn. Therefore,, the design sstrength of th he splice is φRn = 435 kiips

Figuree 4.24 Block k Shear Checck for Exam mple 4.13 EX XAMPLE 4.13b 4 Teension Stren ngth of Sppliced Memb bers by AS SD

Go oal:

W-shapes. Deteermine the deesign strengthh of a splice bbetween two W

Giiven:

Two o W14×43 A992 wide flannges are splicced by flangee plates, as shhown in Figu ure 4.23, witth 7/8 in. diaameter bolts arranged as shown in standard holees. The ASD available streength of a grooup of six bollts in a flangee is 151 member kipss. The plates will be seleccted such thaat they do noot limit the m stren ngth.

SO OLUTION

Sttep 1:

W14×43. Deteermine the deesign strengthh for the limit state of yieldding of the W From m Manual Taable 1-1, Ag = 12.6 in.2

Pn = Fy Ag = 50 (122.6 ) = 630 kipps Pn Ω = 630 1.67 = 377 kips

Sttep 2:

The area to bbe deducted ffor each Deteermine the net area of thee W14×43. T flan nge, with tf = 0.530 0 in., is 2 ( 7 8 + 1 8 )( 0.530 ) = 1.06 in.2 two flanges frrom the gross area Thu us, with the deeduction for tw An = 122.6 − 2 (1.06 ) = 10.5 in.2

Sttep 3:

×43 is treatedd as two tee seections, Deteermine the sh hear lag factoor. The W14× for each WT is found in M Manual Tablee 1-8 as each h a WT7×21 1.5. The x fo 1.31 1 in. With L = 2(3.0) = 6.0 in., 1.31 = 00.782 U =1− 6.0

Chapter 4

Tension Members

109

Specification Table D3.1 provides that for this case, if b f ≥ 2d 3 , a value of U = 0.90 may be used and if b f < 2 d 3 U = 0.85. From Manual Table 1-1, bf = 8.0 in. and d = 13.7 in. thus: b f = 8.0 < 2 (13.7 ) 3 = 9.13 in. Therefore use U = 0.85 Step 4:

Determine the allowable strength for the limit state of rupture. Ae = UAn = 0.85 (10.5 ) = 8.93 in.2

Pn = Fu Ae = 65 ( 8.93) = 580 kips Pn Ω = 580 2.00 = 290 kips

Step 5:

Determine the allowable block shear strength of the flanges. The block shear limit state must be checked for tear-out of the flanges, as shown in Figure 4.24. The calculations will be carried out for one block as shown in the figure and the total obtained by adding the results for all four flange sections. Rupture on the tension plane: Fu Ant = 65 ( 2.0 − (1 2 )( 7 8 + 1 8 ) ) ( 0.530 ) = 51.7 kips Yield on the shear plane: 0.6Fy Agv = 0.6 ( 50)( 8.00)( 0.530) = 127 kips Rupture on the shear plane: 0.6 Fu Anv = 0.6 ( 65 ) ( 8.00 − 2.5 ( 7 8 + 1 8 ) ) ( 0.530 ) = 114 kips Because shear rupture is less than shear yield, the allowable strength for a single block shear element is, with Ubs = 1.0, Rn = (114 + 1.0 ( 51.7 ) ) = 166 kips Rn Ω = 166 2.00 = 83.0 kips

and the block shear strength of the W14×43 is Rn Ω = 4 ( 83.0) = 332 kips Step 6:

Compare the design strength for each limit state. Bolt design shear strength (given, 2 flanges ×151 kips/(bolt group)) Yielding of member Rupture of the member Block shear for the member

Step 7:

302 kips 377 kips 290 kips 332 kips

Rupture of the member controls the design. Therefore, the design strength of the splice is

110 Chapter 4

Tension Members Rn Ω = 290 kips

EXAMPLE 4.14a Tension Strength of an Angle by LRFD

Goal:

Determine the design strength of one of a pair of angles in a tension member.

Given:

The truss diagonal member in Figure 4.25 consists of a pair of angles L4×3×3/8 that are loaded in tension. The bolts to be used are 3/4 in. and the steel is A36. The bolt design shear strength for one angle in this connection is 53.7 kips.

SOLUTION

Step 1:

Determine the single angle design strength for the limit state of yielding. Ag = 2.49 in.2

Pn = Fy Ag = 36 ( 2.49) = 89.6 kips

φPn = 0.90 ( 89.6 ) = 80.6 kips Step 2:

Determine the single angle design strength for the limit state of rupture. An = 2.49 − ( 3 4 + 1 8 )( 3 8 ) = 2.16 in.2 The shear lag coefficient is U =1−

x 0.775 =1− = 0.871 6.0 l

and the effective net area is Ae = AnU = 2.16 ( 0.871) = 1.88 in.2 The nominal strength is Pn = Fu Ae = 58 (1.88) = 109 kips Therefore the design strength is φPn = 0.75 (109) = 81.8 kips Step 3:

Determine the single angle design strength in block shear. Rupture on the tension plane: Fu Ant = 58 (1.5 − (1 2 )( 3 4 + 1 8 ) ) ( 3 8 ) = 23.1 kips Yield on the shear plane: 0.6Fy Agv = 0.6 ( 36)( 7.25)( 3 8) = 58.7 kips Rupture on the shear plane: 0.6 Fu Anv = 0.6 ( 58 ) ( 7.25 − 2.5 ( 3 4 + 1 8 ) ) ( 3 8 ) = 66.1 kips Because shear yield is less than shear rupture, the design strength for a single block shear element is, with Ubs = 1.0, Rn = ( 58.7 + 1.0 ( 23.1) ) = 81.8 kips

φRn = 0.75 ( 81.8) = 61.4 kips

C Chapter 4

Step 4:

T Tension Mem mbers

Compare C the single angle ddesign strengtth for each lim mit state. 53.7 kipps 80.6 kipps 81.8 kipps 61.4 kipps

Bolt B design sttrength Yielding Y of th he member Rupture R of thee member Block B shear for fo the membeer

Step 5:

111

The T bolt desig gn strength controls the deesign. Thereffore, the desiggn strength of o one angle is i φRn = 53. 7 kips The T design strrength for thhe pair of anglles is φRn = 2 ( 53.7 ) =107 kips

Figure 4.225 Truss Diiagonal Mem mber for Exampple 4.14

EXAMPLE E 4.14b Tension Sttrength of an Angle by ASD

Goal:

Determine D the allowable sstrength of oone of a pair of angles inn a tension member. m

Given:

The T truss diaagonal membber in Figuree 4.25 consissts of a pair of angles L4×3×3/8 L thaat are loaded iin tension. Thhe bolts to be used are 3/4 in. and the steel s is A36. The bolt aallowable sheear strength for one anggle in this connection c is 35.7 kips.

SOLUTION N

Step 1:

Determine D thee single anglee allowable sttrength for thee limit state oof yielding. Ag = 2.49 in.2

Pn = Fy Ag = 366 ( 2.49 ) = 89.66 kips Pn Ω = 89.6 1. 67 = 53.7 kipps

Step 2:

Determine D thee single anglee allowable sttrength for thee limit state oof rupture. An = 2.449 − ( 3 4 + 1 8 )( 3 8 ) = 2.166 in.2

112 Chapter 4

Tension Members The shear lag coefficient is U =1−

x 0.775 =1− = 0.871 l 6.0

and the effective net area is Ae = AnU = 2.16 ( 0.871) = 1.88 in.2 The nominal strength is Pn = Fu Ae = 58 (1.88) = 109 kips Therefore the allowable strength is Pn Ω = 109 2.00 = 54.5 kips Step 3:

Determine the single angle allowable strength in block shear. Rupture on the tension plane: Fu Ant = 58 (1.5 − (1 2 )( 3 4 + 1 8 ) ) ( 3 8 ) = 23.1 kips Yield on the shear plane: 0.6Fy Agv = 0.6 ( 36)( 7.25)( 3 8) = 58.7 kips Rupture on the shear plane: 0.6 Fu Anv = 0.6 ( 58 ) ( 7.25 − 2.5 ( 3 4 + 1 8 ) ) ( 3 8 ) = 66.1 kips Because shear yield is less than shear rupture, the allowable strength for a single block shear element is, with Ubs = 1.0, Rn = ( 58.7 + 1.0 ( 23.1) ) = 81.8 kips Rn Ω = 81.8 2.00 = 40.9 kips

Step 4:

Compare the single angle allowable strength for each limit state. Bolt allowable strength Yielding of the member Rupture of the member Block shear for the member

Step 5:

35.7 kips 53.7 kips 54.5 kips 40.9 kips

The bolt allowable strength controls the design. Therefore, the allowable strength of one angle is Rn Ω = 35.7 kips The design strength for the pair of angles is Rn Ω = 2 ( 35.7 ) =71.4 kips

4.8

PIN-CONNECTED MEMBERS When a pin connection is to be made in a tension member, a hole is cut in both the member and the parts to which it is to be attached. A pin is inserted in the hole and a mechanical means is

C Chapter 4

T Tension Mem mbers

113

found to keep the eleements togeth her. This typee of connectioon is the clossest to a true frictionless F 4.26 shows s the endd of a pin-coonnected mem mber and the dimensions pin as caan be made. Figure needed to o determine its i strength. These T membeers are not paarticularly com mmon in builldings; they are used mainly for special applications, such aas hangers inn suspension structures or connecting b structu ures. links in bridge Specification S Section D5 identifies thee limit statess for which pin-connected members must be designed. Th hese are (1) teension on the effective nett area, (2) sheear on the efffective area, ng on the pro ojected area of the pin, andd (4) yieldingg on the gross section. The strength of (3) bearin the pin-cconnected tension member is taken as thhe lowest streength predicteed by each off these limit states. For F tension on n the effectivee net area, acttually a ruptuure limit state Pn = Fu ( 2tbe ) (A AISC D5-1)

φt = 0.75(LRFD) Ωt = 2.000 (ASD) where t is i the thickness of the plate and be is thhe effective w width of the pllate, taken as (2t + 0.63) in inchess, but not morre than the acctual distancee from the edgge of the holee to the edge of the part, measured d perpendicullar to the direction of the fo force, be ≤ b . For F shear on the t effective area, a again a rrupture limit state Pn = 0.6 Fu Asf (A AISC D5-2) φsf = 0..75(LRFD)

Ωsf = 2.000 (ASD)

where Asf = 2t(a + d/2 2), a is the sh hortest distancce from the eddge of the pinn hole to the edge of the member measured parrallel to the diirection of thee force, and d is the pin diameter. For F bearing on n the projecteed area of the pin, from Seection J7, Pn = 1.8Fy Apb ((AISC J7-1)

φt = 0.75 (LRFD))

Ωt = 2.000 (ASD)

where Apb ojected area of o the pin. p = td, the pro For F yielding in the gross seection,

Pn = Fy Ag φt = 0.90 (LRFD))

(A AISC D2-1)

Ωt = 1.67 (ASD)

Figure 4.226 Pin-Connnected Tensioon Member

EXAMPLE E 4.15a Pin-Conneected Member Design by LRFD

Goal:

Design D a pin-connected meember using L LRFD.

Given:

A dead load of o 30 kips annd a live loadd of 70 kips aare to be suppported. The steel s is A572 Gr. 50 with a yield stress of 50 ksi andd an ultimate strength of 65 6 ksi. Use a 3/4 in. plate w with a 4 in. piin.

114 Chapter 4

Tension Members

SOLUTION

Step 1:

Determine the required strength. Pu = 1.2 ( 30) + 1.6 ( 70) = 148 kips

Step 2:

Determine the minimum required effective net width for the limit state of rupture based on Equation D5-1, P 148 = 2.02 in. ( be )min = u = φFu 2t 0.75(65)(2)(0.750) and the maximum be from Section D5.1 is ( be )max = 2t + 0.63 = 2 ( 0.750 ) + 0.63 = 2.13 in. Therefore, try a 9.0 in. plate, which will give an actual distance to the edge of the plate equal to 2.5 in. which is greater than (be)min. Thus, (be)max is used to calculate the rupture strength of the plate.

Step 3:

Determine the design strength of this 9×3/4 plate for the limit state of tension rupture using Equation D5-1. φPn = φFu ( 2tbe ) = 0.75 ( 65 ) ( 2 ( 0.750 ) ) ( 2.13 ) = 156 > 148 kips

Step 4:

Determine the design strength for the limit state of shear rupture using Equation D5-2. For a 9 in. plate and a 4 in. pin, a = b = 2.5 in. Asf = 2t ( a + d 2 ) = 2 ( 0.750 )( 2.5 + 4.0 2 ) = 6.75 in.2 φPn = φ0.6 Fu Asf = 0.75 ( 0.6 ( 65 ) ) ( 6.75 ) = 197 > 148 kips

Step 5:

Determine the design strength for the limit state of bearing on the projected area of the 4 in. pin using Equation J7-1. Apb = td = 0.750 ( 4.0 ) = 3.0 in.2

φPn = φ1.8Fy Apb = 0.75 (1.8)( 50)( 3.0) = 203 > 148 kips Step 6:

Determine the design strength for the limit state of yielding on the gross area of the member using Equation D2-1. φPn = φFy Ag = 0.9 ( 50)( 0.750)( 9.00 ) = 304 > 148 kips

Step 7:

Since each limit state has design strength greater than the required strength, the 9×3/4 pin-connected member with a 4 in. pin will be sufficient to carry the applied load.

EXAMPLE 4.15b Pin-Connected Member Design by ASD

Goal:

Design a pin-connected member using ASD.

Given:

A dead load of 30 kips and a live load of 70 kips are to be supported. The steel is A572 Gr. 50 with a yield stress of 50 ksi and an ultimate strength of 65 ksi. Use a 3/4 in. plate with a 4 in. pin.

SOLUTION

Step 1:

Determine the required strength.

Chapter 4

Tension Members

115

Pa = 30 + 70 = 100 kips

Step 2:

Determine the minimum required effective net width for the limit state of rupture based on Equation D5-1, 100 ( 2.00 ) PΩ = 2.05 in. ( be )min = a = Fu 2t 65(2)(0.750) and the maximum be from Section D5.1 is ( be )max = 2t + 0.63 = 2 ( 0.750 ) + 0.63 = 2.13 in. Therefore, try a 9.0 in. plate, which will give an actual distance to the edge of the plate equal to 2.5 in. which is greater than (be)min. Thus, (be)max is used to calculate the rupture strength of the plate.

Step 3:

Determine the allowable strength of this 9×3/4 plate for the limit state of tension rupture using Equation D5-1. Pn Ω = Fu ( 2tbe ) Ω = 65 ( 2 ( 0.750 ) ) ( 2.13 ) 2.00 = 104 > 100 kips

Step 4:

Determine the allowable strength for the limit state of shear rupture using Equation D5-2. For a 9 in. plate and a 4 in. pin, a = b = 2.5 in. Asf = 2t ( a + d 2 ) = 2 ( 0.750 )( 2.5 + 4.0 2 ) = 6.75 in.2

Pn Ω = 0.6Fu Asf Ω = 0.6 ( 65)( 6.75) 2.00 = 132 > 100 kips Step 5:

Determine the allowable strength for the limit state of bearing on the projected area of the 4 in. pin using Equation J7-1. Apb = td = 0.750 ( 4.0 ) = 3.0 in.2

Pn Ω = 1.8Fy Apb Ω = 1.8 ( 50)( 3.0 ) 2.00 = 135 > 100 kips Step 6:

Determine the allowable strength for the limit state of yielding on the gross area of the member using Equation D2-1. Pn Ω = Fy Ag Ω = 50 ( 0.750)( 9.00) 1.67 = 202 > 100 kips

Step 7:

Since each limit state has allowable strength greater than the required strength, the 9×3/4 pin-connected member with a 4 in. pin will be sufficient to carry the applied load.

4.9 EYEBARS AND RODS Eyebar tension members are not commonly used in new construction but may be found in special applications where the objective is a design that has some historical context. Thus, the provisions for the design of these members are still found in Section D6 of the Specification. Historically, they were commonly used as tension members in trusses and as links forming the main tension member in suspension bridges. Eyebars are designed only for the limit state of yielding on the

1116 Chapter 4

Tension Members M gross section n because thee dimensionall requirementts preclude thhe possibility of failure at aany load below that leevel. Figure 4.27 4 shows a schematic off an eyebar annd Figure 4.4 shows an eyeebar in a building app plication. Rod ds are commo only used forr tension mem mbers in situuations wheree the requiredd tensile strength is small. Thesee tension meembers are ggenerally connsidered secoondary membbers and include sag rods, hangerss, and tie rodss, although thhere are cases where rods pplay a significcant role p forcees. Rods may y also be usedd as part of thhe lateral braacing system in walls in carrying primary and roofs. hough it is po ossible to con nnect rods by welding to thhe structure, tthreading andd bolting Alth is the most common c meaans of connecttion. Rods caan be threadedd in two wayss. Standard roods have threads that reduce the cross-sectiona c al area througgh the removval of materiaal. The upset rod has ds, with the threads t reducing that area to somethingg larger than the gross areea of the enlarged end rod. The streength of the rod depends on o the mannerr in which thee threads are aapplied. For a standard th hreaded rod, th he nominal sttrength is given in Specificcation Sectionn J3.6 as Fn = 0.75Fu over the areaa of the unthreeaded body oof the rod, which gives Pn = 0.755Fu Ab (4-7) and for desig gn, φt = 0.75 (LRFD) ( Ωt = 2.000 (ASD)

4.110

BUILT T-UP TENSION MEMB BERS o the Specifiication allowss tension mem mbers that aree fabricated frrom a combinnation of Section D4 of shapes and plates. p Their strength is determined in the same waay as the strenngth for single-shape tension mem mbers. Howev ver, the desig gner must rem member that in bolted buiilt-up membeers, bolts are usually placed p along the member length to tie the various sshapes togethher. These bollts result in holes alon ng the membeer length, nott just at the ennds, so that ruupture on thee effective nett section may becomee the controllling limit state at a locatioon other than the member eend. Welded built-up members wiill not normallly experiencee rupture on thhe effective nnet area. Perfforated coverr plates or tie plates can be used to tie the separrate shapes ttogether. Limitations on the spacin ng of these elements are aalso providedd in Section D D4, and requiirements for the placeement of boltss can be found in Section JJ3.5.

Figure 4.277 Eyebar Geeometry

4.111

TRUSS S MEMBER RS ommon tensio on members found in buillding structurres are the tennsion web annd chord The most co members off trusses. Tru usses are norrmally foundd as roof struuctures and aas transfer sttructures within a buiilding. Depen nding on the particular p loadd patterns thaat a truss mighht experiencee, a truss member mig ght be called d upon to alw ways resist teension or to resist tensionn in some caases and

C Chapter 4

T Tension Mem mbers

117

compresssion in otherrs. In cases in which a member is required to carry both ttension and compresssion, it will need n to be sizeed accordinglly. Because thhe compressioon strength oof a member is normaally significan ntly less than the tension strength of thhat same mem mber, as willl be seen in Chapter 5, 5 compressio on may actuallly control thee design. The T typical trruss member can be compposed of either single shappes or a com mbination of shapes. When W composed of a com mbination of shhapes, the reqquirements ddiscussed in S Section 4.10 must be included. Ottherwise, trusss members aare designed the same way as any otther tension member discussed in this chapter. Examples 4 .10, 4.11, andd 4.14 showeed the applicaation of the p to several s truss tension t membbers. tension provisions

4.12

BRA ACING ME EMBERS t memberrs, members used u to providde lateral loaad resistance ffor a buildingg might also As with truss be called d upon to carrry tension un nder some coonditions and compressionn under others. They too would neeed to be desiigned to resisst both loads. However, it is often morre economicall to provide twice as many tension n members an nd to assume tthat if a tensioon member w were called uppon to resist d buckle and d therefore caarry no load. This would permit all brracing to be compresssion, it would designed d as tension-o only memberrs and almosst certainly ppermit them tto have a sm maller cross section than t if they were w required d to resist coompression. A An additionall simplificatioon that this assumptiion entails is the eliminattion of potenntial compression memberrs from the aanalysis for member forces. This may result in the structuure being a determinate structure rathher than an minate one, thu us simplifying g the analysiss. indeterm

F igure 4.28 Brraced Frame for Example 4.16

EXAMPLE E 4.16a Tension Member M Design by LRFD L

Goal:

Select S an apprropriate tensi on member fo for use as a brrace in an X-bbraced frame. f

Given:

The T member is i a tension-oonly diagonal as shown in F Figure 4.28 aand must resist r a force resulting r from m the horizonntal wind loadd, W = 120 kipps. Use a pair p of equal leg l angles or a threaded rood of A36 steeel.

SOLUTION N

Step 1:

Determine D thee force in the brace labeledd AC if the brrace labeled B BD is assumed a not to t act since it would be in compression for this loadiing. The most m critical load l combinaation would bee LRFD Loadd Combinatioon 5, 0.9 Dead D + 1.0 Wind. W Using siimilar trianglees, ⎛ 72.1 ⎞ Pu = 1.0 (120 ) ⎜ ⎟ = 144 kips ⎝ 60.0 ⎠

118 Chapter 4

Tension Members Step 2:

Determine the minimum required gross area based on the limit state of yielding. Ag min = 144 ( 0.90 ( 36 ) ) = 4.44 in.2

Step 3:

Based on this minimum gross area, from Manual Table 1-15, select 2L2-1/2×2-1/2×1/2 with Ag = 4.52 in.2

Step 4:

Determine the minimum effective net area based on the limit state of rupture. Ae min = 144 ( 0.75 ( 58 ) ) = 3.31 in.2

Step 5:

Thus, the combination of holes and shear lag may not reduce the area of this pair of angles by more than Ae Ag = 3.31 4.52 = 0.732

Step 6:

Determine the slenderness ratio of the selected double angles and compare it with the recommended maximum of 300. From Manual Table 1-15, rx = 0.735 in. and L 72.1(12 ) = = 1180 > 300 rx 0.735 A check of Manual Table 1-15 shows that no pair of double angles will satisfy the slenderness limit of 300.

Step 7:

Select a threaded rod to meet the strength requirement. The maximum slenderness limit does not apply to rods. For yielding, the minimum gross area is the same as for the double angles. For rupture of a threaded rod, it is reasonable to assume that the effective net area will not be less than 0.75Ag. Thus, using the minimum effective net area from step 3, Ae = 3.31, the corresponding minimum gross area is 3.31 Ag = = 4.41 0.75 Thus, the limit state of yielding controls and the minimum required area is 4.44 in.2. Therefore, select a 2½ in. diameter threaded rod with A = 4.91 in.2.

EXAMPLE 4.16b Tension Member Design by ASD

Goal:

Select an appropriate tension member for use as a brace in an X-braced frame.

Given:

The member is a tension-only diagonal as shown in Figure 4.28 and must resist a force resulting from the horizontal wind load, W = 120 kips. Use a pair of equal leg angles or a threaded rod of A36 steel.

Chapter 4

SOLUTION

Tension Members

119

Step 1:

Determine the force in the brace labeled AC if the brace labeled BD is assumed not to act since it would be in compression for this loading. The most critical load combination would be ASD Load Combination 5, Dead + 0.6Wind. Using similar triangles, ⎛ 72.1 ⎞ Pa = 0.6 (120 ) ⎜ ⎟ = 86.5 kips ⎝ 60.0 ⎠

Step 2:

Determine the minimum required gross area based on the limit state of yielding. Ag min = 86.5 ( 36 1.67 ) = 4.01 in.2

Step 3:

Based on this minimum gross area, from Manual Table 1-15, select 2L3-1/2×3-1/2×5/16 with Ag = 4.20 in.2

Step 4:

Determine the minimum effective net area based on the limit state of rupture. Ae min = 86.5 ( 58 2.00 ) = 2.98 in.2

Step 5:

Thus, the combination of holes and shear lag may not reduce the area of this pair of angles by more than Ae Ag = 2.98 4.20 = 0.710

Step 6:

Determine the slenderness ratio of the selected double angles and compare it with the recommended maximum of 300. From Manual Table 1-15, rx = 1.08 in. and L 72.1(12 ) = = 801 > 300 rx 1.08 A check of Manual Table 1-15 shows that no pair of double angles will satisfy the slenderness limit of 300.

Step 7:

Select a threaded rod to meet the strength requirement. The maximum slenderness limit does not apply to rods. For yielding, the minimum gross area is the same as for the double angles. For rupture of a threaded rod, it is reasonable to assume that the effective net area will not be less than 0.75Ag. Thus, using the minimum effective net area from step 3, Ae = 2.98, the corresponding minimum gross area is 2.98 Ag = = 3.97 0.75 Thus, the limit state of yielding controls and the minimum required area is 4.01 in.2. Therefore, select a 2-3/8 in. diameter threaded rod with A = 4.43 in.2.

1120 Chapterr 4 44.13

Tension n Members

PROB BLEMS

11. Determine th he gross and neet areas for an 8×3/4 in. platee w with a single lin ne of standard holes for 7/8 in n. bolts. 22. Determine th he gross and neet areas for a 10×1/2 in. platee w with a single lin ne of standard holes for 3/4 in n. bolts. 33. Determine th he gross and net n areas for a 6×5/8 in. platee w with a single lin ne of standard holes for 1 in. bolts.

15. D Determine the gross and net areas for a C115×40 with four lines of standaard holes for 33/4 in. bolts. E Each flange will ccontain one linne of bolts and the web will ccontain two lines of bolts. 16. D Determine the gross and net areas for a C110×25 with two llines of standarrd holes in the web for 7/8 inn. bolts. 17. D Determine the net width forr a 10×1/2 in. plate with 3/4 iin. bolts placeed in three linnes as shown in Figure P4.177.

44. Determine th he gross and neet areas for a 10×1/2 in. platee w with two lines of o standard holles for 7/8 in. bolts. b 55. Determine th he gross and neet areas for a 12×5/8 in. platee w with two lines of o standard holles for 3/4 in. bolts. b 66. Determine th he gross and net n areas for a 14×1 in. platee w with three liness of standard ho oles for 3/4 in. bolts. 77. Determine the t gross and net areas forr an L4×4×1/2 2 w with two lines,, one in each leg, of standard d holes for 3/4 4 inn. bolts. 88. Determine the t gross and net areas forr an L5×5×5/8 8 w with two lines,, one in each leg, of standard d holes for 7/8 8 inn. bolts. 99. Determine the t gross and net areas for an L8×4×9/16 6 w with three liness of standard holes, h two in th he 8 in. leg and d oone in the 4 in. leg, for 7/8 in.. bolts.

P4.177 18. D Determine the net width forr a 12×1/2 in. plate with 3/4 iin. bolts placeed in three linnes as shown in Figure P4.188.

110. Determine the gross and net n areas for a WT8×20 with h thhree lines off standard holles for 3/4 in n. bolts. Each h eelement of the WT W will be atttached to the co onnection. 111. Determine the gross and d net areas fo or a WT12×31 1 w with three lines of standard holes for 7/8 in. i bolts. Each h eelement of the WT W will be atttached to the co onnection. 112. Determine the gross and net n areas for a WT7×41 with h thhree lines off standard holles for 3/4 in n. bolts. Each h eelement of the WT W will be atttached to the co onnection.

P4.188 19. D Determine the net area for thhe L6×4×5/8 w with 7/8 in. boltss shown in Figuure P4.19.

113. Determine the gross and net n areas for a WT9×25 with h ffour lines of staandard holes fo or 3/4 in. bolts. Each elementt oof the WT will be attached to o the connection (two lines off bbolts in the stem m). 114. Determine the gross and net areas for a C15×50 with h ffive lines of staandard holes for f 7/8 in. boltts. Each flangee w will contain on ne line of boltts and the weeb will contain n thhree lines of bo olts. P4.199

Chhapter 4

Teension Membbers

121

220. Determine the net area fo or the L6×4×5/8 with 7/8 in. bbolts shown in Figure P4.20.

P4.233

P P4.20 221. Determine the gross and net areas for a double-anglee tension membeer composed of o two L4×4× ×1/2 shapes ass sshown in Figu ure P4.21 wiith holes for 3/4 in. boltss sstaggered in each leg.

24. F For a WT7×34 attached throuugh a flange to a 10×1 in. platee with six 3/4 iin. bolts at a sppacing of 3 in.., placed in two rows of threee bolts as shown in Figuure P4.24, deterrmine the sheaar lag factor (consider Case 2 and Case 7) annd effective nett area of the W WT.

P4.244 P P4.21 222. Determine the gross and net areas for a double-anglee tension membeer composed of o two L6×6× ×5/8 shapes ass sshown in Figu ure P4.22 wiith holes for 3/4 in. boltss sstaggered in each leg.

25. A single L6×6× ×1 is used as a tension brace in a multistoryy building. Onee leg of the anggle is attached to a gusset platee with a single line of three 77/8 in. bolts att a spacing of 3 in. Determinee the shear lagg factor (considder Case 2 and C Case 8) and efffective net areaa. 26. A single L7×44×3/4 is used as a tension brace in a multii-story buildingg. The 7 in. legg of the angle is attached to a ggusset plate wiith a single linee of four 7/8 inn. bolts at a spaciing of 3 in. Deetermine the sshear lag factorr (consider Casee 2 and Case 8)) and effective net area.

P P4.22 223. For a WT8 8×50 attached through t a flang ge to a 12×3/4 4 inn. plate with eight e 7/8 in. bo olts at a spacin ng of 3 in. and d pplaced in two rows r as shown n in Figure P4 4.23, determinee thhe shear lag factor f (consideer Case 2 and d Case 7) and d eeffective net area of the WT.

27. A single L8×66×1/2 is used as a tension brace in a multii-story buildingg. The 8 in. legg of the angle is attached to a gusset plate w with two lines of four 3/4 in. bolts at a spaciing of 3 in. Deetermine the sshear lag factorr (consider Casee 2 and Case 8)) and effective net area. 28. T The WT8×50 oof Problem 23 is welded aloong the tips of thhe flange for a length of 12 in. on eaach flange. Deterrmine the sheaar lag factor (C Case 2) and efffective net area for the WT. 29. D Determine the available strenngth of a 12×11/2 in. A36 platee connected to two 12 in. plates, as shownn in Figure P4.299, with two lines of 3/4 in. bolts. C Considering

1122 Chapterr 4

Tension n Members

yyielding and neet section ruptu ure, determine the (a) design n sstrength by LRF FD and (b) allo owable strengtth by ASD.

36. D Determine the available strenngth of a WT9×20, A992 steel,, with the flangges welded to a 1/2 in. gussett plate with uneqqual length lonngitudinal wellds as shown in Figure 4.20bb. Consideringg yielding annd net sectioon rupture, deterrmine the (a) design strenngth by LRFD D and (b) allow wable strength bby ASD. 37. D Determine the available strenngth of a WT7×15, A992 steel,, with the flangges welded to a 1/2 in. gussett plate with uneqqual length lonngitudinal weldds, a 5 in. weldd and a 11 in. w weld. Considerring yielding and net sectioon rupture, deterrmine the (a) design strenngth by LRFD D and (b) allow wable strength bby ASD.

P P4.29 330. Determine the availablee strength of an L6×4×3/4 4 aattached throug gh the long leg to a gusset plaate with ten 3/4 4 inn. bolts, in two o lines of five bolts, at a 3 in n. spacing. Usee A A36 steel. Con nsidering yield ding and net seection rupture,, ddetermine the (a) design strength s by LRFD L and (b)) aallowable stren ngth by ASD. 331. Determinee the available strength of o a C8×11.5 5 aattached throug gh the web to a gusset plate with eight 7/8 8 inn. bolts, in two o lines of four bolts, at a 3 in n. spacing. Usee A A36 steel. Con nsidering yield ding and net seection rupture,, ddetermine the (a) design strength s by LRFD L and (b)) aallowable stren ngth by ASD. 332. Determine the t available sttrength of a C1 12×30 attached d thhrough the weeb to a gusset plate p with ten 7/8 7 in. bolts, in n tw wo lines of fiv ve bolts, at a 3 in. spacing. Use U A36 steel. C Considering yiielding and neet section ruptu ure, determinee thhe (a) design n strength by LRFD and (b) allowablee sstrength by ASD D. 333. Determine the available strength s of an 8×1/2 8 in. A572 2 G Gr. 50 plate co onnected with three lines off 7/8 in. bolts. C Considering yiielding and neet section ruptu ure, determinee thhe (a) design n strength by LRFD and (b) allowablee sstrength by ASD D. 334. Determine the available strength s of a WT7×15, W A992 2 ssteel, with the flanges f welded d to a 1/2 in. gu usset plate by a 110 in. weld along each sidee of the flangee. Considering g yyielding and neet section ruptu ure, determine the (a) design n sstrength by LRF FD and (b) allo owable strengtth by ASD. 335. Determine the available strength s of a WT9×20, W A992 2 ssteel, with the flanges f welded d to a 1/2 in. gu usset plate by a 112 in. weld along each sidee of the flangee. Considering g yyielding and neet section ruptu ure, determine the (a) design n sstrength by LRF FD and (b) allo owable strengtth by ASD.

38. D Design a 12 ftt long, single-aangle tension m member to suppoort a live loadd of 45 kips annd a dead load of 15 kips (L/D = 3). The meember is to bee connected thhrough one leg. Estimate threee bolts in a sinngle line. Usee A36 steel and llimit the slendderness ratio too 300. Considerr yield and net seection rupture,, and design byy (a) LRFD andd (b) ASD. 39. D Design a 12 ft llong, single-anngle tension meember as in Probllem 38 with the same totaal service loadd, 60 kips. Usingg a live load oof 6.7 kips andd a dead load oof 53.3 kips (L/D = 0.126), (a)) design by L LRFD and (b) design by ASD D. 40. D Design a 12 ft llong, single-anngle tension meember as in Probllem 38 with thhe same servicce load using a live load of 500 kips and a dead load of 10 kkips (L/D = 5). Design by (a) L LRFD and (b) A ASD. 41. D Design a 33 fft long WT teension wind bbrace for a multii-story buildinng to resist a wind force off 290 kips. Use A A992 steel annd 7/8 in. boltss connected to the flange only.. Assume two lines of bolts with at least three bolts per lline. Limit thee slenderness rratio to 300. Consider yieldd and net sectioon rupture, andd design by (a) LRFD and (b) A ASD. Design a W12 A992 tensionn member for a truss that 42. D will carry a dead lload of 70 kipps and a live looad of 210 kips. The flanges w will be bolted to the conneccting plates with 7/8 in. bolts llocated so thatt four bolts wiill occur in any net section. A Assume at leaast three boltss per line. Conssider yield andd net section ruupture, and dessign by (a) LRFD D and (b) ASD D. 43. D Design a W14 A992 tensionn member for a truss that will ccarry a dead looad of 120 kipps and a live looad of 360 kips. The flanges w will be bolted to the conneccting plates with 7/8 in. bolts llocated so thatt four bolts wiill occur in any net section. A Assume at leaast three boltss per line. Conssider yield andd net section ruupture, and dessign by (a) LRFD D and (b) ASD D.

Chhapter 4 444. An L4×3×3 3/4 is attached d to a gusset plate with threee 33/4 in. bolts spaaced at 3 in. with w an end and d edge distancee oof 1.5 in. as shown in Fiigure P4.44. Determine D thee aavailable block k shear strength of the A36 6 angle by (a)) L LRFD and (b) ASD. A

Teension Membbers

123

48. A A36 steel and 33/4 in. bolts aree used in the boolted splice show wn in Figure P4.48. Conssider yield, nnet section ruptuure, and block shear rupture. Determine thee (a) design strenngth by LRFD aand (b) the alloowable strengthh by ASD.

P P4.44 445. A 7×3/4 in. i A36 plate is shown in Figure P4.45,, w where the holles are for 3//4 in. bolts. Determine D thee aavailable block k shear strength h by (a) LRFD and (b) ASD.

P4.488 49. A A572 Grade 500 steel and 7/88 in. bolts are uused in the bolteed splice shownn in Figure P44.48. Consider yield, net sectioon rupture, annd block shear rupture. Dettermine the (a) ddesign strengtth by LRFD and (b) the allowable strenngth by ASD.

P P4.45 446. Determine the available block shear sttrength for thee A A992 steel WT T6×20, attach hed through th he flange with h eeight 3/4 in. bo olts as shown in n Figure P4.46 6, by (a) LRFD D aand (b) ASD.

P P4.46 447. Determine the available block shear sttrength for thee A A992 steel WT T12×34, attacched through th he flange with h eeight 7/8 in. bo olts as shown in n Figure P4.46 6, by (a) LRFD D aand (b) ASD.

50. IIntegrated Dessign Project. L Lateral load reesistance in the nnorth-south direection is proviided by a chevvron braced framee as shown in Figure 1.24. B Before the forcces in these braciing members ccan be determ mined, the speccified wind load must be deterrmined. At thiis stage in thee design, a simpplified approachh to wind loadd calculation m might yield the fo following loadss at each level. Roof Fourth floor Third floor Second floor Total wind loaad

54.0 kipss 102.0 kipps 93.0 kipss 88.0 kipss 337.0 kipps

It is perm missible to trreat the chevrron braced framees as verticaal trusses witth pinned joiints at all interssections. Usinng the two bbraced framess to share equallly the givenn wind loadinng, analyze thhe vertical truss es for the wiind in one diirection only. It will be windward bracce is in tensioon and the apparrent that the w leewaard brace is inn compression. When the winnd blows in the oother direction, the brace forces reverse. Design alll the braces as if the tenssion forces contrrol. Although it is usually compression fforces that contrrol brace desiggn, in this chaapter we are foocusing on tensioon. It would bbe useful to coonsider the usee of several typess of members, e.g., angles, cchannels, and W-shapes, to ressist these forcees. .

C Chapterr 5

C Comprression n M Members

Ruttgers Universitty School of Buusiness, Piscataaway, NJ Photo coourtesy of WSP Parsons Brinnckerhoff

5.11

COMPR RESSION MEMBERS M IN STRUC CTURES Compression n members arre structural elements e subjjected to axiaal forces that ttend to push the ends of the mem mbers toward d each other. The most ccommon com mpression meember in a bbuilding structure is a column. Columns are vertical v membbers that suppport the horiizontal elemeents of a roof or floor system. Sev veral columns can be seenn in Figure 55.1 as part off a building sttructure. They are th he primary ellements that provide p the vvertical spacee to form ann occupiable volume. Other comp pression mem mbers are fou und in trussess as chord annd web mem mbers and as bracing members in floors and walls. w Other naames often ussed to identify fy compressioon members aare struts and posts. Throughout T this t chapter the t terms coompression m member and ccolumn will be used interchangeaably. The compression n members diiscussed in thhis chapter exxperience onlyy axial forcess. In real structures, additional a load d effects are often exertedd on a compreession member that wouldd tend to combine ben nding with th he axial force. These combbined force m members are ccalled beam-ccolumns and are disscussed in Chapter C 8. Th he majority oof the provisions that appply to comp mpression members aree located in Chapter C E of th he Specificatiion. Table 5.1 lists th he sections off the Specificaation and part rts of the Mannual discussed in this chapter.

5.22

CROSS--SECTIONAL SHAPE ES FOR CO OMPRESSIO ON MEMB BERS n members caarry axial forcces, so the prrimary cross-ssectional propperty of intereest is the Compression area. Thus, the t simple rellationship bettween force annd stress, P (5.1) f = A is applicablle. As long as this relationship dictaates compresssion membeer strength, aall cross sections witth the same area will perrform in the same way. IIn real structu tures, howeveer, other factors influ uence the strrength of thee compressioon member, and the disttribution of tthe area becomes imp portant.

1224

Chapter 5

Figure 5.1 5

Comppression Mem mbers

125

Columnss in a Multisto ory Building

Photo cou urtesy Greg Griieco

In I building structures, s thee typical com mpression m member is a column and the typical column is i a rolled wiide-flange meember. Later discussions oof compressioon member sttrength will show thaat the W-shap pe does not have the mostt efficient disttribution of m material for compression memberss. It does, how wever, provid de a compressiion member tthat can easilyy be connecteed to other memberss of the system m such as beaams and otherr columns. Thhis feature siggnificantly innfluences its selection n as an approp priate column cross sectionn. Figure F 5.2 sh hows examplees of rolled aand built-up shapes that aare used as compression memberss. Many of th hese are the saame shapes uused for the teension members discussedd in Chapter 4. This is i reasonablee because thee forces beinng consideredd in these tw wo cases are both axial, although h they act in the t opposite direction. d How wever, other factors that iinfluence the strength of compresssion memberss will dictate additional crriteria for the selection of tthe most efficcient shapes for these members. The T tee and angle a shown in Figure 5.2cc and d are coommonly useed as chords aand webs of trusses. In I these appliications, the geometry g of tthe shapes helps simplify tthe connectioons between memberss. Angles are also used in n pairs as buiilt-up compreession membeers, with the connecting element between b the two t angles as shown in Figgure 5.2h. Thhe channel cann be found in trusses as a single eleement or com mbined with another a channnel as shown in Figure 5.22b, i, l, and m m. Built-up Table 5.1

Sections of o Specification and Parts of Manual Coovered in thiss Chapter Specificationn

B3 B4 E1 E2 E3 E4 E5 E6 E7

Design Baasis Classificattion of Sections for Local B Buckling General Prrovisions Slendernesss Limitationss and Effectivve Length Compressiive Strength for Flexuraal Buckling of Members without Slendeer Elements Compressiive Strength for Torsionall and Flexuraal-Torsional B Buckling of Mem mbers withou ut Slender Eleements Single-Angle Compresssion Memberrs Built-up Members M Members with w Slender Elements Manual

Part 1 Part 4 Part 6

Dimension ns and Properrties Design of Compression n Members Design of Members Sub bject to Combbined Loadinng

1226 Chapter 5

Figgure 5.2

Compresssion Memberss

Ro olled Shapes and Built-up Shapes for Compression M Members columns can n also be foun nd using chan nnels. The holllow structuraal sections (H HSS) shown inn Figure 5.2e, f, and d g are comm monly found as columns in buildings,, particularly one-story sttructures where the connections to o the shape can be simpliffied by carry ing beams ovver the colum mns. The distribution of the materiaal in these shaapes is the moost efficient ffor columns.

5.33

COMPR RESSION MEMBER M STRENGTH S H f were to t impact thee strength of a compressioon member, thhe simple axiial stress If no other factors relationship given in Equation E 5.1 could be ussed to descrribe member strength. Thhus, the maximum fo orce that a compression meember could rresist at yieldd would be Py = Fy Ag (5.2) where Py is the yield load, sometimess called the sqquash load; Fy is the yieldd stress; and Ag is the gross area. This is the response r thatt would be exxpected if a very short sspecimen, onee whose length appro oximates its other two dim mensions, weere to be tessted in comprression. This type of column test specimen, sh hown in Figurre 5.3a, is callled a stub coolumn. Becauuse most comppression members wiill have a len ngth that greaatly exceeds its other dim mensions, lenggth effects caannot be ignored. A more m realistic column is sh hown in a testt frame in Figuure 5.3b.

5.33.1

Euler Column C To address the t impact off length on co ompression m member behavvior, a simplee model, as shown in Figure 5.4, is i used. The Swiss S mathem matician Leonnard Euler firrst presented tthis analysis in 1759. A number of o assumption ns are made in i this colum mn model: (1)) the column ends are fricctionless pins, (2) thee column is peerfectly straiight, (3) the looad is appliedd along the ceentroidal axis,, and (4)

Chapter 5

(a)

Comppression Mem mbers

127

((b)

Figure 5.3 5 Column Testing. T (a) Sttub column. (bb) Long colum umn (a) Photo courtesy Proff. Dr. Mario Fontana, F (b) P Photo courtesyy Mohammed Ali Morovat aand Michael Engelhard dt, University of o Texas at Austin

the materrial behaves elastically. e Baased on thesee assumptionss, this columnn model is usuually called the perfeect column or the pure colu umn. Figure F 5.4a sh hows the perffect column w with an applieed load that w will not causee any lateral displacem ment or yiellding. In this arrangemeent, the loadd can be inccreased with no lateral displacem ment of the column. c How wever, at a paarticular loadd, defined as the critical load or the buckling load, Pcr, the column will displacce laterally as shown inn Figure 5.44b. In this configuraation, the dasshed line reprresents the o riginal positiion of the meember and thhe solid line representts the displacced position. Note N that an axis system iis presented iin the figure, with the zaxis alon ng the membeer length and the y-axis traansverse to thhe member leength. This places the xaxis perp pendicular to the plane of the figure. Thhe x- and y-axxes corresponnd to the centtroidal axes of the cro oss section. A free body diagram d of thee lower portioon of the coluumn in its dispplaced positioon is shown in Figuree 5.4c. If mom ments are takeen about pointt C, equilibriuum requires M z = Pcr y From thee principles of o mechanics and using sm mall displacem ment theory, the differentiial equation relating moment m to cu urvature of thee deflected m member is giveen as d2y M =− z 2 EI x dz Combining these two o equations and a rearrangiing the terms yields the differential eequation of equilibriu um, d 2 y Pcr y=0 + dz 2 EI x

1228 Chapter 5

Compresssion Memberss

onditions for Elastic E Colum mns Figure 5.4 Stability Co If the coeffi ficient of the second term m is taken as k 2 = Pcr EI x , the differenntial equationn for the column beco omes d2y + k2 y = 0 2 dz which is a standard seccond-order liinear ordinarry differentiaal equation. T The solutionn to this equation is given g by y = A sin kz + B cos kz (5.3) where A an nd B are con nstants of integration. To further evalluate this equuation, the bboundary conditions must m be applieed. Because at a z = 0, y = 0 and at z = L, y = 0, we find that B=0 and

A sin kL = 0

For Equation 5.3 to havee a nontrivial solution, (sinn kL) must equal zero. Thhis requires thhat kL = nπ, where n is any integer. Substituting g for k and reearranging yieelds n 2 π 2 EI x (5.4) Pcr = L2 Because n can be taken as a any integer, Equation 5.44 has a minim mum when n = 1. This is caalled the Euler buckliing load or th he critical bucckling load annd is given as π 2 EI x (5.5) Pcr = L2 If values forr B and kL arre substituted into Equatioon 5.3, the shaape of the buuckled columnn can be determined from f z⎞ ⎛ (5.6) y = A sin ⎜ nπ ⎟ ⎝ L⎠

129

Chapter 5

Comppression Mem mbers

Figurre 5.5

Shappe of Buckledd Columns

Because any value for fo A will sattisfy Equatio n 5.6, a uniqque magnitudde for the diisplacement cannot be b determined d; however, it is clear thaat the shape oof the buckleed column is a half sine curve wh hen n = 1. Th his is shown in i Figure 5.55a. For other vvalues of n, ddifferent buckkled shapes will result along with the higher criitical bucklinng load. Whenn n > 1, these shapes are reeferred to as higher mode m shapes. Several S cases are shown inn Figure 5.5b,, c, and d. In aall cases, the basic shape is the sin ne curve. In orrder for thesee higher modees to occur, soome type of pphysical restrraint against buckling is required at a the point where w the buckkled shape crosses the origginal, undefleected shape. This can be accomplisshed with the addition of bbraces, which is discussed later. We W now havee two equation ns to predict tthe column sttrength: Equaation 5.2, whiich does not address length; l and Equation E 5.5,, which does . These two equations arre plotted in Figure 5.6. Because the derivation n of the Eulerr equation waas based on ellastic behavioor and the coluumn cannot carry mo ore load than the t yield load d, there is an uupper limit too the column sstrength. If I the length at which thiss limit occurrs is taken ass Ly, it can bbe determinedd by setting Equation n 5.2 equal to Equation 5.5 and solving ffor length, givving EI x Ly = π Fy Ag To simpllify this equattion, the radiu us of gyrationn, r, will be ussed, where I r= A Because the moment of inertia dep pends on the axis being cconsidered, annd A is the grross area of the sectio on, which is independent of o axis, r will depend on thhe buckling axxis. In the derrivation just developeed, the axis off buckling forr the column oof Figure 5.4 was taken as the x-axis; thhus, E Ly = πrx Fy For this theoretical t deevelopment, a column whoose length is leess than Ly would fail by yyielding and could bee called a sho ort column, whereas w a collumn with a length greateer than Ly woould fail by buckling and be called d a long colum mn.

1330 Chapter 5

Compresssion Memberss

Fiigure 5.6

C Column Strenggth Based on Length

It iss also helpful to write Equ uation 5.5 in tterms of stresss. Dividing bboth sides by the area and substitu uting again forr the radius off gyration yieelds π2 E (5.7) Fcr = 2 ⎛ L⎞ ⎜ ⎟ ⎝r⎠ In this equattion, the radiu us of gyration n is left unsubbscripted so tthat it can be applied to whhichever axis is determined to be the t critical ax xis. A plot of sstress versus L/r would be of the same sshape as the plot of fo orce versus L in Figure 5.6 6. 5.33.2

Other Boundary B Co onditions Derivation of o the bucklin ng equations presented ass Equations 55.5 and 5.7 inncluded the bboundary condition off frictionless pins at both ends. For perrfect columnss with other bboundary connditions, the momentt will not be zero at the ends, e and thiss will result iin a nonhomoogeneous diffferential equation. Solving the resulting diffe ferential equaation and appplying the aappropriate bboundary conditions will w lead to a buckling equation e of a form similaar to the preevious equatiions. To generalize th he buckling equation e for other end condditions, the coolumn length,, L, is replaceed by the column effeective length, KL, where K is the effecctive length ffactor. Thus, the general bbuckling equations beecome π2 EI Pcr = (5.8) 2 ( KL ) and π2 E (5.9) Fcr = 2 ⎛ KL ⎞ ⎜ ⎟ ⎝ r ⎠ Figu ure 5.7 depiccts the origin nal pin-endedd column wiith several examples of ccolumns showing thee influence off different end d conditions. All columns are shown w with the lowerr support fixed againsst lateral tran nslation. Threee of the coluumns have uppper ends thaat are also reestrained from lateral translation, and a three others have upp er ends that aare free to traanslate. The eeffective length can be b visualized d as the lengtth between innflection poinnts, where thhe curvature rreverses. This result is similar to th he original deerivation whenn n was takenn as some inteeger other thaan 1. It is most easily seen in Figure 5.7b and c but b can also bbe seen in Figgure 5.7d by vvisualizing thee

Chapter 5

Figure 5.7 5

Comppression Mem mbers

131

Column Buckled Shap pe for Differeent End Condditions Figure 5.88 Extended S Shape of Buckled Column from F Figure 5.7d extended d buckled shap pe above the column as shhown in Figurre 5.8. In all cases, the buckled curve is a segm ment of the sin ne curve. Thee most importtant thing to oobserve is thaat the columnn with fixed ends in Figure F 5.7b has h an effectiive length off 0.5L, whereas the colum mn in Figure 55.7a has an effective length of L. Thus, the fiixed-end coluumn will havve four times the strength of the pinended co olumn.

5.3.3

Com mbination of Bracing B and End Conditiions The influ uence of interrmediate braccing on the efffective lengthh was touched upon in thee discussion of the hiigher modes of o buckling. In I those casees, the bucklinng resulted inn equal-lengtth segments that refleected the mod de number. Thus, T a colum mn with n = 2 had two equual segments, whereas a

Figure 5.9 5

Buckled Shape for Co olumns with IIntermediate B Braces

1332 Chapter 5

Compresssion Memberss column with h n = 3 buck kled with three equal segm ments. If phyysical braces are used to provide buckling ressistance to thee column, thee effective lenngth will deppend on the loocation of thee braces. Figure 5.9 shows three columns witth pinned endds and interm mediate suppports. The coolumn in Figure 5.9a is the same as the column in Figure 5.55b. The effecttive length is 0.5L, so K = 0.5. The column in Figure F 5.9b sh hows lateral braces b in an unsymmetriccal arrangemeent with one segment equal to L/3 3 and the otheer to 2L/3. Although A the eexact locationn of the inflection point w would be slightly into o the longer segment, s norrmal practice is to take thhe longest unnbraced lengthh as the effective len ngth; thus KL = 2L/3, so K = 2/3. The coolumn in Figuure 5.9c is braaced at two loocations. The longest unbraced len ngth for this case gives an eeffective lenggth KL = 0.5L and a corressponding K = 0.5. A general g rule caan be stated that, t when thee column endds are pinned, the longest uunbraced length is thee effective len ngth for bucklling in that dirrection. Wheen other end d conditions are a present, tthese two influences musst be combinned. The columns of Figure 5.10 illustrate the influence off combinationns of end supports and braacing on the column effective leng gth. The end conditions w would influencce only the effective lengtth of the end segmentt of the colum mn. For the co olumn in Figuure 5.10a, the lower segmeent has L = a, and that segment would buckle with w an effectiv ve length KL = a. The uppper segment hhas L = b but also has a fixed end. Thus, it wou uld buckle witth an effectivve length KL = 0.7b, obtainned by combiining the end conditio ons of Figure 5.7c with th he length, b. T Thus, the relaationship betw ween lengths a and b determine which w end of the t column dictates d the ovverall columnn effective lenngth. As an eexample, the column in Figure 5.10b shows thaat the lowest ssegment wouuld set the collumn effectivve length at 0.35L.

Figure 5.10 Bu uckled Shapee for Columnss with Differeent End Condditions and Inttermediate Brraces

EX XAMPLE 5.1 5 Th heoretical Coolumn Stren ngth

Goa al:

Deterrmine the theeoretical strenngth for a pinn-ended colum mn and whetther it will first f buckle orr yield.

Giv ven:

A W1 10×33, A992,, column withh a length of 220 ft.

SO OLUTION

Step p 1:

Deterrmine the load d that would ccause bucklinng. With no other info ormation, it m must be assum med that this ccolumn will bbuckle aboutt its weak axiis, if it buckl es at all, because the effeective length, KL = 20 ft for both axes.

Chapter 5

Compression Members

133

From Manual Table 1-1, Iy = 36.6 in.4 and Ag = 9.71 in.2. The load that would cause it to buckle is 2 π2 EI y π ( 29,000 )( 36.6 ) Pcr = − = 182 kips 2 2 ( KL ) ( 20 (12 ) ) Step 2:

Determine the load that would cause yielding. Py = Fy Ag = 50 ( 9.71) = 486 kips

Step 3:

Conclusion: Because Pcr < Py, the theoretical column strength is P = 182 kips

and the column would buckle before it could reach its yield stress.

EXAMPLE 5.2 Critical Buckling Load

Goal:

Determine the overall column length that, if exceeded, would theoretically cause the column to buckle elastically before yielding.

Given:

A W8×31 column with fixed supports. Use steels with (a) Fy = 40 ksi and (b) Fy = 100 ksi.

SOLUTION

Step 1:

From Manual Table 1-1, Iy = 37.1 in.4 and Ag = 9.13 in.2.

Part a Step 2:

Step 3:

Determine the force that would cause the column to yield when Fy = 40 ksi. Py = Fy Ag = 40 ( 9.13 ) = 365 kips To determine the length that would cause this same load to be the buckling load for the pinned-pinned case, set this force equal to the buckling load equation and determine the length from 2 π2 EI y π ( 29,000 )( 37.1) 365 kips = 2 = L L2 which gives L=

π2 ( 29,000 )( 37.1) 365

= 171 in.

So the length is L=

Step 4:

171 = 14.3 ft 12

Since a fixed-end column has an effective length equal to one-half the actual length, buckling will not occur if the actual length is less than or equal to:

L = 2 (14.3) = 28.6 ft for a column with Fy = 40 ksi

134 Chapter 5

Compression Members

Part b Step 5:

Part 6:

Determine the force that would cause the column to yield when Fy = 100 ksi. Py = Fy Ag = 100 ( 9.13 ) = 913 kips To determine the length that would cause this same load to be the buckling load for the pinned-pinned case, set this force equal to the buckling force and determine the length from 2 π2 EI y π ( 29,000 )( 37.1) 913 kips = 2 = L L2 which gives L=

π2 ( 29,000 )( 37.1) 913

= 108 in.

So the length is L=

Step 7:

108 = 9.0 ft 12

Since a fixed-end column has an effective length equal to one-half the actual length, buckling will not occur if the actual length is less than or equal to:

L = 2 ( 9.0 ) = 18.0 ft for a column with Fy = 100 ksi

5.3.4

Real Column

Physical testing of specimens that effectively model columns found in real building structures, like that seen in Figure 5.3b, has shown that column strength was not as great as either the buckling load predicted by the Euler buckling equation or the squash load predicted by material yielding. This inability of the theory to predict actual behavior was recognized early, and numerous factors were found to be the cause. Three main factors influence column strength: material inelasticity, column initial out-of-straightness, and end conditions. The influence of column end conditions has already been discussed with respect to effective length determination. Material inelasticity and initial out-of-straightness, which also significantly impact real column strength, are discussed here. Inelastic behavior of a column directly results from built-in or residual stresses in the cross section. These residual stresses are, in turn, the direct result of the manufacturing process. Steel is produced with heat, and heat is also necessary to form the steel into the shapes used in construction. Once the shape is fully formed, it is cooled. During this cooling process residual stresses are developed. Figure 5.11 shows a wide-flange cross section in various stages of cooling. Initially, as shown in Figure 5.11a, the tips of the flanges with the most surface area to give off heat begin to cool. This material contracts as it cools, eventually reaching the ambient temperature. At this point, the fibers in this part of the section reach what is expected to be their final length.

Chapter 5

Comppression Mem mbers

135

Figu ure 5.11 Diistribution of Residual Streesses As A adjacent fibers cool,, they too ccontract. In the processs of contraccting, these subsequeently cooling fibers pull on o the previoously cooled ffibers, placinng the latter uunder some amount of o compressiv ve stress. Figu ure 5.11b shoows a cross seection with addditional flangge elements cooled. When W the preeviously cooled portion oof the cross section proviides enough stiffness to restrain the t contractio on of the subssequently coooling materiaal, a tensile sttress is develloped in the now-cooling materiall because it cannot cont ntract as it w would withouut this restraaint. When completeely cooled, as shown in Fig gure 5.11c, thhe tips of the fflanges and thhe middle of the web are put into compression, c and the flang ge-web junctuure is put intoo tension. Thuus, the first fibbers to cool are in com mpression, whereas w the lasst to cool are in tension. Several S differrent representtations of thee residual streess distributioon have beenn suggested. One distrribution is sh hown in Figurre 5.11c. Thee magnitude of the maxim mum residual stress does not depend on the maaterial yield strength but iss a function oof material thhickness. In aaddition, the compresssive residual stress is of critical c intereest in the connsideration off compressionn members. The mag gnitude of thiss residual streess varies from m 10 ksi to aabout 30 ksi, depending onn the shape. The high her values are found in wid de flanges witth the thickestt flange elemeents. To T understand d the overall impact of thhese residual stresses on ccolumn behavvior, a stub column will w again be investigated. Figure 5.12 sshows the stress-strain relaation for a shhort column, one that will not buck kle but exhib bits the influeence of residuual stresses. A As the colum mn is loaded with an axial a load, thee member sho ortens and thee correspondiing strain andd stress are deeveloped, as if this were w a perfecttly elastic sp pecimen. The response off a perfectly elastic, perfeectly plastic column is i shown by th he dashed lin ne in Figure 5 .12. When thhe applied streess is added too a member with residual compresssive stress, th he stub colum mn begins to shorten at a ggreater rate ass the tips of the flang ge become strressed beyond d the yield str tress. This pooint is identifiied in Figure 5.12 as Fp, the propo ortional limit Thus, the strress-strain currve moves offf the straight dashed line aand follows the curveed solid line. Continuing to add load tto the columnn results in ggreater strain for a given stress, an nd the column n eventually reaches the yyield stress off the perfectly elastic matterial. Thus, the only difference beetween the beh havior of the actual colum mn and the usuual test specim men used to ne the stress-strain relatio onship is thatt the real coolumn behavees inelasticallly as those determin portions of its cross seection with co ompressive reesidual stressees reach the m material yield stress. If I a new term m, the tangentt modulus, ET , is defined aas the slope oof a tangent too the actual stress-strrain curve at any a point sho own in Figuree 5.12, an impproved predicction of colum mn buckling strength can be obtain ned by modify ying the Eulerr buckling equ quation so thatt π2 ET I x Pcr = 2 ( KL )

1336 Chapter 5

Compresssion Memberss

Figu ure 5.12 Stuub Column StrressStrainn Diagrams w with and without Residdual Stress Thus, as the column is loaded beyond its elastic lim mit, ET decreases and the buuckling strenggth does also. This paartially accou unts for the in nability of thee Euler bucklling equation to accuratelyy predict column stren ngth. Ano other factor to t significanttly impact c olumn strenggth is the coolumn initiall out-ofstraightness.. Once again n, the manufaacturing proceess for steel shapes impaacts the abilityy of the column to caarry the prediicted load. In this case, thee problem is rrelated to the fact that no sttructural steel membeer comes out of the producction process perfectly straaight. The AIS SC Code of SStandard Practice perrmits an initiaal out-of-straightness equaal to 1/1000 oof the length between poiints with lateral support. Although h this appearss to be a smaall variation from straighttness, it still impacts column stren ngth. Figu ure 5.13a sh hows a perffectly elastic,, pin-ended column witth an initial out-ofstraightness,, δ. A compaarison of this column c diagrram with thaat used to deriive the Euler column,

Figure 5.13 Influence of o Initial Out--of-Straightneess on Colum mn Strength

Chapter 5

Compression Members

137

Figure 5.4, shows that the moment along the column length will be greater for this initially crooked column in its buckled position than it would have been for an initially straight column. Thus, the solution to the differential equation would be different. In addition, because the applied load works at an eccentricity from the column along its length, even before buckling, a moment is applied to the column that has not yet been accounted for. Figure 5.13b shows the load versus lateral displacement diagram for this initially crooked column compared to that of the initially straight column. This column not only exhibits greater lateral displacement, it also has a lower maximum strength. When these two factors are combined, the Euler equation cannot properly describe column behavior on its own. Thus, the development of curves to predict column behavior has historically been a matter of curve-fitting the test data in an attempt to present a simple representation of column behavior. 5.3.5

AISC Provisions

The compression members discussed thus far have either yielding or overall column buckling as the controlling limit state. Figure 5.14 plots sample column test data compared to the Euler equation and the squash load. The Structural Stability Research Council proposed three equations to predict column behavior. To simplify column design, AISC selected a single curve described using two segments as their representation of column strength. The design basis for ASD and LRFD were presented in Sections 1.9 and 1.10, respectively. The strength equations are repeated here in order to reinforce the relationship between the nominal strength, resistance factor, and safety factor presented throughout the Specification. The requirement for ASD is Rn Ω

(AISC B3.2)

Ru ≤ φRn

(AISC B3.1)

Ra ≤

The requirement for LRFD is

As indicated earlier, the Specification provides the relationship to determine nominal strength and the corresponding resistance factor and safety factor for each limit state to be considered. The provisions for compression members with nonslender elements, i.e., no local buckling, are given in Specification Section E3. The nominal column strength for the limit state of flexural buckling of members with nonslender elements is Pn = Fcr Ag

and

φc = 0.9 ( LRFD )

(AISC E3-1)

Ω c = 1.67 ( ASD )

where Ag is the gross area of the section and Fcr is the flexural buckling stress. (The Euler column derivation in Section 5.3.1 addressed the limit state of flexural buckling.)

1338 Chapter 5

Compresssion Memberss

Figure 5.14 Sample Co olumn Test Data Comparedd to Theoreticcal Column S Strength The Specification n defines Lc as a the effectivve length and shows it equual to KL. Thiss is the same effectiive length factor, K, discusssed earlier. T To capture collumn behavioor when inelasstic buckling dom minates colum mn strength, that t is, wheree residual stresses become important, the Specification n provides thaat when Fy Lc / r ≤ 4.71 E / Fy or ≤ 2.225 Fe Fy ⎡ Fcr = ⎢ 0.6658 Fe ⎢⎣

⎤ ⎥ Fy ⎥⎦

(AIS SC E3-2)

To capture behaavior when in nelastic bucklling is not a factor and iinitial crookedness is hat is when dominant, th Fy Lc / r > 4.71 E / Fy or > 2.225 Fe Fcr = 00.877 Fe SC E3-3) (AIS where Fe is the elastic bu uckling stress; the Euler buuckling stresss previously ppresented as E Equation 5.9 and restaated here is π2 E (AIS SC E3-4) Fe = 2 ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠

Chapter 5

Figure 5.15 5

Comppression Mem mbers

139

Lc/r verrsus Critical Strength S

The T flexural buckling b stressses for three different steeels, A36, A9992, and A5144, versus the slenderneess ratio, Lc/r, are shown in i Figure 5.155. For very sllender columnns, the buckliing stress is independ dent of the maaterial yield. The division between elasstic and inelaastic behaviorr, Equations E3-2 and d E3-3, corressponds to Lc/r values of 1334, 113, and 80.2 for steells with a yieldd of 36, 50, and 100 ksi, k respectiv vely. Early E editionss of the LRF FD Specificattion defined the exponentt of Equationn E3-2 in a slightly different d form m that makes the t presentatioon a bit simpler. If a new tterm is defineed such that 2 Fy ⎛ Lc ⎞ Fy λ c2 = =⎜ ⎟ Fe ⎝ πr ⎠ E then the dividing d poin nt between elaastic and inelaastic behaviorr, where Lc E = 44.71 r Fy becomes

λc =

Lc Fy 4.71 = = 1.55 π πr E

By substituting λ c2 = Fy Fe , the criitical flexurall buckling streess for λ c ≤ 1.5 becomes

(

2

)

Fcr = 0.658λc Fy

(5.10)

and for λ c > 1.5 ,

Fcr =

0.877 Fy λ c2

(5.11)

A plot of o the ratio of critical flexural fl buckl kling stress too yield stresss as a functtion of the slenderneess parameteer, λc, is giveen in Figuree 5.16. Usingg this formullation it is eevident that regardlesss of the steell yield stress, the ratio of fflexural buckkling stress too yield stress is the same when plo otted against the t slenderness parameter,, λc. Table 5.22 provides theese numericall values in a convenieent, usable forrm.

1440 Chapter 5

Compresssion Memberss

Figure 5.16 λc versus Critical C Stresss Earllier editions of o the ASD and a LRFD Sppecifications indicated thaat there shoulld be an upper limit on o the magnittude of the sleenderness rattio at Lc/r = 2000. The intennt for this limiit was to have the eng gineer recogn nize that for veery slender coolumns, the fl flexural bucklling stress waas so low as to make the column very v inefficieent. This lim mit has been rremoved in rrecent editionns of the Specification n because theere are many factors influeencing columnn strength thaat indicate thaat a very slender colu umn might acttually be acceeptable. Sectioon E2 simplyy informs the designer thatt column slenderness should prefeerably b e kep pt to somethinng less than 200. Table 55.3 gives the flexural Table 5.2 Ratio of Critical Stress to Yield Stress λc

Fcr/Fy

λc

Fcr/Fy

λc

Fcr/Fy

λc

Fcrr/Fy

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90

1.000 0.999 0.996 0.991 0.983 0.974 0.963 0.950 0.935 0.919 0.901 0.881 0.860 0.838 0.815 0.790 0.765 0.739 0.712

0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85

0.685 0.658 0.630 0.603 0.575 0.547 0.520 0.493 0.466 0.440 0.415 0.390 0.365 0.343 0.322 0.303 0.286 0.271 0.256

1.90 1.95 2.00 2.05 2.10 2.15 2.20 2.25 2.30 2.35 2.40 2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80

0.243 0.231 0.219 0.209 0.199 0.190 0.181 0.173 0.166 0.159 0.152 0.146 0.140 0.135 0.130 0.125 0.120 0.116 0.112

2.85 2.90 2.95 3.00 3.05 3.10 3.15 3.20 3.25 3.30 3.35 3.40 3.45 3.50 3.55 3.60 3.65 3.70 3.75

0.1108 0.1104 0.1101 0.09974 0.09943 0.09913 0.08884 0.08856 0.08830 0.08805 0.07781 0.07759 0.07737 0.07716 0.06696 0.06677 0.06658 0.06641 0.06624

Chapter 5 Table 5.3

Critical Stress Values for Three Steels Fcr (ksi)

KL/r 0 10 14 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100

Compression Members

Fy = 36 ksi 36.0 35.8 35.6 35.4 35.2 35.1 34.9 34.7 34.5 34.3 34.1 33.9 33.6 33.4 33.1 32.8 32.5 32.2 31.9 31.6 31.2 30.9 30.5 30.2 29.8 29.4 29.0 28.6 28.2 27.8 27.4 27.0 26.6 26.1 25.7 25.3 24.8 24.4 23.9 23.5 23.1 22.6 22.2 21.7 21.3

Fy = 50 ksi 50.0 49.6 49.3 48.8 48.6 48.3 47.9 47.6 47.2 46.8 46.4 45.9 45.5 45.0 44.5 43.9 43.4 42.8 42.2 41.6 41.0 40.4 39.8 39.1 38.4 37.7 37.1 36.4 35.7 34.9 34.2 33.5 32.8 32.0 31.3 30.6 29.8 29.1 28.4 27.7 26.9 26.2 25.5 24.8 24.1

Fcr (ksi) Fy = 100 ksi 100.0 98.5 97.2 95.4 94.3 93.2 91.9 90.6 89.2 87.7 86.1 84.4 82.7 81.0 79.1 77.3 75.3 73.4 71.4 69.4 67.3 65.3 63.2 61.1 59.1 57.0 54.9 52.9 50.9 48.8 46.9 44.9 43.0 41.1 39.2 37.3 35.6 33.9 32.4 31.0 29.7 28.4 27.2 26.1 25.1

KL/r 102 104 106 108 110 112 114 116 118 120 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 152 154 156 158 160 162 164 166 168 170 172 174 176 178 180 182 184 186 190 200

Fy = 36 ksi 20.8 20.4 19.9 19.5 19.0 18.6 18.2 17.7 17.3 16.9 16.4 16.0 15.6 15.2 14.8 14.4 14.0 13.6 13.2 12.8 12.4 12.1 11.8 11.5 11.2 10.9 10.6 10.3 10.1 9.81 9.56 9.33 9.11 8.89 8.69 8.48 8.29 8.10 7.92 7.75 7.58 7.41 7.26 6.95 6.28

Fy = 50 ksi 23.4 22.7 22.0 21.3 20.6 20.0 19.3 18.7 18.0 17.4 16.9 16.3 15.8 15.3 14.9 14.4 14.0 13.6 13.2 12.8 12.4 12.1 11.8 11.5 11.2 10.9 10.6 10.3 10.1 9.81 9.56 9.33 9.11 8.89 8.69 8.48 8.29 8.10 7.92 7.75 7.58 7.41 7.26 6.95 6.28

Fy = 100 ksi 24.1 23.2 22.3 21.5 20.7 20.0 19.3 18.7 18.0 17.4 16.9 16.3 15.8 15.3 14.9 14.4 14.0 13.6 13.2 12.8 12.4 12.1 11.8 11.5 11.2 10.9 10.6 10.3 10.1 9.81 9.56 9.33 9.11 8.89 8.69 8.48 8.29 8.10 7.92 7.75 7.58 7.41 7.26 6.95 6.28

141

1442 Chapter 5

Compresssion Memberss

Figure 5.17 7 Columns for f Examples 5.3, 5.4, and 5.5. buckling strress for valuees of Lc/r fro om 0 to 200 for steels w with three diffferent yield stresses. Manual Tab ble 4-14 prov vides an expanded versionn of this tablee for six diffeerent yield strresses at slenderness ratio incremeents of 1.0 ft. Goa al: EX XAMPLE 5.3 5 Coolumn Stren ngth byy AISC Provvisions Givven:

SO OLUTION

Deterrmine the available columnn strength. A W1 12×79 pin-en nded column w with a lengthh of 10.0 ft. aas shown in F Figure 5.17aa. Use A992 steel.

Step p 1:

From m Manual Table 1-1, rx = 5..34 in., ry = 3..05 in., and A = 23.2 in.2.

Step p 2:

Deterrmine the con ntrolling effecctive slendernness ratio. use the length h is 10.0 ft aand the colum mn has pinned ends, Lc = KL = Becau 10.0 ft f for both thee x-axis and y -axis. Thus,

Lcx 10.0 (12 ) = = 22.5 5.34 rx and

Lcy 10.0 (12 ) = = 39.3 3.05 ry

Sincee

Lcy Lcx > ry rx

Chapter 5

Compression Members

143

the y-axis controls. Step 3:

Determine which column strength equation to use. Since Lc E 29,000 = 39.3 < 4.71 = 4.71 = 113 r Fy 50 use Equation E3-2

Step 4:

Determine the Euler buckling stress. π2 ( 29,000 ) Fe = = 185 ksi 2 ( 39.3)

Step 5:

Determine the critical stress from Equation E3-2. ⎛ Fy ⎞ ⎜ ⎟

⎛ 50 ⎞ ⎜ ⎟

Fcr = 0.658⎝ Fe ⎠ Fy = 0.658⎝ 185 ⎠ ( 50 ) = 44.7 ksi

Step 6: For LRFD Step 7: For ASD Step 7:

Determine the nominal strength. Pn = 44.7 ( 23.2 ) = 1040 kips

Determine the design strength for LRFD. φPn = 0.9 (1040 ) = 936 kips

Determine the allowable strength for ASD. Pn 1040 = = 623 kips Ω 1.67

Goal: EXAMPLE 5.4 Column Strength by AISC Provisions Given:

SOLUTION

Determine the available column strength. A W10×49 column with a length of 20.0 ft, one end pinned and the other end fixed for the y-axis, and both ends pinned for the x-axis, as shown in Figure 5.17b. Use A992 steel.

Step 1:

From Manual Table 1-1, rx = 4.35 in., ry = 2.54 in., and A = 14.4 in.2.

Step 2:

Determine the effective length factors from Figure 5.7. Comparing the columns shown in Figure 5.17b with those shown in Figure 5.7, the effective length factors are Ky = 0.7 and Kx = 1.0.

Step 3:

Determine the x- and y-axis slenderness ratios.

Lcx K x L 1.0 ( 20.0 )(12 ) = = = 55.2 4.35 rx rx

144 Chapter 5

Compression Members

Lcy K y L 0.7 ( 20.0 )(12 ) = = = 66.1 2.54 ry ry Step 4:

Using the larger slenderness ratio, determine which column strength equation to use. Since Lc E 29,000 = 66.1 < 4.71 = 4.71 = 113, r Fy 50 use Equation E3-2

Step 5:

Determine the Euler buckling stress. π2 ( 29,000 ) Fe = = 65.5 ksi 2 ( 66.1)

Step 6:

Determine the critical stress from Equation E3-2. Fcr = 0.658

Step 7: For LRFD Step 8: For ASD Step 8:

⎛ Fy ⎞ ⎜ ⎟ ⎝ Fe ⎠

Fy = 0.658

⎛ 50 ⎞ ⎜ ⎟ ⎝ 65.5 ⎠

( 50 ) = 36.3 ksi

Determine the nominal strength. Pn = 36.3 (14.4 ) = 523 kips

Determine the design strength for LRFD. φPn = 0.9 ( 523 ) = 471 kips

Determine the allowable strength for ASD.

Pn Ω = 523 1.67 = 313 kips

Goal: EXAMPLE 5.5 Column Strength by AISC Provisions Given:

SOLUTION

Determine the available column strength. A W14×53 column with a length of 40.0 ft, both ends fixed for the y-axis, and one end pinned and one end fixed for the x-axis, as shown in Figure 5.17c. Use A992 steel.

Step 1:

From Manual Table 1-1, rx = 5.89 in., ry = 1.92 in., and A = 15.6 in.2.

Step 2:

Determine the effective length factors from Figure 5.7. Comparing the columns shown in Figure 5.17c with those shown in Figure 5.7, the effective length factors are Ky = 0.5 and Kx = 0.7.

Step 3:

Determine the x- and y-axis slenderness ratios. Lcx K x L 0.7 ( 40.0 )(12 ) = = = 57.0 5.89 rx rx

Chapter 5

Compression Members 145

Lcy K y L 0.5 ( 40.0 )(12 ) = = = 125 1.92 ry ry Step 4:

Using the larger slenderness ratio, determine which column strength equation to use. Since Lc E 29,000 = 125 > 4.71 = 4.71 = 113 r Fy 50 use Equation E3-3

Step 5:

Determine the Euler buckling stress. π2 ( 29,000 ) Fe = = 18.3 ksi 2 (125)

Step 6:

Determine the critical stress from Equation E3-3. Fcr = 0.877 Fe = 0.877 (18.3 ) = 16.0 ksi

Step 7:

Determine the nominal strength. Pn = 16.0 (15.6 ) = 250 kips

For LRFD Step 8: For ASD Step 8

Determine the design strength for LRFD. φPn = 0.9 ( 250 ) = 225 kips

Determine the allowable strength for ASD. Pn Ω = 250 1.67 = 150 kips

5.4 ADDITIONAL LIMIT STATES FOR COMPRESSION Two limit states for compression members were discussed in Section 5.3, yielding and flexural buckling. The strength equations provided in Specification Section E3 clearly show that the upper limit for column strength, FyAg, is reached only for the zero-length column. Thus, the provisions are presented in the Specification as applying to the limit state of flexural buckling only, even though they do consider yielding. Singly symmetric, unsymmetric, and certain doubly symmetric members may also be limited by torsional buckling or flexural-torsional buckling. The strength provisions for these limit states are given in Section E4 of the Specification and are discussed here in Section 5.8. For some column profiles, another limit state may actually control overall column strength. The individual elements of a column cross section may buckle locally at a stress below the stress that would cause the overall column to buckle. If this is the case, the column is said to be a column with slender elements. The impact of these slender elements on column strength is determined through the use of an effective area which is smaller than the actual area of the member. The additional provisions for these types of members are presented in Section 5.6.

1446 Chapter 5

Compresssion Memberss

Figure 5.18 Values of Effective E Len ngth Factor, K Copyright © American Insttitute of Steel Construction, C R Reprinted with Permission. A All rights reservved.

5.55

LENGT TH EFFECT TS The effectiv ve lengths thaat have been n discussed w were all relateed to fairly ssimple colum mns with easily defineed end condittions and braccing locationss. Once a collumn is recoggnized as beinng a part of a real stru ucture, determ mining the efffective lengthh becomes m more involved. Moreover, ffor more complex strructures, it might m be sim mpler to dete rmine the buuckling strenngth of the sstructure through anaalysis than thrrough the usee of the effecctive length ffactor, K. Usiing that analyysis, the elastic buck kling stress off the individu ual columns, Fe, can be ddetermined. T This can then be used directly in the t column strength s equaations. Howevver, for this book, colum mn elastic bucckling is determined through a calculation c off effective leength. This approach maay incorporatte some simplificatio ons that wou uld not be maade in an acttual bucklingg analysis annd, dependingg on the approach used to determiine K, may in nclude assumpptions of behaavior that thee actual structture may not satisfy. A fiirst attempt att incorporatin ng some realiistic aspects oof structures iis shown in T Table CA-7.1 of thee Commentary y and here in Figure 5.18. The columnss shown in thiis figure are tthe same as those shown in Figu ure 5.7, and the t same K- factors are sshown and iddentified heree as the theoretical K-values. Wh hat is new herre is the pressentation of rrecommendedd design values when ideal conditiions are apprroximated. Most of these rrecommendedd values are bbased on the fact that perfectly rig gid connection ns are difficu ult to obtain. T Thus, for exaample, a fixedd-end columnn (case a of Figure 5.18) would have h a theoretical K = 0.55, but if the end connectiions were to actually rotate, even just a small amount, a the efffective lengtth would increease. As the eend rotation inncreases toward whatt would occurr for a pin-end ded column, K would apprroach 1.0. Thuus, the recom mmended value of K is 0.65. A sim milar assessm ment of the oother cases w with a fixed end should leaad to an understandin ng of the ideea behind theese recommennded values, each being a bit higher tthan the theoretical value v because actual colu umn end connditions are uunlikely to m match the theeoretical assumptionss.

Chapter 5

Comppression Mem mbers

147

Figurre 5.19 Typpical Moment Frame

When W a colum mn is part of a frame, as shhown in Figuure 5.19, the sstiffness of thhe members framing into i the colum mn impact thee rotation thatt could occur at the columnn ends. As A with the riigid supports discussed forr the columnss in Figure 5..18, these endd conditions permit th he column en nd to rotate. The amountt of this rotaation is someething betweeen the zero rotation of a fixed support s and the t free rotaation of a pinn support. W When the collumn under consideraation is part of a frame where w the ennds of the coolumn are noot permitted to displace laterally relative to eaach other, the frame is calleed a braced fr frame, a sidessway preventeed frame, or a sideswa ay inhibited frame— f shown n as cases a, b, and d in F Figure 5.18. F For a column in a braced frame, th he possible K-factors rangee from 0.5 to 1.0. In framees of this typee, K is often taaken as 1.0, a conserv vative approx ximation thatt simplifies ddesign. In facct, Specification Appendixx 7, Section 7.2.3(a) says that in braced b framees K shall be taken as 1.00 unless analysis shows thhat a lower value is appropriate. When the co olumn under cconsiderationn is in a fram me in which thhe ends are permitted d to move latterally, the fraame is calledd a moment frrame, an unbrraced frame, a sidesway permitted d frame, or a sidesway uniinhibited fram me— shown aas cases c, e, aand f in Figurre 5.18. For the threee cases shown n there, the lo owest value oof K is 1.0. T The other extrreme case, noot shown in Figure 5.18, is a pin--ended colum mn in an unbrraced frame. The effectivve length of tthis column would th heoretically bee infinite. Th hus, K-values for columns in moment fframes range from 1.0 to infinity. The T determinaation of reliab ble effective llength factorss and thus relliable effectivve lengths is a critical aspect of collumn design. Several apprroaches are prresented in thhe literature bbut the most commonly used apprroach is thro ough the aliggnment chartts presented in the Com mmentary to Appendix x 7. The dev velopment off these chartss is based onn a set of asssumptions thaat are often violated in real strucctures; neverttheless, the aalignment chharts are usedd extensivelyy and often modified d in an attemp pt to account for f variations from these aassumptions. These T assump ptions, as giveen in the Com mmentary to A Appendix 7, aare: 1. 1 2. 2 3. 3

Behav vior is purely y elastic. All members m have a constant crross section. All jo oints are rigid d.

148 Chapter 5

Compression Members 4. 5.

6. 7. 8. 9. 10.

For columns in frames with sidesway inhibited, rotations at opposite ends of the restraining beams are equal in magnitude and opposite in direction, producing single curvature bending. For columns in frames with sidesway uninhibited, rotations at opposite ends of the restraining beams are equal in magnitude and direction, producing reverse curvature bending. The stiffness parameter L P /EI of all columns is equal. Joint restraint is distributed to the column above and below the joint in proportion to EI/L for the two columns. All columns buckle simultaneously. No significant axial compression force exists in the girders. Shear deformations are neglected

Using these assumptions, the following equation can be obtained for columns in sidesway inhibited frames. GAGB G +G ⎛ π/K ⎞ 2 tan ( π/ 2Κ ) 2 −1 = 0 (AISC C-A-7-1) ( π/K ) + ⎛⎜ A B ⎞⎟ ⎜⎜1 − ⎟+ 4 2 ( π/Κ ) ⎝ ⎠ ⎝ tan ( π/Κ ) ⎠⎟ For sidesway uninhibited frames, the following equation is obtained.

GAGB ( π/Κ ) − 36 2

6 ( GA + GB )

−

( π/K ) = 0 tan ( π/K )

(AISC C-A-7-2)

In Equations C-A-7-1 and C-A-7-2, the terms GA and GB relate to the relative stiffness of the columns and beams framing into the column at ends A and B, respectively, as given by ∑ ( EI /L )col G= (AISC C-A-7-3) ∑ ( EI /L ) g If the beams and columns behave elastically, as noted in assumption 1, this reduces to ∑ ( I /L )col G= ∑ ( I /L ) g

(5.12)

Equations C-A-7-1 and C-A-7-2 are transcendental equations that do not have a closed-form solution. With the computer methods readily available today, iterative solutions are easily obtained. However, that was not always the case, and a graphical solution was developed in the early 1960s that has become a standard approach for obtaining solutions. Such graphical solutions are called nomographs or alignment charts. Figure 5.20 shows the nomograph for sidesway inhibited frames, and Figure 5.21 gives the chart for sidesway uninhibited frames. Approximate solutions to Equations C-A-7-1 and C-A-7-2 have also been presented in design rules and the literature. The French have used the following equations in their design rules since 1966. For sidesway inhibited, 3GAGB + 1.4 ( GA + GB ) + 0.64 K= (5.13) 3GAGB + 2 ( GA + GB ) + 1.28 For sidesway uninhibited, K=

1.6GAGB + 4 ( G A + GB ) + 7.5 GA + GB + 7.5

(5.14)

Chapter 5

Figure 5.20 5 Alignm ment Chart forr a Braced Frame F (Sidesw way Inhibited d) Copyright © American Institute of Steeel Constructtion, Reprinted d with Permissiion. All rights reseerved.

Comppression Mem mbers

149

Figure 5.211 Alignmentt Chart for ann Unbraced Frame (Sideesway Uninhiibited) Copyright © American Insttitute of Steel Construction,, Reprinted witth Permission. All rights reserved.

These ap pproximate eq quations are said to be aaccurate withiin 2 percent. For design this should easily yieeld results as accurate as th hose obtainedd by reading a value from tthe alignmentt charts. For F the speciial case wheere GA = GB, even simpller equations can be expressed. For sidesway y inhibited, G + 0.4 (5.15) K= G + 0.8 For sidessway uninhibiited, (5.16) K = 0.8G + 1.0 Equation ns 5.15 and 5.16 might be particularly p uuseful for preliminary desiggn.

EXAMPLE E 5.6 Column Efffective Length

Goal: G

Deetermine the column c effecttive length ussing (a) the allignment charrt and (b) Eq quation 5.14.

Given: G

Th he column AB B in a momennt frame is shhown in Figuure 5.22. Asssume that thee column hass its web in thhe plane of thee frame. The beams also hhave their weebs in the plaane of the fraame and thuss beams and columns aree bending ab bout their major, x-axis. It would be verry unusual for a beam in a moment fraame to have the primary bending mom ments about other than thhe x-axis. Ho owever, colum mns may be ooriented for bending aboutt either princippal axis.

SOLUTION N

Part P a Step S 1:

Deetermine mem mber propertiees from Manuual Table 1-1.

150 Chapter 5

Compression Members end A:

W16×36; Igx = 448 in.4 W10×88; Icx = 534 in.4

end B:

W16×77; Igx = 1110 in.4 W10×88; Icx = 534 in.4

Step 2:

Determine the stiffness ratio at each end using Equation 5.12 ⎛ 534 ⎞ 2⎜ ⎟ 14 ⎠ GA = ⎝ = 2.04 ⎛ 448 ⎞ 2⎜ ⎟ ⎝ 24 ⎠ ⎛ 534 ⎞ 2⎜ ⎟ 14 ⎠ GB = ⎝ = 0.825 ⎛ 1110 ⎞ 2⎜ ⎟ ⎝ 24 ⎠

Step 3:

Use the alignment chart shown in Figure 5.21 for a sidesway uninhibited frame. Enter GA and GB on the appropriate scales and construct a straight line between them, as shown in Figure 5.23. The intersection with the scale for K gives the effective length factor, in this case,

K = 1.42 Thus, Part b Step 4:

Lc = KL = 1.42 (14.0 ) = 19.9 ft

Determine K using the stiffness ratios, GA and GB, determined in part (a) Step 2 and Equation 5.14.

K= Thus,

1.6 ( 2.04 )( 0.825) + 4 ( 2.04 + 0.825) + 7.5 2.04 + 0.825 + 7.5

= 1.45

Lc = KL = 1.45 (14.0 ) = 20.3 ft

Note that K determined graphically from the alignment chart and K calculated with Equation 5.14 are very close, as might be expected.

EXAMPLE 5.7 Column Effective Length

Goal:

Determine the column effective length for the column of Example 5.6 using the alignment chart if the column is bending about its weak axis.

Given:

The column AB in a moment frame is shown in Figure 5.22. However, for this example assume that the column has its web perpendicular to the plane of the frame, thus it is bending about its minor or weak axis.

SOLUTION

Step 1:

Find member properties from Manual Table 1-1. end A: W16×36; Igx = 448 in.4 W10×88; Icy = 179 in.4

Chapter 5 en nd B:

Comppression Mem mbers

151

W116×77; Igx = 1110 in.4 W W10×88; Icy = 179 in.4

Step S 2:

Deetermine the stiffness s ratioo at each end uusing Equatioon 5.12. ⎛ 179 ⎞ 2⎜ ⎟ 14 ⎠ GA = ⎝ = 0.685 ⎛ 448 ⎞ 2⎜ ⎟ ⎝ 24 ⎠ ⎛ 179 ⎞ 2⎜ ⎟ 14 ⎠ GB = ⎝ = 0.276 ⎛ 1110 ⎞ 2⎜ ⎟ ⎝ 24 ⎠

Step S 3:

Usse the alignm ment chart shoown in Figurre 5.21 for a sidesway unninhibited fraame. Enter th he values for G A and GB onn the appropriate scales annd draw a strraight line beetween them. The line’s intersection with the scaale for K giv ves the effecttive length facctor—in this case, K = 1.16 Th hus,

Step S 4:

5 Figure 5.22

Lc = K KL = 1.16 (144.0 ) = 16.2 ft

No ote that the reeduction in m moment of inerrtia of the collumns results in the beeams providin ng more end rrestraint, reduucing the effecctive length ffactor for thee column and d thus reducinng the columnn effective lenngth.

Multi-sstory Frame fo or Examples 55.6 and 5.7

1552 Chapter 5

Compresssion Memberss

Figure 5.23 Alignment Chart for Example 5.66 Copyright © A American Instiitute of Steel Construction, Inc. Reprintedd with Permissiion. All rights reseerved.

5.55.1

Effectiv ve Length forr Inelastic Co olumns The assump ption of elastiic behavior for f all membeers of a fram me is regularlly violated. W We have already seen n the role thatt residual streesses play in determining column strenngth through inelastic behavior. Th hus, it is useeful to accom mmodate this inelastic behhavior in the determinatioon of Kfactors. Thee assumption n of elastic behavior is important iin the calcuulation of G as the simplificatio on is made to o move from Equation C-A A-7-3 to Equuation 5.12. R Returning to E Equation C-A-7-3 an nd assuming that all collumns framinng into a jooint have thee same moddulus of elasticity—w which is equaal to the tangeent modulus, ET shown inn Figure 5.12— —and that thhe beams behave elasttically, the deefinition of G for inelastic bbehavior becoomes E (∑( I /L)c ) Ginnelastic = T (5.17) E (∑( I /L) g ) If G for elasstic behavior is i taken as Gelastic , then Gineelastic can be foormulated as e ⎛E ⎞ Ginelastic = ⎜ T ⎟ Gelastic ⎝ E ⎠

(5.18)

Thus, includ ding inelasticc column beh havior simplyy results in a modificationn of G. The ratio of tangent mod dulus to elasttic modulus iss always lesss than 1, so thhe assumptioon of elastic bbehavior for this application leadss to a conserv vative estimaate, as can bee seen by entering the nom mograph with lower G-values an nd determinin ng the correesponding K--factor. Before a straighttforward approach to including in nelastic effects in the deterrmination of effective length can be prroposed, the relationsship between the tangent modulus m and thhe elastic moodulus must be established. The Commentary y to Appendix x 7 of the Speecification inndicates that τb = ET /E, as given in Chapter C for fo the direct analysis a meth hod, should bee used to account for coluumn inelasticity in the effective len ngth method. Thus, if αPr /Pns ≤ 0.5

τb = 1.0

(AISC C C2-2a)

Chapter 5

Compression Members 153

and if αPr /Pns > 0.5

⎛ αP τb = 4 ⎜ r ⎝ Pns

⎞ ⎡ ⎛ αPr ⎟ ⎢1 − ⎜ P ⎠ ⎣ ⎝ ns

⎞⎤ ⎟⎥ ⎠⎦

(AISC C2-2b)

where α = 1.0 for LRFD and α = 1.6 for ASD. Pns is the cross section compression strength. For members without slender elements Pns = Py. For compression members with slender elements, Pns = FyAe which is addressed in the section 5.6. Manual Table 4-13 provides values for τb based on the required strength, Pu/Ag and Pa/Ag. The use of Table 4-13 assumes that the column is loaded to its full available strength. If it is not, the table provides a conservative assessment of the inelastic stiffness reduction factor and the effective length.

EXAMPLE 5.8 Inelastic Column Effective Length

Goal:

Determine the inelastic column effective length using the alignment chart.

Given:

Determine the inelastic effective length for the column in Example 5.6. The column has an LRFD required strength of Pu = 950 kips and an ASD required strength of Pa = 633 kips. Use Equation 5.14 in place of the alignment chart. The column is A992 steel.

SOLUTION

Step 1:

From Manual Table 1-1, for a W10×88 A = 26.0 in.2, and from Example 5.6, the elastic stiffness ratios are GA = 2.04 and GB = 0.825.

For LRFD Step 2:

Determine the required stress based on the required strength. Pu 950 = = 36.5 ksi A 26.0

Step 3:

Determine the stiffness reduction factor from Manual Table 4-13, interpolating between 36 and 37 ksi. τb = 0.788

Step 4:

Determine the inelastic stiffness ratios by multiplying the elastic stiffness ratios by the stiffness reduction factor. GiA = 0.788 ( 2.04 ) = 1.61

GiB = 0.788 ( 0.825) = 0.650 Step 5:

Determine K from Equation 5.14. 1.6 (1.61)( 0.650 ) + 4 (1.61 + 0.650 ) + 7.5 K= = 1.37 1.61 + 0.650 + 7.5 Thus, Lc = KL = 1.37 (14.0 ) = 19.2 ft Note that the effective length factor and thus the effective length is less than that determined in Example 5.6, as expected.

For ASD Step 2:

Determine the required stress based on the required strength. Pa 633 = = 24.3 ksi A 26.0

1554 Chapter 5

Compresssion Memberss Step p 3:

Deterrmine the stiffness reduuction factorr from Mannual Table 4-13, interp polating betweeen 24 and 255 ksi. τb = 0.691

Step p 4:

Deterrmine the ineelastic stiffnesss ratios by m multiplying thhe elastic stifffness ratioss by the stiffneess reductionn factor. GiA = 00.691( 2.04 ) = 1.41

GiB = 00.691( 0.825 ) = 0.570 Deterrmine K from Equation 5.114. 1 (1.41)( 0.5770 ) + 4 (1.41 + 0.570 ) + 7.55 1.6 K= = 1.33 1.441 + 0.570 + 77.5 Thus,, Lc = KL = 1.33 (14.0 ) = 18.6 ft

Step p 5:

Note that the effeective length factor and thhus the effecctive length iis less than that t determineed in Examplle 5.6, as expeected. 5.55.2 Effective Length when n Supporting g Gravity On nly Columnss Another con ndition that influences i co olumn bucklinng and thus the effectivee length factoor is the existence off columns th hat carry only y gravity loaad and contrribute nothingg to the lateeral load resistance or o stability off the structurre. Figure 5.224a illustratees a simple sstructure of tthis type where the sttability or lateral load resiisting columnn is the flagpoole column on the left, column A, and the gravity only collumn is the pin p ended coolumn on thee right, colum mn B. The looad P is applied to co olumn A and the load Q is applied to coolumn B.

P

Q

L

P

B

P

A

B

A P+Q

Q

M = PΔ + Q Δ (b)) olumn Providiing Lateral Reestraint for a Gravity Onlyy Column Flagpole Co

(a))

P+Q

L

P Q

Δ

Q

QΔ L

L

A

Figgure 5.24

Δ

M = PΔ + Q Δ (c)

Chapter 5

Compression Members 155

If Q = 0, column A behaves as if column B did not exist since it just goes along for the ride. However, when Q is not zero, buckling of column A leads to lateral displacement, Δ, at the top of columns A and B. Thus, equilibrium requires a lateral force be exerted at the top of column B. This force must be resisted by column A as shown in Figure 5.24b. In the displaced position illustrated in Figure 5.24b, equilibrium of column A requires a resisting moment at the support of M = PΔ + QΔ . Figure 5.24c shows column A with an applied load, ( P + Q ) , which in the displaced position produces a moment at the support of M = PΔ + QΔ . Thus, the column in Figure 5.24c can be thought of as a representation of column A in Figure 5.24b with only slight error. Since these two columns are considered equal, if column A can support the load ( P + Q ) , it should be adequate for column A to support its load, P, and the effect of the load Q on column B. Considering elastic buckling this can be stated as π2 EI P + Q = (5.19) ( ) 2 ( Ko L ) Where Ko is the K-factor for column A. For this example, the theoretical K-factor for the flagpole column is Ko = 2. Another way to approach this problem would be to continue to consider that column A supports only the load P but use a K-factor that accounts for the influence of the load on the gravity only column, Kn. This can be stated as π2 EI P= (5.20) 2 ( Kn L ) Since these two equations represent the same structure, they can be solved for π 2 EI L2 . Thus, from Equation 5.19 π2 EI L2 = K o2 ( P + Q ) and from Equation 5.20 π 2 EI L2 = K n2 P . Setting these equal and solving for Kn yields P+Q Q Kn = Ko = Ko 1 + (5.21) P P Thus, a column that supports a load P and also must provide stability for load Q on gravity only columns may be designed using this modified effective length factor Kn.

EXAMPLE 5.9 Gravity Only Columns and Effective Length

SOLUTION

Goal:

Determine the in-plane nominal strength of the column that is required to carry a concentrated load and provide lateral stability for gravity only columns. Also, determine the strength of the column if there is no load on the gravity only column.

Given:

A W14×90 column shown in Figure 5.25 as Column A is to a) carry an applied load, P, and provide lateral restraint to a gravity only column, Column B, carrying the load Q = 2P and b) carry an applied load, P, with no load on the gravity only column. The W14×90 is oriented so the web is in the plane of the frame. Use A992 steel.

Step 1:

Determine the effective length factor for the W14×90 column without considering the gravity only column. Since this is a flagpole column, from Figure 5.7f, the theoretical K-factor is 2.0

156 Chapter 5

Compression Members Part a Step 2:

Using Equation 5.21, determine the modified effective length factor to account for the gravity only column load, Q = 2P. Thus Q 2P K n = K o 1 + = 2.0 1 + = 3.46 P P

Step 3:

From Manual Table 1-1

Step 4:

Determine which column strength equation to use. Since K n L 3.46 (15 (12 ) ) 29,000 = = 101 < 4.71 = 113 rx 6.14 50

A = 26.5 in.2 and rx = 6.14

use Equation E3-2 Step 5:

Determine the Euler buckling stress π2 ( 29,000 ) = 28.1 ksi Fe = 2 (101)

Step 6:

Determine the critical stress from Equation E3-2 ⎛ Fy ⎞ ⎛ 50 ⎞ ⎛ ⎛ ⎜ ⎟⎞ ⎜ ⎟⎞ Fcr = ⎜ 0.658⎝ Fe ⎠ ⎟ Fy = ⎜ 0.658⎝ 28.1 ⎠ ⎟ 50 = 23.7 ksi ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

Step 7:

Determine the nominal strength Pn = 23.7 ( 26.5 ) = 628 kips

Part b Step 8:

With no load on the gravity only column, Kn = Ko = 2.0. Determine which column strength equation to use. Since 29,000 K n L 2.0 (15 (12 ) ) = = 58.6 < 4.71 = 113 rx 6.14 50 use Equation E3-2

Step 9:

Determine the Euler buckling stress π2 ( 29,000 ) = 83.3 ksi Fe = 2 ( 58.6)

Step 10:

Determine the critical stress from Equation E3-2 ⎛ Fy ⎞ ⎛ 50 ⎞ ⎛ ⎛ ⎜ ⎟⎞ ⎜ ⎟⎞ Fcr = ⎜ 0.658⎝ Fe ⎠ ⎟ Fy = ⎜ 0.658⎝ 83.3 ⎠ ⎟ 50 = 38.9 ksi ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

Step 11:

Determine the nominal strength Pn = 38.9 ( 26.5 ) = 1030 kips

Chapter 5 P

Comppression Mem mbers

157

Q

Fiigure 5.25 Single Storyy Frame forr Example 5.99

5.6

SLEN NDER ELEMENTS IN N COMPRE ESSION As mentioned in Section 5.4, the columns dis cussed thus ffar are controolled by overrall column buckling. For some sh hapes, anotheer form of bucckling may acctually controol column streength: local buckling of the elemeents that makee up the colum mn shape. W Whether the shhape is rolled or built up, it can bee thought of as a being comp posed of a grroup of intercconnected plaates. Dependiing on how these plaates are suppo orted by each other, they ccould buckle aat a stress bellow the criticcal buckling stress off the overall column. c This is local buckkling, also caalled plate buuckling, and iis shown in Figure 5.26. Local bu uckling is desscribed througgh a plate crittical bucklingg equation sim milar to the Euler bucckling equatio on for column ns. The criticaal buckling sttress for an axxially loaded plate is k π2 E (5.22) Fcr = 2 2 ⎛b⎞ 112 (1 − ν ) ⎜ ⎟ ⎝t⎠ where k is a plate bu uckling coeffiicient that deepends on thee plate loadinng, edge condditions, and length-to o-width ratio; ν is Poisson n’s ratio; and b/t is the rattio of the widdth perpendiccular to the compresssion force to the thickness of the platee. The width--to-thickness ratio is calleed the plate slenderneess and is sim milar in functtion to the coolumn slenderrness. This crritical stress pplotted as a function of width-to-th hickness ratio o is shown as the dashed cuurve in Figurre 5.27.

Figure 5.2 26 Column n Tested to Failure through Lo ocal Buckling g Photo courttesy Perry Greeen

Figure 5.277 Plate Streength in Comppression

158 Chapter 5

Compression Members As with overall column buckling, an inelastic transition exists between elastic buckling and element yielding. This transition is due to the existence of residual stresses and imperfections in the element, just as in the case of overall column buckling, and results in the inelastic portion of the curve shown in Figure 5.27. In addition, for plates with low b/t ratios, strain hardening plays a critical role in their behavior, and plates with large b/t ratios have significant postbuckling strength. To ensure that local buckling will not control column strength, the critical plate buckling stress for local buckling is limited to the critical buckling stress for overall column buckling. This approach results in a different maximum plate slenderness value for each corresponding column slenderness value. However, to alert the designer to the need for consideration of plate buckling, an initial check on element slenderness is made assuming that the stress in the plate has reached the yield stress. The development of the Specification provisions starts by finding a plate slenderness that sets the plate buckling stress equal to the column yield stress. Equation 5.22 then becomes b k π2 E = t 12 (1 − ν 2 ) Fy

Taking ν = 0.3, the standard value for steel, this plate slenderness becomes b kE = 0.95 t Fy

(5.23)

(5.24)

which is shown as point O in Figure 5.27. This point is well above the inelastic buckling curve. In order to obtain a b/t that would bring the inelastic buckling stress closer to the yield stress, a somewhat arbitrary slenderness limit is taken as 0.7 times the limit that corresponds to the column yield stress, which gives b kE = 0.665 t Fy This is indicated as point D in Figure 5.27. The remaining factor to be determined is the plate buckling coefficient, k. This factor is a function of the stress distribution, the edge support conditions and the aspect ratio of the plate. For a plate with uniform compression on opposite ends and simply supported on all four sides, the minimum value of k can be shown to be 4.0. For actual columns, with the variety of potential cross section shapes available, the determination of the value of k becomes much more complicated as the actual edge supports, stress distribution and aspect ratio vary. The limiting width-to-thickness ratios are given in Specification Table B4.1a. These limits may be given as E λ r = c3 (5.25) Fy where c3 is given in Table 5.4 for several elements in uniform compression taken from Table B4.1a of the Specification. The apparent plate buckling coefficient, k, used to obtain these values is also given in Table 5.4. For shapes with element slenderness exceeding these λr values plate buckling must be considered. As already shown, these limits are based on the assumption that the column is stressed to Fy. Since columns are rarely stressed to that level, it is very possible that what appears to be a slender element compression member based on Table B4.1a may not actually see its strength limited by the limit state of local buckling.

Compression Members 159

Chapter 5 Table 5.4

Parameters for Consideration of Compression Member Local Buckling

*

Case

λ

c3

k

c4

c5

1

b/t

0.56

0.71

0.834

0.184

b/t

0.45

0.46

0.671

0.148

d/t h/tw

0.75 1.49

1.27 5.0

1.12 1.95

0.246 0.351

b/t

1.40

4.43

1.93

0.386

Flanges of rolled I-shaped sections, plates projecting from rolled Ishaped sections; outstanding legs of pairs of angles connected with continuous contact, flanges of channels, and flanges of tees 3 Legs of single angles, legs of double angles with separators, and all other unstiffened elements 4 Stems of tees 5 Webs of doubly symmetric I-shaped sections and channels 6 Walls of rectangular HSS and boxes of uniform thickness * From Table B4.1a in the Specification.

For W-shapes, Case 1 in Table B4.1a, with Fy = 50 ksi, the flange slenderness limit is λ rf = 0.56 E Fy = 13.5 , and all W-shapes have a flange slenderness less than this limit. For webs of these W-shapes, Case 5 in Table B4.1a, λ rw = 1.49 E Fy = 35.9 , and many available W-shapes have a web slenderness that exceeds this limit and are classified as slender. Design of slender element compression members according to the Specification follows the same requirements as those for compression members without slender elements, with one modification. To account for slender element behavior, the full area of the slender element cannot be used. Thus, a reduced effective area, Ae, is used in place of the gross area, Ag, to determine column strength. For columns with slender elements, Section E7 indicates that column strength is given by Pn = Fcr Ae (AISC E7-1) which is to be used in place of Equation E3-1 but with the same critical stress, Fcr. Once the designer is directed to the slender element provisions of Section E7 it is apparent that the critical stress based on the controlling limit state must first be determined. Then, using that stress, the actual plate element slenderness at the transition from elastic to inelastic buckling can be determined. For the web of a rolled W-shape, Case 5 in Table B4.1a, the limiting width-to-thickness ratio becomes h E E λ r = = c3 = 1.49 tw Fcr Fcr If the width-to-thickness ratio of the web does not exceed this value, the usable width of the web is the actual width and no change in area is required. If the width-to-thickness ratio of the web does exceed this value, a reduced width must be determined through ⎛ F ⎞ F be = b ⎜⎜1 − c1 el ⎟⎟ el (AISC E7-3) Fcr ⎠ Fcr ⎝ where the elastic plate buckling stress from Equation 5.22 is presented as

160 Chapter 5

Compression Members 2

⎛ λ ⎞ Fel = ⎜ c2 r ⎟ Fy (AISC E7-4) ⎝ λ ⎠ The constants c1 and c2 are given in Specification Table E7-1. If Equations 5.25 and E7-4 are substituted into Equation E7-3, the effective width becomes E ⎛ c1c2 c3 E ⎞ be = c2 c3t (5.26) ⎜1 − ⎟ Fcr ⎜⎝ ( b t ) Fcr ⎟⎠ Equation 5.26 may be simplified by substituting c4 = c2c3 and c5 = c1c2c3 which yields

be = c4t

E Fcr

⎛ c5 E ⎞ ⎜⎜1 − ⎟⎟ ⎝ ( b t ) Fcr ⎠

(5.27)

Values for c4 and c5 are given in Table 5.4. Continuing the consideration of W-shape webs by making the appropriate substitutions into Equation 5.27 yields E ⎡ 0.351 E ⎤ be = 1.95t (5.28) ⎢1 − ⎥ Fcr ⎣ (b /t ) Fcr ⎦ Once the effective width of a slender element is obtained, the corresponding effective area of the member can be determined. Since hot-rolled shapes have fillets at the junction of the plate elements, the best approach for determining the effective area is to use the gross area and deduct the appropriate ineffective element area. This will be illustrated in the example.

EXAMPLE 5.10 Strength of Column with Slender Elements

SOLUTION

Goal:

Determine the available strength of a compression member with a slender web.

Given:

Use a W16×26 as a column with Lcy = 5.0 ft.

Note:

In Manual Table 1-1, this shape is identified with footnote c indicating that it must be considered as a slender element member for compression. It is the most slender-web W-shape available and is not normally used as a column.

Step 1:

From Manual Table 1-1, A = 7.68 in.2, h/tw = 56.8, tw = 0.250 in., ry = 1.12 in.

Step 2:

Determine the web slenderness limit from Specification Table B4.1a, Case 5. E 29,000 λ rw = 1.49 = 1.49 = 35.9 Fy 50

Step 3:

Check the slenderness of the web. h = 56.8 > λ rw = 35.9 tw Thus, the shape must be treated as one with a slender web. It has already been established that all W-shapes have nonslender flanges, so that check

Chapter 5

Compression Members 161

will not be made here. Step 4:

Determine the Euler buckling stress, Fe, for Lc = 5.0 ft. π 2 (29,000) Fe = = 99.7 ksi 2 ⎛ 5 (12 ) ⎞ ⎜ ⎟ ⎝ 1.12 ⎠

Step 5:

Determine Fcr

Fe = 99.7 ksi > Fy 2.25 = 50 2.25 = 22.2 ksi

Therefore use Equation E3-2 ⎛ 50 ⎞ ⎜ ⎟

Fcr = 0.658⎝ 99.7 ⎠ ( 50 ) = 40.5 ksi Step 6:

Check the slenderness of the web against the new limit using Fcr in place of Fy. E 29,000 λ rw = 1.49 = 1.49 = 39.9 40.5 Fcr since h = 56.8 > λ rw = 39.9 tw the effective web width must be determined.

Step 7:

Determine the effective width of the web using Equation 5.28 E ⎡ 0.351 E ⎤ be = 1.95t ⎢1 − ⎥ Fcr ⎣ (b /t ) Fcr ⎦

= 1.95(0.250) Step 8:

29,000 40.5

⎡ 0.351 29,000 ⎤ ⎢1 − ⎥ = 10.9 in. 56.8 40.5 ⎦ ⎣

Determine the actual web width. The width of the web plate is given by h. However, a value of h is not specifically available in the Manual, so with h/tw = 56.8 and tw = 0.250, h can be determined as h = ( h tw ) tw = 56.8 ( 0.250 ) = 14.2 in.

Step 9:

Determine the effective area. Because be < h, use be to determine Ae. To properly account for the fillets at the web-flange junction, the area of the ineffective web is deducted from the gross area of the shape; thus, Ae = Ag − ( h − be ) tw = 7.68 − (14.2 − 10.9 )( 0.250 ) = 6.86 in.2

Step 10:

Determine the nominal strength of the column. Pn = Fcr Ae = 40.5 ( 6.86) = 278 kips

162 Chapter 5

Compression Members For LRFD Step 11: For ASD Step 11:

EXAMPLE 5.11 Strength of Column with Slender Elements

SOLUTION

Determine the design strength for this slender web column with Lc = 5.0 ft. φPn = 0.9 ( 278) =250 kips

Determine the design strength for this slender web column with Lc = 5.0 ft. Pn Ω = 278 1.67 =166 kips

Goal:

Determine the available strength of a compression member with a slender web.

Given:

Use the W16×26 column from Example 5.10 but with Lcy = 15.0 ft.

Note:

This shape has already been shown to have a slender web based on Table B4.1a

Step 1:

From Manual Table 1-1, A = 7.68 in.2, h/tw = 56.8, tw = 0.250 in., ry = 1.12 in.

Step 2:

Determine the web slenderness limit from Specification Table B4.1a, Case 5. E 29.000 λ rw = 1.49 = 1.49 = 35.9 Fy 50

Step 3:

Check the slenderness of the web. h = 56.8 > λ rw = 35.9 tw Thus, the shape has a slender web. It has already been established that all W-shapes have nonslender flanges.

Step 4:

Determine the Euler buckling stress, Fe, for Lc = 15.0 ft. π 2 (29,000) Fe = = 11.1 ksi 2 ⎛ 15 (12 ) ⎞ ⎜ ⎟ ⎝ 1.12 ⎠

Step 5:

Determine Fcr .

Fe = 11.1 ksi < Fy 2.25 = 50 2.25 = 22.2 ksi

Therefore use Equation E3-3 Fcr = 0.877 (11.1) = 9.73 ksi Step 6:

Check the slenderness of the web against the new limit using Fcr in place of Fy.

Chapter 5

λ rw = 1.49

Compression Members 163

E 29,000 = 1.49 = 81.3 9.73 Fcr

since h = 56.8 < λ rw = 81.3 tw

the column will not be limited by local buckling of the web Step 7: For LRFD Step 8:

For ASD Step 8:

Determine the nominal strength of the column. Pn = Fcr Ag = 9.73( 7.68) = 74.7 kips

Determine the design strength for this slender web column with Lc = 15.0 ft. φPn = 0.9 ( 74.7 ) =67.2 kips

Determine the design strength for this slender web column with Lc = 15.0 ft. Pn Ω = 74.7 1.67 =44.7 kips

Examples 5.10 and 5.11 illustrate that a W-shape that appears to be a slender element shape based on Table B4.1a may not actually be limited in strength because of that slender element, the limit state of local buckling. It can be shown that the W16×26 considered in these examples will have its strength limited by local buckling for columns with effective length, Lc, up to about 10.8 ft. Above that effective length, the slender elements will not impact overall column strength.

5.7

COLUMN DESIGN TABLES A review of the AISC column equations, E3-2 and E3-3, shows that the only factor other than shape geometry and material strength that influences the determination of column strength is the slenderness ratio. Therefore, it is convenient to tabulate column strength as a function of slenderness. Part 4 of the Manual contains tables for W-shapes, HP-shapes, and HSS and several singly symmetric shapes. Figure 5.28 shows a sample of Manual Table 4-1a for several W14 sections with Fy = 50 ksi. As with all of the available strength tables in the Manual, both allowable strength (ASD) and design strength (LRFD) values are given. Tables 4-1b and 4-1c are provided for selected W-shapes that are commonly available with Fy = 65 and 70 ksi respectively. The values in these column tables are based on the assumption that the column will buckle about its weak axis. For all W-shapes this is the y-axis, so the values in the tables are given in terms of the effective length with respect to the least radius of gyration, ry. Their use is quite straightforward when the critical buckling length is about this axis. An approach that permits the use of these tables when the strong axis controls will be addressed following the example.

1664 Chapter 5

Compresssion Memberss

A Compresssion Figure 5.28 Available Strength in Axial Copyright © American Insttitute of Steel Construction, C R Reprinted with Permission. A All rights reservved.

Chapter 5

Comppression Mem mbers

165

Figurre 5.29 Collumns for Exaamples 5.12 and 5.13

EXAMPLE E 5.12a Column Deesign by LRFD

Goal: G

Deetermine the least-weightt section to carry the looads given uusing the lim mited selectio on available inn Figure 5.28.

Given: G

Th he column is shown in Figgure 5.29a. Itt must resist tthe followingg loads in thee appropriatee combinationns: PD = 56 kkips, PL = 1772 kips, and PW = 176 kip ps. Use A992 2 steel. Assum me the live looad comes frrom a distribuuted load lesss than 100 psf, so that thee LRFD load factor on livee load may bee taken as 0.5 5 for load com mbination 4.

SOLUTION N

Step S 1:

Deetermine thee maximum required sstrength usinng the LRF FD load co ombinations from fr Section 22.4. 1. 1.4PD = 1.4(56) = 788.4 kips 2. 1.2PD + 1.6PL = 1.2((56) + 1.6(172) = 342 kipss 4. 1.2PD + 0.5PL + 1.0P W = 1.2(56) + 0.5(172) + 1.0(176) = 3229 kips 6. 0.9PD + 1.0PW = 0.99 (56) + 1.0(176) = 226 kipps So o the column must m carry Pu = 342 kips.

Step S 2:

Th he column haas the same efffective lengtth about the x- and y-axes,, so enter thee table in Fig gure 5.28 withh Lc = 18 ft. S Scanning acrooss the table aat Lc = 18 ft and checkin ng the LRFD D values, seleect the least--weight shape in this po ortion of the taable that can support this lload. Seelect a W14×6 61 with a desiign compresssion strength φPn = 456 kips

166 Chapter 5

Compression Members

EXAMPLE 5.12b Column Design by ASD

Goal:

Determine the least-weight section to carry the loads given using the limited selection available in Figure 5.28.

Given:

The column is shown in Figure 5.29a. It must resist the following loads in the appropriate combinations: PD = 56 kips, PL = 172 kips, and PW = 176 kips. Use A992 steel.

SOLUTION

Step 1:

Determine the maximum required strength using the ASD load combinations from Section 2.4. 1. PD = 56 kips 2. PD + PL = 56 + 172 = 228 kips 5. PD + 0.60PW = 56 + 0.6(176) = 162 kips 6. PD + 0.75 PL + 0.75(0.6PW ) = 56 + 0.75(172) +0.75(0.6(176)) = 264 kips 7. 0.6 PD +0.6 PW = 0.6(56) + 0.6(176) = 139 kips So the column must carry Pa = 264 kips.

Step 2:

The column has the same effective length about the x- and y-axes, so enter the table in Figure 5.28 with Lc = 18 ft. Scanning across the table at Lc = 18 ft and checking the ASD values, select the least-weight shape in this portion of the table that can support this load Select a W14×61 with an allowable compression strength Pn Ω = 304 kips

If the largest slenderness ratio for a particular column happens to be for x-axis buckling, the tables may not be entered directly with the x-axis effective length because the table effective length is intended to be used in conjunction with the least radius of gyration. However, it is possible to determine a modified effective length that, when used in the table, will result in the correct column strength. When the x-axis controls column strength, the slenderness ratio used in the column equations is Lcx/rx. To use the column tables, an effective length, (Lc)eff, must be determined that, when combined with ry, gives the same slenderness ratio. So ( Lc )eff Lcx = ry rx Solving this equation for (Lc)eff yields Lcx ( Lc )eff = (rx / ry ) With this modified effective length, the tables can be entered and a suitable column selected. There is one difficulty with this process, however. Until a column section is known, the value for rx/ry cannot be determined. To account for this, a quick scan of the column tables should be done to estimate rx/ry. Then, when a section is selected, the assumption can be verified and an adjustment made if necessary.

Chapter 5

EXAMPLE 5.13 Column Design

SOLUTION

Compression Members 167

Goal:

Determine the least-weight section to carry the force given using the limited selection available through Figure 5.28. Design by LRFD and ASD.

Given:

The column is shown in Figure 5.29b. Use the loading from Example 5.12.

Step 1:

Determine the effective length for each axis. Bracing of the y-axis, shown in Figure 5.29b, yields Lcy = 10.0 ft. The unbraced x-axis has Lcx = 30.0 ft.

Step 2:

Determine (Lc)eff for the x-axis. Select a representative rx/ry from Figure 5.28. There are two general possibilities. Assume that the larger shapes might be needed to carry the load, and try rx/ry = 2.44. Thus, L 30.0 ( Lc )eff = cx = = 12.3 ft (rx /ry ) 2.44

Step 3:

Determine the controlling effective length. Because (Lc)eff = 12.3 ft is greater than Lcy = 10.0 ft, enter the table with Lc = 12.3 ft and interpolate between 12 ft and 13 ft.

For LRFD Step 4:

From Example 5.12a the column must have a design strength greater than Pu = 342 kips with Lc = 12.3 ft. Try a W14×43, which happens to be the smallest column available with the limited selection available in Figure 5.28. This column has rx/ry = 3.08.

Step 5:

Determine (Lc)eff with this new rx/ry. Thus, 30.0 = 9.74 ft ( Lc )eff = 3.08

Step 6:

Determine the new controlling effective length. Because (Lc)eff = 9.74 ft is now less than Lcy = 10.0 ft, enter the table with 10.0 ft and note that the W14×43 has a design strength of 422 kips, which is greater than the required strength of 342 kips.

Step 7:

Therefore, use the selected W14×43 Note: The W14×43 is identified in the table by a footnote as slender for Fy = 50 ksi. This is not an issue for our design because the impact of any slender element has already been taken into account in generating the table as stated in the same footnote. Using the full complement of tables available in the Manual results in a smaller W12 section having the ability to carry the given load.

168 Chapter 5

Compression Members For ASD Step 4:

From Example 5.12b the column must have an allowable strength greater than Pa = 264 kips with Lc = 12.3 ft. Try a W14×48. This column has rx/ry = 3.06.

Step 5:

Determine (Lc)eff with this new rx/ry. Thus, 30.0 = 9.80 ft ( Lc )eff = 3.06

Step 6:

Determine the new controlling effective length. Because (Lc)eff = 9.80 ft is now less than Lcy = 10.0 ft, enter the table with 10.0 ft and see that the W14×43 has an allowable strength of 281 kips, which is greater than the required strength of 268 kips and rx/ry = 3.08 which is greater than that for the W14×48 so Lcy = 10.0 ft will still control.

Step 7:

Therefore, use the selected W14×43 Note: The W14×43 is identified in the table by a footnote as slender for Fy = 50 ksi. This is not an issue for our design because the impact of any slender element has already been taken into account in generating the table as stated in the same footnote. Using the full complement of tables available in the Manual results in a smaller W12 section not having the ability to carry the given load.

Table 4-1a in Part 4 of the Manual includes shapes from a W8×31 up to a W14×873. All of the shapes included are considered column shapes and have reasonably similar strengths about the x- and y-axes. That is, the shapes are close to being square and rx/ry is not extremely large, ranging from 1.59 to 3.08. Any of the other available W-shapes may be used for columns if desired, but it must be recognized that the relationship between the x- and y-axes is such that the y-axis will control unless significant bracing is provided. These shapes are generally considered beam shapes. Since beams are intended to be used to carry flexure, the relationship between the xand y-axes is not as critical. For example, for a W16×26 with a length of 24 ft braced at the ends only for the x-axis and rx/ry = 5.59, the y-axis will control unless it is braced at least every 4.29 ft. The W-shape column tables in Part 4 of the Manual for Fy = 50 ksi also exclude the smallest W-shapes in an attempt to direct the design engineer toward using shapes that are more appropriate when considering connections. That does not mean that these smaller shapes are not acceptable for use as columns. The tables in Part 6 of the Manual, which will be discussed in Chapter 8, can be used for the design of columns and they include all of the W-shapes.

Chapter 5

EXAMPLE 5.14 Column Design

SOLUTION

Compression Members 169

Goal:

Determine the least-weight section to carry the force given using the small shapes provided in the column tables in Manual Part 6. Design by LRFD and ASD.

Given:

The column has an effective length for both axes of 10 ft and must carry a concentrated dead load of 8 kips and a concentrated live load of 24 kips.

For LRFD Step 1:

Step 2:

Determine the required strength for the load combination 1.2D + 1.6L. Pu = 1.2(8.0) + 1.6(24.0) = 48.0 kips Using Manual Table 6-2, select the lightest column to support this load. Select the W4×13.

φPn = 60.1 kips For ASD Step 1:

Step 2:

Determine the required strength for the load combination D + L. Pa = 8.0 + 24.0 = 32.0 kips Using Manual Table 6-2, select the lightest column to support this load. Select the W4×13.

Pn = 40.0 kips Ω

5.8 TORSIONAL BUCKLING AND FLEXURAL-TORSIONAL BUCKLING Up to this point, the discussion has addressed the limit states of flexural buckling and local buckling. Two additional limit states for column behavior must be addressed: torsional buckling and flexural-torsional buckling. Doubly symmetric shapes normally fail through flexural buckling, as discussed earlier in this chapter, or through torsional buckling. Singly symmetric and nonsymmetric shapes can fail through flexural, torsional, or flexural-torsional buckling. Because the shapes normally used for steel members are not well suited to resist torsion, except for closed HSS, it is usually desirable to avoid any torsional limit states through proper bracing of the column or by avoiding torsional loading. If either of the torsional limit states must be evaluated, the applicable Specification provisions are found in Section E4, except for the special cases associated with single angles, which are found in Section E5. All members addressed in Section E4 are assumed to buckle torsionally about their shear center. For these shapes an elastic buckling stress, Fe, is determined, which is then used in Equations E3-2 and E3-3 to determine the column critical stress, Fcr. The equations given in Section E4 defining the elastic buckling stress for the limit states of torsional and flexural-torsional buckling are also found in several other books with varying notation, including Buckling Strength of Metal Structures. 1 The equations of Section E3 are used to account for such factors as inelastic buckling, initial out-of-straightness, and residual stresses.

1

Bleich, F. Buckling Strength of Metal Structures. New York: McGraw-Hill, 1952.

1770 Chapter 5

Compresssion Memberss

F Figure 5.30

Column for Example 5.15

The provisions for single an ngles meetinng a specificc set of criteeria take a ddifferent approach. By B limiting th he way that load l is applieed to the endds of a singlle-angle comppression member, an effective slen nderness is esstablished, whhich is then used in Equations E3-2 andd E3-3 to determine th he column critical stress, Fcr. The limit state off torsional bu uckling is not normally connsidered in thhe design of W W-shape columns wh hen the y-ax xis is the co ontrolling axxis for flexurral buckling. Torsional bbuckling generally do oes not goverrn, and when it does, the ccritical load ddiffers very litttle from the strength determined from flexuraal buckling. For other m member typees, such as W WT or doubble-angle compression n members offten used in trrusses, torsionnal limit states are quite im mportant. An additional factor in determ mining strenggth based on these limit sstates is the ttorsional effective len ngth. The Com mmentary reccommends thaat, conservativvely, the torsional effectivve length be taken ass the column length and provides sevveral other possibilities iff greater accuuracy is desired.

EX XAMPLE 5.15 5 Str trength of a WSh hape Column n wiith Torsiona al Bu uckling

Goa al:

Deterrmine the av vailable strenngth of a W W-shape coluumn and connsider torsio onal buckling..

Giv ven:

A W1 14×48 A992 column as shhown in Figuure 5.30 is brraced laterallly and torsio onally at its en nds. At mid-hheight it is braaced to resist buckling aboout the y-axiss, but it canno ot resist torsioonal bucklingg based on thee bracing shoown in Figurre 5.30c.

SO OLUTION

Step p 1:

From m Manual Table 1-1, Ag = 14.1 in.2, Ix = 484 in.4, Iy = 51.4 in.4, rx = 5.85 inn., ry = 1.91 inn., Cw = 2240 in..6, J = 1.45 inn.4, h/tw = 33.66

Step p 2:

Deterrmine the web b slenderness limit. E 299,000 λ rw = 1.49 = 1.49 = 35.9 Fy 50

Chapter 5 Step 3:

Compression Members 171

Check slenderness of the web and flange. h = 33.6 < λ rw = 35.9 tw Therefore, the web is not slender. As previously discussed, all W-shapes have nonslender flanges.

Step 4:

Determine the critical stress for y-axis buckling. ⎛ Lc ⎞ 10 (12 ) = 62.8 ⎜ ⎟ = 1.91 ⎝ r ⎠y

Fe =

π2 E ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠

=

2

π2 ( 29,000 )

( 62.8)

2

= 72.6 ksi >

Fy = 22.2 ksi 2.25

Therefore, use Equation E3-2: ⎛ 50 ⎞

Fcr = ( 0.658)⎜⎝ 72.6 ⎟⎠ ( 50 ) = 37.5 ksi Step 5:

Determine the critical stress for x-axis buckling.

20 (12 ) ⎛ Lc ⎞ = 41.0 ⎜ ⎟ = 5.85 ⎝ r ⎠x

Fe =

π2 E ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠

2

=

π2 ( 29, 000 )

( 41.0 )

2

= 170 ksi >

Fy = 22.2 ksi 2.25

Therefore, use Equation E3-2: ⎛ 50 ⎞

Fcr = ( 0.658)⎜⎝ 171 ⎟⎠ ( 50 ) = 44.2 ksi

Step 6:

Determine the critical stress for z-axis buckling, or twisting about the shear center, using Section E4(a) Equation E4-2. ⎡ π2 ECw ⎤ 1 Fe = ⎢ + GJ ⎥ 2 ⎢⎣ ( Lcz ) ⎥⎦ I x + I y ⎡ π2 ( 29,000 )( 2240 ) ⎤ 1 ⎥ =⎢ + 11, 200 1.45 ( ) 2 ⎢ ⎥ 484 + 51.4 ( 20 (12 ) ) ⎣ ⎦ Fy = 51.1 ksi > = 22.2 ksi 2.25 Therefore use Equation E3-2: ⎛ 50 ⎞

Fcr = ( 0.658)⎜⎝ 51.1 ⎟⎠ ( 50 ) = 33.2 ksi Step 7:

Select the lowest critical stress determined in Steps 4, 5, and 6. Fcr = 33.2 ksi

172 Chapter 5

Compression Members Since the controlling stress comes from Step 6, the strength of the column is controlled by torsional buckling. Step 8:

For LRFD Step 9: For ASD Step 9:

Determine the nominal strength of the column Pn = 33.2 (14.1) = 468 kips Note: Determination of Fcr in steps 4, 5, and 6 could have been delayed until after the controlling, smallest, value of Fe had been determined and then Fcr determined only once. Determine the column design strength. φPn = 0.9 ( 468 ) = 421 kips

Determine the column allowable strength. Pn 468 = = 280 kips Ω 1.67

EXAMPLE 5.16 Strength of a WTShape Column

Goal:

Determine the available strength of a WT-shape column with consideration of flexural, torsional, and flexural-torsional buckling.

Given:

A WT7×34 A992 column is 10.0 torsionally at its ends only.

SOLUTION

Step 1:

From Manual Table 1-8, Ag = 10.0 in.2, Ix = 32.6 in.4, Iy = 60.7 in.4, rx = 1.81 in., ry = 2.46 in., tf = 0.720, Cw = 3.21 in.6, J = 1.50 in.4 , d/tw = 16.9, y = 1.29 in. , b f 2t f = 6.97

Step 2:

Determine the flange and stem slenderness limits from Table B4.1 cases 1 and 4. E 29,000 λ rf = 0.56 = 0.56 = 13.5 Fy 50

λ rw = 0.75 Step 3:

ft long and is braced laterally and

E 29,000 = 0.75 = 18.1 Fy 50

Check slenderness of the flange and stem. bf = 6.97 ≤ λ rf = 13.5 2t f d = 16.9 < λ rw = 18.1 tw Therefore the WT has nonslender flange and stem.

Compression Members 173

Chapter 5 Step 4:

Determine the nominal strength for flexural buckling. Since Lcx = Lcy and the x-axis has the smallest radius of gyration, flexural buckling will be controlled by the x-axis. 29,000 E ⎛ Lc ⎞ 10 (12 ) = 66.3 ≤ 4.71 = 4.71 = 113 ⎜ ⎟ = 1.81 50 Fy ⎝ r ⎠x

Fex =

π2 E ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠

2

=

π2 ( 29,000 )

( 66.3)

2

= 65.1 ksi >

Fy = 22.2 ksi 2.25

Therefore, use Equation E3-2: ⎛ Fy ⎞

⎛ 50 ⎞

Fcr = ( 0.658 )⎜⎝ Fe . ⎟⎠ Fy = ( 0.658 )⎜⎝ 65.1 ⎟⎠ ( 50 ) = 36.3 ksi

and

Step 5:

Pn = 36.3 (10.0 ) = 363 kips

To determine flexural-torsional buckling, the elastic buckling stress for yaxis buckling is required. ⎛ Lc ⎞ 10 (12 ) = 48.8 ⎜ ⎟ = 2.46 ⎝ r ⎠y

Fey =

Step 6:

π2 E ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠

2

=

π2 ( 29,000 )

( 48.8)

2

= 120 ksi

Determine the flexural-torsional elastic buckling stress for z-axis buckling using Equation E4-3. The shear center of a WT-shape is at the stem-flange intersection. Thus, the distance from the centroid to the shear center is tf 0.720 xo = 0, yo = y − = 1.29 − =0.930 in. 2 2 and from Equation E4-9 Ix + I y 32.6 + 60.7 ro2 = xo2 + yo2 + = 0.0 + 0.9302 + = 10.2 Ag 10.0 and Equation E4-8

H =1−

xo2 + yo2 0 + 0.9302 = − = 0.915 1 ro2 10.2

From the user note in Section E4, take Cw = 0 in Equation E4-7. Thus, 11, 200 (1.50 ) GJ Fez = = = 165 ksi 2 Ag ro 10.0 (10.2 ) Step 7:

Determine the flexural-torsional elastic buckling stress for the singly symmetric member using Equation E4-3.

174 Chapter 5

Compression Members 4 Fey Fez H ⎞ ⎞ ⎛⎜ ⎟ 1 − 1 − ⎟⎜ 2 ⎠⎝ ( Fey + Fez ) ⎟⎠ ⎛ 120 + 165 ⎞ ⎛ 4 (120 )(165 )( 0.915 ) ⎞ ⎟ = 105 ksi = ⎜⎜ ⎟⎟ ⎜ 1 − 1 − 2 ⎟ (120 + 165) ⎝ 2 ( 0.915 ) ⎠ ⎜⎝ ⎠

⎛ Fey + Fez Fe = ⎜ ⎝ 2H

Determine the critical stress using the flexural-torsional elastic buckling stress Fy Fe = 105 ksi > = 22.2 ksi 2.25

Step 8:

Therefore, using Equation E3-2 50

Fcr = ( 0.658 )105 ( 50 ) = 41.0 ksi Determine the nominal strength of the column for the limit state of flexuraltorsional buckling. Pn = 41.0 (10.0 ) = 410 kips

Step 9:

For LRFD Step 10:

For ASD Step 10:

5.9

Determine the column design strength. Since the nominal strength for flexural buckling about the x-axis is less than the flexural-torsional buckling strength, φPn = 0.9 ( 363) = 327 kips

Determine the column allowable strength. Since the nominal strength for flexural buckling about the x-axis is less than the flexural-torsional buckling strength, Pn 363 = = 217 kips Ω 1.67

SINGLE-ANGLE COMPRESSION MEMBERS Single-angle compression members would be designed for flexural-torsional buckling according to the provisions in Specification Section E4 except for an exclusion for angles with b t ≤ 0.71 E Fy . All hot rolled, A36 angles satisfy this exclusion limit so they need not be checked for flexural-torsional buckling. Studies show that the compressive strength of single angles can be reasonably predicted using the column equations of Specification Section E3 if a modified effective length is used and the member satisfies the following limiting criteria as found in Specification Section E5. 1. 2. 3. 4.

Members are loaded at their ends in compression through the same one leg. Members are attached by either welding or a connection containing a minimum of two bolts. There are no intermediate transverse loads. Lc r as determined in this section does not exceed 200.

Chapter 5 5.

Compression Members 175

For unequal leg angles the ratio of the long leg width to short leg width is less than 1.7.

Two cases are given for these provisions: (1) angles that are individual members or web members of planar trusses, and (2) angles that are web members in box or space trusses. This distinction is intended to reflect the difference in restraint provided by the elements to which the compression members are attached. The first set of equations is for angles that 1. 2. 3.

Are individual members or web members of planar trusses. Are equal-leg angles or unequal-leg angles connected through the longer leg. Have adjacent web members attached to the same side of a gusset plate or truss chord.

Buckling is assumed to occur about the geometric axis parallel to the attached leg. Since this may be either the x- or y-axis, the Specification uses the subscript a and then defines ra as the radius of gyration about the axis parallel to the attached leg. L If ≤ 80 , ra Lc L (AISC E5-1) = 72 + 0.75 r ra L and if > 80 ra Lc L (AISC E5-2) = 32 + 1.25 r ra These effective lengths must be modified if the unequal-leg angles are attached through the shorter legs. The provisions of Specification Section E5 should be reviewed for these angles as well as for similar angles in box or space trusses. Goal: EXAMPLE 5.17 Strength of SingleAngle Compression Given: Member

SOLUTION

Step 1:

Determine the available strength of a 10.0 ft single-angle compression member using A36 steel and the provisions of Specification Section E5. A 4×4×1/2 angle is a web member in a planar truss. It is attached by two bolts at each end through the same leg. From Manual Table 1-7, A = 3.75 in.2 and rx = 1.21.

Step 2:

Determine the slenderness ratio for the axis parallel to the connected leg, ra = rx. L L 10.0(12) = = = 99.2 ra rx 1.21

Step 3:

Determine which equation will give the effective slenderness ratio. Because L = 99.2 > 80 ra

use Equation E5-2.

176 Chapter 5

Compression Members Step 4:

Determine the effective slenderness ratio from Equation E5.2. Lc = 32 + 1.25 ( 99.2 ) = 156 < 200 r

Step 5:

Determine which column strength equation to use. Because

Lc 29,000 = 156 > 4.71 = 134 r 36 use Equation E3-3. Step 6:

Determine the Euler buckling stress. π2 E π 2 (29,000) Fe = = = 11.8 ksi 2 (156) 2 ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠

Step 7:

Determine the critical stress from Equation E3-3. Fcr = 0.877 Fe = 0.877 (11.8 ) = 10.3 ksi

Step 8:

Determine the nominal strength. Pn = Fcr A = 10.3 ( 3.75 ) = 38.6 kips

For LRFD Step 9: For ASD Step 9:

Determine the design strength. φPn = 0.9 ( 38.6 ) =34.7 kips

Determine the allowable strength. Pn Ω = 38.6 1.67 =23.1 kips

For single angle compression members that do not meet the criteria set forth in Section E5 for use of the modified slenderness ratio equations, the provisions of Sections E3 or E7 must be followed. The provisions in Section E4 for torsional or flexural-torsional buckling do not need to be followed for hot-rolled angles since all currently available hot-rolled A36 angles meet the leg slenderness exclusion of b t ≤ 0.71 E Fy . Thus, the strength for flexural buckling about the principal axes must be assessed. Goal: EXAMPLE 5.18 Strength of SingleAngle Compression Given: Member

SOLUTION

Step 1:

Determine the available strength of a 10.0 ft single-angle compression member using A36 steel. A 4×4×1/2 angle is a web member in a planar truss. It is attached by single bolts at each end through the same leg. From Manual Table 1-7, A = 3.75 in.2 and rx = 1.21, rz = 0.776 in.2.

Chapter 5

Compression Members 177

Step 2:

Determine the slenderness ratio for the minor (weak) principal axis. Lc 10.0(12) = = 155 rz 0.776

Step 3:

Determine which column strength equation to use. Because

Lc 29,000 = 155 > 4.71 = 134 36 rz Use Equation E3-3. Step 4:

Determine the Euler buckling stress. π2 E π 2 (29,000) Fe = = = 11.9 ksi 2 (155) 2 ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠

Step 5:

Determine the critical stress from Equation E3-3. Fcr = 0.877 Fe = 0.877 (11.9 ) = 10.4 ksi

Step 6:

Determine the nominal strength. Pn = Fcr A = 10.4 ( 3.75 ) = 39.0 kips

For LRFD Step 7: For ASD Step 7:

Determine the design strength. φPn = 0.9 ( 39.0 ) =35.1 kips

Determine the allowable strength. Pn Ω = 39.0 1.67 =23.4 kips

5.10 BUILT-UP MEMBERS Members composed of more than one shape are called built-up members. Several of these were illustrated in Figure 5.2h through n. Built-up members composed of two shapes are covered in Specification Section E6. Compressive strength is addressed by establishing the slenderness ratio and referring to Specification Section E3, E4, or E7 as appropriate. If a built-up section buckles so that the fasteners between the shapes are not stressed in shear but simply “go along for the ride,” the only requirement is that the slenderness ratio of the shape between fasteners be no greater than 0.75 times the controlling slenderness ratio of the built-up shape. If overall buckling would put the fasteners into shear, then the controlling slenderness ratio will be somewhat greater than the slenderness ratio of the built-up shape. This modified slenderness ratio is used to account for the effect of shearing deformations through the connectors. Thus, the effective slenderness ratio for a built-up member with snug-tight connectors will be greater than the same member with pre-tensioned or welded connectors. In addition, the spacing of the intermediate connectors will influence the modified slenderness ratio.

178 Chapter 5

Compression Members For intermediate connectors that are bolted snug-tight, the modified slenderness ratio is always greater than the slenderness ratio of the built-up member acting as a unit and is specified as 2

⎛ Lc ⎞ ⎛ Lc ⎞ ⎛ a ⎞ ⎜ ⎟ = ⎜ ⎟ +⎜ ⎟ ⎝ r ⎠m ⎝ r ⎠o ⎝ ri ⎠

2

(AISC E6-1)

If the intermediate connectors are welded or pre-tensioned bolted, the shearing deformation in the connectors is significantly less than for snug-tight connectors and the modified slenderness ratio may be equal to the slenderness ratio of the built-up member acting as a unit. For this case, the modified slenderness ratio is specified as, a when ≤ 40 , ri

⎛ Lc ⎞ ⎛ Lc ⎞ ⎜ ⎟ =⎜ ⎟ r ⎝ ⎠ m ⎝ r ⎠o and when

(AISC E6-2a)

a > 40 , ri 2

⎛ Ki a ⎞ ⎛ Lc ⎞ ⎛ Lc ⎞ ⎜ ⎟ = ⎜ ⎟ +⎜ ⎟ ⎝ r ⎠m ⎝ r ⎠o ⎝ ri ⎠

2

(AISC E6-2b)

where ⎛ Lc ⎞ ⎜ ⎟ = column slenderness of built-up member acting as a unit ⎝ r ⎠o K i = 0.5 for angles back-to-back = 0.75 for channels back-to-back = 0.86 for all other shapes a = distance between connectors ri = minimum radius of gyration of individual component

The remaining provisions in Specification Section E6 address dimensions and detailing requirements. These provisions are based on judgment and experience and are provided to ensure that the built-up member behaves in a way consistent with the strength provisions already discussed. The ends of built-up compression members must be either welded or pre-tensioned bolted in order to ensure that the member can work together as a unit. Even the smallest amount of slip in the end connections could mean that the built-up member is unable to carry any more load than the components individually. Along the length of built-up members, the longitudinal spacing of connectors must be sufficient to provide for transfer of the required shear force in the buckled member. The Commentary of the Specification gives guidance on how to determine the magnitude of that requirement. A built-up compression member with connectors spaced so that the slenderness ratio of the shape between fasteners is no greater than 0.75 times the controlling slenderness ratio of the built-up shape will not automatically satisfy this strength requirement. The Manual provides tables of properties for double angles, double channels, and Ishapes with cap channels in Part 1 and tables of compressive strength for double-angle compression members in Part 4.

Chapter 5 Goal: EXAMPLE 5.19 Strength of a Builtup Double-Angle Given: Compression Member

SOLUTION

Step 1:

Compression Members 179

Determine the available strength of a 10.0 ft double-angle compression member using A36 steel. Two 5×3×1/4 angles, long legs back-to-back with a 3/8 in. gap are used as a chord member in a planar truss. The angles are welded at each end to a gusset plate and along the length at two intermediate points with a spacing of 40 in. From Manual Table 1-15 for double angles A = 3.88 in.2 , rx = 1.62 in., ry = 1.19 in., ro = 2.52 in. and H = 0.638 in. From Manual Table 1-7 for single angles rz = 0.652 in. and J = 0.0438 in.4

Step 2:

Determine the slenderness ratio for each axis if the member works as a unit. L 10.0(12) = = 74.1 rx 1.62 L 10.0(12) = = 101 ry 1.19

Step 3:

Determine the effective slenderness ratio for buckling about the y-axis, the axis that will put the connectors in shear. Since the intermediate connectors are spaced at 40 in. a a 40 = = = 61.3 > 40 ri rz 0.652 Therefore use Equation E6.2b 2

2 ⎛ Lc ⎞ ⎛ Lc ⎞ ⎛ Ki a ⎞ = 1012 + ( 0.5 ( 61.3) ) = 106 ⎜ ⎟ = ⎜ ⎟ +⎜ ⎟ ⎝ r ⎠m ⎝ r ⎠o ⎝ ri ⎠

Step 4:

Check the maximum permitted slenderness ratio between connectors a = 61.3 < 0.75 (106 ) = 79.5 ri

Step 5:

Determine the elastic buckling stress for flexural buckling using the modified slenderness ratio π 2 ( 29,000 ) π2 E Fey = = = 25.5 ksi 2 2 (106 ) ⎛ Lc ⎞ ⎜ ⎟ ⎝ r ⎠m

Step 6:

Determine the elastic buckling stress for torsional buckling using Equation E4-7 with Cw = 0 based on the user note. 11, 200 ( 2 ( 0.0438 ) ) GJ Fez = = = 39.8 ksi 2 2 Ag ro 3.88 ( 2.52 )

Step 7:

Determine the elastic buckling stress for flexural-torsional buckling using Equation E4-3.

180 Chapter 5

Compression Members

4 Fey Fez H ⎞ ⎞ ⎛⎜ ⎟ ⎟ ⎜1 − 1 − 2 ⎠⎝ ( Fey + Fez ) ⎟⎠ ⎛ 25.5 + 39.8 ⎞ ⎡ 4 ( 25.5)( 39.8 )( 0.638 ) ⎤ ⎢ ⎥ = 19.1 ksi = ⎜⎜ − − 1 1 ⎟⎟ 2 ⎥⎦ ( 25.5 + 39.8) ⎝ 2 ( 0.638 ) ⎠ ⎢⎣

⎛ Fey + Fez Fe = ⎜ ⎝ 2H

Step 8:

Determine the critical stress. Since the elastic buckling stress for flexural-torsional buckling is less than that for flexural buckling, use that to determine the critical stress. Fy 36 = = 1.88 < 2.25 Fe 19.1 Therefore use Equation E3-2 Fy ⎛ Fcr = ⎜ 0.658 Fe ⎜ ⎝

Step 9:

⎞ ⎟ Fy = ( 0.6581.88 ) ( 36 ) = 16.4 ksi ⎟ ⎠

Determine if the local buckling must be included. For the short leg b t = 3.0 0.25 = 12.0 and for the long leg b t = 5.0 0.25 = 20.0 From Table B4.1a case 3 E 29,000 λ r = 0.45 = 0.45 = 18.9 Fcr 16.4 Since b t = 20.0 > λ r = 18.9 the effective area for the long leg must be determined.

Step 10:

Determine the effective area using Equation 5-27 and the values of c4 = 0.671 and c5 = 0.148 from Table 5.4. E ⎡ 0.148 E ⎤ be = 0.671t ⎢1 − ⎥ Fcr ⎣ bt Fcr ⎦ 29,000 ⎡ 0.148 29,000 ⎤ = 0.671( 0.25 ) ⎢1 − ⎥ = 4.86 in. 16.4 ⎣ 20 16.4 ⎦

So the effective area of the two angles is Ae = 3.88 − 2 ( 5.0 − 4.86 )( 0.25) = 3.81 in.2 Step 11: For LRFD Step 12:

Determine the nominal strength. Pn = Fcr Ae = 16.4 ( 3.81) = 62.5 kips

Determine the design strength.

Chapter 5

Comppression Mem mbers

181

φPn = 0.9 ( 62.5) = =56.3 kips For F ASD A Step S 12:

5.11

Deetermine the allowable a streength. Pn Ω = 62.5 1.677 =37.4 kips

CO OLUMN BASE PLATES When co olumns are su upported on material otheer than steel, such as conncrete or maasonry, it is necessary y to distributte their load over an are a significantlly larger thaan the gross area of the column. In these situaations, a colum mn base platee similar to thaat shown in F Figure 5.31 is used. Column C base plates may be attached to the column iin the shop, ass shown in Fiigure 5.31a, or shippeed separately to the site and a attached iin the field. C Columns are normally weelded to the plate butt may be attaached with an ngles when llarge plates m must be shippped separately. In either case, the selection of the t dimension ns and thickneess of the plaate follows thee same rules. Column C base plates are no ormally attachhed to a footiing or pier w with anchor roods, and the space beetween the plate and the support is ffilled with a non-shrink ggrout. A leveling plate, leveling nuts, or shim ms (as shown n in Figure 5..31b) are useed to level thhe column baase plate. In cases wh here the colum mn supports an n axial comprression only, anchor rods aare not designned to resist a specific force. How wever, all collumn base pllates must bee anchored w with a minim mum of four anchor ro ods according g to the Occup pational Safeety and Healthh Administrattion (OSHA) regulations in OSHA A 29 CFR 192 26 Subpart R Safety Standdards for Steeel Erection. Figure 5.32 iillustrates a column with w base platte in plan (Fig gure 5.32a) annd elevation (Figure 5.32bb), including four anchor rods. To T determine the area of bearing b that iss required, thee strength of the material uupon which Section J8 off the Specificcation gives the base plate is bearring must be evaluated. Foor concrete, S provision ns identical to t those given in the conncrete code, A ACI 318. Whhen the beariing plate is covering the full area of the concreete support, thhe nominal beearing strengtth is

Figure 5.31 5

Exam mple of a W-Shape Columnn and Base Pllate

Photos co ourtesy Douglas Steel Fabricaating Corporatiion

1882 Chapter 5

Compresssion Memberss

Figure 5.32

Column and Base Platte Section andd Plan Includding Anchor R Rods Pn = Pp = 0.85 f c′A1

(AIS SC J8-1)

where f c' is the specified d concrete co ompressive sstrength and A1 is the arrea of the pllate and concrete. If the plate doess not cover alll of the concrrete, there will be an increase in strengtth due to the spread of the load as it progresses down throughh the concretee. In this casee the nominall bearing strength is given g as (AIS SC J8-2) Pn = Pp = 0.85 f c′A1 A2 A1 ≤ 1.77 f c′A1 Here A2 is th he maximum m area of conccrete with thee same shape as the bearinng plate. The limit on the right sid de of the equaation imposes a maximum ratio of areass of 4:1. If thhe supporting element is designed based on thee bearing streength of the ssoil, it will bee relatively eeasy to determ mine the extent to wh hich the base plate covers the t concrete ffoundation orr pier. In all ccases, φ = 0.665 and Ω = 2.31. The thickness off a column baase plate is a function of the bending strength of thhe plate. Since bendin ng has not yet been covereed in this bookk, this topic is deferred to Section 11.111. Those wishing to address a base plate p design further f shouldd proceed to Section 11.11 and to the eexample problems giv ven there as well w as AISC Design Guiddes 1, 7 and 100.

55.12

PROB BLEMS

11. Determine the theoreticaal buckling sttrength (Eulerr bbuckling load)) for a W8×58 A992 collumn with an n eeffective lengtth of 20 ft. Will W the theorretical column n bbuckle or yield at this length??

4. D Determine the theoretical bbuckling strength (Euler buckkling load) forr an HSS 100×5×3/8 A5000 Grade C colum mn with an effective lenggth of 20 ft. Will the theorretical column buckle or yieldd at this lengthh?

22. Determine the theoreticaal buckling sttrength (Eulerr bbuckling load)) for a W16×77 A36 collumn with an n eeffective lengtth of 12 ft. Will W the theorretical column n bbuckle or yield at this length??

5. Foor a W12×72 A992 column, determine thhe effective lengtth at which thhe theoretical bbuckling strenngth (Euler buckkling load) will equal the yieldd strength.

33. Determine the theoreticaal buckling sttrength (Eulerr bbuckling load)) for a W24×370 A992 co olumn with an n eeffective lengtth of 15 ft. Will W the theorretical column n bbuckle or yield at this length??

6. Foor a W6×16 A A992 column, determine the effective lengtth at which thhe theoretical bbuckling strenngth (Euler buckkling load) will equal the yieldd strength. 7. Foor an HP8×36 A572 Grade 550 column, dettermine the effecctive length at which the theooretical bucklinng strength (Euleer buckling loaad) will equal tthe yield strenggth.

Chapteer 5 88. A W14×120 0 column has an a effective len ngth for y-axiss bbuckling equal to 24 ft. Deterrmine the effecctive length forr thhe x-axis that will provide th he same theoreetical buckling g sstrength (Euler buckling load)). 99. A W14×53 column has an effective len ngth for x-axiss bbuckling equal to 20 ft. Deterrmine the effecctive length forr thhe y-axis that will provide th he same theoreetical buckling g sstrength (Euler buckling load)). 110. An HSS12×6×1/2 column n has an effecctive length forr xx-axis buckling g equal to 18 ft. Determinee the effectivee length for the y-axis y that will provide the saame theoreticall bbuckling streng gth (Euler buck kling load). 111. A W14×13 32 A992 colum mn has an effecctive length off 336 ft about both axes. Determine the availablee ccompressive sttrength for thee column. Deteermine the (a)) ddesign strength h by LRFD an nd (b) allowab ble strength by y A ASD. Is this an n elastic or inelastic buckling condition? 112. Determine the available compressive strength for a W W12×152 A99 92 column witth an effectivee length aboutt bboth axes of 40 ft. Determin ne the (a) desig gn strength by y L LRFD and (b)) allowable sttrength by AS SD. Is this an n eelastic or inelasstic buckling co ondition? 113. A W6×15 A992 A column has h an effectivee length of 8 ft aabout both ax xes. Determinee the availablee compressivee sstrength for thee column. Deteermine the (a) design d strength h bby LRFD and (b) allowable strength by ASD. A Is this an n eelastic or inelasstic buckling co ondition?

Comprression Membbers

183

Deterrmine the (a)) design strenngth by LRFD D and (b) allow wable strength bby ASD. 18. D Determine the available com mpressive strenngth for an HSS 6×4×3/8 A5000 Grade C collumn where thhe effective lengtth is 10 ft abouut the y-axis annd 15 ft about the x-axis. Deterrmine the (a)) design strenngth by LRFD D and (b) allow wable strength by ASD. Is thhis an elastic oor inelastic buckkling condition?? 19. A round HSS 16.000×0.3755 A500 Grade C column has aan effective leength of 18 ft. Determine thee available comppressive strenggth and indicatte whether this is due to elastiic or inelasticc buckling. D Determine the (a) design strenngth by LRFD aand (b) allowab able strength byy ASD. 20. A W8×40 is ussed as a 12 ft ccolumn in a braaced frame with W16×26 beam ms at the top and bottom ass shown in Figurre P5.20. The ccolumns abovee and below aree also 12 ft W8× ×40s. The beam ms provide m moment restraiint at each colum mn end. Deteermine the efffective length using the alignnment chart annd the available compressivve strength, and tthe (a) designn strength by L LRFD and (b)) allowable strenngth by ASD. Assume that tthe columns arre oriented for ((i) buckling abbout the weakk axis and (iii) buckling abouut the strong axxis. All steel is A992. 21. If the structuure described in Problem 20 is an unbraaced frame, determine the effective length and comppressive strenggth as requestedd in Problem 20.

114. Determine the available compressive strength s for an n M M4×6 A36 collumn with an effective leng gth about both h aaxes of 7 ft. Determine D the (a) ( design stren ngth by LRFD D aand (b) allowaable strength by ASD. Is this an elastic orr innelastic buckliing condition? 115. A W14×21 11 A992 colum mn has an effecctive length off 440 ft about both axes. Determine the availablee ccompressive sttrength for thee column. Deteermine the (a)) ddesign strength h by LRFD an nd (b) allowab ble strength by y A ASD. Is this an n elastic or inelastic buckling condition? 116. Determine the available compressive strength for a W W12×65 A992 column when n the effective length is 20 ft aabout the y-axis and 40 ft abo out the x-axis. Determine thee ((a) design stren ngth by LRFD D and (b) allow wable strength h bby ASD. Is thiss an elastic or inelastic bucklling condition?? D Describe a com mmon conditio on where the effective length h is different about the differentt axes. 117. A W8×24 A992 column n has an effecctive length off 112.5 ft about the y-axis an nd 28 ft abo out the x-axis. D Determine the available a comp pressive strength and indicatee w whether this is due to elaastic or inelaastic buckling.

P5.200 22. A W12×136 ccolumn is show wn in Figure P P5.22 with end cconditions thaat approximatee ideal conditioons. Using the rrecommended approximate vvalues from Coommentary

1184 Chapterr 5

Compreession Membeers

T Table C-A-7.1,, determine thee effective len ngths for the y-aaxis and the x-axis. Which effective lengtth will controll thhe column streength?

and a compressivee live load of 1100 kips by (a) LRFD and ((b) ASD. An A992 W122×53 is used as a column in a building 26. A with an effective leength for each axis of 15 ft. Determine whetther the columnn will carry a ccompressive deead load of 85 kkips and a com mpressive live load of 255 kkips by (a) LRFD D and (b) ASD D. 27. A An A992 W8× ×67 is used in a structure too support a dead load of 60 kkips and a livee load of 100 kips. The colum mn has an eeffective lengtth of 24 ft. Determine whetther the colum mn will supporrt the load by (a) LRFD and ((b) ASD.

P P5.22 223. A W12×87 column is sho own in Figure P5.23 P with end d cconditions thatt approximate ideal conditio ons. Using thee rrecommended approximate values from Commentary y T Table C-A-7.1,, determine thee effective len ngths for the y-aaxis and the x-axis. Which effective lengtth will controll thhe column streength?

28. A An A992 W100×54 is used as a column in a building with an effective length of 30 ftt. Determine w whether the colum mn will carry a compressive dead load of 224 kips and a com mpressive livee load of 72 kiips by (a) LRF FD and (b) ASD D. 29. A An A992 W16× ×77 is used as a column in a building to suppoort a dead loaad of 130 kipss and a live looad of 200 kips. The column effective lengtth is 20 ft forr the y-axis and 330 ft for the xx-axis. Determ mine whether tthe column will ssupport the loaad by (a) LRFD D and (b) ASD. 30. A An A992 W21×111 is used aas a column in a building to suupport a dead lload of 120 kipps and a live load of 300 kips. The column effective lengtth is 22 ft forr the y-axis and 333 ft for the xx-axis. Determ mine whether tthe column will ssupport the loaad by (a) LRFD D and (b) ASD. 31. A An A992 W24×131 is used aas a column in a building to suupport a dead lload of 245 kipps and a live load of 500 kips. The column hhas an effectivve length aboutt the y-axis of 188 ft and an efffective length aabout the x-axxis of 36 ft. Deterrmine whetherr the column w will support thhe load by (a) L LRFD and (b) A ASD.

P P5.23 224. A W10×60 0 column with an effective length l of 30 ft ffor both axes is called upon to carry a com mpressive dead d looad of 123 kip ps and a comprressive live loaad of 170 kips. D Determine wheether the colum mn will suppo ort the load by y ((a) LRFD and (b) ASD. Evalluate the streng gth for (i) Fy = 550 ksi and (ii) Fy = 70 ksi. 225. A W14×25 57 A992 colum mn in a building g has effectivee lengths of 16 ft for both ax xes. Determin ne whether thee ccolumn will caarry a compressive dead loaad of 800 kipss

An A300 Gr. C HSS7×7×1/22 is used as a column to 32. A suppoort a dead loaad of 175 kipss and a live looad of 100 kips. The columnn has an effeective length of 10 ft. will support thhe load by Deterrmine whetherr the column w (a) L LRFD and (b) A ASD. 33. For the W100×88 column with bracingg and end condditions shownn in Figure P5.33, deterrmine the theorretical effectivve length for eaach axis and identify the axis tthat will limit tthe column strength. ×28 column with bracingg and end 34. For the W8× condditions shownn in Figure P5.34, deterrmine the theorretical effectivve length for eaach axis and identify the axis tthat will limit tthe column strength.

Chapteer 5

Comprression Membbers

185

36. A W14×176 coolumn with ennd conditions aand bracing is shoown in Figuree P5.36. Determ mine the least theoretical braciing and its locaation about thee y-axis, in ordder that the y-axiis not control thhe strength of tthe column. 37. A W12×72 coolumn is an exxterior 2nd stoory column (griddline A) with sstrong axis bucckling in the plane of the framee in an unbrraced multi-stoory frame. Thhe column below w is also a W12×72. Beam ms and dimensiions are as show wn in Figure P P5.37. Determ mine the effective length for tthis conditionn and the corrresponding coompressive strenngth by (a) LRF FD and (b) ASD D. All steel is A992. P P5.33

P P5.34

P5.366

335. A W10×100 column with h end conditions and bracing g is shown in Fig gure P5.35. Deetermine the leeast theoreticall bbracing and its location abou ut the y-axis, in n order that thee yy-axis not contrrol the strength h of the column n.

P5.377 38. A W12×72 coluumn is an exteerior 2nd story column (at gridliine C) with sttrong axis buckkling in the pllane of the framee in an unbrraced multi-stoory frame. Thhe column below w is also a W12×72. Beam ms and dimensiions are as show wn in Figure P P5.37. Determ mine the effective length for tthis conditionn and the corrresponding coompressive strenngth by (a) LRF FD and (b) ASD D. All steel is A992. P P5.35

186 Chapter 5

Compression Members

39. A W12×72 column is an interior 2nd story column (at gridline B) with strong axis buckling in the plane of the frame in an unbraced multi-story frame. The column below is also a W12×72. Beams and dimensions are as shown in Figure P5.37. Determine the effective length for this condition and the corresponding compressive strength by (a) LRFD and (b) ASD. All steel is A992. 40. A W12×50 column is an interior column with strong axis buckling in the plane of the frame in an unbraced multi-story frame. The columns above and below are also W12×50. The beams framing in at the top are W16×31 and those at the bottom are W16×40. The columns are 14 ft and the beam span is 25 ft. The column carries a dead load of 75 kips and a live load of 150 kips. Determine the inelastic effective length for this condition and the corresponding compressive strength by (a) LRFD and (b) ASD. All steel is A992. 41. Select the least-weight W12 A992 column to carry a live load of 130 kips and a dead load of 100 kips with an effective length about both axes of 14 ft by (a) LRFD and (b) ASD. 42. Select the least-weight W14 A992 column to carry a dead load of 200 kips and a live load of 600 kips if the effective length about both axes is 20 ft by (a) LRFD and (b) ASD. 43. Select the least-weight W10 A992 column to carry a dead load of 80 kips and a live load of 280 kips with an effective length about both axes of 15 ft by (a) LRFD and (b) ASD. 44. Select the least-weight W8 A992 column to carry a dead load of 20 kips and a live load of 50 kips with an effective length about both axes of 30 ft by (a) LRFD and (b) ASD. 45. Select the least-weight W6 A992 column to carry a dead load of 12 kips and a live load of 36 kips with an effective length about both axes of 8 ft by (a) LRFD and (b) ASD.

48. If the column in Problem 44 had an effective length of 32 ft, select the lightest-weight W12 to support this load by (a) LRFD and (b) ASD. 49. A W14 A992 column must support a dead load of 80 kips and a live load of 300 kips. The column is 22 ft long and has end conditions that approximate the ideal conditions of a fixed support at one end and a pin support at the other. Select the lightest-weight W14 to support this load by (a) LRFD and (b) ASD. 50. Select the least-weight W8 A992 column to support a dead load of 170 kips with an effective length of 16 ft by (a) LRFD and (b) ASD. 51. A column with an effective length of 21 ft must support a dead load of 120 kips, a live load of 175 kips, and a wind load of 84 kips. Select the lightest W14 A992 member to support the load by (a) LRFD and (b) ASD. 52. A W14×99 A992 column is 20 ft long, pinned at each end, and braced at mid-height to prevent lateral movement for buckling about the y-axis. However, the yaxis bracing is not adequate to resist torsion. Considering flexural and torsional buckling, determine the nominal strength of this compression member. 53. An A36 single-angle compression web member of a truss is 10 ft long and attached to gusset plates through the same leg at each end with a minimum of two bolts. The member must carry a dead load of 8 kips and a live load of 10 kips. Select the least-weight equal leg angle to carry this load by (a) LRFD and (b) ASD. 54. If the compression web member of Problem 53 were loaded concentrically, determine the least-weight single angle to carry the load by (a) LRFD and (b) ASD. 55. A W16×26 A992 compression member has a slender web when used in uniform compression. Determine the available strength by (a) LRFD and (b) ASD when the effective length is (i) 6 ft and (ii) 12 ft.

46. Select the least-weight W8 A992 column to carry a dead load of 13 kips and a live load of 39 kips with an effective length about both axes of 14 ft by (a) LRFD and (b) ASD.

56. The W14×43 is the only A992 column shown in the Manual column tables that has a slender web. Determine the available strength for this column if the effective length is 5 ft, and show whether the slender web impacts that strength by (a) LRFD and (b) ASD.

47. A column with pin ends for both axes must be selected to carry a compressive dead load of 95 kips and a compressive live load of 285 kips. The column is 16 ft long and is in a braced frame. Select the lightest-weight W12 to support this load by (a) LRFD and (b) ASD.

57. Determine the available strength of a WT6×22.5 A992 steel compression chord of a truss with effective length Lc = 14 ft. Consider the member braced laterally and torsionally at its ends only. Determine by (a) LRFD and (b) ASD. 58. Determine the available strength of a WT12×125 A992 steel column with effective length Lc = 18 ft.

Chapteer 5 C Consider the member m braced d laterally and d torsionally att itts ends only. Determine D by (aa) LRFD and (b b) ASD. 559. Determine the t available sttrength of a C9 9×20 A36 steell ccompression ch hord of a trusss with effectiv ve length, Lc = 116 ft. Consid der the mem mber braced laterally and d toorsionally at itts ends only. Determine D by (a) LRFD and d ((b) ASD. 660. Determine the availablee strength of an a A36, 20 ft loong, 2L6×4×5 5/8 double-angle compression n member in a pplanar truss. The angles are long l legs back--to-back with a 33/8 in. gap. Thee angles are welded w at each end e to a gussett pplate and along g the length at two intermediaate points with h a spacing of 80 0 in. Determin ne the (a) desiign strength by y L LRFD and (b) allowable a stren ngth by ASD. 661. Determine the availablee strength of an a A36, 15 ft loong, 2L6×3-1//2×1/2 double-angle compreession memberr inn a planar trusss. The angles are long legs back-to-back k w with a 3/8 in n. gap. The angles are co onnected with h ppretensioned bo olts at each end to a gusset plate p and along g thhe length at tw wo intermediatee points with snug-tight boltss aat a spacing off 60 in. Deterrmine the (a) design d strength h bby LRFD and (b) ( allowable sttrength by ASD D. M Multi-Chapterr Problems 662. Using thee framing plaan shown in Figure P5.62 2 ((presented earllier as Figuree 2.7), design n the columnss m marked 4 and 5. 5 This is the saame structure used u in Section n 22.5, where loaad calculationss with live lo oad reductionss w were discussed d. Those calcu ulations can be reused here. L Load case 2 fo or dead plus liive load is to be considered. T The building is an office build ding with a nom minal live load d oof 50 pounds per p square foott (psf) and a calculated c dead d looad of 70 psf. 4: Inteerior column D-2 D regardless of o deck span directiion 5: Exterior column D-4 D regardless of deck span directiion D Design for colu umn length L = 14 ft and K = 1.0 using (a)) L LRFD and (b) ASD. A

P P5.62

Comprression Membbers

187

63. T The framing pplan shown in F Figure P5.63 iis the same as thhe one shown in Figure P2.220 for an 18-sstory office buildding. It must suupport a dead load of 80 psff and a live load of 50 psf. IIn all cases, tthe decking sspans in a direcction from linne A toward line E. Deteermine the requiired axial streength for the columns and design the colum mns as requireed below for ((a) design by LRFD and (b) ddesign by ASD.. The required axial load strengths were deterrmined in Prooblem 20 of Chapter 2. Use a story heighht of 13.5 ft in a braced framee so that K = 1.0. iv: The coolumn at the coorner on lines 1 and E that supports eeight levels. v: The collumn on the eddge at the inteersection of lines 4 andd A that supporrts eight levelss. vi: The interior colum mn at the interrsection of line 4 andd the point bettween lines C and D that supports thhree levels.

P5.633 64. IIntegrated Dessign Project. U Using the gravvity column loadss determined iin Chapter 2, design the grravity-only colum mns. Design columns as singgle-story mem mbers. (It is oftenn more econoomical to usse multi-storyy columns becauuse of construcction costs.) Seelect the final ccolumns so that tthey are two-sstory from bellow grade to tthe second floorr, two-story froom the second floor to the foourth floor, and tthen use a singgle-story colum mn to support thhe roof. Design the columns in the braced fraame for the graviity loads deterrmined in Chappter 2 and the wind load deterrmined in Chaapter 4. Remem mber that the wind load mustt be considered to act in tw wo directions, so use the largeest compressionn forces from w wind to combinne with the graviity loads. Using thee wind load aanalysis from Chapter 4, desiggn all the bracces as compresssion memberss. Compare the teension design w with the comppression designn and select the ap appropriate finaal members.

C Chapterr 6

B Bendin ng Memberss

Massachusetts M C College of Art aand Design, Deesign and Meddia Center Photograph P by Fadi Asmar/LERA Consultinng Structural E Engineers

6.11

BENDIN NG MEMBERS IN STRUCTURE ES A bending member carrries load applied normal to its longituudinal axis aand transfers it to its support poin nts through beending momeents and sheaars. In buildinng constructioon, the most ccommon application of bending members m is to o provide suppport for flooors or roofs. These beamss can be either simplle span or co ontinuous, and d normally trransfer their load to otherr structural m members such as collumns, girders, and wallls. Although the terms bbeam and giirder are oftten used interchangeaably because both are ben nding memberrs, beam norm mally refers tto a bending member that directly y supports an n applied load d whereas girrder usually refers to a bbending mem mber that supports a beam. b The disstinction is no ot important fo for design beccause the sam me criteria appply to all bending mem mbers. The most comm monly used sh hapes for bennding memberrs are the I-sshaped cross sections and, of thesse, the W-shaape is domin nant. Howeveer, there are numerous situations wheere other shapes are used u as bendiing members. L-shapes arre commonly used as linteels over opennings, Tshapes are found f as chorrds of trusses that may be called upon to resist bendding along w with axial forces, and C-shapes C may y coexist with h W-shapes inn floor system ms. In addition a to thee use of the sttandard shapees, engineers often find it necessary to develop their own sh hapes by com mbining shapees and/or plattes. Several eexamples of tthese built-upp shapes are shown in Figure 6.1. 6 Although h the use off these builtt-up shapes is permitted by the Specification n, they may not n be econom mical becausee of the laborr costs associaated with fabrication. The compleexity that ressults from thee wide varieety of possiblle shapes is the reason sso many separate pro ovisions are in ncluded in Ch hapter F of thee Specificatioon. The most comm mon and econo omical bendiing members are those thhat can reach the full material yield strength without w being limited by buuckling of anny cross-sectional elements. These members aree referred to as a compact members m and aare addressedd first. Table 6.1 lists the sections of the Specificaation and parts of the Manuual that are discussed in this chaptter.

1888

Chapter 6

Bending Mem mbers 189

Figure 6.1 Built-up Beaams

6.2

STRE ENGTH OF F BEAMS i applied to a bending meember, resultinng in a bending moment, sstresses are developed in As load is the crosss section. For a load at or below b the noominal level, tthe load magnnitude establiished in the building code, it is reeasonable to expect e the enntire beam crooss section too behave elastically. The stresses and a strains aree distributed as shown in F Figure 6.2a. T This elastic beehavior occurrs whenever the materrial’s behavio or correspond ds to the initiaal straight-linne portion of tthe stress-straain curve of Figure 3..5. From F the basic principles of strength oof materials, the relationshhip between the applied moment and resulting g stresses is giiven by the faamiliar flexuree formula: My (6.1) fy = I where M = any y applied mom ment that stressses the sectioon in the elasttic range y = disttance from th he neutral axiss to the point where the strress is to be determined I = mo oment of inerttia fy = resu ulting bendin ng stress at loccation y Table 6.1

Sections of o Specification and Parts of Manual Coovered in Thiis Chapter Specifi fication

B3 B4 F1 F2 F3 F6 F9 F10 Chapter G H1 J10 Chapter L Appendix x1

Design n Basis Classification of Seections for Loocal Bucklingg Generaal Provisions Doublly Symmetricc Compact II-Shaped Meembers and C Channels Beent about Th heir Major Ax xis Doublly Symmetricc I-Shaped M Members withh Compact W Webs and Nonncompact or Slender Flan nges Bent aboout Their Major Axis I-Shap ped Members and Channells Bent about Their Minor Axis Tees and a Double Angles A Loadedd in the Planee of Symmetryy Singlee Angles Design n of Memberss for Shear Doublly and Singly Symmetric M Members Subjject to Flexurre and Axial F Force Flangees and Webs with w Concenttrated Forces Design n for Serviceaability Design n by Advanceed Analysis Mannual

Part 1 Part 3

Dimen nsions and Pro operties Design n of Flexural Members

1990 Chapter 6

Bending Members M

Figure 6.2 Cross-Sectiional Bendin ng Stresses aand Strains: (a) elastic; ((b) yield; (c)) partial plastic; (d) plastic p Norrmally the strress at the extreme fiber, tthat is, the fiiber most distant from thee neutral axis, is of in nterest because the greatesst stress occuurs at this poiint. The distaance from thee neutral axis to the extreme fiber may m be taken n as c and the flexure formuula becomes M Mc M (6.2) = fb = I S where S = section modu ulus fb = extreme fiberr bending streess The moment thatt causes the extreme e fiber to reach the yield stress, Fy, is called tthe yield moment, My. The corresp ponding stresss and strain diiagrams are shhown in Figuure 6.2b. If the load is increased beeyond the yieeld moment, the t strain in thhe extreme fiiber increasess; however, thhe stress remains at Fy because th hese fibers arre behaving aas depicted bby the plateauu on the stress-strain diagram, sho own previoussly in Figuress 3.5 and 3.6. The stress att some pointss on the crosss section closer to thee neutral axis also reach th he yield stresss while those even closer, from the neuutral axis to the locatio on dimension ned as y in Fig gure 6.2c, rem main elastic. As the t moment continues to increase, thee portion of tthe cross secttion experienncing the yield stress continues to increase untill the entire seection experieences the yielld stress, as sshown in Figure 6.2d.. Note that th here is no corrresponding sstrain diagram m given for thhis loading condition since the strrain would bee unlimited over o the entirre cross sectioon. The mom ment that cauuses this stress distrib bution is calleed the plastic moment, Mp. Since the strrain is unlimiited, the crosss section undergoes unlimited u rotaation and the condition is referred to aas a plastic hiinge. A crosss section that is capab ble of attainin ng this stresss distribution and correspoonding momeent is referredd to as a compact secction. At eveery stage of lo oading, equiliibrium of thee cross sectionn requires, at all

Chapter 6

Figure 6.3 6

Bending Mem mbers 191

Equilibriium in a Doub bly Symmetriical Wide-Flaange Shape

times, th hat the total in nternal tensio on force be eequal to the ttotal internal compressionn force. The basic principles of strrength of matterials are adddressed in nuumerous textts, such as M Mechanics of Materialls.1 For F the doubly symmetricc wide flangee shape show wn in Figure 6.3, equilibriium for the plastic moment m occurrs when the portion p of the shape abovee the elastic nneutral axis iss stressed to the yield d stress in com mpression wh hile the portioon below the elastic neutrral axis is streessed to the yield streess in tension n. For a nonssymmetric shaape, the areaa above the elastic neutrall axis is not equal to the area below w the elastic neutral axis. Thus, a new axis that gives equal areaas in tension and comp pression musst be defined. This new axxis dividing thhe section intto two equal areas is the plastic neutral axis (P PNA). For sym mmetric shappes the elasticc and plastic nneutral axes ccoincide, as was the case for the wide flange. For nonsym mmetric shapees these neuttral axes are at different locationss. Because B equiilibrium meaans that the tension and compressionn forces are equal and opposite,, they form a force couple.. Although m moments can bbe taken abouut any reference point for this case,, it is common practice to take momentts about the P PNA. The mom ment that corrresponds to this fully y yielded stresss distribution n, Mp, is deterrmined from (6.3) M p = Fy ( Ac yc ) + Fy ( At yt ) where At and Ac are the t equal tenssion and com mpression areaas, respectiveely, and yc annd yt are the distancess from the centroid c of th he area to thhe PNA for the tension and compresssion areas, respectiv vely. Equation n 6.3 may be simplified to ⎛ A⎞ M p = Fy ⎜ ⎟ ( yc + yt ) (6.4) ⎝2⎠ The T two term ms multiplied by the yieldd stress are fuunctions onlyy of the geom metry of the cross secction and aree normally co ombined and called the pplastic sectionn modulus, Z Z. Thus, the plastic moment m is giveen as (6.5) M p = Fy Z The plasttic section mo odulus is tabu ulated for all aavailable shappes in Part 1 oof the Manuaal. Chapter C F off the Specifica ation containns the provisiions for design of flexuraal members subject to o bending. Fo or a given beaam to attain itts full plastic moment strenngth, it must be compact and satissfy the laterall support criteeria establisheed in Sectionn F2. If these criteria are nnot met, the strength is defined as something leess than Mp. T The criteria too be satisfied are defined bby two limit 1

Philpot, Tim mothy. Mechan nics of Materia als: An Integra ated Learning SSystem. Wiley, 2013.

1992 Chapter 6

Bending Members M states in add dition to yield ding: local bu uckling (a nonncompact shaape) and lateral torsional bbuckling. These limit states and theeir impact on beam strength th are discusseed in Sections 6.4 and 6.5..

EX XAMPLE 6.1 6 Pllastic Momeent Str trength for a Syymmetric Sh hape

Goa al:

Deterrmine the plaastic moment strength of a W-shape uusing the moddel of three rectangular plates p and com mpare the callculated plastiic section moodulus to thee value determ mined from thhe Manual.

Giv ven:

A W2 24×192 is mo odeled as show wn in Figure 66.4. Assume Fy = 50 ksi.

SO OLUTION

Step p 1:

Deterrmine the locaation of the pllastic neutral axis. Becau use the shape is symmetricc, the plastic nneutral axis iss located on thhe axis of o symmetry.

Step p 2:

Deterrmine the plaastic section m modulus as tthe sum of thhe moments oof the areas about the plaastic neutral aaxis. A A ⎛ ⎞ Z = ( yc + yt ) = 2 ⎜ Af y f + w yw ⎟ 2 2 ⎠ ⎝

⎡ ⎛ 22.58 11.46 ⎞ 22.588 ( 0.810 ) ⎛ 22.58 ⎞ ⎤ 3 + Z = 2 ⎢13.0 (1.4 46 ) ⎜ ⎟+ ⎜ ⎟ ⎥ = 560 inn. 2 2 2 4 ⎠ ⎝ ⎝ ⎠ ⎦ ⎣ Step p 3:

Deterrmine the plasstic moment sstrength as thhe plastic secttion modulus times the yiield stress using Equation 66.5 M p = Fy Z = 550 ( 560 ) = 288, 000 in.-kipss or M p = 28,,000 12 = 23330 ft-kips

Step p 4:

Comp pare the calcculated plasttic section m modulus valuue with that from Manu ual Table 1-1.. From m the table, Zx = 559 in.3. This shows s that th he impact of tthe simplificaation in usingg rectangular plates and ig gnoring the fiillets at the flaange-web junnction is smalll.

Figure F 6.4 W24×192 M Model for Exam mple 6.1

Chapter 6

Figure 6.5

EXAMPLE E 6.2 Plastic Secction Modulus fo or a Nonsymmeetric Shape SOLUTION N

Goal: G Given: G

Step S 1:

Bending Mem mbers 193

T-Beam Moodel for Exam mple 6.2

Lo ocate the plastic neutral axxis and determ mine the plastiic section moodulus for a WT. W A WT12×51.5 modeled as ttwo plates is shown in Figgure 6.5. Asssume that Fy = 50 ksi. Deetermine the area a of the T--shape. Aflange = 9.00 ( 0.980 ) = 8.82 in.2

Astem = 0.550 (12.3 − 00.980 ) = 6.23 iin.2 Atotal = 8.82 + 6.23 = 15.1 in.2 Step S 2:

Deetermine one--half of the arrea, because one-half of thhe area must be above thee plastic neuttral axis and oone-half mustt be below. Atotal 15.1 = = 77.55 in.2 2 2

Step S 3:

Deetermine wheether the plaastic neutral aaxis is in thee flange or tthe stem. Beecause half of o the area iss less than tthe area of thhe flange, thhe plastic neeutral axis is in i the flange aand 7.55 yp = = 0 .839 in. 9.0 wiith the plasticc neutral axis measured froom the top of the flange.

Step S 4:

Deetermine the plastic sectioon modulus aas the sum oof the momennts of the areeas about the plastic neutraal axis. ⎛ 0.839 ⎞ ⎛ 0.980 − 0.839 ⎞ Z = 7.55 ⎜ ⎟ ⎟ + ( 8.82 − 77.55 ) ⎜ 2 ⎝ ⎠ ⎝ 2 ⎠ 12.3 − 0.980 ⎞ ⎛ + 6.23 ⎜ 0.9980 − 0.839 + ⎟ 2 ⎠ ⎝

= 3.17 + 0.08895 + 36.1 = 339.4 in.3 Step S 5:

Co ompare these values with tthe values in M Manual Tablee 1-8. y p = 0.841 in. and Z = 39.2 in..3 Th his shows thaat the impact of the simplification in ussing rectangullar plates an nd ignoring th he fillets at thee flange-web junction is sm mall.

1994 Chapter 6

Bending Members M

Ano other useful term is the shape factorr defined ass the ratio off the plastic section modulus to o the elastic section s modu ulus, shape factoor =

Z S

Values for the shape factor f are no ot given in thhe Manual bbut are readdily determinned and their signifi ficance will be b discussed d later. 6.33

DESIGN N OF COMPACT LAT TERALLY SUPPORTE S ED WIDE-F FLANGE B BEAMS Section F1(b b) of the Speecification requires that aall beams be restrained at the supportss against twist about their longitud dinal axis. Fo or a compactt beam to attaain its plasticc moment streength, it must also bee laterally sup pported at som me specified length along its compresssion flange. T This may be accompliished through h the attachmeent of intermeediate beams or by attachinng decking att regular intervals as shown in Fiigure 6.6. When W this is tthe case the beam is saidd to have fulll lateral support. Secction 6.2 sho owed that thee nominal strrength of a ccompact mem mber with fulll lateral support—that is, in whicch no bucklin ng limit statees control—iss determined by the limit state of yielding. For this limit staate, Specification Section F F2 provides tthat SC F2-1) (AIS M n = M p = Fy Z Speccification Secction F1 also indicates i thatt for all flexurral limit statess, design strenngth and allowab ble strength arre to be determ mined using φ = 0.900 ( LRFD ) Ω = 1.67 ( AS SD ) The design basis from Section ns B3.1 and B B3.2, discusseed in Chapterr 1, are repeatted here. The requirem ment for ASD D is R (AIS SC B3-2) Ra ≤ n Ω ment for LRF FD is The requirem Ru ≤ φRn (AIS SC B3-1) ure 6.7 showss a portion off Manual Tabble 3-2, whichh lists W-shappes in order oof major Figu axis plastic section mod dulus. Shapes are groupedd so the leastt weight mem mber with thee largest plastic sectio on modulus is i given in bo old at the top of each grouup. This faciliitates selectioon of the most econom mical W-shap pe for beams controlled byy the limit staate of yieldingg. This figuree will be used in the following f exaample to aid in n the design oof a compact,, laterally suppported beam.

Figure 6.66 Lateral Suupport of Beaam Using Steeel Deck

Chapter 6

Figure 6.7 6

W-Shapees: Selection by Zx

Bending Mem mbers 195

Copyrightt © American Institute I of Steeel Constructioon. Reprinted w with Permissionn. All rights resserved.

196 Chapter 6

Bending Members

EXAMPLE 6.3a Beam Design by LRFD

Goal:

Select the least-weight wide flange member for the conditions given.

Given:

An A992 beam, simply supported at both ends, spans 20 ft and is loaded at midspan with a dead load of 8.0 kips and a live load of 24.0 kips, as shown in Figure 6.8, in addition to its self-weight. Assume full lateral support and a compact section.

SOLUTION

Step 1:

Determine the required strength using the governing LRFD load combination from Section 2.4. Pu = 1.2 PD + 1.6 PL = 1.2 ( 8.0 ) + 1.6 ( 24.0 ) = 48.0 kips

Mu = Step 2:

Pu L 48 ( 20 ) = = 240 ft-kips 4 4

Determine the required plastic section modulus. For a compact, fully braced section, M n = M p = Fy Z Thus, because Specification Section B3.1 provides that the required moment be less than or equal to the available moment, M u ≤ φM n = φFy Z , and 240 (12 ) M Z req = u = = 64.0 in.3 φFy 0.90 ( 50 )

Step 3:

Using the required plastic section modulus, select the minimum-weight Wshape from the plastic section modulus economy table shown in Figure 6.7. Start at the bottom of the Zx column and move up until a shape in bold with at least Zx = 64.0 in.3 is found. Select W18×35 with Z=66.5 in.3 This is the most economical W-shape, based on the section weight, which provides the required plastic section modulus.

Step 4:

An alternative approach, also using Figure 6.7, would be to enter the table with the required moment, Mu = 240 ft-kips, and proceed up the φMn column. The same section will be selected with this approach.

Step 5:

Determine the additional required strength based on the actual weight of the chosen beam. The beam weighs 35 lb/ft, which gives an additional moment of ⎛ 0.035 ( 20 )2 ⎞ ⎟ = 1.2 (1.75 ) = 2.10 ft-kips M u ( self -weight ) = 1.2 ⎜ ⎜ ⎟ 8 ⎝ ⎠

Step 6:

Since the maximum moment for both the applied load and the self-weight occur at the same point, combine this moment with the moment due to superimposed load to determine the new required strength. M u = 240 + 2.10 = 242 ft-kips

Chapter 6

Bending Members 197

Step 7:

Determine the new required plastic section modulus. ( 242 )(12 ) = 64.5 in.3 M Z req = u = φFy 0.90 ( 50 )

Step 8:

Make the final selection. This required plastic section modulus is less than that provided by the W18×35 already chosen. Therefore, select the W18×35 Note that this is a compact section.

EXAMPLE 6.3b Beam Design By ASD

Goal:

Select the least-weight wide flange member for the conditions given.

Given:

An A992 beam, simply supported at both ends, spans 20 ft and is loaded at mid-span with a dead load of 8.0 kips and a live load of 24.0 kips, as shown in Figure 6.8. Assume full lateral support and a compact section.

SOLUTION

Step 1:

Determine the required strength using the ASD load combinations from Section 2.4. Pa = PD + PL = ( 8.0 ) + ( 24.0 ) = 32.0 kips

Ma = Step 2:

Determine the required plastic section modulus. For a compact, fully braced section, M n = M p = Fy Z Thus, because Specification Section B3.2 provides that the required moment be less than or equal to the available moment, M a ≤ M n /Ω = Fy Z /Ω, and Z req =

Step 3:

Pa L 32.0 ( 20 ) = = 160 ft-kips 4 4

160 (12 ) 160 (12 ) Ma = = = 64.0 in.3 Fy /Ω ( 50 / 1.67 ) 30

Using the required plastic section modulus, select the minimum-weight W-shape from the plastic section modulus economy table shown in Figure 6.7. Start at the bottom of the Zx column and move up until a shape in bold with at least Zx = 64.0 in.3 is found. Select W18×35 with Z = 66.5 in.3 This is the most economical W-shape, based on section weight, which provides the required plastic section modulus.

Step 4:

An alternative approach, again using Figure 6.7, would be to enter the table with the required moment, Ma = 160 ft-kips, and proceed up the Mn/Ω column. The same section will be selected with this approach.

Step 5:

Determine the additional required strength based on the actual weight of the chosen beam. The beam weighs 35 lb/ft, which gives an additional

1998 Chapter 6

Bending Members M momeent of M a ( self -weight ) =

0.035 ( 20 ) 8

2

= 1.75 ft-kips

Step p 6:

Sincee the maximum m moment foor both the appplied load annd the self-weeight occurr at the same point, combbine this mom ment with thee moment duue to superrimposed load d to determinee the new requ quired strengthh. M a = 1600 + 1.75 = 1622 ft-kips

Step p 7:

Deterrmine the new w required plaastic section m modulus. 162 12 ( )( ) Ma Z req = = 64.8 in.3 = Fy /Ω 30

Step p 8:

Makee the final seleection. This reequired plastiic section moodulus is less tthan that provided p by th he W18×35 allready chosenn. Therefore, select the W18×35 Note that this is a compact c sectiion

Note that since the t live-to-deead load ratioo in Examplee 6.3 is 3, beefore inclusion of the self-weight, the required plastic sectio on modulus iin Examples 6.3a and 6.3bb are the sam me. Once the self-weig ght is added to t the problem m, the live-to--dead load raatio is no longger 3 and the required plastic sectio on modulus iss seen to be slightly differeent between tthe LRFD andd ASD exampples.

Figure 6.8 Beam Used in Example 6.3a 6

Chapter 6

Figure 6..9

EXAMPLE E 6.4a Beam Desiign by LRFD

SOLUTION N

Bending Mem mbers 199

Framing Plan for Exam mple 6.4.

Goal: G

Deesign a W-sh hape floor beaam for the inntermediate beam marked A on the flo oor plan show wn in Figure 66.9.

Given: G

Th he beam is loaded uniform mly from the ffloor with a live load of 80 pounds peer square foott (psf) and a ddead load in aaddition to thhe beam self-w weight of 60 0 psf. The beaam will have full lateral suupport providded by the flooor deck, an nd a compact section will bbe selected. U Use A992 steeel.

Step S 1:

Deetermine the required r loadd and momentt. wu = (1.2 wD + 1.6 wL ) Ltrib = (1.2 ( 600 ) + 1.6 ( 80 ) ) (10 ) = 2000 1bb/ft wu L2 2.0 kips/ft ( 26 ft ) = = 169 ft-kips 8 8 2

Mu =

Step S 2:

Deetermine the required r plasttic section moodulus. Fo or a compactt, fully braceed beam, Mn = Mp = FyZ.. Section B3.1 of the Sp pecification reequires that M u ≤ φM n = φFy Z Th herefore (169 )(112 ) = 45.1 in..3 M Z req = u = φFy 0.90 ( 550 )

Step S 3:

Ussing the plasttic section m modulus econoomy table in Figure 6.7, sselect the mo ost economiccal W-shape bbased on leastt weight. W1 4×30 with Z = 47.3 in.3

Step S 4:

Deetermine the additional a reqquired strengtth based on thhe actual weigght of the ch hosen beam. The T beam weiighs 30 lb/ft, which gives an additionall moment off

200 Chapter 6

Bending Members M u ( self -weight )

Step 5:

⎛ 0.030 ( 26 )2 = 1.2 ⎜ ⎜ 8 ⎝

⎞ ⎟ = 1.2 ( 2.54 ) = 3.04 ft-kips ⎟ ⎠

Since the maximum moment for both the applied load and the self-weight occur at the same point, combine this moment with the moment due to superimposed load to determine the new required strength.

M u = 169 + 3.04 = 172 ft-kips Step 6:

Determine the new required plastic section modulus. (172 )(12 ) = 45.9 in.3 M Z req = u = φFy 0.90 ( 50 )

Step 7:

Make the final selection. This required plastic section modulus is less than that provided by the W14×30 already chosen. Therefore, select the W14×30

Step 8:

As shown in Example 6.3, an alternative approach is to use the required moment, Mu = 172 ft-kips, and enter the φM n column to determine the same W-shape.

EXAMPLE 6.4b Beam Design by ASD

Goal:

Design a W-shape floor beam for the intermediate beam marked A on the floor plan shown in Figure 6.9.

Given:

The beam is loaded uniformly from the floor with a live load of 80 psf and a dead load in addition to the beam self-weight of 60 psf. The beam will have full lateral support provided by the floor deck, and a compact section will be selected. Use A992 steel.

SOLUTION

Step 1:

Determine the required load and moment. wa = ( wD + wL ) Ltrib = ( 60 + 80 )(10 ) = 1400 1b/ft

w L2 1.40 kips/ft ( 26 ft ) Ma = a = = 118 ft-kips 8 8 Determine the required plastic section modulus. 2

Step 2:

For a compact, fully braced beam, Mn = Mp = FyZ. Section B3.2 of the Specification requires that Fy Z M Ma ≤ n = Ω Ω Therefore, (118 )(12 ) = 47.2 in.3 Ma Z req = = 30 Fy /Ω

Chapter 6

Bending Members 201

Step 3:

Using the plastic section modulus economy table in Figure 6.7, select the most economical W-shape based on least weight. W14×30 with Z = 47.3 in.3

Step 4:

Determine the additional required strength based on the actual weight of the chosen beam. The beam weighs 30 lb/ft, which gives an additional moment of 2 0.030 ( 26 ) M a ( self -weight ) = = 2.54 ft-kips 8

Step 5:

Since the maximum moment for both the applied load and the self-weight occur at the same point, combine this moment with the moment due to superimposed load to determine the new required strength. M a = 118 + 2.54 = 121 ft-kips

Step 6:

Determine the new required plastic section modulus. (121)(12 ) = 48.4 in.3 Ma Z req = = Fy /Ω 30

Step 7:

Make the final selection. This required plastic section modulus is more than that provided by the W14×30. Therefore, select the next most economical shape from the table W16×31 By inspection, this additional 1 lb/ft is acceptable.

Step 8:

As shown in Example 6.3, an alternative approach is to use the required moment, Ma = 121 ft-kips, and enter the Mn /Ω column to determine the same W-shape.

An alternative approach for including beam self-weight is to start with a preliminary estimate of the beam weight. That way it will already be included when the first trial shape is selected. Then the assumed weight is compared to the weight of the selected shape. If the assumed weight is greater than the selected weight, the beam is adequate. If there is a significant discrepancy, either over or under, a second trial shape can be selected.

6.4 DESIGN OF COMPACT LATERALLY UNSUPPORTED WIDE-FLANGE BEAMS 6.4.1

Lateral-Torsional Buckling The compression region of a bending member cross section has a tendency to buckle similarly to how a pure compression member buckles. The major difference is that the bending tension region helps to resist that buckling. The upper half of the wide flange member in bending acts as a T in pure compression. This T is fully braced about its horizontal axis by the web so it will not buckle in that direction, but it can be unbraced for some distance for buckling about its vertical axis. Thus, it will have a tendency to buckle laterally. Because the tension region also tends to restrain the lateral buckling, the shape actually buckles in a combined lateral and torsional mode. The beam midspan deflects in the plane down and buckles laterally, causing it to twist. This is illustrated in Figure 6.10 by a beam in a laboratory test that has failed due to lateral-torsional

2002 Chapter 6

Bending Members M

Figure 6.10

Example of o Lateral-Torsional Bucklling of a Beam m in a Test Frrame

Photo courtessy of Donald W. W White

buckling. Th he beam appeears to have a tendency to fall over on itts weak axis. In order to reesist this tendency, Sp pecification Sections S B3.4 and F1(b) re quire that all bending mem mbers be restrrained at their supporrt points again nst rotation ab bout their longgitudinal axiss. If the beam m has sufficiennt lateral and/or torsio onal support along its len ngth, a comppact cross secction can devvelop the yielld stress before buckling. This is the t case that was discusseed in Sectionss 6.2 and 6.3.. If the beam tends to buckle before the yield stress is reacched, the nom minal momennt strength is less than thee plastic moment. To ensure e that a beam cross section s can deevelop its fulll plastic mom ment strength without lateral-torsio onal buckling g, Specificatio on Section F2 .2 limits the sslenderness too Lb E (6.6) ≤ 1.76 ry Fy where Lb = unbraced len ngth of the co ompression flaange ry = radius of gyrration for the shape about tthe y-axis The practicaal application n of this limiitation is to uuse the unbrraced length alone, rather than in combination n with the rad dius of gyrattion in the foorm of a slennderness ratioo. This resultts in the requirementt for attaining g the full plaastic momennt strength giiven the Speccification as Lb ≤ L p where E SC F2-5) (AIS L p = 1.76 ry Fy t maximum m unbraced len ngth that wouuld permit thee shape to reach its plastic moment Thus, Lp is the strength. Th his value is tabulated t for each W-shappe and can bbe found in M Manual Tablle 3-2, a portion of which w is shown in Figure 6.7, and severaal other locatiions in the M Manual. Since A992 is the preferred d specificatio on for W-shap pes, the tabullated values iin Table 3-2 aare based on Fy = 50 ksi. Wheen the unbraaced length of o a beam exxceeds Lp, itts strength is reduced duee to the tendency off the memberr to buckle laaterally at a l oad level bellow what woould cause thee plastic moment to be b reached. For F simply sup pported doubbly symmetricc members, a closed-form solution

Chapter 6

Bending Members 203

is well established in the literature. The Guide to Stability Design Criteria for Metal Structures2 shows the development of the Specification equations based on the closed-form solution. The elastic lateral-torsional buckling (LTB) strength of a W-shape is given in Specification Section F2.2 as (AISC F2-3) M n = Fcr S x ≤ M p where Fcr =

Cb π2 E ⎛ Lb ⎞ ⎜r ⎟ ⎝ ts ⎠

2

1 + 0.078

Jc ⎛ Lb ⎞ S x ho ⎜⎝ rts ⎟⎠

2

(AISC F2-4)

which can be combined to give the nominal moment strength for elastic lateral-torsional buckling as Mn =

Cb π2 ES x ⎛ Lb ⎞ ⎜r ⎟ ⎝ ts ⎠

2

1 + 0.078

Jc ⎛ Lb ⎞ S x ho ⎜⎝ rts ⎟⎠

2

(6.7)

A beam buckles elastically if the actual stress in the member at buckling does not exceed Fy at any point. Because all hot rolled shapes have built-in residual stresses as discussed for columns in Section 5.3.4, there is a practical limit to the usefulness of this elastic LTB equation. The Specification sets the level of the residual stress at 0.3Fy so that only 0.7Fy is available to resist a bending moment from applied load while still behaving elastically. This limit results in the maximum elastic moment, M rLTB = 0.7 Fy S x . If Equation F2-4 is set equal to 0.7Fy, a limiting unbraced length, Lr, beyond which the member buckles elastically may be determined. This limit as provided in Specification Section F2 is

Lr = 1.95rts

E 0.7 Fy

2

Jc ⎛ Jc ⎞ ⎛ 0.7 Fy ⎞ + ⎜ + 6.76 ⎜ ⎟ ⎟ S x ho ⎝ E ⎠ ⎝ S x ho ⎠

2

(AISC F2-6)

With an unbraced length between Lp and Lr, the beam behaves inelastically. In this range, the nominal moment, Mn, is reasonably well predicted by a straight-line equation when compared to the results of laboratory tests. The Specification equation for the nominal moment strength is ⎡ ⎛ Lb − Lp ⎞ ⎤ M n = Cb ⎢ M p − ( M p − 0.7 Fy S x ) ⎜ (AISC F2-2) ⎟⎥ ⎝ Lr − Lp ⎠ ⎦⎥ ⎣⎢ Equation F2-2 represents interpolation between the end points of a straight line. These end points correspond to the plastic moment, Mp given by Equation F2-1, at an unbraced length of Lp and an elastic moment, M rLTB = 0.7 Fy S x , at an unbraced length Lr. Although the determination of Fcr and Lr from Equations F2-4 and F2-6 may look somewhat daunting, the Manual has extensive tables that permit their determination with little effort, and the equations also can be readily automated using spreadsheet calculations. The complete picture of the nominal moment strength of a beam as a function of unbraced length is presented in Figure 6.11, where the curve segments are labeled according to the appropriate strength equations. Curves similar to these are available in Manual Table 3-10 for each W-shape and Table 3-11 for C- and MC-shapes. For all of these curves, the moment gradient factor, Cb, is taken as 1.0. Examples of these curves are given in Figure 6.12. 2

Ziemian, R. D. Guide to Stability Design Criteria for Metal Structures, 6th ed. New York: John Wiley & Sons, 2010.

2004 Chapter 6

Bending Members M

Figure 6.11 Lateral-To orsional Buckling

EX XAMPLE 6.5a 6 Beeam Design by LR RFD

Goa al:

Select a W-shape beam b consideering the unbrraced length oof the compreession flangee using Manu ual Table 3-100.

Giv ven:

The beam b spans 30.0 3 ft and haas a requiredd strength of 2282 ft-kips. IIt will have lateral supporrt at its ends aand at midspaan. Use A9922 steel.

SO OLUTION

Step p 1:

Deterrmine the req quired momeent strength aand unbracedd length. From m the given n information,, M u = 282 ft-kipps

Lb = 15.0 ft Step p 2:

Select the least-w weight W-shappe from the portion of M Manual Tablee 3-10 given n in Figure 6.1 12. Enterr Table 3-10 at a a design strrength of ϕMn = 282 ft-kipps and an unbbraced length h of Lb = 15.0 ft. The firsst solid line tto the upper right indicatees the least-weight shapee sufficient to carry this mooment at this unbraced lenggth.

Step p 3:

Thereefore, select a W18×55

EX XAMPLE 6.5b 6 Beeam Design by AS SD

Goa al:

Select a W-shape beam b consideering the unbrraced length oof the compreession flangee using Manu ual Table 3-100.

Giv ven:

The beam b spans 30.0 3 ft and haas a requiredd strength of 188 ft-kips. IIt will have lateral supporrt at its ends aand at midspaan. Use A9922 steel.

Chapter 6

Figure 6.12 6

Bending Mem mbers 205

W-Shap pes: Available Moment veersus Unbraceed Length

Copyrightt © American Institute I of Steeel Constructioon. Reprinted w with Permissionn. All rights resserved.

206 Chapter 6 SOLUTION

Bending Members Step 1:

Determine the required moment strength and unbraced length. From the given information, M a = 188 ft-kips

Lb = 15.0 ft Step 2:

Select the least-weight W-shape from the portion of Manual Table 3-10 given in Figure 6.12. Enter Table 3-10 at an allowable strength of Mn/Ω = 188 ft-kips, and an unbraced length of Lb = 15 ft. The first solid line to the upper right indicates the least-weight shape sufficient to carry this moment at this unbraced length.

Step 3:

Therefore, select a W18×55

When Mn is to be determined through a calculation, an additional simplification can be applied to the straight-line portion of the curve. From Equation F2-2, the ratio ⎛ M p − 0.7 Fy S x ⎞ ⎛ M p − M rLTB ⎞ ⎜ ⎟=⎜ ⎟ Lr − Lp ⎝ ⎠ ⎝ Lr − Lp ⎠ is a constant for each beam shape and is defined as the variable BF when written in terms of nominal strength. This constant is tabulated in Manual Table 3-2, and shown in Figure 6.7, as φBF and BF/Ω for LRFD and ASD, respectively. Thus, for nominal strength, with Cb again taken as 1.0, Equation F2-2 can be rewritten as (6.8a) M n = M p − BF ( Lb − L p ) and for LRFD as (6.8b) φM n = φM p − φBF ( Lb − L p ) and for ASD as M n M p BF = − ( Lb − Lp ) (6.8c) Ω Ω Ω Further inspection of Manual Table 3-2 shows that two additional moment values are given, φb M rx and M rx Ωb . These values are the LRFD and ASD values for M rLTB = 0.7 Fy S x . 6.4.2

Moment Gradient The nominal strength of a beam limited by lateral-torsional buckling as defined when Cb = 1.0 assumes that the moment is uniform across the entire length of the beam as shown in Figure 6.13a. For the limit state of lateral-torsional buckling this is the most severe loading case possible, because it stresses the entire length of the beam to its maximum level, just as for a column. For any other loading pattern, and resulting moment diagram, the compressive force in the beam would vary with the moment diagram. Thus, the reduced stresses along the member length would result in a reduced tendency for lateral-torsional buckling and an increase in strength. The variation in moment over a particular unbraced segment of the beam is called the moment gradient, which describes how the moment varies along a specific length.

Chapter 6

Figure 6.13 6

Bending Mem mbers 207

Resistaance to the Maaximum Mom ment under Thhree Differentt Loading Coonditions

For F the normaal case of loaading that prooduces a mom ment diagram m that is not constant, the nominal moment stren ngth calculateed through Eqquations F2-22, F2-3, and F F2-4 will be iincreased to account for the mom ment gradient when Cb is not taken ass 1.0. The laateral-torsional buckling modificaation factor, Cb, accounts for f nonuniform m moment diiagrams overr the unsuppoorted length. There haave been many y proposed eq quations for thhe determinat ation of Cb in the literature.. The Guide to Stabillity Design Criteria C for Metal M Structurres (see footnnote 2) provides a good ssummary of these pro oposals. The equation pressented in Speccification Secction F1 is 122.5M max Cb = (F1-1) 2.5M max + 3M A + 4M B + 3M C where Mmax = absolu ute value of maximum m mom ment in the unnbraced segm ment MA = absolutee value of moment at quartter point of thhe unbraced seegment MB = absolutee value of moment at centeerline of the uunbraced segm ment MC = absolutee value of mo oment at threee-quarters poinnt of the unbrraced segmennt Cb = 1.0 for a uniform m moment an nd can be connservatively ttaken as 1.0 ffor other casees. In doing so, howeever, the desig gner may be sacrificing s siggnificant econnomy. Figuree 6.14 providees examples of loadin ng conditions,, bracing locaations, and thee correspondiing Cb valuess found in Maanual Table 3-1.

2008 Chapter 6

Bending Members M

Figure 6.14 Values forr Cb for Simplly Supported Beams

Copyright © American Insttitute of Steel Construction. C R Reprinted with Permission. A All rights reservved.

The effect of thee moment graadient factor, Cb, is to alteer the nominnal moment–uunbraced length relatiionship by a constant, c as sh hown in Figuure 6.15. The shaded area sshows the inccrease in moment cap pacity as a ressult of the usee of Cb. Regaardless of how w small the unnbraced lengtth might be, the nom minal moment strength of the t member ccan never excceed the plasstic moment sstrength. Thus, the up pper portion of o the curve in n Figure 6.15 is terminatedd at Mp.

Figuree 6.15 Effecct of Moment Gradiennt

Chapter 6

Bending Mem mbers 209

Figure 6.116 Beam U Used in Exampple 6.6 Goal: G EXAMPLE E 6.6a Beam Stren ngth and Design byy LRFD Considerin ng Moment Gradient G

SOLUTION N

Deetermine wheether the W144×34 beam shhown in Figuure 6.16 will carry the giv ven load. Co onsider the m moment gradiient (a) Cb = 1.0 and (b) Cb from Eq quation F1-1, and then (c)) determine thhe least-weighht section to carry the loaad using the correct c Cb .

Given: G

Figure 6.16 shows a beam that is fixedd at one suppport and pinned at the oth her. The beam m has a conceentrated deadd load of 8 kiips and a concentrated liv ve load of 24 4 kips at midsspan. A lateraal brace is loocated at eachh support an nd at the load point.

Step S 1:

Deetermine the required r strenngth. For the lload combinaation of 1.2D + 1.6L, Pu = 1.2((8.0) + 1.6(244.0) = 48.0 kipps

Step S 2:

Deetermine thee maximum moment frrom an elasstic analysiss of the ind determinate beam. b At the fixed end thee moment is shown in Figgure 6.16 as M u = 180 ftt-kips

Step S 3:

Deetermine the needed valuues from Ma Manual Table 3-2 in ordeer to use Eq quation 6.8b. For a W14×334, Z = 54.6 in.3, Lp = 5.400 ft, Lr = 15.66 ft, φMp = 2055 ft-kips, andd φBF = 7.55 kkips

210 Chapter 6

Bending Members Part a Step 4:

Cb = 1.0

Determine the design moment strength for lateral bracing of the compression flange at the supports and the load, Lb = 10 ft. Because

Lb = 10 ft > L p < Lr φM n = Cb ( φM p − φBF ( Lb − L p ) ) ≤ φM p φM n = 1.0 ( 205 − 7.55(10.0 − 5.40) ) = 170 ft-kips < 205 ft-kips

As an alternative approach, Manual Table 3-10 can be entered with an unbraced length of 10 ft and the design strength of the W14×34 determined to be 170 ft-kips. Therefore, since φM n = 170 ft-kips < M u = 180 ft-kips

The W14×34 beam will not work if Cb = 1.0 Part b Step 5:

Use the Calculated Value of Cb Determine the correct Cb for the two unbraced segments of the beam. For the unbraced segment BC, the first beam in Figure 6.14 can be used to obtain Cb = 1.67 since the moment gradient for the segment from the load to the support in Figure 6.14 is the same as the moment gradient for segment BC of the beam being considered here. This Cb corresponds to the maximum moment of 150 ft-kips at point B on the beam. The W14×34 can resist this moment without consideration of Cb, as shown in part (a) Step 4 above. For the unbraced segment AB, Figure 6.14 cannot be used and Cb must be calculated. Using Equation F1-1 and the moment values given in Figure 6.16 at the quarter points of the segment AB, 12.5 (180 ) Cb = = 2.24 2.5 (180 ) + 3 ( 97.5 ) + 4 (15.0 ) + 3 ( 67.5 )

Step 6:

Determine the design moment strength using the calculated value of Cb and the design moment strength determined from part (a), with Equation 6.8b amplified by Cb and limited to φMp. φM n = Cb ( φM p − φBF ( Lb − L p ) ) ≤ φM p

= 2.24 ( 205 − 7.55(10.0 − 5.40) ) = 2.24(170) = 381 ft-kips > φM p = 205 ft-kips Therefore, the limiting strength of the beam is

Chapter 6

Bending Members 211

φM n = 205 ft-kips > 180 ft-kips So the W14×43 is adequate for bending. Part c Step 7:

Considering that Cb = 2.24 and 1.67 for the two segments, a smaller section can be tried. Assuming the beam can be treated as fully braced as a result of including the influence of the moment gradient, Cb not equal to 1.0, select from Manual Table 3-2 a W16×31 which is the lightest W-shape to carry the given moment, Mu = 180 ft-kips. Determine the values needed to evaluate the shape from Manual Table 3-2. φM p = 203 ft-kips, L p = 4.13 ft, Lr = 11.8 ft, φBF = 10.3 kips

Step 8:

First consider unbraced segment BC. As determined in Step 5 above, Cb = 1.67. Because Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8 ft, use Equation 6.8b with Cb. φM n = Cb ( φM p − φBF ( Lb − Lp ) ) ≤ φM p

= 1.67 ( 203 − 10.3(10.0 − 4.13) ) = 1.67(143) = 239 ft-kips > φM p = 203 ft-kips Note that for this unbraced segment, without the use of Cb, φM n = 143 ft-kips < 150 ft-kips Thus, with Cb included φM n = 239 ft-kips > φM p = 203 ft-kips Therefore the design strength is φM n = 203 ft-kips > 150 ft-kips Step 9:

For unbraced segment AB, again Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8 ft, and again use Equation 6.8b with Cb.= 2.24 as determined in Step 5.

φM n = Cb ( φM p − φBF ( Lb − Lp ) )) ≤ φM p = 2.24 ( 203 − 10.3(10.0 − 4.13) ) = 2.24(143) = 320 ft-kips > φM p = 203 ft-kips where φM n = 320 ft-kips > φM p = 203 ft-kips

Step 10:

Thus, the design strength is φM n = 203 ft-kips > 180 ft-kips So the W16×31 will also work.

212 Chapter 6

Bending Members

EXAMPLE 6.6b Beam Strength and Design by ASD Considering Moment Gradient

Goal:

Determine whether the W14×34 beam shown in Figure 6.16 will carry the given load. Consider the moment gradient (a) Cb = 1.0 and (b) Cb from Equation F1-1, and then (c) determine the least-weight section to carry the load using the correct Cb.

Given:

Figure 6.16 shows a beam that is fixed at one support and pinned at the other. The beam has a concentrated dead load of 8 kips and a concentrated live load of 24 kips at midspan. A lateral brace is located at each support and at the load point.

SOLUTION

Step 1:

Determine the required strength. For the load combination of D + L, Pa = 8.0 + 24.0 = 32.0 kips

Step 2:

Determine the maximum moment from an elastic analysis of the indeterminate beam. At the fixed end the moment is shown in Figure 6.16 as M a = 120 ft-kips

Step 3:

Determine the needed values from Manual Table 3-2 in order to use Equation 6.8b. For a W14×34, Z = 54.6 in.3, Lp = 5.40 ft, Lr = 15.6 ft, Mp/Ω =136 ft-kips, and BF/Ω = 5.01kips

Part a Step 4:

Cb = 1.0

Determine the allowable moment strength for lateral bracing of the compression flange at the supports and the load, Lb = 10 ft. Because

Lb = 10 ft > L p < Lr M n Ω = ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω = (136 − 5.01(10.0 − 5.40) ) = 113 ft-kips < 136 ft-kips

As an alternative approach, Manual Table 3-10 can be entered with an unbraced length of 10 ft and the design strength of the W14×34 determined to be 113 ft-kips. Therefore, since M n Ω = 113 ft-kips < M a = 120 ft-kips

The W14×34 beam will not work if Cb = 1.0 Part b Step 5:

Use the Calculated Value of Cb Determine the correct Cb for the two unbraced segments of the beam. For the unbraced segment BC, the first beam in Figure 6.14 can be used to obtain Cb = 1.67 since the moment gradient for the segment from the load to the support in Figure 6.14 is the same as the moment gradient for

Chapter 6

Bending Members 213

segment BC of the beam being considered here. This Cb corresponds to the maximum moment of 100 ft-kips at point B on the beam. The W14×34 can resist this moment without consideration of Cb, as shown in part (a) Step 4 above. For the unbraced segment AB, Figure 6.14 cannot be used and Cb must be calculated. Using Equation F1-1 and the moment values given in Figure 6.16 at the quarter points of the segment AB, Cb =

Step 6:

12.5 (120 )

2.5 (120 ) + 3 ( 65.0 ) + 4 (10.0 ) + 3 ( 45.0 )

= 2.24

Determine the allowable moment strength using the calculated value of Cb and the allowable moment strength determined from part (a), with Equation 6.8b amplified by Cb and limited to Mp/Ω. M n Ω = Cb ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω = 2.24 (136 − 5.01(10.0 − 5.40) ) = 2.24(113) = 253 ft-kips > M p Ω = 136 ft-kips

Therefore, the limiting strength of the beam is M n Ω = 136 ft-kips > 120 ft-kips So the W14×43 is adequate for bending. Part c Step 7:

Considering that Cb = 2.24 and 1.67 for the two segments, a smaller section can be tried. Assuming the beam can be treated as fully braced as a result of including the influence of the moment gradient, Cb not equal to 1.0, select from Manual Table 3-2 a W16×31 which is the lightest W-shape to carry the given moment, Ma = 120 ft-kips. Determine the values needed to evaluate the shape from Manual Table 3-2. M p Ω = 135 ft-kips, L p = 4.13 ft, Lr = 11.8 ft, BF Ω = 6.86 kips

Step 8:

First consider unbraced segment BC. As determined in Step 5 above, Cb = 1.67. Because Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8 ft, use Equation 6.8b with Cb. M n Ω = Cb ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω = 1.67 (135 − 6.86 (10.0 − 4.13) ) = 1.67(94.7) = 158 ft-kips > M p Ω = 135 ft-kips Note that for this unbraced segment, without the use of Cb, M n Ω = 94.7 ft-kips < 100 ft-kips Thus, with Cb included M n Ω = 158 ft-kips > M p Ω = 135 ft-kips

214 Chapter 6

Bending Members Therefore the allowable strength is M n Ω = 135 ft-kips > 100 ft-kips Step 9:

For unbraced segment AB, again Lb = 10 ft > Lp = 4.13 ft, and < Lr = 11.8 ft, and again use Equation 6.8b with Cb.= 2.24 as determined in Step 5. M n Ω = Cb ( M p Ω − ( BF Ω ) ( Lb − L p ) ) ≤ M p Ω = 2.24 (135 − 6.86 (10.0 − 4.13 ) ) = 2.24(94.7) = 212 ft-kips > M p Ω = 135 ft-kips

where M n Ω = 212 ft-kips > M p Ω = 135 ft-kips

Step 10:

Thus, the allowable strength is M n Ω = 135 ft-kips > 120 ft-kips So the W16×31 will also work.

Manual Table 3-10 was shown in Example 6.5 to be an efficient aid in design considering unbraced length. When design is to consider moment gradient as well as unbraced length, the charts in Table 3-10 can be equally useful. Since the curves in Table 3-10 are based on Cb = 1.0, some modifications to the required moment strength must be made before entering the table. Starting with Equation F2-2 and dividing both sides by Cb gives M n Cb equal to the nominal strength based on Cb = 1.0. Thus, if Table 3-10 is entered with M n Cb a member can be selected. Of course the limit on Equation F2-2 must still be addressed as will be shown in Example 6.7.

EXAMPLE 6.7a Beam Design by LRFD

Goal:

Select a W-shape beam, and consider the unbraced length of the compression flange and the moment gradient, using Manual Table 3-10.

Given

The beam spans 30.0 ft and has a required strength of 282 ft-kips as in Example 6.5a. It is loaded with a concentrated load at the midspan and will have lateral support at its ends and at midspan. Use A992 steel.

SOLUTION

Step 1:

Determine the required moment strength and unbraced length. From the given information, M u = 282 ft-kips

Lb = 15.0 ft Step 2:

Determine the appropriate Cb from Figure 6.14. For a beam with a concentrated load at midspan and lateral supports at the load and at the ends, Cb = 1.67.

Step 3:

Determine the moment to use for entering Manual Table 3-10. To account for the increased strength due to the moment gradient, the required moment is divided by Cb, and that value is used to enter the table along with the given unbraced length.

Chapter 6

Bending Members 215

Thus, enter the table at the design moment strength φM n 282 = =169 ft-kips 1.67 Cb and unbraced length Lb = 15.0 ft. Step 4:

The first solid line above and to the right indicates the least-weight shape sufficient to carry this moment at this unbraced length. Since this is not found in Figure 6.12, use the actual Manual Table 3-10 and select a W12×40. This member’s strength curve appears to be right on the intersection of the required moment strength and given unbraced length.

Step 5:

Determine the maximum strength of the W12×40. Remember that in no case can ϕMn be greater than ϕMp . From Manual Table 3-2, φM p = 214 ft-kips

Since this is less than the required strength, Mu = 282 ft-kips , this W-shape will not be adequate even with the use of Cb. Step 6:

Select a shape with φM p ≥ 282 ft-kips from Manual Table 3-2. Thus , select W18×40, φM p = 294 ft-kips

Step 7:

Confirm in Manual Table 3-10 that the line, either solid or dashed, for the W18×40 is above and to the right of the intersection of φM n = 169 ft-kips and Lb = 15 ft. Thus, select the W18×40

EXAMPLE 6.7b Beam Design by ASD

Goal:

Select a W-shape beam, and consider the unbraced length of the compression flange and the moment gradient, using Manual Table 3-10.

Given:

The beam spans 30.0 ft and has a required strength of 188 ft-kips as in Example 6.5b. It is loaded with a concentrated load at the midspan and will have lateral support at its ends and at midspan. Use A992 steel.

SOLUTION

Step 1:

Determine the required moment strength and unbraced length. From the given information, M a = 188 ft-kips

Lb = 15.0 ft Step 2:

Determine the appropriate Cb from Figure 6.14. For a beam with a concentrated load at midspan and lateral supports at the load and at the ends, Cb = 1.67.

216 Chapter 6

Bending Members Step 3:

Determine the moment to use for entering Manual Table 3-10. To account for the increased strength due to the moment gradient, the required moment is divided by Cb, and that value is used to enter the table along with the given unbraced length. Thus, enter the table at allowable moment strength ( M n Ω ) Cb = 188 1.67 =113 ft-kips and unbraced length Lb = 15.0 ft.

Step 4:

The first solid line above and to the right indicates the least-weight shape sufficient to carry this moment at this unbraced length. Since this is not found in Figure 6.12, use the actual Manual Table 3-10 and select a W12×40. This member’s strength curve appears to be right on the intersection of the required moment strength and given unbraced length.

Step 5:

Determine the maximum strength of the W12×40. Remember that in no case can M n Ω be greater than M p Ω From Manual Table 3-2, M p Ω = 142 ft-kips Since this is less than the required strength, M a = 188 ft-kips , this W-shape will not be adequate even with the use of Cb.

Step 6:

Select a shape with M p Ω ≥ 188 ft-kips from Manual Table 3-2. Thus, select a W18×40 with M p Ω = 196 ft-kips

Step 7:

Confirm in Manual Table 3-10 that the line, either solid or dashed, for the W18×40 is above and to the right of the intersection of M n Ω = 113 ft-kips and Lb = 15 ft. Thus, select the W18×40

6.5 DESIGN OF NONCOMPACT BEAMS 6.5.1

Local Buckling

Local buckling occurs when a compression element of a cross section buckles under load before it reaches the yield stress. Because this buckling occurs at a stress lower than the yield stress, the shape is not capable of reaching the plastic moment. Thus, the shape is not compact and the nominal strength of the member is something less than Mp. Buckling of the flange and web

Chapter 6

Figure 6.17 6

mbers 217 Bending Mem

Examplle of Flange Local L Bucklinng

Photo cou urtesy of Donalld W. White

elementss, and lateral-torsional bucckling of the ssection, do nnot occur in issolation, so itt is difficult to illustraate them indiv vidually. Figu ure 6.17 prim marily illustrattes local buckkling of the ccompression flange off a wide-flang ge beam duriing loading inn an experim mental test. Thhese failures ooccur when the flang ge or web aree slender, and d they can bee predicted thhrough the uuse of the plaate buckling equation discussed in Chapter 5. The T projectingg flange of a w wide-flange m member is considered an unstiffen ned element because the weeb supports oonly one edgee, while the otther edge is uunsupported and free to rotate. Th he wide-flang ge web is coonnected at bboth its edgess to the flangges, so it is considereed a “stiffened” element. Table T B4.1b of the Speciffication proviides the limitting width-to-thickness rattios, λp, for the flang ge and web to ensure that the t full plasticc moment strrength can be reached. Whhen both the flange an nd web have slenderness ratios (b/tf annd h/tw, respeectively) lesss than or equal to the λp values giiven in the taable, the shap pes are calledd compact shhapes. If eithher element eexceeds this value, the shape canno ot be called compact c and tthe nominal sstrength must be reduced. T These latter shapes arre discussed here. h For F the flang ge of a W-sh hape to be ccompact (case 10 in Table B4.1b), itts width-tothicknesss ratio must saatisfy the folllowing limit: b E (6.9) λ f = ≤ λ ppf = 0.38 t Fy where b /t / = bf/2tf. Forr the web to be b compact (ccase 15 in Tabble B4.1b), thhe limiting rattio is h E (6.10) λ w = ≤ λ pw = 3.76 Fy tw Using thee common A9 992 steel with h Fy = 50 ksi, these limits bbecome: for a com mpact flange bf ≤ λ pf = 9.15 2t f and for a compact web b h ≤ λ pw = 90.6 tw A compaarison of these limits with the data giveen in Manual Table 1-1 shhows that the majority of the W-sh hapes have co ompact flangees and all havee compact weebs.

2118 Chapter 6

Bending Members M

S frrom Specificaation Table B4.1 Figure 6.18 Definition of Element Slenderness Copyright © American Insttitute of Steel Construction. C R Reprinted with Permission. A All rights reservved.

Figu ure 6.18 illusstrates these dimensions d foor several com mmonly usedd sections aloong with the slendern ness limits (w width-to-thick kness ratios) as found in Specificationn Table B4.1bb. Other shapes can also a be found in Table B4.1b. 6.55.2

Flange Local Buckliing The full range of nominaal moment strrength, Mn, off a cross sectiion can be exxpressed as a function of flange sleenderness, λf; this relationsship is shownn in Figure 6.19. The threee regions in thhe figure identify threee types of beehavior. The first f region reppresents plasttic behavior, in which the shape is capable of attaining a its full fu plastic mo oment strengtth. This strenngth was disccussed in Secttion 6.2. Shapes that fall into this region r are callled compact.. The behavioor exhibited inn the middle rregion is inelastic, an nd shapes thaat fit this cateegory are callled noncomppact. Shapes that fall into the last region exhiibit elastic buckling b and d are called slender shap apes. The prrovisions for doubly symmetric I-shaped meembers that exhibit thesee last two fforms of behhavior are ggiven in n Section F3. Specification For I-shaped secctions, the div viding line b etween comppact and nonccompact flannges was given in Equ uation 6.9. Th he division beetween noncoompact and slender flange sections is a function of the residu ual stresses present p in thee hot rolled m member. As w was the case with lateral-ttorsional buckling, th he Specificatiion assumes that elastic bbehavior cont ntinues up to the point whhere the elastic mom ment, MrFLB = 0.7FySx. As with w lateral-toorsional buckkling, this limit is a functioon of the residual streess set at a value v of 0.3F Fy. This corrresponds to a flange slenderness, as ffound in Specification n Table B4.1b b, of E (6.11) λ rf = 1.00 Fy

Chapter 6

Figure 6.19 6

Bending Mem mbers 219

Flange Local Buckliing Strength.

The stren ngth at the jun nction of com mpact behavioor and noncom mpact behaviior is the plasstic moment strength M n = M p = Fy Z At the ju unction of non ncompact and d slender behaavior, the mom ment is an elaastic moment defined as M rFLB = 0.7 Fy S x The stren ngth for nonco ompact shapees is representted by a straigght line betweeen these poinnts. Thus, ⎡ ⎛ λ − λ pf ⎞ ⎤ M n = ⎢ M p − ( M p − M rFLB ) ⎜ (6.12) ⎟⎥ ⎝ λ rf − λ pf ⎠ ⎦⎥ ⎣⎢ or as giveen in the Speccification ⎡ ⎛ λ − λ pf ⎞ ⎤ M n = ⎢ M p − ( M p − 0.7 Fy S x ) ⎜ (A AISC F3-1) ⎟⎥ ⎝ λ rf − λ pf ⎠ ⎦⎥ ⎣⎢ For A992 2 steel with Fy = 50 ksi, Equation E 6.11 provides an upper limit too the noncom mpact flange of E 29, 000 = 11.0 λ rf = 1.0 = 24.1 Fy 50 A review of Manual M Tablee 1-1 for bf /22tf shows thatt there are noo W-shapes w with flanges that exceeed this limit. Thus, all wid de-flange shappes have either compact orr noncompactt flanges. A further reeview of the tables showss that only 100 W-shapes hhave noncomppact flanges for Fy = 50 ksi. 6.5.3

Web b Local Buck kling A compaarison of the slenderness criteria c for weeb local buckkling given inn Equation 6.10 with the data avaiilable in Manual Table 1-1 1 for h/tw indiicates that all W-shapes haave compact w webs for Fy = 50 ksi. Thus there is no need to o address weeb local buckkling for W-shapes. Noncoompact and slender webs, w howeveer, are addressed for built--up members in Chapter 7, where platee girders are treated.

220 Chapter 6

Bending Members

EXAMPLE 6.8 Bending Strength of Noncompact Beam

Goal:

Determine the nominal moment strength and then the design moment strength (LRFD) and allowable moment strength (ASD) for a W-shape with noncompact flange.

Given:

A simply supported W6×15 spans 10 ft. It is braced at the ends and at the midspan (Lb = 5 ft). The steel is A992.

SOLUTION

Step 1:

Determine the nominal strength. Check the limits for flange local buckling. For the flange, from Manual Table 1-1, bf = 11.5 > λ pf = 0.38 E Fy = 9.15 2t f

Therefore, the flange is not compact. Checking for a slender flange, even though our previous review of the Manual data indicated that no W-shapes exceeded this requirement, bf = 11.5 < λ rf = 1.0 E Fy = 24.1 2t f

Because λ pf < b f /2t f < λ rf , the shape has a noncompact flange. Step 2:

Check the limit states for web local buckling. For the web, from Manual Table 1-1, h = 21.6 < λ pw = 3.76 E Fy = 90.6 tw So the web is compact, as expected from our earlier evaluation of all Wshapes.

Step 3:

Because the shape is noncompact (flange), determine the nominal moment strength by Equation F3-1. From Manual Table 1-1 Zx = 10.8 in.3 and Sx= 9.72 in.3 and M p = Fy Z x = 50 (10.8 ) = 540 in.-kips Thus, for flange local buckling ⎡ ⎛ λ − λ pf ⎞ ⎤ M n = ⎢ M p − ( M p − 0.7 Fy S x ) ⎜ ⎟⎥ ⎢⎣ ⎝ λ rf − λ pf ⎠ ⎥⎦ ⎡ ⎛ 11.5 − 9.15 ⎞ ⎤ = ⎢540 − ( 540 − 0.7 ( 50 ) 9.72 ) ⎜ ⎟ ⎥ = 509 in.-kips ⎝ 24.1 − 9.15 ⎠ ⎦ ⎣ or 509 in.-kip Mn = = 42.4 ft-kips 12 in./ft

Step 4:

Check for the limit state of lateral-torsional buckling. For this shape, from Manual Table 1-1 ry = 1.45 in. and

Chapter 6

Bending Members 221

L p = 1.76 ry E Fy = 1.76 (1.45 ) 29, 000 50 = 61.5 in.

Thus, Lp = 5.13 ft, which is greater than Lb = 5.0 ft, so the beam is adequately braced to resist the plastic moment. For lateral-torsional buckling, M n = M p = Fy Z = 50 (10.8 ) = 540 in.-kips or M n = 540 /12 = 45.0 ft-kips Step 5:

For LRFD Step 6: For ASD Step 6:

Because the moment based on flange local buckling, 42.4 ft-kips, is less than the moment based on lateral-torsional buckling, 45.0 ft-kips, local buckling controls and M n = 42.4 ft-kips

Determine the design moment for LRFD.

φMn = 0.9 ( 42.4) =38.2 ft-kips

Determine the allowable moment for ASD M n Ω = 42.4 1.67 =25.4 ft-kips

EXAMPLE 6.9 Bending Strength of a Noncompact Beam

Goal:

Determine the (a) design moment strength (LRFD) and (b) allowable moment strength (ASD) for a W-shape with noncompact flange using Table 3-2.

Given:

A simply supported W14×90 spans 15 ft and carries a uniformly distributed load. It is braced only at the supports. The steel is A992. Note in Table 3-2 that this W-shape is identified with a superscript f indicating in a footnote that the flange is not compact.

SOLUTION

Step 1:

Check the unbraced length to determine if lateral-torsional buckling is a limit state that must be considered with Lb = 15.0 ft. From Manual Table 3-2, Lp = 15.1 > Lb = 15.0 ft Thus, lateral-torsional buckling does not influence the beam strength.

Step 2:

Knowing that the shape is not compact, check the limits for flange local buckling. From Manual Table 1-1, bf = 10.2 > λ pf = 0.38 E Fy = 9.15 2t f Therefore, the flange is not compact. We know that no W-shapes have

222 Chapter 6

Bending Members slender flanges but we need that limit. Thus, λ rf = 1.0 E Fy = 24.1

Step 3:

Determine the nominal moment strength using Equation F3-1. From Manual Table 1-1 Zx = 157 in.3 and Sx = 143 in.3 and M p = Fy Z x = 50 (157 ) = 7850 in.-kips and ⎡ ⎛ λ − λ pf ⎞ ⎤ M n = ⎢ M p − ( M p − 0.7 Fy S x ) ⎜ ⎟⎥ ⎢⎣ ⎝ λ rf − λ pf ⎠ ⎥⎦

⎡ ⎛ 10.2 − 9.15 ⎞ ⎤ = ⎢7850 − ( 7850 − 0.7 ( 50 )(143) ) ⎜ ⎟ ⎥ = 7650 in.-kips ⎝ 24.1 − 9.15 ⎠ ⎦ ⎣ Thus,

Mn = For LRFD Step 4:

7650 = 638 ft-kips 12

Determine the design strength and compare it to φb M px from Table 3-2 φM n = 0.9 ( 638 ) = 574 ft-kips

which is identical to the value from Table 3-2 For ASD Step 4:

Determine the allowable strength and compare it to M px Ωb from Table 3-2 M n 1.67 = 638 1.67 = 382 ft-kips which is identical to the value from Table 3-2

Step 5:

Conclusion: Manual Table 3-2 includes consideration of the flange compactness for noncompact flange W-shapes.

6.6 DESIGN OF BEAMS FOR WEAK AXIS BENDING Up to this point, I-shaped beams have been assumed to be bending about an axis parallel to their flanges, called the x-axis. A quick scan of the shape property tables in the Manual shows that the section modulus and plastic section modulus about the x-axis are larger than the corresponding values about the other orthogonal axis, the y-axis. Thus, bending about the x-axis is called strong axis or major axis bending, whereas bending about the y-axis is called weak axis or minor axis bending. Although beams are not normally oriented for bending about this weak axis, a situation may arise when it is necessary to determine the strength of a beam in this orientation. Design of I-shaped beams for weak axis bending is relatively easy. Section F6 of the Specification applies to I-shaped members and channels bent about their minor axis. Two limit states are identified: yielding and flange local buckling. The flange and web referred to here are the same elements as for the case when the shape is bending about its major axis. Thus, the limits on flange slenderness are the same as discussed earlier. For those few W-shapes with noncompact flanges, an equation similar to that used previously for noncompact flanges is required.

Chapter 6

Bending Members 223

For the limit state of yielding, M n = M p = Fy Z y ≤ 1.6 Fy S y

(AISC F6-1)

An I-shaped member bending about its weak axis has properties close to those of a rectangle. For the rectangle, the ratio of the plastic moment to the elastic yield moment, called the shape factor, equals 1.5. The addition of the web alters the elastic section modulus and plastic section modulus so that the shape factor for these weak axis bending members exceeds 1.5. To ensure an appropriate level of rotational capacity at the plastic limit state, the shape factor for weak axis bending is limited to 1.6. All but five W-shapes meet this limitation. Although I-shaped members are not often called upon to carry moment about the y-axis as pure bending members, they are called upon to participate in combined bending as discussed in Section 6.12 and combined with axial load as discussed in Chapter 8.

EXAMPLE 6.10 Weak Axis Bending Strength of Beam

Goal:

Determine the nominal moment strength and then the design moment strength (LRFD) and the allowable moment strength (ASD) for bending about the y-axis.

Given:

A simply supported W10×30, A992 steel, is loaded to cause bending about its weak axis.

SOLUTION

Step 1:

Determine the nominal strength. Check the limits for flange local buckling. For the flange, from Manual Table 1-1, bf = 5.70 < λ pf = 0.38 E Fy = 9.15 2t f

Therefore, the flange is compact. Step 2:

For the limit state of yielding, from Equation F6-1 with Zy = 8.8 in.3 and Sy = 5.75 in.3 from Manual Table 1-1 M n = Fy Z y = 50 ( 8.84 ) = 442 in.-kips

≤ 1.6 Fy S y = 1.6 ( 50 )( 5.75) = 460 in.-kips Thus the strength is controlled by the plastic moment and 442 Mn = = 36.8 ft-kips 12 For LRFD Step 3: For ASD Step 3:

For LRFD, determine the design moment. φM n = 0.9 ( 36.8 ) =33.1 ft-kips

For ASD, determine the allowable moment. M n Ω = 36.8 1.67 =22.0 ft-kips

224 Chapter 6

Bending Members

EXAMPLE 6.11 Weak Axis Bending Strength of Beam

Goal:

Determine the nominal moment strength, design moment strength (LRFD), and allowable moment strength (ASD) for bending about the y-axis.

Given:

A simply supported W40×392, A992 steel, is loaded to cause bending about its weak axis.

SOLUTION

Step 1:

Determine the nominal moment strength. Check the limits for flange local buckling. For the flange, from Manual Table 1-1, bf = 2.45 < λ pf = 0.38 E Fy = 9.15 2t f Therefore, the flange is compact.

Step 2:

For the limit state of yielding, from Equation F6-1 with Zy = 212 in.3 and Sy = 130 in.3 from Manual Table 1-1 M n = Fy Z y = 50 ( 212 ) = 10,600 in.-kips

≤ 1.6 Fy S y = 1.6 ( 50 )(130 ) = 10, 400 in.-kips Thus the strength is controlled by the upper limit and 10, 400 Mn = = 867 ft-kips 12 For LRFD Step 3: For ASD Step 3:

For LRFD, determine the design moment. φM n = 0.9 ( 867 ) = 780 ft-kips

For ASD, determine the allowable moment. M n Ω = 867 1.67 = 519 ft-kips

6.7 DESIGN OF BEAMS FOR SHEAR Chapter G of the Specification establishes the requirements for beam shear. Although shear failures are uncommon with rolled sections, the strength must still be confirmed. A beam can fail in shear by yielding or buckling. Beam webs also need to be checked for shear rupture on the net area of the web when bolt holes are present. Shear rupture is addressed in the discussion of connections in Chapter 10. The nominal shear yielding strength is based on the von Mises criterion,3 which states that for an unreinforced beam web that is stocky enough not to fail by buckling, the shear strength can be taken as Fy 3 = 0.58Fy . The Specification rounds this stress to 0.6Fy and provides, in Section G2, the shear strength as (AISC G2-1) Vn = 0.6 Fy Aw Cv1

3

Mendelson, A. Plasticity: Theory and Application, New York: Macmillan, 1968.

Chapter 6

Bending Members 225

where Aw is the area of the web, taken as the total depth times the web thickness. The web shear coefficient, Cv1, is used to account for shear web buckling. Thus, if the web is capable of reaching yield, Cv1 = 1.0. To ensure that the beam web is capable of reaching yield before buckling, the Specification sets the limit on web slenderness at h k E ≤ 1.10 v tw Fy where kv = 5.34 for unstiffened webs. All current ASTM A6 rolled I-shaped members have webs that meet the criteria for kv = 5.34, and all A992 W-shapes meet the criteria for web yielding, Cv1 = 1.0. Thus, the nominal shear strength of all rolled W-shapes, using Equation G2-1 with Cv1 = 1.0, can be taken as (6.13) Vn = 0.6 Fy Aw Determining the shear design strength or allowable strength is complicated by a variation in resistance and safety factors. To keep the beam shear strength provisions the same in the 2016 Specification as in the allowable stress specifications prior to 2005, the resistance and safety factors for a particular set of rolled I-shapes was liberalized. Thus, for webs of rolled I-shapes with h /t w ≤ 2.24 E / Fy , φ = 1.00

( LRFD )

Ω = 1.50

( ASD )

φ = 0.90

( LRFD )

Ω = 1.67

( ASD )

For all other shapes,

EXAMPLE 6.12 Shear Strength of Beam

SOLUTION

Goal:

Determine the nominal shear strength, design shear strength (LRFD), and allowable shear strength (ASD) when the beam is bending about the x-axis and compare to the shear strength given in Manual Table 3-2.

Given:

A W16×31, A992 steel member is loaded to cause shear in its web.

Step 1:

Determine the nominal strength. Check the limits for yielding of the web. From the user note at the end of Section G2.1 we know that all W-shapes will yield in shear. Thus, Cv1 = 1.0 and from Manual Table 1-1 d = 15.9 in., tw = 0.275 in., and h/tw = 51.6

Step 2:

For the limit state of yielding,

Vn = 0.6Fy AwCv1 = 0.6 ( 50)(15.9)( 0.275)(1.0) = 131 kips

Step 3:

Determine the resistance factor and the safety factor. h E 29, 000 = 51.6 < 2.24 = 2.24 = 53.9 tw Fy 50 Therefore, φv = 1.00 and Ωv = 1.50

226 Chapter 6

Bending Members For LRFD Step 4:

For LRFD, determine the design shear strength. φVn = 1.0 (131) = 131 kips From Manual Table 3-2 φVn = 131 kips . Thus the answers are the same.

For ASD Step 4:

For ASD, determine the allowable shear strength. Vn Ω = 131 1.50 = 87.3 kips From Manual Table 3-2 Vn Ω = 87.5 kips . Thus the results are close. The difference is due to round off.

EXAMPLE 6.13 Shear Strength of Beam

SOLUTION

Goal:

Determine the nominal shear strength, design shear strength (LRFD), and allowable shear strength (ASD) when the beam is bending about the x-axis and compare to the shear strength given in Manual Table 3-2.

Given:

A W16×26, A992 steel member is loaded to cause shear in its web.

Step 1:

Determine the nominal strength. Check the limits for yielding of the web. From the user note at the end of Section G2.1 we know that all W-shapes will yield in shear. Thus, Cv1 = 1.0 and from Manual Table 1-1 d = 15.7 in., tw = 0.250 in., and h/tw = 56.8

Step 2:

For the limit state of yielding, Vn = 0.6Fy AwCv1

= 0.6 ( 50)(15.7 )( 0.250)(1.0) = 118 kips Step 3:

For LRFD Step 4:

Determine the resistance factor and the safety factor. h E 29, 000 = 56.8 > 2.24 = 2.24 = 53.9 tw Fy 50 Therefore, φv = 0.90 and Ωv = 1.67

For LRFD, determine the design shear strength. φVn = 0.90 (118 ) = 106 kips From Manual Table 3-2 φVn = 106 kips . Thus the answers are the same

For ASD Step 4:

For ASD, determine the allowable shear strength.

Chapter 6

Bending Members 227

Vn Ω = 118 1.67 = 70.7 kips

From Manual Table 3-2 Vn Ω = 70.5 kips . Thus the results are close. The difference is due to round off.

6.8 CONTINUOUS BEAMS Beams that span over more than two supports are called continuous beams. Unlike simple beams, continuous beams are indeterminate and must be analyzed by applying more than the three basic equations of equilibrium. Although indeterminate analysis is not within the scope of this book, a few aspects should be addressed, even if only briefly. The Manual includes shears, moments, and deflections for several continuous beams with various uniform load patterns in Manual Table 3-23. These results come from an elastic indeterminate analysis and can be used for the design of any beams that fit the support and loading conditions. It has long been known that material ductility permits steel members to redistribute load. When one section of a member becomes overloaded, it can redistribute a portion of its load to a less loaded section. This redistribution can be accounted for through an analysis method called plastic analysis or through a number of more modern methods capable of modeling the real behavior of the members. These methods may collectively be called advanced analysis and may be used in structural steel design through the provisions of Appendix 1 of the Specification. This appendix also permits use of the simplified plastic analysis approach for continuous beams through Appendix Section 1.3. To allow the designer to take advantage of some of the redistribution that is accounted for in plastic analysis, Section B3.3 gives provisions for moment redistribution in beams. Design of beams and girders that are compact and have sufficiently braced compression flanges may take advantage of this simplified redistribution approach. The compact criteria are those already discussed, whereas the unbraced length criteria are different. To use the simplified redistribution, for doubly symmetric I-shaped beams the unbraced length of the compression flange, Lb, must be less than that given in Section F13.5 as ⎡ ⎛ M ⎞⎤ ⎛ E ⎞ Lm = ⎢0.12 + 0.076 ⎜ 1 ⎟ ⎥ ⎜ ⎟ ry (AISC F13-8) ⎝ M 2 ⎠ ⎦ ⎝ Fy ⎠ ⎣ where M1 is the smaller and M2 is the larger moment at the ends of the unbraced length. The moment ratio is positive when the moments cause reverse curvature and negative when they cause single curvature. When these criteria are satisfied, the beam can be proportioned for 0.9 times the negative moments at points of support. This redistribution is permitted only for gravity-loading cases and moments determined through an elastic analysis. When this reduction in negative moment is used, the positive moment must be increased to maintain equilibrium. This can be accomplished simply by adding to the maximum positive moment 0.1 times the average original negative moments.

EXAMPLE 6.14a Continuous Beam Design by LRFD

Goal:

Select a compact, fully braced section for use as a continuous beam.

Given:

The beam is continuous over three spans of 30 ft each. It supports a live load of 2.5 kip/ft and a dead load of 1.8 kip/ft. Use A992 steel.

SOLUTION

Step 1:

Determine the required strength.

228 Chapter 6

Bending Members The design load is wu = 1.2(1.8) + 1.6(2.5) = 6.16 kip/ ft. From the beam shear, moment, and deflection diagrams in Manual Table 3-23, case 39, the critical span is the exterior span, where the negative moment is 2 − M BA = 0.100 wl 2 = 0.100 ( 6.16 )( 30 ) = 554 ft-kips and the positive moment is 2 + M BA = 0.0800 wl 2 = 0.0800 ( 6.16 )( 30 ) = 444 ft-kips Step 2:

Consider redistribution of moments according to Specification Section B3.3. A design could be carried out for a maximum moment of 554 ft-kips, but with redistribution this moment may be reduced to M BA = 0.9 ( 554 ) = 499 ft-kips provided that the positive moment is increased by the average negative moment reduction. Thus, 0 + 0.1( 554 ) M AB = 444 + = 472 ft-kips 2

Step 3:

Determine the required plastic section modulus. Even with the increase in positive moment, the negative moment is still the maximum moment, so for a moment of 499 ft-kips, 499 (12 ) Z req = = 133 in.3 0.9 ( 50 )

Step 4:

. EXAMPLE 6.14b Goal: Continuous Beam Given: Design by ASD

SOLUTION

Step 1:

Select the least-weight W-shape from Manual Table 3-2. W24×55 with Z = 134 in.3

Select a compact, fully braced section for use as a continuous beam. The beam must be continuous over three spans of 30 ft each. It must support a live load of 2.5 kip/ft and a dead load of 1.8 kip/ft. Use A992 steel. Determine the required strength. The design load is wa = (1.8) + (2.5) = 4.3 kip/ft. From the beam shear, moment, and deflection diagrams in Manual Table 323, case 39, the critical span is the exterior span, where the negative moment is − M BA = 0.100 wl 2 = 0.100 (4.3)(30) 2 = 387 ft-kips

Chapter 6

Bending Members 229

and the positive moment is + MBA = 0.0800 wl 2 = 0.0800(4.3)(30) 2 = 310 ft-kips Step 2:

Consider redistribution of moments according to Specification Section B3.3. A design could be carried out for a maximum moment of 387 ft-kips, but with redistribution this moment may be reduced to MBA = 0.9(387) = 348 ft-kips provided that the positive moment is increased by the average negative moment reduction. Thus, 0 + 0.1(387) MAB = 310 + = 329 ft-kips 2

Step 3:

Determine the required plastic section modulus. Even with this increase in the positive moment, the negative moment is still the maximum moment, so for a moment of 348 ft-kips 348 (12 ) Z req = = 139 in.3 ( 50 / 1.67 )

Step 4:

Select the least-weight W-shape from Manual Table 3-2. W21×62 with Z = 144 in.3

Note that in the preceding example, the reduction in negative moment leads to selection of a smaller W-shape than would have resulted from design for the original negative moment.

6.9 PLASTIC ANALYSIS AND DESIGN OF CONTINUOUS BEAMS Up to this point, it has been assumed that the plastic moment strength of a bending member could be compared to the maximum elastic moment on a beam to satisfy the strength requirements of the Specification. This is accurate for determinate members in which the occurrence of the plastic moment at the single point of maximum moment results in the development of a single plastic hinge, which would lead to member collapse. However, for indeterminate structures, such as continuous beams, more than one plastic hinge must form before the beam would actually collapse, and this provides some additional capacity that an elastic analysis cannot capture. The formation of plastic hinges in the appropriate locations causes a collapse, and the geometry of this collapse is called a failure or collapse mechanism. This is the approach referred to as plastic analysis, permitted by Appendix Section 1.3 of the Specification for use with LRFD only. The Specification only addresses beam design by plastic analysis. Use beyond beams, such as for frames, is up to the designer; proper consideration of second-order effects on stability is required. Thus, only the determination of plastic collapse mechanisms for beams is considered here. The formation of a beam failure mechanism may best be understood by following the load history of a fixed-end beam with a uniformly distributed load. The beam and moment diagrams that result from an elastic indeterminate analysis are given in Figure 6.20a. The largest moments occur at the fixed ends and are given by wL2/12. If the load on the beam is increased, the beam behaves elastically until the moments on the ends equal the plastic moment strength of

2330 Chapter 6

Bending Members M

Fiigure 6.20 B Beam and Mooment Diagraams fo r the Developpment of a Plaastic Mechannism the memberr, as shown in i Figure 6.2 20b. Because the applicatiion of additional load cauuses the member to rotate r at its ends e while maaintaining thee plastic mom ment, these points behave as pins. These pins are a called pla astic hinges. In this case, the load is designated as w1. The mem mber can continue to accept load beyond b this w1, functioninng as a simplee beam, untill a third plasttic hinge forms at thee beam midspan. The forrmation of thhis third hingge makes the beam unstabble, thus forming the collapse mecchanism. Thee mechanism and correspoonding momeent diagram aare given in Figure 6.2 20c. For the collapse mechanism just describeed, equilibriuum requires thhat the simple beam moment, wuL2/8, equal tw wice the plastiic moment; thhus, M p = wu L2 /16 (6.14) Had d this beam been b designeed based on an elastic annalysis, it woould have reqquired a moment cap pacity greaterr than or equ ual to wuL2/112. Using a pplastic analyssis, a smallerr plastic moment streength, equal to t wuL2/16, must m be proviided for in thhe design. Thuus, in this caase of an indeterminatte beam, plaastic analysiss has the pootential to result in a sm maller membeer being required to carry c this sam me load. An additional ad dvantage to the t use of pllastic analysiis for indeterrminate beam ms is the simplicity of o the analysiss. By observaation, regardlless of the ovverall geomettry of the conntinuous beam, each segment betw ween supportss can be evaluuated indepenndently of the other segmennts. This means that any beam seg gment, contin nuous at eachh end and loaaded with a uuniformly disstributed load, exhibits the same collapse c mech hanism. Thus , the relation between the applied loadd and the plastic mom ment will be as a given in Equation 6.14.. Plastic analyysis results ffor additional loading s are given in and beam configuration c i Figure 6.221. Additionaal examples, as well as tthe

Chapter 6

Figure 6.21 6

Bending Mem mbers 231

Loading and Beam Configuration C ns Resulting ffrom Plastic A Analysis

developm ment of these relations thrrough applicaation of energgy principles, can be foundd in several books inccluding, Appllied Plastic Design D in Stee l.4 To T ensure thaat a given beeam cross secction can unddergo the necessary rotattion at each plastic hiinge, the Speccification requ uires that Fy ≤ 65 ksi , thaat the section bbe compact, aand that the compresssion flange be braced such h that the unbbraced lengthh in the area of the hinge is less than Lpd given n as Equatio on A-1-5 in Specificationn Appendix Section 1.3. If these lim mits are not satisfied,, the member design must be b based on aan elastic anallysis.

4

Disque, R. O. O Applied Dessign in Steel. New N York: Van n Nostrand Reinnhold, 1971.

232 Chapter 6

Bending Members

Goal: EXAMPLE 6.15 Beam Design using Plastic Analysis Given: (LRFD only)

SOLUTION

Design a beam using plastic analysis and A992 steel. Plastic analysis is applicable only for LRFD load combinations. A beam is simply supported at one end and fixed at the other, similar to that shown in Figure 6.16 for Example 6.6. It spans 20 ft and is loaded at its midspan with a dead load of 16.0 kips and a live load of 48.0 kips. It is assumed that the final section will be compact and adequately braced.

Step 1:

Determine the required strength. Pu = 1.2 (16.0 ) + 1.6 ( 48.0 ) = 96.0 kips Using the plastic analysis results from Figure 6.21c, 96.0 (10.0 )(10.0 ) Pab φM p req = = = 320 ft-kips ( a + 2b ) 10.0 + 2 (10.0 )

Step 2:

Select the required W-shape from Manual Table 3-2. W21× 44, φM p = 358 ft-kips

Step 3:

Check the initial assumptions on compactness and lateral bracing. A check of the compact flange and web criteria shows that this shape is compact. With continuous bracing, the beam can be designed through plastic analysis. Thus, select a

W 21× 44

6.10 T-SHAPED MEMBERS IN BENDING T-shaped members are normally cut from I-shaped members by splitting the I-shape down the longitudinal axis. Thus they are often referred to as split tees. When they are made from Wshapes they are called WT-shapes. Similarly, when made from M-shapes as MT-shapes and Sshapes as ST-shapes. Provisions for beams formed by combining a pair of angles to form a T and for beams made from a split I-shape and loaded in the axis of symmetry are both found in Section F9 of the Specification. These singly symmetric members may be loaded with the stem in tension or compression. Four limit states must be considered in the design of these T-shaped members: yielding, lateral-torsional buckling, flange local buckling, and stem local buckling. The Specification combines the provisions for tees and double angles, and those for stem or web legs in tension or compression. This section addresses tees while double angles will be treated after the discussion of single angle bending members. 6.10.1

Yielding of Tees

For the limit state of yielding Mn = M p

(AISC F9-1)

and Mp is limited, depending on the orientation of the section. For the stem in tension (AISC F9-2) M p = Fy Z x ≤ 1.6 M y

Chapter 6

Bending Members 233

and for the stem in compression Mp = My

(AISC F9-4)

where M y = Fy S x . The limit on Equation F9-2 is necessary to ensure that the member is capable of rotating sufficiently to attain the plastic moment strength without the extreme fibers of the shape reaching into the strain-hardening region. This is the same limit that was discussed for I-shaped members bending about their weak axis and can be thought of as limiting the section shape factor. Limiting the plastic moment to the elastic moment, as in Equation F9-4 when the stem is in compression, is a conservative assumption that reflects the limited knowledge available to predict what moment the tee in this orientation could actually attain. 6.10.2

Lateral-Torsional Buckling of Tees

Lateral-torsional buckling also must account for the orientation of the shape. For stems in tension, the limiting unbraced length for the limit state of yielding is the same as it was for doubly symmetric I-shapes. As given in Section F9 it is E Lp = 1.76ry (AISC F9-8) Fy The limiting unbraced length for inelastic lateral-torsional buckling is ⎛ E ⎞ IyJ ⎛ Fy ⎞ dS Lr = 1.95 ⎜ ⎟ 2.36 ⎜ ⎟ x + 1 ⎝E⎠ J ⎝ Fy ⎠ S x

(AISC F9-9)

If the unbraced length is less than Lp, the limit state of lateral-torsional buckling does not apply. If the unbraced length is between Lp and Lr, the nominal strength, when the stem is in tension, is given by a straight line as ⎡ Lb − Lp ⎤ Mn = M p − ( M p − M y ) ⎢ (AISC F9-6) ⎥ ⎣ Lr − Lp ⎦ It should be noted that this is similar to the straight line equations used for other types of members but when Lb = Lr , Mn is set to My. When the unbraced length is greater than Lr and the stem is in tension, the nominal strength is given by Mn = Mcr (AISC F9-7) and

M cr =

(

1.95E I y J B + 1 + B2 Lb

)

(AISC F9-10)

where ⎛ d ⎞ Iy B = +2.3 ⎜ ⎟ ⎝ Lb ⎠ J

(AISC F9-11)

For the stem in compression at any point along the span, whenever the unbraced length is greater than Lp Mn = Mcr (AISC F9-7)

234 Chapter 6

Bending Members and again

M cr =

(

1.95E I y J B + 1 + B2 Lb

)

(AISC F9-10)

but the sign on B is changed to negative. Thus, ⎛ d ⎞ Iy B = −2.3 ⎜ ⎟ ⎝ Lb ⎠ J

6.10.3

(AISC F9-11)

Flange Local Buckling of Tees

The limit state of flange local buckling for tees reflects the same behavior as for the I-shapes from which they are cut. The limiting width-to-thickness ratios are the same as discussed earlier, and the nominal strength equation is the same, except that it is limited to 1.6My. For compact flanges, the limit state of flange local buckling does not apply. For noncompact flanges, λ p < λ = b f 2t f ≤ λ r ,

⎡ ⎛ λ − λ pf M n = ⎢ M p − ( M p − 0.7 Fy S xc ) ⎜ ⎢⎣ ⎝ λ rf − λ pf

⎞⎤ ⎟ ⎥ ≤ 1.6M y ⎠ ⎥⎦

(AISC F9-14)

and for slender flanges, λr < λ, Mn =

0.7 ES xc ⎛ bf ⎞ ⎜ ⎟ ⎝ 2t f ⎠

(AISC F9-15)

2

Sxc is the section modulus referred to the compression flange. If the stem is in compression, this limit state does not apply. 6.10.4

Stem Local Buckling of Tees

When the stem is in flexural compression, the nominal strength for the limit state of stem local buckling is given by M n = Fcr S x (AISC F9-16) Sx is the elastic section modulus and Fcr is the critical stress, which is dependent on the stem slenderness, d tw . Thus, when d E ≤ 0.84 tw Fy

the stem will not buckle and Fcr = Fy

When 0.84

E d E < ≤ 1.52 Fy t w Fy

⎡ d Fcr = ⎢1.43 − 0.515 tw ⎢⎣

Fy E

⎤ ⎥ Fy ⎥⎦

(AISC F9-18)

Chapter 6

Bending Mem mbers 235

Figure 6.22 6 T-Beam Orientation for f Example 66.16

and when n

d E > 1.52 Fy tw

Fcr =

1.52 E ⎛d⎞ ⎜t ⎟ ⎝ w⎠

2

(A AISC F9-19)

If the stem is in tensio on, this limit state s does nott apply.

EXAMPLE E 6.16 Bending Sttrength of WT-Shap ape

Goal: G

Deetermine the nominal mom ment strengthh for the giveen WT membber if the steem is in (a) teension (Figuree 6.22a) and ((b) compressiion (Figure 6..22b).

Given: G

A WT9×17.5 is i used as a beam to suppport gravity loads and haas lateral su upport provideed at 5 ft interrvals.

SOLUTION N

Step S 1:

Deetermine the section propeerties for the WT-shapes ffrom Manual Table 18. Zx = 11 1.2 in.3, Sx = 6.21 in.3, d = 8.85 in., tw = 0.300 in., Iy = 7.67 inn.4, J = 0.252 in.4, ry = 1.222 in.

Part P a

Deetermine the nominal n mom ment strength for the stem in tension. Thhe WT is oriented as show wn in Figure 6.22a.

Step S 2:

Deetermine the nominal n mom ment strength for the limit state of yieldiing using Eq quation F9-2. M y = Fy S x = 50 ( 6.21) = 311 in.-kipps

M p = Fy Z x ≤ 1.6 M y = 50 (11.2 ) = 5560 in.-kips ≤ 1.6 ( 311) = 4498 in.-kips Th hus, Mn = 498 in.-kkips for the liimit state of yyielding Step S 3:

Deetermine the nominal mooment strenggth for the llimit state off lateraltorrsional buckliing. From Eqquation F9-8

Lp = 1.76ry an nd Equation F9-9 F

E 299,000 = 1..76 (1.22) = 51.7 in. ⇒ 4.31 ft Fy 50

236 Chapter 6

Bending Members ⎛ E ⎞ IyJ Lr = 1.95 ⎜ ⎟ ⎝ Fy ⎠ S x

⎛ Fy ⎞ dS 2.36 ⎜ ⎟ x + 1 ⎝ E ⎠ J

⎛ 50 ⎞ 8.85 ( 6.21) ⎛ 29,000 ⎞ 7.67 ( 0.252 ) = 1.95 ⎜ +1 2.36 ⎜ ⎟ ⎟ 6.21 ⎝ 50 ⎠ ⎝ 29,000 ⎠ 0.252 = 348 in. ⇒ 29.0 ft

Thus, since L p < Lb = 5.0 ft < Lr the beam will buckle inelastically and using Equation F9-6

⎡ Lb − Lp ⎤ Mn = M p − (M p − M y ) ⎢ ⎥ ⎣ Lr − Lp ⎦ ⎡ 5.0 − 4.31 ⎤ = 498 − ( 498 − 311) ⎢ ⎥ = 493 in.-kips ⎣ 29.0 − 4.31 ⎦ Step 4:

Consider the limit state of flange local buckling. The limit state of flange local buckling does not apply to the WT9×17.5 because the flange is compact.

Step 5:

The controlling limit state for the WT with the stem in tension is the smaller strength given by the limit states of yielding and lateral-torsional buckling. Thus, for the limit state of lateral-torsional buckling M n = 493 12 = 41.1 ft-kips

Part b

Determine the nominal moment strength for the stem in compression. The WT is oriented as shown in Figure 6.22b.

Step 6:

Determine the nominal moment strength for the limit state of yielding using Equation F9-4 Mp is limited to My so that, from Step 2, M n = M p = M y = 311 in.-kips

Step 7:

Determine the nominal moment strength for the limit state of lateraltorsional buckling for this orientation with Lb = 5.0 ft. Determine B from Equation F9-11 using the negative sign since the stem is in tension. Thus, ⎛ 8.85 ⎞ 7.67 ⎛ d ⎞ Iy B = −2.3 ⎜ ⎟ = −2.3 ⎜⎜ = −1.87 ⎟⎟ 5 12 0.252 ( ) ⎝ Lb ⎠ J ⎝ ⎠ and from Equation F9-10

M cr =

(

1.95E I y J B + 1 + B2 Lb

)

Chapter 6

M n = M cr =

1.95 ( 29,000 ) 5.0 ( 2.0 )

Bending Members 237

2 7.67 ( 0.252 ) ⎡ −1.87 + 1 + ( −1.87 ) ⎤ ⎥⎦ ⎣⎢

= 328 in.-kips Step 8:

Determine the nominal moment strength for the limit state of stem local buckling. The stem slenderness is d 8.85 = = 29.5 > 0.84 E Fy = 20.2 tw 0.300 < 1.52 E Fy = 36.6 Therefore, from Equation F9-18 ⎡ d Fy ⎤ Fcr = ⎢1.43 − 0.515 ⎥ Fy t w E ⎥⎦ ⎢⎣ ⎡ 8.85 50 ⎤ = ⎢1.43 − 0.515 ⎥ ( 50 ) = 40.0 ksi 0.300 29,000 ⎦ ⎣ and M n = Fcr S x = 40.0 ( 6.21) = 248 in.-kips

Step 9:

Determine the controlling limit state strength for the WT with the stem in compression. The nominal moment strength is the smallest of the limit states checked; thus, for the limit state of stem local buckling M n = 248 12 = 20.7 ft-kips Note: This example shows that using a WT-shape with the stem in compression significantly compromises the strength of the member. Even so, beams with this orientation are often easier to construct for such applications as lintels in masonry walls where WTs are used in this orientation.

6.11 SINGLE-ANGLE BENDING MEMBERS When single angles are used as bending members, they can be bending about one of the geometric axes, parallel to the legs, or about one of the principal axes. They are often used as lintels over openings in masonry walls, where they are bending about the geometric axes. Unfortunately, this most useful orientation of the single-angle bending member is also the most complex orientation for the determination of strength. Figures 6.23a and b show a single angle oriented for bending about the geometric axis, and Figures 6.23c and d show the angle oriented for bending about the minor and major principal axes, respectively. For the doubly symmetric I-shapes and singly symmetric tees discussed to this point, the shear center is located on an axis of symmetry. Typically, loading is applied symmetrically through the shear center. The WT-shapes considered in Section 6.10 were only discussed for loading on the axis of symmetry. For single angles this is not the case. An angle bending about a geometric axis and loaded through the shear center, an axis through the center of the leg

2338 Chapter 6

Bending Members M

gle Bending about a Geome tric Axis andd Principal Axxis Figure 6.23 Single-Ang w experience single axis bending. Unffortunately itt is quite com mmon for anglles to be thickness, will loaded throu ugh a point other o than the shear centerr, in which caase the angle will experiennce both bending and d torsion. Th he treatment of o torsion is beyond the scope of thiss book althouugh it is addressed in n Specification n Chapter H. Speccification Secction F10 giv ves the proviisions for sinngle-angle bennding membeers. The limit states to t be checked d for these meembers are yiielding, lateraal-torsional buuckling, and lleg local buckling. 6.111.1

Yieldin ng The ratio off the plastic section s modu ulus to the elaastic section modulus, shaape factor, foor angles can easily become b quite large. Thus, in order to be sure that the angle is not strained into the strain–hardeening region, the nominal moment m for thhe limit state of yielding iss taken as C F10-1) (AISC M n = 11.5 M y where M y = Fy S , and S is taken as the t lowest secction moduluus about the aaxis of bendinng. This applies for either e bending g about a prin ncipal axis or a geometric aaxis.

6.111.2

Leg Lo ocal Buckling Legs of angles in compreession have th he same tenddency to buckkle as other coompression ellements. Specification n Table B4-1b b defines the limiting slen derness, b/t, iin case 12, as E λ p = 0.544 Fy and E λ r = 0.911 Fy The bending streength of a sing mber as a funnction of leg slenderness iis shown gle-angle mem in Figure 6.24. 6 In the region of no oncompact bbehavior, λ p < b t ≤ λ r , thhe nominal moment strength is given g by the sttraight-line ass ⎛ ⎛ b ⎞ Fy ⎞ M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟ (AISC C F10-6) ⎟⎟ ⎜ t E ⎝ ⎠ ⎠ ⎝

Chapter 6

Figure 6.24 6

Bending Mem mbers 239

Strengtth of a Single Angle as a Fuunction of Leeg Slendernesss

and the nominal n mom ment strength for the regionn of slender bbehavior is giiven as a com mbination of Equation ns F10-7 and F10-8 F as 0.71ESc Mn = 2 ⎛b⎞ ⎜ ⎟ ⎝t⎠ where Sc is the elasticc section mod dulus to the tooe in compression, relativee to the axis of bending, either geeometric or prrincipal axis. For bending about one off the geometrric axes for aan equal-leg angle with no lateral-torsional resttraint, Sc musst be taken ass 0.8 times the geometric aaxis section modulus to reflect thee effect of actu ual bending sstresses aboutt an inclined aaxis that resullt due to the absence of lateral-torssional restrain nt. The largeest width-to-thhickness ratioos for each leeg width of angles giiven in Manual Table 1-7 are shown in Figure 6.24. The angle leg with the larrgest widthto-thickn ness ratio is b t = 5 0.25 = 20 2 . It can be sseen from thee figure that n none of these angles will be consid dered slender. 6.11.3

Latteral-Torsion nal Buckling The limit state of lateeral-torsional buckling is a function off the axis of bbending and w whether the toe of the angle is at a maximum stress in tenssion or comprression. For bbending abouut the minor principall axis and forr angles with full lateral suupport, the lim mit state of laateral-torsionnal buckling does not apply. Thus, provisions arre given only for bending aabout the majjor principal aaxis and for the speciial case of ben nding about th he geometric axis of equall leg angles. For F all cases of lateral-torsional buckliing, the nomiinal moment strength is a function of the elastiic lateral-torssional bucklin ng moment, M cr, which iss determined for major principal axis bending or geometric axis bending of equal leg angles such tthat for M y M cr ≤ 1.0

⎛ My ⎞ M n = ⎜1.92 − 1.117 ⎟ M y ≤ 1.5M y ⎜ M cr ⎟⎠ ⎝

(A AISC F10-2)

2440 Chapter 6

Bending Members M

Figure 6.25

Strength of a Single Angle A as a Funnction of Lateeral-Torsionaal Buckling

and for M y M cr > 1.0

⎛ 00.71M cr ⎞ M n = ⎜ 0.92 − ⎟ M cr My ⎠ ⎝

(AISC C F10-3)

The nominaal bending streength divided d by the yieldd moment forr the limit staate of lateral-ttorsional buckling is given in Figu ure 6.25 as a function of thhe yield mom ment divided bby the elasticc lateraltorsional buckling momeent. It can be seen that as tthe yield mom ment divided by the elasticc lateraltorsional buckling momeent increases, the nominal bbending strenngth of the meember divideed by the yield momen nt decreases. For bending abo out the majorr principal axxis, Figure 66.23d, the elaastic lateral-ttorsional buckling mo oment is 2 9 EA Arz tCb ⎛⎜ β w rz ⎞ βw rz ⎞⎟ ⎛ + M cr = 1 + ⎜ 4.4 4.4 (AISC C F10-4) 8Lb ⎜ Lb t ⎟⎠ Lb t ⎟ ⎝ ⎝ ⎠ The variablees in Equation n F10-4 are the t same as ppreviously defined except for the new vvariable, βw, which is a section pro operty for sing gle angles bennt about theirr major principal axis. It is zero for equal leg an ngles, positivee for angles with w their shoort leg in com mpression andd negative foor angles with their lo ong leg in com mpression. Values V of βw aare given in C Commentary Table C-F100.1. Note that βw is in ndependent off angle thickn ness. All uneequal leg anggles must be evaluated forr lateraltorsional bu uckling about the major prrincipal axis, even if the aactual appliedd load is bending the angle about one of its geo ometric axes. Thus, for thi s condition, th the applied geeometric axis bending moment must be converteed to momentts about both principal axees. For geometric axis a bending the strength of equal legg angles mayy be determiined for conditions of no lateraal-torsional reestraint or ffor lateral-torrsional restraaint at the ppoint of maximum moment m only y. For any otther conditioon of lateral--torsional resstraint, they m must be evaluated fo or bending abo out the princiipal axes. Wheen there is no o lateral restraaint, the yieldd moment My, is taken as 0.80FyS wheree S is the geometric section modullus. The elasttic lateral-torrsional buckliing moment for loading tthat puts m stress as co ompression att the toe is the maximum

Chapter 6

M cr =

Bending Members 241

2 ⎞ 0.58Eb 4tCb ⎛⎜ ⎛ Lb t ⎞ ⎟ 1 0.88 1 + − ⎜ 2 ⎟ ⎜ ⎟ L2b b ⎠ ⎝ ⎝ ⎠

(AISC F10-5a)

and when the maximum stress is tension at the toe

M cr =

2 ⎞ 0.58Eb 4tCb ⎛⎜ ⎛ Lb t ⎞ ⎟ 1 0.88 1 + + ⎜ 2 ⎟ ⎜ ⎟ L2b b ⎠ ⎝ ⎝ ⎠

(AISC F10-5b)

Note that when the toe is at a maximum stress in tension, the elastic lateral-torsional buckling moment is greater than when the maximum stress is compression at the toe. This means that the nominal moment strength for lateral-torsional buckling will be greater when the maximum stress is tension at the toe. For the case of lateral-torsional restraint at the point of maximum moment, Mcr is taken as 1.25 times the value determined from Equations F10-5 and My is taken as 1.0FyS.

EXAMPLE 6.17 Bending Strength of a Single Angle

Goal:

Determine the available bending strength of an equal-leg angle loaded about the geometric axis through the shear center.

Given:

A 6×6×5/16 A36 angle is required to span 8.0 ft on a simple span with lateral supports at the ends only. The vertical leg is up so that it is stressed in compression due to the uniform gravity load.

SOLUTION

Step 1:

Determine the required properties from Manual Table 1-7 S x = S y = 2.95 in.3 b = 6.0 in. t = 0.3125 in.

Step 2:

Determine the nominal moment strength for the limit state of yielding. Use Equation F10.1. M n = 1.5 Fy S x = 1.5 ( 36 ( 2.95 ) ) = 159 in.-kips

Step 3:

Determine the nominal moment strength for the limit state of lateraltorsional buckling, first determine the elastic lateral-torsional buckling moment. For an equal-leg angle with maximum compression in the toe (leg up and top in compression), use Equation F10-5a. 2 ⎞ 0.58 Eb 4tCb ⎛⎜ ⎛ Lb t ⎞ ⎟ M cr = 1 + 0.88 − 1 ⎜ ⎟ ⎜ ⎟ L2b b2 ⎠ ⎝ ⎝ ⎠ 2 ⎛ ⎞ 4 ⎛ 8.0 (12 )( 0.3125 ) ⎞ 0.58 ( 29,000 )( 6.0 ) ( 0.3125 )(1.0 ) ⎜ ⎟ − 1⎟⎟ = 2 2 ⎜ 1 + 0.88 ⎜⎜ ⎟ ( 6.0 ) (8.0 (12 ) ) ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = 199 in.-kips

Step 4:

Determine the yield moment to be used when there is no lateral-torsional restraint. This accounts for the presence of some torsion by using 0.8Sx. Thus, M y = Fy ( 0.8 S x ) = 36 ( 0.8 ( 2.95 ) ) = 85.0 in.-kips

242 Chapter 6

Bending Members Step 5:

Since M y M cr = 85.0 199 = 0.427 ≤ 1.0 , use Equation F10-2 to determine the nominal moment strength for the limit state of lateral-torsional buckling. ⎛ My ⎞ M n = ⎜ 1.92 − 1.17 ⎟⎟ M y ≤ 1.5M y ⎜ M cr ⎝ ⎠

⎛ 85.0 ⎞ = ⎜⎜ 1.92 − 1.17 ⎟ 85.0 199 ⎟⎠ ⎝ = 98.2 in.-kips ≤ 1.5M y = 1.5 ( 85.0 ) = 128 in.-kips Thus, M n = 98.2 in.-kips

Step 6:

Check the leg slenderness to determine if the strength limit state of leg local buckling must be determined. For the 6×6×5/16 angle, b t = 6.0 0.3125 = 19.2 . From Specification Table B4.1b, the slenderness limits are E 29, 000 λ p = 0.54 = 0.54 = 15.3 Fy 36 λ r = 0.91

E 29, 000 = 0.91 = 25.8 Fy 36

Therefore, the angle is noncompact so the strength for the limit state of leg local buckling must be determined. Step 7:

As with lateral-torsional buckling, the reduced section modulus must be used when bending is about the geometric axis. Thus, using Equation F10-6 ⎛ ⎛ b ⎞ Fy ⎞ M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟ ⎟ ⎜ ⎝ t ⎠ E ⎟⎠ ⎝

⎛ 36 ⎞ ⎛ 6.0 ⎞ = 36 ( 0.8 ( 2.95) ) ⎜⎜ 2.43 − 1.72 ⎜ ⎟ = 108 in.-kips ⎟ ⎝ 0.3125 ⎠ 29,000 ⎟⎠ ⎝ Step 8:

For LRFD Step 9: For ASD Step 9:

The nominal bending strength is the lowest value based on the three limit states checked. Thus, lateral-torsional buckling controls and M n = 98.2 in.-kips

Determine the design strength φM n = 0.9 ( 98.2 ) =88.4 in.-kips

Determine the allowable strength M n 98.2 = =58.8 in.-kips Ω 1.67

Chapter 6

Bending Members 243

6.12 DOUBLE ANGLE MEMBERS IN BENDING T-shaped bending members were discussed in Section 6.10 and it was pointed out that tees and double angles are both treated in Specification Section F9. The same four limit states considered for tees must be considered for double angle members: yielding, lateral-torsional buckling, flange local buckling, and stem local buckling. As with tees, double angles must be loaded in the plane of symmetry. Again, these limit states are treated differently if the web legs are in compression or tension. The following describes the application of Specification Section F9 to double angle bending members. 6.12.1

Yielding of Double Angles

For the limit state of yielding Mn = M p

(AISC F9-1)

and Mp is limited, depending on the orientation of the section. For the web legs in tension (AISC F9-2) M p = Fy Z x ≤ 1.6 M y and for web legs in compression M p = 1.5 M y

(AISC F9-5)

where M y = Fy S x

The limit on Equation F9-2 is required for double angles just as it was for tees, to ensure that the member is capable of rotating sufficiently to attain the plastic moment strength without the extreme fibers of the shape reaching into the strain-hardening region. Limiting the plastic moment to 1.5 M y is slightly more restrictive than the limit in Equation F9-2 but is significantly greater than it was for tees. 6.12.2

Lateral-Torsional Buckling of Double Angles

Lateral-torsional buckling also must account for the orientation of the shape. For web legs in tension, the provisions are the same as they were for tees. The limiting unbraced lengths are E Lp = 1.76ry (AISC F9-8) Fy and

⎛ E ⎞ IyJ Lr = 1.95 ⎜ ⎟ ⎝ Fy ⎠ S x

⎛ Fy ⎞ dS 2.36 ⎜ ⎟ x + 1 ⎝E⎠ J

(AISC F9-9)

As usual, if the unbraced length is less than Lp, the limit state of lateral-torsional buckling does not apply. If the unbraced length is between Lp and Lr, the nominal strength, when the web legs are in tension, is given by the same straight-line that was used for tees,

244 Chapter 6

Bending Members

⎡ Lb − Lp ⎤ Mn = M p − ( M p − M y ) ⎢ ⎥ ⎣ Lr − Lp ⎦

(AISC F9-6)

When the unbraced length is greater than Lr and the stem is in tension, the nominal strength is given by Mn = Mcr (AISC F9-7) and

M cr =

(

1.95E I y J B + 1 + B2 Lb

)

(AISC F9-10)

where ⎛ d ⎞ Iy B = +2.3 ⎜ ⎟ ⎝ Lb ⎠ J

(AISC F9-11)

For web legs in compression at any point along the span, whenever the unbraced length is greater than Lp, the nominal moment strength is determined as it was for single angles, through Equations F10-2 and F10-3. For M y M cr ≤ 1.0

⎛ My ⎞ M n = ⎜1.92 − 1.17 ⎟ M y ≤ 1.5M y ⎜ M cr ⎟⎠ ⎝

(AISC F10-2)

and for M y M cr > 1.0

⎛ 0.71M cr ⎞ M n = ⎜ 0.92 − ⎟ M cr My ⎠ ⎝ The elastic lateral-torsional buckling moment is determined as 1.95E M cr = I y J B + 1 + B2 Lb

(

(AISC F10-3)

)

(AISC F9-10)

with ⎛ d ⎞ Iy B = −2.3 ⎜ ⎟ ⎝ Lb ⎠ J

and M y = Fy S x

6.12.3

Leg Local Buckling of Double Angles

Leg local buckling of double angles is treated the same as leg local buckling of single angles by reference from Specification Section F9 to Section F10. As discussed for single angles, the limiting slenderness, b/t, in case 12, is E λ p = 0.54 Fy and

Chapter 6 λ r = 0.91

Bending Members 245

E Fy

For noncompact angles, λ p < b t ≤ λ r , the nominal moment strength is given by

⎛ ⎛ b ⎞ Fy ⎞ M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟ ⎟ ⎜ ⎝ t ⎠ E ⎟⎠ ⎝

(AISC F10-6)

For slender angles the nominal moment strength is given as a combination of Equations F10-7 and F10-8 as 0.71ESc Mn = 2 ⎛b⎞ ⎜ ⎟ ⎝t⎠ For both, Sc is the elastic section modulus of the double angle member for the toe in compression.

EXAMPLE 6.18 Bending Strength of a Double Angle

Goal:

Determine the available bending strength of a double equal-leg angle loaded in the plane of symmetry.

Given:

Two 6×6×5/16 A36 angles are required to span 8.0 ft on a simple span with lateral supports at the ends only. The vertical legs are in contact and are oriented up so that they are stressed in compression due to the uniform gravity load.

SOLUTION

Step 1:

Determine the required properties from Manual Table 1-7 for a single angle S x = 2.95 in.3 I x = 13.0 in.4 b = 6.0 in. t = 0.3125 in. From Manual Table 1-15 for double angles, ry = 2.47 in. and 2S x = 2 ( 2.95) = 5.90 in.3

Step 2:

Determine the nominal moment strength for the limit state of yielding. Use Equation F9.5. M n = 1.5 Fy S x = 1.5 ( 36 ( 5.90 ) ) = 319 in.-kips

Step 3:

Determine the nominal moment strength for the limit state of lateraltorsional buckling. First determine the unbraced length beyond which lateral-torsional buckling must be considered, Lp, based on Equation F9-8. E 29,000 Lp = 1.76ry = 1.76 ( 2.47 ) = 123 in. Fy 36 Since Lb = 8.0 ft = 96.0 in.< 123 in. lateral torsional buckling does not need to be checked.

Step 6:

Check the leg slenderness to determine if the strength limit state of leg local buckling must be determined using Section F10.3 by reference from Section F9.4. For two 6×6×5/16 angles with the back-to-back legs in flexural compression, b t = 6.0 0.3125 = 19.2 . From Specification Table B4.1b, the slenderness limits are

2446 Chapter 6

Bending Members M λ p = 0.54

E 299, 000 = 15.3 = 0.54 Fy 36

λ r = 0.91

E 299, 000 = 25.8 = 0.91 Fy 36

Thereefore, the an ngles are nonncompact juust as for the single anggle of Exam mple 6.17 so the t strength ffor the limit sstate of leg loocal bucklingg must be dettermined. Step p 7:

Unlik ke for the sing gle angle, thee reduced secttion moduluss is not used. Thus, using Equation F10-6 ⎛ ⎛ b ⎞ Fy ⎞ M n = Fy Sc ⎜ 2.43 − 1.72 ⎜ ⎟ ⎟ ⎜ ⎝ t ⎠ E ⎟⎠ ⎝

⎛ 36 ⎞ ⎛ 6.0 ⎞ = 36 ( 5.990 ) ⎜⎜ 2.43 − 1.772 ⎜ ⎟ = 2269 in.-kips ⎟ ⎝ 0.3125 ⎠ 29,000 ⎟⎠ ⎝ Step p 8:

Forr LRF FD Step p 9: Forr ASD D Step p 9:

The nominal n bend ding strength is the lowestt value based on the limit states check ked. Thus, leg g local bucklinng controls annd M n = 269 in.-kiips

Deterrmine the desiign strength φM n = 00.9 ( 269 ) =2422 in.-kips

Deterrmine the allo owable strengtth M n Ω = 2269 1.67 =16 1 in.-kips

h of the doublle angle beam m in Example 6.18 is more than twice thhat of the singgle angle The strength beam in Exaample 6.17, ev ven with the same limit staate, leg local buckling, conntrolling the sstrength. Because of symmetry about the planee of loading, tthe reduced ssection moduulus that was used for a is not used in determining the local bucklinng strength ffor the double angle the single angle member. In addition, lateeral-torsional buckling doees not even nneed to be chhecked for thee double angle beam where it poteentially could d have controllled for the siingle angle beeam. Clearly some of the modes of failure for a single angle beam and a ddouble angle beam can be quite differennt.

F Figure 6.26 Biaxial Bennding of IS Shaped Beam m

Chapter 6

Bending Mem mbers 247

Figure 66.27 Simple Linear Interaaction Diagram for Biaxial B Bending

6.13

MEM MBERS IN N BIAXIAL BENDING Bending members arre often callled upon to resist forcess that result in bending about two orthogon nal axes. Exam mples of thiss member typpe are crane ggirders and rroof purlins in industrial buildingss. Regardless of the actual orientation of an appliedd moment, it is possible tto break the moment into compon nents about th he two princ ipal axes, as shown in Fiigure 6.26. O Once this is accompliished, the ab bility of the section to reesist the com mbined momeents can be determined through the t interaction n equation. Chapter H addresses thee interaction of forces. F Specification S For the com mbination of momentss, a simple lin near interactio on equation iss used, as shoown in Figure 6.27. This is taken from the equattion provided d in Specifica ation Section H1 for combbined axial looad and mom ment. When the axial load is zero, Equation H1-1b reduces too M rx M ry + ≤ 1.0 Mcx M cy where th he moment terrms relate to the x- and yy-axes, the nuumerator is thhe required sttrength, and the denominator is thee available sttrength, deterrmined as thoough the mem mber were bennding about only onee axis at a tim me. Thus, if th he required x--axis momennt is 79 percennt of the x-axxis strength, only 21 percent p of thee y-axis streng gth is availabble to resist m moment. Moree attention is ggiven to the use of intteraction equaations when axial a load is ccombined withh the bendingg moment in C Chapter 8.

6.14

SER RVICEABIL LITY CRIT TERIA FOR R BEAMS There are several serv viceability co onsiderations that the desiigner must adddress. A genneral set of provision ns are found in i Specificatio on Chapter L L. Although faailure to satisfy these criteria may not impact th he strength off the memberr or overall sttructure, it m may lead to thee first signs oof difficulty for succeessful compleetion of a projject. The speccific criteria sshould be disscussed in dettail with the designer’’s client so th hat the quality y of the final product is coonsistent withh the expectaations of the owner. Although A exceessive floor deflection d mayy not be a saafety issue, it may lead to cracking of finishes and an unaccceptable appeearance. It m may also be aan indicator that other seerviceability issues arre likely, such h as vibration n. Experiencee may indicaate that an annnoying amouunt of floor vibration n may be pressent at first, but b occupantss become useed to it with time. The cliient may be unwilling g to deal witth this period d of dissatisffaction and innsist that thee system be ddesigned to prevent vibration v com mplaints. Thiss must be knnown at the bbeginning off a project, not after the occupantts move in and a find the floor responnse objectionaable. The enngineer must be sure to

248 Chapter 6

Bending Members identify these considerations for the owner so that the decisions made are appropriate to achieve the expected outcome. Beams generally have two serviceability issues that must be addressed directly, deflection and vibration. With today’s high strength steels, deflection criteria may often control designs for typical building spans and loads. A third issue is overall building drift, where the beams in moment frames play a significant role in the response.

6.14.1

Deflection

Deflection is the normal response of a beam to its imposed load. It is impossible to erect a beam with zero deflection under load, but the designer will be able to limit that deflection with proper attention to this limit state. Deflections must be addressed for a variety of loading cases. Deflection under dead load is critical because it impacts the construction process, including the amount of concrete fill needed to form a flat and level floor. Live load deflection is critical because it impacts the finishes of elements attached to the floor, such as ceilings and walls, and may be visible to the occupants. Experience has demonstrated that live load deflection is not a problem if it is limited to 1/360 of the span. Dead load deflection limitations are a function of the particular structural element and loading. AISC Design Guide 3: Serviceability Design Considerations for Steel Buildings covers deflection and other serviceability design criteria. 6.14.2

Vibration

Although vibration of floor systems is not a safety consideration, it can be a very annoying response and very difficult to correct after the building is erected. The most common problem occurs with wide-open spaces with very little damping, such as the jewelry department in a department store. To reduce the risk of annoyance, a general rule is to space the beams or joists sufficiently far apart so that the slab thickness is large enough to provide the needed stiffness and damping. AISC Design Guide 11: Vibration of Steel Framed Structural Systems Due to Human Activity covers the design of steel-framed floor systems for human comfort. 6.14.3

Drift

Under lateral loading, a building will sway sideways. This lateral displacement is called drift. As with deflection and vibration, drift is usually not a safety consideration, but it can be annoying and have a negative impact on nonstructural elements, causing cracks in finishes. Beams and girders are important in reducing the drift, and their final size might actually be determined by drift considerations. However, the impact of drift considerations on beams cannot be determined without also evaluating the other parts of the lateral load–resisting system. This serviceability limit state is treated in Chapter 8. Drift is also discussed in Design Guide 3. Because beam deflection is a serviceability consideration, calculations are carried out using the specific loads under which the serviceability considerations are to be checked. This can be live load, deal load, or some combination of loads, but normally does not include any load factors. Thus, regardless of whether a design is completed using LRFD or ASD, serviceability considerations are checked for the same loads. Numerous elastic analysis techniques are available to determine the maximum deflection of a given beam and loading. Some common loading conditions with their corresponding maximum deflections are shown in Figure 6.28. These and many others are given in Manual Table 3-23.

Chapter 6

Figure 6.28 6 Deflectio ons

Bending Mem mbers 249

Some Common C Loading Conditioons with Theiir Correspondding Maximum m

EXAMPLE E 6.19 Live Load Deflection

Goal: G

Ch heck the live load deflectioon of a previoously designedd beam.

Given: G

Usse the inform mation from Example 6.33, where a W W18×35 was selected. Liimit the live lo oad deflectionn for an accepptable designn to 1/360 of thhe span.

SOLUTION N

Step S 1:

Co ollect the requ uired informaation from Exxample 6.3. Fo or the W18×3 35, I = 510 inn.4. The live looad is 24 kips applied at thhe center off a 20 ft span.

Step S 2:

Deetermine the live load defflection. Usinng the deflecttion equation found in Figure 6.28 for case (b),

24 ( 20.0) (12) PL3 = 0.4667 in. = 48EI 48 ( 29,000))( 510) 3

Δ= Step S 3:

3

Co ompare the caalculated defllection to the given limit. Th he deflection limit is

20.0(12) = 0.667in . 360 Beecause the callculated defleection is less tthan the defleection limit, Δ = 0.4467 in. < Δ mmax = 0.667 in. Δ maax =

Th he deflection satisfies the sset criteria

250 Chapter 6

Bending Members

EXAMPLE 6.20 Beam Design through Deflection Limit

Goal:

Select a W-shape to satisfy a live load deflection limit.

Given:

Use the data from Example 6.19, except that the deflection limit is set to a more severe level of 1/1000 of the span. If the selected member does not meet the established criteria, select a W-shape that satisfies the limitation.

SOLUTION

Step 1:

Check the new deflection limit.

Δmax =

20.0 (12) 1000

= 0.240 in.

From Example 6.19 we know already that the given beam deflects too much. Step 2

Determine the minimum acceptable moment of inertia, Imin, necessary to ensure that the deflection does not exceed the given limit. Rearranging the maximum deflection for a concentrated load at mid-span equation to solve for Imin,

24 ( 20.0) (12) PL3 = = = 993 in.4 48E Δmax 48 ( 29,000)( 0.240) 3

I min Step 3:

3

Select a beam with I ≥ 993 in.4 and, from Example 6.3, one that satisfies the strength limit, Z ≥ 64.0 in.3 . From the moment of inertia tables, Manual Table 3-3, select a W21×55, I = 1140 in.4 and Z = 126 in.3 This is the lightest-weight W-shape that will satisfy the required moment of inertia.

6.15

CONCENTRATED FORCES ON BEAMS Before a beam can be called upon to carry a given load, that load must be transferred to the beam through some type of connection. In a similar manner, the beam reactions must be carried to the beam’s supporting structure through some type of connection. Although the majority of beams are loaded through connections to their webs, some may be loaded by applying a concentrated force to the top flange, and some will have their reactions resisted by bearing on a supporting element. In these cases, a check must be made to establish that the beam web has sufficient strength to resist the applied forces. Four limit states determine the load carrying strength of the web to resist these concentrated forces: web local yielding, web local crippling, web sidesway buckling, and flange local bending. These limit states are all described in Section J10 of the Specification. Although it is possible to select a beam with a web sufficiently thick that these limit states do not control, it is normally more economical to add bearing stiffeners under the concentrated loads to provide the necessary strength. For the limit states of web local yielding and web local crippling, it may be easiest to increase the length of bearing, which would eliminate the requirement for bearing stiffeners. However, if the web sidesway buckling limit state is exceeded, stiffeners are required. In applications where W-shapes are used, this limit state is particularly critical for continuous

Chapter 6

Bending Mem mbers 251

Figure 6.29 6 Single Concentrated C Force Applieed to a Beam beams in n the negativee moment region where tthe bottom flange is the ccompression flange. The limit statte of flange lo ocal bending is i a concern oonly when tennsion loads arre applied to beams. The design of stiffeners iss covered in Section 7.4 oof this book uunder the discussion of pllate girders, because they are mucch more comm monly foundd in that appliication. The llimit states of web local yielding and web locaal crippling will w be discusssed here withh an eye towaard using beaaring length as the controlling factor in resisting g these limit sstates. 6.15.1 Web Local Yieldiing When a single conceentrated forcee is applied tto a beam ass shown in F Figure 6.29, tthe force is assumed to be deliverred to the beaam over a lenggth of bearingg, lb. It is thenn distributed through the flange an nd into the web. w The narro owest portionn of the web is the criticaal section. This occurs at the toe of o the web-to--flange fillet, dimension k in Figure 6.229a. There arre two variablles listed in Manual Table 1-1 forr this dimenssion, kdes and kdet. In this aapplication kdes is to be used since it representts the lower bound b on the actual k dim mension foundd in the produuction of W-sshapes. The distributiion of the forrce along thee web takes pplace at a sloope of 1:2.5. Thus, when the critical section iss reached, thee force has beeen distribute d over a lenggth of lb plus 22.5k in each ddirection. If the conceentrated forcee is applied so that the forrce is distribuuted along thee web in bothh directions, this distrribution increeases the bearring length bby 5k as show wn in Figure 6.29b. If thee bearing is close to the end of the member, distributionn takes placee only in onne direction, toward the midspan.. The Specificcation definess “close to thee member ennd” as being w within the meember depth from the end. Thus, th he available leength of the w web is (lb + 2..5k), as shownn in Figure 6.29c. The T nominal strength of the beam weeb when the concentratedd force to bee resisted is applied at a a distance from f the mem mber end that iis greater thann the depth off the memberr, d, is (A AISC J10-2) Rn = Fyw t w ( 5k + lb ) When the concentrateed force to bee resisted is aapplied at a ddistance from m the memberr end that is less than or equal to th he depth of th he member, d,, the nominal strength is (A AISC J10-3) Rn = Fyw t w ( 2.5k + lb )

252 Chapter 6

Bending Members where Fyw lb k tw

= yield stress of the web = length of bearing = kdes= distance from the outer face of the flange to the web toe of the fillet weld = web thickness

For web local yielding φ = 1.0 (LRFD)

Ω = 1.50 (ASD)

6.15.2 Web local crippling

The criteria for the limit state of web local crippling also depend on the location of the force with respect to the end of the member and the length of bearing, lb. The Specification equations apply to both I-shaped members and HSS. When applied to I-shaped members, Qf = 1.0. When the concentrated compressive force is applied at a distance from the member end that is greater than or equal to d/2, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f 2 Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf (AISC J10-4) tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ When the force is applied at a distance less than d/2 and lb/d ≤ 0.2, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f 2 Rn = 0.40tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣

(AISC J10-5a)

and when lb/d > 0.2, 1.5 ⎡ ⎛ 4l ⎞ ⎛ tw ⎞ ⎤ EFywt f b Rn = 0.40t ⎢1 + ⎜ − 0.2 ⎟ ⎜ ⎟ ⎥ Qf tw ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ ⎝ d 2 w

(AISC J10-5b)

For web local crippling φ = 0.75 (LRFD)

Ω = 2.0 (ASD)

When the beam reaction is transferred to the supporting material, it is often necessary to use a beam bearing plate to spread the load over a sufficiently large area to avoid failure of the supporting material. These beam bearing plates are addressed in Section 11.11.

EXAMPLE 6.21a Beam Bearing Strength by LRFD

SOLUTION

Goal:

Determine the required length of bearing for a W-shape to support (a) a midspan concentrated load and (b) concentrated reactions at the end of a simple span.

Given:

A W18×35 A992 beam was selected in Example 6.3a as a simply supported member spanning 20 ft and carrying a midspan dead load of 8.0 kips and live load of 24.0 kips.

Step 1:

Determine section properties from Manual Table 1-1 d = 17.7 in., tw = 0.300 in., tf = 0.425 in., kdes = 0.827 in.

Step 2:

Determine the required bearing strength at the load point and at the reactions using the LRFD load combinations from Section 2.4.

Chapter 6

Bending Members 253

Pu = 1.2 PD + 1.6 PL = 1.2 ( 8.0 ) + 1.6 ( 24.0 ) = 48.0 kips Ru =

Part a Step 3:

Pu 48 = = 24 kips 2 2

At the load For the limit state of yielding, determine the required length within the web to resist the applied load. The required length, based on Equation J10-2 where N = ( 5k + lb ) , is then

( 5k + lb ) = Step 4:

Pn P φ 48 1.0 = u = = 3.20 in. Fywtw Fywtw 50 ( 0.300 )

Determine the minimum length of bearing, lb. Since this force is located so that distribution takes place in both directions, lb = 3.20 − 5 ( 0.827 ) = −0.935 in. A negative bearing length in this calculation means that the distribution, 5k, is more than enough to carry the required load. Thus, the minimum bearing length for normal practice of 3.0 in. is adequate.

Step 5:

For the limit state of web local crippling, determine the nominal strength if the bearing length is taken as a practical minimum of 3.0 in. From Equation J10-4, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f 2 Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ 1.5 2 ⎡ ⎛ 3.00 ⎞⎛ 0.300 ⎞ ⎤ 29,000(50)(0.425) = 0.80 ( 0.300 ) ⎢1 + 3 ⎜ (1.0 ) ⎟⎜ ⎟ ⎥ 0.300 ⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎦⎥ ⎣⎢

= 134 kips Step 6:

Determine the design strength for web local crippling, φRn = 0.75(134) = 101 kips

Step 7:

For the final selection at mid-span, since Ru = 48 kips < φRn = 101 kips a 3.0 in. bearing length will be sufficient.

Part b Step 8:

At the end reaction Since the end reactions are less than the concentrated load at mid-span, the 3.0 in. minimum bearing length will likely be sufficient for the end reactions also. Thus, check a 3.0 in. bearing. Rn = Fywtw ( 2.5k + lb ) = 50 ( 0.300 ) ( 2.5 ( 0.827 ) + 3.0 ) = 76.0 kips

φRn = 1.0 ( 76.0 ) = 76.0 kips Step 9:

Check the limit state of web local crippling. Since the reaction is at the end of the member at a distance less than d 2 and lb d = 3.0 17.7 = 0.17 ≤ 0.2

254 Chapter 6

Bending Members use Equation J10-5a. 1.5 ⎡ ⎛ l ⎞ ⎛ t ⎞ ⎤ EFywt f Rn = 0.40tw2 ⎢1 + 3 ⎜ b ⎟ ⎜ w ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ 1.5 ⎡ ⎛ 3.0 ⎞⎛ 0.300 ⎞ ⎤ 29,000 ( 50 )( 0.425) = 0.40 ( 0.300 ) ⎢1 + 3 ⎜ (1.0) ⎟⎜ ⎟ ⎥ 0.300 ⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎦⎥ ⎣⎢ 2

= 67.2 kips So,

φRn = 0.75 ( 67.2 ) = 50.4 kips

Step 10:

Determine the strength for the end reaction. Since web local crippling controls, φRn = 50.4 ≥ 24.0 kips

Step 11:

Final selection. Use a minimum bearing plate length along the beam web of 3.0 in. at the concentrated load and at the end reactions.

EXAMPLE 6.21b Beam Bearing Strength by ASD

SOLUTION

Goal:

Determine the required length of bearing for a W-shape to support (a) a midspan concentrated load and (b) concentrated reactions at the end of a simple span.

Given:

A W18×35 A992 beam was selected in Example 6.3b, as a simply supported member spanning 20 ft and carrying a midspan dead load of 8.0 kips and live load of 24.0 kips.

Step 1:

Determine section properties from Manual Table 1-1 d = 17.7 in., tw = 0.300 in., tf = 0.425 in., kdes = 0.827 in.

Step 2:

Determine the required bearing strength at the load point and at the reactions using the ASD load combinations from Section 2.4. Pa = PD + PL = 8.0 + 24.0 = 32.0 kips P 32 = 16 kips Ra = a = 2 2

Part a Step 3:

At the load For the limit state of yielding, determine the required length within the web to resist the applied load. The required length, based on Equation J10-2 where N = ( 5k + lb ) , is then

( 5k + lb ) =

1.5 ( 32 ) Pn ΩPa = = = 3.20 in. Fywt w Fy t w 50 ( 0.300 )

Chapter 6 Step 4:

Bending Members 255

Determine the minimum length of bearing, lb. Since this force is located so that distribution takes place in both directions, lb = 3.20 − 5 ( 0.827 ) = −0.935 in. A negative bearing length in this calculation means that the distribution, 5k, is more than enough to carry the required load. Thus, the minimum bearing length for normal practice of 3.0 in. is adequate.

Step 5:

For the limit state of web local crippling, determine the nominal strength if the bearing length is taken as a practical minimum of 3.0 in. From Equation J10-4, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFy t f 2 Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ 1.5 ⎡ ⎛ 3.00 ⎞⎛ 0.300 ⎞ ⎤ 29,000(50)(0.425) = 0.80 ( 0.300 ) ⎢1 + 3 ⎜ (1.0 ) ⎟⎜ ⎟ ⎥ 0.300 ⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎥⎦ ⎢⎣ = 134 kips Determine the allowable strength for web local crippling, Rn 134 = = 67.0 kips Ω 2.00 2

Step 6:

Step 7:

For the final selection at mid-span, since Ra = 32 kips < Rn Ω = 67 kips a 3.0 in. bearing length will be sufficient.

Part b Step 8:

Step 9:

At the end reaction Since the end reactions are less than the concentrated load at mid-span, the 3.0 in. minimum bearing length will likely be sufficient for the end reactions also. Thus, check a 3.0 in. bearing. Rn = Fywtw ( 2.5k + lb ) = 50 ( 0.300 ) ( 2.5 ( 0.827 ) + 3.0 ) = 76.0 kips Rn Ω = 76.0 1.5 = 50.7 kips Check the limit state of web local crippling. Since the reaction is at the end of the member at a distance less than d 2 and lb d = 3.0 17.7 = 0.17 ≤ 0.2 use Equation J10-5a. 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f Rn = 0.40t ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ 2 w

1.5 2 ⎡ ⎛ 3.0 ⎞⎛ 0.300 ⎞ ⎤ 29,000 ( 50 )( 0.425) = 0.40 ( 0.300 ) ⎢1 + 3 ⎜ (1.0) ⎟⎜ ⎟ ⎥ 0.300 ⎝ 17.7 ⎠⎝ 0.425 ⎠ ⎥⎦ ⎢⎣ = 67.2 kips

So, Rn Ω = 67.2 2.0 = 33.6 kips

Step 10:

Determine the strength for the end reaction. Since web local crippling

2556 Chapter 6

Bending Members M contro ols, Rn Ω = 33.6 >16.00 kips

Step p 11:

Final selection Use a minimum bearing b plate length along the beam weeb of 3.0 in. at the conceentrated load and at the endd reactions.

6.116

OPEN N WEB STE EEL JOISTS S AND JOIS ST GIRDER RS The term op pen web steel joist j refers to o a building pproduct made according too the design sttandards of the Steel Joist Institutte (SJI). They y are manufacctured trussess regularly ussed for buildiing floor and roof sysstems. An ex xample of their use in a buuilding is shoown in Figuree 6.30. As a bbuilding product, thee structural en ngineer is nott involved inn actually dessigning the truuss but ratherr selects the product from a table,, much like seelecting a parrticular W-shhape. Althouggh these mem mbers are made from steel, they are a not desig gned accordinng to any A AISC standarrds but ratherr to the standards fo ound in the SJJI publication n Standard Sppecifications,, Load Tabless, and Weighht Tables for Steel Joiists and Joist Girders (2010). Som me advantagess of open weeb steel joistss include the fact that theyy are lightweight and can easily span s long disstances, theirr open webs can easily aaccommodatee passing mechanical systems thro ough the stru ucture, and in n some appli cations they are more ecoonomical thaan rolled steel shapess. Disadvantaages include a lower loaad carrying ccapacity thann rolled shappes, thus requiring mu uch closer sp pacing; the in nability to eassily accommoodate concenttrated loads aat points other than trruss panel points; and poteential vibratioon issues wheen they are ussed for floor ssystems. In all appliccations, the economic e adv vantages or ddisadvantages of using an open web stteel joist must be asseessed for the specific s cond ditions in quesstion. Fou ur types of op pen web steell joists are ddefined in SJII standards: K K-series, KCS S-series, LH-series an nd DLH-seriees. K-seeries joists arre perhaps th he most comm monly used jjoists for flooor and roof ssystems. They are available in dep pths from 10 in. to 30 in., and design taables are available coverinng spans up to 60 ft. ft The standard designatiion for K-seeries joists iss 16K6, wheere the first number represents th he depth, 16 in.; the letterr indicates thhe series desiignation, K; aand the third number represents place p within th he series, 6. The T series deesignation ideentifies the deetails of mannufacture of the trusss, including the t sizes of the t elements that make uup the truss. It is here thhat each manufactureer has the opp portunity to deesign its own specific joistt. For a 16K joist there are seven

n of Open Web b Steel Joists Figure 6.30 Application Photo courtessy Douglas Steeel Fabricating Corporation

Chapter 6

Bending Mem mbers 257

Figure 6.31 6 LRFD Economy E Tab ble for K-Seri es Open Webb Steel Joists Standard Specifications,, Load Tables, and Weight Ta Tables for Steel Joists and Joisst Girders (20110), © Steel Jo oist Institute. Reprinted R with Permission. A All rights reservved.

258 Chapter 6

Bending Members designations within the series: 2, 3, 4, 5, 6, 7, and 9. Each increase in designation indicates an increase in weight. For example, the seven designations of the 16K indicate joists that weigh approximately 5.5, 6.3, 7.0, 7.5, 8.1, 8.6, and 10.0 lb/ft. Weight is given as an approximate value since each manufacturer has the opportunity to produce joists with different cross section elements. The one constant, however, is that whatever joist is produced, it must have the strength indicated in the SJI standard. Figure 6.31 shows an example of the selection tables published by SJI. K-series joists are designed to carry a uniformly distributed load. Thus, the bending strength can vary along the span from a maximum at midspan to zero at the support, approximating a parabolic moment diagram, and the shear strength can vary from zero at midspan to a maximum at the support, approximating a triangular shear diagram. Design tables like those in Figure 6.31 give two numbers for each joist and span combination. Joist strength in terms of load per foot of span is given as the upper number, and load per foot of span to cause a deflection of 1 360 of the span is given as the lower number. KCS-series joists are identified in the same fashion as the K-series joists. Thus, a 10KCS3 is a 10 in. deep KCS-series joist with a series designation of 3. The difference between K- and KCS-series joists is directly related to the moment and shear diagrams. KCS-series joists are designed for a diagram for uniform moment over all interior panel points and a constant shear. These joists are particularly useful for supporting loads that combine uniformly distributed and concentrated forces. Unlike the K-series joists, the design tables for the KCS-series give only the moment and shear capacities. LH-series joists are long-span joists. Their strengths are tabulated for spans from 25 ft to 96 ft. LH-series joists have depths from 18 in. to 48 in. A typical designation would be 28LH05. LH-series joists are used for floor and roof systems, and, like K-series joists, they are designed for a uniformly distributed load. DLH-series joists are deep long-span joists intended primarily to support roof decks. DLH-series joists start at a depth of 52 in. and go up to 120 in. The design tables indicate spans from 62 ft up to 240 ft. The designation system is the same as for other joists; a 68DLH17 is a 68 in. deep DLH-series joist with a series designation of 17. Another product that is designed according to the SJI standards is the joist girder. Joist girders are essentially pre-engineered trusses that are intended to support concentrated loads at the panel points. They are used to support open web steel joists that are evenly spaced and introduce the same load at each panel point. A possible designation for a joist girder is 44G8N12K. The first number indicates a 44 in. deep member and the G indicates a joist girder. The 8N indicates that there are eight joist spaces, which is the same as saying there are seven concentrated loads on the girder. The final number indicates the load in kips. If the load is for an LRFD design the last letter would be F, indicating a factored load, and the load magnitude would be the LRFD required strength.

EXAMPLE 6.22 Open Web Steel Joist Selection by LRFD

Goal:

Select a K-series open web steel joist from the limited selection available in Figure 6.31 to satisfy strength and deflection criteria.

Given:

Joists supporting a roof deck span 30 ft and are spaced 6.0 ft on center. The load is a uniformly distributed dead load of 20 psf and live load of 30 psf. Select the lightest 18 in. K-series open web steel joist available in Figure 6.31 to support this load. Limit deflection to 1/360 of the span.

SOLUTION

Step 1:

Determine the required load in pounds per foot.

Step 2:

Select a K-series joist for a span of 30 ft.

wu = (1.2wD + 1.6wL ) Ltrib = (1.2 ( 20) + 1.6 ( 30) ) ( 6) = 432 1b/ft

Chapter 6

Bending Members 259

An 18K6 will support a factored load of 451 lb/ft. This is the upper number in the table. The other number in the table is 175 lb/ft. This is the serviceability load that will produce a deflection of 1/360 of the span. Step 3:

Check the deflection of the 18K6 for the serviceability live load of 30 psf. For the 6.0 ft joist spacing, the joist carries a serviceability live load of 6(30)=180 lb/ft. Since this exceeds the load given in the table, the next larger joist should be selected if this deflection limit is to be satisfied.

Step 4:

Check the 18K7. For strength the capacity is 502 lb/ft > 432 lb/ft. For deflection the capacity is 194 lb/ft > 180 lb/ft.

Step 5:

Check to be sure the joist can carry its self-weight. The joist weighs approximately 9.0 lb/ft. The factored dead load due to self-weight is wself = 1.2(9.0) = 10.8 lb/ft Thus, the total load the joist must carry is

wu = 432 + 10.8 = 443 lb/ft < 502 lb/ft Step 6:

Final selection: 18K7

6.17

PROBLEMS

1. Determine the elastic section modulus and the plastic section modulus for a W40×199 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual.

5. Determine the elastic section modulus and the plastic section modulus for a W18×35 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual.

2. Determine the elastic section modulus and the plastic section modulus for a W36×330 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual.

6. Determine the plastic section modulus for a W44×230 modeled as three rectangles forming the flanges and the web. Compare the calculated value to that given in the Manual.

3. Determine the elastic section modulus and the plastic section modulus for a W33×130 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual.

7. Determine the plastic section modulus for a W27×84 modeled as three rectangles forming the flanges and the web. Compare the calculated value to that given in the Manual.

4. Determine the elastic section modulus and the plastic section modulus for a W24×55 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual.

8. Determine the plastic section modulus for a W16×31 modeled as three rectangles forming the flanges and the web. Compare the calculated value to that given in the Manual.

260 Chapter 6

Bending Members

9. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a WT15×45 modeled as two rectangles forming the flange and the stem. Compare the calculated values to those given in the Manual.

18. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a MC13×50 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual.

10. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a WT8×50 modeled as two rectangles forming the flange and the stem. Compare the calculated values to those given in the Manual.

19. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus, all about the geometric axis, for a L4×4×1/2 modeled as two rectangles. Compare the calculated values to those given in the Manual.

11. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a WT5×22.5 modeled as two rectangles forming the flange and the stem. Compare the calculated values to those given in the Manual.

20. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus, all about the geometric axis, for a L8×8×7/8 modeled as two rectangles. Compare the calculated values to those given in the Manual.

12. Determine the plastic section modulus for an HSS8×6×1/2 modeled as four rectangles forming the flanges and webs. Remember to use the design wall thickness for the plate thickness and ignore the corner radius. Compare the calculated value to that given in the Manual.

21. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus, all about the geometric axis, for a L3×3×1/4 modeled as two rectangles. Compare the calculated values to those given in the Manual.

13. Determine the plastic section modulus for an HSS12×8×5/8 modeled as four rectangles forming the flanges and webs. Remember to use the design wall thickness for the plate thickness and ignore the corner radius. Compare the calculated value to that given in the Manual.

22. A simply supported beam spans 25 ft and carries a uniformly distributed dead load of 0.6 kip/ft, including the beam self-weight, and a live load of 2.1 kip/ft. Determine the minimum required plastic section modulus and select the lightest-weight W-shape to carry the moment. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD.

14. Determine the plastic section modulus for an HSS8×8×1/2 modeled as four rectangles forming the flanges and webs. Remember to use the design wall thickness for the plate thickness and ignore the corner radius. Compare the calculated value to that given in the Manual.

23. For flexure only, determine the lightest-weight Wshape to carry a uniform dead load of 1.2 kip/ft, including the beam self-weight, and a live load of 3.2 kip/ft on a simple span of 20 ft. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD.

15. Determine the plastic section modulus for a round HSS10×0.500. Remember to use the design wall thickness in your calculations. Compare the calculated value to that given in the Manual.

24. A beam is required to carry a uniform dead load of 0.80 kip/ft, including its self-weight, and a concentrated live load of 14 kips at the center of a 30 ft span. For bending only, determine the least-weight W-shape to carry the load. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD.

16. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a C15×40 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual. 17. Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a C8×11.5 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual.

25. A simply supported beam spans 32 ft and carries a uniformly distributed dead load of 1.8 kip/ft plus the beam self-weight and a live load of 5.4 kip/ft. Determine the minimum required plastic section modulus and select the lightest-weight W-shape to carry the moment. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD.

Chapter 6 26. For bending only, determine the lightest-weight Wshape to carry a uniform dead load of 4.5 kip/ft plus the beam self-weight and a live load of 3.5 kip/ft on a simple span of 24 ft. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD.

Bending Members 261

and at the load point. Use A992 steel and Cb = 1.0. Design for flexure by (a) LRFD and (b) ASD.

27. A beam is required to carry a uniform dead load of 3.4 kip/ft plus its self-weight, and a concentrated dead load of 12 kips and a concentrated live load of 20 kips, both at the center of a 40 ft span. For bending only, determine the least-weight W-shape to carry the load. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD.

34. A 36 ft simply supported beam is loaded with concentrated loads 16 ft in from each support. On one end, the dead load is 8.0 kips and the live load is 18.0 kips. At the other end, the dead load is 4.0 kips and the live load is 9.0 kips. Include the self-weight of the beam in the design. Lateral supports are provided at the supports and the load points. Considering only bending, determine the least-weight W-shape to carry the load. Use A992 steel and Cb = 1.0. Design by (a) LRFD and (b) ASD.

28. Considering only bending, determine the lightestweight W-shape to carry the following loads: a uniform dead load of 0.6 kip/ft plus self-weight, a concentrated dead load of 2.1 kips, and a concentrated live load of 6.4 kips, located at the center of a 16-ft span. Assume full lateral support and A992 steel. Design by (a) LRFD and (b) ASD.

35. An 18 ft simple span beam is loaded with a uniform dead load of 1.4 kip/ft, including the beam self-weight, and a uniform live load of 2.3 kip/ft. The lateral supports are located at the supports and at 6.0 ft intervals. Considering only bending, determine the least-weight Wshape to carry the load. Use A992 steel and Cb = 1.0. Design by (a) LRFD and (b) ASD.

29. Considering only bending, determine the lightest Wshape to carry a uniform dead load of 4.0 kip/ft plus the self-weight and a uniform live load of 2.3 kip/ft on a simple span of 10.0 ft. Assume full lateral support and A992 steel. Design by (a) LRFD and (b) ASD.

36. A 40 ft simple span beam is loaded with a uniform dead load of 2.2 kip/ft plus the beam self-weight and a uniform live load of 3.6 kip/ft. The lateral supports are located at the supports and at the midpoint of the span. Considering only bending, determine the least-weight Wshape to carry the load. Use A992 steel and Cb = 1.0. Design by (a) LRFD and (b) ASD.

30. A 24 ft simple span laterally supported beam is required to carry a total uniformly distributed service load of 8.0 k/ft. Determine the lightest, A992, W-shape to carry this load, considering only flexure and the limit state of yielding, if the load is broken down as follows. Use LRFD. a. Live load = 1.0 k/ft; dead load = 7.0 kips/ft b. Live load = 3.0 k/ft; dead load = 5.0 kips/ft c. Live load = 5.0 k/ft; dead load = 3.0 kips/ft d. Live load = 7.0 k/ft; dead load = 1.0 kips/ft. 31. Repeat the designs specified in Problem 30 using ASD. 32. A 30 ft simply supported beam is loaded at the third points of the span with concentrated dead loads of 10.0 kips and live loads of 15.0 kips. Lateral supports are provided at the supports and at the load points. The selfweight of the beam will be ignored for this problem. Considering only flexure, determine the least-weight Wshape to carry the load. Use A992 steel and Cb = 1.0. Design by (a) LRFD and (b) ASD. 33. Determine the least-weight W-shape to support a concentrated dead load of 24 kips plus the beam selfweight and a concentrated live load of 15 kips on a 30 ft span. The concentrated loads are located at the midpoint of the span. Lateral supports are provided at the supports

37. An A992 W18×60 is used on a 36 ft simple span to carry a uniformly distributed load. Using Cb = 1.0, determine the locations of lateral supports in order to provide just enough strength to carry (a) a design moment of 435 ft-kips and (b) an allowable moment of 290 ft-kips. 38. An A992 W16×31 is used on an 18 ft simple span to carry a uniformly distributed dead load of 0.95 kips/ft plus self-weight and a live load of 1.9 kips/ft. Using Cb = 1.0, determine the locations of lateral supports in order to provide just enough strength to carry this loading by (a) LRFD and (b) ASD. 39. Redesign the beam of Problem 33 using the correct value of Cb. 40. Redesign the beam of Problem 34 using the correct value of Cb. 41. Redesign the beam of Problem 35 using the correct value of Cb. 42. There are no tables in the Manual for the design of beams with Fy other than 50 ksi. Construct the moment strength vs. unbraced length curve, similar to those in Manual Table 3-10, for a W21×68 A572 Grade 65 steel. Show the curve up to an unbraced length of 40 ft.

262 Chapter 6

Bending Members

43. Since the chart developed for Problem 42 was similar to those in Manual Table 3-10, it used Cb = 1.0. Redo Problem 42 with values of Cb = 1.0, 1.14, 1.32, and 1.67. 44. Plot the nominal moment strength vs. unbraced length for a W27×84 using A992 and A913 Grade 65 steel. Compare the nominal moment strengths for an unbraced length of 18 ft. 45. Plot the nominal moment strength vs. unbraced length for a W16×31 using A36 and A992 steel. Compare the nominal moment strengths for an unbraced length of 7.0 ft. 46. Determine the available shear strength of a W24×62 A992 steel member by (a) LRFD and (b) ASD. 47. Determine the available shear strength of a W30×90 A992 steel member by (a) LRFD and (b) ASD. 48. A girder that carries a uniformly distributed dead load of 1.7 k/ft plus its self-weight and three concentrated live loads of 15 kips at the quarter points of the 36 ft span is to be sized. Using A992 steel, determine the lightest Wshape to carry the load with lateral supports provided at the supports and load points. Use the correct Cb, check shear, and limit deflection to 1/360 of span. Design by (a) LRFD and (b) ASD. 49. A 32 ft simple span beam carries a uniform dead load of 2.4 k/ft plus its self-weight and a uniform live load of 3.0 k/ft. The beam is laterally supported at the supports only. Determine the minimum-weight W-shape to carry the load using A992 steel. Use the correct Cb, check shear, and limit live load deflection to 1/360 of span. Design by (a) LRFD and (b) ASD. 50. A 36 ft simple span beam carries a uniformly distributed dead load of 3.4 kip/ft plus its self-weight and a uniformly distributed live load of 2.4 kip/ft. Determine the least-weight W-shape to carry the load. Use the correct Cb, check shear, and limit live load deflection to 1/360 of the span. Use A992 steel and assume full lateral support. Design by (a) LRFD and (b) ASD. 51. A simple span beam with a uniformly distributed dead load of 1.1 k/ft, including the self-weight, and concentrated dead loads of 3.4 kips and live loads of 6.0 kips at the third points of a 24-ft span, is to be designed with lateral supports at the third points and live load deflection limited to 1/360 of the span. Use the correct Cb, and be sure to check shear. Determine the least-weight Wshape to carry the loads. Use A992 steel. Design by (a) LRFD and (b) ASD. 52. A fixed-end beam on a 28 ft span is required to carry a total ultimate uniformly distributed load of 32.0 kips.

Using plastic analysis and A992 steel, determine the design moment and select the lightest W-shape. Assume (a) full lateral support and (b) lateral support at the ends and center line. 53. A beam is fixed at one support and simply supported at the other. A concentrated ultimate load of 32.0 kips is applied at the center of the 40 ft span. Using plastic analysis and A992 steel, determine the lightest W-shape to carry the load when the nominal depth of the beam is limited to 18 in. Assume (a) full lateral support and (b) lateral supports at the ends and the load. 54. A fixed-end beam on a 40 ft span is required to carry a total ultimate uniformly distributed load of 72.5 kips. Using plastic analysis and A992 steel, determine the lightest-weight W-shape to carry the load. Assume full lateral support. 55. A three-span continuous beam is to be selected to carry a uniformly distributed dead load of 4.7 kip/ft, including its self-weight, and a uniformly distributed live load of 10.5 kip/ft. Be sure to check the shear strength of this beam. Use A992 steel and assume full lateral support. Design using moment redistribution by (a) LRFD and (b) ASD. 56. Determine the available minor axis bending strength of a W18×130 of A992 steel. Determine by (a) LRFD and (b) ASD. 57. Determine the available minor axis bending strength of a W10×12 of A992 steel. Determine by (a) LRFD and (b) ASD. 58. Determine the available minor axis bending strength of a W40×392 of A992 steel. Determine by (a) LRFD and (b) ASD. 59. Determine the available bending strength of a WT8×25, A992 steel, if the stem is in compression and the unbraced length is 10 ft. Determine by (a) LRFD and (b) ASD. 60. Determine the available bending strength of a WT12×52, A992 steel, if the stem is in tension and the unbraced length is 8 ft. Determine by (a) LRFD and (b) ASD. 61. For the beam designed in Problem 34, determine the minimum required lengths of bearing (at the concentrated loads and at the supports) considering the limit state of web yielding and the limit state of web local crippling. Determine by (a) LRFD and (b) ASD. 62. For the girder designed in Problem 48, determine the minimum required lengths of bearing (at the concentrated

C Chapter 6 looads and at th he supports) considering thee limit state off w web yielding and a the limit state s of web lo ocal crippling. D Determine by (a) LRFD and (b) ( ASD.

663. Obtain via the Internet a set of open web w steel joistt ddesign tables from any manufacturerr. Select thee sshallowest K-series joist to span 35 ft and support a rooff ssystem that carries a dead loaad of 25 psf and d a live load off 335 psf. The joiists will be spaaced (i) 6.0 ft on center, (ii)) 88.0 ft on centerr, and (iii) 10.0 0 ft on center. Limit L live load d ddeflection to sp pan/360. Design n by (a) LRFD D and (b) ASD. M Multi-Chapterr Questions 664. Using thee framing plaan shown in Figure P6.64 4 ((presented earliier as Figure 2..7), design the beams marked d 11, 2, and 3. Thiis is the same structure s used in Section 2.5,, w where load calculations with h live load reeductions weree ddiscussed. Tho ose calculation ns can be reussed here. Load d ccase 2 for deaad plus live load is to be co onsidered. Thee bbuilding is an office o building g with a nomin nal live load off 550 pounds per square s foot (pssf) and a calcullated dead load d oof 70 psf. 1: Girderr AB on line 2-2 if the flo oor deck spanss from m line 1-1 to 2-2 2 to 3-3: 2: Floor beam 2-3 on line D-D if the t floor deck k span ns from line C-C to D-D to E--E: 3: Floor beam 2-3 on line C-C if the t floor deck k span ns from line B-B to C-C to D--D: D Design by (a) LRFD L and (b) ASD. A

P P6.64 665. The framin ng plan shown in Figure P6.6 65 is the samee aas that shown n in Figure P2 2.20 for an 18-story 1 officee bbuilding. It mu ust support a deead load of 80 0 psf and a livee looad of 50 psff. In all casess, the decking g spans in thee ddirection from m line A toward line E. Determine D thee rrequired momeent and shear strength for the t beams and d ddesign the beaams as requireed below for (a) design by y L LRFD and (b) design by ASD D. The requireed moment and d sshear strength were w determined in Problem 20 of Chapterr 22.

Bending Membbers 263

i. The beam b etween columnn lines 2 and 3 along line D. iii. The girder oon column line 3 between coluumn line E and midwayy between liness D and C. iiii. The beam oon the line bettween lines C and D and column liness 3 and 4.

P6.655 66. IIntegrated Deesign Project. Design the beeams in the graviity-only system m for the loadds determined in Chapter 2. Bee sure to indiicate all assum mptions regardding lateral suppoort. Reconsideer the proposedd framing planns given in Figurres 1.20 and 1.21. For the panel bounded by column lines A, C, 6, and 77, span the beaams in the 30 fft direction and rredesign. Com mpare the resullts with the deesigns done abovve. For open w web steel joists, manufactureer literature incluudes load tablees. These may bbe found on thhe Internet. Selecct open web stteel joists for thhe roof using tthe loading deterrmined in Chappter 2. Use spacing differentt from that sugg ested in Figuure 1.22 if thhat appears too be more approopriate.

C Chapterr 7

P Plate Girder G rs

N New York Univversity Kimm mel Center Photograph P by Fadi Asmar/LERA Consultinng Structural E Engineers

7.11

BACKG GROUND A plate gird der is a bendiing member composed c of individual stteel plates. A Although platee girders are normally y the memberr of choice forr situations inn which the avvailable rolledd shapes are nnot deep or large enough to carry the intended load, there iss no requirem ment that they will always bbe at the deep or larg ge end of the spectrum s of member m sizess. Beams fabrricated from individual steel plates to meet a specific requireement are gen nerally identiffied in the fielld as plate girrders. Platte girders are usually used d in building structures foor special situuations such as those involving veery long span ns or very laarge loads. Peerhaps their m most commoon applicationn is as a transfer gird der, which iss a bending member thatt supports a structure above and perm mits the column spaccing to be chaanged below. They are alsso very comm mon in industrrial structuress for use as crane gird ders and as support for larrge pieces off equipment. IIn commerciaal buildings, they are often used to o span large open o areas to meet particuular architectuural requiremeents; becausee of their normally greater depth and a resulting stiffness, theyy tend to defflect less thann other potenttial long span solutions. An examp ple of a buildiing applicatioon of the platee girder is shoown in Figuree 7.1. The cross section n of a typical plate girder is shown in F Figure 7.2 annd a large platte girder being transp ported on a truck t is show wn in Figure 7.3. Althouggh it is possible to combiine steel plates into numerous n geo ometries, the plate girderss addressed hhere are thosse formed froom three plates, one for f the web and a two for th he flanges. B ecause the w web and flangges of the platte girder are fabricateed from indiv vidual plates, they can be designed witth the web annd flange plattes from the same grade of steel (a ( homogeneo ous plate girdder) or from ddifferent graddes of steel (aa hybrid plate girder)). For hybrid d girders, the flanges are uusually fabriccated with a hhigher grade of steel than that useed in the web b. This takes advantage a of tthe higher strresses that cann be developeed in the flanges, whiich are locateed a greater distance d from m the neutral axis than thee web, resultting in a higher mom ment strengtth contributiion. Hybrid girders aree relatively common inn bridge construction n, though theey are rarely used in builldings. In thee past, hybriid girders haave been included in the AISC Speecification, bu ut they are noot specificallyy addressed iin the currentt edition. y, they are nott discussed in n this book. Accordingly

2664

Chapterr 7

7 Figure 7.1

Applicattion of a Platee Girder

Figure 7.2 7

Typical Plate P Girder Definitions D

Figure 7.3 7

Plate Girrder Being Trransported

Photo cou urtesy LeJeune Steel Co.

Plate Giirders 265

2666 Chapter 7

Plate Gird ders

Fig gure 7.4

Siingly Symmettric Plate Girdder

Ano other type of plate girder is the singly ssymmetric girrder, one withh flanges thatt are not of the same size, as seen in Figure 7.4 4. Although ssingly symmeetric plate girrders are addrressed in the AISC Sp pecification, they are not particularly common in bbuildings andd are not speecifically addressed heere. Howeverr, the principlles for all of tthese plate girrders are the same, and thee careful application of o the Specifiication provissions will leadd to an econoomical and saafe design forr each of them. Buillt-up I-shapeed members with compacct webs are designed acccording to thhe same provisions as a rolled I-shaped memb bers presenteed in Specifiication Sectioons F2 and F3 and discussed in n Chapter 6 off this book. The discussionn of plate girdders in this chhapter addressses these built-up I-sh hapes with noncompact or slender webss. Table 7.1 lists th he sections off the Specificaation and part rts of the Mannual discussed in this chapter.

7.22

HOMOG GENEOUS PLATE GIIRDERS IN N BENDING G The behavio or of plate girrders will be addressed herre by consideering flexure and shear sepparately. In flexure, a plate girder is considered d in this bookk as either nooncompact or slender accoording to the proportiions of the web. w Flanges can be com mpact, noncom mpact, or slender, and thee flange slenderness is treated thee same as thaat discussed inn Chapter 6. Thus, it is possible, for eexample, for a noncom mpact web plate girder to have h a slendeer flange, poteentially contrrolling the cappacity of the memberr. The design n rules for each type of girder, noncompact web or slender w web, are considered separately. s With h our discusssion limited to o doubly sym mmetric plate girders, the limit states thhat must be considered are compression flang ge yielding, ccompression flange local buckling, web local buckling, an nd lateral-torssional bucklin ng. These are the same lim mit states considered for thhe rolled I-shapes in Chapter C 6 witth the addition n of web locaal buckling. R Remember thaat all W-shapees have Table 7.1 Sections of Specification S and a Parts of M Manual Coveered in This C Chapter Specificatioon B4 F4

Classification of Sections for Lo ocal Bucklingg Otherr I-Shaped Meembers with Compact C or N Noncompact W Webs, Bent abbout Their Major M Axis F5 Doub bly Symmetricc and Singly Symmetric S I- Shaped Mem mbers with Sleender Webs Bent B about Their Major Axis G Desig gn of Memberrs for Shear J4.4 Streng gth of Elemen nts in Compreession J10 Flang ges and Webs with Concen ntrated Forcess

Chapterr 7

Figuree 7.5 Platte Flexurral Strength

Plate Giirders 267

Girder

Nominal

mit state wass not considdered for thoose bending compact webs so the web local buckling lim onal limit statte of tension fflange yieldinng found in Sp Specification S Sections F4 memberss. The additio and F5 can be ignored d, because thee compressionn flange alwaays controls oover the tensioon flange in ymmetric hom mogeneous members. m Platte girders witth noncompacct webs are aaddressed in doubly sy Specifica ation Section F4, and thosse with slendder webs in S Section F5. T The nominal strength of plate gird ders, for all limit l states, can c be describbed as shownn in Figure 7.5. As with tthe rolled Ishaped members m disccussed in Chaapter 6, the bbehavior is pllastic, inelastiic, or elastic. Figure 7.5 shows th hat the plastiic behavior corresponds c to an area oof the figuree described aas compact. Inelastic behavior co orresponds to o the area iidentified as noncompact, and elastiic behavior correspon nds to the areea identified as a slender. Applying A the same figuree to the lateraal-torsional bbuckling limiit state, the ffully braced region co orresponds to o plastic behaavior, whereass the partiallyy braced region corresponnds to either inelastic or elastic bucckling. The T flexural design d strengtth (LRFD) annd allowable sstrength (ASD D) are determ mined just as they were for the flexu ural memberss discussed inn Chapter 6 w with φ = 0.9 0 (LRFD) Ω = 1.67 (ASD) The requ uirement for ASD A is Ra ≤

The requ uirement for LRFD L is

7.2.1

Rn Ω

Ru ≤ φRn

(A AISC B3-2)

(A AISC B3-1)

Nonccompact Weeb Plate Gird ders uence of web slenderness on o the strengtth of plate girrders is not trreated as a separate limit The influ state to be b assessed th hrough its ow wn set of requuirements. Ratther, web slennderness is trreated as an effect on n the flange yielding y or flaange local buuckling strenggth and the laateral-torsionnal buckling strength as provided in n Specificatio on Section F44. The T slenderneess parameterr of the web iis defined as λw = hc/tw. Foor the doublyy symmetric plate gird der, case 15 in i Specification Table B4..1b, this can bbe simplifiedd to λw = h/tw, where h is the clear distance betw ween the flan nges. Throughhout this chap apter, hc in alll Specification equations can be reeplaced by h.. Similarly Sxc can be repl aced by S , and Myc can bbe replaced byy My. For a x x a plate gird der to be nonccompact, the following reqquirements arre set:

2668 Chapter 7

Plate Gird ders λ pw < λ w ≤ λ rw where, accorrding to Table B4.1b of the Specificatioon,

λ pw = 3.766

E Fy

λ rw = 5.700

E Fy

and

The influence off the noncom mpact web iss characterizeed through thhe web plastification factor, Rpc. This T factor is used to assess the abilityy of the sectioon to reach itss full plastic capacity and modifiees the flange yielding y or lo ocal buckling and lateral toorsional buckkling limit staates. The web plastificcation factor for a noncom mpact web is ggiven in the Sp Specification aas ⎡ Mp ⎛ Mp ⎞⎛ λ − λ pw ⎞ ⎤ M p −⎜ − 1 ⎟⎜ R pc = ⎢ C F4-9b) (AISC ⎟⎥ ≤ ⎢⎣ M yc ⎝ M yc ⎠⎝ λ rw − λ pw ⎠ ⎦⎥ M yc where M p = Fy Z x ≤ 1.6 Fy S x

and M yc = Fy S xc

Equ uation F4-9b is i shown in Figure F 7.6 foor two values of Mp/Myc, oone with a m maximum value of 1.6 6 to account for f the upper limit on Mp, and one arbittrarily taken aat 1.2 as an eexample. The ratio Mp/Myc is the shape factor that was disscussed in Chhapter 6. As was the casee in that discussion, it i must be lim mited to 1.6 in order to ennsure that the necessary rootation can takke place before strain n hardening occurs o as the section underrgoes plastic deformation. The minimuum Rpc is seen to be 1.0, 1 regardless of Mp/Myc. Thus, a consservative apprroach would be to take Rppc = 1.0. This is what the user note in Secttion F4 is im mplying wheen it says thhat Section F5 may conservativeely be used fo or shapes thatt actually fall under the reqquirements off Section F4. Because a plate gird der with a weeb that is on nly slightly nnoncompact w would have ssignificant addditional strength refl flected throug gh the use off Rpc, and beccause the calcculation of Rpc is not parrticularly difficult, th here is no ad dvantage to this simplifiication. Thuss, Equation F4-9b will bbe used throughout this t chapter, as a appropriatee.

Figure 7.6 Factor

Web Plastificcation

Chapterr 7

Figure 7.7 7

Plate Giirders 269

Nominall Flexural Streength Based oon Flange Loocal Buckling

Flange Local L Bucklin ng The requ uirements for compression n flange local buckling aree given in Section F4.3. T The strength of a nonccompact web plate girder as a a function of flange slennderness is shhown in Figure 7.7.For a compact flange girderr, the impact of the nonco mpact web iss to modify thhe strength byy the factor Rpc, as giiven in Equatiion F4-1. (A AISC F4-1) M n = R pc M yc At A the lower limit for a noncompact web, when the web is jjust barely nnoncompact, λ w = λ pww , then R pc =

Mp M yc

And the nominal n stren ngth for a com mpact flange ggirder is Mp Mn = M yc = M p M yyc

(7.1)

(7.2)

At A the upper limit for a no oncompact w web, that is thhe web is veryy close to beiing slender, nd the nominaal strength forr a compact fl flange girder bbecomes λ w = λ pr , Rpc = 1.0 an M n = 1. 0 M yc = Fy S xcc

(7.3)

For F a noncom mpact flange girder, at thhe juncture bbetween the nnoncompact aand slender flange, λ f = λ rf , the nominal n mom ment strength iis given as M n = FL S xc = 0.7 Fy S xc

(7.4)

where the use of FL = 0.7 Fy , for double symmeetric memberss, accounts foor the stress abbove which the mem mber inelastic buckling app plies. This is the same ass the residual stress assum med to have occurred in the hot rollled shapes, even e though thhis is a weldeed shape. The T moment strength for a noncompacct flange platte girder is ffound throughh the linear interpolaation between the end points, as shown iin Figure 7.7,, and is givenn by Equation F4-13. ⎛ λ − λ pf ⎞ M n = R pc M yc − ( R pc M yc − FL S xxc ) ⎜ (A AISC F4-13) ⎟ ⎝ λ rf − λ pf ⎠

2770 Chapter 7

Plate Gird ders The influence off the noncom mpact web is diminished as the flangee becomes m more and more nonco ompact. Oncee the flange becomes b slendder, Rpc is noo longer needded, and flannge local buckling con ntrols the streength of the girder. For a slender flaange girder, the t behavior is an elastic buckling phenomena deppicted in Figure 7.7 and given as 0.99 Ekc S xc (AISC C F4-14) Mn = λ2 where λ = b f 2t f and th he plate buckliing factor is

kc =

4

(7-5) h / tw This plate buckling b facttor must be taken in the range from 0.35 to 0.766 for the purrpose of calculation, even though it may actually be outside that range. kling Lateral-Torrsional Buck Lateral-torsiional bucklin ng for noncom mpact web ggirders is adddressed in Seection F4.2. Lateraltorsional bu uckling behav vior for noncompact web plate girderss is, in princiiple, the sam me as for rolled beam ms. However, the equations in the Sppecification aare slightly aaltered, and tthe web plastification n factor mustt be included. As when connsidering the noncompact flange, the innfluence of the noncompact web is diminished d as the laterral-torsional bbuckling respponse becom mes more dominant. Figure F 7.8 sh hows the stren ngth of a nooncompact weeb plate girdder when connsidering lateral-torsio onal buckling g. For a plate girdeer with lateraal supports att a spacing no greater thaan Lp, Equatioon F4-1, which inclu udes the influ uence of Rpc, again definnes the girderr strength. T This girder w would be considered to t have full laateral supportt. The definittion of this limiting unbraaced length is slightly different thaan it was for a compact web b member. Thhis differencee is slight, butt has been useed in the Specification n, because itt gives more accurate resuults when ussed for singlyy symmetric girders. Thus, for no oncompact weeb girders E L p = 1.11rt SC F4-7) (AIS Fy For I-shapess with a rectaangular comp pression flangge, the effecttive radius off gyration forr lateraltorsional bu uckling, rt, iss the radius of o gyration off the compresssion flange plus one thirrd of the compression n portion of th he web, given n as

Figure 7.8 Nominal Fleexural Streng gth Based on U Unbraced Lenngth

Chapter 7

rt =

Plate Girders 271

b fc

(AISC F4-11)

⎛ 1 ⎞ 12 ⎜ 1 + aw ⎟ ⎝ 6 ⎠

where

aw =

hc t w b fc t fc

(AISC F4-12)

The strength of a section undergoing elastic lateral-torsional buckling, Lb > Lr, can be obtained through plate buckling theory. The Specification uses a combination of Equations F4-3 and F4-5 to determine the nominal moment strength as

Mn =

Cb π2 ES xc ⎛ Lb ⎞ ⎜r ⎟ ⎝ t ⎠

2

2

1 + 0.078

J ⎛ Lb ⎞ ≤ R pc M yc S xc ho ⎜⎝ rt ⎟⎠

(7.6)

This is essentially the same equation as was used for compact web girders as given through Equations F2-3 and F2-4 but with rt replacing rts. Because built-in stresses occur in plate girders just as residual stresses do for rolled shapes, elastic buckling cannot occur if the applied stress pushes the actual stress on the shape beyond the yield stress. With the built-in stress taken as 0.3Fy, the available elastic stress is again taken as 0.7Fy. Thus, Equation 7.4 again gives the limiting strength, this time for elastic lateraltorsional buckling. Using this strength with Equation 7.6, the unbraced length that defines the limit of elastic lateral-torsional buckling is obtained as

Lr = 1.95rt

E FL

2

J ⎛ J ⎞ ⎛F ⎞ + ⎜ + 6.76 ⎜ L ⎟ ⎟ S xc ho ⎝ E⎠ ⎝ S xc ho ⎠

2

(AISC F4-8)

If Equation F4-8 is modified to reflect only the doubly symmetric girders considered in this chapter, and FL is taken as 0.7Fy, it becomes Lr = 1.95rt

E 0.7 Fy

2

J ⎛ J ⎞ ⎛ 0.7 Fy ⎞ + ⎜ + 6.76 ⎜ ⎟ ⎟ S x ho ⎝ E ⎠ ⎝ S x ho ⎠

2

(7.7)

The lateral-torsional buckling strength when the member has an unbraced length between Lp and Lr is given by the same straight-line equation used previously. This time, however, it accounts for the noncompact web by including Rpc at the upper limit; thus, with FL = 0.7 Fy and

S xc = S x , Equation F4-2 becomes ⎡ ⎛ Lb − Lp ⎞ ⎤ M n = Cb ⎢ R pc M yc – ( R pc M yc – 0.7Fy S x ) ⎜ ⎟ ⎥ ≤ R pc M yc ⎝ Lr − L p ⎠ ⎥⎦ ⎣⎢ 7.2.2

(7.8)

Slender Web Plate Girders

Slender web plate girders are covered in Specification Section F5. They are those built-up members with web slenderness, λ w = hc tw , exceeding the limit λ rw = 5.70

E Fy

2772 Chapter 7

Plate Gird ders

Figgure 7.9 Beending Strenggth Redduction Factoor T B4.1b, case 15. as given in Table As was w the case for noncomp pact web mem mbers, slendeer web membbers are evaluuated by assessing ho ow the web sllenderness influences the oother applicaable limit statees. The impacct of the slender web b on member strength is characterized through the bbending strenngth reductionn factor, Rpg. The ben nding strength h reduction faactor is given by Equation F5-6 and is shown in Figuure 7.9. ⎛ hc aw E ⎞ R pg = 1 – (AIS SC F5-6) ⎜⎜ – 5.7 ⎟ ≤ 1.0 1200 + 300aw ⎝ tw Fy ⎟⎠ where aw is as defined by y Equation F4-12 F but in tthis applicatioon is limited to a value noo greater than 10. Speccification Secction F13.2 places p limitss on the propportions of m members thatt can be designed acccording to itss provisions. The T web-to-fflange area raatio, aw, is lim mited to 10 to prevent the designerr from using these t provisio ons for membbers that are essentially w webs with flannges that are little mo ore than smalll stiffeners. In n addition, thee web slenderness is limitted so that h/ttw ≤ 260. This ensuress that the gird der is not so slender that tthe stated proovisions do noot properly reeflect its behavior. The bending streength reductio on factor reduuces the strenngth of the giirder uniform mly in all ranges of flaange local bu uckling and laateral-torsionaal buckling. T Thus, it couldd simply be hheld as a final reductiion, as shown n in the Specification, or ccould be usedd within the primary equaations as shown below w. Because itt is uniformly y applied to aall of the limiit states, the influence of a slender web on flexu ural strength is easily visuaalized. Flange Local Buckling The strength h of a plate girder g as a fu unction of flaange slendernness is shownn in Figure 7..10. The slenderness parameters for fo the flange are defined iin Table B4.11b and are the same as thoose used c flangge plate girdeer, λ f ≤ λ pf , in Chapter 6 and Section 7.2.1. For a compact M n = R pg Fy S xc

(AIS SC F5-1)

Because Rpgg will not exceed 1.0, thee maximum bbending strenngth of the sslender web ggirder is limited to th he yield momeent. At the t juncture between the noncompact and slender flange, λ f = λ rf , the strrength is limited to ellastic behavio or, after accou unting for buillt-in stresses. Thus (7.9) M n = R pg ( 0.7 Fy S xc )

Chapterr 7

7 Figure 7.10

Plate Giirders 273

Nominaal Flexural Sttrength Basedd on Flange Slenderness

minal moment strength for the t slender w web–noncomppact flange plaate girder is ggiven by the The nom linear traansition which h is a combinaation of Equaations F5-7 annd F5-8 as ⎡ ⎛ λ – λ pf ⎞ ⎤ (7.10) M n = R pg S xc ⎢ Fy – (0.3Fy ) ⎜ ⎟⎥ ⎢⎣ ⎝ λ rf – λ pf ⎠ ⎦⎥ For F the slend der web–slend der flange meember, the sttrength is thee same as it w was for the noncomp pact web–slen nder flange member m and ggiven in Equaation F4-14, eexcept for thee use of Rpg. Thus, thee combination n of Equation ns F5-7 and F55-9 gives 0. 9Ekc S xc R pg Mn = (7.11) λ2 The platee buckling facctor, kc, is as previously p deefined. Lateral-Torsional Bu uckling Lateral-to orsional buck kling for thee slender weeb girder apppears quite similar to thhat for the noncomp pact web gird der and is sho own in Figuree 7.11. However, some diffferences musst be noted. For a meember to be considered c ass having full llateral support, its unbraced length is llimited to a spacing not n greater thaan Lp, where

Figure 7.11 7

Nominaal Flexural Sttrength Basedd on Unbracedd Length

2774 Chapter 7

Plate Gird ders

L p = 1.1rt

E Fy

(AIS SC F4-7)

This is the same limit used u for non ncompact webb girders, buut is differennt than that uused for compact web members. The elastic lateraal-torsional buckling b strenngth of the sllender web ggirder is giveen in the Specification n as a combin nation of Equations F5-1 aand F5-4 as Cb π 2 ESS xc R pg (7.12) Mn = ≤ R pg Fy S xc 2 ⎛ Lb ⎞ ⎜r ⎟ ⎝ t ⎠ Wheen this streng gth is set equaal to the correesponding streength limit giiven by Equattion 7.9, the limiting unbraced len ngth, Lr, becom mes E Lr = πrt (AIS SC F5-5) 0.7Fy This limit for fo elastic lateeral-torsional buckling is nnot the samee as was usedd for the nonccompact web girder. Thus, Lr is different fo or each of thhe three typees of plate ggirder: compaact web, noncompactt web, and sleender web. For the inelasticc lateral-torsiional bucklinng region, thhe strength iis given by a linear equation sim milar to thosse used prev viously, with the additionn of the Rpg multiplier; tthus the combination n of Equationss F5-2 and F5 5-3 becomes ⎡ ⎛ Lb – L p ⎞ ⎤ (7.13) M n = Cb R pg S xc ⎢ Fy – (0.33Fy ) ⎜ x ⎟ ⎥ ≤ R pg Fy S xc ⎢⎣ ⎝ Lr – Lp ⎠ ⎦⎥

Figuree 7.12 Platee Girder for E Example 7.1

Chapter 7

EXAMPLE 7.1 Plate Girder Flexural Strength

SOLUTION

Plate Girders 275

Goal:

Determine the available flexural strength for two different plate girder designs, a web thickness of (a) 3/8 in. (noncompact) and (b) 1/4 in. (slender).

Given:

The cross section of a homogeneous A36 plate girder is shown in Figure 7.12. The span is 120 ft and the unbraced length of the compression flange is 20 ft. Assume Cb = 1.0.

Step 1:

Determine the section properties for both plate girders. (a) 3/8 in. web (b) 1/4 in. web tw = 0.375 in. tw = 0.25 in. A = 63.5 in.2 A = 57.5 in.2 4 Ix = 30,600 in. Ix = 29,500 in.4 4 Iy = 2560 in.4 Iy = 2560 in. 3 Sx = 1230 in. Sx = 1190 in.3 Zx = 1330 in.3 Zx = 1260 in.3 ry = 6.35 in. ry = 6.67 in.

Part (a) Step 2:

For the plate girder with a 3/8 in. web plate Check the web slenderness in order to determine which sections of the Specification must be followed. Thus, this is a noncompact web girder, and the provisions of Section F4 must be followed. The web plastification factor must be determined.

Step 3:

Determine the shape factor and calculate Rpc. M p Z x 1330 = = = 1.08 < 1.6 M y c S x 1230 Therefore, use 1.08 in the calculation of Rpc, in Equation F4-9b. ⎡ Mp ⎛ Mp ⎞⎛ λ − λ pw ⎞ ⎤ M p Rpc = ⎢ −⎜ − 1⎟⎜ ⎟⎥ ≤ ⎢⎣ M yc ⎝ M yc ⎠⎝ λ rw − λ pw ⎠ ⎥⎦ M yc

⎡ ⎛ 128 – 107 ⎞ ⎤ = ⎢1.08 – (1.08 – 1.0 ) ⎜ ⎟ ⎥ = 1.05 ≤ 1.08 ⎝ 162 – 107 ⎠ ⎦ ⎣ Step 4:

Determine the nominal bending strength for the limit state of compression flange yielding. M n = R pc Fy S xc

= 1.05 ( 36 )(1230 ) = 46,500in.-kips = Step 5:

46,500 = 3880ft-kips 12

Check the unbraced length for lateral torsional buckling, with Lb = 20 ft or 240 in. First determine the effective radius of gyration.

276 Chapter 7

Plate Girders aw =

48 ( 0.375 ) hc t w = = 0.791 b fc t fc 26.0 ( 0.875 )

and from Equation F4-11 b fc 26.0 rt = = = 7.05 in. ⎛ aw ⎞ ⎛ 0.791 ⎞ 12 ⎜ 1 + ⎟ 12 ⎜1 + ⎟ 6 ⎠ 6 ⎠ ⎝ ⎝ From Equation F4-7 L p = 1.1rt

E Fy 29,000 = 220 in. 36

= 1.1(7.05) =

220 = 18.3ft 12

And from Equation F4-8 2

J ⎛ J ⎞ ⎛ E⎞ ⎛F ⎞ + ⎜ + 6.76 ⎜ L ⎟ Lr = 1.95rt ⎜ ⎟ ⎟ ⎝ E⎠ ⎝ FL ⎠ S xc ho ⎝ S xc ho ⎠

2

⎛ 29,000 ⎞ ⎛ 0.7 ( 36 ) ⎞ ⎛ 12.5 ⎞ 12.5 = 1.95(7.05) ⎜⎜ + ⎜ ⎟⎟ ⎟ ⎟ + 6.76 ⎜ ⎝ 1230(48.9) ⎠ ⎝ 29,000 ⎠ ⎝ 0.7 ( 36 ) ⎠ 1230(48.9) 787 = 787 in. = = 65.6 ft 12 2

where

⎛ 26(0.875)3 ⎞ 48 ( 0.375 ) 1 J = ∑ bt 3 = 2 ⎜ = 12.5 in.4 ⎟+ 3 3 3 ⎝ ⎠ 3

Step 6:

Determine the nominal strength based on the limit state of lateraltorsional buckling. Because the unbraced length is between Lp and Lr, the straight line equation, Equation 7.8, is used; thus ⎡ ⎛ Lb − Lp ⎞ ⎤ M n = Cb ⎢ R pc M yc – ( R pc M yc – 0.7Fy S x ) ⎜ ⎟ ⎥ ≤ R pc M yc ⎢⎣ ⎝ Lr − Lp ⎠ ⎥⎦

⎡ 0.7 ( 36 )(1230 ) ⎞ ⎛ 20.0 – 18.3 ⎞ ⎤ ⎛ = 1.0 ⎢3880 – ⎜ 3880 – ⎟⎜ ⎟ ⎥ = 3830 ft-kips 12 ⎝ ⎠ ⎝ 65.6 – 18.3 ⎠ ⎦⎥ ⎣⎢ Step 7:

Check for the limit state of compression flange local buckling. bf 26 = = 14.9 λf = 2t f 2 ( 0.875 ) and the limits are

2

Chapter 7 λ pf = 0.38 and, with kc =

E 29,000 = 0.38 = 10.8 Fy 36

4 = 0.354 > 0.35 , 48 / 0.375

λ rf = 0.95

Step 8:

Plate Girders 277

0.354 ( 29,000 ) kc E = 0.95 = 19.2 0.7 ( 36 ) FL

Determine the nominal moment strength for the limit state of flange local buckling. Because λpf < λf < λrf, the shape has a noncompact flange, so that, from Equation F4-13, ⎛ λ − λ pf ⎞ M n = R pc M yc − ( R pc M yc − FL S xc ) ⎜ ⎟ ⎝ λ rf − λ pf ⎠ 0.7(36)(1230) ⎞⎛ 14.9 – 10.8 ⎞ ⎛ = 3880 – ⎜ 3880 – ⎟⎜ ⎟ = 3250 ft-kips 12 ⎝ ⎠⎝ 19.2 – 10.8 ⎠

Step 9:

Determine the lowest nominal moment for the limit states checked. For compression flange local buckling, M n = 3250 ft-kips

For LRFD Step 10:

φM n = 0.9 ( 3250) = 2930 ft-kips

For ASD Step 10:

M n 3250 = = 1950 ft-kips Ω 1.67

Part (b) Step 11:

For the plate girder with a 1/4 in. web plate Check the web slenderness in order to determine which sections of the Specification must be followed. From step 2, λ rw = 162 and h 48 λw = c = = 192 > λ rw = 162 tw 0.25 Thus, this is a slender web plate girder and the provisions of Section F5 must be followed.

Step 12:

Determine the bending strength reduction factor, Equation F5-6. 48.0 ( 0.25 ) ht aw = c w = = 0.527 Lb = 8 ftt 12

Thus lateral torsion nal buckling iis not a factorr.

Step p 3:

Check k the slendern ness limits forr flange locall buckling. b bf 14.0 λf = = = 28.0 = t 2t f 2 ( 0.2550 )

Chapter 7

Plate Girders 281

The limiting slenderness from Table B4.1b case 11 is E 29,000 = 0.38 = 9.15 λ pf = 0.38 Fy 50 and with

kc = λ rf = 0.95

Step 4:

4 4 = = 0.548 h tw 20.0 / 0.375 0.548 ( 29,000 ) kc E = 0.95 = 20.2 0.7 Fy 0.7 ( 50 )

Determine the nominal strength for the limit state of flange local buckling. Since λf = 28.0 is greater than λrf = 20.2, the flange is slender. Thus, for flange local buckling, using Equation F3-2, 0.9Ekc S x Mn = λ 2f =

Step 5:

0.9 (29,000)(0.548)(94.4) = 144 ft-kip (28.0) 2 (12)

Determine the lowest nominal moment for the limit states checked. Since web local buckling and lateral torsional buckling are not factors for this member, the only limit state considered is flange local buckling. This is the controlling limit state. Thus M n = 144 ft-kips

For LRFD Step 6: For ASD Step 6:

φM n = 0.9 (144) = 130 ft-kips M n 144 = = 86.2 ft-kips Ω 1.67

7.3 HOMOGENEOUS PLATE GIRDERS IN SHEAR Shear is an important factor in the behavior and design of plate girders because the webs have the potential to be relatively thin. Two design procedures are available for shear design of plate girders. One accounts for the postbuckling strength available through tension field action, whereas the other includes postbuckling buckling strength of the web, without relying on tension field action. Transverse stiffeners can be used to increase web shear strength but are not required unless the shear strength of the web is less than the required strength. Sometimes, a thicker web may be required even if stiffeners are used. The limit states for shear are web yielding and web buckling. If tension field action is not considered, postbuckling strength is determined using a model that accounts for web stress redistribution in a member with or without transverse stiffeners. These provisions are covered in Specification Section G2.1. Under certain circumstances it is possible to take advantage of the postbuckling strength of the girder web through an approach called tension field action. Research

2882 Chapter 7

Plate Gird ders

Figure 7.14 Web Buck kling Showing g Tension Fieeld Action Photo courtessy Donald Whiite

has demonsttrated that a plate p girder with w transversse stiffeners aand a thin webb can act sim milar to a Pratt truss once o the web buckles, thu us providing additional poostbuckling sttrength. The buckled web of a plaate girder is sh hown in Figu ure 7.14, and tthe truss moddel behavior is illustrated inn Figure 7.15, where the buckled panel of thee girder simullates the tenssion diagonall of the truss and the stiffener rep presents the vertical web member. Thhe designer m must decide whether to use this tension field d action or to design witho out tension fieeld action. Onne anomaly w with the provissions for shear occurs when the tension t field action approoach actuallyy produces leess strength tthan the approach wiithout tension n field action n. This is the result of the Specificationn using two ddifferent theoretical models m to prredict shear strength. s It iss an acceptabble situation that simply must be checked wheen using tension field actio on if the maxiimum shear sstrength is dessired. It will w be seen th hat web yieldiing controls tthe maximum m strength of tthe girder weeb. If the size of the girder g web peermits web yielding, there will be no addvantage to considering sttiffeners, with or with hout tension fiield action. Avaailable shear strength s for webs w of platee girders, regaardless of weeb width-to-thhickness ratio, is deteermined using g φ = 0.9 and Ω = 1.67.

7.33.1

Nontension Field Acction The provisions for sheaar strength off a plate girdder without ttension field action are ggiven in Specification n Section G2 2.1(b). The no ominal shear strength is a function of thhe slendernesss of the web defined d as λwv = h/tw. The single limit used to ddefine the rannges of behavior is kE λ wvp = 1.100 v Fy

Chapterr 7

Plate Giirders 283

Figure F 7.15 Plate Girderr Showing Tennsion Field A Action

Figure F 7.16 Limitations on Plate Girdder to Permit Tension Fieldd Action b plate buckling coefficien nt, kv, for unsstiffened webs of I-shapedd members thhat meet the The web proportio oning criteria of the Speciffication—thatt is, λwv < 2600—is taken ass kv = 5.34. For stiffened webs, wh here stiffenerss are spaced at a a distance aa, as shown inn Figure 7.16, 5 kv = 5 + ( a /h ) 2 but is tak ken as 5.34 wh hen a/h > 3.0 0. The T nominal shear s strength h of a girder w without tensioon field actionn is given by AISC G2-1) (A Vn = 00.6 Fy AwCv1 where Aw = the overaall depth timees the web thiickness Cv1, the web shear s coefficieent, is a functtion of web shhear slendernness For F λ wv ≤ 1.100 kv E /Fy ,

Cv1 = 1.0 For F λ wv > 1.100 kv E /Fy ,

(A AISC G2-3)

2884 Chapter 7

Plate Gird ders

Figure 7.17 Web Sheaar Coefficientt as a Functioon of Web Shear Slenderneess

C v1 =

1..10 kv E / Fy h tw

(AISC G2-4)

The web shear co oefficient is sh hown in Figuure 7.17 for tw wo cases of thhe web plate bbuckling coefficient, kv = 5.34 and d kv = 10.0. For F a web witth Cv1 = 1.0 tthe web reachhes its yield sstrength. w shear sleenderness succh that Cv1 is less than 1.0,, the web bucckles. Compaaring the For a web with two curves in i Figure 7.17 shows the impact i of addding stiffenerrs to a girder without tensiion field action; a girrder with no stiffeners, s kv = 5.34; and a girder with sstiffeners spacced so that thhe panels are square, a/h a = 1.0 and kv = 10.0.

7.33.2

Tension n Field Action Although th he Specification does not require that tension field action be coonsidered, a ddesigner may take ad dvantage of tension field acction when stiiffeners are prresent. The im mpact of tensiion field action is gen nerally to incrrease the web b shear strenggth, with the nnominal shearr strength dettermined as a combin nation of web buckling streength and webb postbucklinng strength. B Both of these strength components are functionss of stiffener spacing. Ten nsion field acttion is addresssed in Speciffication Sectiion G2.2. To include tensiion field action in th he strength caalculation, th he plate girdeer must meett certain lim mitations. Figuure 7.16 illustrates th hese four limittations as desscribed here.

No Tension n Field Action n in End Pan nels Figure 7.16aa shows a plate girder wiith a potentiaally buckled w web. The diaagonal tensionn that is developed in i the web brings two orthogonal ccomponents of force to the flange-stiffener intersection.. The verticaal stiffener resists the veertical compoonent, and thhe flange ressists the horizontal component, c ju ust as for thee Pratt truss. The end pannel has no addjacent panell to help resist the ho orizontal comp ponent, so thiis last panel m must resist thhe shear forcee through beam m shear, without consideration of tension field action. Everyy stiffened plaate girder hass end panels thhat must be designed d without tenssion field acttion. This usuually results iin narrower ppanels at the ends of tension field d girders.

Chapter 7

Plate Girders 285

Proportions of Panels The Specification limits the proportions of stiffened panels such that tension field action may not be considered if a >3 h

Figure 7.16b shows a portion of a stiffened plate girder with stiffeners placed at the limit of a/h = 3. The panel is quite elongated, and its effectiveness at resisting vertical forces is significantly reduced in comparison to that of a panel with a smaller aspect ratio, such as that shown in Figure 7.16a. Proportions of Web to Flange Section G2.2(b) provides two equations for shear strength with tension field action, depending on the relative proportions of the web and flange. The first, when applied to doubly symmetric plate girders, is that the ratio of web area to flange area not exceeding 2.5 and web height to flange width not exceeding 6. If these limits are exceeded, the flanges are not sufficient to fully resist the developed diagonal tension forces. Figures 7.16c and d show plate girders that meet these limits. In this case, the nominal shear strength is give as ⎛ ⎞ 1 – Cv 2 ⎜ ⎟ Vn = 0.6Fy Aw Cv 2 + (AISC G2-7) 2 ⎟ ⎜ 1.15 1 a / h + ( ) ⎝ ⎠

If these limits are not met, the nominal shear strength is given as ⎛ ⎜ 1 – Cv 2 Vn = 0.6Fy Aw ⎜ Cv 2 + 2 ⎜ 1.15 a h + 1 + ( a /h ) ⎝

(

)

⎞ ⎟ ⎟ ⎟ ⎠

(AISC G2-8)

The web shear buckling coefficient, Cv2, is a function of the web width-to-thickness ratio. It is based on a model of shear buckling that takes no postbuckling strength into account so it is somewhat different from the coefficient, Cv1, given in Specification Section G2.1 when tension field action is not considered. When h tw ≤ λ wvp = 1.10 kv E Fy

Cv 2 = 1.0

(AISC G2-9)

when λ wvp = 1.10 kv E Fy < h tw ≤ λ wvr = 1.37 kv E Fy Cv 2 =

1.10 kv E Fy h tw

(AISC G2-10)

and when h tw > λ wvr = 1.37 kv E Fy

Cv 2 =

1.51kv E

( h tw )

2

Fy

(AISC G2-11)

Figure 7.18 shows the web shear strength in terms of Vn/(0.6FyAw) for a girder with and without tension field action. These girders meet the more stringent web to flange proportion limits that permit the use of Equation G2-7. Strength is given as a function of web shear slenderness for three panel sizes. Equation G2-7 can be rewritten to show that the strength due to tension

2886 Chapter 7

Plate Gird ders

Figure 7.18 Web Shearr Strength forr Girder with and without T Tension Fieldd Action n of the prebuuckling strenngth and the ppostbuckling strength field action is simply thee combination as ⎤ ⎡ 1 – Cv 2 ⎥ Vn = 0.6Fy AwCv 2 + 0.6Fy Aw ⎢ ⎢1.15 1 + a /h 2 ⎥ ( ) ⎦ ⎣ The prebuckling strength can be seen in Fiigure 7.18 as the strength of the girder without tension field d action. The addition of th he postbucklinng strength ussing the tension field actioon model shifts the cu urves for each particular a/h h shown in thhe figure. The end panel in n a tension field f plate girrder must bee especially rrigid in orderr for the remainder of the web to properly p funcction as a Praatt truss; thus the stiffener spacing for thhe panel next to the support musst be less thaan that withinn the span, aand shear in the end pannel must t rules for a girder witho out tension fieeld action. conform to the

7.44

STIFFE ENERS FOR R PLATE GIRDERS G When stiffeeners are required for a plate p girder, they can be either interm mediate stiffe feners or bearing stiff ffeners. The purpose p of in ntermediate sstiffeners is tto increase ggirder shear sstrength, either by controlling c th he buckling strength of the girder w web or by ppermitting addditional postbuckling g strength to be achieved. These stiffenners are distriibuted along the girder lenngth and result in pan nel sizes with h aspect ratio os, a/h, that im mpact girder shear strengtth. Bearing stiffeners usually occu ur at the locations of con ncentrated looads or reactiions. They permit the traansfer of concentrated d forces that could c not alreeady be transfferred throughh direct bearinng on the girdder web.

7.44.1

Intermeediate Stiffen ners The Specificcation requireements for inttermediate stiiffeners are prrescriptive inn nature. Therre are no forces for which w these stiffeners mu ust be sized;; they are siimply sized tto meet the specific limitations provided p in Section S G2.3.. There are tw wo requiremeents for stiffeener sizing. F First is a width-to-thickness limit and a second is a minimum m moment of innertia. For transverse stiiffeners, the width-to-thick w kness ratio, ( b t ) st , is limitted such that E ⎛b⎞ ⎜ ⎟ ≤ 0.566 Fyst ⎝ t ⎠ st

C G2-12) (AISC

Chapterr 7

Plate Giirders 287

where Fyst d strength of the t stiffener. Since interm mediate stiffenners are not ddesigned for y is the yield strength, it is not un ncommon for the stiffenerr and the webb to have diffferent strenggths when a higher strrength materiial is used forr the web. Transverse T sttiffeners used d to developp the available web shearr strength m must have a moment of inertia, Istt, about an axis a in the w web center forr stiffener paairs or about the face in w the web plate for sing gle stiffeners as shown inn Figure 7.19,, such that the minimum contact with moment of inertia bassed on consideeration of no postbucklingg strength is ⎡ 2.5 ⎤ I stt 2 = ⎢ − 2 ⎥ bp tw3 ≥ 0.55bp tw3 (AIISC G2-15) 2 ⎢⎣ ( a h ) ⎥⎦ mum of a and h. h where bp is the minim The T minimum m moment of inertia requirred when the full postbuckkling strengthh or tension field actiion strength iss included is defined d in thee Specificationn as 1.5

⎛ Fyw ⎞ h 4ρ1.3 st I st1 = ⎜ ⎟ 440 ⎝ E ⎠

(AIISC G2-14)

where ρst = the larger of Fyw/Fyst and 11.0 o the web Fyw = yield stress of quired shear strength s is less than the ful l available shhear strength, a linear interrpolation, as If the req indicated d by Equation n G2-13, may y be used to ddetermine thee actual requiired stiffener moment of inertia. Thus, T (AIISC G2-13) I st ≥ I st 2 + ( I st1 − I st 2 ) ρ w which acccounts for the t amount of o postbucklinng strength, either with oor without teension field action, in ncluded in the shear streng gth calculatio n, where

Figure 7.19 7

Web Sttiffener Minim mum Momentt of Inertia

2888 Chapter 7

Plate Gird ders

Figure 7.20 Detailing Requirement R for Intermediiate Stiffenerss Ist Ist1 Ist2 Vr Vc1 Vc2 ρw

= moment m of inerrtia of the tran nsverse stiffe ner = minimum m mom ment of inertiaa of the transvverse stiffenerr required forr the developm ment of thee full shear po ostbuckling sttrength, Vc1 = minimum m mom ment of inertiaa of the transvverse stiffenerr required to ddevelop buckkling streength without considering any postbuckkling strengthh, Vc2 = reequired shear strength in th he panel beingg considered = av vailable postb buckling shearr strength as ddefined in Section G2.1 orr G2.2 = av vailable shearr strength wheen consideringg no postbuckkling strengthh, using Cv2 inn Eq quation G2-1 in i place of Cv1 v . = th he maximum shear s ratio wiithin the paneels on each sidde of the transsverse stiffenner, ⎡ Vr − Vc 2 ⎤ ⎢V − V ⎥ ≥ 0 ⎣ c1 c 2 ⎦

provides detaailing requireements for inntermediate sttiffeners. In addition, a the Specification S They can be stopped sh hort of the teension flange and, when uused in pairss, do not neeed to be T weld byy which they are attachedd to the web is to be attached to the compresssion flange. The b fourr and six tim mes the web tthickness from m the near tooe to the webb-flange terminated between weld, as sho own in Figuree 7.20, but theere is no speccific requirem ment for sizingg that weld. N Normally it would be sized based on the platee thickness. W When single stiffeners aree used, they must be attached to the compresssion flange, if it consistts of a rectaangular plate,, to resist anny uplift ue to torsion n in the flang ge. Because intermediatee stiffeners pprovide a connvenient tendency du mechanism to transfer brracing forces to the girder,, these stiffenners also musst be connecteed to the n flange and must m be capab ble of transmiitting 1 percennt of the totall flange force. compression

7.44.2

Bearing g Stiffeners quired when the strength oof the girder web is not ssufficient to rresist the Bearing stifffeners are req concentrated d forces exertted on it. Alth hough bearingg stiffeners ccan be requireed for rolled II-shaped members an nd were discu ussed in Section 6.14, theyy are much m more likely to be required ffor plate girders, partticularly at th he girder sup pports. Speciffication Sectiion J10 addreesses the apppropriate limit states. Normally th he forces to be b resisted arre compressivve in nature. For those caases, the w local criippling, and web sideswaay buckling m must be limit states of web locaal yielding, web checked. When the appliied load is tensile, web loocal yielding and flange loocal bending must be considered. If the strengtth of the web is insufficiennt to resist thhe applied forrce, bearing stiffeners can be used..

Chapterr 7

Figure 7.21 7

Plate Giirders 289

Single Concentrated C d Force Applieed to Beam

The T relationsh hip between available str ength and noominal strenggth varies forr each limit state asssociated with web strengtth. Thus, eithher design strengths or aallowable strrengths, not nominal strengths, mu ust be comparred to determiine the minim mum web strength. The apppropriate with the follow wing discussiion of the corrresponding resistance factors and safety factorrs are given w limit stattes.

Web Loccal Yielding When a single concentrated forcee, tension or compression,, is applied tto a girder, aas shown in t be delivereed to the girdder over a lenngth of bearinng, lb, and is Figure 7..21, the forcee is assumed to then disttributed throu ugh the flangee and into thee web. For a plate girder,, the web just below the web-flan nge weld, dim mensioned as k in Figure 7..21a, is the crritical locatioon. For plate ggirders, this dimensio on, k, is taken n as the thick kness of the fl flange plus thhe dimension of the weld. As was the case for the W-shape,, the distributtion takes plaace along a liine with a sloope of 1:2.5. T Thus, when the criticcal section is reached, the force has beeen distributedd over a lengtth of lb plus 22.5k in each direction n. If the conceentrated forcee is applied sso that the forrce distributees along the w web in both direction ns, this distrib bution increasses the bearinng length by 5k, as shownn in Figure 77.21b. If the bearing is i close to thee end of the member, m distr tribution takes place only in one directtion, toward the midspan. The Speecification deefines “close to the membber end” as being within tthe member depth fro om the end. Thus, the availlable length oof the web is ((lb + 2.5k), as shown in Figgure 7.21c. The T nominal strength of the t girder weeb when the concentratedd force to bee resisted is applied at a a distance from f the mem mber end that iis greater thann the depth off the memberr, d, is (A AISC J10-2) Rn = Fyyw t w ( 5k + lb ) When W the con ncentrated forrce to be resissted is applieed at a distancce from the m member end that is lesss than or equ ual to the deptth of the mem mber, d, the noominal strenggth is AISC J10-3) (A Rn = Fywwt w ( 2.5k + lb ) where Fyw = yield stress of the web w lb = length h of bearing k = distan nce from the outer o face of tthe flange to the web toe oof the fillet weeld tw = web thickness t

290 Chapter 7

Plate Girders

For web local yielding, φ = 1.0 (LRFD)

Ω = 1.50 (ASD)

Web local crippling The criteria for the limit state of web local crippling of a plate girder is the same as it was for rolled W-shapes. It again depends on the location of the force with respect to the end of the girder and Qf =1.0 for non HSS members. When the concentrated compressive force is applied at a distance from the member end that is greater than or equal to d/2, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f 2 Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf (AISC J10-4) tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣

When the force is applied at a distance less than d/2 and lb/d ≤ 0.2, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFywt f 2 Rn = 0.40tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ and when lb/d > 0.2, 1.5 ⎡ ⎛ 4l ⎞ ⎛ tw ⎞ ⎤ EFywt f b 2 − 0.2 ⎟ ⎜ ⎟ ⎥ Rn = 0.40tw ⎢1 + ⎜ Qf tw ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ ⎝ d For web local crippling, φ = 0.75 (LRFD) Ω = 2.0 (ASD)

(AISC J10-5a)

AISC J10-5b)

Web Sidesway Buckling The web of a plate girder is generally a slender element, as has already been discussed. If the tension and compression flanges of the girder are not prevented from displacing laterally with respect to each other at the point of a compressive load, web sidesway buckling must be assessed. Two provisions are given for web sidesway buckling: (1) if the compression flange is restrained against rotation, such as when it is attached to a slab, and (2) if it is not. When the compression flange is restrained against rotation and the ratio of web slenderness to lateral buckling slenderness, ( h tw ) ( Lb b f ) ≤ 2.3 , the nominal strength is given as 3 ⎛ h /t w ⎞ ⎤ Cr tw3 t f ⎡ ⎢1 + 0.4 ⎜ Rn = ⎟ ⎥ h2 ⎢ Lb /b f ⎠ ⎥ ⎝ ⎣ ⎦

(AISC J10-6)

If ( h tw ) ( Lb b f ) > 2.3 , the limit state of web sidesway buckling does not apply. When the compression flange is not restrained against rotation and the ratio of web slenderness to lateral buckling slenderness, ( h tw ) ( Lb b f ) ≤ 1.7 , the nominal strength is given as 3 Cr tw3 t f ⎡ ⎛ h /tw ⎞ ⎤ ⎢0.4 ⎜ Rn = (AISC J10-7) ⎟ ⎥ h 2 ⎢ ⎝ Lb /b f ⎠ ⎥ ⎣ ⎦ and if ( h tw ) ( Lb b f ) > 1.7 , the limit state of web sidesway buckling does not apply. In the above equations, = largest laterally unbraced length along either flange at the point of load Lb

Chapterr 7

Plate Giirders 291

Figure 7.222 Flange Loocal Bending foor an Applied Tension Load Cr

=

960,000 kssi when, at thhe location off the force, Mu < My or 1.5M Ma < My; 480,000 ksi when, at thhe location off the force, Mu ≥ My or 1.5M Ma ≥ My

For F web sidessway buckling g, φ = 0.85 0 (LRFD)

Ω = 1.766 (ASD)

Flange Local L Bendin ng This limiit state appliees when a sin ngle tensile c oncentrated fforce is appliied to the flannge and the length off loading acro oss the memb ber flange is ggreater than oor equal to 0.15bf, as show wn in Figure 7.22. Thee nominal streength is Rn = 6.25 Fyf t 2f (A AISC J10-1) If I the force is applied at a distance less than 10tf from m the membeer end, the strrength must be reduceed by 50 perccent. For F flange loccal bending, Ω= 1.67 (ASD) φ = 0.9 (LRFD)

7.4.3

Bearring Stiffenerr Design a checked, a decision is made regarding the need for bearing Once thee appropriate limit states are stiffenerss. Although itt is possible to t select a weeb plate that would not reequire bearingg stiffeners, this is no ot usually the most econom mical approacch, even thouggh the additioon of stiffenerrs is a highlabor, an nd thus high-ccost, activity.. When bearinng stiffeners are to be sizzed, Section JJ10.8 of the Specifica ation requiress that they be b sized accoording to the provisions ffor tension m members or compresssion memberss, as appropriiate. Stiffeners S dessigned to resisst tensile forc es must be deesigned accorrding to the reequirements of Sectio on J4.1 which h refer to Chaapter D for thee difference bbetween the rrequired strenngth and the minimum m available lim mit state stren ngth. The stifffener must bbe welded to the flange annd web, and these welds must be siized to resist the force beinng transferredd to the stiffenners. Stiffeners S req quired to ressist compresssive forces m must be designed accordding to the provision ns of Chapterr E, except fo or stiffeners w with KL/r ≤ 25, which mayy be designedd with Fcr = Fy, accorrding to Sectiion J4.4, for the t differencee between thee required strrength and the minimum availablee limit state strength. s Theese stiffeners must either bear on or bbe welded to the loaded flange an nd welded to the web, and d these welds must be sized to resist thee force being transferred to the stiffeners.

2992 Chapter 7

Plate Gird ders

Fig gure 7.23 Pllate Girder foor Example 7.3

EX XAMPLE 7.3 7 Pllate Girder Fllexural and Sh hear Strengtth

Goa al:

Deterrmine the av vailable mom ment and sheaar strength uusing tensionn field action n.

Giv ven:

A bu uilt-up membeer is shown in Figure 7..23. The beaam is continuuously lateraally braced. Use U A572 Gr.. 50 for the m member platees and A36 fo for the stiffen ners. Sx = 464 4 in.3.

SO OLUTION

Step p 1:

Check k the web sleenderness to ddetermine whhich section oof the Specificcation must be used. Usin ng Table B4.11b case 15 h 40 29,000 λw = = = 160 > λ rw = 5.7 = 137 t w 0.25 50 Thus,, this is a slender web plaate girder, annd the provisiions of Sectioon F5 must be applied.

Step p 2:

Deterrmine the ben nding strengthh reduction factor. The ratio of web arrea to flange area is given as 40.0(0.250) aw = = 1.00 10.0(1.00) f Equation n F5-6 And from

Rpg = 1 – Step p 3:

⎛ 40.0 1.00 29,000 ⎞ – 5.7 ⎜⎜ ⎟ = 0.985 50 ⎟⎠ 1200 + 300(1..00) ⎝ 0.250

Deterrmine the nom minal momentt strength for the limit state of yielding.. 0.985(50)(4664) M n = R pg Fy S xc = = 1900 ft-kips 12

Chapter 7 Step 4:

Plate Girders 293

Determine the nominal moment strength for the limit state of lateraltorsional buckling. The compression flange is fully braced, so this limit state does not apply.

Step 5:

Determine the nominal moment strength for the limit state of flange local buckling. The flange slenderness is, from Table B4.1b case 11 b bf 10.0 29,000 = = 5.0 < λ pf = 0.38 = 9.15 λf = = t 2t f 2 (1.0 ) 50 Therefore, the flange is compact, and there is no reduction in strength.

Step 6:

Determine the lowest nominal moment strength for the limit states considered, yielding, lateral-torsional buckling and flange local buckling. For the limit state of yielding, M n = 1900 ft-kips

For LRFD Step 7:

Determine the design moment strength. φM n = 0.9 (1900) =1710 ft-kips

For ASD Step 7:

Determine the allowable moment strength.

Step 8:

Determine the shear strength with tension field action. First check the girder proportions against the prescriptive requirements of Section G2.2.

M n Ω = 1900 1.67 = 1140 ft-kips

Proportions of panel: a 60.0 = = 1.50 < 3 h 40.0

Proportions of web to flange: Aw 42.0(0.250) = = 1.05 < 2. 5 10.0(1.00) Af Proportions of web height to flange width: h 40.0 = = 4.00 < 6.0 b f 10.0 Because the criteria have been satisfied, the tension field action provisions of Section G2.2(b)(1) may be used for all but the end panel. Step 9:

Check the web shear slenderness limits for determination of the shear strength coefficient. The web shear buckling coefficient from Equation G25 is

294 Chapter 7

Plate Girders kv = 5 +

5 5 =5+ = 7.22 2 ( a /h ) (60.0/40.0) 2

Thus the limits for determining Cv2 are kE 7.22(29,000) λ wvp = 1.1 v = 1.1 = 71.2 50 Fy and λ wvr = 1.37

kv E 7.22 (29,000) = 1.37 = 88.7 50 Fy

Because λ wv = h tw = 40.0 0.250 = 160 is greater than λwvp and λwvr, the web plate will buckle elastically. Step 10:

Determine the shear strength coefficient from Equation G2-11. 1.51(7.22)(29,000) Cv 2 = = 0.247 (160) 2 (50) The nominal shear strength with tension field action from Equation G2-7 is ⎛ ⎡ 1 – 0.247 ⎤⎞ Vn = 0.6(50)(42.0)(0.250) ⎜ (0.247 + ⎢ ⎥ ⎟ = 192 kips 2 ⎟ ⎜ ⎢ ⎥⎦ ⎠ 1.15 1 (1.50) + ⎣ ⎝

For LRFD Step 11:

Determine the design shear strength. φVn = 0.9 (192) = 173 kips

For ASD Step 11:

Determine the allowable shear strength.

Step 12:

Check the intermediate stiffener size for meeting the criteria.

Vn Ω = 192 1.67 = 115 kips

Check the width-to-thickness ratio of the stiffener from Equation G2-12 4.5 29,000 ⎛b⎞ = 12.0 ≤ 0.56 = 15.9 ⎜ ⎟ = 36 ⎝ t ⎠ st 0.375 So the stiffener satisfies this requirement For the single plate stiffener, b h3 0.375(4.50)3 = 11.4 in.4 I st = st = 3 3 Step 13:

Check the minimum stiffener moment of inertia when tension field action is considered. Since the required shear strength is not given, the stiffeners will be checked to see that they are capable of providing the full tension field strength of the web using Equation G2-14.

Chapter 7

1.5

h 4 ρ1.3 st ⎛ Fyw ⎞ 40 ⎜⎝ E ⎟⎠

1.3

50 1.5 ( 40.0 ) ⎛⎜ ⎞⎟ ⎝ 36 ⎠ ⎛ 50 ⎞ = 7.02 in.4 = ⎜ 29,000 ⎟ 40 ⎝ ⎠ 4

I st1 =

Plate Girders 295

and I st1 = 7.02 < 11.4 in.4

Thus, these stiffeners are adequate to permit full use of the available shear strength with tension field action, as given in Step 11.

7.5 PROBLEMS 1. Determine the available moment strength of an A36 plate girder with a 50×1/2 in. web plate and equal flange plates of 12×1 in. Assume there is full lateral support. Determine by (a) LRFD and (b) ASD. 2. Determine the available moment strength of an A572 Gr. 50 plate girder with an 80×1/2 in. web plate and equal flange plates of 16×1/2 in. Assume there is full lateral support. Determine by (a) LRFD and (b) ASD. 3. Determine the available moment strength of an A572 Gr. 50 plate girder with a web plate of 40×1/2 in. and equal flange plates of 10×1 in. Assume there is full lateral support. Determine by (a) LRFD and (b) ASD. 4. Determine the available moment strength of an A36 plate girder with a web plate of 75×3/8 in. and equal flange plates of 14×1-1/4 in. Assume there is full lateral support. Determine by (a) LRFD and (b) ASD. 5. For a plate girder spanning 80 ft with lateral supports at the supports and midspan, determine the nominal moment strength. The girder is an A36 member with a web plate of 45×3/4 in. and flange plates of 9×3/4 in. 6. For the plate girder of Problem 1, determine the available moment strength if the girder spans 120 ft and has lateral supports at the ends and the third points of the span. Determine by (a) LRFD and (b) ASD. 7. For the plate girder of Problem 2, determine the available moment strength if the girder spans 90 ft and has lateral supports at the ends and the quarter points of the span. Determine by (a) LRFD and (b) ASD. 8. For the plate girder of Problem 3, determine the available moment strength if the girder spans 70 ft and has lateral supports at the ends only. Determine by (a) LRFD and (b) ASD. 9. For the plate girder of Problem 4, determine the available moment strength if the girder spans 160 ft and

has lateral supports at the ends and the third points of the span. Determine by (a) LRFD and (b) ASD. 10. Determine the available moment strength of an A36 plate girder with a 50×1/2 in. web plate and equal flange plates of 12×1/2 in. Assume there is full lateral support. Determine by (a) LRFD and (b) ASD. 11. Determine the available moment strength of an A572 Gr. 50 plate girder with a web plate of 40×1/2 in. and equal flange plates of 12×1/2 in. Assume there is full lateral support. Determine by (a) LRFD and (b) ASD. 12. Determine the available moment strength of an A572 Gr. 50 plate girder with a web plate of 45×3/4 in. and flange plates of 14×1/2 in. Assume there is full lateral support. Determine by (a) LRFD and (b) ASD. 13. Determine the available shear strength of an A572 Gr. 50 plate girder without transverse stiffeners. The web plate is 100×3/4 in. and the flange plates are 15×1-1/2 in. Determine by (a) LRFD and (b) ASD. 14. Determine the available shear strength of the girder in Problem 1 if there are no intermediate stiffeners. Determine by (a) LRFD and (b) ASD. 15. Determine the available shear strength of an A572 Gr. 50 plate girder with transverse stiffeners spaced every 100 in. The web plate is 100×3/4 in. and the flange plates are 15×1-1/2 in. Determine by (a) LRFD and (b) ASD. 16. Determine the available moment and available shear strength of an A36 plate girder on a 100 ft span with stiffeners at the ends and 20 in. from the ends, along with intermediate stiffeners spaced at 40 in. on center within the span. Assume there is full lateral support. The girder has a web plate of 40×1/4 in. and flange plates of 8×3/4 in. Determine by (a) LRFD and (b) ASD.

296 Chapter 7

Plate Girders

17. Consider the plate girder from Problem 16 and the intermediate stiffener between a 20 in. end panel and an adjacent 40 in. panel. Does a pair of 3×1/4 in. stiffener plates satisfy all intermediate transverse stiffener requirements for this location? Evaluate by (a) LRFD and (b) ASD. 18. Consider the plate girder from Problem 16 and the intermediate stiffener between a 20 in. end panel and an adjacent 40 in. panel. Does a single 3-1/2×3/8 in. stiffener plate satisfy all intermediate transverse stiffener requirements for this location? Evaluate by (a) LRFD and (b) ASD. 19. A girder is to be designed to span 125 ft and carry a uniformly distributed live load of 2.0 kip/ft and a uniformly distributed dead load of 1.9 kip/ft including the girder weight. Assume that the compression flange has full lateral support, and determine the web and flange plates necessary if the web plate is limited to a depth of 50 in. Determine any stiffener requirements for a girder without tension field action. Use A572 Gr. 50 steel. Design by (a) LRFD and (b) ASD. 20. Design a girder for the conditions of Problem 19 if the compression flange is braced at 1/3 points.

Chapter

8

Beam m-Colu umns and Frame F e Behav vior

KFC Yuum! Center, Loouisville, KY photo © Bob Perzel

8.1

INTR RODUCTIO ON Beam-co olumns are meembers subjected to axial forces and beending momeents simultaneeously; thus their beh havior falls somewhere betw ween that of aan axially loaaded column aand that of a bbeam under pure ben nding. It is thu us possible to o consider thee beam or axxially loaded m member as sppecial cases of the beeam-column. Practical applications off the beam-coolumn are nuumerous. Theey occur as chord meembers in tru usses, as elem ments of rigidlly connected frameworks, and as membbers of pinconnecteed structures with transverrse or eccenttric loads. It is not alwayys possible too look at a member and determin ne whether it is a beam-co lumn or not; some knowleedge of the acctual forces being caarried by the member is required r to ccategorize it as a beam-column. However, many structural members are a subjected to these coombined forcees, and the bbeam-columnn is a very common element in building structtures. The T manner in i which the combined looads are trannsferred to a particular beeam-column significan ntly impacts the ability of o the membeer to resist thhose loads. S Starting with the axially loaded column, bendiing moments can occur fr from various sources. Lateeral load can be applied directly to t the membeer, as is the caase for a trusss top chord orr a column suupporting the lateral load from a wall. w Alternativ vely, the axiaal force can bee applied at ssome eccentriicity from the centroid of the colum mn as a resu ult of the speecific connecctions. In adddition, the m member can rreceive end momentss from its con nnection to other memberrs of the struccture, such ass in a rigid frrame. In all cases, th he relation off the beam-co olumn to thee other elemeents of the sstructure is im mportant in determin ning both the applied a forcess and the strenngth of the m member. To T understan nd the behavior of beam--columns, it is common practice to llook at the response predicted by y an interactio on equation. T The response of a beam-coolumn to an aaxial load P, major ax xis moment Mx, and minor axis momentt My is presennted on the thrree-dimensionnal diagram shown in n Figure 8.1. Each axis in this diagram m represents thhe capacity oof the memberr when it is subjected d to loading of o one type on nly, whereas the curves reepresent the combination oof two types of loadin ng. The surfacce formed by y connecting tthe three curvves representss the interactiion of axial load and biaxial bendiing. This interraction surfacce is of intereest to the desiggner.

297

2998

Chapter 8

Beam-C Columns and Frame F Behav vior

Figuree 8.1 Ultimaate Interaction Surface forr a mn. Stockyy Beam-Colum

The end points of o the curves shown in Fiigure 8.1 deppend on the sstrength of thhe beammembers (Chaapter 6). columns as described forr compression n members (C Chapter 5) annd bending m The shape of o the curvess between theese end pointts depends oon the properrties of the particular member as well w as the pro operties of otther memberss of the structuure. rts of the Mannual discussed in this Table 8.1 lists th he sections off the Specificaation and part chapter.

8.22

SECON ND-ORDER EFFECTS The single most m-column is w what are m compliccating factor in the analy sis and desiggn of a beam effects. Seco known as second-order s ond-order eff ffects are thee changes in member forrces and mmonly usedd elastic moments ass the direct result r of stru uctural deform mations. Beccause the com methods of structural anaalysis assumee that all defoormations aree small, and bbecause the eqquations of equilibriu um are written n using the un ndeformed coonfiguration oof the structurre, these methhods are not able to capture the additional second-order s effects that occur in reaal structures without adjustment. The results of o that type of o analysis arre called firstt-order effectss—that is, firrst-order forces, first--order momeents, and firstt-order displaacements. To account for the influencce of the deformation ns, an addition nal analysis must m be perforrmed. The ressults of this aadditional anaalysis are referred to as a the second--order effects.. Several approach hes are availlable for inclluding secondd-order effeccts in an anaalysis. A nelastic analyssis would takke into accounnt the actual ddeformation oof the complete seccond-order in Table 8.1 Sections of Specification S and a Parts of M Manual Coveered in This C Chapter Speecification B3 Design Basis C Design for Stability S H Combined Forrces and Torsion Design of Members M for C Appendix 6 Stability Braacing for Coluumns and Beaams Appendix 7 Design for Staability Alternative Methods M of D Appendix 8 Approximatee Second-Ordder Analysis Manual M Part 1 Dimensions and Propertiees Part 3 Design of Fllexural Membbers Part 4 Members Design of Co ompression M Part 6 Design of Members M Subjeect to Combinned Loading

Beam m-Columns annd Frame Behhavior

Chappter 8

299

structure and the resu ulting forces,, as well as tthe sequencee of loading and the behaavior of the structure after any of its compon nents are streessed beyondd the elastic limit. This aapproach to analysis is generally more m complex x than is neceessary for noormal design. A similar appproach that includes the actual defformations bu ut that does nnot include ineelastic behaviior is usually sufficient. An A approach that is consistent with noormal design office practiice and with how beamcolumns have been handled h for many m years usses a first-ordder elastic annalysis and am mplification factors to o approximatee the second-order effects.. This approaach applies theese amplificaation factors as multip pliers to the reesults of the first-order f anaalysis to obtaiin the second--order effectss. Two T differentt deflection components c tthat could occcur in a beaam-column innfluence the momentss in that beam m-column. Th he first, illusstrated in Figgure 8.2a, is tthe deflectionn along the length off the memberr that results from the mom ment along thhe member. IIn this case, tthe member ends musst remain in th heir original position p relatiive to each otther; thus, no sway is conssidered. The moment created by the t load, P, acting at ann eccentricityy δ2 from thee deformed m member, is superimp posed on the moment m resullting from thee applied endd moments. Because the m magnitude of this additional momen nt depends on n the propertiies of the collumn itself, thhis is called tthe member effect. When W the beaam-column is part of a struucture that is permitted to sway, the dissplacements of the ov verall structure also influeence the mom ments in the member. Forr a beam-coluumn that is permitted d to sway an amount a Δ2, ass shown in Fiigure 8.2b, thhe additional m moment is givven by PΔ2. Because the lateral diisplacement of o a given meember is a fuunction of thee properties oof all of the memberss in a given sttory, this is caalled the struccture effect. To T understand d the magnitu ude of the pootential increease in momeents on a coluumn due to second-o order effects, two simple calculations will be carriied out. The first is for a 20 ft long column similar s to thatt shown in Fig gure 8.2a. A W W12×96 meember is used to carry an axxial load Pu = 400 kip ps and equal end e momentss of Mu = 200 ft-kips bendiing the membber in single ccurvature. A first-ordeer analysis yieelds an axial force f in the c olumn of 4000 kips and a bbending momeent at every point alo ong the colum mn length of 200 ft-kips. The maximuum deflectionn of the mem mber at mid height du ue to the mom ment is

M u L2 200 ( 200 ) (1728 ) = = 00.715 in. 8 EI 8 ( 29,0000 )( 833) 2

δ=

With h the occurren nce of this defflection, the appplied load of 400 kips is now at an ecccentricity

Figu ure 8.2 Coluumn Dispplacements forr SecondOrdeer Effects.

300

Chapter 8

Beam-Columns and Frame Behavior

from the member in its displaced position. Thus, an additional moment is induced into the member equal to 400 ( 0.715 ) M additional = = 23.8 ft-kips 12

The addition of this additional moment to the original internal moment of 200 ft-kips yields the second-order moment, M 2 nd = 200 + 23.8 = 224 ft-kips Thus, there is an amplification of the moment by 224/200 = 1.12. If this were the final case, second-order analysis would be fairly simple. Unfortunately, the additional moment just determined also causes additional deflection, which, in turn, causes additional moment. This process continues until equilibrium is reached. The process is an iterative one, and is nonlinear. A second example is a column similar to that shown in Figure 8.2b. The same W12×96 member is used, and the axial force is again Pu = 400 kips. In this case, the column is a cantilever with a moment of Mu = 200 ft-kips applied at the top. This moment will cause a horizontal deflection at the top of the column of

M L2 200 ( 20 ) (1728 ) Δ= u = = 2.86 in. 2 EI 2 ( 29,000 )( 833) 2

In this displaced position, the 400 kip load is now at an eccentricity from the fixed support, which induces an additional moment 400 ( 2.86 ) M additional = = 95.3 ft-kips 12 The addition of this additional moment to the original support moment of 200 ft-kips yields the second-order moment M 2 nd = 200 + 95.3 = 295 ft-kips

which is an increase of 1.48 times the first-order moment. Again, this is not the end of the required calculations; this additional moment causes additional deflections and additional moments. Both of these second-order effects are significant in real structures and must be accounted for in the design of beam-columns according to Section C1 of the Specification. Procedures for incorporating these effects will be addressed once an overall approach to beam-column design is established.

8.3

INTERACTION PRINCIPLES The interaction of axial load and bending within the elastic response range of a beam-column can be investigated through the straightforward techniques of superposition. This is the approach normally considered in elementary strength of materials in which the normal stress due to an axial force is added to the normal stress due to a bending moment. Although the superposition of individual stress effects is both simple and correct for elastic stresses, there are significant limitations when applying this approach to the limit states of real structures. These include: 1. Superposition of stress is correct only for behavior within the elastic range, and only for similar stress types.

Beam-Columns and Frame Behavior

Chapter 8

301

Superposition of strain can be extended only into the inelastic range when deformations are small. 3. Superposition cannot account for member deformations or stability effects such as local buckling. 4. Superposition cannot account for structural deflections and system stability. With these limitations in mind, it is desirable to develop interaction equations that will reflect the true limit states behavior of beam-columns. Any limit state interaction equation must reflect the following characteristics: 2.

Axial Load 1.

Maximum column strength

2.

Individual column slenderness

Bending Moment 1.

Lateral support conditions

2.

Sidesway conditions

3.

Member second-order effects

4.

Structure second-order effects

5.

Moment variation along the member

The resulting equations must also provide a close correlation with test results and theoretical analyses for beam-columns, including the two limiting cases of pure bending and pure compression. Application of the resulting interaction equations can be regarded as a process of determining available axial strength in the presence of a given bending moment or determining the available moment strength in the presence of a given axial load. An applied bending moment consumes a portion of the column strength, leaving a reduced axial load strength. When the two actions are added together, the resulting total load must not exceed the total column strength. Conversely, the axial load can be regarded as consuming a fraction of the moment strength. This fraction, plus the applied moments, must not exceed the maximum beam strength.

8.4 INTERACTION EQUATIONS A simple form of the three-dimensional interaction equation is Pr M rx M ry + + ≤ 1.0 Pc M cx M cy

(8.1)

where the terms with the subscript r represent the required strength and those with the subscript c represent the available strength. This interaction equation is plotted in Figure 8.3. The figure shows that this results in a straight line representation of the interaction between any two of the load components. The

3002

Chapter 8

Beam-C Columns and Frame Behav vior

Figure 8.3

Simpliffied Interactioon Surface.

Figuree 8.4a Interacction Diagram m for Stub W W14×82 Colum mn.

Figurre 8.4b Norm malized Interacction Diagraam for Stub W14× ×82 Column.

Beam-Columns and Frame Behavior

Chapter 8

303

horizontal plane of Figure 8.3 represents the interaction of moments in the two principal axis directions, called biaxial bending, whereas the vertical planes represent the interaction of axial compression plus either major or minor axis bending. It should also be apparent that the threedimensional aspect is represented by a plane with intercepts given by the straight lines on the three coordinate planes. The interaction equations in Chapter H of the Specification result from fitting interaction equations that are similar to the form of Equation 8.1 to a set of data developed from an analysis of forces and moments for various plastic stress distributions on a stub column. Figure 8.4a shows the actual analysis results for a W14×82 stub column. Figure 8.4b shows the same data plotted as functions of the normalized axial strength Py and flexural strength Mp. In both cases, the influence of length on the axial or flexural strength is not included. Using curves of this type, developed for a wide variety of steel beam-column shapes, two equations were developed that are conservative and accurate for x-axis bending. When applied to y-axis bending, they are significantly more conservative; however, simplicity of design and the infrequent use of weak axis bending justify this extra level of conservatism. An additional modification to these equations is required to account for length effects. Rather than normalizing the curves on the yield load and the plastic moment as was done in Figure 8.4b, the equations were developed around the nominal strength of the column and the nominal strength of the beam. The resulting equations are Equations H1-1a and H1-1b in the Specification and are plotted in Figure 8.5. The equations shown here consider bending about both principal axes, whereas the plot in Figure 8.5 is for single-axis bending. P For r ≥ 0.2 , Pc

For

Pr 8 ⎛ M rx M ry ⎞ + ⎜ + ⎟ ≤ 1.0 Pc 9 ⎝ M cx M cy ⎠

(AISC H1-1a)

Pr ⎛ M rx M ry ⎞ +⎜ + ⎟ ≤ 1.0 2 Pc ⎝ M cx M cy ⎠

(AISC H1-1b)

Pr < 0.2 , Pc

where Pr = required compressive strength, kips Pc = available compressive strength, kips Mr = required flexural strength, ft-kips Mc = available flexural strength, ft-kips x = subscript relating symbol to strong axis bending y = subscript relating symbol to weak axis bending It is important to note that 1. The available column strength, Pc, is based on the axis of the column with the largest slenderness ratio. This is not necessarily the axis about which bending takes place. 2. The available bending strength, Mc, is based on the bending strength of the beam without axial load, including the influence of all the beam limit states. 3. The required compressive strength, Pr, is the force in the member, including secondorder effects.

3004

Chapter 8

Beam-C Columns and Frame F Behav vior

Figgure 8.5 Intteraction Equuations H11-1a and H1-11b.

4.

The required flexural strength, Mr, is the bending m moment in thhe member, inncluding second-ordeer effects.

Second-ordeer forces and moments can n be determineed through a second-orderr analysis or bby a modification n of the resu ults of a first-order analyysis using am mplification ffactors as meentioned earlier. Thesse amplificatiion factors wiill be discusseed as they rellate to bracedd frames (Secttion 8.5) and momentt frames (Secttion 8.6). Add ditional provissions are avaiilable for casees where the axial strengthh limit state iss out-ofplane buckling and the flexural f stren ngth limit staate is lateral-ttorsional bucckling for bennding in plane. Equaations H1-1a and b are co onservative foor this situatiion, but an addditional appproach is available. Sp pecification Section S H1.3 provides thaat (1) for the limit state off in-plane insstability, Equations H1-1a H and H1 1-1b should be b used wherre the comprressive strenggth is determ mined for buckling in the plane of bending an nd Mcx = Mp , and (2) forr the limit sttates of out-of-plane buckling and d lateral-torsiional buckling g

Pr ⎛ Pr ⎜1..5 − 0.5 Pcy ⎝ Pcy

2

⎞ ⎛ M rx ⎞ ⎟+⎜ ⎟ ≤ 1.0 ⎠ ⎝ Cb M cx ⎠

(AISC H1-3)

where Pcy = available co ompressive sttrength out off the plane of bending Mcx = available laateral-torsional buckling sttrength for strrong axis bennding with Cb = 1.0 g factor discuussed in Chappter 6 Cb = lateral-torsiional buckling If there is significant s biaxial bending g, meaning th that the requiired-moment--to-available--moment ratio for y-ax xis bending iss greater than n or equal to 00.05, then thiss option is noot available. A Although this optionall approach caan provide a more m economiical solution in some casess, it is not useed in the examples orr problems in this book.

Beam m-Columns annd Frame Behhavior

Figgure 8.6

Chappter 8

305

Braaced Frame.

Figgure 8.7 Ann Axially Loaaded Coolumn with Equal and Oppposite End Mooments.

8.5

BRAC CED FRAM MES A frame is considered d to be braced d if a positivee system—thaat is, an actuaal system suchh as a shear wall (maasonry, concrrete, steel, orr other materrial) or diagoonal steel meember—as illlustrated in Figure 8.6, serves to resist the latteral loads, sttabilize the fframe under ggravity loadss, and resist lateral diisplacements. In these casees, columns arre consideredd braced againnst lateral trannslation and the in-pllane K-factorr can be takeen as 1.0, acccording to A Appendix Seection 7.2.3(aa), unless a rational analysis a indiccates that a lo ower value iss appropriate.. This is the ttype of colum mn that was

306

Chapter 8

Beam-Columns and Frame Behavior

discussed in Chapter 6. Later in this chapter the requirements for bracing to ensure that a structure can be considered a braced frame, as found in Appendix 6, are discussed. If the column in a braced frame is rigidly connected to a girder, bending moments result from the application of the gravity loads to the girder. These moments can be determined through a first-order elastic analysis. The additional second-order moments resulting from the displacement along the column length can be determined through the application of an amplification factor. The full derivation of the amplification factor has been presented by various authors.1,2 Although this derivation is quite complex, a somewhat simplified derivation is presented here to help establish the background. An axially loaded column with equal and opposite end moments is shown in Figure 8.7a. This is the same column that was discussed in Section 8.2. The resulting moment diagram is shown in Figure 8.7b where the moments from both the end moments and the secondary effects are given. The maximum moment occurring at the mid-height of the column, Mr, is shown to be M r = M 1 + Pδ 2

The amplification factor is defined as AF =

M r M 1 + Pδ 2 = M1 M1

Rearranging terms yields

AF =

1 Pδ 2 1– M 1 + Pδ 2

Two simplifying assumptions will be made. The first is based on the assumption that δ is sufficiently small that δ2 δ ≈ 1 M 1 + Pδ 2 M 1 and the second, using the beam deflection, δ1 = M1 L2/8EI, assumes that M1 8EI π 2 EI = 2 ≈ 2 = Pe L L δ1 Because these simplifying assumptions are in error in opposite directions, they tend to be offsetting. This results in a fairly accurate prediction of the amplification. Thus, 1 (8.2) AF = 1 – P /Pe A comparison between the actual amplification and that given by Equation 8.2 is shown in Figure 8.8. The discussion so far has assumed that the moments at each end of the column are equal and opposite, and that the resulting moment diagram is uniform. This is the most severe loading case for a beam-column braced against translation. If the moment is not uniformly distributed, the

1

Galambos, T. V., Structural Members and Frames. Englewood Cliffs, NJ: Prentice Hall, Inc., 1968. Johnson, B. G., Ed., Guide to Stability Design Criteria for Metal Structures, 3rd ed., SSRC, New York: Wiley, 1976.

2

Beam m-Columns annd Frame Behhavior

Chappter 8

307

F Figure 8.8 A Amplified Mooment: E Exact and Approximate. displacem ment along th he member is less than ppreviously considered andd the resultingg amplified moment is less than in ndicated. It haas been custoomary in desiggn practice too use the casee of uniform moment as a base an nd to provid de for other moment disttributions by converting tthem to an equivalen nt uniform mo oment throug gh the use of aan additional factor, Cm. Numerous N stu udies have sh hown that a reasonably aaccurate correection resultss for beamcolumns braced again nst translation and not subjeect to transveerse loading bbetween their supports, if the mom ment is reduced d through its multiplication m n by Cm, wheere

Cm = 0.6 − 0.4 ( M 1 M 2 )

(A AISC A-8-4)

t smaller to o larger momeents at the ennds of the mem mber unbraceed length in M1/M2 is the ratio of the the planee of bending g. M1/M2 is positive p wheen the membber is bent inn reverse currvature and negative when bent in n single curvaature. For F beam-colu umns in braced frames whhere the mem mber is subjectted to transveerse loading between supports, Cm may be taken n from Comm mentary Tablee C-A-8.1, orr conservativeely taken as 1.0. The T combinattion of the am mplification ffactor, AF, annd the equivaalent momentt factor, Cm, accounts for the total member seco ondary effectss. This combiined factor is given as B1 inn Appendix 8 of the Specification S as Cm B1 = ≥ 1.0 (A AISC A-8-3) α Pr 1– Pe1 where α = 1.6 forr ASD and 1.0 1 for LRFD D to accounnt for the nonlinear behavvior of the structurre at its ultim mate strength Pr = required strength, which w may be taken as the ffirst-order reqquired strengtth, Pnt + Plt, when used u in moment frames Pe1 = Euler buckling b load for the colum mn in the planne of bendingg with an effecctive length factor, K = 1.0 Thus, thee value of Mr in Equations H1-1a and H H1-1b is takenn as

M r = B1M nt nt on the beaam-column. T The subscript nt indicates tthat for this where Mnt is the maxiimum momen case, the column doess not undergo o any lateral translation oof its ends. It is possible fo for Cm to be less than n 1.0 and for Equation A-8 8-3 to give ann amplificatioon factor less than 1.0. Thhis indicates

3008

Chapter 8

Beam-C Columns and Frame F Behav vior

Figure 8.9 Three-Dimeensional Braced Frame forr a Single-Story Structure. that the com mbination of the t Pδ effectss and the nonuuniform mom ment results inn a moment lless than the maximu um moment on the beam m-column frrom a first-oorder analysiis. In this case, the amplification factor B1 = 1.0.

EX XAMPLE 8.1a 8 Brraced Framee Coolumn Desig gn for Coombined Ax xial an nd Bending by b LR RFD

Goa al:

Desig gn column A1 A in Figure 8.9 for the given loadss using the L LRFD provisions and thee second-ordeer amplificatioon factor provvided in Apppendix 8 of th he Specificatiion.

Giv ven:

The three-dimensional braced fframe for a siingle-story sttructure is givven in Figurre 8.9. Rigid connections are providedd at the roof level for collumns A1, B1, B A4, and B4. B All other column connnections are ppinned. Dead Load = 50 psf, p Snow Lo oad = 20 psf, Roof Live Looad = 10 psf,, and Wind L Load = 20 psf horizontal. Use U A992 steeel. Assume thhat the X-braacing is sufficciently stifferr than the rigiid frames to rresist all lateraal load.

Beam-Columns and Frame Behavior Chapter 8

SOLUTION

Step 1:

309

Determine the appropriate load combinations. From ASCE 7, Section 2.3, the following two combinations are considered. ASCE 7 load combination 3 1.2D + 1.6(Lr or S or R) + (0.5L or 0.5W) ASCE 7 load combination 4 1.2D + 1.0W + 0.5L + 0.5(Lr or S or R)

Step 2:

Determine the factored roof gravity loads for each load combination. For load combination 3 1.2(50) + 1.6(20) = 92 psf and for load combination 4 1.2(50) + 0.5(20) = 70 psf Because column A1 does not participate in the lateral load resistance, the worst case loading will use the uniformly distributed roof load of 92 psf.

Step 3:

Carry out a preliminary first-order analysis. Because the structure is indeterminate, a number of approaches can be taken. If an arbitrary 6:1 ratio of moment of inertia for beams to columns is assumed, a moment distribution analysis yields the moment and force given in Figure 8.9b. Thus, the column will be designed to carry Pu = 29.1 kips and Mu = 37.7 ft-kips

Step 4:

Select a trial size for column A1 and determine its compressive strength and bending strength. Try W10×33. (Section 8.8 addresses trial section selection.) From Manual Table 1-1 A = 9.71 in.2, rx = 4.19 in., ry = 1.94 in., Ix = 171 in.4, rx/ry = 2.16 The column is oriented so that bending is about the x-axis of the column. It is braced against sidesway by the diagonal braces in panel A2–A3 and is pinned at the bottom and rigidly connected at the top in the plane of bending. The column is also braced out of the plane of bending by the brace in panel A1–B1. Because this column is part of a braced frame, K = 1.0 can be used. Although the Specification permits the use of a lower K-factor if justified by analysis, this is not recommended because it would likely require significantly more stiffness in the braced panel. From Manual Table 4-1a, for y-axis buckling φPn = 214 kips for Lc = 16.0 ft From Manual Table 3-10 φMn = 113 ft-kips for Lb = 16.0 ft

310

Chapter 8

Beam-Columns and Frame Behavior Step 5:

Check the W10×33 for combined axial load and bending in-plane. For an unbraced length of 16 ft, the Euler load is π 2 EI π 2 (29,000)(171) Pe1 = 2 = = 1330 kips Lc1 (16.0(12)) 2 The column is bent in single curvature between bracing points, the end points, and the moment at the base is zero, so M1/M2 = 0.0. Thus Cm = 0.6 – 0.4(0.0) = 0.6 Therefore, the amplification factor, with α = 1.0, becomes Cm 0.6 B1 = = = 0.613 ≤ 1.0 αP 1.0 ( 29.1) 1− r 1– Pe1 1330 The Specification requires that B1 not be less than 1.0. Therefore, taking B1 = 1.0, Mrx = B1(Mx) = 1.0(37.7) = 37.7 ft-kips To determine which equation to use, calculate Pu 29.1 = = 0.136 < 0.2 φPn 214 Therefore, use Equation H1-1b Pu Mu + ≤ 1.0 2φPn φM n 37.7 0.5(0.136) + = 0.402 < 1.0 113 Thus, the W10×33 will easily carry the given loads. The solution to Equation H1-1b indicates that there is a fairly wide extra margin of safety. It would be appropriate to consider a smaller column for a more economical design.

EXAMPLE 8.1b Braced Frame Column Design for Combined Axial and Bending by ASD

Goal:

Design column A1 in Figure 8.9 for the given loads using the ASD provisions and the second-order amplification factor provided in Appendix 8 of the Specification.

Given:

The three-dimensional braced frame for a single-story structure is given in Figure 8.9. Rigid connections are provided at the roof level for columns A1, B1, A4, and B4. All other column connections are pinned. Dead Load = 50 psf, Snow Load = 20 psf, Roof Live Load = 10 psf, and Wind Load = 20 psf horizontal. Use A992 steel. Assume that the X-bracing is sufficiently stiffer than the rigid frames to resist all lateral load.

Beam-Columns and Frame Behavior Chapter 8

SOLUTION

Step 1:

311

Determine the appropriate load combinations. From ASCE 7, Section 2.3, the following two combinations are considered. ASCE 7 load combination 3 D + (Lr or S or R) ASCE 7 load combination 6 D + 0.75(0.6W) + 0.75(Lr or S or R)

Step 2:

Determine the factored roof gravity loads for each load combination. For load combination 3 50 + 20 = 70 psf and for load combination 6 50 + 0.75(20) = 65 psf Because column A1 does not participate in the lateral load resistance, the worst case loading will use the uniformly distributed roof load of 70 psf.

Step 3:

Carry out a preliminary first-order analysis. Because the structure is indeterminate, a number of approaches can be taken. If an arbitrary 6:1 ratio of moment of inertia for beams to columns is assumed, a moment distribution analysis yields the moment and force given in Figure 8.9c. Thus, the column will be designed to carry Pa = 22.1 kips and Ma = 28.7 ft-kips

Step 4:

Select a trial size for column A1 and determine its compressive strength and bending strength. Try W10×33. (Section 8.8 addresses trial section selection.) From Manual Table 1-1 A = 9.71 in.2, rx = 4.19 in., ry = 1.94 in., Ix = 171 in.4, rx/ry = 2.16 The column is oriented so that bending is about the x-axis of the column. It is braced against sidesway by the diagonal braces in panel A2–A3 and is pinned at the bottom and rigidly connected at the top in the plane of bending. The column is also braced out of the plane of bending by the brace in panel A1–B1. Because this column is part of a braced frame, K = 1.0 can be used. Although the Specification permits the use of a lower K-factor if justified by analysis, this is not recommended because it would likely require significantly more stiffness in the braced panel. From Manual Table 4-1a for y-axis buckling Pn Ω = 142 kips for Lc = 16.0 ft From Manual Table 3-10 M n Ω = 74.9 ft-kips for Lb = 16.0 ft

312

Chapter 8

Beam-Columns and Frame Behavior Step 5:

Check the W10×33 for combined axial load and bending in-plane. For an unbraced length of 16 ft, the Euler load is π 2 EI π 2 (29,000)(171) Pe1 = 2 = = 1330 kips Lc1 (16.0(12)) 2 The column is bent in single curvature between bracing points, the end points, and the moment at the base is zero, so M1/M2 = 0.0. Thus Cm = 0.6 – 0.4(0.0) = 0.6 Therefore, the amplification factor, with α = 1.6. becomes Cm 0.6 B1 = = = 0.616 ≤ 1.0 αP 1.6 ( 22.1) 1− r 1– Pe1 1330 The Specification requires that B1 not be less than 1.0. Therefore, taking B1 = 1.0, Mrx = B1(Mx) = 1.0(28.7) = 28.7 ft-kips To determine which equation to use, calculate Pu 22.1 = = 0.156 < 0.2 Pn Ω 142 Therefore, use Equation H1-1b Pa Ma + ≤ 1.0 2 Pn Ω M n Ω 28.7 0.5(0.156) + = 0.461 < 1.0 74.9 Thus, the W10×33 will easily carry the given loads. The solution to Equation H1-1b indicates that there is a fairly wide extra margin of safety. It would be appropriate to consider a smaller column for a more economical design.

8.6

Moment Frames A moment frame depends on the stiffness of the beams and columns that make up the frame for stability under gravity loads and under combined gravity and lateral loads. Unlike braced frames, there is no external structure to lean against for stability. Columns in moment frames are subjected to both axial load and moment and experience lateral translation. The same interaction equations, Equations H1-1a and H1-1b, are used to design beamcolumns in moment frames as were previously used for braced frames. However, in addition to the member second-order effects discussed in Section 8.5, there is the additional second-order effect that results from the sway or lateral displacement of the frame.

Beam m-Columns annd Frame Behhavior

8 Figure 8.10

Chappter 8

313

Structu ure Second-Orrder Effect: S Sway.

Figure F 8.10 sh hows a cantileever or flag ppole column uunder the actioon of an axial load and a lateral lo oad. Figure 8.10a is the column as viewed for a first-orderr elastic anallysis where equilibriu um requires a moment at the bottom, M lt = HL. Thhe deflection that results aat the top of the colum mn, Δ1, is the elastic deflecction of a canttilever, so H HL3 Δ1 = (8.3) 3EI A second-ord der analysis yields y the forcces and displlacements as shown in Figure 8.10b. The displacement, Δ2, is the total displacement d t, including seecond-order eeffects, and thhe moment, including g second-ordeer effects, is (8.4) B2 M lt = HL + P Δ 2 An A equivalen nt lateral load d can be dettermined thatt results in thhe same mom ment at the bottom of o the column n as in the seecond-order aanalysis. Thiss load is H + PΔ2/L and iis shown in Figure 8..10c. It I may be assu umed, with only slight erroor, that the diisplacements at the top of the column for the caases in Figurees 8.10b and c are the samee. Thus, usingg the equivaleent lateral loaad ( H + PΔ2 /L) L3 HL3 ⎛ PΔ2 ⎞ ⎛ PΔ2 ⎞ = Δ2 = (8.5) ⎜1 + ⎟ = Δ1 ⎜1 + ⎟ HL ⎠ HL ⎠ 3EII 3EI ⎝ ⎝ Equation n 8.5 can now w be solved forr Δ2, where Δ2 =

Δ1 P Δ1 1− HL

(8.6)

and the result r substitu uted into Equ uation 8.4. Soolving the ressulting equatiion for the am mplification factor, B2, and simpliffying yields

314

Chapter 8

Beam-Columns and Frame Behavior B2 =

Δ2 1 = Δ1 1 – PΔ1 HL

(8.7)

Considering that the typical beam-column will be part of some larger structure, this equation must be modified to include the effect of the multistory and multibay characteristics of the actual structure. This is easily accomplished by summing the total gravity load on the columns in the story and the total lateral load in the story. Thus, Equation 8.7 becomes 1 (8.8) B2 = ΣPΔ1 1– ΣHL This amplification factor is essentially that given in Appendix 8 of the Specification as Equation A-8-6, when combined with Equation A-8-7 1 1 (AISC A-8-6) B2 = = ≥ 1.0 αPstory αPstory Δ H 1– 1– Pe story RM HL where Pstory = total gravity load on the story Pe story =

measure of lateral strength of the structure = RM

HL ΔH

(AISC A-8-7)

ΔH = story drift from a first-order analysis due to the lateral load, H α = 1.0 for LRFD and 1.6 for ASD to account for the nonlinear behavior of the structure at its ultimate strength

⎛ Pmf ⎞ RM = 1 − 0.15 ⎜ (AISC A-8-8) ⎟ ⎝ Pstory ⎠ Pmf = the total vertical load in columns that are part of the lateral load resisting system The variable RM accounts for the influence of the member effect on the sidesway displacement that could not be accounted for in the simplified derivation above. If all the columns are moment frame columns, Pmf/Pstory = 1.0 and RM = 0.85. For braced frames, Pmf = 0 and RM = 1.0. For frames with a combination of columns resisting lateral load through bending and gravity only or leaning columns, the value of RM will be between these limits. It is often desirable to limit the lateral displacement, or drift, of a structure during the design phase. ASCE 7 Appendix C Commentary provides some general guidance. This limit can be defined using a drift index, which is the story drift divided by the story height, ΔH/L. The design then proceeds by selecting members so that the final structure performs as desired. This is similar to beam design, where deflection is the serviceability criterion. Because a limit on the drift index can be established without knowing member sizes, it can be used in Equation A-8-6; thus an analysis with assumed member sizes is unnecessary. With this amplification for sidesway, the moment, Mr, to be used in Equations AISC H11a and AISC H1-1b, can be evaluated. Mr must include both the member and structure secondorder effects. Thus, a first-order analysis without sidesway is carried out, yielding moments, Mnt, that is without translation, to be amplified by B1. Next, a first-order analysis including lateral loads and permitting translation is carried out. This yields moments, Mlt, with translation, to be amplified by B2. The resulting second-order moment is M r = B1M nt + B2 M lt (AISC A8-1)

Beam-Columns and Frame Behavior

Chapter 8

315

where B1 is given by Equation A-8-3 B2 is given by Equation A-8-6 Mnt = first-order moments when the structure is not permitted to translate laterally Mlt = first-order moments that result from just the lateral translation Mlt could include moments that result from unsymmetrical frame properties or loading as well as from lateral loads. In most real structures, however, moments resulting from this lack of symmetry are usually small and are thus often ignored. The second-order force is Pr = Pnt + B2 Plt (AISC A-8-2) The sum of Pnt and Plt for the entire structure will equal the total gravity load on the structure, since the sum of Plt will be zero. For the individual column, however, it is important to amplify the portion of the individual column force that comes from the lateral load. For situations where there is no lateral load on the structure, it may be necessary to incorporate a minimum lateral load in order to capture the second-order effects of the gravity loads. This is covered in Section 8.7 where the three methods provided in the Specification for treating stability analysis and design are discussed.

EXAMPLE 8.2a Moment Frame Strength Check for Combined Compression and Bending by LRFD

Goal:

Using the LRFD provisions, determine whether the W14×90, A992 column shown in Figure 8.11 is adequate to carry the imposed loading.

Given:

An exterior column from an intermediate level of a multi-story moment frame is shown in Figure 8.11. The column is part of a braced frame out of the plane of the figure. Figure 8.11a shows the elevation of the frame with the member to be checked labeled AB. The same column section will be used for the level above and below the column AB. A first-order analysis of the frame for gravity loads plus the minimum lateral load (the minimum lateral load will be discussed in Section 8.7) results in the forces shown in Figure 8.11b, whereas the results for gravity plus wind are shown in Figure 8.11c. Assume that the frame drift under service loads is limited to height/300 for a story shear, H = 148 kips.

SOLUTION

Step 1:

Determine the column effective length factor in the plane of bending. Using the effective length alignment chart introduced in Chapter 5 and given in Commentary Figure C-A-7.2, determine the effective length for buckling in the plane of the moment frame. At each joint there are two columns and one beam framing in. Thus, ⎛ 999 ⎞ 2⎜ ⎟ Σ ( I L )c 12.5 ⎠ G A = GB = = ⎝ = 2.28 Σ ( I L )g ⎛ 2100 ⎞ ⎜ ⎟ ⎝ 30.0 ⎠ Thus, from Figure 5.20, K = 1.66.

Chapter 8

Beam-C Columns and Frame F Behav vior

W24×76 7

A

B

8 at 12.5 ft

4

4

Ix = 21 100 in.

Ix = 999 in.

W14×90

316

A B

7 W24×76 4

Ix = 2100 in.

4 at 300.0 ft (a)

Figure 8.1 11 Exterior Column From m an Intermed diate Level off a Multistoryy Rigid Framee (Example 8.2).

Beam-Columns and Frame Behavior Step 2:

Chapter 8

317

Determine the controlling effective length. With rx/ry = 1.66 for the W14 × 90,

( Lcx )eff = ( KL )eff

=

( KL ) x rx ry

=

1.66 (12.5 ) 1.66

= 12.5 ft

Lcy = KLy = 1.0(12.5) = 12.5 ft Step 3:

Since the effective length about each axis is 12.5 ft, determine the column design axial strength using Lc = 12.5. From the column tables, Manual Table 4-1a, for Lc = 12.5 ft, φPn = 1060 kips

Step 4:

Determine the first-order moments and forces for the loading combination that includes wind, 1.2D +0.5L + 1.0W. The column end moments given in Figure 8.11c are a combination of moments resulting from a nonsway gravity load analysis and a wind analysis: Moment for end A: Moment for end B:

Mnt = 96.7 ft-kips Mlt = 154 ft-kips Mnt = 48.3 ft-kips Mlt = 154 ft-kips

Compression:

Pnt = 354 kips Plt = 99.0 kips Step 5:

Determine the second-order moments by amplifying the first-order moments. No-translation amplification: The no-translation moments must be amplified by B1. From Figure 8.11c it is seen that the end moments bend the column in reverse curvature: M 1 48.3 = = 0.50 M 2 96.7

Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4 Pe1 =

π2 E π 2 (29,000)(999) = = 12,700 kips L2c1 (1.0(12.5)(12)) 2

Thus, with α = 1.0 for LRFD and Pr = 354 + 99 = 453 kips, Equation A-8-3 yields

318

Chapter 8

Beam-Columns and Frame Behavior

B1 =

Cm 0.4 = = 0.415 < 1.0 (1.0)(453) αPr 1− 1– Pe1 12,700

Therefore, B1 = 1.0. Translation amplification: The translation forces and moments must be amplified by B2. The design drift limit of height/300 and Equation A-8-6 are used to determine B2. The total service lateral load on this story is given as H = 148 kips Additional given information is that the total gravity load for this load combination in Figure 8.11c is Pstory = 2110 kips The drift limit under the service lateral load of 148 kips is ΔH = L/300 = 12.5(12)/300 = 0.50 in. Remember that in the calculation of B2, H can be taken as any convenient magnitude, as long as ΔH is the corresponding displacement. This is because it is the ratio of H to ΔH that is used in the determination of Pe story. Thus, with α = 1.0 for LRFD and RM = 0.85 assuming all columns are moment frame columns, Equation A-8-7 gives Pe story =

RM HL 0.85(148)(12.5)(12) = = 37,700 kips ΔH 0.50

and Equation A-8-6 gives 1 1 B2 = = = 1.06>1.0 αPstory (1.0)2110 ⎞ ⎛ 1− 1– ⎜ ⎟ Pe story ⎝ 37,700 ⎠ Thus, the second-order compressive force and moment are Pr = Pnt + B2(Plt) = 354 + 1.06(99) = 459 kips Mr = B1(Mnt) + B2(Mlt) = 1.0(96.7) + 1.06(154) = 260 ft-kips These represent the required strength for this load combination. Step 6:

Determine whether the W14×90 will provide the required strength based on the appropriate interaction equation. The unbraced length of the compression flange for pure bending is 12.5 ft, which is less than Lp = 15.1 ft for this section, taking into account that its flange is noncompact. Thus, from Manual Table 3-2, the design moment strength of the section is

Beam-Columns and Frame Behavior

Chapter 8

319

φMn = 574 ft-kips Determine the appropriate interaction equation. From Step 3, φPn = 1060 kips; Pu 459 = = 0.433 > 0.2 φPn 1060 so use Equation H1-1a, which yields Pu 8 ⎛ M u + ⎜ φPn 9 ⎝ φM n

⎞ ⎟ ≤ 1.0 ⎠

8 ⎛ 260 ⎞ 0.433 + ⎜ ⎟ = 0.836 < 1.0 9 ⎝ 574 ⎠

Thus, the W14×90 is adequate for this load combination. Step 7:

Check the section for the gravity-only load combination, 1.2D + 1.6L. Because this is a gravity-only load combination, Specification Appendix Section 7.2.2, by reference to Section C2.2b, requires that the analysis include a minimum lateral load of 0.002 times the gravity load. This will be further discussed in Section 8.7. For this load combination, the total story gravity load must also be known and is given in Figure 8.11b as Pstory = 2430 kips. Thus, for this frame the minimum lateral load is 0.002Pstory = 0.002(2430) = 4.86 kips at this level. The forces and moments given in Figure 8.11b include the effects of this minimum lateral load. The magnitude of the lateral translation effect is small in this case. Since both the moment due to the minimum lateral load and the amplification factor, B2, are expected to be small, the forces and moments used for this check will be assumed to come from a no-translation case, with little error. If the minimum lateral load would produce large moments or the amplification factor, B2, calculated in Step 5, were large, this would not be a good assumption. Therefore, at end A, Mnt = 142 ftkips, at end B Mnt = 71.0 ft-kips, and Pnt = 522 kips. A quick review of the determination of B1 from the first part of this solution shows that the only change is in the magnitude of the axial force and the member end moments; thus M 1 71.0 = = 0.50 M 2 142 Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4

Pe1 =

π2 EI π 2 (29,000)(999) = = 12,700 kips L2c1 (1.0(12.5)(12)) 2

Thus, with α = 1.0 for LRFD, Cm 0.4 B1 = = = 0.417 < 1.0 (1.0)(522) αP 1− r 1 – Pe1 12,700

320

Chapter 8

Beam-Columns and Frame Behavior Note that B1 is again 1.0. With the assumption that there is no lateral translation, Mlt = 0.0 and B2 is unnecessary,thus

Pr = 522 kips, M r = 1.0(142) = 142 ft-kips Again using Equation H1-1a, Pu 8 ⎛ M u ⎞ + ⎜ ⎟ ≤ 1.0 φPn 8 ⎝ φM n ⎠ 522 8 ⎛ 142 ⎞ + ⎜ ⎟ = 0.712 < 1.0 1060 9 ⎝ 574 ⎠ Thus, the W14×90 is adequate for both load combinations.

EXAMPLE 8.2b Moment Frame Strength Check for Combined Compression and Bending by ASD

Goal:

Using the ASD provisions, determine whether the W14×90, A992 column shown in Figure 8.11 is adequate to carry the imposed loading.

Given:

An exterior column from an intermediate level of a multi-story moment frame is shown in Figure 8.11. The column is part of a braced frame out of the plane of the figure. Figure 8.11a shows the elevation of the frame with the member to be checked labeled AB. The same column section will be used for the level above and below the column AB. A first-order analysis of the frame for gravity loads plus the minimum lateral load (the minimum lateral load will be discussed in Section 8.7) results in the forces shown in Figure 8.11d, whereas the results for gravity plus wind are shown in Figure 8.11e. Assume that the frame drift under service loads is limited to height/300 for a story shear, H = 148 kips.

SOLUTION

Step 1:

Determine the column effective length factor in the plane of bending. Using the effective length alignment chart introduced in Chapter 5 and given in Commentary Figure C-A-7.2, determine the effective length for buckling in the plane of the moment frame. At each joint there are two columns and one beam framing in. Thus, ⎛ 999 ⎞ 2⎜ ⎟ Σ ( I L )c 12.5 ⎠ G A = GB = = ⎝ = 2.28 Σ ( I L )g ⎛ 2100 ⎞ ⎜ ⎟ ⎝ 30.0 ⎠ Thus, from Figure 5.20, K = 1.66.

Step 2:

Determine the controlling effective length. With rx/ry = 1.66 for the W14 × 90,

( Lcx )eff = ( KL )eff

=

( KL ) x rx ry

=

1.66 (12.5 ) 1.66

= 12.5 ft

Beam-Columns and Frame Behavior

Chapter 8

321

Lcy = KLy = 1.0(12.5) = 12.5 ft Step 3:

Since the effective length about each axis is 12.5 ft, determine the column allowable axial strength using Lc = 12.5. From the column tables, Manual Table 4-1a, for Lc = 12.5 ft, Pn Ω = 703 kips

Step 4:

Determine the first-order moments and forces for the loading combination that includes wind, D +0.75L + 0.75(0.6W). The column end moments given in Figure 8.11e are a combination of moments resulting from a nonsway gravity load analysis and a wind analysis: Moment for end A: Moment for end B:

Mnt = 78.0 ft-kips Mlt = 96.0 ft-kips Mnt = 39.0 ft-kips Mlt = 96.0 ft-kips

Compression:

Pnt = 280 kips Plt = 62.0 kips Step 5:

Determine the second-order moments by amplifying the first-order moments. No-translation amplification: The no-translation moments must be amplified by B1. From Figure 8.11e it is seen that the end moments bend the column in reverse curvature: M 1 39.0 = = 0.50 M 2 78.0

Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4 Pe1 =

π2 E π 2 (29,000)(999) = = 12,700 kips L2c1 (1.0(12.5)(12)) 2

Thus, with α = 1.6 for ASD and Pr = 280 + 62 = 342 kips, Equation A-8-3 yields Cm 0.4 B1 = = = 0.418 < 1.0 (1.6)(342) αP 1− r 1 – Pe1 12,700 Therefore, B1 = 1.0. Translation amplification: The translation forces and moments must be amplified by B2. The design drift limit of height/300 and Equation A-8-6 are used to determine B2.

322

Chapter 8

Beam-Columns and Frame Behavior The total service lateral load on this story is given as H = 148 kips Additional given information is that the total gravity load for this load combination in Figure 8.11e is Pstory = 1670 kips The drift limit under the service lateral load of 148 kips is ΔH = L/300 = 12.5(12)/300 = 0.50 in. Remember that in the calculation of B2, H can be taken as any convenient magnitude, as long as ΔH is the corresponding displacement. This is because it is the ratio of H to ΔH that is used in the determination of Pe story. Thus, with α = 1.6 for ASD and RM = 0.85 assuming all columns are moment frame columns, Equation A-8-7 gives Pe story =

RM HL 0.85(148)(12.5)(12) = = 37,700 kips ΔH 0.50

and Equation A-8-6 gives 1 1 B2 = = = 1.08>1.0 αPstory ⎛ (1.6)1670 ⎞ 1− 1– ⎜ ⎟ Pe story ⎝ 37,700 ⎠

Thus, the second-order compressive force and moment are Pr = Pnt + B2(Plt) = 280 + 1.08(62.0) = 347 kips Mr = B1(Mnt) + B2(Mlt) = 1.0(78.0) + 1.08(96.0) = 182 ft-kips These represent the required strength for this load combination. Step 6:

Determine whether this shape will provide the required strength based on the appropriate interaction equation. The unbraced length of the compression flange for pure bending is 12.5 ft, which is less than Lp = 15.1 ft for this section, taking into account that its flange is noncompact. Thus, from Manual Table 3-2, the allowable moment strength of the section is M n Ω = 382 ft-kips Determine the appropriate interaction equation. From Step 3, φPn = 1060 kips; Pa 347 = = 0.494 > 0.2 Pn Ω 703 so use Equation H1-1a, which yields

Beam-Columns and Frame Behavior

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323

Pa 8 ⎛ Ma ⎞ + ⎜ ⎟ ≤ 1.0 Pn Ω 9 ⎝ M n Ω ⎠ 8 ⎛ 182 ⎞ 0.494 + ⎜ ⎟ = 0.918 < 1.0 9 ⎝ 382 ⎠

Thus, the W14×90 is adequate for this load combination. Step 7:

Check the section for the gravity-only load combination, D + L. Because this is a gravity-only load combination, Specification Appendix Section 7.2.2, by reference to Section C2.2b, requires that the analysis include a minimum lateral load of 0.002 times the gravity load. This will be further discussed in Section 8.7. For this load combination, the total story gravity load must also be known and is given in Figure 8.11d as Pstory = 1690 kips. Thus, for this frame the minimum lateral load is 0.002Pstory = 0.002(1690) = 3.38 kips at this level. The forces and moments given in Figure 8.11d include the effects of this minimum lateral load. The magnitude of the lateral translation effect is small in this case. Since both the moment due to the minimum lateral load and the amplification factor, B2, are expected to be small, the forces and moments used for this check will be assumed to come from a no-translation case, with little error. If the minimum lateral load would produce large moments or the amplification factor, B2, calculated in Step 5, were large, this would not be a good assumption. Therefore, at end A, Mnt = 95.0 ftkips, at end B Mnt = 47.0 ft-kips, and Pnt = 348 kips. A quick review of the determination of B1 from the first part of this solution shows that the only change is in the magnitude of the axial force and the member end moments; thus M 1 47.0 = = 0.50 M 2 95.0

Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4(0.50) = 0.4 Pe1 =

π2 EI π 2 (29,000)(999) = = 12,700 kips L2c1 (1.0(12.5)(12)) 2

Thus, with α = 1.6 for ASD, Cm 0.4 B1 = = = 0.418 < 1.0 αPr (1.6)(348) 1− 1– Pe1 12,700 Note that B1 is again 1.0. With the assumption that there is no lateral translation, Mlt = 0.0 and B2 is unnecessary, thus Pr = 348 kips, M r = 1.0(95.0) = 95.0 ft-kips Again using Equation H1-1a,

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Chapter 8

Beam-C Columns and Frame F Behav vior Pa 8 ⎛ Ma ⎞ + ⎜ ⎟ ≤ 1.0 Pn Ω 8 ⎝ M n Ω ⎠ 348 8 ⎛ 95.0 ⎞ + ⎜ ⎟ = 0.716 < 1.0 703 9 ⎝ 382 ⎠

Thus,, the W1 14×90 is adeqquate for bothh load combinnations. The moments in the beams an nd the beam-ccolumn conneections must also be ampllified for the critical case c to accou unt for the seccond-order efffects. This iss done by connsidering equuilibrium of the beam m-column join nt. The ampliffied momentss in the colum mn above andd below the jjoint are added togeth her and this sum distributeed to the beam ms which fram me into the jooint accordingg to their stiffnesses. These T momen nts then establish the conneection designn moments.

8.77

SPECIF FICATION PROVISIO ONS FOR ST TABILITY Y ANALYSIIS AND DES SIGN Up to this po oint, the discu ussion of the interaction off compressionn and bendingg has concenttrated on the develop pment of the interaction equations annd one approoach to incorrporate seconnd-order effects. The Specification n actually provides three appproaches to deal with theese two closelly linked issues. The most m direct ap pproach is to use a generaal second-ordeer analysis in conjunction with the Direct Analy ysis Method described d in Chapter C C. A general g second d-order analy ysis yields forrces and mom ments that caan be used dirrectly in the interaction equations of Chapter H without thee need to resoort to amplifiication factorrs as just described. The T disadvan ntage to this approach is that, since tthe extremelyy useful prinnciple of superpositio on cannot be used (since the t structurall response is nonlinear), a complete nnonlinear analysis mu ust be carried d out for each h load combiination. A ddiscussion of general, or rrigorous, methods of second-order analysis is beyond the scoope of this boook. Thus, in the remaindeer of this book, if seccond-order efffects have no ot already beeen included in the analysis results given, the amplified first-order anaalysis approacch will be ussed to obtain the requiredd second-ordeer forces and momentts.

Figure 8.12 8 Comp parison of thee Effectivee Length Metthod and the Direct Analysis A Meth hod

Beam-Columns and Frame Behavior

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325

8.7.1 Direct Analysis Method

The advantage of the direct analysis method of Specification Chapter C is that for the design of compression members, the effective length factor is taken as 1.0. Since for braced frames, K can always be taken as 1.0 based on Section 7.2.3(a), the direct analysis method is particularly useful for moment frames. The Specification requires that the stiffness of all elements contributing to the lateral load resistance of the structure be reduced. Thus, rather than using EA for the axial stiffness of the members, the modified stiffness, EA* = 0.8EA, is used. Similarly for flexural stiffness, the modified stiffness EI* = 0.8τbEI, where τb accounts for the influence of residual stresses on second-order effects, is used. It should be remembered that the influence of residual stresses on the strength of compression and bending members was already discussed in Chapters 5 and 6. The use of τb in this instance is to capture the influence of those same residual stresses on displacements and thus on second-order effects. This is the same τb used with the alignment charts in the determination of the effective length in Chapter 5. Figure 8.12 shows a comparison between the effective length method already presented and the direct analysis method for a simple structure. Equations H1-1a and H1-1b are plotted for the effective length method and labeled with K = 2.66. This indicates that the compressive strength of the member has been determined using K = 2.66. The nonlinear load-moment curve is identified with EI and EA to indicate that the nominal stiffnesses are used to determine this behavior. The intersection of these two curves indicates that this load and moment combination satisfy the interaction equation. Equations H1-1a and H1-1b are plotted for the direct analysis method and identified with K = 1.0. This indicates that the compressive strength of the member has been determined with K = 1.0. Note that regardless of which approach is selected to determine the compressive strength, the flexural strength is the same for both methods. The nonlinear load-moment curve is identified with EI* and EA* to indicate that the reduced stiffnesses were used to determine this behavior. The intersection of these two curves indicates that this load and moment combination satisfy the interaction equation. Next, note that the axial load magnitude for both of these intersections is nearly the same. Thus, the load that satisfies the interaction equation is the same regardless of which method is used. Since the direct analysis method did not require the determination of K, it is a significantly simpler method than the effective length method. Another consideration that has only briefly been mentioned to this point is the requirement in Section C1 that the influence of geometric imperfections be considered. As with residual stresses, the influence of geometric imperfection on the strength of compression members has already been addressed through the Specification column strength equations. The requirement here is to consider the influence of out-of-plumbness on the stability of the structure. This may be accomplished by modeling the structure in its out-of-plumb condition or through the use of notional loads to simulate the out-of-plumbness. These notional loads will be discussed later in this section. It should be noted that this is not a requirement of the direct analysis method alone but a general requirement for determining required strength. In addition to the direct analysis method, two other design methods are given in the Specification. They are found in Appendix 7. The limitations on the application of these methods are based on the direct analysis method. 8.7.2 Effective Length Method

Appendix 7.2 provides the requirements for the effective length method. This is the approach already described earlier in this chapter for braced and moment frames. It is valid so long as the ratio of second-order deflection to first-order deflection, Δ2/Δ1, is equal to or less than 1.5. Another way to state this requirement is to remember that Δ2/Δ1 = B2, so the effective length method is valid as long as B2 ≤ 1.5. Although this check was not made in Example 8.2, it can now be seen that it was acceptable to use the effective length method in that example, since for both

326

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Beam-Columns and Frame Behavior

LRFD and ASD, B2 ≤ 1.5. A special case occurs when B2 ≤ 1.1. In this case, columns in moment frames can be designed using K = 1.0. The effective length method is essentially the same method used in past practice with the addition of the requirement of a minimum lateral load to be applied in gravity-only load combinations. This is the notional load discussed above to account for initial out-of-plumbness. It is the same as the minimum lateral load used in Example 8.2 and will be discussed later in this section. 8.7.3 First-Order Analysis Method

A third method is given in the Appendix 7.3, the first-order analysis method. This approach permits design without direct consideration of second-order effects except through the application of additional notional lateral loads that account for structure out-of-plumbness and second-order effects. This is possible because of the limits placed on the implementation of this method. As with the effective length method, the structure must support gravity loads primarily through vertical columns, walls or frames and the ratio of the second-order drift to first-order drift must be less than or equal to 1.5. Additionally, compression members that participate in lateral load resistance must behave elastically according to αPr ≤ 0.5Pns (AISC A-7-1) With the foregoing limitations and the application of the notional load given by N i = 2.1α ( Δ L ) Yi ≥ 0.0042Yi (AISC A-7-2) compression members may be designed using K = 1.0. Story gravity load, Yi, is defined in the following. 8.7.4 Geometric Imperfections

Now, consider in more detail the requirement to consider geometric imperfections. Specification Section C1 requires that geometric imperfections be considered in the analysis and design of structures. There are two types of geometric imperfections that must be considered: initial out-ofstraightness and initial out-of-plumbness. The strength equations of Specification Chapter E already account for initial out-of-straightness. To consider out-of-plumbness, the structure may be modeled in the out-of-plumb position or notional loads, as given in Specification Section C2.2b, may be used. The AISC Code of Standard Practice permits columns to be built with an out-of-plumbness tolerance of height/500. Figure 8.13a shows the upper two stories of a column that is out-of-plumb, L/500 = 0.002L. For level 1, with a load P1 applied, a horizontal force of 0.002P1 is required for equilibrium as shown. The next story down, level 2, is also out-of-plumb by the same amount and the load from above is added to the load introduced at that level so that the column must carry P1 + P2. Since this load is also applied at the eccentricity of 0.002L, equilibrium requires that the column be restrained by a force of 0002(P1 + P2) as shown for level 2. When these two columns are put together, it can be seen in Figure 8.13b that the horizontal force at the intersection becomes 0.002 P2. Thus, Figure 8.13b shows for this two story out-of-plumb column, the horizontal forces that are required to keep it in equilibrium. The same effect could be accomplished if the column was modeled plumb and the restraining forces from Figure 8.13b were applied as loads as shown in Figure 8.13c. Thus, the out-of-plumbness can be modeled with a lateral load equal to 0.2 percent of the gravity load introduced at each level of the structure. This lateral load is called a notional load and is taken as Ni = 0.002αYi where Yi is the total gravity load on story i and α is 1.0 for LRFD and 1.6 for ASD, as discussed earlier. The Specification includes α as described here, but when the amplified first-order analysis method is used to obtain second-order effects, this is not

Beam m-Columns annd Frame Behhavior

Lev vel 1, Top

Chappter 8

327

Level 2, Nexxt level down

Figure F 8.13 Notional Lo oad Model forr Geometric Im mperfections. necessary y, since α is included i in th he B1 and B2 calculations. This notionaal load is the ssame as the “minimu um lateral load d” used in Ex xample 8.2.

8.7.5 Comparison of Meethods The three methods of frame stabiility analysis just describeed will be coompared usinng a simple determin nate structure.. Figure 8.14 shows a onee-bay unbraceed frame withh an LRFD ggravity plus lateral lo oad combinattion. Column A is a flag pole columnn and providees all of the lateral load resistance while colum mn B is a graavity only coolumn. Gravitty only colum mns will be ddiscussed in pth in Section n 8.10. Colum mn A is a W1 4×90 bendingg about its strrong axis, collumn B can more dep be any siize sufficientt to support th he gravity loaad since it ddoes not contrribute to the lateral load resistance, and the beaam is assumed to be a rigidd element.

3228

Chapter 8

Beam-C Columns and Frame F Behav vior

One-bay Unnbraced Fram Figgure 8.14 me for Coomparison of A Analysis Metthods

Effective Leength Method d: First the efffective lengthh method of A Appendix Secction 7.2 will be used along with the t B1 – B2 amplification a for second-oorder effects. If the structuure is preventted from swaying, thee nt analysis produces, forr column A, P nt = 200 kipps and Mnt = 0 ft-kips. Thhe lateral translation analysis a produ uces, for colu umn A, Plt = 0 kips and Mnt = 300 ft-kips. Since theere is no moment in the t nt analysis, there are no o P-δ effects (member effe fects) and no nneed to determ mine B1. To assess th he P-Δ effects (sway effectts), B2 will be determined. The 20 kip laateral load prooduces a drift calculated as for a caantilevered beeam,

20 (15 (112 ) ) HL3 = 1.344 in. Δ= = 3EII 3 ( 29,000 )( 999 ) 3

The total graavity load on the structure is 400 kips. Half of this lload is on thee lateral load rresisting column A an nd half is on the t gravity on nly column. T Thus, ⎛ Pmf ⎞ ⎛ 200 ⎞ RM = 1 − 0.1 15 ⎜ ⎟ = 1 − 0.15 ⎜ ⎟ = 0.925 ⎝ 400 ⎠ ⎝ Ptory ⎠ and 200 (15 (12 ) ) H HL = 22490 kips Pe story = RM = 0.925 ΔH 1.34 Thus, 1 1 = B2 = = 11.19 αPstory 1.0 ( 400 ) 1− 1− Pe story 2490 xt, consider the t limitation ns on use off the effectivve length meethod. This sstructure Nex supports graavity loads th hrough verticaal columns soo it meets thee first limitattion in Sectioon 7.2.1. The second limitation reequires that th he second-ordder amplificaation, B2, be less than or equal to 1.5. Since B2 = 1.19 ≤ 1.5 the effective length methood may be useed for this fraame. The required streength, includiing second-o rder effects iis found throuugh Equationns A-8-1 and A-8-2. M r = B1M nt + B2 M lt = 0 + 11.19 ( 300 ) = 357 ft-kips and Pr = Pnt + B2 Plt = 200 + 1.19 ( 0 ) = 2000 kips n the effectiv ve length meethod is deterrmination off the effectivee length The next step in factor. The effective len ngth factor for fo the flag ppole column alone is Kx = 2.0. How wever, as

Beam-Columns and Frame Behavior

Chapter 8

329

discussed in Chapter 5, the inclusion of the gravity only column with load will increase the effective length of column A. Using the approach presented in Chapter 5, with the load on the moment frame column, Pmf = 200 kips and the load on the gravity only column, Pgrav only = 200 kips, the effective length factor is K x* = K x 1 + Pgrav only Pmf = 2.0 1 + 200 200 = 2.83 Assuming that the frame is braced out of the plane of the frame, Ky = 1.0. The available strength of the W14×90 column can be determined from Table 6-2 for an unbraced length of the compression flange Lb = 15 ft < Lp = 15.1 ft, φM n = 574 ft-kips . The controlling effective length is for x-axis buckling, thus

( Lcx )eff

= 2.83(15) 1.66 = 25.6 ft and

φPn = 720 kips. With the required strength and available strength determined, the interaction equation can be checked. First determine which interaction equation should be used. Since Pr φPn = 200 720 = 0.278 > 0.2 use Equation H1-1a, thus

Pr 8 ⎛ M r ⎞ 200 8 ⎛ 357 ⎞ + = + ⎜ ⎟ = 0.278 + 0.553 = 0.831 < 1.0 Pc 9 ⎜⎝ M c ⎟⎠ 720 9 ⎝ 574 ⎠ So the W14×90 is shown to be adequate by the effective length method. First-Order Analysis Method: The first-order analysis method of Appendix Section 7.3 may be used for those structures that meet the limitations of Section 7.3.1. These limitations are the same as for the effective length method with the addition of the requirement that the columns behave elastically such that αPr ≤ 0.5Pns (AISC A-7-1)

Since the W14×90 column does not have slender elements for compression, Pns = Py = Fy Ag = 50 ( 26.5 ) = 1330 kips and for the frame of Figure 8.14 αPr = 200 kips . Thus

200 ≤ 0.5 (1330 ) = 666 kips and the first-order analysis method may be used. The required strength for the first-order analysis method is determined from a first-order analysis that includes a notional load defined by Equation A-7-2 added to the lateral load in all load combinations. This notional load accounts for both the initial out-of-plumbness of the structure and second-order effects. Thus, N i = 2.1α ( Δ L ) Yi ≥ 0.0042Yi (AISC A-7-2) For our structure, Δ = 1.34 in. as before and Yi = 400 kips so the notional load for this load combination is

(

)

Ni = 2.1(1.0) 1.34 (15 (12) ) ( 400) = 6.25 kips ≥ 0.0042 ( 400) = 1.68 kips

Thus, the lateral load in the analysis will be increased from 20 kips to 26.3 kips. The results of the first-order analysis for the determinate structure are Pu = 200 kips and Mu =26.3(15) = 395 ftkips. Although this is called the first-order analysis method, it does require that the moment be amplified by B1 found using Cm B1 = ≥ 1.0 (AISC A-8-3) 1 − αPr Pe1

330

Chapter 8

Beam-Columns and Frame Behavior

This amplification addresses the member effect and is influenced by the buckling strength of the column as a pin ended column in a no sway condition, Pe1, and the equivalent uniform moment factor, Cm. Thus, Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 − 0.4 ( 0 395 ) = 0.6 and with EI* = EI, π2 EI * π2 (29,000)(999) Pe1 = 2 = = 8830 kips 2 Lc1 (15 (12) ) which gives

B1 =

0.6 = 0.614 < 1.0 1 − 200 8830

Therefore there is no amplification needed so Pr = Pu = 200 kips and M r = Mu = 395 ft-kips . The available moment strength of the W14×90 column determined previously from Table 6-2 is unchanged, thus φM n = 574 ft-kips . The controlling effective length is for y-axis buckling, thus Lcy = 15.0 ft and φPn = 1000 kips . With the required strength and available strength determined, the interaction equation can be checked. First determine which interaction equation should be used. Since Pr φPn = 200 1000 = 0.20 ≤ 0.2 use Equation H1-1a, thus Pr 8 ⎛ M r ⎞ 200 8 ⎛ 395 ⎞ + = + ⎜ ⎟ = 0.200 + 0.612 = 0.812 < 1.0 Pc 9 ⎜⎝ M c ⎟⎠ 1000 9 ⎝ 574 ⎠ So the W14×90 is shown to be adequate by the first order analysis method. Direct Analysis Method: The third method to be considered is the direct analysis method of Chapter C. There are no limitations on the use of the direct analysis method like there are on the effective length or first-order analysis methods and second-order effects and initial out-ofplumbness must be accounted for as they were for the effective length method. The only new requirement is that the stiffness of all members that contribute to the lateral load resistance be * reduced in the analysis to EI * = 0.8τb EI and EA = 0.8EA . It is this stiffness reduction that permits the use of an effective length factor equal to one when using the direct analysis method. From the discussion of the effective length method it was seen that B2 was less than 1.5 when using the unreduced stiffness thus the notional load to account for out-of-plumbness does not need to be added to the lateral load. Thus, from a first order analysis of the determinate structure, Pu = 200 kips and M u = 20.0 (15 ) = 300 ft-kips . As for the effective length method, τb = 1.0 so * that the flexural stiffness of column A will be taken as EI = 0.8EI . Thus the 20 kip lateral load produces a drift calculated as for a cantilevered beam,

20 (15 (12 ) ) HL3 Δ= = = 1.68 in. * 3EI 3( 0.8)( 29,000)( 999 ) 3

The total gravity load on the structure is 400 kips. Half of this load is on the lateral load resisting column A and half is on the gravity only column. Thus again, ⎛ Pmf ⎞ ⎛ 200 ⎞ RM = 1 − 0.15 ⎜ ⎟ = 1 − 0.15 ⎜ ⎟ = 0.925 ⎝ 400 ⎠ ⎝ Ptory ⎠ and

Beam-Columns and Frame Behavior Pe story = RM

Chapter 8

331

20 (15 (12 ) ) HL = 0.925 = 1980 kips 1.68 ΔH

Thus, 1 1 = = 1.25 αPstory 1.0 ( 400 ) 1− 1− Pe story 1980 Note that the drift increased from what was calculated for the effective length method and therefore the second-order amplification increased. The required strength, including second-order effects is found through Equations A-8-1 and A-8-2. M r = B1M nt + B2 M lt = 0 + 1.25 ( 300 ) = 375 ft-kips and Pr = Pnt + B2 Plt = 200 + 1.25 ( 0 ) = 200 kips B2 =

The available moment strength of the W14×90 column determined previously from Table 6-2 is unchanged, thus φM n = 574 ft-kips . The controlling effective length is for y-axis buckling, thus Lcy = 15.0 ft and φPn = 1000 kips . With the required strength and available strength determined, the interaction equation can be checked. First determine which interaction equation should be used. Since Pr φPn = 200 1000 = 0.20 ≤ 0.2 use Equation H1-1a, thus

Pr 8 ⎛ M r ⎞ 200 8 ⎛ 375 ⎞ + = + ⎜ ⎟ = 0.200 + 0.581 = 0.781 < 1.0 Pc 9 ⎜⎝ M c ⎟⎠ 1000 9 ⎝ 574 ⎠ So the W14×90 is adequate by the direct analysis method. Note that based on the results of the interaction equation, this approach is less conservative than the other two methods. Since the only new requirement of the direct analysis method is to use a reduced stiffness in calculating secondorder effects and this permits the use of an effective length factor in the lateral load resisting direction of one, this is clearly the simplest and most direct method available. Three methods of analysis are available and all three have their place in design. It is up to the user to determine when to use each approach most efficiently.

EXAMPLE 8.3a Direct Analysis Method for Column Design by LRFD

Goal:

Using the LRFD provisions and the results from a second-order direct analysis, determine if a W14×132, A992 member is adequate to carry the given loads and moments.

Given:

The column has a length of 16 ft and is braced at the ends only. The results of the second-order direct analysis are Pu = 800 kips, Mux = 300 ft kips, and Muy = 76 ft kips.

SOLUTION

Step 1:

Determine the required strength. Since the given results are from a second-order analysis, there is no need to amplify forces and moments; thus Pr = Pu = 800 kips, M rx = M ux = 300 ft kips,

M ry = M uy = 76 ft kips

332

Chapter 8

Beam-Columns and Frame Behavior Step 2:

Determine the available compressive strength of the column. Since the given results are from a direct analysis, K = 1.0; thus, from Manual Table 4-1 with Lc = 16.0 ft, φPn = 1440 kips

Step 3:

Determine the available strength for bending about the x-axis. With an unbraced length Lb = 16 ft, from Manual Table 3-2,

φM p = 878 ft kips, L p = 13.3 ft, φBF = 7.74 kips

and

Step 4:

φM nx = φM p − φBF ( Lb − Lp ) = 878 − 7.74 (16.0 − 13.3) = 857 ft kips

Determine the available strength for bending about the y-axis. From Manual Table 3-4, φM ny = 424 ft kips

Step 5:

Check the W14×132 for combined axial load and bending. To determine which equation to use, check Pu 800 = = 0.556 ≥ 0.2 φPn 1440 Therefore, use Equation H1-1a. M ry Pr M rx + + ≤ 1.0 φPn φM nx φM ny 0.556 +

300 76 + = 1.09 > 1.0 857 424

Thus, the W14×132 will not carry the given load.

EXAMPLE 8.3a Goal: Direct Analysis Method for Column Design Given: by ASD

SOLUTION

Step 1:

Using the ASD provisions and the results from a second-order direct analysis, determine if a W14×132, A992 member is adequate to carry the given loads and moments. The column has a length of 16 ft and is braced at the ends only. The results of the second-order direct analysis are Pa = 530 kips, Max = 200 ft kips, and May = 52 ft kips. Determine the required strength. Since the given results are from a second-order analysis, there is no need to amplify forces and moments. Thus, Pr = Pa = 530 kips, M rx = M ax = 200 ft kips,

M ry = M ay = 52 ft kips

Beam-Columns and Frame Behavior Step 2:

Chapter 8

333

Determine the available compressive strength of the column. Since the given results are from a direct analysis, K = 1.0. Thus, from Manual Table 4-1 with Lc = 16.0 ft, Pn = 960 kips Ω

Step 3:

Determine the available strength for bending about the x-axis. With an unbraced length Lb = 16 ft, from Manual Table 3-2, Mp BF = 584 ft kips, Lp = 13.3 ft, = 5.15 kips Ω Ω And M nx M p BF = − ( Lb − Lp ) = 584 − 5.15 (16.0 − 13.3) = 570 ft kips Ω Ω Ω

Step 4:

Determine the available strength for bending about the y-axis. From Manual Table 3-4 M ny = 282 ft kips Ω

Step 5:

Check the W14×132 for combined axial load and bending. To determine which equation to use, check Pu 530 = = 0.552 ≥ 0.2 φPn 960 Therefore, use Equation H1-1a. M ry Pr M rx + + ≤ 1.0 Pn Ω M nx Ω M ny Ω 0.552 +

200 52 + = 1.09 > 1.0 570 282

Thus, the W14×132 will not carry the given load.

8.8 INITIAL BEAM-COLUMN SELECTION Beam-column design is a trial-and-error process that requires that the beam-column section be known before any of the critical parameters can be determined for use in the appropriate interaction equations. There are numerous approaches to determining a preliminary beam-column size. Each incorporates its own level of sophistication and results in its own level of accuracy. Regardless of the approach used to select the trial section, one factor remains—the trial section must ultimately satisfy the appropriate interaction equation. To establish a simple, yet useful, approach to selecting a trial section, Equation H1-1a is modified by multiplying each term by Pc which yields 8 M rx Pc 8 M ry Pc Pr + + ≤ Pc (8.9) 9 M cx 9 M cy Then multiplying the third term by Mcx/Mcx, letting

334

Chapter 8

Beam-Columns and Frame Behavior m=

8Pc 9M cx

and

U=

M cx M cy

and substituting into Equation 8.9 yields Pr + mM rx + mUM ry ≤ Pc

(8.10)

Because Equation 8.10 calls for the comparison of the left side of the equation to the column strength, Pc, Equation 8.10 can be thought of as an effective axial load; thus Peff = Pr + mM rx + mUM ry ≤ Pc (8.11) The accuracy used in the evaluation of m and U dictates the accuracy with which Equation 8.11 represents the strength of the column being selected. Because at this point in a design the actual column section is not known, exact values of m and U cannot be determined. Past editions of the AISC Manual have presented numerous approaches to the evaluation of these multipliers. A simpler approach however, is more useful for preliminary design. If the influence of the length—that is, all buckling influence on Pc and Mcx—is neglected, the ratio, Pc/Mcx, becomes A/Zx, and m = 8A/9Zx. Evaluation of this m for all W6 to W14 shapes with the inclusion of a units correction factor of 12 results in the average m values given in Table 8.2. If the relationship between the area, A, and the plastic section modulus, Zx, is established using an approximate internal moment arm of 0.89d, where d is the nominal depth of the member in inches, then m = 24/d. This value is also presented in Table 8.2. This new m is close enough to the average m that it may be readily used for preliminary design. When bending occurs about the y-axis, U must be evaluated. A review of the same W6 to W14 shapes results in the average U values given in Table 8.2. However, an in-depth review of the U values for these sections shows that only the smallest sections for each nominal depth have U values appreciably larger than 3. Thus, a reasonable value of U = 3.0 can be used for the first trial. More accurate evaluations of these multipliers, including length effects, have been conducted, but there does not appear to be a need for this additional accuracy in a preliminary design. Once the initial section is selected, however, the actual Specification provisions must be satisfied. Table 8.2 Shape W6 W8 W10 W12 W14

EXAMPLE 8.4a Initial Trial Section Selection by LRFD SOLUTION

Simplified Bending Factors mavg m = 24/d 4.41 4.00 3.25 3.00 2.62 2.40 2.08 2.00 1.71 1.71

Uavg 3.01 3.11 3.62 3.47 2.81

Goal:

Determine the initial trial section for a column.

Given:

The loadings of Figure 8.11c are to be used. Assume the column is a W14 and use A992 steel. Also, use the simplified values of Table 8.2, m = 24/d.

Step 1:

Obtain the required strength from Figure 8.11c. Use the first-order analysis

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335

results. Pu = 453 kips M u = 251 ft-kips Step 2:

Determine the effective load by combining the axial force and the bending moment. For a W14, m = 1.71, so

Peff = 453 + 1.71( 251) = 882 kips Step 3:

Select a trial column size to carry the required force, Peff. Using an effective length Lc = 12.5 ft, from Manual Table 4-1, the lightest W14 to carry this load is W14 × 90 with φPn = 1060 kips Example 8.2a showed that this column adequately carries the imposed load. Because the approach used here is expected to be conservative, it would be appropriate to consider the next smaller selection, a W14×82, and check it against the appropriate interaction equations.

EXAMPLE 8.4b Initial Trial Section Selection by ASD SOLUTION

Goal:

Determine the initial trial section for a column.

Given:

The loadings of Figure 8.11e are to be used. Assume the column is a W14, and use A992 steel. Also, use the simplified values of Table 8.2, m = 24/d.

Step 1:

Obtain the required strength from Figure 8.11e. Use the first-order analysis results. Pa = 342 kips M a = 174 ft-kips

Step 2:

Determine the effective load by combining the axial force and the bending moment. For a W14, m = 1.71; thus Peff = 342 + 1.71(174 ) = 640 kips

Step 3:

Select a trial column size to carry the required force, Peff. Using an effective length Lc = 12.5 ft, from Manual Table 4-1, the lightest W14 to carry this load is W14 × 90 with Pn Ω = 703 kips Example 8.2b showed that this column adequately carries the imposed load. Because the approach used here is expected to be conservative, it would be appropriate to consider the next smaller selection, a W14×82, and check it against the appropriate interaction equations.

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Every column section selected must be checked through the appropriate interaction equations for the second-order forces and moments. Thus, the process for the initial selection should be quick and reasonable. The experienced designer will rapidly learn to rely on that experience rather than these simplified approaches.

8.9

BEAM-COLUMN DESIGN USING MANUAL PART 6 Manual Part 6, Design of Members Subject to Combined Loading contains Table 6-2 which includes the axial and flexural strength for all W-shapes. Although these tables are presented here as they relate to combined loading, they can also be used for compression only, bending only, tension only and shear. There is no information found in Table 6-2 that is not already included in other Parts of the Manual already discussed. The advantage for combined loading is that all of the available strength values needed are found in one location. Figure 8.15 is a portion of Manual Table 6-2. It shows that the compressive strength for a given section is a function of the effective length about the weak axis of the member. The effective length is tabulated in the center of the table with the compressive strengths shown on the left portion of the table. This portion of the table is used in exactly the same way as the column tables in Part 4 of the Manual. The strong axis bending strength is a function of the unbraced length of the compression flange of the beam. Previously, this information was available only through the beam curves in Part 3 of the Manual. In Table 6-2 it is tabulated on the right portion of the table with the same column of lengths now defined as the unbraced length of the compression flange. Weak axis bending is not a function of length, so only one value is given for each shape. Although not used for beam-columns, when tension is combined with bending, the table also provides tension yield and rupture strength.

Goal: EXAMPLE 8.5a Combined Strength Check Using Manual Part 6 and Given: LRFD

Check the strength of a beam-column using Manual Part 6 and compare to the results of Example 8.2a.

Step 1:

Determine the values needed from Manual Table 6-2 (Figure 8.15). The column is required to carry a compressive force with an effective length about the y-axis of 12.5 ft and an x-axis moment with an unbraced length of 12.5 ft. Thus, from Figure 8.15, φPn = 1060 kips

SOLUTION

It has already been shown that the W14×90 column of Example 8.2a is adequate by LRFD. Use the required strength values given in Example 8.2a and recheck this shape using the values found in Figure 8.15 or Manual Table 6-2.

φM n = 574 ft-kips Step 2:

Determine which interaction equation to use. Pr 459 = = 0.433 > 0.2 φPn 1060 Therefore, use Equation H1-1a. 200 8 ⎛ 260 ⎞ + ⎜ ⎟ = 0.433 + 0.403 = 0.836 < 1.0 1060 9 ⎝ 574 ⎠ Therefore, as previously determined in Example 8.2a, the shape is adequate for this column and this load combination. The results from Manual Tables

Beam-Columns and Frame Behavior

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337

6-2 are exactly the same as those determined from Table 4-1 for compression and Table 3-2 for bending.

EXAMPLE 8.5b Combined Strength Check Using Part 6 and ASD

Goal:

Check the strength of a beam-column using Manual Part 6 and compare to the results of Example 8.2b.

Given:

It has already been shown that the W14×90 column of Example 8.2b is adequate by ASD. Use the required strength values given in Example 8.2b and recheck this shape using the values found in Figure 8.15 or Manual Table 6-2.

SOLUTION

Step 1:

Determine the values needed from Manual Table 6-2 (Figure 8.15). The column is required to carry a compressive force with an effective length about the y-axis of 12.5 ft and an x-axis moment with an unbraced length of 12.5 ft. Thus, from Figure 8.15, Pn Ω = 703 kips M n Ω = 382 ft-kips

Step 2:

Determine which interaction equation to use. Pr 347 = = 0.494 > 0.2 Pn Ω 703 Thus, use Equation H1-1a. 347 8 ⎛ 182 ⎞ + ⎜ ⎟ = 0.494 + 0.424 = 0.918 < 1.0 703 9 ⎝ 382 ⎠ Thus, as previously determined in Example 8.2b, the shape is adequate for this column and this load combination. The results from Manual Tables 6-2 are exactly the same as those determined from Table 4-1 for compression and Table 3-2 for bending.

8.10 COMBINED SIMPLE AND MOMENT FRAMES The practical design of steel structures often results in frames that combine segments of rigidly connected elements with segments that are pin connected as was the case in the example frame used in Section 8.7.5. If these structures rely on the moment frame to resist lateral load and to provide the overall stability of the structure, the rigidly connected columns are called upon to carry more load than appears to be directly applied to them. In these combined simple and moment frames, the simple columns “lean” on the moment frames in order to maintain their stability and thus are often called leaning columns. They are also called gravity columns which is a more appropriate term since they participate only in carrying gravity loads. These columns can be designed with an effective length factor, K = 1.0, regardless of the approach to analysis that has been taken. Because these gravity only columns have no lateral stability of their own, the moment frame columns must be designed to provide the lateral stability for the full frame. Although this combination of framing types makes design of a structure more complicated, it can also be economically advantageous, because the combination can reduce the number of moment connections for the full structure and thereby reduce overall cost.

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Beam-C Columns and Frame F Behav vior

Figure 8.15 Comb bined Axial an nd Bending Strength S for W W-Shapes. Copyrig ght © Americcan Institute of o Steel Consttruction, Inc. Reprinted wiith Permissionn. All rights reserved.

Beam m-Columns annd Frame Behhavior

Chappter 8

339

Fig gure 8.16 Piinned Base U Unbraced Fram me. Numerous N deesign approacches have beeen proposed ffor consideraation of the ggravity only column and a associateed moment frrame design.33,4 Yura propposes to desiggn columns thhat provide lateral sttability for th he total load on the fram me at the storry in questionn, whereas L LeMessurier presents a modified effective e lengtth factor thatt accounts forr the full fram me stability. P Perhaps the a is th hat presented by Yura, as ffollows. most straaightforward approach The T two-colu umn frame sh hown in Figuure 8.16a is a moment fframe with ppinned base columns and a rigidly y connected beam. b The coolumn sizes are selected so that, undeer the loads shown, they buckle simultaneou usly in a siidesway modde, because their load is directly proportio onal to the sttiffness of thee members. E Equilibrium iin the displacced position iis shown in Figure 8.16b. The latteral displacement of the frame, Δ, ressults in a mooment at the ttop of each column equal e to the lo oad applied on o the columnn times the diisplacement, as shown. Thhese are the second-o order effects discussed d in Section 8.6. The total loaad on the fraame is 700 kiips, and the total PΔ moment is 70 00Δ, divided between b the ttwo columns based on the load that eachh carries. If I the load on n the right-han nd column iss reduced to 5500 kips, the column doess not buckle sidewayss, because thee moment at the t top is now w less than 6000Δ. To reachh the bucklingg condition, a horizon ntal force musst be applied at the top of tthe column, aas shown in Figure 8.17b. T This force

Fiigure 8.17 C Columns from m Unbraced F Frame wiith Revised L Loading.

3

Yura, J. A., “The “ Effectivee Length of Collumns in Unbraced Frames,” Engineering JJournal, AISC,, Vol. 8, No. 2,, 1971, pp. 37–42. 4 LeMessurierr, W. J., “A Praactical Method of Second Ord der Analysis,” Engineering JJournal, AISC, Vol. 14, No. 22, 1977, pp. 49–67.

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Chapter 8

Beam-C Columns and Frame F Behav vior

can result on nly from action on the lefft column thatt is transmitteed through thhe beam. Equuilibrium of the left co olumn, shown n in Figure 8..17a, requiress that an addiitional columnn load of 1000 kips be applied to th hat column in n order for th he load on thhe frame to bbe in equilibriium in this ddisplaced position. Th he total framee capacity iss still 700 kipps and the tootal second-oorder momennt is still 700Δ. oad that an in ndividual collumn can resiist is limited to that perm mitted for The maximum lo the column in i a braced frrame for whicch K = 1.0. Inn this examplee case, the lefft column couuld resist 400 kips and d the right co olumn 2400 kips. k This is aan increase off four times tthe load origiinally on the column, because the effective e leng gth factor for eeach column would be redduced from 2.0 to 1.0. nal capacity of o the left column is only with respectt to the bendiing axis. The column The addition would have the same cap pacity about th he other axis as it did prioor to reducingg the load on tthe right column. ne column to carry increassed load wheen another coolumn in the frame is The ability of on called upon to carry lesss than its critiical load for lateral buckliing is an impportant characcteristic. d column to lean on a m moment framee column, provided that tthe total This allows a pin-ended d on the framee can be carrieed by the rigidd frame. gravity load

Fiigure 8.18 Frame F Used iin Example 88.6.

Figure 8.19 Nomiinal Wind Loaad, Snow Loaad, and Dead Load (Exampple 8.6).

Beam-Columns and Frame Behavior

EXAMPLE 8.6a Moment Frame Strength and Stability by LRFD

Chapter 8

341

Goal:

Determine whether the structure shown in Figure 8.18 has sufficient strength and stability to carry the imposed loads.

Given:

The frame shown in Figures 8.18 and 8.19 is similar to that in Example 8.1, except that the in-plane stability and lateral load resistance is provided by the moment frame action at the four corners. The exterior columns are W8×40, and the roof girder is assumed to be rigid. Out-of-plane stability and lateral load resistance is provided by X-bracing along column lines 1 and 4. The loading is the same as that for Example 8.1: Dead Load = 50 psf, Snow Load = 20 psf, Roof Live Load = 10 psf, and Wind Load = 20 psf horizontal. Use A992 steel.

SOLUTION

Step 1:

The analysis of the frame for gravity loads as given for Example 8.1 will be used. Because different load combinations may be critical, however, the analysis results for nominal Snow and nominal Dead Load are given in Figure 8.19b. The analysis results for nominal Wind Load acting to the left are given in Figure 8.19c.

Step 2:

Determine the first-order forces and moments for the column on lines A-1. For ASCE 7 load combination 3: Pu = 1.2(15.8) + 1.6(6.33) + 0.5(0.710) = 29.1 + 0.355 = 29.5 kips Mu = 1.2(20.5) + 1.6(8.20) + 0.5(32.0) = 37.7 + 16.0 = 53.7 ft-kips For ASCE 7 load combination 4: Pu = 1.2(15.8) + 0.5(6.33) + 1.0(0.710) = 22.1 + 0.710 = 22.8 kips Mu = 1.2(20.5) + 0.5(8.20) + 1.0(32.0) = 28.7 + 32.0 = 60.7 ft-kips

Step 3:

Determine the total story gravity load acting on one frame. Dead = 0.05 ksf (90 ft)(50 ft)/2 frames = 113 kips Snow = 0.02 ksf (90 ft)(50 ft)/2 frames = 45.0 kips

Step 4:

Determine the second-order forces and moments for load combination 3. Gravity loads will be assumed to yield the no-translation effects, and wind load to yield the lateral translation effects. From Step 2, Pnt = 29.1 kips, Plt = 0.355 kips, Mnt = 37.7 ft-kips, Mlt = 16.0 ft-kips For the W8×40, A = 11.7 in.2, Ix = 146 in.4, rx = 3.53 in., rx/ry = 1.73 In the plane of the frame,

⎛ 0 ⎞ Cm = 0.6 = 0.4 ( M 1 M 2 ) = 0.6 – 0.4 ⎜ ⎟ = 0.6 ⎝ 37.7 ⎠ π2 EI π2 (29,000)(146) Pe1 = 2 x = = 1130 kips (16.0(12)) 2 Lc1

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Chapter 8

Beam-Columns and Frame Behavior and with Pr = Pnt + Plt = 29.1 + 0.355 = 29.5 Cm 0.6 B1 = = = 0.616 < 1.0 29.5 αPr 1− 1– Pe1 1130 Therefore, use B1 = 1.0. To determine the sway amplification, the total gravity load on the frame for this load combination from Step 3 is Pstory = 1.2(113) + 1.6(45.0) = 208 kips A serviceability drift index of 0.003 is maintained under the actual wind loads. Therefore, H = 4.0 kips, and Δ/L = 0.003 is used to determine the sway amplification factor. If this limit is not met at the completion of the design, the second-order effects must be recalculated. The sway amplification is given by

B2 =

1 ⎛ αP 1 − ⎜ story ⎝ Pe story

⎞ ⎟ ⎠

> 1.0

(AISC A-8-6)

and Pe story = RM

HL ΔH

(AISC A-8-7)

Since one third of the load is on the moment frame corner columns, Equation A-8-8 gives ⎛1⎞ RM = 1 − 0.15 ⎜ ⎟ = 0.95 ⎝ 3⎠ Thus, with α = 1.0 for LRFD, Equation A-8-6 becomes 1 1 B2 = = = 1.20 1.0 ( 208) ⎛ αPstory ⎛ Δ ⎞ ⎞ 1− ⎜ (0.003) ⎜ ⎟⎟ 1− 0.95 ( 4.0 ) ⎝ RM H ⎝ L ⎠ ⎠ Thus, the second-order force and moment are Mr = 1.0(37.7) + 1.20(16.0) = 56.9 ft-kips Pr = 29.1 + 1.20(0.355) = 29.5 kips Step 5:

Determine whether the column satisfies the interaction equation. Because the roof beam is assumed to be rigid in this example, use the recommended design value of K = 2.0 from Figure 5.17 case f in the plane of the frame, Lcx = 2.0(16.0) = 32.0 ft. Out of the plane of the frame, this is a braced frame where K = 1.0; thus, Lcy = 16.0 ft. Determine the critical buckling axis.

Beam-Columns and Frame Behavior

( Lcx )eff

=

Chapter 8

343

Lcx 32.0 = = 18.5 ft > Lcy = 16.0 ft rx / ry 1.73

Thus, from Manual Table 6-2, using Lc =( Lcx) eff = 18.5 ft, φPn = 222 kips and from Manual Table 6-2 with an unbraced length of Lb = 16 ft φMnx = 128 ft-kips Determine the appropriate interaction equation to use. Pr 29.5 = = 0.133 < 0.2 φPn 222 Therefore, use Equation H1-1b. Pu M + u ≤ 1.0 2φPn φM n 29.5 56.9 + = 0.511 < 1.0 2(222) 128 Thus, the column is adequate for this load combination. Step 6:

Determine the first-order forces and moments for load combination 4 with the same assumption as to translation and no-translation effects. From Step 2. Pnt = 22.1 kips, Plt = 0.710 kips, Mnt = 28.7 ft-kips, Mlt = 32.0 ft-kips

Step 7:

Determine the second-order forces and moments. In the plane of the frame, as in Step 4,

⎛ 0 ⎞ Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4 ⎜ ⎟ = 0.6 ⎝ 28.7 ⎠ π2 EI π2 (29, 000)(146) Pe1 = 2 x = = 1130 kips (16(12)) 2 Lc1 and B1 =

Cm 0.6 = = 0.612 < 1.0 22.8 αP 1− r 1– Pe1 1130

Therefore, use B1 = 1.0. To determine the sway amplification, the total gravity load on the frame is Pu = 1.2(113) + 0.5(45.0) = 158 kips Again, a serviceability drift index of 0.003 is maintained under the actual wind loads. Therefore, H = 4.0 kips, and Δ/L = 0.003 is used to determine the sway amplification factor. As before, RM = 0.95 so

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Beam-Columns and Frame Behavior

B2 =

1 1 = = 1.14 1.0 (158) ⎛ αPstory ⎛ Δ ⎞ ⎞ 1– ⎜ (0.003) ⎜ ⎟⎟ 1– 0.95 ( 4.0 ) ⎝ RM H ⎝ L ⎠ ⎠

Thus, the second-order force and moment are Mr = 1.0(28.7) + 1.14(32.0) = 65.2 ft-kips Pr = 22.1 + 1.14(0.710) = 22.9 kips Step 8:

Determine whether the column satisfies the interaction equation. Using the same strength values found in Step 5, determine the appropriate interaction equation. Pr 22.9 = = 0.103 < 0.2 φPn 222 Therefore, use Equation H1-1b. Pu M + u ≤ 1.0 2φPn φM n 22.9 65.2 + = 0.561 < 1.0 2(222) 128 Thus, the column is adequate for this load combination also.

Step 9:

The W8×40 is shown to be adequate for gravity and wind loads in combination. Now, check to see that these columns have sufficient capacity to brace the interior pinned columns for load combination 3, which will put the greatest load on the gravity only columns.

Step 10:

For stability in the plane of the frame, using the Yura approach discussed in Section 8.10, the total load on the structure is to be resisted by the four corner columns; thus Dead Load = 0.05 ksf (50 ft)(90 ft)/4 columns = 56.3 kips/column Snow Load = 0.02 ksf (50 ft)(90 ft)/4 columns = 22.5 kips/column Thus, for load combination 3 Pu = 1.2(56.3) + 1.6(22.5) +0.5(0.710) = 104 kips Mu = 1.2(20.5) + 1.6(8.20) + 0.5(32.0) = 53.7 ft-kips

Step 11:

Determine the second-order amplification. As before, for the length Lx = 16.0 ft, Pe1 = 1130 kips, and Cm = 0.6, the second-order amplification for member effect is Cm 0.6 B1 = = = 0.66 < 1.0 104 αPr 1− 1– Pe1 1130 Therefore, use B1 = 1.0 and Pr = Pu = 104 kips. Sway amplification will be the same as determined in step 4, since the gravity load is the same; thus B2 = 1.20. Therefore Mr = Mu = 1.0(37.7) + 1.20(16.0) = 56.9 ft-kips.

Beam-Columns and Frame Behavior Step 12:

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345

Check the corner columns for interaction under these forces and moments. As determined in Step 5 for in-plane buckling, φPnx = 222 kips φMnx = 128 ft-kips Checking for the appropriate interaction equation, Pu 104 = = 0.468 > 0.2 φPn 222 Thus, use Equation H1-1a. Pu 8 ⎛ M u x ⎞ + ⎜ ⎟ ≤ 1.0 φPn 9 ⎝ φM n x ⎠ 104 8 ⎛ 56.9 ⎞ + ⎜ ⎟ = 0.864 < 1.0 222 9 ⎝ 128 ⎠

Thus, the W8×40 is adequate for both strength under combined load and stability for supporting the gravity only columns.

EXAMPLE 8.6b Moment Frame Strength and Stability by ASD

Goal:

Determine whether the structure shown in Figure 8.18 has sufficient strength and stability to carry the imposed loads.

Given:

The frame shown in Figures 8.18 and 8.19 is similar to that in Example 8.1 except that the in-plane stability and lateral load resistance is provided by the rigid frame action at the four corners. The exterior columns are W8×40, and the roof girder is assumed to be rigid. Out-of-plane stability and lateral load resistance is provided by X-bracing along column lines 1 and 4. The loading is the same as that for Example 8.1: Dead Load = 50 psf, Snow Load = 20 psf, Roof Live Load = 10 psf, and Wind Load = 20 psf horizontal. Use A992 steel.

SOLUTION

Step 1:

The analysis of the frame for gravity loads as given for Example 8.1 will be used. Because different load combinations may be critical, however, the analysis results for nominal Snow and nominal Dead Load are given in Figure 8.19b. The analysis results for nominal Wind Load acting to the left are given in Figure 8.19c.

Step 2:

Determine the first-order forces and moments for the column on lines A-1. For ASCE 7 load combination 3: Pa = (15.8) + (6.33) = 22.1 kips Ma = (20.5) + (8.20) = 28.7 ft-kips For ASCE 7 load combination 6: Pa = (15.8) + 0.75(6.33) + 0.75(0.6(0.710)) = 20.9 kips Ma = (20.5) + 0.75(8.20) + 0.75(0.6(32.0)) = 41.1 ft-kips

Step 3:

Determine the total story gravity load acting on one frame. Dead = 0.05 ksf (90 ft)(50 ft)/2 frames = 113 kips Snow = 0.02 ksf (90 ft)(50 ft)/2 frames = 45.0 kips

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Chapter 8

Beam-Columns and Frame Behavior Step 4:

Determine the second-order forces and moments for load combination 3. Gravity loads will be assumed to yield the no-translation effects. With no wind load, there will be no lateral translation effects; thus From Step 2: Pnt = 22.1 kips, Plt = 0 kips, Mnt = 28.7 ft-kips, Mlt = 0 ft-kips For the W8×40: A = 11.7 in.2, Ix = 146 in.4, rx = 3.53 in., rx/ry = 1.73 In the plane of the frame:

⎛ 0 ⎞ Cm = 0.6 − 0.4 ( M 1 M 2 ) = 0.6 – 0.4 ⎜ ⎟ = 0.6 ⎝ 28.7 ⎠ π2 EI π2 (29,000)(146) Pe1 = 2 x = = 1130 kips (16.0(12)) 2 Lc1 and Cm 0.6 = = 0.619 < 1.0 1.6(22.1) αPr 1− 1– Pe1 1130 Therefore, use B1 = 1.0. B1 =

To determine the sway amplification, even though there are no lateral translation forces or moments for this combination, the total gravity load on the frame for this load combination from Step 3 is Pstory = (113) + (45.0) = 158 kips A serviceability drift index of 0.003 is maintained under the actual wind loads. Therefore, H = 4.0 kips, and Δ/L = 0.003 is used to determine the sway amplification factor. If this limit is not met at the completion of the design, the second-order effects must be recalculated. The sway amplification is given by

B2 =

1 ⎛ αP 1 − ⎜ story ⎝ Pe story

⎞ ⎟ ⎠

> 1.0

(AISC A-8-6)

and Pe story = RM

HL ΔH

(AISC A-8-7)

Since one third of the load is on the moment frame corner columns, Equation A-8-8 gives ⎛1⎞ RM = 1 − 0.15 ⎜ ⎟ = 0.95 ⎝ 3⎠ Thus, with α = 1.6 for ASD, Equation A-8-6 becomes

Beam-Columns and Frame Behavior

B2 =

Chapter 8

347

1 1 = = 1.25 ⎛ αPstory ⎛ Δ ⎞ ⎞ 1 – 1.6(158) (0.003) 1– ⎜ ⎜ ⎟⎟ 0.95 ( 4.0 ) ⎝ RM H ⎝ L ⎠ ⎠

Thus, the second-order force and moment are Mr = 1.0(28.7) + 1.25(0) = 28.7 ft-kips Pr = 22.1 + 1.25(0) = 22.1 kips Step 5:

Determine whether the column satisfies the interaction equation. Because the roof beam is assumed to be rigid in this example, use the recommended design value of K = 2.0 from Figure 5.17 case f in the plane of the frame, Lcx = 2(16.0) = 32.0 ft. Out of the plane of the frame, this is a braced frame where K = 1.0; thus, Lcy = 16.0 ft. Determining the critical buckling axis. L 32 = 18.5 ft > Lcy = 16.0 ft ( Lcx )eff = cx = rx /ry 1.73 Thus, from Manual Table 6-2, using Lc = (Lcx)eff = 18.5 ft, Pn/Ω = 148 kips and from Manual Table 6-2, with an unbraced length of Lb = 16 ft, Mnx/Ω = 84.9 ft-kips Determine the appropriate interaction equation to use. Pr 22.1 = = 0.149 < 0.2 Pn /Ω 148 Therefore, use Equation H1-1b. Pa Ma + ≤ 1.0 2 ( Pn Ω ) ( M n Ω )

22.1 28.7 + = 0.413 < 1.0 2(148) 84.9 Thus, the column is adequate for this load combination. Step 6:

Determine the first-order forces and moments for load combination 6. Gravity loads will be assumed to yield the no-translation effects, and wind load will yield the lateral translation effects. From Step 2. Pnt = 20.5 kips, Plt = 0.320 kips, Mnt = 26.7 ft-kips, Mlt = 14.4 ft-kips

Step 7:

Determine the second-order forces and moments. In the plane of the frame, as in Step 4,

348

Chapter 8

Beam-Columns and Frame Behavior ⎛ 0 ⎞ Cm = 0.6 − 0.4 ( M 1 M 2 ) 0.6 – 0.4 ⎜ ⎟ = 0.6 ⎝ 26.7 ⎠ π2 EI π2 (29,000)(146) = 1130 kips Pe1 = 2 x = Lc1 (16(12)) 2

and Cm 0.6 = = 0.618 < 1.0 1.6(20.8) αP 1− r 1– Pe1 1130 Therefore, use B1 = 1.0. B1 =

To determine the sway amplification, the total gravity load on the frame is Pstory = (113) + 0.75(45.0) = 147 kips Again, a serviceability drift index of 0.003 is maintained under the actual wind loads. Therefore, H = 4.0 kips and Δ/L = 0.003 is used to determine the sway amplification factor. As before, RM = 0.95 so 1 1 B2 = = = 1.23 ⎛ αPstory ⎛ Δ ⎞ ⎞ 1 – 1.6(147) (0.003) 1– ⎜ ⎜ ⎟⎟ 0.95 ( 4.0 ) ⎝ RM H ⎝ L ⎠ ⎠ Thus, the second-order force and moment are Mr = 1.0(26.7) + 1.23(14.4) = 44.4 ft-kips and, adding in the lateral load effect amplified by B2, Pr = 20.5 + 1.23(0.320) = 20.9 kips Step 8:

Determine whether the column satisfies the interaction equation. Using the same values found in Step 5, determine the appropriate interaction equation. Pr 20.9 = = 0.141 < 0.2 Pn /Ω 148 Therefore, use Equation H1-1b. Pa Ma + ≤ 1.0 2 ( Pn Ω ) ( M n Ω )

20.9 44.4 + = 0.594 < 1.0 2(148) 84.9 Thus, the column is adequate for this load combination also. Step 9:

The W8×40 is shown to be adequate for gravity and wind loads in combination. Now, check to see that these columns have sufficient capacity to brace the interior pinned columns for gravity load only. This load combination puts the greatest load in the gravity only columns.

Beam-Columns and Frame Behavior Step 10:

Chapter 8

349

For stability in the plane of the frame, using the Yura approach discussed in Section 8.10, the total load on the structure is to be resisted by the four corner columns; thus Dead Load = 0.05 ksf (50 ft)(90 ft)/4 columns = 56.3 kips Snow Load = 0.02 ksf (50 ft)(90 ft)/4 columns = 22.5 kips Thus, for load combination 3, Pa = (56.3) + (22.5) = 78.8 kips Ma = (20.5) + (8.20) = 28.7 ft-kips

Step 11:

Determine the second-order amplification As before, for the length Lx = 16.0 ft, Pe1 = 1130 kips and Cm = 0.6, the second-order amplification for member effect is Cm 0.6 B1 = = = 0.675 < 1.0 1.6(78.8) αP 1− r 1– P1 1130 Therefore, use B1 = 1.0 and Pr = Pa = 78.8 kips Sway amplification will be the same as determined in step 4, since the gravity load is the same; thus, B2 =1.25. Therefore

M r = 1.0 ( 28.7 ) + 1.25 ( 0 ) = 28.7 ft-kips Step 12:

Check the corner columns for interaction under this force and moment. As determined in Step 5 for in-plane buckling, Pn/Ω = 148 kips Mnx/Ω = 84.9 ft-kips Checking for the appropriate interaction equation, Pu 78.8 = = 0.532 > 0.2 Pn /Ω 148 Thus, use Equation H1-1a. Pa 8 ⎛ Max ⎞ + ⎜ ≤ .0 Pn /Ω 9 ⎝ M n x /Ω ⎟⎠ 78.8 8 ⎛ 28.7 ⎞ + ⎜ ⎟ = 0.833 < 1.0 148 9 ⎝ 84.9 ⎠

Thus, the W8×40 is adequate for both strength under combined load and stability for supporting the gravity only columns.

8.11 PARTIALLY RESTRAINED FRAMES The beams and columns in the frames considered up to this point have all been connected with moment-resisting fully restrained (FR) connections or simple pinned connections. These latter simple connections are defined in Specification Section B3.4a. Partially restrained connections,

3550

Chapter 8

Beam-C Columns and Frame F Behav vior

Figure 8.20 Moment Rotation R Curves for Uniform mly Loaded B Beam and Tyypical Connecctions. defined in Specification Sp Section S B3.4b b along with F FR connectioons, have histoorically been referred to as semirig gid connectio ons. When theese PR conneections are inccluded as the connecting eelements in a structural frame, they y influence bo oth the strenggth and stabiliity of the struccture. fore considerin ng the partiallly restrained frame, it willl be helpful too look at the ppartially Befo restrained beam. b The rellationship between the ennd moment aand end rotatiion for a sym mmetric, uniformly lo oaded prismattic beam can be obtained ffrom the welll-known sloppe deflection eequation as E EI θ WL + M = –2 (8.12) L 12 on is plotted in n Figure 8.20 0a and labeledd as the beam line. This equatio

F Figure 8.21 Beam Line aand C Connection Cuurves.

Beam m-Columns annd Frame Behhavior

Chappter 8

351

Fiigure 8.22 IInfluence of tthe PR Coonnection on the Maximum m Positive annd Negative M Moments of a Beam. All A PR connections exhibitt some rotatioon as a result of an appliedd moment. Thhe momentrotation characteristic c s of these con nnections are the key to deetermining thee type of connnection and thus the behavior off the structuree. Moment-rootation curvees for three ggeneric connnections are shown in n Figure 8.20 0b and are laabeled rigid, ssimple, and P PR. Numerouus research sttudies have been con nducted in an n effort to id dentify the m moment-rotatioon curves forr real connecctions. Two compilations of these curves have been b publisheed.5,6 The T relationsh hip between the t moment-rrotation charaacteristics of a connection and a beam can be seeen by plottin ng the beam line and connnection curvve together, aas shown in F Figure 8.21. Equilibriium is obtained when th he beam linne and the cconnection ccurve intersect. Normal engineeriing practice treats t connecctions capablee of resistingg at least 90 ppercent of the fixed-end moment (FEM) as rig gid and those capable of reesisting no m more than 20 percent of thhe fixed-end moment as simple. All A connection ns that exhibitt an ability too resist momeent between tthese limits must be treated as partially restraained connecctions, accounnting for theeir true moment-rotation characterristics. The T influencee of the PR co onnection onn the maximum m positive annd negative m moments on the beam m is seen in Figure F 8.22. Here, H the ratioo of positive or negative m moment to thhe fixed-end moment (FEM) is plotted against th he ratio of beeam stiffness, EI/L, to a linnear connectioon stiffness, M/θ. Thee moment forr which the beam b must bbe designed raanges from 00.75 times the fixed-end moment to 1.5 times the t fixed-end moment, deppending on the stiffness off the connectioon. When W PR con nnections aree used to connnect beams aand columns to form PR frames, the analysis becomes mu uch more com mplex. The reesults of num merous studiess dealing withh this issue have beeen reported. Although A som me practical deesigns have bbeen carried oout, widespreaad practical design of PR frames is not comm mon. In addittion to the problems assoociated with modeling a particular connection,, the question n of loading ssequence arisses. Because real, partiallyy restrained connectio ons behave nonlinearly, th he sequence oof applied loaads influencess the structuraal response. The apprroach to load application may m have morre significancce than the acccuracy of the connection model ussed in the anaalysis. 5

Goverdhan, A. A V., A Collecction of Experiimental Momen nt Rotation Cuurves and Evaluuation of Preddiction Equationns for SemiRigid Connecctions, Master of o Science Theesis, Vanderbilt University, N Nashville, TN, 1983. 6 Kishi, N., an nd Chen, W. F,. Data Base off Steel Beam-to o-Column Connnections, CE-S STR-86-26, Weest Lafayette, IIN: Purdue University, School of Engin neering, 1986.

3552

Chapter 8

Beam-C Columns and Frame F Behav vior

Alth hough a comp plete, theoretiical analysis oof a partially restrained fraame may currrently be beyond the scope of norrmal engineerring practice,, a simplifiedd approach exxists that is nnot only well within the scope of practice, p but also commonnly carried ouut in everydayy design and hhas been for over half a century. This T approach h can be referrred to as Fleexible Momennt Connectionns. It has historically been called Type 2 with Wind. The F Flexible Mom ment Connecction approacch relies heavily on the nonlinearr moment-rottation behaviior of the PR R connectionn although thhe actual curve is not used. In addiition, it relies on a phenom menon called sshake-down, which showss that the connection, although ex xhibiting nonlinear behaviior initially, behaves lineearly after a limited number of applications a of lateral load.7 The moment-rotaation curve for fo a typical P PR connectioon is shown inn Figure 8.233a along with the beaam line for a uniformly lo oaded beam. T The point labbeled 0 represents equilibrrium for the applied gravity loads. The applicaation of wind load producees moments aat the beam eends that add to the gravity g momeent at the leew ward end of tthe beam andd subtract from m the windw ward end. Because mo oment at the windward w end d is being rem moved, the connnection behaaves elasticallly with a stiffness clo ose to the oriiginal connecction stiffnesss, whereas att the leewardd end, the connnection continues to o move along the nonlineaar connection curve. Pointss labeled 1 annd 1′ in Figure 8.23b represent eq quilibrium und der the first ap pplication of wind to the fr frame. Wheen the wind load l is remov ved, the connnection movess from pointss 1 and 1′ to points 2 and 2′, as sh hown in Figu ure 8.23c. Th he next applic ation of a winnd load that iss larger than the first

Figure 8.233 Moment-R Rotation Curves Show wing Shake-D Down

7

G Geschwindner, L. L F., and Disq que, R. O., “Fleexible Momentt Connections ffor Unbraced F Frames Subject to Lateral Forrces—A Reeturn to Simpliccity.” Engineerring Journal, AISC, A Vol. 42, No. 2, 2005, ppp 99–112.

Beam-Columns and Frame Behavior

Chapter 8

353

and in the opposite direction will see the connection behavior move to points 3 and 3′. Note that on the windward side, the magnitude of this applied wind moment dictates whether the connection behaves linearly or follows the nonlinear curve, as shown in Figure 8.23d. Removal of this wind load causes the connection on one end to unload and on the other end to load, both linearly as shown in Figure 8.23e. Any further application of wind load, less than the maximum already applied, will see the connection behave linearly. In addition, the maximum moment on the connection is still close to that applied originally from the gravity load. Thus, the condition described in Figure 8.23f shows that shake-down has taken place and the connection now behaves linearly for both loading and unloading. The design procedure used to account for this shake-down is straight forward. All beams are designed as simple beams using the appropriate load combinations. This assures that the beams are adequate, regardless of the actual connection stiffness, as was seen in Figure 8.22. Wind load moments are determined through a modified portal analysis where the leeward column is assumed not to participate in the lateral load resistance. Connections are sized to resist the resulting moments, again for the appropriate load combinations. In addition, it is particularly important to provide connections that have sufficient ductility to accommodate the large rotations that will occur, without overloading the bolts or welds under combined gravity and wind. Columns must be designed to provide frame stability under gravity loads as well as gravity plus wind. The columns may be designed using the approach that was presented for columns in moment frames, but with two essential differences from the conventional rigid frame design: 1. Because the gravity load is likely to load the connection to its plastic moment capacity, the column can be restrained only by a girder on one side and this girder will act as if it is pinned at its far end. Therefore, in computing the girder stiffness rotation factor, Ig/Lg, for use in the effective length alignment chart, the girder length should be doubled. 2. One of the external columns, the leeward column for the wind loading case, cannot participate in frame stability, because it will be attached to a connection that is at its plastic moment capacity. The stability of the frame may be assured, however, by designing the remaining columns to support the total frame load. For the exterior column, the moment in the beam to column joint is equal to the capacity of the connection. It is sufficiently accurate to assume that this moment is distributed one-half to the upper column and one-half to the lower column. For interior columns, the greatest realistically possible difference in moments resulting from the girders framing into the column should be distributed equally to the columns above and below the joint.

EXAMPLE 8.7a Column Design with Flexible Wind Connections by LRFD

Goal:

Select girders and columns for a building with flexible wind connections.

Given:

An intermediate story of a three-story building is given in Figure 8.24. Story height is 12 ft. The frame is braced in the direction normal to that shown. Use the LRFD provisions and A992 steel.

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Chapter 8

Beam-C Columns and Frame F Behav vior

Figure 8.24 Intermediaate Story of a Three-Story Building (Exxample 8.7).

SO OLUTION

Step p 1:

Deterrmine the requ uired forces aand moments for the load ccombination 1.2D D + 0.5L + 1..0W The loads shown in i Figure 8.244 are the codde-specified nnominal loadss. The requirred forces aree calculated uusing tributaryy areas as folllows. Graviity loads on exterior colum mns. 1.2D = 1.2 2(25 kips + 0..75 kips/ft (155 ft)) = 43.5 kkips 0.5L = 0.5 5 (75 kips + 2 .25 kips/ft (155 ft)) = 54.4 kkips Total =97.9 kkips Graviity loads on in nterior colum mns 1.2D = 1.2 2 (50 kips + 00.75 kips/ft (30 ft)) = 87.0 kkips 0.5L = 0.5 5 (150 kips + 22.25 kips/ft (330 ft)) = 109 kips Total= 196 kips

Step p 2:

Graviity load on girrders for the w worst case 1.22D + 1.6L: 1.2D = 1.2 2 (0.75 kips/ftt (30 ft)) = 277.0 kips 1.6L = 1.6 6 (2.25 kips/ftt (30 ft)) = 1008 kips Total = 1335 kips Desig gn the girder for f the simplee beam momeent assuming full lateral suupport using Manual Table 3-2 or 6-2. M u = WL 8 = 135 ( 30.0 ) 8 = 506 ft-kipss

Beam-Columns and Frame Behavior Therefore use

Step 3:

Chapter 8

355

W21×62 (φMn = 540 ft-kips, Ix = 1330 in.4)

Design the columns for the gravity load on the interior column using Manual Table 4-1. For buckling out of the plane in a braced frame, K = 1.0 and Lcy = 12.0 Thus, with Pu = 196 kips ,try W14×43, (φPn = 371 kips, Ix = 428 in.4, rx/ry = 3.08)

Step 4:

To check the column for stability in the plane, determine the effective length factor from the alignment chart with ⎛ 428 ⎞ 2⎜ ⎟ Σ ( I L )c 12.0 ⎠ Gtop = Gbottom = = ⎝ = 3.21 Σ ( I L ) g ⎛ 1330 ⎞ ⎜ ⎟ ⎝ 2(30.0) ⎠

Note that only one beam is capable of restraining the column and that the beam is pinned at its far end; thus the effective beam length is taken as twice its actual length. Considering the stress in the column under load, the stiffness reduction factor can be determined. Pu A = 196 12.6 = 15.6 ksi Thus, since Pu/A = 15.6 < 0.5Fy, the stiffness reduction factor from Manual Table 4-13 is τb = 1.00. The stiffness ratio then remains Gtop = Gbottom = 3.21 which yields, from the alignment chart, Figure 5.20 K = 1.87 Step 5:

Determine the effective length in the plane of bending. L 1.87(12.0) = 7.29 ft ( Lcx )eff = cx = rx /ry 3.08

Step 6

Determine the column compressive strength from Manual Table 4-1 or 6-2 with Lc = 7.29 ft. φPn = 484 kips

Step 7:

Determine the second-order moment. The applied wind moment is Mu = 1.0(6.0)(12.0) = 72.0 ft-kips and the applied force is Pu = 196 kips. Considering all the moment as a translation moment and using Commentary equation C-A-8-1

356

Chapter 8

Beam-Columns and Frame Behavior Pe story =

π2 EI

( K2 L )

2

=

π2 (29,000)(428) = 1690 kips (1.87(12.0)(12)) 2

Therefore, for all three columns, 1 1 B2 = = = 1.13 1.0 ( 3(196) ) ⎛ αPstory ⎞ 1− ⎜ ⎟ 1 – 3(1690) ⎝ Pe story ⎠ and Mr = 1.13 (72.0) = 81.4 ft-kips Step 8:

Determine whether the column satisfies the interaction equation Pu 196 = = 0.405 > 0.2 φPn 484 Therefore, use Equation H1-1a, φMn = 222, from Manual Table 3-10 or 62, which results in Pu 8 ⎛ M u ⎞ + ⎜ ⎟ ≤ 1.0 φPn 9 ⎝ φM n ⎠ 8 ⎛ 81.4 ⎞ 0.405 + ⎜ ⎟ = 0.731 < 1.0 9 ⎝ 222 ⎠

This indicates that the W14×43 is adequate for stability. The members can then be used as a starting point in a more rigorous analysis.

EXAMPLE 8.7b Column Design with Flexible Wind Connections by ASD

Goal:

Select girders and columns for a building with flexible wind connections.

Given:

An intermediate story of a three-story building is given in Figure 8.24. Story height is 12 ft. The frame is braced in the direction normal to that shown. Use the ASD provisions and A992 steel.

Step 1:

Determine the required forces and moments for the load combination D + 0.75L +0.75(0.6W) The loads shown in Figure 8.24 are the code-specified nominal loads. The required forces are calculated using tributary areas as follows. Gravity loads on exterior columns D = (25 k + 0.75 k/ft (15 ft)) = 36.3 kips 0.75L = 0.75 (75 k + 2.25 k/ft (15 ft)) = 81.6 kips Total = 118 kips Gravity loads on interior columns D = (50 k + 0.75 k/ft (30 ft)) = 72.5 kips 0.75L = 0.75 (150 k + 2.25 k/ft (30 ft)) = 163 kips Total = 236 kips Gravity load on girders for the worst case, D + L D = (0.75 k/ft (30 ft)) = 22.5 kips L = (2.25 k/ft (30 ft)) = 67.5 kips Total = 90.0 kips

Beam-Columns and Frame Behavior Step 2:

357

Design the girder for the simple beam moment assuming full lateral support using Manual Table 3-2 or 6-2. M a = WL 8 = 90.0 ( 30.0 ) 8 = 338 ft-kips Therefore use

Step 3:

Chapter 8

W21×62 (Mn/Ω = 359 ft-kips, Ix = 1330 in.4)

Design the columns for gravity load on the interior column using Manual Table 4-1 or 6-2. For buckling out of the plane in a braced frame, K = 1.0 and Lcy = 12.0 Thus, with Pa = 236 kips try W14×43 (Pn/Ω = 247 kips, Ix = 428 in.4, rx/ry = 3.08)

Step 4:

To check the column for stability in the plane, determine the effective length factor from the alignment chart with ⎛ 428 ⎞ 2⎜ ⎟ Σ ( I L )c 12.0 ⎠ = ⎝ = 3.21 Gtop = Gbottom = Σ ( I L ) g ⎛ 1330 ⎞ ⎜ ⎟ ⎝ 2(30.0) ⎠ Note that only one beam is capable of restraining the column and that that beam is pinned at its far end; thus the effective beam length is taken as twice its actual length. Considering the stress in the column under load, the stiffness reduction factor can be determined. Pa 236 = = 18.7 ksi A 12.6 Thus, from the Manual Table 4-13, the stiffness reduction factor τb = 0.960. The inelastic stiffness ratio then becomes Gtop = Gbottom = 0.960(3.21) = 3.08 which yields, from the alignment chart, Figure 5.20 K = 1.84

Step 5:

Determine the effective length in the plane of bending. L 1.84(12.0) = 7.17 ft ( Lcx )eff = cx = rx /ry 3.08

Step 6:

Determine the column compressive strength from Manual Table 4-1 or 6-2 with Lc = 7.17 ft. Pn/Ω = 324 kips

358

Chapter 8

Beam-Columns and Frame Behavior Step 7:

Determine the second-order moment. The applied wind moment is Ma = 0.75(0.6)(6.0(12.0)) = 32.4 ft-kips and the applied force is Pa = 236 kips. Considering all the moment as a translation moment and using Commentary equation C-A-8-1 π2 EI π2 (29,000)(428) Pe story = = = 1740 kips 2 2 ( K 2 L ) (1.84(12)(12)) αPa = 1.6(236) = 378 kips Therefore, for all three columns, 1 1 B2 = = = 1.28 3(378) αPstory 1– 1− 3(1740) Pe story and Mr = 1.28(32.4) = 41.5 ft-kips

Step 8:

Determine whether the column satisfies the interaction equation Pr 236 = = 0.728 < 0.2 Pn /Ω 324 Therefore, use Equation H1-1a, Mn/Ω = 148, from Manual Table 3-10 or 62, which results in Pa 8 ⎛ Ma ⎞ + ⎜⎜ ⎟ ≤ 1.0 ( Pn Ω ) 9 ⎝ ( M n Ω ) ⎟⎠ 8 ⎛ 41.5 ⎞ 0.728 + ⎜ ⎟ = 0.98 < 1.0 9 ⎝ 148 ⎠ This indicates that the W14×43 is adequate for stability. These members can then be used as a starting point in a more rigorous analysis.

After an acceptable column is selected, the lateral displacement of the structure must be checked. Coverage of drift in wind moment frames is beyond the treatment intended here, but is covered in the Geschwindner and Disque paper already referenced.

8.12 STABILITY BRACING DESIGN Braces in steel structures are used to reduce the effective length of columns, reduce the unbraced length of beams, and provide overall structural stability. The discussion of columns in Chapter 5 showed how braces could be effective in reducing effective length and thereby increasing column strength. Chapter 6 demonstrated how the unbraced length of a beam influenced its strength, and earlier in this chapter the influence of sway on the stability of a structure was discussed. Every case assumed that the given bracing requirements were satisfied, but nothing was said about the

Beam m-Columns annd Frame Behhavior

Chappter 8

359

Figure 8.25 8 Definitiions of Bracin ng Types. Coppyright © Ameerican Institutee of Steel Consstruction, Inc. Repriinted with Perm mission. All rig ghts reserved.

strength or stiffness of the required d braces. For cases when bbraces are nott specifically included in the secon nd-order anaalysis, design n of braces w will follow thhe provisionss of Appendiix 6 of the Specifica ation. Append dix 6 treats bracing for coolumns and beams similarlly, although tthe specific requirem ments are diffeerent. Two typ pes of braces are defined: ppoint bracingg and panel brracing Point P bracing controls the movement oof a point on the member without interraction with any adjacent braced points. p Thesee braces woulld be attachedd to the mem mber and thenn to a fixed support, such as the ab butment show wn in Figure 88.25b. Panel P bracing g relies on oth her braced poiints of the struucture to provvide support. A diagonal brace wiithin a frame would be a panel brace, as shown in Figure 8.25aa. In this casse, the axial deformattion of the diiagonal bracee is a functioon of the dispplacement att each end off the brace. Because the horizonttal strut is ussually a part of a very sttiff floor systtem that has significant strength in its plane, th he strength an nd stiffness oof the diagonaal element usuually controlss the overall behaviorr of this braced system. The T brace req quirements of the Specificcation are inttended to enaable the mem mbers being designed d to reach theeir maximum m strength bassed on the leength betweenn bracing points and an effective length factorr, K = 1.0. A brace b has twoo requirementts: strength annd stiffness. A brace that is inadeq quate in eitheer of these reespects is nott sufficient too enable the member it iss bracing to perform as it was desiigned.

8.12.1

Collumn Bracin ng For colum mn panel braccing, the requ uired shear strrength of the bbracing systeem is

Vbr = 0.005Pr and the required r shearr stiffness is 1 ⎛ 2P βbr = ⎜ r φ ⎝ Lbrb

⎞ ⎟ (LRFD) ⎠ φ = 0.75 5 (LRFD)

⎛ 2P ⎞ βbr = Ω ⎜ r ⎟ (ASD) ⎝ Lbr ⎠ Ω = 2..00 (ASD)

AISC A-6-1) (A (A AISC A-6-2)

360

Chapter 8

Beam-Columns and Frame Behavior

where Lbr = unbraced length of the panel under consideration Pr = required strength of the column within the panel under consideration for ASD or LRFD as appropriate for the design method being used. For a column point brace, the required brace strength is Pbr = 0.01 Pr and the required brace stiffness is 1 ⎛ 8P βbr = ⎜ r φ ⎝ Lbr

⎞ ⎟ (LRFD) ⎠ φ = 0.75 (LRFD)

(AISC A-6-3)

⎛ 8P ⎞ βbr = Ω ⎜ r ⎟ (ASD) ⎝ Lbr ⎠ Ω = 2.00 (ASD)

(AISC A-6-4)

where Lbr = laterally unbraced length adjacent to the point brace Pr = required strength for ASD or LRFD as appropriate for the design method being used. It should be noted that the requirements for point braces are significantly greater than those for panel braces. Thus, if a panel bracing system can be developed, it has the potential to be the more economical approach. 8.12.2

Beam Bracing

For a beam panel brace, the required shear strength of the bracing system is

⎛M C ⎞ Vbr = 0.01⎜ r d ⎟ ⎝ ho ⎠ and the required panel brace stiffness is 1 ⎛ 4M r Cd ⎞ βbr = ⎜ (LRFD) φ ⎝ Lbr ho ⎟⎠ φ = 0.75 (LRFD)

⎛ 4M r Cd βbr = Ω ⎜ ⎝ Lbr ho Ω = 2.00 (ASD)

(AISC A-6-5)

⎞ ⎟ (ASD) ⎠

(AISC A-6-6)

where ho = distance between flange centroids Cd = 1.0 for single curvature bending and 2.0 for the brace closest to the inflection point for double curvature bending Lbr = laterally unbraced length within the panel under consideration Mr = the largest required flexural strength of the beam within the unbraced lengths adjacent to the point being braced For a beam point brace, the required strength of the brace is Pbr = 0.02MrCd/ho and the required brace stiffness is

1 ⎛ 10M r Cd ⎞ βbr = ⎜ (LRFD) φ ⎝ Lbr ho ⎟⎠ φ = 0.75 (LRFD)

⎛ 10M r Cd βbr = Ω ⎜ ⎝ Lbr ho Ω = 2.00 (ASD)

(AISC A-6-7)

⎞ ⎟ (ASD) ⎠

where ho = distance between flange centroids Cd = 1.0 for single curvature and 2.0 for double curvature as above

(AISC A-6-8)

Beam-Columns and Frame Behavior

Chapter 8

361

Lbr = laterally unbraced length adjacent to the point brace Mr = the largest required flexural strength of the beam within the unbraced lengths adjacent to the point being braced As with column bracing, the requirements for point braces are greater than those for panel braces. 8.12.3

Frame Bracing

Frame bracing and column bracing are accomplished by the same panel and point braces and may be designed using the same stiffness and strength equations. However, the most direct approach to bracing design for frames is to include the braces in the model when a second-order analysis is carried out. When that is the case, the provisions of Appendix 6 do not need to be checked.

EXAMPLE 8.8a Bracing Design by LRFD

Goal:

Determine the required bracing for a braced frame to provide stability for the gravity load.

Given:

Using the LRFD requirements, select a rod to provide the point bracing shown in the center panel of the three-bay frame of Figure 8.9a to provide stability for a total gravity dead load of 113 kips and live load of 45 kips.

SOLUTION

Step 1:

Determine the required brace stiffness for gravity load. For the gravity load, the required brace stiffness is based on 1.2D + 1.6L. Pr = 1.2 (113) + 1.6 ( 45.0 ) = 208 kips and from Equation A-6-4 1 ⎛ 8P βbr = ⎜ r φ ⎝ Lbr

Step 2:

1 ⎛ 8(208) ⎞ ⎞ ⎟ = 0.75 ⎜ 16.0 ⎟ = 139 kips/ft ⎝ ⎠ ⎠

Determine the required brace area based on required stiffness and accounting for the angle of the brace. Based on the geometry of the brace from Figure 8.9, where θ is the angle of the brace with the horizontal and Lr = 34.0 ft is the length of the brace, A E βbr = br cos 2 θ = 139 kips/ft Lr This results in a required brace area β L 139(34.0) = 0.209 in.2 Abr = br 2r = 2 E cos θ 30 ⎛ ⎞ 29,000 ⎜ ⎟ ⎝ 34 ⎠

Step 3:

Determine the required brace force for gravity load. The required horizontal brace force for a point brace given by Equation A-6-3 is Pbr = 0.01Pr = 0.01( 208) = 2.08 kips which gives a force in the member of Pbr ( angle ) = 2.08 ( 34 30 ) = 2.36 kips

362

Chapter 8

Beam-Columns and Frame Behavior and a required area, assuming Fy = 36 ksi for a rod, of Pbr ( angle) 2.36 Abr = = = 0.0728 in.2 φFy 0.9(36) Step 4:

For the dead plus live load case, Amin = 0.209 in.2

Step 5:

Select a rod to meet the required area for the controlling case of stiffness for the dead plus live load case where Amin = 0.209 in.2 use a 5/8-in. rod with A = 0.307 in.2

EXAMPLE 8.8b Bracing Design by ASD

Goal:

Determine the required bracing for a braced frame to provide stability for the gravity load.

Given:

Using the ASD requirements, select a rod to provide the point bracing shown in the center panel of the three-bay frame of Figure 8.9a to provide stability for a total gravity dead load of 113 kips and live load of 45 kips.

SOLUTION

Step 1:

Determine the required brace stiffness for gravity load. For the gravity load, the required brace stiffness is based on D + L. Pr = 113 + 45.0 = 158 kips and from Equation A-6-4

⎛ 8P βbr = Ω ⎜ r ⎝ Lbr Step 2:

⎞ ⎛ 8(158) ⎞ ⎟ = 2.00 ⎜ 16.0 ⎟ = 158 kips/ft ⎝ ⎠ ⎠

Determine the required brace area based on required stiffness and accounting for the angle of the brace. Based on the geometry of the brace from Figure 8.9, where θ is the angle of the brace with the horizontal and Lr =34.0 ft is the length of the brace. A E βbr = br cos 2 θ = 158 kips/ft Lr This results in a required brace area β L 158(34.0) Abr = br 2r = = 0.238 in.2 2 E cos θ ⎛ 30 ⎞ 29,000 ⎜ ⎟ ⎝ 34 ⎠

Step 3:

Determine the required brace force for gravity load. The required horizontal brace force for a point brace given by Equation A6-3 is Pbr = 0.01Pr = 0.01(158 ) = 1.58 kips

Beam-Columns and Frame Behavior

Step 4:

Step 5:

Chapter 8

363

which gives a force in the member of Pbr ( angle ) = 1.58 ( 34 30 ) = 1.79 kips and a required area, assuming Fy = 36 ksi for a rod, of Pbr ( angle ) 1.79 Ab r = = = 0.0830 in.2 Fy Ω (36 1.67) For the dead plus live load case, Amin = 0.238 in.2 Select a rod to meet the required area for the controlling case of stiffness for the dead plus live load case, Amin = 0.238 in.2. use a 5/8-in. rod with A = 0.307 in.2

8.13 TENSION PLUS BENDING Throughout this chapter, the case of combined compression plus bending has been treated. That is the most common case of combined loading in typical building structures. However, the Specification also has provisions, in Section H1.2, for combining flexure and tension. The addition of a tension force to a member already undergoing bending may be beneficial. The interaction equations for combined tension and flexure are the same as those already discussed and given as Equations H1-1a and H1-1b. However, if the flexural strength is controlled by the limit state of lateral-torsional buckling, the addition of a tension force can increase bending strength. This is accounted for in the Specification by the introduction of a modification factor to be applied to Cb. Thus, for doubly symmetric members, Cb in Chapter F can be multiplied by 1 + αPr Pey for axial tension that acts concurrently with flexure, where

Pey = π2 EI y L2b and α = 1.0 for LRFD and 1.6 for ASD, as before. The limit that Mn cannot exceed Mp still must be satisfied as it was for beam design discussed in Chapter 6.

EXAMPLE 8.9a Combined Tension and Bending by LRFD

Goal:

Check the given W-shape beam for combined tension and bending

Given:

A W16×77 beam spans 25 ft and carries a uniform dead load of 0.92 kips/ft and a uniform live load of 2.79 kips/ft. It also carries a tension live load of 62.5 kips. The member is braced at the ends only for lateral-torsional buckling. Use A992 steel.

SOLUTION

Step 1:

Determine the required moment strength

wu = 1.2 ( 0.92 ) + 1.6 ( 2.79 ) = 5.57 kips/ft

Mu =

Step 2:

Step 3:

5.57 ( 25 )

2

= 435 ft-kips 8 Determine the required tension strength Tu = 1.6 ( 62.5 ) = 100 kips

Determine the available moment strength. With Lb = 25 ft and Cb = 1.14, from Manual Table 6-2 φM n = 1.14 ( 382 ) = 435 ft-kips < φM p = 563 ft-kips

364

Chapter 8

Beam-Columns and Frame Behavior Step 4:

Determine the available tension strength for the limit state of yielding. Connections at the end of the member are at a location of zero moment so tension rupture will not be a factor for interaction with bending. From Table 6-2

φTn = 1020 kips Step 5:

Determine the increase to be applied to Cb when tension is applied in conjunction with moment strength determined for the lateral-torsional buckling limit state. 2 π2 EI π ( 29,000 )(138 ) Pey = 2 = = 439 kips 2 Lb ( 25 (12) ) 1+

1.0 (100 ) αPr = 1+ = 1.11 Pey 439

Step 6:

Moment strength when considered in combination with tension

Step 7:

Determine the interaction equation to use Pr 100 = = 0.098 < 0.2 Pc 1020

Step 8:

Use Equation H1-1b Pr M 0.098 435 + r = + = 0.049 + 0.901 = 0.95 < 1.0 2 Pc M c 2 483

φM n = 1.11( 435 ) = 483 ft-kips < φM p = 563 ft-kips

So the beam is adequate to carry the bending moment and tension force. Step 9:

Check the beam for bending alone, in case the tension force were not there. Use the available moment strength from Step 3. M r 435 = = 1.0 ≤ 1.0 M c 435 So the beam would just be adequate. In cases where the application of the tension force increases interaction strength and that force may not actually occur, it is important to check the member for flexure alone. The W16×77 is adequate to carry the applied loads.

EXAMPLE 8.9b Combined Tension and Bending by ASD

Goal:

Check the given W-shape beam for combined tension and bending

Given:

A W16×77 beam spans 25 ft and carries a uniform dead load of 0.92 kips/ft and a uniform live load of 2.79 kips/ft. It also carries a tension live load of 62.5 kips. The member is braced at the ends only for lateral-torsional buckling. Use A992 steel.

SOLUTION

Step 1:

Determine the required moment strength

Beam-Columns and Frame Behavior

Chapter 8

365

wa = 0.92 + 2.79 = 3.71 kips/ft Ma =

Step 2:

3.71( 25 )

2

8 Determine the required tension strength

= 290 ft-kips

Ta = 62.5 kips Step 3:

Determine the available moment strength. With Lb = 25 ft and Cb = 1.14, from Manual Table 6-2 Mp Mn = 1.14 ( 254 ) = 290 ft-kips < = 374 ft-kips Ω Ω

Step 4:

Determine the available tension strength for the limit state of yielding. Connections at the end of the member are at a location of zero moment so tension rupture will not be a factor for interaction with bending. From Table 6-2 Tn = 677 kips Ω

Step 5:

Determine the increase to be applied to Cb when tension is applied in conjunction with moment strength determined for the lateral-torsional buckling limit state.

Pey =

2 π2 EI π ( 29,000 )(138 ) = = 439 kips 2 L2b ( 25 (12) )

1+

1.6 ( 62.5 ) αPr = 1+ = 1.11 439 Pey

Step 6:

Moment strength when considered in combination with tension Mp Mn = 1.11( 290 ) = 322 ft-kips < = 374 ft-kips Ω Ω

Step 7:

Determine the interaction equation to use Pr 62.5 = = 0.092 < 0.2 Pc 677

Step 8:

Use Equation H1-1b Pr M 0.092 290 + r = + = 0.046 + 0.901 = 0.95 < 1.0 2 Pc M c 2 322 So the beam is adequate to carry the bending moment and tension force.

Step 9:

Check the beam for bending alone, in case the tension force were not there. Use the available moment strength from Step 3. M r 290 = = 1.0 ≤ 1.0 M c 290 So the beam would just be adequate. In cases where the application of the

366

Chapter 8

Beam-Columns and Frame Behavior tension force increases interaction strength and that force may not actually occur, it is important to check the member for flexure alone. The W16×77 is adequate to carry the applied loads.

8.14

PROBLEMS

Unless noted otherwise, all columns should be considered pinned in a braced frame out of the plane being considered in the problem with bending about the strong axis. 1. Determine whether a W14×90, A992 column with a length of 12.5 ft is adequate in a braced frame to carry the following loads from a first-order analysis: a compressive dead load of 100 kips and live load of 300 kips, a dead load moment of 30 ft-kips and live load moment of 70 ftkips at one end, and a dead load moment of 15 ft-kips and a live load moment of 35 ft-kips at the other. The member is bending in reverse curvature about the strong axis. Determine by (a) LRFD and (b) ASD. 2. An A992 W12×58 is used as a 14 ft column in a braced frame to carry a compressive dead load of 60 kips and live load of 120 kips. Will this column be adequate to carry a dead load moment of 30 ft-kips and live load moment of 60 ft-kips at each end, bending the column in single curvature about the strong axis? The analysis results are from a first-order analysis. Determine by (a) LRFD and (b) ASD. 3. Determine whether a W12×190, A992 column with a length of 22 ft is adequate in a braced frame to carry the following loads from a first-order analysis: a compressive dead load of 300 kips and live load of 500 kips, a dead load moment of 50 ft-kips and live load moment of 100 ft-kips at one end, and a dead load moment of 25 ft-kips and a live load moment of 50 ft-kips at the other. The member is bending in reverse curvature about the strong axis. Determine by (a) LRFD and (b) ASD. 4. An A992 W10×60 is used as a 13 ft column in a braced frame to carry a compressive dead load of 74 kips and live load of 120 kips. Will this column be adequate to carry a dead load moment of 30 ft-kips and live load moment of 45 ft-kips at each end, bending the column in single curvature about the strong axis? The analysis results are from a first-order analysis. Determine by (a) LRFD and (b) ASD. 5. Given a W14×500, A992 42 ft column in a braced frame with a compressive dead load of 90 kips and live load of 270 kips. Maintaining a live load to dead load ratio of 3, determine the maximum live and dead load second-order moments that can be applied about the

strong axis on the upper end when the lower end is pinned by (a) LRFD and (b) ASD. 6. Given a W14×132, A992 15 ft column in a braced frame with a compressive dead load of 350 kips and live load of 350 kips, and maintaining a live load to dead load ratio of 1, determine the maximum live and dead load second-order moments that can be applied about the strong axis on the upper end when the lower end is pinned by (a) LRFD and (b) ASD. 7. Reconsider the column and loadings in Problem 1 if that column were bent in single curvature by (a) LRFD and (b) ASD. 8. Reconsider the column and loadings in Problem 2 if that column were bent in reverse curvature by (a) LRFD and (b) ASD. 9. Reconsider the column and loadings in Problem 3 if that column were bent in single curvature by (a) LRFD and (b) ASD. 10. Reconsider the column and loadings in Problem 4 if that column were bent in reverse curvature by (a) LRFD and (b) ASD. 11. A 14 ft pin-ended column in a braced frame must carry a compressive dead load of 85 kips and live load of 280 kips, along with a uniformly distributed transverse dead load of 0.4 kips/ft and live load of 1.3 kips/ft. Will a W14×68, A992 member be adequate if the transverse load is applied to put bending about the strong axis? Determine by (a) LRFD and (b) ASD. 12. A pin-ended chord of a truss is treated as a member in a braced frame. Its length is 12 ft. It must carry a compressive dead load of 90 kips and live load of 170 kips, along with a uniformly distributed transverse dead load of 1.1 kips/ft and live load of 2.3 kips/ft. Will a W8×58, A992 member be adequate if the transverse load is applied to put bending about the strong axis? Determine by (a) LRFD and (b) ASD. 13. A moment frame is designed so that under a service lateral load H = 150 kips, the frame drifts no more than L/400. There are a total of 15 columns in this frame, so Pstory is 15 times the load on this column. A 13 ft,

Beam-C Columns and Frame Behavvior W W14×120, A9 992 column iss to be checcked. Analysiss rresults are fro om a first-order analysis. The T column iss ccalled upon to carry a comprressive dead load of 100 kipss aand live load of 300 kips. This load willl be taken ass ccoming from a no-translatio on analysis. The T top of thee ccolumn is loadeed with no-tran nslation dead lo oad moment off 225 ft-kips and a no-translation live load mo oment of 80 ft-kkips. The translation momentts applied to th hat column end d aare a dead loaad moment off 35 ft-kips an nd a live load d m moment of 100 0 ft-kips. The lo ower end of thee column feels hhalf of these moments. m The column c is bend ding in reversee ccurvature aboutt the strong ax xis. Will the W14×120, W A992 2 m member be ad dequate to caarry this load ding? Analysiss sshows that thee effective len ngth factor in n the plane off bbending is 1.66 6. Determine by y (a) LRFD and (b) ASD.

Chapteer 8

367

17. A two-story single bay fraame is shownn in Figure P8.177. The uniforrm live and ddead loads aree indicated alongg with the windd load. A first--order elastic aanalysis has yieldded the results shown in the ffigure for the ggiven loads and the appropriatte notional looads. Assuminng that the storyy drift is limiteed to height/3000 under the ggiven wind loadss, determine whether the first- and seecond-story colum mns are adequuate. The gravitty loads will pproduce the no-trranslation resullts and the winnd load will pproduce the transslation results. The members are shown aand are all A9922 steel. Determ mine by (a) LRF FD and (b) AS SD.

114. A W14×1 193, A992 mem mber is propossed for use as a 112.5 ft colum mn in an mom ment frame. The frame iss ddesigned so thaat under a serviice lateral load d H = 120 kips,, thhe frame driftss no more than n L/500. The to otal story load,, Pstory, is 20 tim mes the indiviidual column load. Analysiss rresults are from m a first-orderr analysis. Willl this memberr bbe adequate to o carry a no-trranslation com mpressive dead d looad of 160 kip ps and live load d of 490 kips? The top of thee ccolumn is loaded with a no-trranslation dead d load momentt oof 15 ft-kips an nd a no-translaation live load moment of 30 0 fft-kips. The traanslation mom ments applied to t that column n eend are a dead load moment of 80 ft-kips and a a live load d m moment of 250 0 ft-kips. The column c is ben nding about thee sstrong axis and d, the lower end d of the column n is considered d ppinned, and th he effective length factor is taken as 1.5. D Determine by (a) LRFD and (b) ( ASD. 115. Will an A992W14×48 A be b adequate as a 14 ft column n inn a moment frame fr with a compressive c deead load of 35 5 kkips and livee load of 80 0 kips? One half of thiss ccompressive load is taken as a no-translation load and onee hhalf as a transslation load. The T top and bottom b of thee ccolumn are lo oaded with a no-translatio on dead load d m moment of 20 0 ft-kips and a no-translattion live load d m moment of 55 ft-kips. f The traanslation mom ments applied to o thhe column end ds are a dead lo oad moment off 10 ft-kips and d a live load mo oment of 50 ft-kips. Analy ysis results aree ffrom a first-ord der analysis. The T frame is deesigned so thatt uunder a servicee lateral load H = 50 kips, th he frame driftss nno more than L/300. L The tottal story load, Pstory, is eightt tiimes the indiv vidual column load. The colu umn is bent in n rreverse curvatu ure about the strong axis, and a Kx = 1.3. D Determine by (a) LRFD and (b) ( ASD. 116. Determine whether a 10 ft braced frrame W14×43,, A A992 column can carry a co ompressive deead load of 35 5 kkips and live load of 80 kiips along with h a dead load d m moment of 20 ft-kips and live load momen nt of 40 ft-kips. O One half of theese moments are a applied at the other end,, bbending it in sin ngle curvature.

P8.177 18. Determine w whether the coolumns of thhe two-bay unbraaced frame shhown in Figurre P8.18 are aadequate to suppoort the givenn loading. Results for the first-order analyysis are providded. The gravitty loads will pproduce the

3368

Chapter 8

Beam--Columns and d Frame Behaavior

nno-translation results r and thee wind load wiill produce thee trranslation resu ults. Assume th hat the lateral drift under thee ggiven wind loaad will be limited to a maxim mum of 0.5 in. A All members are a A992 steel and the sizes are as shown. D Determine by (a) LRFD and (b) ( ASD.

20. A 14 ft coluumn in a mom ment frame muust carry a comppressive load of 540 kips aand a momentt about the stronng axis of 1355 ft-kips from an LRFD second-order direcct analysis. W Will a W14× ×74, A992 m member be adequuate?

119. A nonsy ymmetric two o-bay unbracced frame iss rrequired to supp port the live an nd dead loads given g in Figuree P P8.19. Using the results from f the firstt-order elasticc aanalysis provid ded, assuming the axial forcees provided aree ffrom the no-traanslation analyssis and the lateeral drift due to o a 5 kip force iss limited to 0.4 4 in., determinee whether each h ccolumn will be adequate. All members are A992 A steel and d thhe sizes are as shown. Deteermine by (a) LRFD and (b)) A ASD.

21. A 14 ft coluumn in a mom ment frame muust carry a comppressive load of 360 kips aand a momentt about the stronng axis of 90 ftt-kips from an ASD second-oorder direct analyysis. Will a W114×74, A992 m member be adequate? 22. A 28 ft coluumn in a mom ment frame muust carry a comppressive load of 110 kips and moment about the stronng axis of 1000 ft-kips from an LRFD second-order direcct analysis. W Will a W10× ×60, A992 m member be adequuate? 23. A 28 ft coluumn in a mom ment frame muust carry a comppressive load oof 73 kips and moment aboutt the strong axis of 67 ft-kipss from an LR RFD second-orrder direct analyysis. Will a W110×60, A992 m member be adequate? 24a. Select a W-sshape for a collumn with a leength of 15 ft. Thhe results of a second-orderr direct analyssis indicate that the member m must carry a fforce of 700 kkips and a stronng axis momennt of 350 ft-kipss. Design by L LRFD. 24b. Select a W-shape for a column with a leength of 15 ft. Thhe results of a second-orderr direct analyssis indicate that the member m must carry a fforce of 467 kkips and a stronng axis momennt of 230 ft-kipss. Design by A ASD. 25a. Select a W-sshape for a collumn with a leength of 28 ft. Thhe results of a second-orderr direct analyssis indicate that tthe member m must carry a fo force of 1100 kips and a stronng axis momennt of 170 ft-kipss. Design by LRFD. 25b. Select a W-shape for a column with a leength of 28 ft. Thhe results of a second-orderr direct analyssis indicate that the member m must carry a fforce of 730 kkips and a stronng axis momennt of 110 ft-kipss. Design by A ASD. 26a. Select a W-sshape for a collumn with a leength of 14 ft. T The results of a second-ordeer direct analyssis indicate that the member m must carry a fforce of 350 kkips and a stronng axis momennt of 470 ft-kipss. Design by LRFD. 26b. Select a W-shape for a column with a leength of 14 ft. T The results of a second-ordeer direct analyssis indicate that the member m must carry a fforce of 230 kkips and a ASD. stronng axis momennt of 320 ft-kipss. Design by A

P P8.18

27a. Select a W-sshape for a collumn with a leength of 16 ft. T The results of a second-ordeer direct analyssis indicate that tthe member m must carry a fo force of 1250 kips and a stronng axis momennt of 450 ft-kipss. Design by L LRFD.

Beam-C Columns and Frame Behavvior

Chapteer 8

369

P8.19 227b. Select a W-shape for a column with a length of 16 6 fft. The resultss of a second-o order direct an nalysis indicatee thhat the memb ber must carry y a force of 830 kips and a sstrong axis mom ment of 300 ft--kips. Design by b ASD. 228. The two-b bay moment frame fr shown in n Figure P8.28 8 ccontains a sing gle leaning collumn. The results of a first-oorder elastic an nalysis for eacch load are giv ven. Determinee w whether the ex xterior column ns are adequaate to providee sstability for thee frame under dead and livee load. All W-sshapes, orienteed for bending g about the sttrong axis, aree ggiven and the steel s is A992. Determine by (a) LRFD and d ((b) ASD.

Select an A A36 rod to prrovide the poiint bracing 31. wn in the centerr panel of the tthree-bay frame of Figure show 8.9a to provide staability for a tootal gravity deead load of 150 kkips and live load of 60 kipps. Design by (a) LRFD and ((b) ASD. Select an A 32. A36 rod to prrovide the poiint bracing show wn in the centerr panel of the tthree-bay frame of Figure 8.9a to provide staability for a tootal gravity deead load of 180 kkips and live load of 95 kipps. Design by (a) LRFD and ((b) ASD.

229. The two-story frame sh hown in Figurre P8.29 reliess oon the left-hand d columns ben nding about thee strong axis to o pprovide stabiliity. Using thee first-order an nalysis resultss sshown, determiine whether th he given structu ure is adequatee if the steel is A992. A Determin ne by (a) LRFD D and (b) ASD..

33. A 30 ft simpply supported W W24×68 beam m has point braciing at third pooints. The beaam supports a uniformly distriibuted dead lload of 1.2 kkips/ft and a uniformly distriibuted live looad of 2.0 kkips/ft. Deteermine the requiired point bracce strength andd the required ppoint brace stiffnness by (a) LRF FD and (b) AS SD.

330. The two--bay, two-storry frame shown in Figuree P P8.30 is to be designed. d Usin ng the Live, Deead, Snow, and d W Wind Loads giiven in the fig gure, design thee columns and d bbeams to provid de the required d strength and stability by (a)) L LRFD and (b) ASD. A

34. A 30 ft simpply supported W W24×68 beam m has panel braciing in three eqqual panels ovver its length. The beam suppoorts a uniform mly distributed dead load of 1.2 kips/ft and a uniformly distributed livve load of 22.0 kips/ft. Deterrmine the reqquired shear strength of thhe bracing

3370

Chapter 8

Beam--Columns and d Frame Behaavior

ssystem and the required panell brace stiffnesss by (a) LRFD D aand (b) ASD.

P P8.28

P8.299

335. A simply supported W1 18×86 beam spans s 30 ft and d ccarries a uniforrmly distributed d dead load of 0.8 kips/ft and d a uniformly diistributed live load of 2.4 kips/ft. k It also o ccarries a tension live load of 18 kips. The member m is fully y bbraced for latteral-torsional buckling. Use A992 steel. D Determine if th he W18×86 iss adequate for the combined d tension and ben nding by (a) LR RFD and (b) ASD. A

36. A simply suppported W27×884 beam spans 30 ft and carriees a uniformlyy distributed deead load of 1.2 kips/ft and a uniiformly distribbuted live loadd of 2.5 kips//ft. It also carriees a tension llive load of 330 kips. The m member is braceed at third poiints for laterall-torsional bucckling. Use A9922 steel. Determ mine if the W227×84 is adequuate for the combbined tension aand bending byy (a) LRFD andd (b) ASD.

Beam-C Columns and Frame Behavvior

Chapteer 8

371

Desiggn the colum mns and beamss for the resuulting load effeccts and redo thee analysis to chheck the strenggth of these new members and the drift of thhe structure. Coonfirm that the e ffective lengthh method may bbe used. 38. In ntegrated Dessign Project – Direct Analyssis Meth hod Laterral load resisstance in thee east-west diirection is proviided by two pperimeter mom ment frames aas seen in Figurre 1.24. Beforre the forces inn these membbers can be deterrmined, the speecified wind looad must be ddetermined. At thhis stage in thee design, a sim mplified approaach to wind load calculation sim milar to that uused in Chapter 4 might yieldd the followingg loads at each llevel: Roof 4th Floorr 3rd Floorr 2nd Floorr Total Winnd Load P P8.30 337. Integrated Design Project – Effective Length M Method L Lateral load resistance r in the east-west direction iss pprovided by tw wo perimeter moment fram mes as seen in n F Figure 1.24. Beefore the forces in these meembers can bee ddetermined the specified wind d load must be determined. A At this stage in n the design, a simplified app proach to wind d looad calculation n similar to th hat used in Ch hapter 4 mightt yyield the follow wing loads at eaach level: Rooff 4th Floor 3rd Floor F 2nd Floor F Totall Wind Load

32.0 0 kips 59.0 0 kips 54.0 0 kips 50.0 0 kips 195.0 0 kips

T The moment frrames will shaare equally in carrying thesee looads. They will be designed using the efffective length h m method of App pendix 7.2, and d second-order effects will bee inncorporated using u the am mplified first-o order analysiss m method of Appendix 8; thus, superposition may m be used. B Before an analysis may be b carried ou ut, preliminary y m member sizes must m be obtain ned. Using thee gravity loadss ccalculated in Chapter C 2, select preliminarry column and d bbeam sizes with hout concern for f the frame behavior b of thee sstructure. W With these mem mber sizes, thee analysis is to be carried outt ffor dead load d, live load, roof load, an nd wind load. M Members are to o be selected for fo the gravity plus p wind load d ccombination, so s there shou uld be no need to includee nnotional loads.

32.0 kips 59.0 kips 54.0 kips 50.0 kips 195.0 kips

The moment framees will share eequally in carrrying these loadss. They will bbe designed ussing the Direcct Analysis Methhod from Chappter C. Befoore an analysis may be ccarried out, ppreliminary mem mber sizes musst be obtained. Using the graavity loads calcuulated in Chappter 2, select preliminary coolumn and beam m sizes withoutt concern for tthe frame behaavior of the struccture. Withh these memberr sizes, the anaalysis is to be carried out for ddead load, livve load, rooff load, and w wind load, follow wing the general analysis requirements of Section C2. Members are tto be selected for the gravityy plus wind load combination, so there shoulld be no need to include notioonal loads. Desiggn the colum mns and beamss for the resuulting load effeccts and redo thee analysis to chheck the strenggth of these new m members and tthe drift of the structure.

C Chapterr 9

C Compo osite Constr C ruction n

7 Brryant Park, Neew York Photo courteesy Thornton T Tomasetti

9.11

INTROD DUCTION Any structurral member in n which two or o more mateerials having ddifferent stresss-strain relattionships are combineed and called d upon to work w as a sinngle member may be connsidered a coomposite member. Man ny different types t of mem mbers have beeen used thatt could be callled composite. Such members, ass shown in Figure 9.1, incclude (a) a re inforced conccrete beam, ((b) a precast cconcrete beam with a cast-in-placce slab, (c) a “flitch” girdder combiningg wood side members andd a steel plate, (d) a stressed skin panel where plywood is ccombined witth solid woodd members, aand (e) a steel shape combined c witth concrete. The last type off member, an nd similar meembers, are what are norrmally thougght of as composite beams b in build ding applicatiions. Specificaation Chapterr I provides rrules for desiggn of the composite members m illusstrated in Figu ure 9.2. Thesee members arre (a) steel beeams fully enncased in concrete, (b) steel beamss with flat soffit concrete slabs, (c) steel beams com mbined with cconcrete slabs on forrmed steel deeck, (d) steell columns fuully encased in concrete, and (e) holloow steel shapes filled d with concrette. Encased beams and filled columns, c shoown in Figurre 9.2a and e, may not require mechanical anchorage between the steel s and conncrete, other than the natuural bond thaat exists between thee two materiaals; however,, the flexurall and compreession membbers shown inn Figure 9.2b, c, and d always requ uire some forrm of mechannical shear connnection. Reg gardless of thee type of mecchanical shearr device provided, it must connect the ssteel and concrete to form a unit and permit the t two materrials to workk together to resist the loaad. This considerably y increases th he strength off the bare steeel shape. Coomposite beam ms were firstt used in bridge desig gn in the United States arou und 1935. Unntil the inventtion of the shear stud, the cconcrete floor slab was connected to the stringeer beams by m means of wiree spirals or chhannels weldeed to the top flange of the beam, ass shown in Fiigure 9.3.

3772

Chaptter 9

Compposite Construuction 373

Figure 9.1 Composite Members M

S Beams and a Columns Figure 9.2 Composite Steel In I the l940s the t Nelson Sttud Companyy invented thee shear stud, a headed rodd welded to the steel beam by meaans of a speciial device or ggun, as shownn in Figure 9.4. The comppany did not enforce its i patent butt instead enco ouraged nonpproprietary usse of the system, assuminng correctly that the company c wou uld get its shaare of businesss if it becam me popular. Inn a very short time, studs replaced spirals and channels, c and d today studss are used alm most exclusivvely in compposite beam constructtion. In I 1952 AISC C adopted com mposite desiggn rules for eencased beam ms in its speciification for building design, and in l956 it ex xtended them m to beams w with flat soffiits. Although the design procedurre was based on the ultimaate strength oof the compossite section, tthe rules weree written in terms off an allowable stress form mulation, as w was commonn for the timee. Since thenn, allowable stress design for com mposite beamss has often beeen criticizedd as being connvoluted and difficult to understan nd.

Figure 9.3 9

Composite Beam Usin ng a Spiral Shhear Connecttor

3774 Chapter 9

Composiite Construction

Figure 9.4 Installation of a Shear Sttud with a Stuud Gun Photo Courtesy W. Samuel Easterling

In the t current Specification, Sp whether forr ASD or LR RFD, the rulees for the deesign of composite beams b are straightforward and surprisinngly simple. The ultimatee flexural streength of the composiite member is i based on plastic p stress distribution, with the ducctile shear coonnector transferring shear betweeen the steel section and thee concrete slabb. Speccification Secction I1.3 giv ves limitationns on materiaal properties for use in coomposite concrete meembers. Concrrete is limited d to fc′ betweeen 3 ksi and 10 ksi for norrmal-weight cconcrete and between n 3 ksi and 6 ksi k for lightw weight concrette. The speciffied minimum m yield strengtth of the structural stteel used in calculating c th he strength off composite m members is nnot to exceedd 75 ksi. Similarly, fo or reinforcing g steel the lim mit for calculattions is 80 ksii. Thiss chapter disscusses the design d of botth compositee beams and composite ccolumns. Table 9.1 lissts the section ns of the Speccification and parts of the M Manual discuussed in this chhapter. Table 9.1 Sections of Specification S and a Parts of M Manual Coveered in This C Chapter Specificationn I1 I2 I3 I5 I6 I8

Generral Provisionss Axiall Force Flexu ure Comb bined Axial Force F and Flexxure Load Transfer Steel Anchors Manual

Part 3 Part 4

Desig gn of Flexurall Members Desig gn of Compression Membeers

Chapter 9

Composite Construction 375

9.2 ADVANTAGES AND DISADVANTAGES OF COMPOSITE BEAM CONSTRUCTION One feature of composite construction makes it particularly advantageous for use in building structures. The typical building floor system is composed of two main parts: a floor structure that carries load to supporting members, usually a concrete slab or a slab on a metal deck; and the supporting members that span the space between girders, usually steel beams or joists. The advantage of a composite floor system stems from the concrete slab doing “double duty”. The concrete slab, with or without metal deck, is first designed to span between beams. Then, since the concrete is there for that purpose in one direction, it can be used to advantage in the other direction, that of the beam span. All other factors that could be identified as advantages of this type of construction can be traced back to this single feature. A composite beam takes the existing concrete slab and enables it to work with the steel beam to carry the load to the girders. Thus, the resulting system has a greater strength than would be available from the steel beam alone. The composite beam is stronger and stiffer than the noncomposite beam. The advantage of this factor lies in the reduced weight and/or shallower member depth necessary to carry a given load compared to the bare steel beam. Because the concrete slab is in compression and the majority of the steel is in tension, the two materials are working to their best advantage. In addition, the effective beam depth has been increased from just the depth of the steel to the total distance from the top of the slab to the bottom of the steel, increasing the overall efficiency of the member. With regard to stiffness, the composite section has an increased elastic moment of inertia compared to the bare steel beam. Although the actual calculation of the stiffness of the composite section may be approximate in many cases, the impact of the increased stiffness profoundly affects the static deflection. The only disadvantage of composite construction is the added cost of the required shear connectors, referred to as steel headed stud anchors in the Specification. Because the increased strength, or reduction in required steel weight, is normally sufficient to offset the added cost of providing and installing the shear connectors, this increased cost is usually not a true disadvantage.

9.3

SHORED VERSUS UNSHORED CONSTRUCTION Two methods of construction are available for composite beams: shored and unshored construction. Each has advantages and disadvantages, which will be discussed briefly. The difference between these two approaches to the construction of a composite beam is how the selfweight of the wet concrete is carried. When the steel shape alone is called upon to carry the concrete weight, the beam is considered to be unshored. In this case, the steel is stressed and it deflects. This is the simplest approach to constructing the composite beam because the formwork and/or decking is supported directly on the steel beam. Unshored construction may, however, lead to a deflection problem during the construction phase because as the wet concrete is placed, the steel beam deflects. To obtain a level slab, more concrete is placed where the beam deflection is greatest. This means that the contractor must place more concrete than required based on the specified slab thickness, and the designer needs to provide more strength than would be needed if the slab had remained of uniform thickness. For shored construction, temporary supports called shores are placed under the steel beam to carry the wet concrete weight. In this case, the composite section carries the entire load after the shores are removed. No load is carried by the bare steel beam alone, and thus, no deflection occurs during concrete placement. Two factors must be considered in the selection of shored construction: (1) the additional cost, in terms of both time and money, of placing and

3776 Chapter 9

Composiite Construction

Figure 9.5 Effective Fllange Width removing th he temporary shoring; and (2) the potenntial increase in long-term,, dead load deeflection due to creep p in the concrrete, which no ow must partiicipate in carrrying the perm manent weighht of the slab. hough elasticc stress distriibution and deflection unnder service load conditiions are Alth influenced by b whether th he compositee beam is shoored or unshoored, researchh has shown that the ultimate streength of the composite section is indepeendent of the shoring situaation. Thus, thhe use of shoring is en ntirely a serv viceability and d constructabbility questionn that must be considered by both the designer and constrructor. Wheth her ASD or LRFD provvisions are uused, the shoored and unshored sy ystems have the same nominal n streength, as dettermined thrrough applicaation of Specification n Section I3.

9.44

EFFECT TIVE FLAN NGE A cross secttion through a series of typ pical compossite beams is shown in Figgure 9.5. Because the concrete slaab is normallly part of the transverse spanning flooor system, iits thickness and the spacing of th he steel beam ms are usually established pprior to the deesign of the coomposite beam ms. Because the abillity of the slaab to particippate in load ccarrying decreeases as the distance from the beaam centerline increases, so ome limit musst be establishhed to determ mine the portioon of the slab that can n be used in the calculations to determ mine the strenngth of the ccomposite beaam. The Specification n provides tw wo criteria forr determiningg the effectivee width of thee concrete slaab for an interior beam m and an add ditional criterion for an eddge beam in S Section I3.1a.. As shown inn Figure 9.5, the effeective width, beff, is the su um of b′ valuues on each sside of the centerline of tthe steel section. For an interior beeam, b′ is the lesser of spaan b′ ≤ 8 1 b′ ≤ distance d to thee adjacent beaam 2 For an edge beam, the ad dditional criterrion is b′ ≤ disstance to the edge of the sllab The entire th hickness of th he concrete slab is availabble to carry a compressivee force. Howeever, the depth of thee concrete used in calcullations is onnly that requiired to provide sufficient area in compression n to balance the t force transsferred by thee shear conneectors to the ssteel shape. T This may be significan ntly less than n the entire slaab thickness. It should be noted that thhe slab thickness does not influence the effective width of thee slab as it dooes for reinforrced concretee T-beam desiign.

Chaptter 9

9.5

Compposite Construuction 377

STRE ENGTH OF F COMPOS SITE BEAM MS AND SL AB Flat soffi fit composite beams (Figurre 9.2b) are cconstructed ussing formworrk that is set at the same elevation n as the top of o the steel section. The cooncrete slab iss placed direcctly on the stteel section, resulting in a flat surfface at the lev vel of the top of the steel. C Composite beeams with a fformed steel deck (Fig gure 9.2c) aree constructed d with the steeel deck restinng on top of tthe steel beam m or girder. The conccrete is placed on top of th he deck so thhat the concreete ribs and vvoids alternatte. Provided that the portion p of con ncrete requireed to balance tthe tension foorce in the steeel is availablle above the tops of the t ribs, the ultimate streength is deterrmined similaarly for the ttwo types off composite beams. Although A thee steel memb ber may be either shoreed or unshored, the strenngth of the compositte member is independentt of the use oof shores, andd the design rrules are indeependent of the method of constru uction. The T flexural strength s of a composite bbeam under ppositive momeent where cooncrete is in compresssion is preseented in Secction I3.2a oof the Speciffication. In tthis section, strength is developeed for flat sofffit beams. Th he required m modifications to account fo for the use of metal deck are pressented in th he next secttion. For stteel shapes with a webb slendernesss ratio of which is the case c for all rollled W-, S-, aand HP-shapees), the nominnal moment, h / t w ≤ 3 .76 E / Fy (w Mn, is determined from m the plastic distribution d oof stress on thhe composite ssection, and φb = 0.90 (LRFD)) Ωb = 1.677 (ASD) wn in Figuree 9.6 with threee possible plastic stress A composite beam cross section s is show distributiions. Regardlless of the strress distributiion considereed, equilibrium m requires thhat the total tension force f equals the t total com mpression forcce, T = C. Inn Figure 9.6a,, the plastic nneutral axis (PNA) iss located at th he top of the steel shape. T The compresssion force deeveloped usinng all of the concrete is exactly eq qual to the ten nsion force deeveloped usinng all of the ssteel. For the distribution of Figuree 9.6b, the PN NA is located within the steeel shape. In this case, all of the concreete is taking compresssion, but this is not sufficiient to balancce the tensionn force that thhe full steel sshape could provide. Thus, some of the steel shape s is in coompression inn order to saatisfy T = C. The plastic stress disstribution sho own in Figure 9.6c is what occurs whenn less than thee full amount of concrete is needed d to balance th he tensile forcce developedd in the steel sshape. Here thhe PNA is loccated within the concrrete and that portion of th he concrete beelow the PNA A is not usedd because it w would be in tension and a concrete is i not effectiv ve in resisting tension. In I all three caases, equilibrrium of the crross section rrequires that the shear connnectors be capable of o transferring g the force caarried by the cconcrete into the steel. Forr the cases in Figure 9.6a and b, th his is the full strength of th he concrete. F For the case in Figure 9.66c, this is the strength of the steel shape. Becau use the shear connectors c arre carrying thee full amountt of shear forcce required

Figure 9.6 9

Plastic Stress S Distribu ution

378 Chapter 9

Composite Construction

to provide equilibrium using the maximum capacity of one of the elements, this is called a fully composite beam. It is also possible to design a composite beam when the shear force that can be transferred by the shear connectors is less than this amount. In this case, the beam is called a partially composite beam. Although it has less strength than the fully composite member, it is often the most economical solution. The Specification indicates that the plastic stress distribution in the concrete shall be taken as a uniform stress at a magnitude of 0.85fc′. This is the same distribution specified by ACI 318 Building Code Requirements for Reinforced Concrete. In addition, the distribution of stress in the steel is taken as a uniform Fy, as was the case for determining the plastic moment strength of a steel shape. The Specification also provides for the use of a strain compatibility method and an elastic stress distribution method for determining the strength of a composite section. Strain compatibility should be considered when a section is of unusual geometry and elastic stress distribution should be used when the web is not compact. 9.5.1

Fully Composite Beams Establishing which stress distribution is in effect for a particular combination of steel and concrete requires calculating the minimum compressive force as controlled by the three components of the composite beam: concrete, steel, and shear connectors. If all of the concrete were working in compression, (AISC I3-1a) Vc′ = V ′ = 0.85 f c′beff t

If all of the steel shape were working in tension, Vs′ = V ′ = Fy As

(AISC I3-1b)

If the shear studs were carrying their full capacity, each resisting Qn, Vq′ = V ′ = ∑ Qn

(AISC I3-1c)

Because fully composite action is being considered first, Vq′ will not control and is not considered further. If Vs′ ≤ Vc′, the steel is fully stressed and all or only a portion of the concrete is stressed. This is the distribution given in either Figure 9.6a or c. If Vc′ < Vs′, the concrete is fully stressed and the steel is called upon to carry both tension and compression to ensure equilibrium. This results in the distribution shown in Figure 9.6b. Once the proper stress distribution is known, the corresponding forces can be determined and their point of application found. With this information, the nominal moment, Mn, can be determined by taking moments about some reference point. Because the internal forces are equivalent to a force couple, any point of reference can be used for taking moments; however, it is convenient to use a consistent reference point. These calculations use the top of the steel as the point about which moments are taken. Determination of the PNA for the cases in Figure 9.6a and c is quite straightforward. In both cases the steel is fully stressed in tension, so it is said that the steel controls and it is known that the concrete must carry a compressive force equal to Vs′. Only that portion of the concrete required to resist this force will be used, so that force is defined as Cc = 0.85 fc′beff a, where a defines the depth of the concrete stressed to its ultimate. Setting Vs′ = Cc and solving for a yields Fy As Vs′ = a= (9.1) 0.85 f c′beff 0.85 f c′beff

Chapter 9

Composite Construction 379

For the special case where Vs′ is exactly equal to Vc′, the value of a thus obtained is equal to the actual slab thickness, t. This is the case shown in Figure 9.6a. For all other values of a, the distribution of Figure 9.6c results. The nominal flexural strength can then be obtained by taking moments about the top of the steel so that M n = Ts ( d 2 ) + Cc ( t − a 2 ) (9.2) where Ts = Fy As and Cc = Ts because the system must be in equilibrium. When the concrete controls, Vc′ < Vs′, the determination of the PNA is a bit more complex. It is best to consider this case as two separate subcases: (1) the PNA occurring within the steel flange and (2) the PNA occurring within the web. Once it is determined that Vc′ controls, and thus Cc = Vc′, the next step is to determine the force in the steel flange and web respectively from (9.3) T f = Fy b f t f Tw = Ts − 2T f

(9.4)

A comparison between the force in the concrete and the force in the bottom flange plus the web shows whether the PNA is in the top flange or web. If Cc > Tw + Tf, more tension is needed for equilibrium and the PNA must be in the top flange. If Cc < Tw + Tf, less tension is needed for equilibrium and the PNA is in the web. In either case, the difference between the concrete force, Cc, and the available steel force, Ts, must be divided evenly between tension and compression in order to obtain equilibrium. This allows determination of the PNA location and the nominal moment strength. Thus, with As–c = area of steel in compression and As = total area of steel equilibrium is given by Cc + Fy As − c = Ts − Fy As − c

Solving for the area of steel in compression yields T − Cc As −c = s 2Fy

(9.5)

(9.6)

For the case where the PNA is in the flange, the distance from the top of the flange to the PNA is given by x, where A x = s −c (9.7) bf and for the case where the PNA is in the web, As − c − b f t f x= + tf tw

(9.8)

Equation 9.8 can be more easily understood if it is related to the areas being considered. The area of the web in compression is the area of steel in compression less the flange area. This web compression area is divided by the web thickness, and the result is the location of the PNA

380 Chapter 9

Composite Construction

measured from the underside of the flange. Thus, x is simply the thickness of the flange plus the depth of the web in compression.

EXAMPLE 9.1 Fully Composite Beam Strength

SOLUTION

Goal:

Determine the nominal moment strength for the interior composite beam shown as Beam A in Figure 9.7. Also determine the design moment and the allowable moment.

Given:

The section is a W21×44 and supports a 4.5 in. concrete slab. The dimensions are as shown. Fy = 50 ksi and fc′ = 4 ksi. Assume full composite action.

Step 1:

Determine the effective flange width, the minimum of beff = 30.0(12 in./ft)/4 = 90.0 in. and beff = (10.0 + 10.0)(12 in./ft)/2 = 120 in. Therefore use beff = 90.0 in.

Step 2:

Determine the controlling compression force using Equations I3-1a and I31b. Vc′ = 0.85 f c' Ac = 0.85 ( 4.0 )( 90.0 )( 4.5 ) = 1380 kips Vs′ = Fy As = 50.0 (13.0 ) = 650 kips

Assuming full composite action, the shear connectors must carry the smaller of Vc′ and Vs′; thus, Vq′ = 650 kips Because Vs′ is less than Vc′, the PNA is in the concrete. Step 3:

Determine the depth of concrete in compression using Equation 9.1. Fy As 650 a= = = 2.12 in. 0.85 f c′beff 0.85(4)(90.0) The resulting plastic stress distribution is shown in Figure 9.7c.

Step 4:

Determine the nominal moment strength using Equation 9.2, M n = Ts ( d 2 ) + Cc ( t − a 2 ) 2.12 ⎞ ⎛ 20.7 ⎞ ⎛ = 650 ⎜ ⎟ + 650 ⎜ 4.50 − ⎟ = 9000 in.-kips 2 ⎠ ⎝ 2 ⎠ ⎝

⎛ 9000 ⎞ Mn = ⎜ ⎟ = 750 ft-kips ⎝ 12 ⎠ For LRFD Step 5:

The design moment is

φM n = 0.9 ( 750 ) = 675 ft-kips

Chaptter 9 For F ASD A Step S 5:

Compposite Construuction 381

Th he allowable moment m is M n Ω = 750 1.67 = 449 ft-kips

Figure 9.7 Interior Com mposite Beam m (Examples 9.1 9 and 9.2)

EXAMPLE E 9.2 Fully Comp posite Beam Stren ngth

SOLUTION N

Goal: G

Deetermine the nominal mom ment strengthh for the inteerior compossite beam sh hown as Beam m A in Figure 9.7 using a llarger W-shappe. Also deterrmine the wable momennt. deesign momentt and the allow

Given: G

Usse a W21×111 as shown inn Figure 9.8 for the steel m member and the same me full composite action. maaterials as in Example 9.1.. Again, assum

Step S 1:

Deetermine the effective e flannge width. Th will remain thhe same; thus, he effective fllange width w beff = 90.00 in.

Step S 2:

Deetermine the controlling coompression fo force using Eqquations I3-1a and I31b b. Vc′ = 0.85 f c' Ac = 0.85 ( 4.0 )( 990.0 )( 4.5 ) = 11380 kips . Vs′ = Fy As = 500.0 ( 32.6 ) = 16630 kips Asssuming full composite c acttion, Vq′ = 1380 kkips Beecause Vc′ is less l than Vs′, tthe PNA is inn the steel.

Step S 3:

Deetermine wheether the PNA A is in the steeel flange or w web. T f = Fy b f t f = 50 (12.3)( 0.875 ) = 538 kips

Tw = Fy As − 2T f = 1630 − 2 ( 538 ) = 5544 kips Th hus,

Cc = Vc′ = 13800 > T f + Tw = 5538 + 554 = 10090 kips

382 Chapter 9

Composite Construction Because additional tension is required to balance the compression in the concrete, the PNA is in the flange. Step 4:

Determine the area of steel in compression. Use Equation 9.6.

As −c = Step 5:

Ts − Cc 1630 − 1380 = = 2.50 in.2 2Fy 2(50)

Determine the location of the PNA in the flange. The PNA is located down from the top of the steel by a distance x as given by Equation 9.7. A 2.50 x = s −c = = 0.203 in. bf 12.3 The stress distribution for this PNA location is shown in Figure 9.8b.

Step 6:

Determine the nominal moment strength of the composite beam. Moments could be taken about any point to determine the nominal moment; however, a simplified mathematical model shown in Figure 9.8c makes the analysis quicker. In this case, the full area of steel is shown in tension, and the portion in compression is first removed (130 kips on the compression side) and then added in compression (another 130 kips on the compression side), shown as 2(130) = 260 kips. This results in only three forces and moment arms entering the moment equation. Thus ⎛d ⎞ ⎛t⎞ ⎛ x⎞ M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎛ 21.5 ⎞ ⎛ 4.5 ⎞ ⎛ 0.203 ⎞ = 1630 ⎜ ⎟ + 1380 ⎜ ⎟ − 2(2.50) ( 50 ) ⎜ ⎟ = 20,600 in.-kips ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 20,600 Mn = = 1720 ft-kips 12

Step 7:

For LRFD, the design moment is φM n = 0.9 (1720 ) = 1550 ft-kips

Step 7:

For ASD, the allowable moment is M n 1720 = = 1030 ft-kips Ω 1.67

Chaptter 9

Compposite Construuction 383

Figure 9.8 Interior Com mposite Beam m (Example 9.2)

9.5.2

PAR RTIALLY COMPOSIT C TE BEAMS S The com mposite members considereed thus far havve been fully composite. T This means thhat the shear connecto ors have been n assumed to o be capable of transferriing whatever force was rrequired for equilibriu um when eitther the conccrete or steel were fully sstressed. Theere are manyy conditions where th he required strength s of th he compositee beam is lesss than what would resullt from full compositte action. In particular, p theere are cases where the sizze of the steeel member is dictated by factors other o than thee strength off the composiite section. B Because shearr connectors compose a significan nt part of th he cost of a composite c beeam, econom mies can resuult if the low wer required flexural strength s can be b translated to t a reduced nnumber of shhear connectors when the ssteel section and conccrete geometry y are already given. If I the composite section is i viewed unnder elastic sttress distribuutions, partiall composite action caan be more eaasily understo ood. Figure 9..9 shows elasstic stress disttributions for three cases of combiined steel an nd concrete. The T first casee, Figure 9.9aa, is what reesults when thhe concrete simply rests on the steel s with no shear transffer between tthe two mateerials. The reesult is two independ dent memberss that slip paast each otheer at the inteerface. If the two materials are fully connecteed, the elasticc stress distriibution is as shown in Fiigure 9.9c annd the materiials are not permitted d to slip at all. a If some limited amounnt of slip is permitted beetween the stteel and the concrete,, the resulting g elastic stresss distributionn is similar tto that shownn in Figure 9.9b. This is how the partially p com mposite beam would w behavee in the elastic region. The T plastic moment m streng gth for a parrtially compoosite member is the result of a stress distributiion similar to that shown in i Figure 9.100. The PNA w will be in the steel and thee magnitude of the compression c force in thee concrete w will be controlled by thee strength off the shear connecto ors.

9 Figure 9.9

Levels of o Composite Action for Ellastic Behavioor

3884 Chapter 9

Composiite Construction

Figure 9.10 Plastic Stress Distrribution for Partially Compposite Action Reg gardless of the final location of the PN NA, the forcee in the concrrete is limitedd by the strength of the t shear stud ds. Thus, an approach a com mbining thosee taken for thhe three cases of fully composite sections s is ussed for the partially p compposite membber. By the ddefinition of ppartially composite members, m Cc = Vq′ = ΣQn and the deptth of the conccrete acting in n compressionn is given by ∑ Qn a= 0.85 f c′beff

(9.9)

Equations 9.3 through 9.8 can then bee used to deteermine the loccation of the PNA within tthe steel and the nom minal moment can be obtain ned as before . Parttial compositee action is geenerally desccribed in term ms of a perceent of full coomposite action. Thuss, a beam witth shear stud strength equaal to the tensiile strength off the W-shape is said to be 100% composite or o fully comp posite. A beam m with shearr stud strengthh equal to 3//4 of the tensile stren ngth of the W-shape W is saaid to be 75% % compositee or partially composite. T The slip mposite actioon requires a certain between thee steel and th he concrete that t allows fo for partial com amount of ductile d behaviior or deform mation capacitty of the sheaar studs. Although there arre no set requirementts in the Speccification for determining d a minimum leevel of ductillity, the Com mmentary provides gu uidelines to in nsure that thee partially coomposite beaams will behaave as designned. For configuratio ons of compo osite beams in normal ddesign practice, three speecific guideliines are provided thaat ensure suffficient ductility. Thus, if ppartially compposite beams meet one or more of the followin ng condition ns, shear con nnectors aree not subject to failure due to insuufficient deformation n capacity: 1. Beams with w span not exceeding 30 0 ft; 2. Beams with w a degree of compositee action of at least 50%; orr 3. Beams with w an averaage nominal shear connectoor capacity of at least 16 kkips per ft aloong their span. Th his correspond ds to a single 3/4 in. shear stud placed aat 12 in. spaciing on averagge. Beams that do not meet these criteriaa may still peerform adequuately but theey should be checked through a deetailed analysis that captures the actual deformation ccharacteristiccs of the shearr studs.

Chaptter 9

Figure 9.11 9

Compposite Construuction 385

Stress Distribution D and a Forces Ussed in Exampple 9.3

EXAMPLE E 9.3 Partially Composite Beam Strength

Goal: G

Deetermine the nominal mooment strengtth of a partiaally composiite beam. Allso determinee the design m moment and thhe allowable m moment.

Given: G

Co onsider the co oncrete and ssteel beam giiven in Exam mple 9.1 and sshown in Figure 9.7. In this case, hoowever, assuume that the shear conneectors are caapable of transsferring only Vq′ = 500 kipps.

SOLUTION N

Step S 1:

Deetermine the effective e flannge width. Th his is the same as determinned for Exampple 9.1. beff = 90.00 in.

Step S 2:

Deetermine the controlling c coompression foorce. Frrom Example 9.1, Frrom the given n data

Vc′ = 1380 kips Vs′ = 650 kkips Vq′ = 500 kips

Beecause the low west value off the compresssive force is ggiven by Vq′, tthis is a paartially compo osite memberr and Cc = Vq′ = 500 kips. Since Vq′ Vs′ = 500 nd should 5 650 = 0.777 ≥ 0.5 , the bbeam is 77% composite an no ot be subject to t insufficientt connector duuctility. Step S 3:

Deetermine the depth d of the cconcrete workking in comprression from Equation 9.9 9. ΣQn 5000 a= = = 1.633 in. ′ 0.85 f c beff 0.85(44)(90.0)

Step S 4:

Deetermine the area a of steel iin compressioon from Equaation 9.6. T − Cc 650 − 500 As − c = s = = 1.50 in.2 2 Fy 2(550) Beecause this is less than thee area of the fflange, 6.50(00.450) = 2.933 in.2, the PN NA is in the flange. fl

386 Chapter 9

Composite Construction Step 5:

Determine the location of the PNA from Equation 9.7. A 1.50 x = s −c = = 0.231 in. bf 6.50

Step 6:

Determine the nominal moment strength by taking moments about the top of the steel shape using the three forces shown in Figure 9.11. ⎛d ⎞ ⎛t⎞ ⎛x⎞ M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠ 1.63 ⎞ ⎛ 20.7 ⎞ ⎛ ⎛ 0.231 ⎞ = 650 ⎜ ⎟ + 500 ⎜ 4.50 − ⎟ − 2(1.50) ( 50 ) ⎜ ⎟ = 8550 in.-kips 2 ⎠ ⎝ 2 ⎠ ⎝ ⎝ 2 ⎠ 8550 Mn = = 713 ft-kips 12

Step 7:

For LRFD, the design moment is φM n = 0.9 ( 713) = 642 ft-kips

Step 7:

For ASD, the allowable moment is M n 713 = = 427 ft-kips Ω 1.67

The nominal moment strength decreased from 750 ft-kips for the full composite action of Example 9.1 to 713 ft-kips for the level of partial composite action given in Example 9.3. This is approximately a 5 percent reduction in strength, corresponding to more than a 23 percent reduction in shear connector strength. In both cases, the strength of the composite beam is significantly greater than that of the bare steel beam, where the plastic moment strength of the bare steel beam is Mp = 398 ft-kips. It is acceptable to make comparisons at the nominal strength level because for the bare steel beam and the composite beam, the resistance factors and safety factors are the same. 9.5.3

Composite Beam Design Tables

The force transferred between the steel and concrete governs the strength of the composite beams. The shear studs transfer that force to the concrete, so design can be linked to the total shear force, ΣQn. Design tables have been developed that use the shear stud strength in combination with an infinite selection of concrete areas and strengths to determine the flexural strength of the composite beam. These are given in Manual Table 3-19, an example of which is shown here as Figure 9.12. The variables used in Manual Table 3-19 are defined in Figure 9.13. The beam is divided into seven PNA locations; five are in the flange and two are in the web. When the PNA is at the top of the flange, in position 1, the entire steel section is in tension. This is a fully composite beam. When the PNA is in the web at location 7, 25 percent of the potential steel section force is transferred to the concrete through the studs. As shown in Figure 9.13, the flange has five PNA locations and the stud strength for location 6 is one-half the difference between that at locations 5 and 7. These seven PNA locations establish corresponding stud strengths, ΣQn, which are also given in the tables. Designs using PNA location 7 are clearly of concern regarding the deformation capacity of the shear studs since this location represents only 25% composite action.

Chaptter 9

Compposite Construuction 387

Figu ure 9.12 Co omposite W-S Shapes: Availlable Strengthh in Flexure Copy yright © Amerrican Institute of o Steel Constrruction. Reprinnted with Perm mission. All righhts reserved.

3888 Chapter 9

Composiite Construction

Figure 9.12 (Conttinued) Copyrig ght © American n Institute of Steel S Constructiion. Reprinted with Permissioon. All rights rreserved.

Chaptter 9

Figure 9.13 9

Compposite Construuction 389

Definitiion of Variab bles for Use w with Compositte Beam Desiign Tables

Copyrightt © American Institute I of Steeel Constructioon. Reprinted w with Permissionn. All rights resserved. The location where w 50% composite actiion is used w will vary, som metimes beingg in the web and someetimes in the flange, and sh hould be checcked as part oof the design pprocess. Finding F the contribution of o the concrette to the beam m strength reqquires knowlledge of the location of the concrrete compresssive force. A As already disscussed, the force in the concrete is equal to the force in the studs, ΣQ Qn. The mom ment arm for tthat force is defined as Y Y2 in Figure 9.13. It is a function n of the conccrete strengthh and concreete geometry.. These tablees are quite flexible and accomm modate any permitted cooncrete strenngth and effe fective slab w width. The thicknesss of the slab is limited only y by the maxiimum momennt arm given iin the table. Although A thesse tables are of o most valuee in the desiggn of a compoosite beam, thhey can also be used to t check a parrticular comb bination. Sele ction of a com mposite beam m is illustratedd in Section 9.9.

390 Chapter 9

Composite Construction

EXAMPLE 9.4 Composite Beam Strength Using Tables

Goal:

Determine the design flexural strength and allowable flexural strength for the fully composite W16×26.

Given

The W16×26 beam is used with the metal deck and slab shown in Figure 9.14. The effective flange width is given, beff = 60.0 in. Fy = 50 ksi and fc′ = 4 ksi. Use the portion of Table 3-19 from Figure 9.12.

SOLUTION

Step 1:

Determine the controlling compression force using Equations I3-1a and I31b. Vc′ = 0.85 f c'tbeff = 0.85 ( 4.0 )( 3.0 )( 60.0 ) = 612 kips Vs′ = Fy As = 50.0 ( 7.68 ) = 384 kips

Because Vs′ < Vc′, the steel controls and PNA is at or above the top of the flange. In Figure 9.12 this PNA location confirms that the ΣQn = 384 kips. Step 2:

Determine the depth of the concrete acting in compression that is needed to balance this stud force using Equation 9.9. ΣQn 384 = = 1.88 in. a= 0.85 f c′beff 0.85(4)(60.0)

Step 3:

Determine the moment arm of the compressive force from the top of the steel. a 1.88 Y 2 = tc − = 6.0 − = 5.06 in. 2 2

For LRFD Step 4:

Determine the design moment strength from Figure 9.12 Enter the table with Y2 = 5.0, which will be slightly conservative for an actual Y2 = 5.06, and ΣQn = 384. For LRFD, the design moment is φM n = 370 ft-kips

For ASD Step 4:

Determine the allowable moment strength from Figure 9.12 (ASD). Enter the table with Y2 = 5.0, which will be slightly conservative for an actual Y2 = 5.06, and ΣQn = 384. For ASD the allowable moment is M n Ω = 246 ft-kips

Chaptter 9

Compposite Construuction 391

Figu ure 9.14 Co omposite Beam for Examp ple 9.4

9.5.4

Neg gative Mom ment Strengtth Accordin ng to Specification Section n I3.2b, the neegative flexurral strength caan be taken eiither as that for the bare b steel beaam according g to the provvisions of Chhapter F or, iif the composite section satisfies the criteria, as a a compositte section usiing a plastic sstress distribuution. For thee composite section, the t concrete in tension is ignored andd reinforcing steel is placeed in the tenssion region. The resisstance factor and safety factor f are thee same as forr the positivee moment caase, and the nominal flexural stren ngth is calculaated assumingg a plastic strress distribution similar to that for the positive moment m case. The limitatio ons of the Speecification staate that 1. 1 2. 2 3. 3

9.6

The steel s beam mu ust be compacct and adequaately braced aaccording to C Chapter F. Shearr connectors (steel ( headed stud or steel channel anchhors) are provvided in the negattive moment region. r The required r longitudinal reinfforcing bars aare placed wiithin the effeective width of thee slab and are properly devveloped.

SHEA AR STUD STRENGTH S H uction industtry as shear cconnectors aree called steell anchors in What aree recognized in the constru the Specification. Tw wo different ty ypes of shearr connectors: steel studs, called steel hheaded stud anchors, and channelss, called steel channel anchhors are recoggnized by the Specificationn. The shear connecto or used almosst exclusively y in building construction is the steel hheaded stud aanchor. The diameter of the studs must be 3/4 4 in. diameterr or less exceept when useed in a flat sooffit slab in which caase they may be up to 7/8 in. diameterr. In addition the stud diam meter may bee no greater than 2.5 times the flan nge thicknesss of the beam m to which thhey are welded if they are not located directly over o the beam m web. The T nominal strength of a single steell stud, Qn, ussed in a com mposite beam is given in Specifica ation Section I8.2a as

Qn = 0.5 0 Asa f c′ Ec ≤ Rg Rp Asa Fu

((AISC I8-1)

392 Chapter 9 Table 9.2

Composite Construction

Nominal Shear Strength, Qn, for One Stud with Rp = Rg = 1.0 (kips)

Stud diameter (in.) 3/8 1/2 5/8 3/4

Normal-weight concrete

Lightweight concrete

wc = 145 pcf

wc = 110 pcf

fc′ = 3 ksi 5.26 9.35 14.6 21.0

fc′ = 4 ksi 6.53 11.6 18.1 26.1

fc′ = 3 ksi

fc′ = 4 ksi

4.28 7.60 11.9 17.1

5.31 9.43 14.7 21.2

Based on Fu = 65 ksi 7.18 12.8 19.9 28.7

where Asa = f c′ =

cross-sectional area of the shank of the stud, in.2 specified compressive strength of the concrete, ksi

Fu =

minimum specified tensile strength of the stud, ksi

Ec =

Rg , Rp =

modulus of elasticity of the concrete, wc1.5 fc′ , ksi, where wc is the unit weight of the concrete in pounds per cubic foot and fc′ is in ksi. Although this is somewhat different from the equation used by ACI 318, it provides sufficiently accurate results in this instance. reduction factors used to bring the predicted stud strength into agreement with test results. When used in a flat soffit slab, Rg = 1.0 and Rp = 0.75.

The nominal strength of typical 3/4 in. shear studs is given, along with other results, in Table 9.2. Values are given for normal and lightweight concrete with f c′ = 3 ksi and 4 ksi. Values are also given for the stud strength based on the tensile strength of the stud material with Rg = Rp = 1.0. The reductions to be applied when these studs are incorporated into a slab on a metal deck are addressed later. Although not normally used in practice today, channel shear connectors are also permitted by the Specification. The nominal strength of a channel shear connector, Qn, is given as

Qn = 0.3(t f + 0.5tw )la f c′ Ec

(AISC I8-2)

where tf = thickness of channel flange, in. tw = thickness of channel web, in. la = length of channel, in. The strength of the channel shear connector must be developed by welding the channel to the beam flange for the force Qn with appropriate consideration of the eccentricity of the force on the connector. 9.6.1

Number and Placement of Shear Studs

Although a shear stud serves to transfer load between the steel beam and the concrete slab, it is not necessary to place the studs in accordance with the shear diagram of the loaded beam. Tests have demonstrated that the studs have sufficient ductility to redistribute the shear load under the ultimate load condition. Therefore, in design, it is assumed that the studs share the load equally and the total shear force determined according to Section 9.5 can be transferred over the distance

Chaptter 9

9 Figure 9.15

Compposite Construuction 393

Stud Pllacement for Concentrated C d Load

between maximum moment m and zero z moment using studs ddistributed allong this disttance. For a uniform load this ressults in Vq′/Q Qn connectorss on each sidde of the maaximum mom ment at the centerline of the beam m span. In th he case of cooncentrated looads placed aat the third pooints of the beam, thee same number of studs wo ould be used on each side of the beam bbetween the lload and the support, and a minimu um number of o studs woul d be requiredd between thee loads. This is shown in Figure 9..15. Shear S studs must m be placed d so that they have a minim mum of 1 in. oof lateral concrete cover, unless ussed in ribs of formed steel deck. In the direction of tthe shear forcce the minimuum distance from the center of thee stud to the free fr edge is 8 in. for normaal weight conncrete and 10 in. for light weight concrete. For studs s not placced in steel ddeck ribs, the minimum ceenter-to-centerr spacing is six stud diameters alo ong the mem mber and fourr stud diametters transverse to the mem mber. When placed in n metal deck ribs, the spaccing is to be no less than four diameteers in any dirrection. The maximum m center-to-ceenter stud spaacing is eight times the totaal slab thicknness or 36 in.

EXAMPLE E 9.5 Shear Stud d Determinattion

Goal: G

Deetermine the number of 33/4 in. shear studs required over the complete beeam span.

Given: G

SOLUTION N

Step S 1:

Usse the fully composite bbeam of Exam mple 9.1. Assume normaal-weight co oncrete and th he values of E Example 9.1. Deetermine the strength s of a ssingle shear sstud. Frrom Table 9.2 2, based on thhe concrete,

Qn = 0.5 Asa fc′ Ec = 26.1 kips an nd based on th he stud withouut the reduction factors Qn = Asa Fu = 228.7 kips Fo or the slab in Example 9.1 , the stud is w welded directtly to the top flange of thee beam in a fllat soffit slab,, so Rg = 1.00

3994 Chapter 9

Composiite Construction and

Rp = 0.75

Qn = Rg R p Asa Fu = 1.0 ( 0.75 )( 228.7 ) = 21.5 kkips

Use th he lowest Qn, so Step p 2:

Q n = 21.5 kipss

Deterrmine the num mber of studs required. From m Example 9.1

Vq′ = ΣQn = 650 kkips Thus,, the number required r is 650//21.5 = 30.2 sstuds Step p 3:

Deterrmine the totaal number of sstuds requiredd for the beam m. Place thirty-one 3//4 in. shear sstuds on eachh side of the beam betweeen the maxim mum momentt and the zeroo moment. Thhus, Use 62 studss for the entirre beam span LRFD Note that these caalculations arre independennt of the use of ASD or L n the shear fo orce to be trransferred is the same, ass in this exaample, when becau use the calculaations are carrried out at thee nominal strrength level.

9.77

COMPO OSITE BEA AMS WITH H FORMED METAL D DECK The combin nation of form med steel deck k and compossite design is considered tooday to be onne of the most econom mical method ds of floor co onstruction. T The steel deckk is a stay-inn-place formw work for the concretee slab. Cells, which w can be formed by ennclosing the sspace below tthe deck and bbetween the ribs, caan be used to distributee the electriccal and elecctronic system ms of the bbuilding, contributing g greatly to the overall econ nomy of the ssystem.

S Deck Prrofiles Figure 9.16 Common Steel

Chaptter 9

Figure 9.17 9

Compposite Construuction 395

Beam with w Formed Metal M Deck

The T Specifica ation providess rules in Secttion I3.2c forr steel decks w with nominall rib heights of up to 3 in. and averrage rib width hs of 2 in. or more. For a deck that hass ribs narroweer at the top than at th he interface with w the beam m, the width oof the rib usedd in calculatiions must be taken as no more thaan the width at the narrow w portion. A deck section with this proofile is show wn in Figure 9.16 alon ng with other common decck profiles. Sttuds must exttend at least 11-1/2 in. abovve the top of the steel deck. The co oncrete slab th hickness musst be specifiedd to provide 1/2 in. of covver over the top of the installed stu ud. The slab thickness t aboove the metal deck must bee at least 2.0 in., and the deck mu ust be anchorred to the sup pporting beam am with a coombination off puddle welds or other fastenerss and studs at a spacing nott to exceed 188 in. 9.7.1

Deck k Ribs Perpeendicular to Steel S Beam For beam ms supporting g the steel deck the ribs ruun perpendicuular to the beeam, as show wn in Figure 9.17. Thee space below w the top of th he rib containns concrete onnly in the alteernating spacces, so there is no opp portunity to transfer t force at this level.. Thus, the onnly concrete available for calculating the full concrete c forcee is above the top of the deeck. Table 9.3 3 Shear Stu ud Strength Adjustment Faactors Copyrightt © American Institute I of Steeel Constructioon. Reprinted w with Permissionn. All rights reserved.

Condition

Rg

Rp

No deckiing Decking oriented paraallel to the steeel shape

1.0

0.75

1.0

0.75

0.85**

0.75

wr ≥ 1.5 hr wr < 1.5 hr

Decking oriented perp pendicular to the steel shappe Number of studs occu upying the sam me decking riib 1 1.0 0.6† 2 0.6† 0.85 3 or more 0.7 m 0.6† hr = nom minal rib heigh ht, in. wr = averrage width off concrete rib or haunch (ass defined in S Section I3.2c), in. ** For a siingle stud. † This vallue may be in ncreased to 0.7 75 when emid-hht ≥ 2 in.

3996 Chapter 9

Composiite Construction

d Weak Shearr Stud Locatioons Figure 9.18 Strong and The right-hand siide of the ineequality of Eqquation I8-1, without Rp orr Rg, accounts for the tensile stren ngth of the stu ud material. This T strength must be reduuced to accouunt for the loccation of the stud within the concrrete rib, becau use there is a difference inn strength forr a stud placeed closer to the rib wall w in the dirrection of forrce versus onne placed closser to the ribb wall away ffrom the force. The stud s strength value v specifieed on the righht-hand side of the inequaality of Equattion I8-1 includes two o multipliers, Rg and Rp. A simplified ttable of the vvalues for these adjustmennt factors is given in Table T 9.3, and d the strength h of the stud ccontrolled by Fu, with Rp = Rg = 1.0, is given in Table 9.2. Rg is used to acccount for the number of sttuds in a giveen concrete riib. If a rib coontains a single stud, Rg = 1.0; if the t rib contain ns two studs,, Rg = 0.85; aand if a rib coontains three or more studs, Rg = 0.7. 0 Rp is used to acco ount for the lo ocation of thee stud in the rrib in either thhe strong or tthe weak position. Fig gure 9.18 sho ows the stron ng and weakk locations off a stud in reelation to the applied force. Becau use it is diffficult to ensu ure that the sstuds are loccated in the sstrong positioon, it is recommendeed that Rp = 0.6 be used unless the nneed for the sslight strengtth increase iss critical enough to warrant w the extra e effort in i the field tto ensure thaat studs are placed in thee strong position. Wh hen studs aree placed in th he strong posiition, which is at least 2.00 in. from thee loaded side of the rib r edge at miid-height of th he rib, as sho wn in Figure 9.18, Rp can be increasedd to 0.75, which is the same value that t was used for the flat sooffit slab. The maximum stud s spacing is specified aas 36 in., whhich is conveenient becausse many in multiples decks have a rib spacing of o 6 in. 9.77.2

Deck Ribs R Parallel to t Steel Beam m For girders supporting beams b that caarry steel deckk, the ribs ruun parallel too the steel secction, as shown in Figure F 9.19. Concrete bellow the top of the deckk can be useed in calculaating the composite section s propeerties and must m be usedd in shear sttud calculatioons. For callculation purposes, co oncrete below w the top of th he steel deck can be negleccted unless itt is needed to balance the shear stu ud strength. The T design pro ocedure descrribed for flat ssoffit beams aapplies here aas well,

Figure 9.19 Girder with h Formed Meetal Deck

Chaptter 9

Figure 9.20 9

Compposite Construuction 397

Examplle 9.6, case (aa)

provided d sufficient co oncrete is av vailable abovve the top off the steel deeck as determ mined using Equation n 9.1 or 9.9. If I the concrette below the top of the steeel deck musst be used to balance the stud or steel s beam strrength, the on nly differencee in the deterrmination of sstrength is reelated to the changed geometry fro om inclusion of o a portion oof the ribs. When W the dep pth of the steeel deck is 1 -1/2 in. or ggreater, the avverage width, wr, of the haunch or o rib is not to t be less thaan 2 in. for tthe first stud in the transvverse row pluus four stud diameters for each ad dditional stud d. If the deck rib is too naarrow, the deeck can be spplit over the beam and d spaced in su uch a way as to allow for tthe necessary rib width witthout adverseely affecting member strength. For F this deck orientation, Rg is used too account for the width-to--height ratio of the deck rib. When wr /hr ≥ 1.5, Rg = 1.0. Wh hen wr /hr < 11.5, Rg = 0.85.. Rp is taken as 0.75 in all caases where thhe deck ribs aare parallel to the supportinng member, just as fo or the flat sofffit slab.

EXAMPLE E 9.6 Composite Beam Design

Goal: G

Caalculate the design momennt and allowabble moment aand determinee the stud req quirements for fo a compossite section. C Carry out thee calculations for the following threee cases. (a) Full composite action (F igure 9.20). (b) Partial com mposite actionn with ΣQn = 3387 kips , whiich results in the PNA at the centerr of the top flaange of the stteel beam (Figgure 9.21). (c)) Partial comp posite action with ΣQn = 2260 kips , whiich results in the PNA at the bottom m of the top fflange of the ssteel beam (Figure 9.22).

SOLUTION N

Given: G

Usse a W18×35 5 with a 6-in. slab on a 3 iin. metal deckk perpendicular to the m spacing is 12 ft and beeam with the profile p shownn in Figure 9.16c. The beam thee beam span is 40 ft. fc′ = 4 ksi and Fy = 50 ksi. Froom Table 1-1 d = 17.7 in. and tf = 0.42 25 in.

Step S 1:

Deetermine the effective e flannge width. beff = 440.0/4 = 10.00 ft (governs) beff = beam spacinng = 12.0 ft

398 Chapter 9

Composite Construction Step 2:

Determine the compression force using the full concrete and full steel areas using Equations I3-1a and I3-1b. Vc′ = 0.85 f c′tbeff = 0.85 ( 4 )( 3.0 )(120 ) = 1220 kips Vs′ = Fy As = 50 (10.3) = 515 kips

Part a Step 3:

Full Composite Action (Figure 9.20) Determine the controlling concrete force. For full composite action, Cc is the smaller of Vc′ and Vs′; thus, Cc = 515 kips

Step 4:

Calculate the effective depth of the concrete using Equation 9.1. Cc 515 a= = = 1.26 in. ′ 0.85 f c beff 0.85(4)(120) Because a is less than the 3 in. available in the concrete above the deck, the procedures for a flat soffit beam can be used.

Step 5:

Determine the nominal moment strength. ⎛d ⎞ ⎛ a⎞ M n = Ts ⎜ ⎟ + Cc ⎜ t − ⎟ ⎝2⎠ ⎝ 2⎠ 1.26 ⎞ ⎛ 17.7 ⎞ ⎛ = 515 ⎜ ⎟ + 515 ⎜ 6.0 − ⎟ = 7320 in.-kips 2 ⎠ ⎝ 2 ⎠ ⎝ 7320 = = 610 ft-kips 12

For LRFD Step 6:

The design moment is

For ASD Step 6:

The allowable moment is

Step 7:

Determine the strength of a single stud.

φM n = 0.9 ( 610 ) = 549 ft-kips

M n 610 = = 365 ft-kips Ω 1.67

From Table 9.2, the value of a single 3/4 in. stud with normal-weight concrete, fc′ = 4 ksi, is 26.1 kips based on the concrete. However, because the studs are used in conjunction with the metal deck, a check for any required reduction must be made. For the deck profile given, the ribs are spaced at 12.0 in. on center which would permit 39 studs per beam if only one stud were placed in each rib. Thus, assuming two studs per rib with the studs placed in the weakest location, from Tables 9.2 and 9.3, Rg = 0.85 Rp = 0.6 Asc Fu = 28.7 kips

Chapter 9

Composite Construction 399

Rg R p Asa Fu = 0.85 ( 0.6 )( 28.7 ) = 14.6 ≤ 26.1 kips The stud strength is the lower value, based on the stud placement in the metal deck. Step 8:

Determine the number of studs required on each side of the maximum moment. The shear that is to be transferred is 515 kips. Therefore, Number of studs = 515/14.6 = 35.3

Step 9:

Determine the total number of studs required for the beam. Use 36 studs on each half span or 72 studs for the full beam span With this deck profile, studs can be placed in pairs every 12 in. This will nicely accommodate the 72 studs on the 40 ft span with two studs placed in each rib.

Part b Step 10:

Partial Composite Action (Figure 9.21) Determine the controlling concrete force. Because the value of ΣQn = 387 kips given is less than Vc′ and Vs′ as determined in part a, ΣQn controls and this is a partially composite beam. With ΣQn Vs′ = 387 515 = 0.75 stud ductility will not be a concern.

Step 11:

Calculate the effective depth of the concrete using Equation 9.9 with Cc = ΣQn = 387 kips . Cc 387 a= = = 0.949 in. ′ 0.85 f c beff 0.85(4)(120) Because a < 3.0 in., sufficient concrete is available above the metal deck as in part (a).

Step 12:

Determine the area of steel in compression. V ′ − Cc 515 − 387 As −c = s = = 1.28 in.2 2 Fy 2(50)

Step 13:

Determine the location of the PNA in the steel. Assume the PNA is in the flange. A 1.28 = 0.213 in. < t f = 0.425 in. x = s −c = bf 6.00 Therefore, the PNA is in the flange.

Step 14:

Determine the nominal moment strength.

400 Chapter 9

Composite Construction

⎛d ⎞ ⎛t⎞ ⎛ x⎞ M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠ 0.949 ⎞ ⎛ 17.7 ⎞ ⎛ ⎛ 0.213 ⎞ = 515 ⎜ ⎟ + 387 ⎜ 6.0 − ⎟ − 2 (1.28)( 50 ) ⎜ ⎟ = 6680 in.-kips 2 ⎠ ⎝ 2 ⎠ ⎝ ⎝ 2 ⎠ 6680 = = 557 ft-kips 12 For LRFD Step 15:

The design moment is

For ASD Step 15:

The allowable moment is M n ⎛ 557 ⎞ =⎜ ⎟ = 334 ft-kips Ω ⎝ 1.67 ⎠

Step 16:

Determine the stud requirements.

φM n = 0.9 ( 557 ) = 501 ft-kips

The shear to be transferred is 387 kips. The value of stud strength previously determined in part (a) for two studs per rib is Qn = 14.6 kips Step 17:

Determine the required number of shear studs. 387/14.6 = 26.5 studs

Step 18:

Determine the total number of studs required for the beam. Use 27 studs in each half span or 54 studs for the full beam span. With 54 studs symmetrically placed in 39 available ribs, 12 ribs will remain without studs. It is possible to recalculate the stud requirements to see if the higher strength when only one stud per rib is used would work for this beam. Since Rg = 1.0 for a single stud per rib, from part (a) Rg R p Asa Fu = 1.0 ( 0.6 )( 28.7 ) = 17.2 ≤ 26.1 kips and the number of studs required for the half span is 387 17.2 = 22.5 Thus, 46 studs are required and this would still require two studs per rib for 8 ribs. For constructability reasons, 54 studs placed symmetrically, two studs per rib will be selected.

Part c Step 19:

Partial Composite Action (Figure 9.22) Determine the controlling concrete force. Because the value of ΣQn = 260 kips given is less than Vc′ and Vs′, ΣQn

Chapter 9

Composite Construction 401

controls and this is a partially composite beam. With ΣQn Vs′ = 260 515 = 0.50 stud ductility will not be a concern. Step 20:

Calculate the effective depth of concrete using Equation 9.9. Cc 260 a= = = 0.637 in. ′ 0.85 f c beff 0.85(4)(120)

Step 21:

Determine the area of steel in compression. V ′ − Cc 515 − 260 As −c = s = = 2.55 in.2 2 Fy 2(50)

Step 22:

Determine the location of the PNA in the steel. Assume the PNA is in the flange. A 2.55 x = s −c = = 0.425 in. bf 6.00 which is the flange thickness, as expected.

Step 23:

Determine the nominal moment strength. ⎛d ⎞ ⎛t⎞ ⎛ x⎞ M n = Ts ⎜ ⎟ + Cc ⎜ ⎟ − 2 As −c Fy ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠

0.637 ⎞ ⎛ 17.7 ⎞ ⎛ ⎛ 0.425 ⎞ = 515 ⎜ ⎟ + 260 ⎜ 6.0 − ⎟ − 2 ( 2.55 )( 50 ) ⎜ ⎟ = 5980 in.-kips 2 ⎠ ⎝ 2 ⎠ ⎝ ⎝ 2 ⎠ 5980 = = 498 ft-kips 12 For LRFD Step 24: For ASD Step 24:

Step 25:

The design moment is

φM n = 0.9 ( 498 ) = 448 ft-kips

The allowable moment is M n ⎛ 498 ⎞ =⎜ ⎟ = 298 ft kips Ω ⎝ 1.67 ⎠ Determine the stud requirements. The shear to be transferred is 260 kips. As before, the single stud strength is Qn = 14.6 kips

Step 26:

Determine the required number of shear studs. 260/14.6 = 17.8 studs.

Step 27:

Determine the total number of studs required for the beam.

4002 Chapter 9

Composiite Construction Use 18 1 studs in eacch half span oor 36 studs foor the full beaam span. Howeever, this requ uires only a ssingle stud inn each rib, so the Rg = 0.855 does not neeed to be applied. Thereforre, as from paart (b) Qn = 1.0 ( 0.6 )( 28.7 ) = 17.22 < 26.1 kips and 260//17.2 = 15.1 sstuds Thus,, use a total of 32 sstuds which h will also be accommodatted with one sstud per rib.

Figure 9.21 Exam mple 9.6, case (b)

Figure 9.22 Exam mple 9.6, case (c) A co omparison off the results frrom Examplee 9.6 shows thhat the magniitude of the reeduction in moment strength s that results r from a reduction inn the shear foorce transferdd by the shearr studs is less than th he magnitudee of shear strrength reducttion. For parrt (b) of Exaample 9.6, thhe shear strength wass reduced app proximately 25 2 percent buut it producedd a moment strength reduuction of only approx ximately 9 perrcent. Similarrly, for part (cc), the 50 perrcent reduction in stud streength led to an 18 perrcent reductio on in moment strength. Thiis reinforces the idea that a partially coomposite

Chaptter 9

Compposite Construuction 403

beam maay be more economical e th han a fully ccomposite beaam and sugggests why theey are quite common in building design. d

9.8

FULL LY ENCAS SED STEEL L BEAMS Steel beaams fully enccased in conccrete that conntributes to tthe strength oof the final m member are called en ncased beamss. Such beam ms can be deesigned throuugh one of tw wo procedurees given in Specifica ation Section I3.3. The fllexural strenggth can be ccalculated froom the superrposition of elastic sttresses, considering the efffects of shorring. Or, wheen shear connnectors are prrovided, the flexural strength can be based on the plastic sttress distributtion or strainn compatibilitty approach for the composite c secction. Alternaatively, the sstrength can bbe calculatedd from the pllastic stress distributiion on the steeel section allone. In all ccases, φb = 00.9 and Ωb = 1.67. Althouugh encased beams arre not particu ularly common in buildinng construction today, theese provisionns will also apply wh hen determining the flexuraal strength off composite beeam-columnss.

9.9

SELE ECTING A SECTION The desiign of a com mposite beam is somewhatt of a trial-annd-error proccedure, as aree numerous other dessign situation ns. The materrial presentedd thus far in tthis chapter hhas been direccted toward the deterrmination of section s streng gth when the cross section and concretee dimensions are known. This section addressees the prelim minary selectiion of the steeel shape to go along w with a given concrete slab. This procedure p is followed by a discussionn of the design tables foound in the Manual. With W an estim mate of beam depth, the w weight of the bbeam can be estimated. Thhis is based on the asssumption thaat the PNA is within the cooncrete so thaat the full steeel section is aat yield. The resulting dimensions are a given in Figure F 9.23. The T moment arm between the tension fforce in the stteel and the ccompression force in the concrete is given by d ⎛ a⎞ (9.10) moment arm = + ⎜ t − ⎟ 2 ⎝ 2⎠ If the nominal momen nt strength is divided by thhe moment aarm, the requiired tension fforce can be determin ned. If that fo orce is divideed by the steeel yield streess, the requiired area is ddetermined. Multiplying the requirred area by weight w of steell, 3.4 lb/ft forr each squaree inch, yields an estimate of the beam weight. Thus, T 3.4 M n (9.11) beam m weight = d a⎞ ⎛ ⎜ + t − ⎟ Fy 2⎠ ⎝2

9 Figure 9.23

Momen nt Arm for Preeliminary Weeight Determiination

404 Chapter 9

Composite Construction

To determine the beam weight by this approach, the depth of the beam must be estimated. Several approaches have been suggested for this. One simple approach is to take the span in feet and divide it by 2 to get the depth, d, in inches. Another approach to determine the depth of the total composite section (d + t) is to take the span in feet and divide by 4/3 to obtain the total depth in inches. Any reasonable approach gives a starting point. Because the thickness of the slab is determined from the design in the transverse direction, only the effective depth of the concrete is left to be determined. It is generally sufficient to assume that the effective depth of the concrete is 1 in.; therefore, a/2 = 0.5 in. Although this approach to finding a starting point for composite beam design might be helpful, the design tables in the Manual make this much effort unnecessary. Manual Table 3-19 (Figure 9.12) has already been discussed in the context of determining the strength of a given combination of steel and concrete. It will now be approached from the perspective of selecting a section through the use of an example.

EXAMPLE 9.7a Composite Beam Design by LRFD

Goal:

Select the most economical W-shape to be used as a composite beam.

Given:

The composite beam spans 30.0 ft and is spaced at 10.0 ft from adjacent beams. It supports a 5 in. slab on a 2 in. formed steel deck with a profile similar to that shown in Figure 9.16a. The beam must carry a dead load moment of 50.0 ft-kips and a live load moment of 150 ft-kips. In addition, the bare steel beam must be checked for the dead load plus a construction live load moment of 40.0 ft-kips. Fy = 50 ksi and fc′ = 3 ksi.

SOLUTION

Step 1:

Determine the required moment for the composite beam. Mu = 1.2(50.0) + 1.6(150) = 300 ft-kips

Step 2:

Determine the starting moment arm for the concrete from the top of the steel. Manual Table 3-19 is most effective when entered with a value for Y2. In order to start the design process, the moment arm for the compressive force in the concrete must be estimated. It is almost always adequate to assume, as a starting point, that a = 1.0 in. Thus 1.0 Y2 = 5 – = 4.5 in. 2

Step 3:

Select potential W-shapes from Figure 9.12 and Manual Table 3.19. Since the beam span is 30 ft it will be acceptable to use partial composite action lower than 50%. Enter the column in Figure 9.12 with Y2 = 4.5 and proceed down to identify the potential section that will carry the design moment of 300 ft-kips. W16×26 with ϕMn = 306 ft kips and Σ Qn = 242 kips, PNA location 4 Using portions of Table 3.19 from the Manual yields additional possibilities: W16×31 with ϕMn = 319 ft-kips and ΣQn = 164 kips, PNA location 6 W14×34 with ϕMn = 309 ft-kips and ΣQn = 159 kips, PNA location 6 W14×30 with ϕMn = 311 ft-kips and ΣQn = 248 kips, PNA location 4

Chapter 9

Composite Construction 405

W14×26 with ϕMn = 312 ft-kips and ΣQn = 332 kips, PNA location 2 Step 4:

Determine the effective flange width. 30.0(12) beff ≤ = 90.0 in. 4 beff ≤ 10.0(12) = 120 in. Therefore beff = 90.0 in.

Step 5:

Using the W16×31, determine the depth of concrete needed to balance the force in the shear studs. For the beam span and spacing given using Cc = ΣQn = 164 kips and the effective flange width already determined. Cc 164 a= = = 0.715 in. 0.85 f c′beff 0.85(3)(90.0) Because this is less than the value a = 1.0 in. that was assumed to start the problem, the assumption was conservative. Design could continue with the determination of a more accurate required stud strength, or this conservative solution could be used. The required number of studs would be determined as before, accounting for the presence of any formed steel deck and its influence on the individual stud strength.

Step 6:

Determine the required strength of the bare steel beam under dead load plus construction live load. Mu = 1.2(50.0) + 1.6 (40.0) = 124 ft-kips

Step 7:

Check to verify that the bare steel beam will support the required strength. Since this beam has the steel deck attached to its compression flange, it will be considered as fully braced for lateral-torsional buckling under the weight of the construction loads. From Manual Table 3-19, the W16×31 has a design strength of φMp = 203 ft-kips > 124 ft-kips Therefore, the W16×31 is an acceptable selection for strength.

Step 8:

To show what happens when the assumption for a is not quite as good, the W14×26 is considered. Again using the ΣQn determined from the table and the effective flange width, Cc 332 a= = = 1.45 in. 0.85 f c′beff 0.85(3)(90.0) This is significantly greater than the assumed value. To consider this section further, determine a new Y2 such that 1.45 Y2 = 5 − = 4.28 in. 2

406 Chapter 9

Composite Construction Entering Manual Table 3-19 with Y2 = 4.0 as a conservative number, Mu = 300 ft-kips is determined and it corresponds to the same required shear stud strength. Thus, this section also meets the strength requirements, and the design can proceed with stud selection.

EXAMPLE 9.7b Composite Beam Design by ASD

Goal:

Select the most economical W-shape to be used as a composite beam.

Given:

The composite beam spans 30.0 ft and is spaced at 10.0 ft from adjacent beams. It supports a 5 in. slab on a 2 in. formed steel deck with a profile similar to that shown in Figure 9.16a. The beam must carry a dead load moment of 50.0 ft-kips and a live load moment of 150 ft-kips. In addition, the bare steel beam must be checked for the dead load plus a construction live load moment of 40.0 ft-kips. Fy = 50 ksi and fc′ = 3 ksi.

SOLUTION

Step 1:

Determine the required moment for the composite beam. Ma = 50.0 + 150 = 200 ft-kips

Step 2:

Determine the starting moment arm for the concrete from the top of the steel. Manual Table 3-19 is most effective when entered with a value for Y2. In order to start the design process, the moment arm for the compressive force in the concrete must be estimated. It is almost always adequate to assume, as a starting point, that a = 1.0 in. Thus, 1.0 Y2 = 5− = 4.5 in. 2

Step 3:

Select potential W-shapes from Figure 9.12 and Manual Table 3.19. Since the beam span is 30 ft it will be acceptable to use partial composite action lower than 50%. Enter the column in Figure 9.12 with Y2 = 4.5 and proceed down to identify the potential section that will carry the design moment of 200 ft-kips. W16×26 with Mn/Ω = 204 ft kips and ΣQn = 242 kips, PNA location 4 Using portions of Table 3-19 from the Manual yields additional possibilities: W16×31 with Mn/Ω = 212 ft-kips and ΣQn = 164 kips, PNA location 6 W14×34 with Mn/Ω = 205 ft-kips and ΣQn = 159 kips, PNA location 6 W14×30 with Mn/Ω = 207 ft-kips and ΣQn = 248 kips, PNA location 4 W14×26 with Mn/Ω = 208 ft-kips and ΣQn = 332 kips, PNA location 2

Step 4:

Determine the effective flange width. 30.0(12) beff ≤ = 90.0 in. 4 beff ≤ 10.0(12) = 120 in.

Chapter 9

Composite Construction 407

Therefore, beff = 90.0 in. Step 5:

Using the W16×31, determine the depth of concrete needed to balance the force in the shear studs. For the beam span and spacing given using Cc = ΣQn = 164 kips and the effective flange width already determined. Cc 164 a= = = 0.715 in. 0.85 f c′beff 0.85(3)(90.0) Because this is less than the value a = 1.0 in. that was assumed to start the problem, the assumption was conservative. Design could continue with the determination of a more accurate required stud strength, or this conservative solution could be used. The required number of studs would be determined as before, accounting for the presence of any formed steel deck and its influence on the individual stud strength.

Step 6:

Determine the required strength of the bare steel beam under dead load plus construction live load. Ma = 50.0 + 40.0 = 90.0 ft-kips

Step 7:

Check to verify that the bare steel beam will support the required strength. Since this beam has the steel deck attached to its compression flange, it will be considered as fully braced for lateral-torsional buckling under the weight of the construction loads. From Manual Table 3-19, the W16×31 has an allowable strength of M p Ω = 135 ft-kips > 90.0 ft-kips Therefore, the W16×31 is an acceptable selection for strength.

Step 8:

To show what happens when the assumption for a is not quite as good, the W14×26 is considered. Again using the ΣQn determined from the table and the effective flange width, Cc 332 a= = = 1.45 in. 0.85 f c′beff 0.85(3)(90.0) This is significantly greater than the assumed value. To consider this section further, determine a new Y2 such that 1.45 Y2 = 5− = 4.28 in. 2 Entering Manual Table 3-19 with Y2 = 4.0 as a conservative number, Ma = 210 ft-kips is determined and it corresponds to an increased required shear stud strength. Thus, this section meets the strength requirements and the design can proceed with stud selection.

408 Chapter 9 Table 9.4

Shape W16×31 W16×26 W14×34 W14×30 W14×26

Composite Construction Shapes Selected for Example 9.7a (LRFD) Percent composite Weight of action steel Σ Qn Mu 319 164 35.9 930 306 242 63.0 780 309 159 31.8 1020 311 248 56.0 900 312 332 86.2 780

Number of studs 16 24 16 24 32

Equivalent weight of studs 160 240 160 240 320

Equivalent total weight of beam 1090 1020 1180 1140 1100

Which of the many possible sections should be chosen as the final design depends on the overall economics of the situation. One way to compare several choices is to look at the total weight of the steel sections combined with the total quantity of studs required. To make this comparison it is often effective to assume that an installed single shear stud has the equivalent cost of 10 pounds of steel. To make this type of comparison, the five potential sections found initially in Example 9.7a are presented in Table 9.4. Here, it was assumed that Qn = 21.0 kips. This means that no consideration was taken for metal deck reduction. In addition, no check was made for the assumed versus actual a dimension. This table is simply to help determine which of the potential shapes should be considered further. Based on this table, it could be said that the W16×26 with an equivalent weight of 1020 lbs should be investigated further.

9.10 SERVICEABILITY CONSIDERATIONS Three important serviceability considerations are associated with the design of composite floor systems: deflection during construction, vibration under service loads, and live load deflection under service loads. 9.10.1

Deflection During Construction

As discussed in Section 9.3, the Specification permits either shored or unshored construction. With unshored construction, the Specification requires that the steel section alone have adequate strength to support all loads applied prior to the concrete’s attaining 75 percent of its specified strength. The bare steel beam, under the weight of these loads, deflects as an elastic member. Because of this deflection of the beam under the weight of the wet concrete, cambering of the steel beam is often specified. Cambering is the imposition of a permanent upward deflection of the beam in its unloaded state so that, under load, the downward deflection results in a beam without excessive final deflection from the original horizontal position. Predicting the necessary camber is difficult because of the varying methods and sequences of concrete placement used by different contractors as well as such factors as the end restraint provided by the beam connection. Even with camber it is often prudent for the designer to add a little extra concrete load into the design dead load and for the contractor to allow for a little extra concrete in the quantity estimate. In the case of shored construction, deflection during construction is usually not a concern, because the shores are not removed until the concrete has achieved some strength and composite action can be counted upon. The deflection under the wet concrete for shored construction is at a minimum. On the other hand, long-term deflection due to creep of the concrete may have to be investigated because the concrete is stressed under the self-weight as a permanent load along with the sustained service loads.

Chapter 9 9.10.2

Composite Construction 409

Vibration Under Service Loads

Composite construction usually is shallower than comparable noncomposite construction and, therefore, may be more susceptible to perceived vibrations. Because vibration calculations assume that the beam behaves compositely, even when it is not a composite beam, the additional stiffness of a composite beam does not improve its vibration characteristics. If problems occur, they usually occur in applications with long spans and little damping. For instance, a large area of a department store containing only a light jewelry display and with no partition walls connecting one floor to the next might exhibit vibrations that would be perceptible to some customers. On the other hand, an office building constructed with the same floor system could contain full or partial height partitions that would provide sufficient damping to obviate any perceived vibration. Because of wide differences in human perception of vibration and many other factors, vibration problems do not lend themselves to simple solutions. AISC Design Guide 11, Vibrations of Steel Framed Structural Systems Due to Human Activity, provides more information and gives the designer an approach to vibration acceptance criteria, damping, and rational design techniques. 9.10.3

Live Load Deflections

Live load deflections can be a critical design consideration for many applications. Excessive deflection could cause problems with the proper fit of partitions, doors, and equipment and may also result in an unacceptable appearance, including cracking of finishes and other visible evidence of distress. Therefore, a live load deflection calculation should be carried out for most situations. As discussed in Chapter 6, this calculation is made with the service loads for which deflection is of interest, usually the nominal live loads. Because beam deflections are a function of the stiffness of the beam, the modulus of elasticity and moment of inertia of the composite section must be determined. The true moment of inertia at the service load level for which deflections are to be calculated is not easily determined. In addition, the modulus of elasticity of the composite section must account for the interaction of steel and concrete. The normal approach is to transform the concrete into a material that behaves like steel, with the same modulus of elasticity. The moment of inertia of the new transformed section can then be determined. Transformation of the concrete into steel is accomplished by dividing the concrete area by the modular ratio, n = Es /Ec, and using the same thickness. Although this seems like a fairly straightforward process, the problem is determining the thickness of the concrete that is actually participating in resisting the deflection. One approach is to assume that the only concrete participating in deflection resistance is also that which is providing strength. Thus, whether it is a fully composite or partially composite member, a moment of inertia can be determined using the known value of a from the strength calculations. Because the nominal strength is calculated at the ultimate load level, the amount of concrete actually participating for service loads could be significantly more than that used in the strength calculations. Thus, a moment of inertia determined by this approach is less than what might actually be available and is called a lower bound moment of inertia, ILB. Figure 9.24 is a sample from Manual Table 3-20, which gives the lower bound moment of inertia in a format that parallels the strength tables already discussed. Use of these ILB values results in a conservative estimate of service load deflections.

4110 Chapter 9

Composiite Construction

Figure 9.24 Lower-Bound Elasstic Moment of o Inertia, ILB, for Plastic C Composite Secctions

Copyrig ght © American n Institute of Steel S Constructiion. Reprinted with Permissioon. All rights rreserved.

Chapter 9

EXAMPLE 9.8 Deflection

SOLUTION

Composite Construction 411

Goal:

Determine the construction load deflection of the bare steel beam and the service load deflection for the composite beam of Example 9.7

Given:

Consider the W16×26 addressed in Example 9.7. Check the beam for a dead load of 0.45 kips/ft, a construction live load of 0.36 kips/ft, and an inservice live load of 1.35 kips/ft. Compare the construction load deflection to span/360. For the live load deflection, use the lower bound moment of inertia from Figure 9.24 and compare the calculated deflection to span/360 as a design limit.

Step 1:

Determine the total construction load for deflection calculations. w = wD + wL–const. = 0.45 + 0.36 = 0.81 kips/ft

Step 2:

Determine the moment of inertia of the W16×26 from Manual Table 1-1 or Table 3-20. Ix = 301 in.4

Step 3:

Calculate the construction load deflection and compare to span/360. 5(0.81)(30.0) 4 (1728) 30(12) Δ LL = = 1.69 in. > = 1.0 in. 384(29,000)(301) 360 Because the deflection exceeds our limit, cambering of the beam or shoring during construction would be required. Shoring has a significant impact on cost as well as on scheduling; therefore, it is likely that the beam would be cambered or a larger section used. Selecting camber as the appropriate solution, if we must continue to use this W16×26 , the beam should be cambered to approximately 75 percent of the dead load deflection. Thus, 0.75(1.69) = 1.27 in., so specify a camber of 1.25 in.

Step 4:

Assuming that the construction load deflection issue is resolved, determine the live load deflection under the in-service live load. From Example 9.7, the W16×26 was selected using Y2 = 4.5 and the resulting shear stud force was ΣQn = 242 kips at PNA location 4 for both LRFD and ASD.

Step 5:

Determine the lower bound moment of inertia. Using Manual Table 3-20 with the values given in step 4, select ILB = 754 in.4

Step 6:

Determine the live load deflection. 5(1.35)(30.0) 4 (1728) Δ LL = = 1.13 in. 384(29,000)(754)

412 Chapter 9

Composite Construction Step 7:

Compare the calculated deflection with the given limit. span 30(12) Δ LL = 1.13 > = = 1.0 in. 360 360 Because the calculated deflection is greater than the limiting value, the live load deflection is not acceptable based on the given criteria. This result, combined with the construction load deflection issue, would likely lead the designer to select a larger section for this situation, as was actually done in Example 9.7. Deflection calculations are carried out under service loads and are independent of design by LRFD or ASD.

9.11 COMPOSITE COLUMNS Composite columns in building construction have been much slower to gain acceptance than composite beams. Specification provisions were first provided in the 1986 LRFD Specification and for ASD in the 2005 Specification. Although the use of composite columns in buildings is still quite limited, attention to hardening of structures against blast forces will likely bring them more to the forefront. Specification Section I2 provides for two types of composite columns: open shapes encased by concrete and hollow shapes filled with concrete. Composite columns exist at the interface between specification provisions for steel and those for concrete. For a member to qualify as a composite column under the Specification, it must meet the following limitations: 1. 2.

3.

4. 5. 6.

The cross-sectional area of the steel member must constitute at least 1 percent of the gross area. For an encased member the concrete encasement must be reinforced with continuous longitudinal steel as well as lateral ties or spirals. The longitudinal steel area must be at least 0.004 times the gross area, and the tie area must be at least equivalent to No. 3 bars at 12 in. spacing. Tie spacing must not exceed onehalf the smallest dimension of the column. For filled hollow sections, strength is a function of local buckling of the walls of the hollow section. To be classified as a compact section, HSS must have a minimum wall thickness such that b t ≤ 2.26 E Fy for rectangular HSS and D t ≤ 0.15 ( E Fy ) for round HSS. Noncompact and slender wall hollow sections may be used for filled composite columns provided the appropriate Specification provisions are met. For these shapes, the compressive strength will be limited by the limit state of local buckling. The concrete strength, fc′, must be between 3 ksi and 10 ksi for normal-weight concrete and 3 ksi and 6 ksi for lightweight concrete as for beams. The maximum value of Fy for structural steel shapes to be used in calculating strength is 75 ksi as for beams. The maximum value of Fy for reinforcing bars to be used in calculating strength is 80 ksi as for beams.

Although these requirements are usually readily satisfied, for situations where they are not, ACI 318 should also be consulted. To account for the effects of column slenderness on the nominal strength of a composite column, the equations found in Chapter E for steel columns are used with slight modification.

Chapter 9

Composite Construction 413

Because of the combination of two dissimilar materials and the general uncertainties of composite column behavior, the resistance and safety factors are taken as φc = 0.75 (LRFD) Ωc = 2.00 (ASD) To convert the column equations, Equations E3-2 and E3-3, for use with a composite column, multiply each occurrence of stress; the critical stress, Fcr, the yield stress, Fy, and the elastic buckling stress, Fe, by the area of the column and make the following substitutions; Fcr A = Pn , Fy A = Pno and Fe A = Pe . This is presented in Specification Section I2.1 for encased composite columns where P for no ≤ 2.25 Pe ⎛ Pno ⎞ ⎡ ⎜ ⎟⎤ Pn = Pno ⎢ 0.658⎝ Pe ⎠ ⎥ ⎢⎣ ⎥⎦

and for

(AISC I2-2)

Pno > 2.25 Pe

Pn = 0.877 Pe

(AISC I2-3)

Where Pno for the bare steel column was the yield strength of the shape, for a composite column this strength must also include the contribution of the reinforcing steel and the concrete stressed to their ultimate strength. Thus, to account for the reinforcing steel, FysrAsr is added and to account for the concrete, 0.85f’cAc is added, Thus, (AISC I2-4) Pno = Fy As + Fysr Asr + 0.85 f c′Ac Where Pe for the bare steel column was the elastic buckling strength based on the stiffness, EI, of the steel shape, for the composite column it must be based on an effective stiffness, EIeff, which incorporates the contribution of the reinforcing steel and the concrete. Thus, π 2 EI eff (AISC I2-5) Pe = ( Lc )2 and the effective stiffness is a direct summation of the contribution of the steel shape, the reinforcing steel and an appropriate portion of the concrete, thus (AISC I2-6) EI eff = Es I s + E s I sr + C1 Ec I c Recent research has shown that the contribution to stiffness from the concrete is a function of the ratio of the total steel area to the gross area of the composite member, up to 70% of the stiffness of the uncracked concrete section, given by ⎛ A + Asr ⎞ C1 = 0.25 + 3⎜ s (AISC I2-7) ⎟ ≤ 0.7 ⎝ Ag ⎠ In these equations, the s subscript refers to the steel section, the sr subscript refers to the longitudinal reinforcing steel, and the c subscript refers to the concrete. With these variables defined, Equations I2-2 and I2-3 are then used to determine the nominal strength of the encased composite column. For filled columns with compact HSS or box sections, the equivalent yield strength must account for the behavior that these columns exhibit. Thus, as for encased composite columns, the steel section is at yield and the concrete is at 0.85f’c for rectangular sections and 0.95f’c for round sections. This increase to 0.95f’c reflects the increase in strength of the concrete due to the

414 Chapter 9

Composite Construction

confinement of the circular steel section just as for spiral columns designed according to ACI 318. The reinforcing steel is transformed by the modular ratio, Es Ec , into a material that behaves like concrete. Thus ⎛ ⎛ E ⎞⎞ Pno = Pp = Fy As + C2 f c' ⎜ Ac + Asr ⎜ s ⎟ ⎟ (AISC I2-9a,b) ⎝ Ec ⎠ ⎠ ⎝ where C2 = 0.85 for rectangular sections = 0.95 for round sections The effective stiffness of the filled hollow section also accounts for the confinement provided by the rectangular and round steel shape. In this case, research has shown that the contribution to the effective stiffness of the concrete is up to 90% of the stiffness of the uncracked concrete section. Thus, (AISC I2-12) EI eff = Es I s + Es I sr + C3 Ec I c and ⎛ A + Asr ⎞ C3 = 0.45 + 3⎜ s (AISC I2-13) ⎟ ≤ 0.9 A g ⎝ ⎠ Equations I2-2 and I2-3 are then used to determine the nominal strength for the filled compact wall column. For what is called a noncompact wall composite column, the concrete contribution to Pno is reduced from 0.85f’c for rectangular sections and 0.95f’c for round sections to 0.7f’c for both as the wall slenderness increases from λp to λr. For slender wall filled composite members, the contribution of the concrete to Pno remains unchanged but the contribution of the steel shape becomes a function of the critical buckling stress of the wall. These requirements are given in Specification Section I2.2b(b) and I2.2b(c). There are four square ASTM A500 Grade C HSS that are noncompact and one that is slender; six round HSS are noncompact and none are slender.

EXAMPLE 9.9 Composite Column Strength

Goal:

Determine the nominal strength of an encased composite column. Then determine the design strength and the allowable strength.

Given:

The column is composed of a W14×53 encased in 18 in.×22 in. of concrete as shown in Figure 9.25. Additional given information is as follows: Column effective length = 15 ft. Steel shape: Fy = 50 ksi Reinforcing: four #9 bars, Gr. 60, Fy = 60 ksi. Area of one bar = 1.0 in.2 Concrete strength: fc′ = 5 ksi Ec = 1451.5 5.0 = 3900 ksi

SOLUTION

Step 1:

Determine the areas of the components. As = 15.6 in.2 from Manual Table 1-1 Asr = 4 (1.0 ) = 4.0 in.2 Ac = bh − As − Asr = 18.0 ( 22.0 ) − 15.6 − 4.0 = 376 in.2 Ag = bh = 18.0 ( 22.0 ) = 396 in.2

Chapter 9 Step 2:

Composite Construction 415

Check the minimum steel ratios. A 15.6 ρs = s = = 0.0394 > 0.01 Ag 22.0(18.0)

ρsr =

Asr 4.0 = = 0.0101 > 0.004 Ag 22.0(18.0)

So the specified minimums are satisfied. Step 3:

Determine Pno and Pe. From Equation I2-7 ⎛ 15.6 + 4.0 ⎞ C1 = 0.25 + 3 ⎜ ⎟ = 0.398 < 0.7 ⎝ 396 ⎠ By inspection, the y-axis will be the critical buckling axis since it is the weak axis for both the steel shape and the concrete encasement. I s = I y = 57.7 in.4 from Manual Table 1-1

I sr = ΣAd 2 = 4 (1.0 )( 6.625 ) = 176 in.4 2

hb3 − I s − I sr 12 22.0(18.0)3 = − 57.7 − 176 = 10,500 in.4 12

Ic =

From Equation I2-4 Pno = Fy As + Fysr Asr + 0.85 f c′Ac

= 50 (15.6 ) + 60 ( 4.0 ) + 0.85 ( 5 )( 376 ) = 2620 kips and from Equation I2-6 EI eff = Es I s + Es I sr + C3 Ec I c

= 29,000 ( 57.7 ) + 29,000 (176 ) + 0.398 ( 3900 )(10,500 ) = 23.1 × 106 kip-in.2 and from Equation I2-5 π 2 EI eff π 2 (23.1 × 106 ) Pe = = = 7040 kips ( Lc ) 2 (15.0(12)) 2 Step 4:

Determine the controlling column strength equation. Pno 2620 = = 0.372 < 2.25 Pe 7040 Therefore, use Equation I2-2.

Step 5:

For LRFD

Determine the nominal compressive strength. Pn = Pno (0.658) Pno Pe = 2620(0.658) 0.372 = 2240 kips

The design compressive strength is

4116 Chapter 9

Composiite Construction

Step p 6:

φPn = 0.775 ( 2240 ) = 16680 kips

Forr ASD D Step p 6:

The allowable a com mpressing streength is Pn Ω = 22240 2.00 = 1 120 kips

Figure 9.25

Composite Column (Example 9.9)

The Commentary y indicates th hat the tablees found in A AISC Designn Guide 6, Looad and Resistance Factor Desig gn of W-Sha apes Encasedd in Concrete, give connservative ressults for umns by folllowing earlier specificatioons. The desiign guide also includes suuggested encased colu details for im mplementing encased com mposite colum mns in a project. Because ffilled HSS reepresents a limited sett of possible geometries, g sttrength tabless have been ddeveloped forr these shapess and are available on n the AISC weeb page at ww ww.aisc.org/m manualresourcces. The tablees are used inn exactly the same waay as the column tables for fo the unfilleed HSS colum mn previouslyy discussed. In cases where the filled HSS wou uld have stren ngth less thann the unfilled H HSS, the com mposite colum mn tables nd the design should revertt to that for ann unfilled HSS column. do not proviide values, an

EX XAMPLE 9.10 9 HS HSS Composiite Coolumn Stren ngth

Goa al:

Deterrmine the avaailable strenggth of a com mposite filled HSS columnn and comp pare it to the strength of an unfilled HSS S.

Giv ven:

The column c is co omposed of aan A500 Graade C HSS7× ×4×1/2 filledd with normaal weight con ncrete. Additioonal given innformation is aas follows: Lc = 15 ft Fy = 50 ksi fc′ = 5 ksi

SO OLUTION

Step p 1:

Deterrmine the walll slendernesss of the HSS ffor the most sslender wall. From Manu ual Table 1-11 1 and the limiit from Tablee I1.1a, h t = 12.1 < 2.26 E Fy = 2.26 29,000 50 = 54.4 so thee column is co ompact

Chapter 9 Step 2:

Composite Construction 417

Determine the needed properties from Manual Table 1-11. As = 8.81 in.2 , I y = 20.7 in.4 , ry = 1.53 in. and Asr = 0 since there is no reinforcing steel Ac = 6.0 ( 3.0 ) = 18 in.2 without considering the reduction in concrete area due to the rounded corners. Confirm that there is sufficient steel to be considered a composite column A 8.81 ρs = s = = 0.315 > 0.01 Ag 7.0(4.0)

Step 3:

Determine the equivalent yield strength of the column using Equations I29a and I2-9b. For the rectangular HSS, C2 = 0.85. E ⎞ ⎛ Pno = Pp = Fy As + C2 f c′ ⎜ Ac + Asr s ⎟ Ec ⎠ ⎝ = 50 ( 8.81) + 0.85 ( 5 )(18 + 0 ) = 517 kips

Step 4:

Determine the effective stiffness. First determine C3 from Equation I2-13. ⎛ A + Asr ⎞ C3 = 0.45 + 3 ⎜ s ⎟ ⎝ Ag ⎠ ⎛ 8.81 + 0 ⎞ = 0.45 + 3 ⎜⎜ ⎟⎟ ⎝ 7 ( 4) ⎠ = 1.39 > 0.9 Therefore C3 = 0.9 and determine Ec Ec = w1.5 f c′ = 1451.5 5 = 3900 ksi Thus, from Equation I2-12 EI eff

⎛ 6.0 ( 3.0 )3 ⎞ ⎟ = 29,000 ( 20.7 ) + 0.9 ( 3900 ) ⎜ ⎜ ⎟ 12 ⎝ ⎠ 2 = 648,000 kip-in.

Step 5:

Determine the elastic buckling strength from Equation I2-5 2 π 2 EI eff π ( 648,000 ) Pe = = = 197 kips 2 L2c (12 (15 ) )

Step 6:

Determine which compressive strength equation to use Pno 517 = = 2.62 > 2.25 Pe 197 Therefore use Equation I2-3

418 Chapter 9

Composite Construction Step 7:

Determine the nominal compressive strength Pn = 0.877 Pe = 0.877 (197 ) = 173 kips

For LRFD Step 8:

Determine the design strength of the composite column φPn = 0.75 (173) = 130 kips

For ASD Step 8:

Determine the allowable strength of the composite column Pn Ω = 173 2.00 = 86.5 kips

Step 9:

To determine the strength of the unfilled HSS column, first determine the elastic buckling stress from Equation E3-4 π2 ( 29,000 ) π2 E Fe = = = 20.7 ksi 2 2 ⎛ Lc ⎞ ⎛ 12 (15 ) ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ry ⎠ ⎝ 1.53 ⎠

Step 10:

Determine which critical stress equation to use Fy 50 = = 2.42 > 2.25 Fe 20.7 Therefore use Equation E3-3

Step 11:

Determine the critical stress Fcr = 0.877 Fe = 0.877 ( 20.7 ) = 18.2 ksi

Step 12:

Determine the nominal strength Pn = 18.2 ( 8.81) = 160 kips

For LRFD Step 13:

Determine the design strength φPn = 0.9 (160 ) = 144 kips

For ASD Step 13:

Determine the allowable strength Pn Ω = 160 1.67 = 95.8 kips

For LRFD Step 14: For ASD Step 14:

Compare the available strength of the composite column and the unfilled HSS. Since the unfilled HSS has a design strength of 144 kips which is greater than 130 kips for the filled HSS, Use the design strength of the unfilled HSS. Thus, φPn = 144 kips Since the unfilled HSS has an allowable strength of 95.8 kips which is greater than 86.5 kips for the filled HSS, Use the allowable strength of the unfilled HSS. Thus, Pn Ω = 95.8 kips

Chapter 9

Composite Construction 419

EXAMPLE 9.11 HSS Composite Column Strength

Goal:

Determine the available strength of a composite filled HSS column and compare it to the strength of an unfilled HSS.

Given:

The column is composed of an A500 Grade C HSS16×8×5/8 filled with normal weight concrete. Additional given information is as follows: Column effective length, Lc = 21 ft Fy = 50 ksi fc′ = 5 ksi

SOLUTION

Step 1:

Determine the wall slenderness of the HSS for the most slender wall. From Manual Table 1-11 and the limit in Table I1.1a, h t = 16.0 0.625 = 24.5 < 2.26 E Fy = 2.26 29,000 50 = 54.4 so the column is compact

Step 2:

Determine the needed properties from Manual Table 1-11. As = 25.7 in.2 , I y = 274 in.4 , ry = 3.27 in. and Asr = 0 since there is no reinforcing steel Ac = 14.75 ( 6.75 ) = 99.6 in.2 without considering the reduction in concrete area due to the rounded corners. Confirm that there is sufficient steel to be considered a composite column A 25.7 ρs = s = = 0.201 > 0.01 Ag 16.0(8.0)

Step 3:

Determine the equivalent yield strength of the column using Equations I29a and I2-9b. For the rectangular HSS, C2 = 0.85. E ⎞ ⎛ Pno = Pp = Fy As + C2 f c′ ⎜ Ac + Asr s ⎟ Ec ⎠ ⎝ = 50 ( 25.7 ) + 0.85 ( 5 )( 99.6 + 0 ) = 1710 kips

Step 4:

Determine the effective stiffness. First determine C3 using Equation I2-13. ⎛ A + Asr ⎞ C3 = 0.45 + 3 ⎜ s ⎟ ⎝ Ag ⎠ ⎛ 25.7 + 0 ⎞ = 0.45 + 3 ⎜⎜ ⎟⎟ ⎝ 16 ( 8 ) ⎠ = 1.05 > 0.9 Therefore C3 = 0.9 and determine Ec Ec = w1.5 f c′ = 1451.5 5 = 3900 ksi Thus, the effective stiffness from Equation I2-12 is

420 Chapter 9

Composite Construction EI eff

⎛ 14.75 ( 6.75 )3 ⎞ ⎟ = 29,000 ( 274 ) + 0.9 ( 3900 ) ⎜ ⎜ ⎟ 12 ⎝ ⎠ 6 2 = 9.27 × 10 kip-in.

Step 5:

Determine the elastic buckling strength from Equation I2-5 2 6 π 2 EI eff π ( 9.27 × 10 ) Pe = = = 1440 kips 2 L2c (12 ( 21) )

Step 6:

Determine which compressive strength equation to use Pno 1710 = = 1.19 < 2.25 Pe 1440 Therefore use Equation I2-2

Step 7:

Determine the nominal compressive strength Pn = 0.6581.19 (1710 ) = 1040 kips

For LRFD Step 8:

Determine the design strength of the composite column φPn = 0.75 (1040 ) = 780 kips

For ASD Step 8:

Determine the allowable strength of the composite column Pn Ω = 1040 2.00 = 520 kips

Step 9:

To determine the strength of the unfilled HSS column, first determine the elastic buckling stress using Equation E3-4 π2 ( 29,000 ) π2 E Fe = = = 48.2 ksi 2 2 ⎛ Lc ⎞ ⎛ 12 ( 21) ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ry ⎠ ⎝ 3.27 ⎠

Step 10:

Determine which critical stress equation to use Fy 50 = = 1.04 < 2.25 Fe 48.2 Therefore use Equation E3-3

Step 11:

Determine the critical stress Fcr = 0.6581.04 ( 50 ) = 32.4 ksi

Step 12:

Determine the nominal strength Pn = 32.4 ( 25.7 ) = 833 kips

For LRFD Step 13:

Determine the design strength φPn = 0.9 ( 833) = 750 kips

Chapter 9 For ASD Step 13:

Composite Construction 421

Determine the allowable strength Pn Ω = 833 1.67 = 499 kips Compare the available strength of the composite column and the unfilled HSS.

For LRFD Step 14:

For ASD Step 14:

Since the unfilled HSS has a design strength of 750 kips which is less than 780 kips for the filled HSS, Use the design strength of the filled HSS. Thus, φPn = 780 kips

Since the unfilled HSS has an allowable strength of 499 kips which is less than 520 kips for the filled HSS, Use the allowable strength of the filled HSS. Thus, Pn Ω = 520 kips

Examples 9.10 and 9.11 illustrate that filling an HSS is not a guarantee that the column strength will increase. Even in the case where the strength does increase, as in Example 9.11, the increase may not be significant enough warrant the extra cost of adding the concrete. Both of these situations result because of the increased variability of a composite column when compared to an unfilled column and the resulting reduction in resistance factor and increase in safety factor.

9.12 COMPOSITE BEAM-COLUMNS Composite beam-columns have the same potential to occur as bare steel beam-columns. Any application where bending moment and axial force are applied simultaneously needs to be addressed according to the provisions of Specification Section I5. For doubly symmetric composite beam-columns, the most common composite beamcolumns found in building construction, the interaction equations of Specification Chapter H can be used conservatively. For a more accurate approach to determining the available strength, the interaction surface can be developed based on plastic stress distributions and the length modification from Specification Section I2 made, as discussed in Section 9.11. Figure 9.26 provides several potentially useful interaction diagrams for a composite beam-column. Curve 1 is the interaction curve based on a strain compatibility approach similar to that used for developing similar diagrams for reinforced concrete columns, without consideration of length effects. Curve 2 represents a segmented straight-line approximation based on plastic stress distributions, again without incorporating any length effects. Curve 3 is a simplification of curve 2, incorporating length effects and using only one intermediate point between pure axial strength and pure bending strength. Curve 4 results from the application of resistance or safety factors to curve 3. Curve 5 is the result of applying the interaction equations of Chapter H. Only curves 4 and 5 in Figure 9.26 account for resistance or safety factors and the effects of length on beam-column strength. The conservatism of the Chapter H approach may not be that great compared to the curve 4 approach. Part 6 of the Manual includes tables for use in determining the needed points for curve 2 of the interaction diagram in Figure 9.26. The available equations address encased W-shapes for

4222 Chapter 9

Composiite Construction

Figure 9.226 Composite Column Interactionn Diagrams bending abo out the majorr and minor axes, compaact rectangulaar HSS bendding about eiither the major or miinor axis, and d compact rou und HSS. Figgure 9.27 is M Manual Tablee 6-3a for an encased W-shape beending aboutt the major axis, a and Figgure 9.28 is Manual Tabble 6-4 for ccompact rectangular HSS bending g about eitheer axis. If thee curve 4 innteraction of Figure 9.26 is to be uation identifi fied in Figure s 9.27 and 9.28 for pointss A, B and C need be constructed,, only the equ evaluated. Then T the lengtth effects must be applied to points A aand C, and thhe resistance oor safety factors appliied to all threee points. Thiis process ressults in the daata for curve 4 which cann then be used to checck the strength h of a compossite beam-collumn.

EX XAMPLE 9.12a 9 HS HSS Composiite Coolumn Stren ngth un nder Combin ned Looading by LR RFD

Goa al:

Check k the interacttion diagram of Chapter H for a givenn rectangularr HSS underr combined ax xial and flexuural loading.

Giv ven:

The column c is com mposed of ann HSS8×8×1/2 filled withh concrete. Itt must carry an axial load d of Pu = 2000 kips and a bbending mom ment of Mu = 84 ftkips. It has an effeective length of 15 ft for bboth axes. The HSS has Fy = 46 a fc′ = 5 ksi. Use tthe filled H HSS column tables founnd at ksi and www.aisc.org/man nualresources .

SO OLUTION

Step p 1:

Deterrmine the desiign axial strenngth. From m the on-line Table, T with K KLy =15 ft, φPn = 480 kip s

Step p 2:

Deterrmine the flex xural strength . ble, the availlable flexural strength caan be found aat the In thee on-line Tab bottom m of the tablee. Note that itt is not a funcction of unbraaced length. φM n = 140 ft-kiips

Step p 3:

Step p 4:

Deterrmine which interaction i eqquation to usee, H1-1a or H1-1b. Pu 200 = = 0.4177 > 0.2 φPn 480 uation H1-1a.. Thereefore, use Equ Check k the interactiion equation.

Chapter 9

Composite Construction 423

8 ⎛ 84.0 ⎞ 0.417 + ⎜ ⎟ = 0.95 < 1.0 9 ⎝ 140 ⎠ Since the interaction equation is less than 1.0, this column will carry the imposed load.

EXAMPLE 9.12b HSS Composite Column Strength under Combined Loading by ASD

Goal:

Check the interaction diagram of Chapter H for a given rectangular HSS under combined axial and flexural loading.

Given:

The column is composed of an HSS8×8×½ filled with concrete. It must carry an axial load of Pu = 133 kips and a bending moment of Mu = 56 ftkips. It has an effective length of 15 ft for both axes. The HSS has Fy = 46 ksi and fc′ = 5 ksi. Use the filled HSS column tables found at www.aisc.org/manualresources.

SOLUTION

Step 1:

Determine the design axial strength. From the on-line Table, with KLy =15 ft, Pn Ω = 320 kips

Step 2:

Determine the flexural strength. In the on-line Table, the available flexural strength can be found at the bottom of the table. Note that it is not a function of unbraced length. M n Ω = 93.0 ft-kips

Step 3:

Determine which interaction equation to use, H1-1a or H1-1b. Pa ( Pn Ω ) = 133 320 = 0.416 > 0.2 Therefore, use Equation H1-1a.

Step 4:

Check the interaction equation.

8 ⎛ 56.0 ⎞ 0.416 + ⎜ ⎟ = 0.95 < 1.0 9 ⎝ 93.0 ⎠ Since the interaction equation is less than 1.0, this column will carry the imposed load.

4224 Chapter 9

Composiite Construction

Figure 9.27 7

Composiite Column In nteraction Diaagram Points:: Encased W--Shape

Copyright © American Institute of Steel Construction. C R Reprinted with Permission. A All rights reservved.

Chaptter 9

Figure 9.28 9

Compposite Construuction 425

Comp posite Column n Interaction Diagram Points: Filled HS SS

Copyrightt © American Institute I of Steeel Constructioon. Reprinted w with Permissionn. All rights resserved.

426 Chapter 9 9.13

Composite Construction

PROBLEMS

1. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W16×26 spanning 24 ft and spaced 8 ft on center, supporting a 6 in. concrete slab. Use fc′ = 4 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 2. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W16×45 spanning 20 ft and spaced 8 ft on center, supporting a 5 in. concrete slab. Use fc′ = 5 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 3. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W18×65 spanning 20 ft and spaced 6 ft on center, supporting a 5 in. concrete slab. Use fc′ = 5 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 4. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W18×71 spanning 18 ft and spaced 5 ft on center, supporting a 4 in. concrete slab. Use fc′ = 4 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 5. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W14×48 spanning 20 ft and spaced 5 ft on center, supporting a 4 in. concrete slab. Use fc′ = 3 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 6. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W14×68 spanning 24 ft and spaced 6 ft on center, supporting a 4 in. concrete slab. Use fc′ = 3 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 7. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W36×330 spanning 50 ft and spaced 20 ft on center, supporting a 12 in. concrete slab. Use fc′ = 6 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 8. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W27×102 spanning 30 ft and spaced 14 ft on center, supporting a 6 in. concrete

slab. Use fc′ = 6 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 9. Determine the location of the plastic neutral axis and the available moment strength for a flat soffit, fully composite beam composed of a W10×12 spanning 12 ft and spaced 4 ft on center, supporting a 5 in. concrete slab. Use fc′ = 4 ksi and A992 steel. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 10. Repeat Problem 1 with the shear stud capacity limited to Vq′ = 250 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 11. Repeat Problem 2 with the shear stud capacity limited to Vq′ = 500 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 12. Repeat Problem 3 with the shear stud capacity limited to Vq′ = 500 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 13. Repeat Problem 4 with the shear stud capacity limited to Vq′ = 400 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 14. Repeat Problem 5 with the shear stud capacity limited to Vq′ = 300 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 15. Repeat Problem 6 with the shear stud capacity limited to Vq′ = 600 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 16. Repeat Problem 7 with the shear stud capacity limited to Vq′ = 2000 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 17. Repeat Problem 8 with the shear stud capacity limited to Vq′ = 900 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 18. Repeat Problem 9 with the shear stud capacity limited to Vq′ = 100 kips. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 19. Repeat Problems 5 and 14 with Vq′ = 450 kips and plot the results of all three problems as a function of Vq′. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 20. Repeat Problems 6 and 15 with Vq′ = 400 kips and plot the results of all three problems as a function of Vq′.

Chapter 9 Determine (a) design strength by LRFD and (b) allowable strength by ASD. 21. A W12 composite beam spaced every 8 ft is used to support a uniform dead load of 1.0 k/ft and live load of 0.9 k/ft on a 20 ft span. Using a 4 in. flat soffit slab with fc′ = 4 ksi, 3/4 in. shear studs, and A992 steel, determine the least-weight shape and the required number of shear connectors to support the load. Design by (a) LRFD and (b) ASD.

Composite Construction 427

28. Determine the available moment strength for a W18×46 A992 member used as a partially composite beam to support 3 in. of concrete on a 3 in. metal deck for a total slab thickness of 6 in. The metal deck is perpendicular to the beam. The beam spans 30 ft and is spaced 11 ft from adjacent beams. Shear stud strength is Vq′ = 400 kips, fc′ = 5 ksi. Determine (a) design strength by LRFD and (b) allowable strength by ASD.

22. A W14 composite beam is to support a uniform dead load of 1.0 k/ft and live load of 1.4 k/ft. The beam spans 24 ft and is spaced 8 ft from adjacent beams. Using a 5 in. flat soffit slab and 3/4 in. shear studs, determine the leastweight shape and the required number of shear connectors to support the load if fc′ = 4 ksi and A992 steel is used. Design by (a) LRFD and (b) ASD.

29. A composite beam is to span 20 ft and support a 4 in. slab including a 1-1/2 in. metal deck. The deck span is 10 ft. The beam must accommodate a uniformly distributed dead load of 75 psf including the slab weight and live load of 100 psf. The deck has 2 in. ribs spaced 6 in. on center, with deck ribs perpendicular to the beam. Determine the required A992 W-shape and number of 3/4 in. shear studs. Use fc′ = 3 ksi. Design by (a) LRFD and (b) ASD.

23. A W16 composite beam spaced every 12 ft is used to support a uniform dead load of 1.2 k/ft and live load of 1.9 k/ft on a 20 ft span. Using a 4 in. flat soffit slab with fc′ = 5 ksi, 3/4 in. shear studs, and A992 steel, determine the least-weight shape and the required number of shear connectors to support the load. Design by (a) LRFD and (b) ASD.

30. Determine the required W-shape and number of 3/4 in. shear studs for a composite girder that spans 30 ft and supports two concentrated dead loads of 12 kips and live loads of 20 kips at the third points. The 1-1/2 in. metal deck with 2 in. ribs spaced at 6 in. on center is parallel to the girder and supports a total slab of 5 in. Use fc′ = 4 ksi and A992 steel. Design by (a) LRFD and (b) ASD.

24. A W16 composite beam is to support a uniform dead load of 1.8 k/ft and live load of 2.2 k/ft. The beam spans 26 ft and is spaced 10 ft from adjacent beams. Using a 5 in. flat soffit slab and 3/4 in. shear studs, determine the least-weight shape and the required number of shear connectors to support the load if fc′ = 4 ksi and A992 steel is used. Design by (a) LRFD and (b) ASD.

31. Determine the live load deflection for a W24×76 A992 composite beam with a 6 in. total thickness slab on a 3 in. metal deck. The beam spans 28 ft and is spaced at 10 ft intervals. The beam is to carry a live load of 3.4 k/ft. Assume Y2 = 5.5 in. and ΣQn=394 kips.

25. Compare the least-weight A992 W16 and W14 members required to support a uniform dead load of 2.4 k/ft and live load of 3.2 k/ft. The beams span 18 ft and are spaced 12 ft on center. They support a 6 in. concrete slab with fc′ = 4 ksi. Design by (a) LRFD and (b) ASD. 26. W16×31 A992 composite beams are spaced at 10 ft intervals and span 24 ft. The beams support a 2-1/2 in. metal deck perpendicular to the beam with a slab whose total thickness is 5 in. Assuming fully composite action, determine the available moment strength and the number of 3/4 in. shear studs required. The deck has 6 in. wide ribs spaced at 12 in. Use fc′ = 4 ksi. Determine by (a) LRFD and (b) ASD. 27. Determine the available moment strength of a W18×35 A992 composite beam supporting a slab with a total thickness of 5 in. on a 3 in. metal deck perpendicular to the beam. The beam spans 28 ft and is spaced 12 ft from adjacent beams. Use fc′ = 5 ksi. Determine (a) design strength by LRFD and (b) allowable strength by ASD.

32. Determine the live load deflection for a W16×26 A992 composite beam supporting a 6 in. slab on a 2-1/2 in. metal deck. The beam spans 24 ft and is spaced at 8 ft on center. The live load is 2.1 k/ft. Assume Y2 = 5.5 in. and ΣQn= 384 kips. 33. Determine the live load deflection for the beam of Problem 21. 34. Determine the live load deflection for the beam of Problem 22. 35. Determine the available compressive strength of a 20 ft effective length 18×18 in. composite column encasing an A992 W10×68 and eight #8, Gr. 60 reinforcing bars, fc′ = 5 ksi. Each face has three bars with their centers located 2.5 in. from the face of the concrete. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 36. Determine the available compressive strength of a 22×22 in. composite column with an effective length of 16 ft. The concrete encases an A992 W12×120 and eight #9, Gr. 60 bars, fc′ = 5 ksi. Each face has three bars with their centers located 2.5 in. from the face of the concrete.

428 Chapter 9

Composite Construction

Determine (a) design strength by LRFD and (b) allowable strength by ASD. 37. Determine the available compressive strength of a 24×24 in. composite column with an effective length of 12 ft. The concrete encases an A992 W14×132 and eight #10, Gr. 60 bars, fc′ = 5 ksi. Each face has three bars with their centers located 2.5 in. from the face of the concrete. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 38. Determine the available compressive strength of an HSS10×5×3/8 composite column with an effective length of 12 ft. The concrete has fc′ = 5 ksi. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 39. Determine the available compressive strength of an HSS12×10×1/2 composite column with an effective length of 16 ft. about the strong axis and 8 ft for the weak axis. The concrete has fc′ = 5 ksi. Determine (a) design strength by LRFD and (b) allowable strength by ASD.

40. Using interaction equations H1-1a and H1-1b, determine if an HSS10×6×3/16 composite column with an effective length of 20 ft will carry an axial compressive live load of 25 kips and dead load of 18 kips, and a live load moment of 10 ft-kips and a dead load moment of 8 ft-kips, both bending the member about its strong axis. Use f’c = 5 ksi. Determine (a) design strength by LRFD and (b) allowable strength by ASD. 41. Integrated Design Project. The framing plans shown in Figures 1.20 and 1.21 were used to design beams and girders in Chapter 6. These same gravity-only beams and girders will now be designed as composite members. Manufacturers’ data on composite metal decks may be found on the Internet. For the designs carried out here, consider a 3 in. deck with a profile similar to that shown in Figure 9.16c. It will be used with 3 in. of concrete with f c′ = 4 ksi for a total slab thickness of 6.0 in.

Chapter

10

Conneection Elements

Marlins Parkk, Miami, FL Photo: © Chhristy Radecic (www.christyrradecic.com)

10.1

INT TRODUCTIION A steel building b structure is essenttially a collecction of indiviidual memberrs attached too each other to form a stable and serviceable wh hole, called thhe frame. Thee assumed beehavior of the connection between any two mem mbers determ mines how thee structure is analyzed to rresist gravityy and lateral loads. Th moments, shhears, and axxial loads forr which the his analysis, in turn, deteermines the m beams, columns, c and d other memb bers are desiggned. It is, ttherefore, esssential that thhe designer understan nd the basic behavior b of co onnections. Members M are attached to each other thhrough a variiety of conneecting elemennts, such as plates, an ngles, and oth her shapes, ussing mechaniccal fasteners or welds. Thee characteristtics of these connectin ng elements and fastenerss must be unnderstood to assess the reesponse of thhe complete connectio on. With each h connection n, the load traansfer mechannism must bee understood so that the applicablle limit states of the joint can c be evaluatted. Table T 10.1 lissts the section ns of the Speccification and parts of the M Manual discuussed in this chapter.

10.2

BAS SIC CONNE ECTIONS The wide variety of potential mem mber geomettries and arraangements avvailable for cconstruction makes a listing of the correspondin ng connectionns a very dem manding exerccise. Every joint between memberss must be analyzed and dessigned accordding to the unnique aspects oof that joint. Figure F 10.1 shows s severall examples o f tension connnections. Thhe connectionns shown in Figure 10 0.1a, b, and c illustrate waays that a tenssion member ccan be spliced. In each casse, the bolts are subjeected to a sh hear force. Th he butt jointt (Figure 10.11a) and the llap joint (Figgure 10.1b) provide a connection between two o members, w whereas the jooint shown inn Figure 10.1cc shows the connectio on of a singlee member to a pair of mem mbers. This tyype of joint can also be coonsidered as a portion n of the butt joint j shown in i Figure 10..1a. The jointt shown in Figure 10.1d rrepresents a hanger co onnected to th he lower flang ge of a beam;; in this case tthe connectioon is accomplished with

429

4330 Chapter 10 1

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Table 10.1 Sections off Specification n and Parts off Manual Covvered in This Chapter Specification S B3 Chapter D Chapter E G2 J1 J2 J3 J4

Design n Basis Design n of Memberss for Tension Design n of Memberss for Compresssion Memb bers with Unsttiffened or Sttiffened Webss Generaal Provisions Welds Bolts and a Threaded d Parts Affected Elements of o Members aand Connecting Elements Manual

Part 7 Part 8 Part 9

Design n Consideratio ons for Bolts Design n Consideratio ons for Weldss Design n of Connecting Elements

a WT-shapee, and the boltts are subjectted to a tensille load. The cconnection off a tension meember to a gusset platte is shown in n Figure 10.1ee. Here againn, the bolts aree subjected too a shear force. All of these examp ples illustratee bolted con nnections. Sim milar connecctions can bee accomplishhed with welds. The connections illustrated in Figure 10.2 aare bracket coonnections. T The connectionn shown in Figure 10 0.2a is a brack ket attached to o the flange oof a column. IIn this case, thhe bolts are suubjected to shear and d moment in the t plane of th he connectionn when loaded as shown. T The bracket sshown in Figure 10.2 2b, when loaaded as show wn, subjects the bolt group to shear in the planee of the connection and a a momen nt out of the pllane that resuults in a tensile force in thee top bolts.

100.3 BEAM M-TO-COLU UMN CONN NECTIONS S Design of beam--to-column connections c iss addressed iin Chapters 111 and 12. T They are discussed briefly b here since s each connection c iss composed of a combinnation of connnection elements. Th he connection n of a beam to t a column ccan also be acccomplished in a variety oof ways. Figure 10.3 3 illustrates several conn nections of W-shape beeams to W-sshape columnns. The classification n of these con nnections is a function of tthe forces being transferreed between thhe

Figure 10.1 Tension Connectio ons

Chappter 10

Figure 10 .2

Connection Elem ments 431

Bracket C Connections

memberss. The connecctions shown n in Figure 100.3a through d are usuallyy called simpple or shear connectio ons and are covered in Chapter 11, whereas thhose in Figurre 10.3e throough h are generally y referred to as a fixed or mo oment connecttions and covvered in Chapter 12. In I normal prractice beam m-to-column connections are classifieed as simplee or fixed, however; these connections actuallly exhibit a w wide range of behaviors, ass discussed inn Chapter 8. havior can be b described through a pplot of the m moment-rotatiion characterristics of a This beh particular connection n. Typical moment-rotaation relatioonships for three beam m-to-column connectio ons are preseented in Figurre 10.4. Whenn a connectioon is very stifff, it deformss very little, even wheen subjected to t large mom ments. This typpe of connectiion is represeented by curvee a in

Figure 10.3 Beam-to-C Column Connections

4332 Chapter 10 1

Connecttion Elementss

Figure 10.4 Beam-to-Column MomeentRotation Currves Figure 10.4.. At the otherr extreme, wh hen a connecttion is quite fflexible, it will rotate conssiderably but will not develop a sig gnificant mom ment, as show wn by curve c in Figure 100.4. Curve b inn Figure 10.4 is reprresentative off any connecction whose moment rotaation behavioor occurs som mewhere between currves a and c. c Despite som me appreciabble stiffness, these conneections still eexhibit a degree of flexibility; th hus, significant rotation will occur along with significant moment resistance. For the purposess of design, co onnections haave usually bbeen assumedd to behave acccording to the simpllified behavio ors representeed by the vert rtical axis of F Figure 10.4 aas a fixed connnection and the horrizontal axis of Figure 10.4 as a sim mple connectiion. Because connectionss do not actually beh have in this way, w those thaat follow curvves a and c aand exhibit a behavior closse to the idealized connection arre called fix xed connecttions and siimple conneections, respeectively. Specification n Section B3 3.4 divides connections c into two cattegories: sim mple connections and moment con nnections. Th he moment co onnection cat ategory includdes fully resttrained (FR) moment connections and partially y restrained (P PR) moment connections.. FR connectiions transfer moment with a negliigible rotation n between thee connected m members, as sshown in curvve a. PR connnections transfer mom ment between the membeers but the rrotation is noot negligible, as demonstrrated by curve b. It iss the designeer’s responsib bility to mattch connectioon behavior with the apppropriate analysis mo odel and to complete c the connection ddesign so thaat the actual connection bbehavior matches thatt used in the analysis. a Ofteen, this requirres experiencee and judgmeent; the state oof the art is such that it is usually not n possible to o accurately ppredict the M M-θ curve for aanything but tthe most basic of con nnections.

100.4

FULLY Y RESTRA AINED CON NNECTION NS The basic assumption fo or frames with h FR connecctions is that the beams annd columns m maintain their originaal geometric relationship over the enttire loading hhistory. This is normally called a rigid or fixeed connection n. The diagraams in Figuree 10.3e throuugh h show examples of bbeam-tocolumn connections that are usually treated t as FR R connectionss. Although tthey may shoow some relative rotation between members, theey have sufficcient stiffnesss to justify ignnoring this rootation. Figu ure 10.3e shows a connecttion with a weeb plate shopp-welded to thhe column flaange and field-bolted to the beam m web. The beeam flanges have been beveled in thee shop and are field-

Chapter 10

Connection Elements 433

welded to the column. Although the beam web is not continuously connected to the column, it has been repeatedly demonstrated that this connection can adequately transfer the full plastic moment of the beam to the column. The moment strength is derived mostly from the flange connections and is equal to the flange force times beam depth. The small amount of moment in the web connection and local strain hardening in the flanges add to the connection’s ability to reach the full plastic moment of the beam. This connection is generally known as the pre-Northridge connection because it was the de-facto standard connection for seismic applications prior to the 1994 Northridge, California, earthquake. Because its performance under the seismic load of that event was below expectations, it is no longer used in seismic resisting frames. However, it is still used to resist moments due to gravity and wind loads. Figure 10.3f is similar to Figure 10.3e except that the beam frames into the web of the column. To ensure that this connection has adequate ductility, it is important to extend the flange connecting plates beyond the column flange and to design these plates to be a little thicker than the beam flange. Extending the connecting plate reduces the possibility of a tri-axial stress condition near the column flange tips. Thickening the plate reduces the average tension stress in the plate. It also facilitates welding to the beam flange. The connection illustrated in Figure 10.3g is a flange-plate connection. As with the connection shown in Figure 10.3e, the web is connected to transfer the beam shear force only. The flange forces are first transferred to the top and bottom plates and then to the column flange. This is shown as a bolted connection, but it is also possible to fabricate this as a welded connection. For fully welded connections, special care should be taken to address support during erection prior to field welding; this usually means there will be some bolts used even in a fully welded connection. Figure 10.3h is an extended end-plate connection. For this connection, a plate is shopwelded to the end of the beam and then bolted to the column flange. Although this connection is very popular with some fabricators, others tend to avoid it. It must be fabricated with special care so that the end plates are parallel to each other. Also, it is not a very forgiving connection and it can make erection difficult and expensive. Fully restrained connections are covered in Chapter 12.

10.5 SIMPLE AND PARTIALLY RESTRAINED CONNECTIONS Analysis of a frame with PR connections must account for the actual moment-rotation characteristics of the connection. These connections are now referred to as partially restrained connections but have historically been called semi-rigid connections. It is typically not possible to determine whether a connection should be classified as PR just by looking at it. Several connections that appear to be simple actually have the potential to resist significant moment. In the simple connection case, the analysis assumes that the connections are pinned and free to rotate. The rotation capacity of the connection must be sufficient to accommodate the simple beam rotation of the beam to which it is connected. There are basically two ways in which this type of frame can be designed to resist lateral loads and to provide stability for gravity loads. In one case, a positive bracing system is provided, such as diagonal steel bracing or a shear wall. In the second case, lateral stability is provided by the limited restraint offered by the connections and members themselves. This type of connection is called a flexible moment connection. Flexible moment connections are designed for a limited amount of moment resistance accompanied by a significant amount of rotation. The connections are flexible enough to rotate under gravity loads so that no gravity moments are transferred to the columns. At the same time they are assumed to have sufficient strength and stiffness to resist the lateral loads and to provide frame stability. This approach to frame design is a simplification of convenience based upon experience, judgement and past usage. It was addressed in Section 8.11.

434 Chapter 10

Connection Elements

Design of these flexible moment connections will follow the same approach as that for other connections to be discussed later. The design of PR connections is more involved than the flexible moment connection approach because it requires that the frame be analyzed considering the true semi-rigid behavior of the connections. In this case, the actual M-θ curve of the connection must be known. The resulting analysis tends to be rather complex because of the nonlinear behavior of the connection. Although there are currently no commercially available computer programs for analysis of frames with PR connections, there are simplified approaches that will aid in the use of these connections. Figure 10.3 shows examples of simple and PR connections. As mentioned earlier, it is not normally possible to tell by visual inspection whether a connection should be treated as a PR connection. Figure 10.3a shows a double-angle connection, also referred to as a clip angle connection. This connection has been used extensively over the years. In fact, it is usually the standard to which other simple connections are compared. Even though it is readily accepted as a simple connection, it has been shown that under certain circumstances it can be relied upon to resist some moment from lateral load. Figure 10.3b shows a single-plate framing connection that is often referred to as a shear tab. Care must be taken when these connections are designed as simple connections to ensure that the elements have sufficient flexibility to accommodate the simple beam rotation. Figure 10.3c shows a seated connection and Figure 10.3d a stiffened seated connection. Either can be bolted or welded, and they are usually used to frame a beam into the web of a Wshaped column section. Although they may appear to be stiffer than the standard double-angle connection, they are designed to rotate sufficiently without transferring a moment to the column, so they can be treated as simple connections. Simple shear connections are treated in Chapter 11.

10.6

MECHANICAL FASTENERS The mechanical fasteners most commonly used today are bolts. The Specification provides for the use of common bolts and high-strength bolts. It also provides some direction for cases where bolts are to be used in conjunction with rivets in new work on historic structures. There are no provisions for rivets in new construction, however, because these connectors are no longer used in new construction of buildings.

10.6.1

Common Bolts Common bolts are manufactured according to the ASTM A307 specification, as discussed in Section 3.6.3. When used, they are usually found in simple connections for such elements as girts, purlins, light floor beams, bracing, and other applications where the loads are relatively small. Although permitted by the Specification, they are not recommended for normal steel-to-steel connections and should not be used where the loads are cyclic or vibratory, or where fatigue may be a factor. Common bolts are also called machine, unfinished, or rough bolts. They have square or hexagonal heads and nuts and are identified by the grade designation 307A or 307B on the heads. They are available in diameters from 1/4 in. to 4 in. These bolts are usually installed using a spud wrench. No specified pre-tension is required. Because no clamping force is assumed, it is necessary only to tighten the nut sufficiently to prevent it from backing off of the bolt. The design shear and tensile strength are given in Specification Section J3, Table J3.2.

Chappter 10 10.6.2

Connection Elem ments 435

Hig gh-Strength Bolts B High-streength bolts arre presented in Specificattion Section JJ3 in three grroups accordding to their material strength. Fou ur Group A high-strength h bolts are currrently permittted in steel sttructures as specified d by ASTM F3125 F Grades A325, A325M M and F18522 and ASTM A354, Gradee BC. Grade A325M is i the metric equivalent e off A325. Thesee bolts all havve the same m minimum tensiile strength, Fu = 120 0 ksi. Four Group G B boltts are permittted as speciffied by ASTM M F3125 Grades A490, A490M and a F2280 an nd ASTM A354, Grade BD D. Grade A4990M is the meetric equivaleent of A490. These bo olts also all have h the samee minimum teensile strength th, Fu = 150 kksi. Two Grooup C bolts, ASTM F3043 F and F3111 with min nimum tensilee strength of 200 ksi, are ppermitted. Usse of Group C bolts iss restricted by y Section J3.1. Details of tthe material aand other properties of theese bolts are described d in Section 3.6.3. 3 Bolt strrength is baseed on the tenssile strength oof the bolt maaterial. The nominal strength of Group G B boltts is 25 perceent greater thhan that of G Group A boltss. Bolts are generally y referred to by b their gradee designationn without repeeating the AS STM F3125 ddesignation. Thus, in this book thaat model will be followed.. All Group A and B boltss can be used for simple, FR, or PR P connection ns and for bo oth static andd dynamic looading. Bolts have alwayss been very popular for field installation. Th heir use in the shop haas increased considerablyy with the introducttion of autom mated equipmeent and tensioon control bollts (F1852 andd F2280). A325, A A325M M and F1852 bolts are avaailable in twoo types. Type 1, manufactuured from a medium--carbon steel,, is the most commonly uused. It is avvailable in sizzes ranging fr from 1/2 in. through 1-1/2 in. in diameter. Ty ype 3 is a weeathering steeel bolt with ccorrosion chaaracteristics similar to o those of AS STM A242, A588, A and A8 47 steels. Typ ype 3 bolts aree also availabble from 1/2 through 1-1/2 1 in. in diiameter. They y all have a m minimum tensiile strength Fu = 120 ksi. A490, A A490M M and F2280 0 bolts are allso available as Type 1 aand Type 3 aand in sizes ranging from f 1/2 in. to o 1-1/2 in. in diameter. Alll have a minim mum tensile sstrength Fu = 150 ksi. A325 A Type 1 bolts are iden ntified by thee mark “A3255” on the bolt head. Type 3 bolts have the desig gnation “A325” underlined d. A490 boltss are similar,, carry the m marking “A490” with the identificaation underlin ned for Type 3. Example bbolt markings are shown iin Figure 10.5. All bolts should allso be marked d with a symb bol designatinng the manufaacturer as shoown in this caase by SL in the figuree. A354 A bolts arre manufacturred as quenchhed and tempeered alloy steeel in two graddes, BC and BD. Any y steel that meets m the mecchanical and cchemical prooperties in thee standard maay be used. Grade BC C bolts have a tensile stren ngth of 125 orr 115 ksi, bassed on diametter, but are inccluded with Group A bolts where a tensile stren ngth of 120 kksi is used. G Grade BD boltts have a tenssile strength of 150 orr 140 ksi, based on diameteer, but are inccluded with G Group B boltss where a tenssile strength of 150 kssi is used. Th hese bolts are available in sizes rangingg from 1/4 in. to 4.0 in. diaameter with diameters over 2-1/2 in. having th he reduced strrength. A354 bolts are maarked with thee grade and may be marked m with 6 radial lines, 60° apart forr grade BD.

Figure 10.5 1 Examplle Bolt Identiification Markkings from Stt. Louis Screw w and Bolt Coompany

4336 Chapter 10 1

Connecttion Elementss

Figure 10.6 Bolt Definitions D Group C bolts aree currently on nly available as a proprietaary product annd will not bee used in this book altthough the baasic concepts of design disccussed will eqqually apply tto them. Figu ure 10.6a sho ows the principal parts aand dimensionns of a highh-strength bollt: head, shank, bolt length, l and th hread length. Figure 10.6bb shows the pprincipal partss of a tensionn control bolt. Both h A325 and A490 A bolts caan be installeed with a spud wrench, shhown in Figurre 10.7a, or in cases where w a clamp ping force is necessary, ann impact wrennch shown inn Figure 10.7bb. F1852 and F2280 bolts b are instaalled with a mechanical m ddevice that sim multaneously holds the boolt shank and nut and rotates them relative to eaach other. Thee end of the bbolt twists offf when the prescribed tensile forcee is reached,, ensuring the required p re-tension. F Figure 10.7c shows a wreench for installation of these ten nsion control bolts. It shhould be noteed that placeement of boolts in a connection must m provide sufficient cleearance for wrrench access. 100.6.3

Bolt Holes H w bolts aree inserted willl impact the strength of thhe bolts in plaace, it is Because thee holes into which important to o address the hole requirem ments at this point. The SSpecification defines four types of bolt holes th hat are permiitted in steel construction:: standard, ovversized, shorrt slot, and loong slot. Table 10.2 shows s the nom minal hole diimensions forr each of thesse types and ffor bolts from m 1/2 in. diameter up. Figure 10.8 shows the fou ur hole sizes for a 3/4 in. bbolt. Stan ndard holes orr short-slotted d holes transvverse to the diirection of loaad are the staandard to be used unlless one of th he other typees is permitteed by the deesigner. This is because thhe other arrangementts can reduce the final bolt strength andd allow largerr deformationns from moveement of bolts in holees under loadiing. A standarrd hole has a maximum diameter that iss 1/16 in. greaater than

(a) Sp pud Wrenchess

(b) Impact Wrrench

Figure 10.7 7 Wrenches used u to installl Bolts Photos courteesy Tone Co., Ltd. L

(c) Tension C Control Wrennch

Chappter 10

Connection Elem ments 437

Figure 10.8 1 Hole Siizes for a 3/4 in. Diameter Bolt the bolt diameter up to 7/8 in. bo olts and 1/8 in. for 1.0 iin. bolts and larger to accommodate placemen nt of the bollt. Short-slottted holes havve this same dimension iin one directiion but are elongated d in the otherr direction to assist in fit-uup of the connnection partss. Any slot loonger than a short slott should be cllassified as a long slot, eveen if it is not the full lengtth of a long slot as given in Table 10.2. Oversized O holles and long--slotted holes are specifiedd when increaased tolerancce is needed to accom mplish the actual a connecction. If a ddesign includdes other thhan standard holes, the requirem ments of Speciification Secttion J3.2 for w washers com me into play. F For the exam mples in this book, on nly standard holes will be used. u In I addition to o prescribing the sizes of bolt holes, thhe Specificattion gives minnimum and maximum m hole spacing and edg ge distances. Figure 10.99 shows a pplate with hoole spacing dimensio oned with thee standard vaariable namees used in thhe Specificatiion. The minnimum hole spacing, s, for standaard, oversized d, or slotted holes must nnot be less thhan 2-2/3 tim mes the bolt diameter. A spacing of o 3 diameterss, 3d, is preferrred. It will be shown laterr that even at a minimum spacing of o 3d, bolt strrength may be b less than w what it could bbe if the spaccing were jusst a little bit greater. The T maximum m spacing of bolts b in a connnection is 244 times the thiickness of thee connected part or 12 in. This maaximum is no ot a strength rrequirement bbut rather onee that is intendded to keep the conneection plies in n contact and prevent any ppotential moiisture build-up up between thee elements. The T minimum m edge distan nces, le, speciffied are intennded to facilittate constructtion and are not stren ngth related. Table 10.3 shows s the miinimums from m Specificatiion Table J3..4. Because these dim mensions willl be shown to t directly im mpact bolt str trength, it is critical to prrovide edge distancess that are com mpatible with h the requiredd strength of the connection. The maxximum edge distance is 12 times the thickness of the connnected part oor 6 in. for elements nott subject to corrosion n, for the sam me reasons as for f the limits on bolt spaciing. Tab ble 10.2 Nom minal Hole Dimensions, D in n. Hoole dimensionns Bolt diameter 1/2 5/8 3/4 7/8 1 ≥1-1/8

Standaard (dia.)) 9/16 11/16 6 13/16 6 15/16 6 1-1/8 8 d + 1/8

Oversize O (dia.) 5/8 13/16 15/16 1-1/16 1-1/4 d + 5/16

Shortt slot (width × length) 9/16 × 11/16 11/16 × 7/8 13/166 × 1 15/16 × 1-1/8 1-1/8 × 1-5/16 (d + 1/8) × (d + 3/8)

Long slot (width × llength) 9/16 × 1-1/4 11/16 × 1-9/16 13/16 × 1-7/8 15/16 × 22-3/16 1-1/8 × 2-1/2 (d + 1/8) × (2.5 × d)

4338 Chapter 10 1

Connecttion Elementss

Figgure 10.9 H Hole Spacing, Gage, and Edgge Distances

100.7

BOLT LIMIT STA ATES Three basic limit states govern the response r of bbolts in bolteed connectionns: shear throough the shank or thrreads of the bolt, b bearing or tearout onn the elementts being connnected, and teension in the bolt. Cases where loaad reversals are a expected, or fatigue iss a factor, havve an additionnal limit state to preevent slip in the connectiion. This lim mit state appllies only to connections that are classified ass slip-criticall connectionss. For bolts iin shear connnections, the nominal connnection strength is th he sum of thee lowest stren ngth for the lim mit states of bbolt shear, beearing, and teaarout for each bolt in the connectio on. Boltts may be insstalled to a sn nug-tight conndition or preetensioned. S Snug-tight insstallation has no requ uired specific level of tenssion. This is commonly aattained after a few impaccts of an impact wren nch or the fulll effort of an ironworker w with an ordinaary spud wrennch. The plies should be in firm contact, a con ndition that means m the pliees are solidly seated againsst each other,, but not necessarily in i continuouss contact. Snu ug-tight bolts are permittedd in connectioons that are classified as bearing-ty ype connectio ons. Prettensioned bollts are installeed to a tensioon specified iin Table J3.1 in the Specif ification. They must be b used in con nnections sub bjected to vibbratory loads where bolt looosening is a concern and in end connections of built-up members. m Preetensioned bolts are requuired for connections designed as slip-critical connections. c Table 10.3 Minimum Edge E Distanceea from Centter of Standarrd b Hole to Edg ge of Connectted Part Bolt diameteer (in.)

Minimum eddge distance (in.)

1/2 3/4 5/8 7/8 3/4 1 7/8 1-1/8 1 1-1/4 1-1/8 1-1/2 11/4 1-5/8 >1-1/4 1--1/4 × d a If necessary, lesser edge e distancces are perm mitted providded the appropriate provisions from fr Sectionss J3.10 and JJ4 are satisfiied, but edge distancces less than one bolt diaameter are noot permitted w without approval fro om the engineeer of record. b For oversizzed or slotted holes, see Speecification Taable J3.5.

Chappter 10

Connection Elem ments 439

Figure 10.10 1 Bolt Failure F Modess 10.7.1

Bollt Shear pplication of bolts b in connnections is to resist shear. Shear throughh the shank The most common ap of the bo olt is the mean ns whereby th he load, P, inn Figure 10.100a is transferrred from one plate to the other. In this case, thee bolt is shearred along onee plane. Thus,, it is said to bbe a bolt in siingle shear. The arran ngement in Figure 10.10b shows two siide plates connnected to a ccentral plate. IIn this case, the load, P, is transferrred from the center plate tto the side plaates and the bbolt is thereforre loaded in double sh hear. A bolt in n double sheaar has twice thhe shear strenngth as a bolt in single sheaar. For F the limit state of bolt shear, the noominal strenggth is based oon the tensile strength of the bolt and a the locatiion of the sheear plane withh respect to thhe bolt threadds. Section J33.6 provides that Rn = Fn Ab ((AISC J3-1) and Ω = 2.000 (ASD) φ = 0.75 (LRFD) where Fn = shear streess, Fnv, from m Specificationn Table J3.2 Ab = area of th he bolt shank Tab ble 10.4 Nom minal Stress of o Fasteners and a Threadedd Parts, ksi

Description of faasteners A30 07 bolts Group A (e.g., A325) bolts, when w threads are a not excluded from m shear planees (N) Group A (e.g., A325) bolts, when w threads are a excluded from shear pllanes (X) Group B (e.g., A4 490) bolts, when threads are a not excluded from m shear planees (N) Group B (e.g., A4 490) bolts, when threads are a excluded from shear pllanes (X) Threeaded parts meeting m the requirements of Section A3..4, when threadss are not exclu uded from sheear planes (N N) Threeaded parts meeting m the requirements of Section A3..4, when threadss are excluded d from shear planes p (X)

Nominnal tensile stresss, Fnt, ksi

Nominal sshear stress in beariing-type connectionns, Fnv, ksi

45 90

227 554

90

668

113

668

113

884

0.75Fu

0.4550Fu

0.75Fu

0.5663Fu

440 Chapter 10

Connection Elements

The information in Table 10.4 is taken from Specification Table J3.2. Each high-strength fastener has two descriptions. The first is for cases where the threads are not excluded from the shear plane, and the second is for when the threads are excluded from the shear plane. Because, in every case, the area of the bolt shank is used to determine the nominal strength, the reduction in area when the shear plane passes through the threads is accounted for by reducing the nominal shear stress. This is done as a convenience in design so the area can be calculated using the known basic bolt diameter without having to calculate the reduction for threading. When threads are excluded from the shear plane, the bolts are called either A325-X or A490-X bolts. In these cases, Fnv = 0.563Fu. When threads are not excluded from the shear plane, the bolts are referred to as either A325-N or A490-N bolts. In these cases, Fnv = 0.45Fu. Only one value is provided for A307 bolts, and that value is based on the assumption that the threads are included in the shear plane. Unless the designer can be sure that the final connection will result in the bolt threads being excluded from the shear plane, it is usually best to design the connection for the worst case of threads included in the shear plane. 10.7.2

Bolt Bearing and Tearout The available strength for the limit state of bearing and tearout at bolt holes is specified in Section J3.10. Because the material strength of a bolt is greater than that of the material it is bearing on, the only bearing check is for bearing on the material of the connected parts. The Specification provision considers two limit states for bearing strength at bolt holes: the limit state based on shear in the material being connected, as shown in Figure 10.10c, and the limit state of material crushing, as shown in Figure 10.10d. When the clear distance from the edge of the hole to the edge of the part or next hole is less than twice the bolt diameter, the limit state of shear in the plate material, also referred to as tearout, will control. In this case, failure occurs by a piece of material tearing out of the end of the connection as shown in Figure 10.10c or by tearing between holes in the direction of force. The nominal strength for this failure mode, Rn, is provided by shear along the two planes. From statics, Rn = (shear strength)(2 planes)(clear distance)(material thickness) Rn = 0.6 Fu ( 2lc ) t = 1.2lc tFu (AISC J3-6c) where 0.6Fu = ultimate shear strength of the connected material, ksi t = thickness of the material, in. lc = clear edge distance, measured from the edge of the hole to the edge of the material or the next hole; unlike net shear and net tension area calculations, no deduction is made at the hole edge for damage in bearing calculations If the clear distance exceeds 2d, bearing on the connected material will be the controlling limit state, as shown in Figure 10.10d. In this case, the limit state is that of hole distortion and the calculated bolt strength will be Rn = 2.4dtFu (AISC J3-6a) where = bolt diameter d = connected part thickness t Fu = tensile strength of the connected part If deformation at the bolt hole is not a design consideration at service loads, both of these limit states may be increased by one-quarter, so that for tearout

Chappter 10

Connection Elem ments 441

Rn = 1.5lc tFu

(A AISC J3-6d)

Rn = 3.0dtFu

(A AISC J3-6b)

and for bearing b

olts are used in a connectio on with long sslots and the fforce is perpeendicular to thhe slot, bolt When bo strength is reduced such that for teaarout Rn = 1.0lc tFu (A AISC J3-6f) b and for bearing Rn = 2.0dtFu (A AISC J3-6e) As A was the caase for bolt sh hear, the resisstance and saafety factors ffor the limit sstate of bolt bearing are a φ = 0.75 0 (LRFD) Ω = 2.000 (ASD) 10.7.3

Strength at Boltt Holes Normal use u of bolts iss to hold two or more piecces of steel toogether. This of course reqquires holes in each piece p to be co onnected. Wh hen bolts are pplaced througgh holes and a shear forcee is applied, their avaailable strength will be th he lowest vallue obtained according too the limit staates of bolt shear, beearing on each h connected element e and teearout at the bolt hole on each connectted element. The nom minal strength of a connectiion in shear iss the sum of tthe nominal sstrength of thee individual bolts. For F a bolt in standard, oveersized, or loong slotted hooles perpendiicular to the ddirection of loading, nominal stren ngth will be the t lowest vallue accordingg to Equationns J3-1, J3-6a,, and J3-6c. It has beeen shown thaat for bolts insstalled in holees with a cleaar distance of 2d or more, bbearing will control over o tearout. It I is also possiible to determ mine a minimuum plate thicckness to insuure that bolt shear willl control over both bearing g and tearout..

Figu ure 10.11a Minimum M Plaate Thickness so Bolt Sheaar Controls forr A325-N bollts and Fy = 336 ksi

4442 Chapter 10 1

Connecttion Elementss

Figure 10.11b Min nimum Plate Thickness so Bolt Shear C Controls for A A325-N bolts and Fy = 50 kksi Figu ure 10.11 sho ows the minim mum plate thi ckness for A3325-N bolts iin plates withh Fy = 36 ksi and Fy = 50 ksi as a function of clear c end disttance dividedd by bolt diam meter, lc d ffor bolts with diametters from 5/8 in. to 1.5 in. diameter. T The curves staart at the minnimum clear distance given in Tab ble J3.4 divided by bolt diaameter and ennd at a clear ddistance of 2 bbolt diameterrs, above which no inccrease in stren ngth can be achieved. a 100.7.4

Bolt Tension T olt tension, sttrength is direectly based oon the tensilee strength of the bolt For the limiit state of bo material. Section J3.6 pro ovides that Rn = Fn Ab (AIS SC J3-1) with Ω = 2.00 ((ASD) φ = 0.75 (L LRFD) and

Fn Ab

= tensile t stress, Fnt, from Speecification Taable J3.2 = area a of the bo olt shank

Table 10.4 shows the no ominal tensilee stress, Fnt, for bolts takeen from Speccification Tabble J3.2. Note that th here is no disstinction for the t location oof the shear plane, becauuse the bolt iss loaded axially and the t limiting sttress occurs over o the net teensile area. To ease calculaations, the areea of the bolt shank iss again used and a the nomin nal tensile streess appropriaately reduced and given as 0.75Fu.

EX XAMPLE 10.1 1 Boolt Shear Str trength

Goa al:

Deterrmine the available bolt sheear strength.

Giv ven:

(a) A single 3/4 in.. A325-N bollt. (b) A single 7/8 in. A490-X bollt.

SO OLUTION

Parrt (a) Step p 1:

For a single 3/4 in.. A325-N bollt, determine tthe bolt shankk area.

Chapter 10

Ab = Step 2:

Connection Elements 443

πd 2 π(0.75)2 = = 0.442 in.2 4 4

Determine the nominal shear stress. For an A325 bolt Fu = 120 ksi and for the threads included (N) Fnv = 0.45Fu = 0.45(120) = 54 ksi

For LRFD Step 3:

For shear, φ = 0.75 and the design strength is φrn = 0.75(54)(0.442) = 17.9 kips

For ASD Step 3:

For shear, Ω = 2.00 and the allowable strength is rn Ω = 54 ( 0.442 ) 2.00 = 11.9 kips

Part (b) Step 4:

Step 5:

For a single 7/8 in. A490-X bolt, determine the bolt shank area. πd 2 π(0.875)2 Ab = = = 0.601 in.2 4 4 Determine the nominal shear stress. For an A490 bolt Fu = 150 ksi and for the threads excluded (X) Fnv = 0.563Fu = 0.563(150) = 84 ksi

For LRFD Step 6: For ASD Step 6:

For shear, φ = 0.75 and the design strength is φrn = 0.75(84)(0.601) = 37.9 kips For shear, Ω = 2.00 and the allowable strength is rn (84)(0.601) = = 25.2 kips 2.00 Ω Manual Table 7-1 provides single-bolt available shear strength values for a wide range of bolt sizes and strengths.

4444 Chapter 10 1

Connecttion Elementss

Figure 10.12 Lap Jointt for Examplee 10.2

EX XAMPLE 10.2 1 Laap Splice Coonnection Str trength

Goa al:

Deterrmine the available strength th for a four-bbolt connectioon.

Giv ven:

A lap p joint using 1/2 1 in. A36 pllates is given in Figure 10.12. Use (a) 77/8 in. A325 5-X bolts and (b) 7/8 in. A A325-N bolts. The plates are labeled pplate 1 and plate p 2 and thee bolt pairs arre labeled bollts A and boltts B. The layoout of the co onnection is such s that eacch bolt will hhave the samee limiting streength. Thus,, the strength of the joint iss 4 times the sstrength of a single bolt.

SO OLUTION

Parrt a Step p 1:

Deterrmine the nom minal shear strrength for 7/88 in. A325-X bolts. Fnv = 0.563Fu = 0.563 (1220 ) = 68 ksi

Ab = 0.601 in..2 rn = Fnv Ab = 68 ( 0.601) = 40.9 kips Step p 2:

Deterrmine the nom minal bearing strength. Unlesss noted otheerwise assum me deformatioon at servicee load is a ddesign consid deration, thuss use Equationn J3-6a rn = 2.4dtFu = 2.44 ( 7 8)( 0.50 ))( 58) = 60.9 kips k

Step p 3:

Deterrmine the tearrout strength tto the plate eddge using lc = lcA1 = lcB2. Find the t clear distaance from thee bolt hole to the end of thee member, ⎛1⎞ lc = 1.5 − ⎜ ⎟ ( 7/8 + 1//16 ) = 1.03 < 2d = 2 ( 7/8 ) = 1.75 in. ⎝2⎠ Sincee the clear disstance is less than 2 bolt diameters, tearout strengthh will be lesss than bearin ng strength. U Using Equationn J3-6c

Chapter 10

Connection Elements 445

rn = 1.2lc tFu = 1.2 (1.03)( 0.50 )( 58) = 35.8 kips Step 4:

Determine the tearout strength between bolt holes using lc = lcA2 = lcB1. Find the clear distance from one bolt hole to the other, lc = 2.5 − ( 7/8 + 1/16 ) = 1.56 < 2d = 2 ( 7/8) = 1.75 in. Since the clear distance is less than 2 bolt diameters, tearout strength will be less than bearing strength. Using Equation J3-6c rn = 1.2lc tFu = 1.2 (1.56 )( 0.50 )( 58) = 54.3 kips

Step 5:

Determine the nominal strength of bolts A Bolt shear 40.9 kips Bearing 60.9 kips Tearout on plate 1 35.8 kips Tearout on plate 2 54.3 kips Thus, these bolts have a nominal strength per bolt rn = 35.8 kips Determine the nominal strength of bolts B Bolt shear 40.9 kips Bearing 60.9 kips Tearout on plate 1 54.3 kips Tearout on plate 2 35.8 kips Thus, these bolts have a nominal strength per bolt rn = 35.8 kips Therefore, all four bolts have the same strength.

Step 6:

Determine the nominal strength of the 4 bolt connection. Since all 4 bolts have the same strength, the controlling limit state is tearout to the edge of the plate, thus Rn = 4rn = 4 ( 35.8) = 143 kips

For LRFD Step 7: For ASD Step 7:

Part b Step 8:

The design strength of the 4 bolt connection is φRn = 0.75 (143) = 107 kips

The allowable strength for the 4 bolt connection is Rn Ω = 143 2.00 = 71.5 kips

Determine the nominal shear strength for 7/8 in. A325-N bolts.

4446 Chapter 10 1

Connecttion Elementss

Fnv = 0.45Fu = 0.45 (1200 ) = 54 ksi Ab = 0.601 in..2 rn = Fnv Ab = 54 ( 0.601) = 32.5 ksi Step p 9:

Deterrmine the nom minal strengthh of the 4 boltt connection. Sincee bearing and tearout are noot influenced by the locatioon of the sheaar plane through the threads, t the nnominal strenggths determinned in Steps 22, 3, and 4 are unchangeed. Thus, the controlling liimit state for each bolt in tthis conneection is bolt shear. Thus, Rn = 4rn = 4 ( 32.5) = 1130 kips

Forr LRF FD Step p 10: Forr ASD D Step p 10:

The design d strengtth of the 4 bollt connection is φRn = 00.75 (130 ) = 977.5 kips

The allowable a strength for the 4 bolt connecttion is Rn Ω = 130 2.00 = 6 5.0 kips

Figure 10.13 Butt Jointt for Examplee 10.3

EX XAMPLE 10.3 1 Bu utt Splice Coonnection Str trength SO OLUTION

Goa al:

Deterrmine the available strength th of the boltss in the butt spplice connectiion shown n in Figure 10 0.13.

Giv ven:

Use 3/4 3 in. A490-N N bolts and A A36 plates.

Step p 1:

Deterrmine bolt nom minal shear sttrength. Fnv = 0.563Fu = 0.536 (1550 ) = 68 ksi

Ab = 0.442 in..2 rn = Fnv Ab = 68 ( 0.442 ) = 30.1 kips The bolts b are in do ouble shear, thhus rn = 2 ( 30.1) = 60.22 kips

Chapter 10

Connection Elements 447

Step 2:

Determine the nominal strength for bearing on the 1/2 in. middle plate using Equation J3-6a, rn = 2.4dtFu = 2.4 ( 3 4 )(1 2 )( 58) = 52.2 kips

Step 3:

Determine the nominal strength for bearing on the 3/8 in. outer plate using Equation J3-6a, rn = 2.4dtFu = 2.4 ( 3 4 )( 3 8)( 58) = 39.2 kips Since there are 2 outer plates, the nominal bearing strength is rn = 2 ( 39.2 ) = 78.4 kips

Step 4:

Determine the tearout strength in the 1/2 in. middle plate For the bolts closest to the end of the plate ⎛1⎞ lc = 1.25 − ⎜ ⎟ ( 3/4 + 1/16 ) = 0.844 2d = 2 ( 3 4 ) = 1.5 in. Since the clear distance is greater than 2 bolt diameters, tearout strength will be greater than bearing strength. To confirm this, using Equation J3-6c rn = 1.2lc tFu = 1.2 ( 2.19 )( 0.50 )( 58) = 76.2 kips

Step 5:

Determine the tearout strength in the 3/8 in outer plate For the bolts closest to the end of the plate ⎛1⎞ lc = 1.25 − ⎜ ⎟ ( 3/4 + 1/16 ) = 0.844 < 2d = 2 ( 3/4 ) = 1.50 in. ⎝2⎠ Since the clear distance is less than 2 bolt diameters, tearout strength will be less than bearing strength. Using Equation J3-6c rn = 1.2lc tFu = 1.2 ( 0.844 )( 3 8)( 58) = 22.0 kips For the two plates,

rn = 2 ( 22.0 ) = 44.0 kips

For the interior bolts, the clear distance between bolts is lc = 3.0 − ( 3 4 + 1 16 ) = 2.19 > 2d = 2 ( 3 4 ) = 1.5 in. Since the clear distance is greater than 2 bolt diameters, tearout strength

448 Chapter 10

Connection Elements will be greater than bearing strength. To confirm this, using Equation J3-6c rn = 1.2lc tFu = 1.2 ( 2.19 )( 3 8)( 58) = 57.2 kips For the two plates,

Step 6:

rn = 2 ( 57.2 ) = 114 kips

Determine the controlling limit state and the nominal strength of each bolt. For the bolts closest to the end of the outer plate, the limit states and strengths are Bolt shear 60.2 kips Bearing on 1/2 in. plate 52.2 kips Bearing on 2 - 3/8 in. plates 78.4 kips Tearout on 1/2 in. plate 76.2 kips Tearout on 2 - 3/8 in. plate 44.0 kips Thus, these bolts have a nominal strength per bolt rn = 44.0 kips For the bolts closest to the end of the inner plate, the limit states and strengths are Bolt shear 60.2 kips Bearing on 1/2 in. plate 52.2 kips Bearing on 2 - 3/8 in. plates 78.4 kips Tearout on 1/2 in. plate 29.4 kips Tearout on 2 - 3/8 in. plate 114.0 kips Thus, these bolts have a nominal strength per bolt rn = 29.4 kips

Step 7: For LRFD Step 8: For ASD Step 8:

10.7.5

Determine the final connection nominal strength Rn = 2 ( 44.0 ) + 2 ( 29.4 ) = 147 kips

The design strength for the connection is φRn = 0.75 (147 ) = 110 kips

The allowable strength for the connection is Rn Ω = 147 2.00 = 73.5 kips

Slip The limit state of slip is associated with connections that are referred to as slip-critical. Slipcritical connections are designed to prevent slip at the required strength. They should be used only when required, such as when the connection is subjected to fatigue or the connection has oversized holes or slots parallel to the direction of load. Often, connections are specified to be designed as slip-critical when these conditions do not apply. Significant connection economies

Chapter 10

Connection Elements 449

are lost, for no gain, when this is the case. Any slip-critical connection design must also include a check for strength as a bearing-type connection by the methods discussed in the previous sections. Bolts in slip-critical connections would be designated as A325-SC or A490-SC. The nominal strength of a single bolt in a slip-critical connection is given in Specification Section J3.8 as (AISC J3-4) Rn = μDu h f Tb ns where μ Du hf

ns Tb

= mean slip coefficient = 0.30 for Class A surfaces = 0.50 for Class B surfaces = 1.13, a multiplier that reflects the ratio of mean installed bolt pretension to specified bolt pretension. = factor for fillers = 1.0 for one filler between connected parts = 0.85 for more than one filler. If multiple fillers are directly connected independently to transfer load, hf also may be taken as 1.0 = number of slip planes = minimum bolt pretension specified in Table J3.1

For connections with standard size and short-slotted holes perpendicular to the direction of load, φ = 1.00 (LRFD) Ω = 1.50 (ASD) For oversized and short-slotted holes parallel to the direction of the load, φ = 0.85 (LRFD) Ω = 1.76 (ASD) For long-slotted holes φ = 0.70 (LRFD)

Ω = 2.14 (ASD)

EXAMPLE 10.4 Slip-Critical Connection Strength

Goal:

Determine the available slip-critical connection strength for the connection shown in Figure 10.13. This connection was treated as a bearing-type connection in Example 10.3.

Given:

Use 3/4 in. A490-SC bolts in standard holes and A36 plates with a Class A surface. There are no fillers.

SOLUTION

Step 1;

Determine bolt nominal slip strength. From Specification Table J 3.1, the minimum bolt tension, Tb = 35 kips. From the given information, μ = 0.30, Du = 1.13, hf = 1.0, ns = 2. Therefore, Rn = μDh h f Tb ns Rn = 0.30 (1.13 )(1.0 )( 35 )( 2 ) = 23.7 kips

For LRFD Step 2:

Determine the design strength. From Example 10.3, the design bearing strength of the connection was found to be φRn = 110 kips For the slip-critical connection,

450 Chapter 10

Connection Elements

φRn = 1.00 ( 4 )( 23.7 ) = 94.8 kips Thus, the design strength of this slip-critical connection is φRn = 94.8 kips For ASD Step 2:

Determine the allowable strength. From Example 10.3, the allowable bearing strength of the connection was found to be Rn Ω = 73.5 kips For the slip-critical connection, Rn Ω = 4 ( 23.7 1.5) = 63.2 kips Thus, the allowable strength of this slip-critical connection is Rn Ω = 63.2 kips

It can be seen through Example 10.4 that a significant amount of connection strength is lost when a connection that otherwise could be designed as a bearing-type connection is forced to be designed as a slip-critical connection. A situation where the connection of Example 10.4 would need to be considered slip-critical is in a fatigue loading. By considering the resistance factor and safety factor for other hole types, it is also seen that significant strength reductions result for those slip-critical connections. 10.7.6

Combined Tension and Shear in Bearing-Type Connections When bolts are subjected to simultaneous shear and tension, the available strength in each is diminished by the presence of the other. Experimental results indicate that the interaction between shear and tension is best represented by an elliptical relationship as shown as the solid curve in Figure 10.14. This elliptical relationship can conveniently be replaced, with little loss of accuracy, by three straight lines as shown by the dashed lines in Figure 10.14. Specification Section J3.7 provides a nominal tensile stress modified to include the effects of shearing stress, Fnt′ , to be used in determining the nominal bolt tensile strength such that Rn = Fnt′ Ab (AISC J3-2) and φ = 0.75 (LRFD) Ω = 2.00 (ASD) As can be seen in Figure 10.14, when the required stress in either shear or tension is less than or equal to 30 percent of the corresponding available stress, the effects of the combined stresses can be ignored. If both required stresses exceed this 30 percent limit, the modified tensile stress is given by Equations J3-3a and J3-3b as F Fnt′ = 1.3 Fnt − nt f rv ≤ Fnt Frv where Fnt = nominal tensile stress when only tension occurs Frv = available shear stress, φFnv for LRFD or Fnv Ω for ASD Fnv = nominal shear stress when only shear occurs

Chappter 10

Connection Elem ments 451

frv = required shear stress, either e for LRF FD or ASD An exam mple will be giiven in Sectio on 11.10. 10.7.7

Com mbined Tenssion and Shear in Slip-Crritical Conneections When bo olts in slip-ccritical conneections are suubjected to ssimultaneous shear and ttension, the availablee slip strengtth is diminish hed as the aapplied tensille force reduuces the clam mping force provided d by the bolt pretension. p The strength oof the connecttion, thereforre, is a linear function of the forcee compressing g the plies. Th his force is thhe initial pre-ttension, Tb, m minus the appllied load, T. The speccified slip-critical shear vaalue is, thereffore, reducedd by the factoor (1 − T Tb ) . The actual reduction n factor, ksc, is i provided in n Specificatioon Equation JJ3-5, again wiith one for AS SD and one for LRFD D, as

Tu RFD) ≥ 0 (LR Du Tb nb 1..5Ta ksc = 1 − ASD) ≥ 0 (A Du Tb nb

ksc = 1 −

(A AISC J3-5a) (A AISC J3-5b)

where Ta = required r tensiion force usinng ASD load ccombinationss Tu = required r tensiion force usinng LRFD loadd combinationns Tb = minimum m boltt pretension Du = 1.13 nb = number n of bollts carrying thhe applied tennsion

Figure 10.144 Interactioon of Bolt Shear and Teension

10.8

WELDS Welding is the proceess of joining g steel by m melting and fuusing additional metal intto the joint between the two piecees to be joineed. The ease w with which vaarious types oof steel can bbe joined by welding, without exh hibiting cracks and other fflaws, is calleed weldability ty. Most strucctural steels used today accept weelding withou ut the occurreence of unwannted defects. The Americaan Welding Society (AWS) ( defin nes weldabilitty as “the cappacity of a m metal to be w welded under fabrication

4552 Chapter 10 1

Connecttion Elementss

conditions im mposed, into a specific, su uitably designned structure aand to perform m satisfactoriily in the intended serrvice.” Welldability depeends primarily y on the chem mical composiition of the stteel and the thhickness of the materrial. The impaact on weldab bility of the vaarious chemiccal elements iin the composition of steel was disscussed in Ch hapter 3. 100.8.1

Weldin ng Processess For structuraal steel, the four fo most pop pular welding processes as designated bby AWS are: shielded metal arc, su ubmerged arcc, gas shielded d metal arc annd flux cored arc. M Arc Wellding Shielded Metal Shielded meetal arc weld ding (SMAW) is one of thhe oldest wellding processses. It is often called manual or stick s welding. Figure 10.15 5a is a schem matic represenntation of thiss welding proocess. A high voltagee is induced between an electrode andd the metal ppieces that aare to be joinned. The electrode is the source of o the metal introduced i innto the joint tto make the w weld. It is caalled the consumable electrode. When W the wellding operatorr strikes an aarc between tthe electrode and the base metal, the resulting flow of curreent melts the eelectrode andd the base mettal adjacent too it. The electrode is coated with a special ceram mic material called flux. T This flux proteects the molteen metal from absorb bing hydrogen n and other im mpurities duriing the weldinng process. W When the metal cools, a permanentt bond exists between the electrode maaterial and thhe base materrial. Because the flux cools at a diifferent rate th han the metall, it separatess from the weeld and is easiily removed ffrom the joint. Submerged d Arc Weldin ng Submerged arc welding (SAW) is an automatic or semi-automaatic process tthat is used pprimarily when long pieces p of platee are to be joined. It is show wn schematiccally in Figurre 10.15b. SA AW

Figure 10.15 (a) Shield ded Metal Arcc Welding annd (b) Submerrged Arc Wellding

Chappter 10

Connection Elem ments 453

Figure 10.16 Fillet and d Groove Wellds in Section n welds must m be madee in the nearr flat or horiizontal positiion. The fluxx is a granullar material introduceed through a flexible tube on top of the electric aarc. It is an economical process for applicatio ons in which h the use of reepetitive and automated fa fabrication proocedures lendd efficiency to the wo ork. Gas Shieelded Metal Arc A Welding g Gas shieelded metal arrc welding (G GMAW) is a pprocess in whhich a continuuous wire is ffed into the joint to be b welded. The T molten metal m is proteccted from thee atmosphere by gas surroounding the wire. In the field, it iss necessary to o ensure that wind does nnot blow the ggas away from m the joint. This is th he method oftten referred to o as MIG weldding for its usse of inert gasses. Flux Corred Arc Welding Flux corred arc weld ding (FCAW)) is also a ccontinuous-wiire process, except that tthe wire is essentiallly a thin holllow tube filled with flux thhat protects tthe metal as tthe wire meltts. It can be arranged as a semi-automatic proceess, and excepptionally highh production rrates can be aattained. 10.8.2

Typ pes of Welds Four bassic types of welds w are used d in steel connstruction: filllet welds, grooove welds, pplug welds, and slot welds. w Fillet and groove welds w are show wn in Figure 10.16. Plug aand slot weldds fill a hole or slot with weld mateerial to attach h one piece to another. Figure F 10.16aa shows a filllet weld. Thee leg of the w weld is measured along thhe interface between the weld mettal and the baase metal. Thhe throat of thhe weld is thee shortest dim mension and the shearr plane of the weld. Becau use most fillett welds are syymmetrical, w with a 45-degree surface, the throaat is 0.707 tim mes the leg diimension as sshown. The siize of a fillett weld is giveen by its leg dimensio on, in incremeents of 1/16 in n. A groove welld can be either a completee joint penetraation (CJP) ggroove weld, aas shown in Figure 10 0.16b, or a paartial joint peenetration (PJJP) groove weeld, as shownn in Figure 100.16c. Both types of groove weldss have been prequalified p bby AWS. Thiis prequalificcation means that certain weld con nfigurations— —including thee root openinng, R, the anggle of preparattion, α, and thhe effective thicknesss, S—are deeemed practicaal to build aand will carryy the intendeed load. AW WS specifies provision ns for prequaalifying any weld w configurration if circuumstances inddicate that it iis practical. These prrequalified co omplete and partial joint penetration ggroove weldss are shown in detail in Manual Table T 8-2. Th he configurations shown inn Figure 10.166 are schemattic representattions.

4554 Chapter 10 1

Connecttion Elementss

Figgure 10.17 Terminology y for Fillet and Groove Weeld Positions Both h fillet weldss and groove welds can b e laid down in a variety oof different ppositions depending on o the orientaation of the pieces p to be jjoined. The tterminology ffor these possitions is given in Fig gure 10.17. Flaat and horizon ntal welds aree preferred ovver vertical annd overhead w welds. Wellds other than n flat and horrizontal requiire a higher ddeposition ratee and thus result in a higher cost final f connectiion. 100.8.3

Weld Sizes S Specification n Section J2 addresses efffective areass and sizes ffor welds. Thhe effective aarea and other limitattions for groo ove welds aree given in Sppecification S Section J2.1 aand Tables J22.1, J2.2 and J2.3. The T effective areas of filllet welds arre given in Specification Section J2.2a. The minimum sizes for fillet welds w in jointts are based oon the thinnerr of the parts bbeing joined as given in Specificattion Table J2..4. The maxim mum size of ffillet welds foor material lesss than 1/4 in.. thick is the thicknesss of the mateerial, whereass for materiall 1/4 in. thickk or greater, the maximum m size is the material thickness lesss 1/16 in. The minimum leength of filleet welds that are designedd on the basiis of strengthh is four times the weeld leg size. For F welds loaaded longituddinally, the m maximum weldd length is 1000 times the weld leg g size for full effectivenesss. Welds long er than this m must use a redduced effectivve length as given in Section S J2.2b.

100.9

WELD D LIMIT ST TATES The only lim mit state to bee considered for f a weld is rupture. Yielding of the w weld metal will occur, but it occurrs over such a short distaance that it iis not a factoor in connecttion behaviorr. Strain hardening occcurs and rup pture takes plaace without exxcessive yieldding deformaation. The ultimate tenssile strength of o an electrodde may vary from 60 to 120 ksi, depennding on the specified d composition n. AWS classsifies electroddes according to the tensilee strength of tthe weld metal and in ndicates electrrode strength h as FEXX. In thhis notation, the E represeents the electrrode and the XX represents the ten nsile strength h. Thus, a typpical electrodde used to weeld A992 steeel would have a strength of 70 ksi and be design nated as an E770 electrode.

Chapter 10

Connection Elements 455

AWS and AISC specify that for a particular grade of structural steel, as indicated by yield strength, there is a matching electrode. The user note table found in Section J2.6 summarizes the AWS provisions for matching filler metals. Both organizations further specify that the steel can be joined by welding only with the matching electrode or one that is no more than one grade higher. This is to encourage yielding in the base metal before it occurs in the weld. 10.9.1

Fillet Weld Strength For a fillet weld as shown in Figure 10.16a, load is transferred by shear through the throat of the weld and the weld rupture strength is a function of the properties of the electrode. Shear strength provisions for welds are found in Specification Section J2.4 and Table J2.5, where Rn = Fnw Awe (AISC J2-3) φ = 0.75 (LRFD) Ω = 2.00 (ASD) and Fnw = nominal strength of the weld metal per unit area Awe = effective area of the weld FEXX = filler metal classification strength, the weld strength For a longitudinally loaded fillet weld,

Fnw = 0.6 FEXX Because the limit state of all fillet welds is one of shear rupture through the throat, the effective area of a symmetrical fillet weld is the width of the weld at the throat, 0.707w, times the length of the weld, l, so that Awe = 0.707wl The resulting nominal weld strength is Rn = 0.60 FEXX ( 0.707 wl ) For the most commonly used weld electrode, FEXX = 70 ksi, the design strength for LRFD can be determined as φRn = 0.75 ( 0.6 ( 70 ) ) ( 0.707 wl ) = 22.27 wl and the allowable strength for ASD can be determined as Rn Ω = (0.6 (70))( 0.707wl ) 2.00 = 14.85wl It is convenient in design to use the fillet weld strength for a fillet weld with a 1/16 in. leg, which gives Design strength for LRFD φRn = 22.27wl = 22.27 (1 16 )(1.0 ) = 1.392 kips per 1 16 in. of weld per in. of length

and Allowable strength for ASD Rn Ω = 14.85wl = 14.85 (1 16 )(1.0 ) = 0.928 kips per 1 16 in. of weld per in. of length Therefore, a 1/4 in. fillet weld has a design strength of 1.392×4 (sixteenths) = 5.57 kips per inch of length and an allowable strength of 0.928×4 (sixteenths) = 3.71 kips per inch of length.

4556 Chapter 10

Connecttion Elementss

Figure 10.18

Normalizzed Strength vs v Normalizeed Deformatioon for Welds Loaded at ann Angle

Copyright © American Insttitute of Steel Construction. C R Reprinted with Permission. A All rights reservved.

Research has sho own that when n load is appllied to a fillett weld at an aangle other thaan along the length of o the weld, more m strength h is availablee than when tthe load is appplied longituudinally. Figure 10.18 shows thatt as the anglee of load inccreases the ulltimate strenggth increasess. It also shows a co orresponding reduction off deformationn capacity. T The vertical lline in Figurre 10.18 locates the maximum m streength of a weeld loaded lonngitudinally. Thus, welds loaded at an angle to the weld length are seen to exhibit an increase of up to 50% in strengtth. The Specification provides thee nominal streess, based on n the angle off the load to the longitudiinal axis of thhe weld. Thus, (AIS SC J2-5) Fnw = 0.6 FEXX (1.00 + 0.5sin1.5 θ ) where θ = angle of load ding measured d from the weeld longitudinnal axis Thiss strength equ uation is inten nded to be ussed for weldss or weld grouups with unifform leg size in whicch all elementts are in line or parallel annd loaded thrrough the cennter of gravityy. When welds with different orieentations are combined in the same joiint, deformatiion of these ddifferent welds, as illlustrated in Fiigure 10.18, must m be accouunted for. Sppecification Seection J2 provvides an approach fo or concentriccally loaded fillet weld ggroups consiisting of elem ments that aare both longitudinall and transverrse to the direection of the applied loadd. For this case, the nominnal weld strength is taken as thee larger of th he simple suum of the w weld strength without connsidering orientation, given by Rn = Rnwl + Rnwt (AISC C J2-6a) or Rn = 0.855Rnwl + 1.5Rnwwt (AISC C J2-6b) where Rnwl = totaal nominal streength of the llongitudinallyy loaded weldd without connsidering the angle of o load Rnwt = totaal nominal strrength of the transversely loaded weldd without connsidering the angle of o load

Chappter 10

Connection Elem ments 457

1 Weldss for Examplee 10.5 Figure 10.19 Equation n J2-6a provid des the streng gth of the welld group as a function of tthe sum of thee lengths of the welds times the sttrength per unit u length. Im mplementationn of Equationn J2-6b requiires that the longitudiinally loaded weld strength h be discounteed by 0.85 in order for thee transverse w welds to be increased d by 1.5. Wh hen welds loaaded at diffeerent angles aare combinedd, it is criticaal that their deformattions are com mpatible. From m Figure 10.118 it can be seeen that a weeld loaded at 990° reaches its ultimaate strength at a a normalizeed deformatioon significanttly less than thhe deformatioon at which a weld lo oaded at 0° reeaches its ultiimate strengtth. The weld loaded at 0° reaches only about 85% of its ultiimate strength h when its deformation is tthe same as thhe weld loadeed at 90° wheen it is at its ultimate strength. Thee approach of o Equation J22-6b will be most beneficcial when thee transverse welds aree longer than the longitudinal welds. The T forgoing discussion addressed a onlly the strength th of the weldd material. T The material that the weld w is attach hed to is referrred to as the base materiall. This materiial must also be checked for streng gth according g to the provissions of Secti on J4 which w will be discusssed in Sectioon 10.10.

EXAMPLE E 10.5 Weld Stren ngth and Load Anglee

Goal: G

Deetermine the available a strength of the thhree welds givven in Figure 10.19.

Given: G

Th he welds are 3/4 3 in. welds,, 8.0 in. long,, and loaded ((a) along the length of thee weld, (b) trransversely too the weld, aand (c) at a 445-degree anggle to the weeld. Use E70 electrodes.

SOLUTION N

Part P a Step S 1:

Weld W Loaded Along A Its Len gth Deetermine the number n of 1/116 units for thhe given weldd using Equattion J2-5. A 3/4 in. weld is twelve 1/1 6 in. units accross the leg.

Step S 2:

Deetermine the strength s of thhe weld when loaded alongg its length. Th he strength values v alreadyy discussed can be used because thee weld is loaaded longitud dinally.

For F LRFD L Step S 3:

Th he design streength is φRn = 8.0 (12 )(1.392 ) = 134 kipss

For F ASD A Step S 3:

Th he allowable strength s is Rn Ω = 8.0(12)(0.9228) = 89.1 kips

Part P b Step S 1:

Weld W Loaded at a 90 Degrees to the Weld L Length Deetermine the nominal weldd strength usiing the equattion to accounnt for the an ngle of load.

4558 Chapter 10 1

Connecttion Elementss Fw = ( 0.60 FEXX ) (1.0 + 0.5sin1.5 θ ) = ( 0.60 FEXX ) (1.0 + 0.5sin1.5 ( 90 ) ) = ( 0.60 FEXX )(1.5 )

Thereefore, the streength of the w weld is increaased by 1.5 ovver what it is when loaded longitudinaally. Forr LRF FD Step p 2: Forr ASD D Step p 2: Parrt c Step p 1:

d strengtth is The design

φRn = 1.5 (134 ) = 2001 kips

The allowable a strength is Rn Ω = 1.5(89.1) = 134 kips Weld d Loaded at 45 5 Degrees to tthe Weld Lenngth Deterrmine the nom Equation J2-55 to account ffor the minal weld strrength using E angle of load. Fw = ( 0.60 FEXX ) (1.0 + 0.55sin1.5 θ ) = ( 0.60 FEXX ) (1.0 + 0.55sin1.5 ( 45 ) ) = ( 0.60 FEXX )(1.30 )

Thereefore, the strength of the w weld is increassed by 1.30 ovver what it is when loaded longitudinaally. Forr LRF FD Step p 2: Forr ASD D Step p 2:

d strengtth is The design

φRn = 1.3 (134 ) = 1774 kips

The allowable a strength is Rn Ω = 1.3(89.1) = 116 kips

Figure 10.20

C-Shaped W Weld for Exam mple 10.6.

Chapter 10

Connection Elements 459

EXAMPLE 10.6a Weld Strength and Load Angle by LRFD

Goal:

Determine the design strength for C-shaped welds.

Given:

A C-shaped weld group is shown in Figure 10.20 to attach a tension plate to a gusset. Use E70 electrodes and a 7/8 in. weld.

SOLUTION

Step 1:

Determine the design strength for the two 4.0 in. welds parallel to the load. φRnwl = 2 ( 4.0 )(14 )(1.392 ) = 156 kips

Step 2:

Determine the design strength for the 6.0 in. weld transverse to the load. φRnwt = 6 (14 )(1.392 ) = 117 kips

Step 3:

Determine the connection design strength by adding the strength based on length of the welds. φRn = Rnwl + Rnwt = 156 + 117 = 273 kips

Step 4:

Determine the design strength considering the added contribution of the transverse welds while reducing the contribution of the longitudinal welds so that φRn = 0.85Rnwl + 1.5Rnwt = 0.85 (156 ) + 1.5 (117 ) = 308 kips

Step 5:

Determine the weld strength by selecting the larger of the results from steps 3 and 4. Thus, accounting for the difference between the longitudinal and transverse welds provides more strength. φRn = 308 kips

EXAMPLE 10.6b Weld Strength and Load Angle by ASD

Goal:

Determine the design strength for C-shaped welds.

Given:

A C-shaped weld group is shown in Figure 10.20 to attach a tension plate to a gusset. Use E70 electrodes and a 7/8 in. weld.

SOLUTION

Step 1:

Determine the allowable strength for the two 4.0 in. welds parallel to the load. Rnwl Ω = 2(4.0)(14)(0.928) = 104 kips

Step 2:

Determine the allowable strength for the 6.0-in. weld transverse to the load. Rnwt Ω = 6.0(14)(0.928) = 78.0 kips

Step 3:

Determine the connection allowable strength by adding the strength based on length of the welds. Rn Ω = 104 + 78.0 = 182 kips

Step 4:

Determine the allowable strength considering the added contribution of the transverse welds while reducing the contribution of the longitudinal welds so that

460 Chapter 10

Connection Elements Rn Ω = 0.85(104) + 1.5(78.0) = 205 kips

Step 5:

Determine the weld strength by selecting the larger of the results from steps 3 and 4. Thus, accounting for the difference between the longitudinal and transverse welds provides more strength. Rn Ω = 205 kips

EXAMPLE 10.7a Weld Strength and Load Angle by LRFD

Goal:

Determine the design strength for C-shaped welds.

Given:

The C-shaped weld group shown in Figure 10.20 is to attach a tension plate to a gusset. However, the welds parallel to the load will be increased to 10 in. in length. Use E70 electrodes and a 7/8 in. weld.

SOLUTION

Step 1:

Determine the design strength for the two 10.0 in. welds parallel to the load. φRnwl = 2 (10.0 )(14 )(1.392 ) = 390 kips

Step 2:

Determine the design strength for the 6.0 in. weld transverse to the load. φRnwt = 6 (14 )(1.392 ) = 117 kips

Step 3:

Determine the connection design strength by adding the strength based on length of the welds. φRn = Rnwl + Rnwt = 390 + 117 = 507 kips

Step 4:

Determine the design strength considering the added contribution of the transverse welds while reducing the contribution of the longitudinal welds so that φRn = 0.85Rnwl + 1.5Rnwt = 0.85 ( 390 ) + 1.5 (117 ) = 507 kips

Step 5:

Determine the weld strength by selecting the larger of the results from steps 3 and 4. Note that both approaches give the same design strength. Thus, φRn = 507 kips If the welds parallel to the load were any longer, considering only weld length would provide the most strength.

EXAMPLE 10.7b Weld Strength and Load Angle by ASD

Goal:

Determine the design strength for C-shaped welds.

Given:

The C-shaped weld group shown in Figure 10.20 is to attach a tension plate to a gusset. However, the welds parallel to the load will be increased to 10 in. in length. Use E70 electrodes and a 7/8 in. weld.

SOLUTION

Step 1:

Determine the allowable strength for the two 10.0 in. welds parallel to the load. Rnwl Ω = 2(10.0)(14)(0.928) = 260 kips

Chapter 10

Connection Elements 461

Step 2:

Determine the allowable strength for the 6.0 in. weld transverse to the load. Rnwt Ω = 6.0(14)(0.928) = 78.0 kips

Step 3:

Determine the connection allowable strength by adding the strength based on length of the welds. Rn Ω = Rnwl + Rnwt = 260 + 78.0 = 338 kips

Step 4:

Determine the allowable strength considering the added contribution of the transverse welds while reducing the contribution of the longitudinal welds so that Rn Ω = 0.85Rnwl + 1.5 Rnwt = 0.85(260) + 1.5(78.0) = 338 kips

Step 5:

Determine the weld strength by selecting the larger of the results from steps 3 and 4. Note that both approaches give the same allowable strength. Thus, Rn Ω = 338 kips If the welds parallel to the load were any longer, considering only weld length would provide the most strength.

10.9.2

Groove Weld Strength A groove weld can be either a complete or partial joint penetration weld as shown in Figure 10.16b and c. The complete joint penetration (CJP) groove weld is not designed in the usual sense because the weld metal is always stronger than the base metal when properly matching electrodes are used. Therefore, the strength of the base metal controls the design. In the case of a CJP groove weld, the nominal strength of the tension joint is the product of the yield strength of the base material and the cross-sectional area of the smallest piece joined. The nominal strength of a partial joint penetration (PJP) groove weld in a tension joint is similar except that the full cross-sectional area of the joined pieces is not effective. In this case, AWS defines an effective throat dimension, S, which is a function of the configuration of the bevel as shown in Figure 10.16c and Manual Table 8-2.

10.10

CONNECTING ELEMENTS The plates, angles, and other elements that go into making up a connection are called connecting elements. They, along with the region of the members actually involved in the connection, are treated in Section J4. There are provisions for tension, compression, flexure, shear, and block shear.

10.10.1

Connecting Elements in Tension Although the Specification addresses tension in connecting elements in Section J4.1, it does not alter the basic tension provisions found in Specification Chapter D. This means that two limit states are to be considered, the limit state of yielding and the limit state of rupture. Again, for tension, the resistance and safety factors are different for the two limit states, so any comparison of strength must be made at the design or allowable strength level. The design strength is given by φRn and the allowable strength by Rn/Ω, as is the case throughout the Specification. For the limit state of yielding of connecting elements (AISC J4-1) Rn = Fy Ag

462 Chapter 10

Connection Elements φ = 0.90 (LRFD)

Ω = 1.67 (ASD)

For the limit state of rupture of connecting elements Rn = Fu Ae φ = 0.75 (LRFD)

(AISC J4-2)

Ω = 2.00 (ASD)

The areas are determined as for the tension members previously considered. 10.10.2

Connecting Elements in Compression Most connecting elements in compression are relatively short and have a fairly low slenderness ratio. In addition, determination of the appropriate effective length factor requires application of significant engineering judgment, usually amounting to making an educated guess as to an appropriate factor. With this in mind, and to simplify connection design somewhat, the Specification provides a simple relation in Section J4.4 for the compressive strength of connecting elements if the slenderness ratio, lc/r, is less than or equal to 25. For this case, (AISC J4-6) Pn = Fy Ag and the resistance and safety factors are the same as for other compression members, φ = 0.90 (LRFD) Ω = 1.67 (ASD) If the slenderness ratio of the compression element is greater than 25, the element must be designed according to the compression member provisions of Specification Chapter E.

10.10.3

Connecting Elements in Flexure Connecting elements that undergo flexure are to be designed with consideration of flexural yielding, local buckling, flexural lateral-torsional buckling, and flexural rupture. These are the same limit states to be considered for flexural members as presented in Chapter 6.

10.10.4

Connecting Elements in Shear Member design for shear, according to Section G2, requires the consideration of the limit states of shear yielding and shear buckling. Connecting elements and the portion of members affected by the connection must be checked for the limit states of shear yielding and shear rupture according to Section J4.2. Shear yielding occurs on the gross area of the element whereas shear rupture occurs on a section containing holes. Thus, for shear yielding of the element (AISC J4-3) Rn = 0.6 Fy Agv φ = 1.00 (LRFD)

Ω = 1.50 (ASD)

where Agv = gross area subject to shear. The resistance and safety factors for this case are the same as those for the special case of rolled I-shaped members given in Specification Section G2.1(a). For the limit state of shear rupture Rn = 0.6 Fu Anv (AISC J4-4) φ = 0.75 (LRFD) Ω = 2.00 (ASD) where Anv = net area subjected to shear. As was the case for tension rupture, the net area is determined by removing the area of holes from the gross area.

Chappter 10

(a)) Figu ure 10.21

Connection Elem ments 463

(b)

Block B Shear Failure: F (a) welded tensionn connection; (b) bolted beam end conneection

Phottos courtesy Ro obert Driver

10.10.5

Bllock Shear Sttrength The limiit state of bllock shear ru upture can occcur on the cconnecting eelements or tthe affected memberss as shown in i Figure 10..21. Figure 110.21a showss a block shhear failure inn a welded connectio on of a tension member to t the web off a W-shape.. Figure 10.21b shows a bblock shear failure in n a bolted end d connection to t a coped W--shape. Block B shear combines c shear and tensioon failures intto a single, coomplex modee of failure. Block sh hear was discu ussed in Sectiion 4.7 as it ppertained to teension membbers because iit is a major factor in determining tension mem mber strength. It can also be a factor in determining tthe strength of a beam m end reactio on, dependin ng on the connnection geom metry. Thus, it is repeatedd here. The nominal strength for the t limit state of block sheaar rupture is ((AISC J4-5) Rn = 0.6 6 Fu Anv + U bs Fu Ant ≤ 0.6 Fy Agv + U bs Fu Ant where Agv Ant Anv Ubs

= gross shear area = net teension area = net sh hear area = 1.0 fo or uniform ten nsion stress diistribution = 0.5 fo or nonuniform m tension stresss

The resisstance and saffety factors fo or the limit staate of block sshear rupture are again φ = 0.75 (LRFD)) Ω = 2.00 (ASD) Figure 10 0.22 shows a single-anglee tension mem mber attachedd to a gusset plate and a ccoped beam end with h the holes loccated in a sing gle line. The tension area and shear areea are identifiied for each and the area a that woulld tear out is shaded. s

4664 Chapter 10

Connecttion Elementss

Figure 10.22 Example Block Shear Failure e show ws that the eexpected failuure mode willl always A reeview of the block shear equation include tenssion rupture, whereas the shear failure mode will be the smallerr of the shearr rupture and the sheaar yield. The tension t stresss distribution factor, Ubs, iss a function oof the variatioon of the tension stresss over the teension area. Figure F 10.23 shows severaal elements annd the corressponding assumed ten nsile stress disstribution. Th he only case iddentified by tthe Commenttary where the tensile stress distrib bution is not uniform is that of a copeed beam withh two rows oof bolts, as shhown in Figure 10.23 3g.

hear Tensile Stress Distribuution Figure 10.23 Block Sh

Chappter 10

Connection Elem ments 465

Figgure 10.24 C Coped Beam End for Exaample 10.8

EXAMPLE E 10.8 Block Shea ar Strength

Goal: G

Deetermine the block shear design strenngth and alloowable strenggth for a co oped beam.

Given: G

A coped W16× ×40 A992 beaam end is shown in Figurre 10.24. Asssume that thee beam has sttandard holes for 5/8 in. boolts.

SOLUTION N

Step S 1:

Deetermine the gross g and nett shear areas aand net tensioon area for thee beam. Reemember from m the discusssion of tensiion members that for net area, an ad dditional 1/16 6 in. must be added to thee hole size to account for any hole daamage from th he punching ooperation. Froom Manual Table 1-1 tw = 0.305 in. Agv = 11.0 ( 0.305 ) = 3.36 in.2

Anv = (11.0 − 33.5 ( 5 8 + 1 8 ) ) ( 0.305 ) = 2.25 in.2 Ant = ( 4.25 − 11.5 ( 5 8 + 1 8 ) ) ( 0.305 ) = 0.953 in.2 Step S 2:

Deetermine the shear yield and rupture strength andd the tensionn rupture strrength. Fo or this geomeetry, the tensiile stress disttribution is nnonuniform; ttherefore, Ubs b = 0.5. Sh hear yield

Sh hear rupture Teensile rupture

0.6 Fy Agvv = 0.6 ( 50 )( 33.36 ) = 101 kipps 0.6 Fu Anv = 0.6 ( 65)( 2.55) = 99.5 kiips U bs Fu Ant = 0.5 ( 65)( 0.9953) = 31.0 kkips

466 Chapter 10

Connection Elements Step 3:

Determine the nominal block shear strength. Because shear rupture is less than shear yield, combine the shear rupture with the tensile rupture. Thus, Rn = 99.5 + 31.0 = 131 kips

For LRFD Step 4: For ASD Step 4: 10.10.6

The design strength is

φRn = 0.75 (131) =98.3 kips

The allowable strength is Rn Ω = 131 2.00 = 65.5 kips

Connecting Element Rupture Strength at Welds The strength of connection elements at welds cannot always be determined directly. In those cases, it is often convenient to determine the minimum thickness of the connecting element required to match the rupture strength of the weld with the rupture strength of the base metal. The rupture strength of a fillet weld was given in Section 10.9.1 as Rn = 0.60 FEXX ( 0.707 wl ) If this is set equal to the rupture strength of the base metal, given as Rn = 0.6 Fu tl and the thickness is taken as tmin, the minimum required thickness for the base metal to match the weld, with E70 electrodes and the weld size taken as D 16 is

0.60 ( 70 ) ( 0.707 ( D 16 ) l ) = 0.6 Fu tmin l Solving for tmin yields

tmin = 3.09 D Fu

(Manual 9-2)

For a case where there are welds on both sides of the base metal, such as on the web of a beam with angles on both sides, the minimum web thickness will be twice that for a weld on one side. Thus, tmin = 6.19 D Fu (Manual 9-3) The application of these minimum thickness calculations will be illustrated in Chapters 11 and 12 as appropriate for the connections being considered.

10.11

PROBLEMS

1. Develop a table showing the nominal shear strength for A325-N bolts for the following sizes: 5/8, 3/4, 7/8, and 1 in.

3. Develop a table showing the nominal shear strength for A490-N bolts for the following sizes: 5/8, 3/4, 7/8, and 1 in.

2. Develop a table showing the nominal shear strength for A325-X bolts for the following sizes: 5/8, 3/4, 7/8, and 1 in.

4. Develop a table showing the nominal shear strength for A490-X bolts for the following sizes: 5/8, 3/4, 7/8, and 1 in.

Chaptter 10 55. Develop a table showing the design sheear strength forr A A325-N, A325 5-X, A490-N, and A490-X bolts for thee ffollowing sizess: 5/8, 3/4, 7/8, and 1 in. 66. Develop a table showing g the allowable shear strength h ffor A325-N, A325-X, A A490-N N, and A490-X X bolts for thee ffollowing sizess: 5/8, 3/4, 7/8, and 1 in.

Connnection Elements 467

with two 1/2 in. A336 plates. Deteermine (a) desiggn strength by LR RFD and (b) aallowable strenggth by ASD. 15. Determine thhe available sstrength of thhe 7/8 in. A4900-N bolts in thhe lap splice shown in Figuure P10.15 with two 1/2 in. A A572 Grade 550 plates. Dettermine (a) desiggn strength byy LRFD and (bb) allowable sstrength by ASD D.

77. Develop a table showing the design sheear strength forr F F1852-N, F185 52-X, F2280-N N, and F2280-X X bolts for thee ffollowing sizess: 5/8, 3/4, 7/8, and 1 in. 88. Develop a table showing g the allowable shear strength h ffor F1852-N, F1852-X, F F228 80-N, and F2280-X bolts forr thhe following sizes: 5/8, 3/4, 7/8, 7 and 1 in. 99. Determine the available strength of thee 3/4 in. A325-N bolts in the lap l splice show wn in Figure P10.9 P with two o 11/2 in. A36 plates. p Determ mine (a) desig gn strength by y L LRFD and (b) allowable a stren ngth by ASD.

P10.115 16. Determine thhe available sstrength of thhe 7/8 in. A4900-X bolts in thhe lap splice shown in Figuure P10.15 with two 1/2 in. A A572 Grade 550 plates. Dettermine (a) desiggn strength byy LRFD and (bb) allowable sstrength by ASD D. 17. Determine thhe available sstrength of thhe 7/8 in. A3255-N bolts in thhe butt splice shown in Figuure P10.17 with two 1/2 in. siide plates andd a 1 in. main plate. Use A36 plates. Determ mine (a) designn strength by LRFD and (b) alllowable strenggth by ASD.

P P10.9 110. Determine the availab ble strength of o the 3/4 in. A A325-X bolts in n the lap splicee shown in Fig gure P10.9 with h tw wo 1/2 in. A36 plates. Deterrmine (a) design strength by y L LRFD and (b) allowable a stren ngth by ASD. 111. Determine the availab ble strength of o the 7/8 in. A A325-N bolts in n the lap splicee shown in Fig gure P10.9 with h tw wo 1/2 in. A36 plates. Deterrmine (a) design strength by y L LRFD and (b) allowable a stren ngth by ASD. 112. Determine the availablee strength of th he 1 in. F2280-N bolts in the lap l splice show wn in Figure P10.9 P with two o 11/2 in. A36 plates. p Determ mine (a) desig gn strength by y L LRFD and (b) allowable a stren ngth by ASD. 113. Determine the availablee strength of th he 1 in. F2280-X bolts in the lap l splice show wn in Figure P10.9 P with two o 11/2 in. A36 plates. p Determ mine (a) desig gn strength by y L LRFD and (b) allowable a stren ngth by ASD. 114. Determine the availablle strength off the 1-1/8 in. F F2280-N bolts in the lap sp plice shown in n Figure P10.9 9

P10.117 18. Determine thhe available sttrength of thee 1-1/8 in. A4900-X bolts in thhe butt splice shown in Figuure P10.17 with two 1/2 in. siide plates andd a 1 in. main plate. Use A36 plates. Determ mine (a) designn strength by LRFD and (b) alllowable strenggth by ASD. 19. Recreate Figgures 10.11a aand 10.11b foor A325-X boltss. 20. Recreate Figgures 10.11a aand 10.11b foor A490-N boltss.

4468 Chapterr 10

Connection Elemen nts

221. Recreate Figures 10.11 1a and 10.11b b for A490-X X bbolts.

26. Determine thhe available strrength of the six 3/4 in. A3255-N bolts in a 7×1/2 in. A366 plate when thhe bolts are placeed as shown inn Figure P10.226. Determine (a) design strenngth by LRFD aand (b) allowab able strength byy ASD.

222. Determine the availab ble strength of o the 7/8 in. A A490-SC boltss in the slip-ccritical lap splice shown in n F Figure P10.15 with two 1//2 in. A572 Gr. G 50 plates,, sstandard holes,, Class B surfface, and no fillers. f (This iss P Problem 15 as a slip-criticall connection.) Determine (a)) ddesign strength h by LRFD an nd (b) allowab ble strength by y A ASD. 223. Determine the availab ble strength of o the 7/8 in. A A325-SC bolts in the butt sp plice shown in Figure P10.17 7 w with two 1/2 in n. side plates and a 1 in. main m plate. Usee A A36 plates, ov versized holess, Class A su urface, and no o ffillers. (This iss Problem 17 as a a slip-criticaal connection.)) D Determine (a) design d strength h by LRFD and d (b) allowablee sstrength by ASD D. 224. Determine the availablee strength of th he four 3/4 in. A A325-N bolts in i the single L3×3×1/2 A36 when w the boltss aare placed as shown in Fig gure P10.24. Determine (a)) ddesign strength h by LRFD an nd (b) allowab ble strength by y A ASD.

P10.226 27. Determine thhe available strrength of the six 7/8 in. A3255-N bolts in a 7×3/4 in. A366 plate when thhe bolts are placeed as shown inn Figure P10.226. Determine (a) design strenngth by LRFD aand (b) allowab able strength byy ASD. 28. Determine thee available streength of the eiight 3/4 in. A4900-N bolts in ann A992 WT6× ×17.5 when thhe bolts are placeed as shown inn Figure P10.228. Determine (a) design strenngth by LRFD aand (b) allowab able strength byy ASD.

P P10.24 225. Determine the availablee strength of th he three 3/4 in. A A325-N bolts in i the single L4 4×3×3/8 A36 when w the boltss aare placed as shown in Fig gure P10.25. Determine (a)) ddesign strength h by LRFD an nd (b) allowab ble strength by y A ASD.

P P10.25

P10.228 29. Determine thee available streength of two 33/8 in. fillet weldds that are loadded parallel too their length, are 10 in. long,, and are madde from E70 eelectrodes. Dettermine (a) desiggn strength byy LRFD and (bb) allowable sstrength by ASD D. 30. Determine thee available streength of two 1/2 in. fillet weldds that are loadded parallel too their length, are 12 in. long,, and are madde from E70 eelectrodes. Dettermine (a) desiggn strength byy LRFD and (bb) allowable sstrength by ASD D.

Chaptter 10

Connnection Elements 469

331. Determine the availablee strength of tw wo 1/4 in. fillett w welds that are loaded paralllel to their len ngth, are 8 in. loong, and are made m from E7 70 electrodes. Determine (a)) ddesign strength h by LRFD an nd (b) allowab ble strength by y A ASD. 332. If the weelds of Probleem 29 were lo oaded at theirr ccentroid and att 90 degrees to o the weld len ngth, determinee ((a) design stren ngth by LRFD D and (b) allow wable strength h bby ASD. 333. If the weelds of Probleem 29 were lo oaded at theirr ccentroid and att 45 degrees to o the weld len ngth, determinee ((a) design stren ngth by LRFD D and (b) allow wable strength h bby ASD. 334. If the weelds of Probleem 31 were lo oaded at theirr ccentroid and att 80 degrees to o the weld len ngth, determinee ((a) design stren ngth by LRFD D and (b) allow wable strength h bby ASD.

P10.440 41. Determine thhe available bllock shear streength for a A992 beam witth holes for 3//4 in. bolts copedd W21×182 A as shhown in Figuree P10.41. Deterrmine (a) desiggn strength by LR RFD and (b) aallowable strenggth by ASD.

335. If the weelds of Probleem 31 were lo oaded at theirr ccentroid and att 35 degrees to o the weld len ngth, determinee ((a) design stren ngth by LRFD D and (b) allow wable strength h bby ASD. 336. Plot a curvee of the strengtth versus anglee of loading forr a 12 in. long 5//16 in. weld maade from E70 electrodes. e Thee pplot should incllude angles fro om 0 degrees to o 90 degrees. 337. Three 5/1 16 in. welds arre grouped to form a C and d aare loaded at th heir centroid. Determine D the available weld d sstrength if the single transverrse weld is 9 in. and the two o loongitudinal welds w are each h 3 in. Use E70 E electrodes. D Determine (a) design d strength h by LRFD and d (b) allowablee sstrength by ASD D. 338. Repeat Prroblem 37 with h the transversse weld at 3 in. aand the two lo ongitudinal welds at 9 in. eaach. Determinee ((a) design stren ngth by LRFD D and (b) allow wable strength h bby ASD.

P10.441 42. Determine thhe available bllock shear streength for a copedd W24×146 A A992 beam witth holes for 7//8 in. bolts as shhown in Figuree P10.42. Deterrmine (a) desiggn strength by LR RFD and (b) aallowable strenggth by ASD.

339. Three 3/8 8 in. welds are grouped to forrm a C and aree looaded at theirr centroid. Deetermine the available a weld d sstrength if the single transverrse weld is 8 in. and the two o loongitudinal welds w are each h 8 in. Use E70 E electrodes. D Determine (a) design d strength h by LRFD and d (b) allowablee sstrength by ASD D. 440. Determine the availablee block shear strength for a ccoped W16×26 6 A992 beam with w holes for 3/4 3 in. bolts ass sshown in Figurre P10.40. Deteermine (a) desiign strength by y L LRFD and (b) allowable a stren ngth by ASD. P10.442

4470 Chapterr 10

Connection Elemen nts

443. Determine the availablee block shear strength for a ccoped W27×19 94 A992 beam m with holes fo or 7/8 in. boltss aas shown in Fig gure P10.43. Determine D (a) design d strength h bby LRFD and (b) ( allowable sttrength by ASD D.

45. Determine thhe available bllock shear streength for a A992 beam witth holes for 3//4 in. bolts copedd W30×261 A as shhown in Figuree P10.45. Deterrmine (a) desiggn strength by LR RFD and (b) aallowable strenggth by ASD.

P P10.43 444. Determine the availablee block shear strength for a ccoped W30×26 61 A992 beam m with holes fo or 7/8 in. boltss aas shown in Fig gure P10.44. Determine D (a) design d strength h bby LRFD and (b) ( allowable sttrength by ASD D.

P P10.44

P10.445 46. Determine thhe available bllock shear streength for a copedd W44×230 A A992 beam w with holes for 12 - 1 in. boltss spaced at 3.00 in., with horrizontal edge ddistance of 2.0 in. and locatted 1-1/2 in. down from the cope. Deterrmine (a) desiggn strength by LRFD and (b)) allowable strenngth by ASD.

Chapter

11

Simple Con nnectiions

Stanford University, Bing Concert Hall Courtesy of Degenkoolb Engineers

11.1

TYP PES OF SIM MPLE CON NNECTION NS This chaapter addressees two types of simple c onnections, bbeam shear cconnections aand bracing connectio ons. Both aree commonly designed as ppinned jointss – simple, inn the terminoology of the Specifica ation. The connecting elem ments and thee connectors required for these connections have already been b discusseed in Chapter 10. The limitt states that ccontrol the coonnection havve also been discussed d individually y, although th heir link to coonnection dessign may not yet be complletely clear. Connectiion design is a combinatio on of elemennt and connecctor selectionn with a checcking of all appropriaate limit stattes. The goal is to obtainn a connectiion with suff fficient strenggth and the appropriaate stiffness to carry the lo oad in a mannner consistentt with the moddel used in thhe structural analysis. In addition to these sim mple shear connnections, beeam bearing plates and coolumn base plates wiill be discusseed. The T limit stattes to be con nsidered for a particular cconnection deepend on the connection elementss, the connecttion geometry y, and the lo ad path. Theey will be ideentified in thee following sections as each conn nection type is consideredd. A summary ry of the poteential limit sttates at this time, how wever, may prove useful. For bolts, tthe limit stattes of tensilee rupture, sheear rupture, bearing and a tearout, as well as sllip, will be cconsidered. F For welds, thhe only limit state to be considereed is shear rupture, r altho ough weld grroup geometrry will add ssome complexxity to that consideraation. For connecting ellements, the limit states are tensionn yielding annd rupture, compresssion buckling g, shear yieldiing and rupturre, and the fuull range of fleexural limit sttates. Table T 11.1 lissts the section ns of the Speccification and parts of the M Manual discuussed in this chapter.

11.2

SIM MPLE SHEA AR CONNE ECTIONS A signifiicant number of potential connection c geeometries aree associated w with the varioous types of memberss to be conn nected. Five of the mostt commonly used simplee shear connnections are described d in the follow wing sectionss with design examples folllowing. These connectionss are shown

471

472 Chapter 11

Simple Connections

Table 11.1 Sections of Specification and Parts of Manual Covered in This Chapter Specification B4.3 J2 J3 J4 J10

Gross and Net Area Determination Welds Bolts and Threaded Parts Affected Elements of Members and Connecting Elements Flanges and Webs with Concentrated Forces Manual

Part 7 Part 8 Part 9 Part 10 Part 14 Part 15

Design Considerations for Bolts Design Considerations for Welds Design of Connecting Elements Design of Simple Shear Connections Design of Beam Bearing Plates, Column Base Plates, Anchor Rods, and Column Splices Design of Hanger Connections, Bracket Plates, and Crane-Rail Connections

in Figure 11.1a through e as double-angle, single-angle, single-plate (commonly called a shear tab), unstiffened seated, and stiffened seated connections. Part 10 of the Manual includes many tables that can simplify connection design; however, the examples presented here show the required calculations when necessary to improve understanding. Once a calculation has been sufficiently demonstrated, the Manual tables are used. Several design considerations apply to all of the shear connections to be discussed, and in some cases to other types of connections. It is helpful to address these before dealing with the specific connection. The first issue to consider is the location of the hinge within the connecting elements. It is critical that this hinge can actually occur in the real connection, because the analytical model of the connection assumes it behaves as a hinge or pin. The location of the hinge determines what forces and moments, if any, the individual elements must be designed for. In all cases, the hinge is located at the most flexible point within the connection. This may be at the face of the supporting member or at some other point within the connection. Several general design guidelines help to ensure that the connection behaves as desired. In most cases, this means that the hinge is located at the face of the supporting member. For double-angle connections, angle thickness should be limited to a maximum of 5/8 in. The bolts in the outstanding legs, those connecting to the supporting member, should be spaced at as wide a gage as possible; for welded outstanding legs, the vertical welds should be spaced as far apart as possible. These characteristics will ensure that the connection behaves as a simple connection through bending of the outstanding legs. For simple beam connections, the permitted tolerance for beam length must be considered. Although this tolerance is not normally a consideration for member design, it becomes important when the details of connecting members are considered. Beam length tolerance is ±1/4 in. To accommodate fit-up, beams are held back 1/2 in. from the face of the supporting member. Then, when considering the edge distance from a bolt hole to the end of the member, the distance used in calculations should be taken as 1/4 in. less than that actually detailed in cases where an under run in beam length might yield a lower strength. For welded connections, the small loss of weld area due to stopping the weld short of the end of the joint by one to two weld sizes is not typically considered in the calculation of weld strength. It is also helpful to remember the considerations for hole sizes. First, standard holes are sized 1/16 in. larger than the bolt to be inserted for bolts up to 7/8 in. and 1/8 in. larger for 1.0 in. and larger bolts. Then, when considering net sections for the limit states of tension rupture or shear rupture, Specification Section B4.3 requires that an additional 1/16 in. be deducted to

Chhapter 11

Figu ure 11.1

Siimple Connecctions 473

Sim mple Shear Connections C

account for any mateerial damage resulting from m the processs of making the hole. W When a clear distance is calculated for the limit state s of tearouut, the actual hole size is uused. These T design consideration ns are includeed in the exam mples to follow w.

11.3

DOU UBLE-ANG GLE CONN NECTIONS:: BOLTED--BOLTED A doublee-angle shear connection, as shown in F Figure 11.1a,, is perhaps thhe most comm mon simple shear con nnection used d in steel con nstruction. It iis a fairly sim mple connectiion to fabricaate and also provides for fairly eassy erection. When W double- angle connecctions are to bbe installed baack to back, there maay be some prroblems, partticularly whenn the supportting member is a column web. In the case of attachment a to a column weeb, the safetyy requirementts of OSHA ccall for speciaal attention. One solu ution is to stag gger the doub ble angles. Thhis connectioon can easily accommodatee variations in beam length l within n acceptable to olerances. The T double-aangle shear connection c m must be checcked for the following llimit states, grouped according to o the elementts that make up the conneection: bolts, beam web, angles, and supportin ng member: 1. 1 Bolts a. Shear ruptture 2. 2 Beam m

4774 Chapter 11 1

Simple Connections C a.

Bolt B bearing and a tearout onn beam web

Figure 11.2 Connection n Geometry for f Example 111.1

3.

4.

b. Shear S yielding g of the web c. Block B shear on o coped beam m web d. Coped C beam flexural f strenngth Angles a. Bolt B bearing and a tearout onn angles b. Shear S rupture c. Shear S yielding g d. Block B shear Supportin ng member a. Bolt B bearing

h of these lim mit states hass been addresssed in this bbook. In the eexamples thatt follow, Each these limit state checks arre combined into i a compleete connectionn design.

EX XAMPLE 11.1a 1 Boolted-Bolted d Doouble-Anglee Sh hear Connecction byy LRFD

Goa al:

Desig gn a bolted-bo olted double-aangle shear coonnection forr a W18×50 bbeam.

Giv ven:

The W18×50 W beaam is conneected to the web of a W W24×207 annd the conneection must provide p a requuired strengthh Ru = 83.0 kkips. The beam m and colum mn are A992 2 and the 3- 1/2×3-1/2×3//8 angles aree A36. The beam flangee is coped 2 in. i Use 7/8 inn. A325-N boolts in standarrd holes in thhe legs on thee beam web and a short slotts on the outsstanding legs. The basic sttarting geom metry is given in Figure 11. 2.

SO OLUTION

Boltt Shear Deterrmine the num mber of bolts required baseed on the sheear rupture strrength Step p 1: of thee bolts. From m Manual Table 7-1, the de sign shear strrength per bollt is φrn = 24.3 kipps Becau use the bolts are in doublee shear, ns = 2 and the dessign shear strrength of a single bolt is ns φrn = 2 ( 24.3) = 488.6 kips

Chapter 11

Simple Connections 475

Values for double shear are also found in Manual Table 7-1. Because of rounding, the number in the table is slightly greater at 48.7 kips. The total number of bolts required is Ru 83.0 N= = = 1.71 ns ( φrn ) 48.6 Therefore, based only on the limit state of bolt shear, try two bolts. Bolt Strength at Holes in Beam Web Determine bolt bearing and tearout strength on the beam web. Step 2: For the W18×50 beam, tw = 0.355 in. For the two-bolt connection, the top bolt is 1.25 in. from the edge of the beam cope and the second bolt is spaced 3.0 in. from the first. For the top bolt, the clear distance is 1 1 lc = lcv − d h = 1.25 − (7/8 + 1/16) = 0.781 < 2(7/8) = 1.75 2 2 Since the clear distance is less than 2db, tearout controls over bearing. The nominal bolt strength for tearout is

Rn = 1.2lc tFu = 1.2 ( 0.781)( 0.355)( 65) = 21.6 kips and the design strength is φRn = 0.75 ( 21.6) = 16.2 kips Since this is less than the design shear strength of the bolt in double shear it will control the strength of this bolt. For the second bolt lc = s − dh = 3.0 − ( 7 8 + 1 16) = 2.06 > 2 ( 7 8) = 1.75 in. Since the clear distance is greater than 2db, bearing controls over tearout. Thus, the nominal bolt bearing strength is Rn = 2.4dtFu = 2.4 ( 7 8)( 0.355)( 65) = 48.5 kips and the design strength for bearing on the beam web is φRn = 0.75 ( 48.5) = 36.4 kips Since this is less than the shear strength of the bolt in double shear, it will control the strength of this bolt. Therefore, the connection design strength, if only considering the beam web, is based on one bolt limited by tearout and one bolt limited by bearing. Thus, φRn = (16.2 + 36.4) = 52.6 < 83.0 kips

476 Chapter 11

Simple Connections and the two-bolt connection cannot support the load. Step 3:

Determine the number of bolts required considering bearing and tearout on the web. Adding a third bolt spaced at 3.0 in., which means that bearing controls over tearout, gives a connection design strength for bolt bearing and tearout of φRn = (16.2 + 2 ( 36.4 ) ) = 89.0 > 83.0 kips Therefore a three bolt connection will be considered further.

Bolt Strength at Holes in Angle Leg Determine bolt bearing and tearout strength on the A36 angle. The angle Step 4: legs on the beam and on the supporting member are identical and each angle takes one half of the load. For the bottom edge bolt, the clear distance is 1 1 lc = lev − d h = 1.25 − ( 7 8 + 1 16 ) = 0.781 < 2 ( 7 8 ) = 1.75 in. 2 2 Since the clear distance is less than 2db, tearout controls over bearing. The nominal bolt strength for tearout is Rn = 1.2lc tFu = 1.2 ( 0.781)( 3 8 )( 58 ) = 20.4 kips

and the design strength is φRn = 0.75 ( 20.4 ) = 15.3 kips

Since this is less than the design shear strength of the bolt in single shear, it will control the strength of this bolt For the second and third bolts, the clear distance is

lc = s − d h = 3.0 − ( 7 8 + 1 16 ) = 2.06 > 2 ( 7 8 ) = 1.75 in.

Since the clear distance is greater than 2db, bearing controls over tearout. The nominal bolt strength for bearing is Rn = 2.4 dtFu = 2.4 ( 7 8 )( 3 8 )( 58 ) = 45.7 kips

and the design strength for bearing is φRn = 0.75 ( 45.7 ) = 34.3 kips Since this is greater than the design shear strength of the bolt in single shear, bolt shear will control the strength of these two bolts. Connection Strength at Beam Determine the controlling strength for each bolt at the beam web Step 5:

Top bolt Bolt double shear

48.6 kips

Chapter 11 Web tearout Angle bearing ×2 Middle bolt Bolt double shear Web bearing Angle bearing ×2 Bottom bolt Bolt double shear Web bearing Angle tearout ×2

Simple Connections 477

16.2 kips 68.6 kips 48.6 kips 36.4 kips 68.6 kips 48.6 kips 36.4 kips 30.6 kips

Therefore, the strength of the connection at the beam web is φRn = 16.2 + 36.4 + 30.6 = 83.2 > 83.0 kips Bolt Strength at Holes in Supporting Member Consider the outstanding legs of the angles attached to the supporting Step 6: member.

In this case, the bolts are in single shear but there are twice as many bolts, so the load per bolt is half of the load for each of the bolts in the beam web. If the supporting member thickness is at least one-half of the beam web thickness and the strengths are the same, the bolts in the supporting member will be satisfactory. For the W24×207, clear distances result in all three bolts being controlled by bearing over tearout. Thus, with tw = 0.870 in., Rn = 2.4dtFu = 2.4 ( 7 8)( 0.870)( 65) = 119 kips and the design strength is φRn = 0.75 (119) = 89.3 kips Since this is greater than the single shear strength of the bolt, φrn = 24.3 kips , connection strength on the supporting member will be controlled by bolt shear. For six bolts in single shear φRn = 6 ( 24.3) = 146 kips which is greater than the required strength of 83.0 kips so the bolts in the supporting member are adequate Connection Strength at Supporting Member Determine the controlling strength for each bolt at the supporting member Step 7: Top bolt Bolt single shear 24.3 kips Web bearing 89.3 kips Angle tearout 15.3 kips Middle bolt Bolt single shear 24.3 kips Web bearing 89.3 kips Angle bearing 34.3 kips

478 Chapter 11

Simple Connections Bottom bolt Bolt single shear Web bearing Angle bearing

24.3 kips 89.3 kips 34.3 kips

Therefore, the strength of the connection at the beam web is φRn = 15.3 + 24.3 + 24.3 = 63.9 > 83.0 2 =41.5 kips Step 8:

Evaluate the minimum depth of the connection. The beam web connection should be at least half the depth of the beam web, measured as the distance between the fillets, T, given in Manual Table 1-1. This requirement is to prevent twisting of the simple supports. For W18×50, T = 15-1/2 in., so the minimum angle depth should be 7-3/4 in. Thus, the 8-1/2 in. long angle will provide an acceptable connection depth and will also fit within the 15.5 in. available flat length.

Remaining Beam Limit States Check shear yield of the beam web. Step 9:

This is a check that should be carried out during the beam design process. At the point of connection design it is too late to find out that the beam will not be adequate. From Manual Table 3-2, φVn = 192 kips > 83.0 kips Step 10:

Check shear yield of the beam web at the cope This check is carried out according to Section J4.2 where the full area of the remaining web is stressed to Fy and ϕ=1.0. Thus φVn = φ0.6 Fy htw = 1.0 ( 0.6 ( 50 ) ) (18.0 − 2.0 )( 0.355 ) = 170 kips > 83.0 kips

Therefore shear yield strength at the cope is adequate Step 11:

Check shear rupture of the beam web at the cope This check is carried out according to Section J4.2 where the effective area, Ae, the full area of the remaining web less the bolt holes, is stressed to Fu and ϕ=0.75. Thus φVn = φ0.6 Fu Ae ⎛ ⎛ 7 1 ⎞⎞ = 0.75 ( 0.6 ( 65 ) ) ⎜ 18.0 − 2.0 − 3 ⎜ + ⎟ ⎟ ( 0.355 ) ⎝ 8 8 ⎠⎠ ⎝ = 135 kips > 83.0 kips Therefore shear rupture at the cope is adequate

Step 12:

Check the beam web for block shear.

Chapter 11

Simple Connections 479

The equations for block shear rupture are found in Specification Section J4.3 and were presented in Section 10.10.5. First calculate the areas, remembering to account for the 1/4 in. beam length under run tolerance in the tension area calculation, so that leh = 2.0 – 0.25 = 1.75 in.: 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜ 1.75 − ( 7 8 + 1 8 ) ⎟ (0.355) = 0.444 in.2 2 ⎝ ⎠ Agv = ltw = 7.25(0.355) = 2.57 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) (0.355) = 1.69 in.2

Determine the tension rupture strength Fu Ant = 65 ( 0.444 ) = 28.9 kips Consider shear yield and shear rupture and select the least nominal strength; thus, 0.6 Fy Agv = 0.6 ( 50 )( 2.57 ) = 77.1 kips 0.6 Fu Anv = 0.6 ( 65 )(1.69 ) = 65.9 kips

Selecting the shear rupture term and combining it with the tension rupture term gives the connection block shear design strength, and recalling that Ubs = 1.0 for the case of uniform tensile stress distribution, we have φRn = 0.75 ( 65.9 + 1.0 ( 28.9 ) ) = 0.75 ( 94.8 ) = 71.1 < 83.0 kips Thus, the given three-bolt connection is not adequate, with block shear being the critical limit state to this point in our calculations. Step 13:

Revise the connection to meet the block shear strength requirements. Consideration could be given to increasing the number of bolts and thereby increasing the length of the connection. However, because bolt shear required only two bolts, this would not be a particularly economical solution. If the connection were to be lowered on the beam end so that the distance from the center of the top bolt to the edge of the cope were 2.5 in., the connection would have more block shear strength. It should also be noted that the clear end distance for the top bolt in the beam web would increase and thus increase the tearout strength of that top bolt. There is no need to consider that increase since at best it could contribute to reducing the number of bolts and this would have a negative impact on the block shear calculations being addressed in this step. Thus, the new shear areas become

480 Chapter 11

Simple Connections Agv = ltw = 8.50(0.355) = 3.02 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = ( 8.50 − 2.5 ( 7 8 + 1 8 ) ) (0.355) = 2.13 in.2 and the nominal shear yield and rupture strengths become 0.6 Fy Agv = 0.6 ( 50 )( 3.02 ) = 90.6 kips 0.6 Fu Anv = 0.6 ( 65 )( 2.13 ) = 83.1 kips

Selecting the shear rupture term and combining it with the tension rupture term gives the resulting block shear design strength as φRn = 0.75 ( 83.1 + 1.0 ( 28.9 ) ) = 0.75 (112 ) = 84.0 > 83.0 kips Step 14:

Check the flexural strength of the coped beam. It is a good idea to check this limit state during the initial design of the beam. It should be anticipated during the beam design stage that a coped connection will be required, and it is at that stage that a change in beam section can most readily be accommodated. Flexural strength of the coped beam is not addressed in the Specification directly but is covered in Part 9 of the Manual. The moment in the coped beam is taken as the shear force times the eccentricity from the face of the support to the edge of the cope, taken as cope + setback = 7.0 + 0.50 = 7.5 in. in this example. Thus,

M u = Ru e = 83.0 ( 7.5) = 623 in.-kips

(Manual 9-5)

To determine the flexural strength of the coped beam, the web slenderness must be checked. This requires determination of the plate buckling coefficient, k, and the buckling adjustment factor, f. The web slenderness of this coped beam is ho t w = 16.0 0.355 = 45.1

The limiting web slenderness is λ p = 0.46

k1 E Fy

(Manual 9-12)

where k1 = fk

(Manual 9-10)

The buckling adjustment factor, f, is a function of the ratio of the cope length, c, to the beam depth, d, and the plate buckling coefficient, k, is a function of the ratio of cope length, c, to the reduced beam depth, ho. For

Chapter 11

Simple Connections 481

this example, with c d = 7.0 18.0 = 0.389 ≤ 1.0 2c 2(7.0) f = = = 0.778 d 18.0

(Manual 9-14a)

and with c ho = 7.0 16.0 = 0.438 ≤ 1.0 1.65

⎛h ⎞ k = 2.2 ⎜ o ⎟ ⎝ c ⎠

Thus,

1.65

⎛ 16.0 ⎞ = 2.2 ⎜ ⎟ ⎝ 7.0 ⎠

= 8.61 (Manual 9-13a)

k1 = fk = 0.778 ( 8.61) = 6.70 ≥ 1.61

(Manual 9-10)

and

λ p = 0.46

6.70 ( 29,000 ) k1 E = 0.46 = 28.7 Fy 50

Since λ is between λp and 2λp the net elastic and net plastic section modulus will be needed. The net elastic section modulus is taken from Manual Table 9-2. With the depth of the cope, dc = 2.0 in., Snet = 23.4 in.3 The net plastic section modulus is not provided in the Manual so it must be determined. Following the procedure discussed in Chapter 6, the net plastic section modulus for the tee shaped beam that remains after the cope is found to be Z net = 42.4 in.3 and from ⎛ λ ⎞ M n = M p − (M p − M y )⎜ − 1⎟ ⎝ λp ⎠

with

(Manual 9-7)

M p = Fy Z net = 50 ( 42.4 ) = 2120 in.-kips M y = Fy S net = 50 ( 23.4 ) = 1170 in-kips

The nominal moment strength at the cope is ⎛ 45.1 ⎞ M n = 2120 − ( 2120 − 1170 ) ⎜ − 1 ⎟ = 1580 in.-kips ⎝ 28.7 ⎠ and the design strength is φM n = 0.9 (1580) = 1420 in.-kips > 623 in.-kips So the coped beam has sufficient flexural strength.

482 Chapter 11

Simple Connections Angles Step 15:

Check the angles for shear rupture using Equation J4-4. The net area of the angle on the vertical shear plane is Anv = ( l − n ( d h + 1 16 ) ) ta = ( 8.5 − 3 ( 7 8 + 1 8 ) ) ( 3 8 ) = 2.06 in.2 and the design strength for two angles is φVn = 2 ( φ0.6 Fu Agv ) = 2 0.75 ( 0.6 ( 58 )( 2.06 ) ) = 108 kips > 83.0 kips

(

)

So the angles are adequate for shear rupture. Step 16:

Check the angles for shear yield using Equation J4-3. The gross area of the angle on the vertical shear plane is Agv = 8.5 ( 3 8 ) = 3.19 in.2

and the design strength for two angles is φVn = 2 ( φ0.6 Fy Agv ) = 2 1.0 ( 0.6 ( 36 )( 3.19 ) ) = 138 kips > 83.0 kips

(

)

So the angles are also adequate for shear yield. Step 17

Check the angles for block shear. The equations for block shear in the angle are the same as those for the web and as presented in Section 10.10.5. First calculate the areas, 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜ 1.0 − ( 7 8 + 1 8 ) ⎟ ( 3 8 ) = 0.188 in.2 2 ⎝ ⎠ Agv = ltw = 7.25 ( 3 8 ) = 2.72 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) ( 3 8 ) = 1.78 in.2 Determine the tension rupture strength Fu Ant = 58 ( 0.188) = 10.9 kips Consider shear yield and shear rupture, and select the least nominal strength; thus, 0.6 Fy Agv = 0.6 ( 36 )( 2.72 ) = 58.8 kips 0.6 Fu Anv = 0.6 ( 58 )(1.78 ) = 61.9 kips

Selecting the shear yield term and combining it with the tension rupture

Chhapter 11

Siimple Connecctions 483

terrm gives a co onnection bloock shear desiign strength oof the doublee angle— ag gain, Ubs = 1.0 0 for this casee of uniform ttensile stress ddistribution— —of φRn = 2 ( 0.75 ( 558.8 + 10.9 ) ) = 105 kips > 83.0 8 kips Step S 18:

Prresent the finaal connection design. Th he three-bolt connection, rrevised as shoown in Figurre 11.3, is adeequate to caarry the impossed load of 833.0 kips.

Figure 11.3 1 Final Connection C Deesign for Exam mple 11.1

EXAMPLE E 11.1b Bolted-Bollted Double-An ngle Shear Connection by ASD

Goal: G

Deesign a bolted d-bolted doubble-angle sheaar connection for a W18×550 beam.

Given: G

Th he W18×50 beam is connnected to tthe web of a W24×207 and the co onnection musst provide a rrequired strenngth Ra = 55.0 kips. The bbeam and co olumn are A9 992 and the 3-1/2×3-1/22×3/8 angles are A36. T The beam flaange is coped d 2 in. Use 7/88 in. A325-N N bolts in stanndard holes inn the legs on n the beam weeb and short slots on the ooutstanding leegs. The basicc starting geeometry is giv ven in Figure 11.2.

SOLUTION N Step S 1:

Bo olt Shear Deetermine the number n of boolts required bbased on the shear rupturee strength off the bolts. Frrom Manual Table T 7-1, thee design shearr strength per bolt is rn Ω = 16.22 kips Beecause the bo olts are in douuble shear, ns = 2 and the design shearr strength off a single bolt is ns rn Ω = 2 (16.2) = 32.4 kips Vaalues for dou uble shear aree also found in Manual T Table 7-1. Beecause of rou unding, the number n in thee table is slighhtly greater aat 32.5 kips. The total nu umber of boltss required is

484 Chapter 11

Simple Connections

Ra 55.0 = = 1.70 ns ( rn Ω ) 32.4 Therefore, based only on the limit state of bolt shear, try two bolts. N=

Bolt Strength at Holes in Beam Web Determine bolt bearing and tearout strength on the beam web. Step 2:

For the W18×50 beam, tw = 0.355 in. For the two-bolt connection, the top bolt is 1.25 in. from the edge of the beam cope and the second bolt is spaced 3.0 in. from the first. For the top bolt, the clear distance is 1 1 lc = lcv − d h = 1.25 − (7/8 + 1/16) = 0.781 < 2(7/8) = 1.75 2 2 Since the clear distance is less than 2db, tearout controls over bearing. The nominal bolt strength for tearout is Rn = 1.2lc tFu = 1.2 ( 0.781)( 0.355)( 65) = 21.6 kips and the allowable strength is Rn Ω = 21.6 2.00 = 10.8 kips Since this is less than the allowable shear strength of the bolt in double shear it will control the strength of this bolt. For the second bolt lc = s − dh = 3.0 − ( 7 8 + 1 16) = 2.06 > 2 ( 7 8) = 1.75 in. Since the clear distance is greater than 2db, bearing controls over tearout. Thus, the nominal bolt bearing strength is Rn = 2.4dtFu = 2.4 ( 7 8)( 0.355)( 65) = 48.5 kips and the allowable strength for bearing on the beam web is Rn Ω = 48.5 2.00 = 24.3 kips Since this is less than the shear strength of the bolt in double shear, it will control the strength of this bolt. Therefore, the connection allowable strength, if only considering the beam web, is based on one bolt limited by tearout and one bolt limited by bearing. Thus, Rn Ω = (10.8 + 24.3) = 35.1 < 55.0 kips and the two-bolt connection cannot support the load. Step 3:

Determine the number of bolts required considering bearing and tearout on the web.

Chapter 11

Simple Connections 485

Adding a third bolt spaced at 3.0 in., which means that bearing controls over tearout, gives a connection allowable strength for bolt bearing and tearout of Rn Ω = (10.8 + 2 ( 24.3) ) = 59.4 > 55.0 kips Therefore a three bolt connection will be considered further. Bolt Strength at Holes in Angle Leg Determine bolt bearing and tearout strength on the A36 angle. The angle Step 4: legs on the beam and on the supporting member are identical and each angle takes one half of the load.

For the bottom edge bolt, the clear distance is 1 1 lc = lev − d h = 1.25 − ( 7 8 + 1 16 ) = 0.781 < 2 ( 7 8 ) = 1.75 in. 2 2 Since the clear distance is less than 2db, tearout controls over bearing. The nominal bolt strength for tearout is Rn = 1.2lc tFu = 1.2 ( 0.781)( 3 8 )( 58 ) = 20.4 kips

and the allowable strength is Rn Ω = 20.4 2.00 = 10.2 kips

Since this is less than the allowable shear strength of the bolt in single shear, it will control the strength of this bolt For the second and third bolts, the clear distance is

lc = s − d h = 3.0 − ( 7 8 + 1 16 ) = 2.06 > 2 ( 7 8 ) = 1.75 in.

Since the clear distance is greater than 2db, bearing controls over tearout. The nominal bolt strength for bearing is Rn = 2.4 dtFu = 2.4 ( 7 8 )( 3 8 )( 58 ) = 45.7 kips

and the allowable strength for bearing is Rn Ω = 45.7 2.00 = 22.9 kips

Since this is greater than the allowable shear strength of the bolt in single shear, bolt shear will control the strength of these two bolts Connection Strength at Beam Determine the controlling strength for each bolt at the beam web Step 5:

Top bolt Bolt double shear Web tearout Angle bearing ×2 Middle bolt Bolt double shear

32.4 kips 10.8 kips 45.8 kips 32.4 kips

486 Chapter 11

Simple Connections Web bearing Angle bearing ×2 Bottom bolt Bolt double shear Web bearing Angle tearout ×2

24.3 kips 45.8 kips 32.4 kips 24.3 kips 20.4 kips

Therefore, the strength of the connection at the beam web is φRn = 10.8 + 24.3 + 20.4 = 55.5 > 55.0 kips Bolt Strength at Holes in Supporting Member Consider the outstanding legs of the angles attached to the supporting Step 6: member.

In this case, the bolts are in single shear but there are twice as many bolts, so the load per bolt is half of the load for each of the bolts in the beam web. If the supporting member thickness is at least one-half of the beam web thickness and the strengths are the same, the bolts in the supporting member will be satisfactory. For the W24×207, clear distances result in all three bolts being controlled by bearing over tearout. Thus, with tw = 0.870 in., Rn = 2.4dtFu = 2.4 ( 7 8)( 0.870)( 65) = 119 kips and the allowable strength is Rn Ω = 119 2.00 = 59.5 kips Since this is greater than the single shear strength of the bolt, rn Ω = 16.2 kips , connection strength on the supporting member will be controlled by bolt shear. For six bolts in single shear Rn Ω = 6 (16.2) = 97.2 kips which is greater than the required strength of 55.0 kips so the bolts in the supporting member are adequate Connection Strength at Supporting Member Determine the controlling strength for each bolt at the supporting member Step 7: Top bolt Bolt single shear 16.2 kips Web bearing 59.5 kips Angle tearout 10.2 kips Middle bolt Bolt single shear 16.2 kips Web bearing 59.5 kips Angle bearing 22.9 kips Bottom bolt Bolt single shear 16.2 kips Web bearing 59.5 kips Angle bearing 22.9 kips

Chapter 11

Simple Connections 487

Therefore, the strength of the connection at the beam web is Rn Ω = 10.2 + 16.2 + 16.2 = 42.6 > 55.0 2 =27.5 kips Step 8:

Evaluate the minimum depth of the connection. The beam web connection should be at least half the depth of the beam web, measured as the distance between the fillets, T, given in Manual Table 1-1. This requirement is to prevent twisting of the simple supports. For W18×50, T = 15-1/2 in., so the minimum angle depth should be 7-3/4 in. Thus, the 8-1/2 in. long angle will provide an acceptable connection depth and will also fit within the 15.5 in. available flat length.

Remaining Beam Limit States Check shear yield of the beam web. Step 9:

This is a check that should be carried out during the beam design process. At the point of connection design it is too late to find out that the beam will not be adequate. From Manual Table 3-2, Vn Ω = 128 kips > 55.0 kips Step 10:

Check shear yield of the beam web at the cope This check is carried out according to Section J4.2 where the full area of the remaining web is stressed to Fy and Ω = 1.5. Thus Vn Ω = 0.6 Fy ht w Ω = ( 0.6 ( 50 ) ) (18.0 − 2.0 )( 0.355 ) 1.50 = 114 kips > 55.0 kips

Therefore shear yield strength at the cope is adequate Step 11:

Check shear rupture of the beam web at the cope This check is carried out according to Section J4.2 where the effective area, Ae, the full area of the remaining web less the bolt holes, is stressed to Fu and Ω = 2.0. Thus Vn Ω = 0.6 Fu Ae Ω ⎛ ⎛ 7 1 ⎞⎞ = ( 0.6 ( 65 ) ) ⎜ 18.0 − 2.0 − 3 ⎜ + ⎟ ⎟ ( 0.355 ) 2.00 ⎝ 8 8 ⎠⎠ ⎝ = 90.0 kips > 55.0 kips

Therefore shear rupture at the cope is adequate Step 12:

Check the beam web for block shear. The equations for block shear rupture are found in Specification Section J4.3 and were presented in Section 10.10.5. First calculate the areas, remembering to account for the 1/4 in. beam

488 Chapter 11

Simple Connections length under run tolerance in the tension area calculation, so that leh = 2.0 – 0.25 = 1.75 in.: 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜ 1.75 − ( 7 8 + 1 8 ) ⎟ (0.355) = 0.444 in.2 2 ⎝ ⎠ Agv = ltw = 7.25(0.355) = 2.57 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) (0.355) = 1.69 in.2

Determine the tension rupture strength Fu Ant = 65 ( 0.444 ) = 28.9 kips Consider shear yield and shear rupture and select the least nominal strength; thus, 0.6 Fy Agv = 0.6 ( 50 )( 2.57 ) = 77.1 kips 0.6 Fu Anv = 0.6 ( 65 )(1.69 ) = 65.9 kips

Selecting the shear rupture term and combining it with the tension rupture term gives the connection block shear allowable strength, and recalling that Ubs = 1.0 for the case of uniform tensile stress distribution, we have Rn Ω = ( 65.9 + 1.0 ( 28.9 ) ) 2.00 = 47.4 < 55.0 kips Thus, the given three-bolt connection is not adequate, with block shear being the critical limit state to this point in our calculations. Step 13:

Revise the connection to meet the block shear strength requirements. Consideration could be given to increasing the number of bolts and thereby increasing the length of the connection. However, because bolt shear required only two bolts, this would not be a particularly economical solution. If the connection were to be lowered on the beam end so that the distance from the center of the top bolt to the edge of the cope were 2.5 in., the connection would have more block shear strength. It should also be noted that the clear end distance for the top bolt in the beam web would increase and thus increase the tearout strength of that top bolt. There is no need to consider that increase since at best it could contribute to reducing the number of bolts and this would have a negative impact on the block shear calculations being addressed in this step. Thus, the new shear areas become Agv = ltw = 8.50(0.355) = 3.02 in.2

Chapter 11

Simple Connections 489

Anv = ( l − ( n − 0.5)( d h + 1 16 ) ) tw = ( 8.50 − 2.5 ( 7 8 + 1 8) ) (0.355) = 2.13 in.2 and the nominal shear yield and rupture strengths become 0.6 Fy Agv = 0.6 ( 50 )( 3.02 ) = 90.6 kips 0.6 Fu Anv = 0.6 ( 65 )( 2.13 ) = 83.1 kips

Selecting the shear rupture term and combining it with the tension rupture term gives the resulting block shear allowable strength as Rn Ω = ( 83.1 + 1.0 ( 28.9 ) ) 2.00 = 112 2.00 = 56.0 > 55.0 kips Step 14:

Check the flexural strength of the coped beam. It is a good idea to check this limit state during the initial design of the beam. It should be anticipated during the beam design stage that a coped connection will be required, and it is at that stage that a change in beam section can most readily be accommodated. Flexural strength of the coped beam is not addressed in the Specification directly but is covered in Part 9 of the Manual. The moment in the coped beam is taken as the shear force times the eccentricity from the face of the support to the edge of the cope, taken as cope + setback = 7.0 + 0.50 = 7.5 in. in this example. Thus, M a = Ra e = 55.0 ( 7.5) = 413 in.-kips (Manual 9-5) To determine the flexural strength of the coped beam, the web slenderness must be checked. This requires determination of the plate buckling coefficient, k, and the buckling adjustment factor, f. The web slenderness of this coped beam is ho t w = 16.0 0.355 = 45.1 The limiting web slenderness is λ p = 0.46

k1 E Fy

(Manual 9-12)

where k1 = fk

(Manual 9-10)

The buckling adjustment factor, f, is a function of the ratio of the cope length, c, to the beam depth, d, and the plate buckling coefficient, k, is a function of the ratio of cope length, c, to the reduced beam depth, ho. For this example, with c d = 7.0 18.0 = 0.389 ≤ 1.0

f =

2c 2(7.0) = = 0.778 18.0 d

(Manual 9-14a)

and with c ho = 7.0 16.0 = 0.438 ≤ 1.0 1.65

⎛h ⎞ k = 2.2 ⎜ o ⎟ ⎝ c ⎠

Thus,

1.65

⎛ 16.0 ⎞ = 2.2 ⎜ ⎟ ⎝ 7.0 ⎠

= 8.61 (Manual 9-13a)

490 Chapter 11

Simple Connections

k1 = fk = 0.778 ( 8.61) = 6.70 ≥ 1.61

(Manual 9-10)

and

λ p = 0.46

6.70 ( 29,000 ) k1 E = 0.46 = 28.7 Fy 50

Since λ is between λp and 2λp the net elastic and net plastic section modulus will be needed. The net elastic section modulus is taken from Manual Table 9-2. With the depth of the cope, dc = 2.0 in., Snet = 23.4 in.3 The net plastic section modulus is not provided in the Manual so it must be determined. Following the procedure discussed in Chapter 6, the net plastic section modulus for the tee shaped beam that remains after the cope is found to be Z net = 42.4 in.3 and from ⎛ λ ⎞ M n = M p − (M p − M y )⎜ − 1⎟ ⎝ λp ⎠

with

(Manual 9-7)

M p = Fy Z net = 50 ( 42.4 ) = 2120 in.-kips M y = Fy S net = 50 ( 23.4 ) = 1170 in-kips

The nominal moment strength at the cope is ⎛ 45.1 ⎞ M n = 2120 − ( 2120 − 1170 ) ⎜ − 1 ⎟ = 1580 in.-kips ⎝ 28.7 ⎠ and the design strength is M n Ω = 1580 1.67 = 946 in.-kips > 413 in.-kips So the coped beam has sufficient flexural strength. Angles Step 15:

Check the angles for shear rupture using Equation J4-4. The net area of the angle on the vertical shear plane is Anv = ( l − n ( d h + 1 16 ) ) ta = ( 8.5 − 3 ( 7 8 + 1 8 ) ) ( 3 8 ) = 2.06 in.2 and the design strength for two angles is Vn Ω = 2 ( 0.6 Fu Agv Ω ) = 2 ( 0.6 ( 58 )( 2.06 ) ) 2.00 = 71.7 kips > 55.0 kips

(

So the angles are adequate for shear rupture.

)

Simple Connections 491

Chapter 11 Step 16:

Check the angles for shear yield using Equation J4-3. The gross area of the angle on the vertical shear plane is Agv = 8.5 ( 3 8 ) = 3.19 in.2

and the design strength for two angles is Vn Ω = 2 ( 0.6 Fy Agv Ω ) = 2 ( 0.6 ( 36 )( 3.19 ) ) 1.5 = 91.9 kips > 55.0 kips

(

)

So the angles are also adequate for shear yield. Step 17

Check the angles for block shear. The equations for block shear in the angle are the same as those for the web and as presented in Section 10.10.5. First calculate the areas, 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜ 1.0 − ( 7 8 + 1 8 ) ⎟ ( 3 8 ) = 0.188 in.2 2 ⎝ ⎠ Agv = ltw = 7.25 ( 3 8 ) = 2.72 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = ( 7.25 − 2.5 ( 7 8 + 1 8 ) ) ( 3 8 ) = 1.78 in.2 Determine the tension rupture strength Fu Ant = 58 ( 0.188) = 10.9 kips Consider shear yield and shear rupture, and select the least nominal strength; thus, 0.6 Fy Agv = 0.6 ( 36 )( 2.72 ) = 58.8 kips 0.6 Fu Anv = 0.6 ( 58 )(1.78 ) = 61.9 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection block shear design strength of the double angle— again, Ubs = 1.0 for this case of uniform tensile stress distribution—of Rn Ω = 2 ( ( 58.8 + 10.9 ) 2.00 ) = 69.7 kips > 55.0 kips Step 18:

Present the final connection design. The three-bolt connection, revised as shown in Figure 11.3, is adequate to carry the imposed load of 55.0 kips.

4992 Chapter 11 1

Simple Connections C

Figure 11.4 Welded-Bo olted Double--Angle Conneection

111.4

DOUBLE-ANGLE E CONNEC CTIONS: WELDED-BO W OLTED The double--angle shear connection c caan also be connstructed by ccombining weelding and boolting. In this case thee angles are welded w to the beam b web, ass shown in Figgure 11.4. The limit states to o be considerred are 1. Bolts a. Shear S rupture 2. Weld a. Rupture R 3. Beam a. Shear S yielding g of the web b. Block B shear on o coped beam m web c. Coped C beam flexural f strenngth d. Web W strength at the weld Angles 4. a. Bolt B bearing and a tearout onn angles b. Shear S rupture c. Shear S yield d. Block B shear Supportin ng member 5. a. Bolt B bearing and a tearout t were nott considered for the bolted-bolted connnection from Section The limit states that 11.3 are those associated d with the welld. These incllude block shhear of the beaam web as a rresult of the welded connection; weld w rupture, which is infl fluenced by thhe eccentricitty of the force on the weld group; and the stren ngth of the beam web at thee weld. Blocck shear for a welded con nnection diffeers only slighhtly from block shear for a bolted connection. The differen nce is in the lack of holess to be deduucted in deterrmining the nnet area. Thus, the neet shear area and a gross sheear area are thhe same. As a result, yieldding is the conntrolling shear term in n the block sh hear equation for this type of welded coonnection. Welld rupture is a much more complex limiit state to incoorporate in thhis type of connnection design. Chapter 10 discu ussed the stren ngth of a welld loaded at iits centroid aand at any angle. The welds in thee double-anglle connection n are loaded pparallel to thhe length on oone side of thhe angle and perpend dicular to theeir lengths on n the other ttwo sides. Unnfortunately, these welds are not loaded throu ugh their centtroid, so the simplified appproach to com mbining them,, previously sshown in Chapter 10, cannot be used. u The Ma anual uses thhe instantaneeous center of rotation meethod to determine weld w strength in cases like this. This appproach accouunts for the looading at an angle to the weld as well as the eccentricity off the load to tthe weld grouup. Figure 111.5 shows a C C-shaped he geometric variables v labeeled. In the tyypical connecttion design, thhe geometry can be weld with th

Chhapter 11

Siimple Connecctions 493

Figuree 11.5 C-Shhaped Weld G Group set and Manual M Tablee 8-8 can be used to deterrmine the welld group strenngth. The appplication of this tablee is shown in Example 11.2 2. The T beam web strength at the weld is aalso a bit diffi ficult to calcullate. The usuaal approach is to deteermine the tottal strength off the weld annd then proporrtion that forcce to the webb based on a one-inch length of weeb and one-incch length of w weld. This, too, is illustrateed in Examplee 11.2.

1 Connecction for Exam mple 11.2 Figure 11.6

EXAMPLE E 11.2 Welded-Bo olted Double-An ngle Shear Connection

Goal: G

Deetermine the available streength of the w welded-boltedd connection shown in Figure 11.6a.

Given: G

Deetermine the design strenngth and alloowable strenggth of the coonnection sh hown in Figu ure 11.6a forr the three nnew limit staates discussedd for the weelded-bolted double-anglee connection. For the givven 3/16 in. w weld use E7 70 electrodes..

494 Chapter 11 SOLUTION

Simple Connections Step 1:

Determine the nominal strength for the limit state of block shear in the beam web. For the tension area, the length is found by taking the 3.0 in. angle leg and subtracting the 1/2 in. setback and the 1/4 in. potential beam under run tolerance. Thus,

Ant = ( la − 0.50 − 0.25) tw = ( 3.0 − 0.75)( 0.255) = 0.574 in.2

For the gross shear area, the angle is 8.5 in. long and set down from the cope 1/2 in. Thus, the length for block shear calculation is l = 8.5 + 0.5 = 9.0 in.

Agv = ltw = 9.0 ( 0.255) = 2.30 in.2

Therefore, with Ubs = 1.0, the nominal block shear strength is Rn = 0.6 Fy Agv + Ubs Fu Ant = 0.6 ( 50 )( 2.30 ) + 1.0 ( 65)( 0.574 ) = 106 kips For LRFD Step 2:

The design block shear strength is φRn = 0.75(106) = 79.5 kips

For ASD Step 2:

The allowable block shear strength is Rn Ω = 106 2.00 = 53.0 kips

Step 3:

Determine the nominal strength for the limit state of weld rupture. The geometry of the weld is given in Figure 11.6b. The minimum angle thickness is the weld plus 1/16 in. Manual Equation 9-2, discussed in Chapter 10, shows that when welding with properly matched electrodes, this minimum angle thickness will always have sufficient strength to match the weld rupture strength. Thus, the 5/16 in. angle has sufficient strength to support the 3/16 in. weld. Using terms as defined for Manual Table 8-8, the angle is 8.5 in. long, so the weld length is l = 8.5 in. The leg of the angle is 3.0 in. and the weld length is kl= 3.0 – 1/2 in. setback – 1/4 in. under run = 2.25 in. Thus, 2.25 k= = 0.265 8.5 From Manual Table 8-8, for the load parallel to the weld along the angle toe, loading angle = 0°, the location of the weld centroid can be determined. Enter the table with k = 0.265 and interpolate for x from the values at the bottom of the table, which yields x = 0.0466. With this, the weld centroid is determined as xl = 0.0466 ( 8.5) = 0.396 in. The eccentricity of the force is then determined as

Chapter 11

a=

Simple Connections 495

ex (3.0 − 0.396) = = 0.306 l 8.5

Using this value for a and the previously determined value for k, the value of C can be determined from the table as C = 2.59. As indicated in the table, the nominal strength of the weld group is then Rn = CC1 Dl

C has been determined above. C1 represents the electrode strength and is 1.0 for the E70XX electrodes used here. D is the number of sixteenths of an inch in the fillet weld size, and l is the defined length of the weld group. Thus, for each angle with this weld Rn = CC1Dl = 2.59 (1.0 )( 3)( 8.5) = 66.0 kips For LRFD Step 4: Step 5:

The design strength of the welds on the double angle connection is φRn = 2 ( φRn ) = 2 ( 0.75 ( 66.0 ) ) = 99.0 kips Determine the design shear rupture strength for the beam web at the weld. Using Manual Equation 9-2 determine the minimum web thickness to balance the 3/16 in. weld on both sides of the web. 6.19 D 6.19 ( 3) tmin = = = 0.286 in. Fu 65 Since the beam web thickness tw < tmin, an approach that determines an effective weld length is used. The design rupture strength of the 3/16 in. weld of unit length on both sides of the web, using the weld design strength determined in Chapter 10, is twice the weld strength of a single weld, thus design weld strength per unit length = 2 ( 3)(1.392) = 8.35 kips/in. Using this strength the effective length of the weld is weld strength divided by strength per inch. Thus, 99.0 effective weld length = = 11.9 in. 8.35 The design shear rupture strength of a unit length of the beam web, using Equation J4-4, is φ0.6 Fu tw = 0.75 ( 0.6 ( 65)( 0.255) ) = 7.46 kips/in.

Step 6:

Therefore, the beam web design rupture strength at the weld, using the weld effective length, is φRn = (11.9)(7.46) = 88.8 kips Determine the controlling limit state for the three limit states considered in this example. The connection is controlled by the limit state of block shear. Therefore, φRn = 79.5 kips

496 Chapter 11

Simple Connections For ASD Step 4: Step 5:

The allowable strength of the welds on the double angle connection is Rn Ω = 2 ( Rn Ω ) = 2 ( 66.0 2.00 ) = 66.0 kips

Determine the allowable shear rupture strength for the beam web at the weld. Using Manual Equation 9-2 determine the minimum web thickness to balance the 3/16 in. weld on both sides of the web. 6.19 D 6.19 ( 3) tmin = = = 0.286 in. Fu 65 Since the beam web thickness tw < tmin, an approach that determines an effective weld length is used. The allowable rupture strength of the 3/16 in. weld of unit length on both sides of the web, using the weld allowable strength determined in Chapter 10, is twice the weld strength of a single weld, thus

allowable weld strength per unit length = 2 ( 3)( 0.928) = 5.57 kips/in. Using this strength, the effective length of the weld is weld strength divided by strength per inch. Thus, 66.0 effective weld length = = 11.9 in. 5.57 The allowable shear rupture strength of a unit length of the beam web, using Equation J4-4, is 0.6 Fu t w Ω = 0.6(65)(0.255) 2.00 = 4.97 kips/in. Therefore, the beam web allowable rupture strength at the weld, using the weld effective length, is Rn Ω = (11.9)(4.97) = 59.1 kips Step 6:

Determine the controlling limit state for the three limit states considered in this example. The connection is controlled by the limit state of block shear. Therefore, Rn Ω = 53.0 kips

11.5 DOUBLE-ANGLE CONNECTIONS: BOLTED-WELDED This connection is shown in Figure 11.7, where the angles are bolted to the beam web and welded to the supporting member. The beam has a cope on the tension flange to permit the beam to be inserted in the space between the double angles, which are shop-welded to the supporting member, like a knife being inserted into its sheath. This connection might be used on a beam-tocolumn flange connection, but it would not be used as a beam-to-girder connection because of the interference of the girder flange.

Chhapter 11

Siimple Connecctions 497

Figure 11.7 1 Doublee-Angle Bolteed-Welded Coonnection for Example 11.3 For F this conn nection, the new limit statee of weld strrength on thee outstanding legs of the angles must m be consid dered. In addition, considerration must bbe given to thee flexural streength of the coped beeam with the tension t flangee removed. The T strength of the weld on o the outstannding leg of the angle is ddetermined byy an elastic method that t assumes a uniform sheear stress on tthe weld and a linearly varrying tension stress over the loweer five-sixths of the angle, with the topp one-sixth oof the angles in compresssion against each oth her through th he beam web b. The resultiing weld streess is then deetermined byy taking the square ro oot of the sum m of the squ uares of the teension and shhear stresses. Part 10 of tthe Manual gives thee resulting weeld strength ass ⎤ ⎡ ⎡ ⎤ ⎥ ⎢ ⎥ ⎢ 1.392 1 Dl ⎥ 0.928 Dl ⎥ Manual 10-1) (LRFD) ( and Rn Ω = 2 ⎢ (ASD) (Ma φRn = 2 ⎢ ⎢ ⎢ 12.96e 2 ⎥ 12.96e 2 ⎥ ⎥ ⎢ 1+ ⎥ ⎢ 1+ l2 l2 ⎦ ⎣ ⎦ ⎣ where l is the length of o the angle an nd e is the wiidth of the outtstanding leg..

EXAMPLE E 11.3 Bolted-Wellded Double-An ngle Shear Connection

Goal: G

Deetermine the available streength of the w welds for a bbolted-weldedd doublean ngle connectio on.

Given: G

Th he bolted-weelded double -angle conneection is shoown in Figuure 11.7. Asssume 1/4 in. welds with a 1/2 in. returnn on top as shhown.

SOLUTION N

For F LRFD L Step S 1:

Deetermine the design d strenggth. With W the given information, D = 4, l= 8.55 in., and e = 33.0 in.; thus, from Eq quation 10-1aa ⎤ ⎡ ⎡ ⎤ ⎥ ⎢ ⎢ ⎥ 1 .392 Dl ⎥ 1.392(4)((8.5) ⎥ ⎢ ⎢ φRn = 2 =2 ⎢ 2 ⎥ ⎢ 12.96e 2 ⎥ ⎢ 1 + 12.966(3) ⎥ ⎢ 1+ ⎥ 2 2 (8.55) ⎥⎦ l ⎢⎣ ⎣ ⎦ = 58.5 kkips

498 Chapter 11

Simple Connections For ASD Step 1:

Determine the allowable strength. With the given information, D = 4, l= 8.5 in., and e = 3.0 in.; thus, from Equation 10-1b ⎡ ⎢ 0.928 Dl Rn Ω = 2 ⎢ ⎢ 2 ⎢ 1 + 12.96e ⎢⎣ l2

⎡ ⎤ ⎤ ⎢ ⎥ ⎥ ⎥ = 2 ⎢ 0.928(4)(8.5) ⎥ ⎢ ⎥ 2 ⎥ ⎢ 1 + 12.96(3) ⎥ ⎥ ⎢ ⎥⎦ (8.5) 2 ⎥⎦ ⎣

= 39.0 kips

11.6 DOUBLE ANGLE CONNECTIONS: WELDED-WELDED The double-angle welded-welded connection is not a particularly common connection because it requires temporary support for field welding, which usually means bolts are used. This tends to favor a bolted field joint over a welded field joint. Nonetheless, if it is desirable for a particular situation, however, the limit states are those that have already been discussed. The procedures are the same and all potential limit states must be checked.

11.7 SINGLE-ANGLE CONNECTIONS The single-angle connections shown in Figure 11.1b represent a bolted-bolted connection and a bolted-welded connection. In both cases, the connection is shown as a beam-to-girder connection. This is a particularly efficient connection because it eliminates erection problems when transverse beams frame into a girder at the same point on opposite sides of the girder web. It is also efficient because it has fewer parts than the double-angle connection. This connection is growing in popularity with both fabricators and erectors. It aids in erection efficiency because the beam can be installed from one side with the angle pre-attached to the girder. The disadvantages of this connection are that the components, such as angles, bolts, and welds, are larger than for the double-angle connection. For this connection, the bolts in the beam web are in single shear and, if the controlling limit state were bolt shear, it would require twice as many bolts than if it were a double-angle connection. Because all of the beam force must pass through only one angle, the angle likely needs to be larger. Greater weld size and weld length are also required. However, the single angle often is still the best choice in many situations, particularly when limit states such as block shear might otherwise control the strength of the connection. The single-angle connection easily behaves as a simple shear connection, as it is modeled in analysis, because it is more flexible than the previously considered double-angle connections. Because of this increased flexibility, however, this connection is not recommended for laterally unsupported beams that rely on their end connections for lateral stability. The limit states to be checked for the single-angle connection are the same as those for the double-angle connection, with some modifications and additions. The major modifications involve the eccentricities induced in the connecting elements. For the supported beam, as long as there is only one row of bolts, no eccentricities are considered, and this portion of the connection is treated the same as for the double-angle connections. For the outstanding leg, the bolts or welds must be designed to account for the connection eccentricity. This eccentricity also adds the limit states of flexural yielding and flexural rupture for the outstanding leg of the angle. Figure 11.8a and b shows how the eccentricity is measured for a bolted and a welded outstanding leg. Note that the eccentricity is measured, in both cases, from the centerline of the supported beam web. Figure 11.8c and d illustrates, with a bold line, the location and cross

Chhapter 11

Siimple Connecctions 499

Figure 11.8 1 Single-Angle Conneection Eccentrricities f the momeent in the ang gle for the lim mit states of flexural yielding (Figure 11.8c) and section for flexural rupture r (Figurre 11.8d). The T limit staate of flexuraal yielding iss calculated bbased on thee plastic mom ment of the 2 element. Thus, the pllastic section modulus forr the angle leeg without hooles, Z = t pl 4 , is used and the nominal n mom ment strength is i given as M n = Rn ea = Fy Z For the flexural fl yieldiing limit statee, φ = 0.9 and Ω = 1.67. For F the limit state of flexu ural rupture, the plastic ssection modulus of the neet section is needed. This T can be either e determined by calcuulation or obttained from M Manual Tablee 15-3. The resulting nominal mom ment strength h is M n = Rn ea = Fu Z net fl rupturre limit state, φ = 0.75 andd Ω = 2.00. For the flexural To T account fo or the effects of load eccenntricity on thee bolt group, an equivalentt number of bolts mu ust be determiined. This can n be accompllished using M Manual Tablee 7-6, and thee procedure is illustraated in Examp ple 11.4. To account for thhe eccentricitty effects on the weld grouup, Manual Table 8-11 can be ussed. This tablle is for an L L-shaped weld group withh the welds oon only two sides. Th he top of the leg is kept frree except forr a small welld return to eensure sufficient rotation capacity.. Application of this table is i also demonnstrated in Exxample 11.4.

Figure 11.9 1 Single-Angle Conneection for Exaample 11.4

500 Chapter 11

Simple Connections

EXAMPLE 11.4a Goal: Single-Angle Shear Connection by Given: LRFD

SOLUTION

Determine the design strength of a bolted-bolted and a bolted-welded single-angle connection. A single-angle connection is shown in Figure 11.9 for the bolted outstanding leg case (Figure 11.9b) and the welded outstanding leg case (Figure 11.9c). The angle is A36 steel, L3-1/2×3-1/2×3/8×1 ft – 0 in. The bolts are 3/4 in. A325-N, the weld is 3/16 in. with E70 electrodes and the beam is a W16×31, A992 steel.

Part (a): Supported Beam Determine the bolt shear rupture strength. From Manual Table 7-1, Step 1: φrn = 17.9 kips Step 2:

Determine the bolt bearing and tearout strength on the angle and compare to bolt shear. The bottom bolt is 1.5 in. from the bottom of the angle, with the remaining bolts spaced at 3.0 in. Determine the clear distances for each of the bolts. For the bottom bolt 1 1 lc = lev − d h = 1.5 − ( 3 4 + 1 16 ) = 1.09 < 2 ( 3 4 ) = 1.5 in. 2 2 Thus, tearout controls over bearing, and the nominal bolt strength is Rn = 1.2lc tFu = 1.2 (1.09 )( 0.375)( 58) = 28.4 kips The design bolt strength for tearout is φRn = 0.75 ( 28.4 ) = 21.3 kips Since this is greater than the design shear strength of the bolt, shear strength controls for this bolt. For the other bolts lc = s − dh = 3.0 − ( 3 4 + 1 16) = 2.19 > 2 ( 3 4) = 1.5 in. Therefore, bearing controls over tearout, and the nominal bolt strength is Rn = 2.4dtFu = 2.4 ( 3 4)( 0.375)( 58) = 39.2 kips The design bolt strength for bearing is φRn = 0.75 ( 39.2 ) = 29.4 kips Since this is greater than the design shear strength of the bolt, shear strength controls for these bolts also. Thus, for the four bolts in the angle the design strength is limited by bolt shear.

Step 3:

Determine the bolt bearing and tearout strength on the supported beam web

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Simple Connections 501

with tw = 0.275 in.. Because the clear distance is greater than two times the bolt diameter for each of the four bolts, the nominal bearing strength is Rn = 2.4dtFu = 2.4 ( 3 4)( 0.275)( 65) = 32.2 kips and the design bearing strength φRn = 0.75 ( 32.2 ) = 24.2 kips Since this is greater than the design shear strength of the bolt, shear strength controls for these bolts also. Thus, for the four bolts in the beam web the design strength is limited by bolt shear. Step 4:

Determine the connection design shear strength based on the bolts. Since all bolts are limited by bolt shear rupture φRn = 4 (17.9 ) = 71.6 kips

Step 5:

Determine the shear yield strength of the angle. The gross shear area of the angle is

Agv = lta = 12.0 ( 0.375) = 4.50 in.2

and the design shear strength is φVn = φ0.6 Fy Agv = 1.0 ( 0.6 ( 36 )( 4.50 ) ) = 97.2 kips Step 6:

Determine the shear rupture strength of the angle. The net shear area of the angle is Anv = ( l − n ( d h + 1 16 ) ) ta = (12.0 − 4 ( 3 4 + 1 8) ) ( 0.375) = 3.19 in.2 and the design shear strength is φVn = φ0.6 Fu Anv = 0.75 ( 0.6 ( 58)( 3.19 ) ) = 83.3 kips

Step 7:

Determine the block shear strength of the angle. First calculate the areas. 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜ 1.25 − ( 3 4 + 1 8 ) ⎟ ( 0.375 ) = 0.305 in.2 2 ⎝ ⎠ Agv = ltw = 10.5 ( 0.375 ) = 3.94 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.5 − 3.5 ( 3 8 + 1 8 ) ) ( 0.375 ) = 2.79 in.2

502 Chapter 11

Simple Connections Determine the tension rupture strength Fu Ant = 58 ( 0.305) = 17.7 kips Consider shear yield and shear rupture and select the one with the least strength; thus, 0.6 Fy Agv = 0.6 ( 36 )( 3.94 ) = 85.1 kips 0.6 Fu Anv = 0.6 ( 58 )( 2.79 ) = 97.1 kips

Selecting the shear yield term and combining it with the tension rupture term, with Ubs = 1.0 for this case of uniform tensile stress distribution, gives φRn = 0.75 ( 85.1 + 1.0 (17.7 ) ) = 77.1 kips Step 8:

Determine the design strength of the leg attached to the supported member. The design strength is controlled by bolt shear, where φRn = 71.6 kips

Part (b): Bolted Outstanding Leg Check the eccentric shear using Manual Table 7-6 to account for the effects Step 9: of eccentricity on the bolt group.

The eccentricity of the load on the line of bolts is calculated as the bolt distance from the angle heel plus one-half of the beam web as shown in Figure 11.9b; thus, 0.275 ex = 2.25 + = 2.39 in. 2 For the four-bolt connection with bolt spacing of 3.0 in., controlled by bolt shear, Manual Table 7-6 with the load at an angle of zero gives, by interpolation, the effective number of bolts as C = 3.12. Therefore, φRn = Cφrn = 3.12 (17.9) = 55.8 kips As expected, the strength of the bolts in the outstanding leg is less than that in the leg on the supported beam because the outstanding leg must accommodate an eccentricity that is not present in the leg on the beam. Step 10:

Determine the flexural yield strength of the outstanding leg. The plastic section modulus is determined for the rectangle formed by the length and thickness of the angle, and the nominal moment strength is determined by multiplying the plastic section modulus by the yield stress; thus, Z =

and

(0.375)(12 2 ) = 13.5 in.3 4

M n = Fy Z = 36 (13.5) = 486 in.-kips

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Simple Connections 503

Because the moment is the shear force times the eccentricity, using the same eccentricity calculated for the bolt group, φM n 0.9(486) φRn = = = 183 kips e 2.39 Step 11:

Determine the flexural rupture strength of the outstanding leg. The net plastic section modulus is determined for the rectangle less the holes. Although this can readily be calculated, it can also be obtained from Manual Table 15-3, where Z net = 9.56 in.3 . Thus, M n = Fu Z net = 58 ( 9.56 ) = 554 in.-kips and φM n 0.75(554) φRn = = = 174 kips e 2.39

Step 12:

Determine the controlling limit state strength for the bolted outstanding leg. For the bolted outstanding leg, the strength is controlled by shear strength of the bolts with eccentricity, where φRn = 58.8 kips Because this is less than the value for the leg attached to the beam, this is the design strength of the bolted-bolted single-angle connection as long as the supporting member has sufficient bearing and tearout strength.

Part (c): Welded Outstanding Leg Step 13: Determine the eccentric weld rupture strength.

Manual Table 8-10 will be used to determine the eccentric weld rupture strength. The weld for the single-angle connection is applied to the vertical toe and the bottom edge of the angle, not the top, except for a small weld return. This ensures that the angle is sufficiently flexible to behave as a simple connection as modeled in the analysis of a simple beam. Manual Table 8-10 for load at an angle of zero degrees shows this weld on the top or the bottom. Thus, the table can be used for either arrangement because the geometry is the same whether the horizontal weld is at the top or bottom of the connection. Based on the dimensions given in Figure 11.9c, l = 12.0, kl = 3.5, and thus k = 0.292. The weld is given as a 3/16 in. weld with E70 electrodes. From Manual Table 8-10, interpolating between k = 0.2 and k = 0.3 yields x = 0.0336 . Therefore, using the definitions given in Table 8-10, the eccentricity is t 0.275 ex = kl + w − xl = 3.5 + − 0.0336 (12.0 ) = 3.23 in. 2 2 and e 3.23 a= x = = 0.269 l 12.0

504 Chapter 11

Simple Connections With a double interpolation between k = 0.2 and 0.3 and a = 0.25 and 0.30, the coefficient C is determined as C = 2.17 Therefore, with C1 = 1.0 for an E70XX electrode, the nominal weld strength is Rn = CC1Dl = 2.17 (1.0)( 3)(12.0) = 78.1 kips and the design strength is φRn = 0.75 ( 78.1) = 58.6 kips Step 14:

Determine the design strength for the limit state of flexural yielding. For the limit state of flexural yielding of the angle, the strength is the moment determined in step 10 divided by the eccentricity, ex = 3.23 in., φRn = 0.9 ( 486 ) 3.23 = 135 kips

Step 15:

Determine the controlling limit state strength for the welded outstanding legs. For the welded outstanding leg, the design strength is controlled by eccentric shear on the weld, where φRn = 58.6 kips Because this is less than the value for the leg attached to the beam, this is the design strength of the bolted-welded single-angle connection as long as the supporting member has sufficient rupture strength. For both bolted and welded outstanding leg arrangements the outstanding leg controlled the strength of the connection.

EXAMPLE 11.4b Goal: Single-Angle Shear Connection by Given: ASD

SOLUTION

Determine the allowable strength of a bolted-bolted and a bolted-welded single-angle connection. A single-angle connection is shown in Figure 11.9 for the bolted outstanding leg case (Figure 11.9b) and the welded outstanding leg case (Figure 11.9c). The angle is A36 steel, L3-1/2×3-1/2×3/8×1 ft – 0 in. The bolts are 3/4 in. A325-N, the weld is 3/16 in. with E70 electrodes and the beam is a W16×31, A992 steel.

Part (a): Supported Beam Determine the bolt shear rupture strength. From Manual Table 7-1, Step 1: rn Ω = 11.9 kips Step 2:

Determine the bolt bearing and tearout strength on the angle and compare to bolt shear. The bottom bolt is 1.5 in. from the bottom of the angle, with the remaining

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Simple Connections 505

bolts spaced at 3.0 in. Determine the clear distances for each of the bolts. For the bottom bolt 1 1 lc = lev − d h = 1.5 − ( 3 4 + 1 16 ) = 1.09 < 2 ( 3 4 ) = 1.5 in. 2 2 Thus, tearout controls over bearing, and the nominal bolt strength is Rn = 1.2lc tFu = 1.2 (1.09 )( 0.375)( 58) = 28.4 kips The allowable bolt strength for tearout is Rn Ω = 28.4 2.00 = 14.2 kips Since this is greater than the allowable shear strength of the bolt, shear strength controls for this bolt. For the other bolts lc = s − dh = 3.0 − ( 3 4 + 1 16) = 2.19 > 2 ( 3 4) = 1.5 in. Therefore, bearing controls over tearout, and the nominal bolt strength is Rn = 2.4dtFu = 2.4 ( 3 4)( 0.375)( 58) = 39.2 kips The allowable bolt strength for bearing is Rn Ω = 39.2 2.00 = 19.6 kips Since this is greater than the allowable shear strength of the bolt, shear strength controls for these bolts also. Thus, for the four bolts in the angle the allowable strength is limited by bolt shear. Step 3:

Determine the bolt bearing and tearout strength on the supported beam web. Because the clear distance is greater than two times the bolt diameter for each of the four bolts, the nominal bearing strength is Rn = 2.4dtFu = 2.4 ( 3 4 )( 0.275 )( 65) = 32.2 kips and the allowable bearing strength Rn Ω = 32.2 200 = 16.1 kips

Since this is greater than the allowable shear strength of the bolt, shear strength controls for these bolts also. Thus, for the four bolts in the beam web the allowable strength is limited by bolt shear. Step 4:

Determine the connection allowable shear strength based on the bolts. Since all bolts are limited by bolt shear rupture Rn Ω = 4 (11.9 ) = 47.6 kips

506 Chapter 11

Simple Connections Step 5:

Determine the shear yield strength of the angle. The gross shear area of the angle is Agv = lta = 12.0 ( 0.375) = 4.50 in.2 and the allowable shear strength is Vn Ω = 0.6 Fy Agv Ω = ( 0.6 ( 36 )( 4.50 ) ) 1.50 = 64.8 kips

Step 6:

Determine the shear rupture strength of the angle. The net shear area of the angle is Anv = ( l − n ( d h + 1 16 ) ) ta = (12.0 − 4 ( 3 4 + 1 8) ) ( 0.375) = 3.19 in.2 and the allowable shear strength is Vn Ω = 0.6 Fu Anv Ω = ( 0.6 ( 58)( 3.19 ) ) 2.00 = 55.5 kips

Step 7:

Determine the block shear strength of the angle. First calculate the areas. 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜1.25 − ( 3 4 + 1 8 ) ⎟ ( 0.375 ) = 0.305 in.2 2 ⎝ ⎠ Agv = ltw = 10.5 ( 0.375 ) = 3.94 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.5 − 3.5 ( 3 4 + 1 8 ) ) ( 0.375 ) = 2.79 in.2

Determine the tension rupture strength Fu Ant = 58 ( 0.305) = 17.7 kips Consider shear yield and shear rupture and select the one with the least strength; thus, 0.6 Fy Agv = 0.6 ( 36 )( 3.94 ) = 85.1 kips 0.6 Fu Anv = 0.6 ( 58 )( 2.79 ) = 97.1 kips

Selecting the shear yield term and combining it with the tension rupture term, with Ubs = 1.0 for this case of uniform tensile stress distribution, gives Rn Ω = ( 85.1 + 1.0 (17.7 ) ) 2.00 = 51.4 kips Step 8:

Determine the allowable strength of the leg attached to the supported member. The allowable strength is controlled by bolt shear, where Rn Ω = 47.6 kips

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Simple Connections 507

Part (b): Bolted Outstanding Leg Check the eccentric shear using Manual Table 7-6 to account for the effects Step 9: of eccentricity on the bolt group.

The eccentricity of the load on the line of bolts is calculated as the bolt distance from the angle heel plus one-half of the beam web as shown in Figure 11.9b; thus, 0.275 ex = 2.25 + = 2.39 in. 2 For the four-bolt connection with bolt spacing of 3.0 in., controlled by bolt shear, Manual Table 7-6 with the load at an angle of zero gives, by interpolation, the effective number of bolts as C = 3.12. Therefore, Rn Ω = C rn Ω = 3.12 (11.9 ) = 37.1 kips As expected, the strength of the bolts in the outstanding leg is less than that in the leg on the supported beam because the outstanding leg must accommodate an eccentricity that is not present in the leg on the beam. Step 10:

Determine the flexural yield strength of the outstanding leg. The plastic section modulus is determined for the rectangle formed by the length and thickness of the angle, and the nominal moment strength is determined by multiplying the plastic section modulus by the yield stress; thus, Z= and

0.375 (12 ) 4

2

= 13.5 in.3

M n = Fy Z = 36 (13.5) = 486 in.-kips

Because the moment is the shear force times the eccentricity, using the same eccentricity calculated for the bolt group, M Ω (486 1.67) = = 122 kips Rn Ω = n e 2.39 Step 11:

Determine the flexural rupture strength of the outstanding leg. The net plastic section modulus is determined for the rectangle less the holes. Although this can readily be calculated, it can also be obtained from Manual Table 15-3, where Z net = 9.56 in.3 . Thus, M n = Fu Z net = 58 ( 9.56 ) = 554 in.-kips and M Ω (554 2.00) Rn Ω = n = = 116 kips e 2.39

508 Chapter 11

Simple Connections Step 12:

Determine the controlling limit state strength for the bolted outstanding legs. For the bolted outstanding leg, the strength is controlled by shear strength of the bolts with eccentricity, where Rn Ω = 37.1 kips Because this is less than the value for the leg attached to the beam, this is the allowable strength of the bolted-bolted single-angle connection as long as the supporting member has sufficient bearing and tearout strength.

Part (c): Welded Outstanding Leg Step 13: Determine the eccentric weld rupture strength.

Manual Table 8-10 will be used to determine the eccentric weld rupture strength. The weld for the single-angle connection is applied to the vertical toe and the bottom edge of the angle, not the top, except for a small weld return. This ensures that the angle is sufficiently flexible to behave as a simple connection as modeled in the analysis of a simple beam. Manual Table 8-10 for load at an angle of zero degrees shows this weld on the top or the bottom. Thus, the table can be used for either arrangement because the geometry is the same whether the horizontal weld is at the top or bottom of the connection. Based on the dimensions given in Figure 11.9c, l = 12.0, kl = 3.5, and thus k = 0.292. The weld is given as a 3/16 in. weld with E70 electrodes. From Manual Table 8-10, interpolating between k = 0.2 and k = 0.3 yields x = 0.0336 . Therefore, using the definitions given in Table 8-10, the eccentricity is t 0.275 ex = kl + w − xl = 3.5 + − 0.0336 (12.0 ) = 3.23 in. 2 2 and e 3.23 a= x = = 0.269 l 12.0 With a double interpolation between k = 0.2 and 0.3 and a = 0.25 and 0.30, the coefficient C is determined as C = 2.17 Therefore, with C1 = 1.0 for an E70XX electrode, the nominal weld strength is Rn = CC1Dl = 2.17 (1.0)( 3)(12.0) = 78.1 kips and the allowable strength is Rn Ω = 78.1 2.00 = 39.1 kips Step 14:

Determine the allowable strength for the limit state of flexural yielding. For the limit state of flexural yielding of the angle, the strength is the

Chapter 11

Simple Connections 509

moment determined in step 10 divided by the eccentricity, ex = 3.23 in., Rn Ω = ( 486 1.67 ) 3.23 = 90.1 kips Step 15:

Determine the controlling limit state strength for the welded outstanding legs. For the welded outstanding leg, the allowable strength is controlled by eccentric shear on the weld, where Rn Ω = 39.1 kips Because this is less than the value for the leg attached to the beam, this is the allowable strength of the bolted-welded single-angle connection as long as the supporting member has sufficient rupture strength. For both bolted and welded outstanding leg arrangements the outstanding leg controlled the strength of the connection.

11.8 SINGLE-PLATE SHEAR CONNECTIONS The single-plate shear connection, also called a shear tab connection, is shown in Figure 11.1c. It consists of a plate that is shop-welded to the support and field-bolted to the beam and is similar to the single-angle connection when it comes to erection. The shear tab consists of only a single plate, which is about as simple as can be expected. It is welded to the supporting member and must be bolted to the supported beam in order to accommodate the required rotation. Even when bolted to the beam, this connection is stiffer than the single- or double-angle connection and requires careful detailing to ensure sufficient flexibility. The behavior of this connection is similar to that of a double-angle connection except that it achieves its rotation capacity through the bending of the tab and deformation of the plate or beam web in bearing at the bolt holes. Because of the complexity of assessing some of the limit states for this connection, AISC has developed two design approaches—including a somewhat prescriptive approach for what is called the conventional configuration, and a detailed limit states checking procedure for all others, referred to as the extended configuration. The limit states that must be checked are the same for either configuration; the difference is that in the conventional configuration, physical limitations have been set so that most of those limit states do not govern. The potential limit states are 1. Bolts a. Shear rupture with eccentricity 2. Beam a. Bearing on the web b. Shear yielding of web 3. Plate a. Bearing and tearout on the plate b. Elastic yield moment c. Flexural rupture d. Shear yielding e. Shear rupture f. Block shear rupture g. Buckling h. Plastic flexural yielding with shear interaction 4. Weld a. Weld rupture with eccentricity

5110 Chapter 11 1

Simple Connections C

Of these 11 limit states, those associated with fleexure and bucckling of the plate are new w to the o simple con nnection desiign, and the weld rupturee limit state is treated a little bit discussion of differently th han those welld limit statess already disc ussed. The conventional configuratio on of the sheaar tab results iin a connectioon that is veryy simple to design. This T is the typ pe of connecction that is tr treated here. F For other connfigurations, detailed procedures are a given in Part P 10 of the Manual. Thhe dimensionnal limitationss of the convventional shear tab leaad to the follo owing requirements: 1. There can be only a siingle vertical row containiing a total of 2 to 12 bolts. 2. The distaance from thee bolt line to tthe weld line cannot exceed 3-1/2 in. 3. Only stan ndard or shorrt slotted holees transverse to the direction of the loadd can be used. The verttical edge disstance, lev, m 4. must satisfy th the Specificattion minimum ms from Table J3.4 and leh sho ould be greateer than or equ qual to 2 timees the diameteer of the bolt for both b the plate and beam weeb. 5. For stand dard holes, either the platee or beam webb must have t ≤ (db/2 + 1/116) for 2 to 5 boltts, or t ≤ (db/2 / – 1/16) foor 6 to 12 boolts to accomm modate simpple beam rotation. These are given in Ma Manual Table 10-9 along with the thhickness ments for shorrt-slotted holees and the reqquired minimuum eccentriciity to be requirem included. Oncce these limiitations are satisfied, thee connectionn need be chhecked only for the following lim mit states usin ng the eccentrricity specifieed in Manuall Table 10-9 a. Eccentricc bolt shear ru upture b. Bolt bearring and tearo out c. Block sh hear rupture d. Plate sheear yielding e. Plate sheear rupture

Figure 11.10 Shear Tab b Connection n for Examplee 11.5

Chapter 11

Simple Connections 511

EXAMPLE 11.5a Shear Tab Conventional Configuration by LRFD

Goal:

Determine the design strength of a conventional configuration shear tab connection.

Given

The shear tab connection is given in Figure 11.10. The beam is a W16×50 A992 framing into the flange of a W14×90 A992 column with an A36 1/4×4-1/2×1’-0 plate. Use four 3/4 in. A325-N bolts in standard holes.

SOLUTION

Step 1:

Determine whether the given shear tab meets the limitations for the conventional configuration. Limitations for the conventional configuration: 1. 4 bolts—is between 2 and 12 2. a = 3.0 in.—does not exceed 3-1/2 in. 3. Standard holes—standard or short-slotted holes are permitted 4. lev = 1.5 in. > 1.0 in. from Table J3.4 leh = 1.5 in. ≥ 2db = 2 ( 3 4 ) = 1.5 in. 5.

Step 2:

⎛d ⎞ ⎛3 4 ⎞ t plate = 1 4 in. < ⎜ b + 1 16 ⎟ = ⎜ + 1 16 ⎟ = 7 16 in. ⎝ 2 ⎠ ⎝ 2 ⎠

Determine the bolt shear strength considering eccentricity. From Manual Table 7-1,

φrn = 17.9 kips/bolt From Manual Table 10-9 the eccentricity to be included is

e = a 2 = 3.0 2 = 1.5 in. From Manual Table 7-6 the equivalent number of bolts to account for eccentricity in the connection can be determined by interpolating between C = 3.75 for ex = 1.0 in. and C = 3.32 for ex = 2.0 in. Thus, C = 3.54 and φRn = Cφrn = 3.54 (17.9) = 63.4 kips Step 3:

Determine the bolt bearing and tearout strength on the plate assuming it is concentrically loaded. For the bottom bolt

1 1 lc = lev − d h = 1.5 − (3/4 + 1/16) = 1.09 < 2(3/4) = 1.5 2 2 Thus, tearout controls over bearing, and the nominal bolt strength is Rn = 1.2lc t p Fu = 1.2 (1.09 )( 0.250 )( 58) = 19.0 kips The design bolt tearout strength is φRn = 0.75 (19.0) = 14.3 kips Since this is less than the bolt shear strength, it will control the strength of this bolt.

512 Chapter 11

Simple Connections For the other bolts lc = s − dh = 3.0 − ( 3 4 + 1 16) = 2.19 > 2 ( 3 4) = 1.5 in. Therefore, bearing controls over tearout, and the nominal bolt strength is Rn = 2.4dt p Fu = 2.4 ( 3 4)( 0.250 )( 58) = 26.1 kips The design bolt bearing strength is φRn = 0.75 ( 26.1) = 19.6 kips Since this is greater than the bolt shear strength, bolt shear strength will control the strength of these remaining bolts. Step 4:

Determine the bolt bearing and tearout strength on the web. For the beam web, the material is A992, the web thickness is 0.380 in. and the member is not coped. Because the bolts will not tear out and both the strength and thickness of the web are greater than the comparable values for the plate, the beam web will not control. However, strength will still be determined for consistency. For each bolt in bearing Rn = 2.4dtw Fu = 2.4 ( 3 4)( 0.380)( 65) = 44.5 kips and the design strength in the web is φRn = 0.75 ( 44.5 ) = 33.4 kips Since this is greater than the bolt shear strength, bolt shear strength will control for all four bolts in the beam web

Step 6:

Determine the block shear strength of the plate. Calculating the areas, 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ t w 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜ 1.5 − ( 3 4 + 1 8 ) ⎟ ( 0.250 ) = 0.266 in.2 2 ⎝ ⎠ Agv = ltw = 10.5 ( 0.250 ) = 2.63 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.5 − 3.5 ( 3 4 + 1 8 ) ) ( 0.250 ) = 1.86 in.2

Determine the tension rupture strength Fu Ant = 58 ( 0.266 ) = 15.4 kips Consider shear yield and shear rupture and select the one with least

Chapter 11 strength; thus,

Simple Connections 513

0.6 Fy Agv = 0.6 ( 36 )( 2.63 ) = 56.8 kips 0.6 Fu Anv = 0.6 ( 58 )(1.86 ) = 64.7 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection allowable block shear strength, with Ubs = 1.0, of φRn = 0.75 ( 56.8 + 1.0 (15.4 ) ) = 54.2 kips Step 5:

Determine the connection strength based on the bolts in holes, The bottom bolt is controlled by tearout in the plate, φRn = 14.3 kips The other 3 bolts are controlled by shear φRn = 17.9 kips Thus, the connection strength based on the bolts in holes is φRn = 14.3 + 3(17.9) =68.0 kips

Step 7:

Determine the design shear yield strength of the plate. The gross shear area is

Agv = lt p = 12.0 ( 0.250) = 3.00 in.2

and Step 8:

Determine the design shear rupture strength of the plate. The net shear area is Anv = ( l − n ( d h + 1 16 ) ) t p = (12.0 − 4 ( 3 4 + 1 8) ) ( 0.250 ) = 2.13 in.2 and

Step 9:

φVn = φ0.6 Fy Agv = 1.0 ( 0.6 ( 36 )( 3.00 ) ) = 64.8 kips

φVn = φ0.6 Fu Anv = 0.75 ( 0.6 ( 58)( 2.13) ) = 55.6 kips

Determine the design weld rupture strength. The conventional configuration requires that the plate be welded to the supporting member through a pair of fillet welds on each side of the plate, with the weld leg width w = ( 5 8 ) t p . Therefore use a 3/16 in. fillet weld. This develops the strength of either a 36 ksi or 50 ksi plate and therefore does not require any further limit state checking. However, this does not check for the strength of the supporting member.

Step 10:

Determine the minimum thickness of the supporting member to match the weld rupture strength. Using the minimum thickness given by Manual Equation 9-2 3.09 D 3.09 ( 3) tmin = = = 0.143 in. Fu 65

514 Chapter 11

Simple Connections Which is significantly less that the W14×90 flange thickness, t f = 0.710 in. Even if the minimum thickness had not been exceeded by the column flange thickness, the fact that the plate is welded to the column flange at the column flange-web junction would provide sufficient base metal strength. Step 11:

Determine the controlling limit state and design strength of the connection. The design strength is controlled by the limit state of block shear rupture of the plate, where φRn = 54.2 kips A check of Manual Table 10-10a shows that this is quite close to the value for the 11.5 in. plate given there, as would be expected for our 12.0 in. plate.

EXAMPLE 11.5b Shear Tab Conventional Configuration by ASD

Goal:

Determine the allowable strength of a conventional configuration shear tab connection.

Given

The shear tab connection is given in Figure 11.10. The beam is a W16×50 A992 framing into the flange of a W14×90 A992 column with an A36 1/4×4-1/2×1’-0 plate. Use four 3/4 in. A325-N bolts in standard holes.

SOLUTION

Step 1:

Determine whether the given shear tab meets the limitations for the conventional configuration. Limitations for the conventional configuration: 1. 4 bolts—is between 2 and 12 2. a = 3.0 in.—does not exceed 3-1/2 in. 3. Standard holes—standard or short-slotted holes are permitted 4. lev = 1.5 in. > 1.0 in. from Table J3.4 leh = 1.5 in. ≥ 2db = 2 ( 3 4 ) = 1.5 in. 5.

Step 2:

⎛d ⎞ ⎛3 4 ⎞ t plate = 1 4 in. < ⎜ b + 1 16 ⎟ = ⎜ + 1 16 ⎟ = 7 16 in. ⎝ 2 ⎠ ⎝ 2 ⎠

Determine the bolt shear strength considering eccentricity. From Manual Table 7-1,

rn Ω = 11.9 kips/bolt From Manual Table 10-9 the eccentricity to be included is

e = a 2 = 3.0 2 = 1.5 in. From Manual Table 7-6 the equivalent number of bolts to account for eccentricity in the connection can be determined by interpolating between C = 3.75 for ex = 1.0 in. and C = 3.32 for ex = 2.0 in. Thus, C = 3.54 and Rn Ω = C rn Ω = 3.54 (11.9) = 42.1 kips

Chapter 11 Step 3:

Simple Connections 515

Determine the bolt bearing and tearout strength on the plate assuming it is concentrically loaded. For the bottom bolt

1 1 lc = lev − d h = 1.5 − (3/4 + 1/16) = 1.09 < 2(3/4) = 1.5 2 2 Thus, tearout controls over bearing, and the nominal bolt strength is Rn = 1.2lc t p Fu = 1.2 (1.09 )( 0.250 )( 58) = 19.0 kips The allowable bolt tearout strength is Rn Ω = 19.0 2.00 = 9.50 kips Since this is less than the bolt shear strength, it will control the strength of this bolt. For the other bolts lc = s − dh = 3.0 − ( 3 4 + 1 16) = 2.19 > 2 ( 3 4) = 1.5 in. Therefore, bearing controls over tearout, and the nominal bolt strength is Rn = 2.4dt p Fu = 2.4 ( 3 4)( 0.250 )( 58) = 26.1 kips The allowable bolt bearing strength is Rn Ω = 26.1 2.00 = 13.1 kips Since this is greater than the bolt shear strength, bolt shear strength will control the strength of these remaining bolts. Step 4:

Determine the bolt bearing and tearout strength on the web. For the beam web, the material is A992, the web thickness is 0.380 in. and the member is not coped. Because the bolts will not tear out and both the strength and thickness of the web are greater than the comparable values for the plate, the beam web will not control. However, strength will still be determined for consistency. For each bolt in bearing Rn = 2.4dtw Fu = 2.4 ( 3 4)( 0.380)( 65) = 44.5 kips and the allowable strength in the web is Rn Ω = 44.5 2.00 = 22.3 kips

Since this is greater than the bolt shear strength, bolt shear strength will control for all four bolts in the beam web.

516 Chapter 11

Simple Connections Step 5:

Determine the connection strength based on the bolts in holes, The bottom bolt is controlled by tearout in the plate, Rn Ω = 9.50 kips The other 3 bolts are controlled by shear Rn Ω = 11.9 kips Thus, the connection strength based on the bolts in holes is Rn Ω = 9.50 + 3(11.9) =45.2 kips

Step 6:

Determine the block shear strength of the plate. Calculating the areas, 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ t w 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜ 1.5 − ( 3 4 + 1 8 ) ⎟ ( 0.250 ) = 0.266 in.2 2 ⎝ ⎠ Agv = ltw = 10.5 ( 0.250 ) = 2.63 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.5 − 3.5 ( 3 4 + 1 8 ) ) ( 0.250 ) = 1.86 in.2

Determine the tension rupture strength Fu Ant = 58 ( 0.266 ) = 15.4 kips Consider shear yield and shear rupture and select the one with least strength; thus, 0.6 Fy Agv = 0.6 ( 36 )( 2.63 ) = 56.8 kips 0.6 Fu Anv = 0.6 ( 58 )(1.86 ) = 64.7 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection allowable block shear strength, with Ubs = 1.0, of Rn Ω = ( 56.8 + 1.0 (15.4 ) ) 2.00 = 36.1 kips Step 7:

Determine the allowable shear yield strength of the plate. The gross shear area is Agv = lt p = 12.0 ( 0.250 ) = 3.00 in.2 and Vn Ω = 0.6 Fy Agv Ω = ( 0.6 ( 36 )( 3.00 ) ) 1.50 = 43.2 kips

Step 8:

Determine the allowable shear rupture strength of the plate. The net shear area is Anv = ( l − n ( d h + 1 16 ) ) t p = (12.0 − 4 ( 3 4 + 1 8 ) ) ( 0.250 ) = 2.13 in.2 and

Chapter 11

Simple Connections 517

Vn Ω = 0.6 Fu Anv Ω = ( 0.6 ( 58 )( 2.13 ) ) 2.00 = 37.1 kips

Step 9:

Determine the allowable weld rupture strength. The conventional configuration requires that the plate be welded to the supporting member through a pair of fillet welds on each side of the plate, with the weld leg width w = ( 5 8 ) t p . Therefore use a 3/16 in. fillet weld. This develops the strength of either a 36 ksi or 50 ksi plate and therefore does not require any further limit states checking. However, this does not check for the strength of the supporting member.

Step 10:

Determine the minimum thickness of the supporting member to match the weld rupture strength. Using the minimum thickness given by Manual Equation 9-2 3.09 D 3.09 ( 3) tmin = = = 0.143 in. Fu 65 Which is significantly less that the W14×90 flange thickness, t f = 0.710 in. Even if the minimum thickness had not been exceeded by the column flange thickness, the fact that the plate is welded to the column flange at the column flange-web junction would provide sufficient base metal strength.

Step 11:

Determine the controlling limit state and allowable strength of the connection. The allowable strength is controlled by the limit state of block shear rupture of the plate, where Rn Ω = 36.1 kips A check of Manual Table 10-10a shows that this is quite close to the value for the 11.5 in. plate given there, as would be expected for our 12.0 in. plate.

11.9 SEATED CONNECTIONS An unstiffened seated connection is shown in Figure 11.1d, and a stiffened seated connection is shown in Figure 11.1e. These connections are typically used to attach a beam to the web of a column. They can also be used to add capacity to other types of existing connections in a retrofit situation. Because of their simplicity, they are fairly easy connections to erect. They have very few parts: a seat angle, a connection to the supporting member through welds or bolts, a limited connection to the supported member, and a top connection to ensure stability of the supported beam. All of the force is transferred through bearing of the beam on the seat and then through the connection of the seat to the supporting member. When the seat lacks sufficient strength in bending of the top leg, it can be stiffened to produce the stiffened seated connection. The seat can be welded or, as is usually the case, bolted to the supporting member. The connection to the supported member is not designed for a specific strength when only a vertical force is being transferred. The seated connection performs excellently as a simple connection. It

518 Chapter 11

Simple Connections

can rotate sufficiently about the bottom of the beam without imposing any significant moment to the supporting member. The simplicity of this connection results in relatively few limit states to be checked. Because the transfer of force between the beam and seat is through the bearing of the beam on the seat, the limit states of beam web yielding and beam web local crippling must be checked. These limit states were introduced in Section 6.14. The outstanding leg of the seat angle must be checked for the limit states of flexural yielding and shear yielding, and the connection to the supporting member, bolts, or welds must be checked for their appropriate limit states. In summary, the potential limit states are 1. Beam a. Web local yielding b. Web local crippling 2. Seat angle a. Flexural yielding b. Shear yielding 3. Connector a. Bolt or weld shear with eccentricity The nominal strength for the limit state of web local yielding was discussed in Section 7.4.2. If the strength equation for a force applied at the end of a member, Equation J10-3 is rewritten to solve for the minimum required bearing length, it can be used for design. Thus, from Specification Section J10.2, for the limit state of web local yielding R (11.1) lb min = n − 2.5k Fy t w For the limit state of web local yielding, φ = 1.00 and Ω = 1.50. Similarly, from Specification Section J10.3, for the limit state of web local crippling, an equation for the minimum required bearing length can be determined. However, there are three different strength equations, with the choice depending on the relationship lb/d. Because it is most likely that lb will be less than d/2 and very likely it will be less than 0.2d, only Specification Equation J10-5a is treated here. The other situations can both be handled in the same way if need be. Thus, rearranging Equation J10-5a to solve for the minimum required bearing length for the limit state of web local crippling yields, for lb d ≤ 0.2 ,

lb min

d ⎛ Rn = ⎜ 3 ⎜⎝ 0.40tw2

⎞ ⎛ t f ⎞1.5 tw − 1⎟ ⎜ ⎟ ⎟ ⎝ tw ⎠ EFy t f ⎠

(11.2)

For the limit state of web local crippling, φ = 0.75 and Ω = 2.00. Because the unstiffened angle is a very flexible connection, the load levels usually considered are quite low. This tends to result in very small minimum required bearing lengths and, in some calculations, a negative minimum required bearing length. To offset this potential problem, the minimum bearing length for seated connections is taken as kdet. A review of Manual Table 1-1 shows two values for k: kdes is a dimension used in design calculations and is the smaller of the two values, and kdet is a dimension normally used in detailing and is used here because it is the larger of these two values. These two values of k are the result of differences in production by different mills. They represent the extremes of the values actually found and the selection is made within a calculation to give a conservative answer. The outstanding leg of the angle must be capable of supporting the beam reaction applied at an eccentricity from the critical section of the angle. This requires checking the limit state of flexural yielding of the leg. The angle shown in Figure 11.11a is an unstiffened seat angle. The critical section for both flexure and shear is taken as 3/8 in. out from the face of the vertical leg,

Chhapter 11

Siimple Connecctions 519

Figure 11.11 1 Seated d Beam Conn nection

Figure 11.12 1 Unstifffened Seated d Beam Conneection for Exaample 11.6 the radiu us of the fillet.. The eccentriicity is measuured from thiss line to the m midpoint of thhe minimum required bearing lengtth of the beam m on the anglee, lb. Thus, thhe eccentricityy is e = lb 2 + 1 2 in. setbaack + 1 4 in. uunder run − ( ta + 3 / 8) = lb 2 + 3 8 − ta (11.3) minal beam reaaction strengtth, for the lim mit state of fleexural yielding, is based onn the plastic The nom section modulus m of thee seat leg of the t angle. Thuus,

520 Chapter 11

Simple Connections Z=

la ta2 4

and

⎛ l t2 ⎞ Fy ⎜ a a ⎟ 4 ⎠ Fy Z M Rn = n = = ⎝ e e e

(11.4)

Because this is based on flexural yielding, φ = 0.9 and Ω = 1.67. Equation 11.4 can then be solved for the thickness of the angle as a function of the required nominal reaction as 4 Rn e (11.5) ta = Fy la For the limit state of shear yielding, the gross area of the angle leg at this same location is used. Because this is beyond the angle fillet, the angle thickness is used to determine the gross area and (11.6) Rn = 0.6 Fy la ta For the limit state of shear yielding, φ = 1.0 and Ω = 1.5.

EXAMPLE 11.6a Seated Connection Design by LRFD

Goal:

Design an unstiffened welded seated connection.

Given:

An unstiffened welded seated connection is shown in Figure 11.12. A W16×36 beam is framing into a W14×90 column. The beam has an LRFD required strength of Ru = 35 kips . The beam and column are A992 and the angle is A36. From Manual Table 1-1, for the W16×36 , kdet = 1-1/8 in. and kdes = 0.832 in.

SOLUTION

Step 1:

Determine the minimum required bearing length for web yielding using Equation 11.1. R 35 / (1.0) lb min = n − 2.5k = − 2.5(0.832) = 0.293in.< kdet = 1 1 8 in. Fy t w (50)(0.295)

Step 2:

Determine the minimum required bearing length for web local crippling using Equation 11.2, assuming that lb d ≤ 0.2 . lb min =

d ⎛ Rn ⎜ 3 ⎜⎝ 0.40tw2

⎞ ⎛ t f ⎞1.5 tw − 1⎟ ⎜ ⎟ ⎟ ⎝ tw ⎠ EFy t f ⎠

15.9 ⎛ 35 / 0.75 = ⎜ 3 ⎜⎝ (0.40)(0.295) 2

⎞ ⎛ 0.430 ⎞1.5 0.295 − 1⎟ ⎜ ⎟ 29,000(50)(0.430) ⎟⎠ ⎝ 0.295 ⎠

= −0.726 < kdet

Thus, lb = kdet = 1.125 in. and lb/d = 1.125/15.3 = 0.0735 ≤ 0.2, so the correct equation has been used and the bearing length is taken as the minimum required length of 1.125 in.

Chapter 11 Step 3:

Simple Connections 521

Determine the eccentricity to be used in calculating angle thickness. Assume an angle thickness of 1/2 in. From Equation 11.3 e = lb 2 + 3 / 8 − ta = 1.125 2 + 3 8 − 1 2 = 0.438 in.

Step 4:

Determine the minimum required angle thickness based on the limit state of flexural yielding. Assume an 8.0 in. long angle so that it extends beyond the beam flange on both sides and use Equation 11.5. Thus

treq =

4Rn e 4(35 / 0.9)(0.438) = = 0.486 in. Fy l (36)(8.0)

Step 5:

Determine the minimum angle thickness for the limit state of shear yielding using Equation 11.6. Thus, Rn (35 / 1.0) tmin = = = 0.203 in. 0.6 Fy L 0.6(36)(8.0)

Step 6:

Check the selected angle thickness. The 1/2 in. angle provides a thickness greater than each of the minimums determined in steps 4 and 5. Thus, the 1/2 in. angle is adequate.

Step 7:

Determine the required weld size. Use Manual Table 8-4 with the angle equal to 0 degrees to account for the eccentricity. The eccentricity for the weld is taken from the center of bearing to the face of the supporting column, which yields 3 1.125 e= + = 1.31 in. 4 2 Assuming an angle with a 4.0 in. outstanding leg, e 1.31 a= = = 0.328 l 4.0 From Manual Table 8-4, the coefficient is determined for k = 0 through interpolation as C = 2.97, so the minimum weld is Ru 35 Dmin = = = 3.93 φCC1l 0.75(1.0(2.97)(4.0)) Therefore, the calculated minimum weld is 1/4 in. and the angle is an L4×4×1/2, as shown in Figure 11.12. Note: In addition to the seat angle, a top clip angle is needed to provide lateral stability. This angle is not normally designed to support any load and is usually a 1/4 in. angle attached with two bolts to the beam and two bolts to the supporting member.

522 Chapter 11

Simple Connections

EXAMPLE 11.6b Seated Connection Design by ASD

Goal:

Design an unstiffened welded seated connection.

Given:

An unstiffened welded seated connection is shown in Figure 11.12. A W16×36 beam is framing into a W14×90 column. The beam has an ASD required strength of Ra = 23 kips . The beam and column are A992 and the angle is A36. From Manual Table 1-1, for the W16×36 , kdet = 1-1/8 in. and kdes = 0.832 in.

SOLUTION

Step 1:

Determine the minimum required bearing length for web yielding using Equation 11.1. 1.5 ( 23) R lb min = n − 2.5k = − 2.5(0.832) = 0.259in.< kdet = 1 − 1 / 8 in. (50)(0.295) Fy tw

Step 2:

Determine the minimum required bearing length for web local crippling using Equation 11.2, assuming that lb d ≤ 0.2 . lb min

d ⎛ Rn = ⎜ 3 ⎜⎝ 0.40tw2 =

⎞ ⎛ t f ⎞1.5 tw − 1⎟ ⎜ ⎟ ⎟ ⎝ tw ⎠ EFy t f ⎠

15.9 ⎛ 2.00 ( 23) ⎜ 3 ⎜⎝ (0.40)(0.295) 2

⎞ ⎛ 0.430 ⎞1.5 0.295 − 1⎟⎟ ⎜ ⎟ 29,000(50)(0.430) ⎠ ⎝ 0.295 ⎠

= −0.849 < kdet

Step 3:

Thus, lb = kdet = 1.125 in. and lb/d = 1.125/15.3 = 0.0735 ≤ 0.2, so the correct equation has been used and the bearing length is taken as the minimum required length of 1.125 in. Determine the eccentricity to be used in calculating angle thickness. Assume an angle thickness of 1/2 in. From Equation 11.3 e = lb 2 + 3 / 8 − ta = 1.125 2 + 3 8 − 1 2 = 0.438 in.

Step 4:

Determine the minimum required angle thickness based on the limit state of flexural yielding. Assume an 8.0 in. long angle so that it extends beyond the beam flange on both sides and use Equation 11.5. Thus

treq =

4 (1.67 ( 23) ) ( 0.438) 4 Rn e = = 0.483 in. 36 ( 8.0 ) Fy l

Step 5:

Determine the minimum angle thickness for the limit state of shear yielding using Equation 11.6. Thus, 1.50 ( 23) Rn tmin = = = 0.200 in. 0.6 Fy L 0.6 ( 36 )( 8.0 )

Step 6:

Check the selected angle thickness. The 1/2 in. angle provides a thickness greater than each of the minimums

Chapter 11

Simple Connections 523

determined in steps 4 and 5. Thus, the 1/2 in. angle is adequate. Step 7:

Determine the required weld size. Use Manual Table 8-4 with the angle equal to 0 degrees to account for the eccentricity. The eccentricity for the weld is taken from the center of bearing to the face of the supporting column, which yields 3 1.125 e= + = 1.31 in. 4 2 Assuming an angle with a 4.0 in. outstanding leg, e 1.31 a= = = 0.328 l 4.0 From Manual Table 8-4, the coefficient is determined for k = 0 through interpolation as C = 2.97, so the minimum weld is 2.00 ( 23 ) ΩRu Dmin = = = 3.87 CC1l 1.0 ( 2.97 )( 4.0 ) Therefore, the calculated minimum weld is 1/4 in. and the angle is an L4×4×1/2, as shown in Figure 11.12. Note: In addition to the seat angle, a top clip angle is needed to provide lateral stability. This angle is not normally designed to support any load and is usually a 1/4 in. angle attached with two bolts to the beam and two bolts to the supporting member.

A stiffened seated connection is shown in Figure 11.1e, and a detail of the stiffened seated connection is shown in Figure 11.11b. This type of connection is used when the loads are too large to be supported by an unstiffened seat. The stiffener can be a single plate, the stem of a tee, or the back-to-back legs of a pair of angles. A plate on top of the stiffener provides the bearing surface and the location of the bolts required to attach the beam flange to the seat. The limit states for this connection are the same as for the unstiffened connection already discussed but also include the limit state of punching shear in the supporting member because of the potential for a large force from the stiffener being carried to the support. Punching shear on a column web will not be critical if the following parameters set in Part 10 of the Manual are met: 1. The simplified approach is applicable to columns in the following depths with weights per foot no less than: W14×43, W12×40, W10×33, W8×24, W6×20, and W5×16. 2. The width of the stiffener W is no greater than 7.0 in. 3. The beam is bolted, not welded, to the bearing plate at a point no greater than W/2 or 2-5/8 in. from the column face. 4. The top angle has a minimum thickness of 1/4 in. The eccentricity of the beam reaction is taken as 0.8W in determining the strength of the weld or bolt group connecting the seat to the supporting member. Part 10 of the Manual provides tables for the design of stiffened seated connections.

5224 Chapter 11 1

Simple Connections C

Figure 11.13 Light Braacing Connections

111.10

LIGH HT BRACIN NG CONNE ECTIONS Bracing con nnections hav ve as many potential vaariations as ddo the simplle shear connnections discussed ab bove. Figure 11.13a show ws a bolted-w welded tension brace connnection, andd Figure 11.13b show ws a welded-b bolted tension n brace conneection. It is aalso permissibble to have a weldedwelded conn nection or a bolted-bolted b connection, c aalthough thesee are not illusstrated here. The limit states for f these conn nections have already beenn addressed. F For the boltedd-welded connection, shown in Fig gure 11.13a, th hey are 1. Angles a. Tension T yield ding b. Tension T ruptu ure c. Bolt B bearing and a tearout d. Block B shear ru upture Bolts 2. a. Shear S rupture 3. Gusset plate a. Tension T yield ding b. Tension T ruptu ure c. Bolt B bearing and a tearout 4. Welds a. Weld W rupture for combinedd tension andd shear ded-bolted co onnection wheere the gussett is replaced w with a tee, Figgure 11.13b, tthe limit For the weld states are 1. Angles a. Tension T yield ding b. Tension T ruptu ure Welds 2. a. Weld W rupture 3. Tee stem m a. Tension T yield ding b. Tension T ruptu ure c. Block B shear d. Shear S yielding g 4. Tee flang ge a. Flange F bendin ng

Chhapter 11

5. 5 6. 6 7. 7

Siimple Connecctions 525

b. Shear yield ding c. Shear ruptture d. Bolt bearin ng and tearouut e. Block sheaar Bolts a. Combined d shear and tennsion Colum mn flange a. nding Flange ben b. Bolt bearin ng and tearouut Colum mn web a. Web local yielding

Although h they appearr to be simplle connectionns, light braciing connectioons require chhecking for quite a number n of diifferent limit states. Thre e of these li mit states haave not previiously been addressed d: (1) tension n rupture and d tension yieldd within the gusset, (2) liimit state of bbolt rupture due to co ombined shear and tension, and (3) highh-strength bollts in tension w with prying aaction.

Figure 11.14 1 Whitm more Section in a Gusset P Plate Figure F 11.14 shows a sin ngle-angle braace attached to a gusset plate with equal length welds alo ong the sides of the angle. Research haas shown thatt the distributtion of stresses from the brace thrrough the welds into the gu usset is such tthat the entiree width of the gusset is nott effective if it exceed ds the width defined d by a 30-degree 3 anngle spreadingg to both sidees from the beginning of the conn nection to the end of the connection. Thhis width is ddefined as thhe Whitmore section and identified d in Figure 11.14 1 as the critical c sectioon. When thee gusset is w wider than thee Whitmore section, only the Wh hitmore sectio on can be connsidered to rresist the forcce. When thee Whitmore section iss wider than the available plate dimenssion, only thee width of thee plate at the connection end can be b considered d to resist the force. If the connection iss bolted ratheer than weldedd as shown, the Whittmore section n distribution n starts at thhe first bolt aand proceedss to the last bolt in the connectio on. This is illu ustrated in Ex xample 11.7.

5226 Chapter 11 1

Simple Connections C

Figure 11.15 Tapered Gusset G Plate for f Example 11.7

EX XAMPLE 11.7 1 Taapered Gussset Pllate Strength h

Goa al:

Deterrmine the avaailable strenggth of a gusseet plate connnected to a dooubleangle brace. Consiider only tenssion yielding aand tension ruupture.

Giv ven:

The gusset g plate portion p of thee connection is shown in Figure 11.155. The brace is a double angle a bolted w with 3/4 in. bolts to a taperred gusset wiith the dimen nsions as show wn. The platee is A36 steel.

SO OLUTION

Step p 1:

Deterrmine the wid dth of the plat e at the locatiion of the lastt bolt. This is i also the loccation of the W Whitmore secction. By propportions, as seeen in Figurre 11.15c, x 7.25 = 1.0 10.25 Thereefore, x = 0.707 in.. and th he width of th he plate at thiss location is l = 5.0 + 2 ( 0.707 ) = 6.41 in.

Chapter 11

Simple Connections 527

Step 2:

Determine the width of the Whitmore section using the geometry shown in Figure 11.15d. lW = 2 ( 6 tan ( 30 ) ) = 6.93 in.

Step 3:

Determine the width to be used to determine the plate strength. Because the actual plate width at this location is less than the Whitmore section, the actual plate width is used to determine the strength of the plate.

Step 4:

Determine the nominal strength for the limit state of plate yielding. The gross area at the critical location is Ag = 6.41(1 2 ) = 3.21 in.2 and the nominal tensile strength is Tn = Fy Ag = 36 ( 3.21) = 116 kips

For LRFD Step 5:

The design tensile strength is φTn = 0.9 (116 ) = 104 kips

For ASD Step 5:

The allowable tensile strength is Tn Ω = 116 1.67 = 69.5 kips

Step 6:

Determine the nominal strength for the limit state of plate rupture. The effective net area at the same location is Ae = An = ( l − ( d h + 1 16 ) ) t p = ( 6.41 − ( 3 4 + 1 8 ) ) (1 2 ) = 2.77 in.2 Therefore, the nominal tensile strength is Tn = Fu Ae = 58 ( 2.77 ) = 161 kips

For LRFD Step 7:

The design tensile strength is φTn = 0.75 (161) = 121 kips

For ASD Step 7:

The allowable tensile strength is Tn Ω = 161 2.00 = 80.5 kips

For LRFD Step 8:

Determine the design tensile strength of the gusset plate based on the controlling limit state. The design tensile strength of the gusset is φTn = 104 kips

528 Chapter 11

Simple Connections based on the limit state of yielding. For ASD Step 8:

Determine the allowable tensile strength of the gusset plate based on the controlling limit state. The allowable tensile strength of the gusset is Tn Ω = 69.5 kips based on the limit state of yielding.

EXAMPLE 11.8 Uniform-Width Gusset Plate Strength

Goal:

Determine the available strength of a gusset plate of uniform width. Consider only tension yielding and tension rupture.

Given:

Determine the available strength of a gusset plate for the same situation as in Example 11.7, except that the plate has a uniform width of 8.0 in. The plate is again A36 steel and the bolts are 3/4 in.

SOLUTION

Step 1:

Determine the controlling width at the critical section. The width of the plate at the location of the last bolt is given as 8.0 in. The width of the Whitmore section is determined using the same geometry as shown in Figure 11.15d, which again yields lW = 2 ( 6 tan ( 30 ) ) = 6.93 in. In this case, the actual plate width is greater than the Whitmore section, so the Whitmore section width is used to determine the strength of the plate.

Step 2:

Determine the nominal tensile strength for the limit state of plate yielding. The gross area at the critical location is Ag = 6.93(1 2) = 3.47 in.2 The nominal tensile strength is Tn = Fy Ag = 36 ( 3.47 ) = 125 kips

For LRFD Step 3:

The design tensile strength for yielding is φTn = 0.9 (125 ) = 113 kips

For ASD Step 3:

The allowable tensile strength for yielding is Tn Ω = 125 1.67 = 74.9 kips

Chhapter 11

Step S 4:

Siimple Connecctions 529

Deetermine the nominal n strenngth for the lim mit state of pplate rupture. Th he effective net area at the same location is Ae = An = ( lW − ( d h + 1 116 ) ) t p = ( 6.933 − ( 3 4 + 1 8 ) ) (1 2 ) = 3.033 in.2 Th herefore,

Tn = Fu Ae = 58 ( 3.033) = 176 kips

For F LRFD L Step S 5:

he design tenssile strength ffor tension ruppture is Th φTn = 0.75 (176 ) = 132 kips

For F ASD A Step S 5:

Th he allowable tensile t strenggth for tensionn rupture is Tn Ω = 176 2.00 = 88.0 kips

For F LRFD L Step S 6:

Deetermine the design tensiile strength of the gusseet plate basedd on the co ontrolling limiit state. Th he design tenssile strength oof the uniform m-width gusseet is φTn = 113 kkips baased on the lim mit state of yiielding.

For F ASD A Step S 6:

Deetermine the allowable a tennsile strength of the gusset plate based oon the co ontrolling limiit state. Th he allowable tensile t strenggth of the unifform-width guusset is Tn Ω = 74.99 kips baased on the lim mit state of yiielding. No ote: There is no n advantage to using a plate wider thann the Whitmoore section, 6.93 in n.

Figure 11.116 Shear-Teension Interacction for Bolts

5330 Chapter 11

Simple Connections C

Figure 11.17 Modiffied Shear-Teension Interactionn for Bolts The next limit state to addrress concernss high-strenggth bolts in combined shhear and tension. Thiis was briefly y addressed in i Sections 110.7.5 and 100.7.6. A bolt loaded in coombined shear and teension has a reduced ability to resist sshear in a beearing-type connection duue to the presence of tension. In a slip-critical connection, tthe tension reeduces the conntact force annd, thus, lowers the shear required d to cause the connection tto slip. Thesee reductions m must be accouunted for in the design n of connectio ons where theese combinedd limit states ooccur. Testts have show wn that the intteraction of sshear and tennsion in a beaaring-type connnection can be fairly y well prediccted through an a elliptical iinteraction cuurve. Howeveer, for simplicity, the Specification n has adopteed three straig ght lines to approximate the ellipse. Both the elliipse and straight lines are shown in i Figure 11.1 16. Section J33.7 gives two equations forr this interacttion, one for ASD an nd one for LRFD, where the nominall tensile stresss including the effects oof sheartension interraction is F (AISC C J3-3a) Fnt′ = 1.3 1 Fnt − nt f rv ≤ Fnt (LR RFD) φFnv ΩFnt (AISC C J3-3b) Fnt′ = 1.3 1 Fnt − f rv ≤ Fnt (AS D) Fnv where nsile stress for tension alonne from Specif ification Tablle J3.2 Fnt = nominal ten Fnv = nominal sh hear stress for shear alone ffrom Specification Table JJ3.2 frv = required sheear stress These two equations e can n be combineed and writtenn in terms off nominal strrength if the rrequired shear stress, frv, is combin ned with φ an nd Ω to give tthe nominal shhear stress inncluding the effects of shear-tension interaction.. Thus f Fnv′ = rv (LRFD) or Fnv′ = Ωf rv (A ASD) φ so that F (11.7) Fnt′ = 1.3 Fnt − nt Fnv' ≤ Fnt Fnv If Equation 11.7 is divideed by Fnt, another form of tthe interaction equation reesults as Fnt′ F′ = 1.3 − nv ≤ 1.0 Fnt Fnv nship is shown n in Figure 11 1.17. This relation

(11.8)

Chhapter 11

Siimple Connecctions 531

Figure 11.18 1 Conneection for Exaample 11.9

EXAMPLE E 11.9a Bolts in Co ombined Shear and Tension in Bearing g Type Connection ns by LRFD

Goal: G

Deetermine the strength of a connection uusing bolts inn combined sshear and ten nsion and com mpare it to thee applied loadd.

Given: G

An n inclined hanger that suppports a dead load of 10 kkips and a live load of 50 0 kips is show wn in Figure 11.18. The cconnection usses four 1 in. A325-N bo olts.

SOLUTION N

Step S 1:

Deetermine the required r strenngth for the apppropriate loaad combinatioon. Ru = 1.2 (10.0 ) + 1.6 ( 50.0 ) = 92.0 kiips

Step S 2:

Deetermine the force f assignedd to each boltt in tension annd shear. ⎛ 92.0 ⎞ Bolt tensionn = sin ( 30 ) ⎜ ⎟ = 11.5 kips ⎝ 4 ⎠ ⎛ 922.0 ⎞ Bolt shear = cos ( 30 ) ⎜ ⎟ = 19.9 kiips ⎝ 4 ⎠ 11.5 Bolt tensionn stress = f rt = = 144.6 ksi 0.785 19.9 = 25.44 ksi Bolt shear sstress = f rv = 0.785

Step S 3:

Deetermine the reduced r nomiinal tensile sttress. Th he nominal sh hear and tenssile stress vaalues are, from m Specification Table J3.2, Fnv = 54 kksi Fnt = 90 kksi nd the nomiinal shear sstress includiing the effeects of tensiion-shear an intteraction is f 25.4 Fnv′ = rv = = 33.9 ksi φ 0.75 Th hus, from Equ uation 11.7

532 Chapter 11

Simple Connections Fnt′ = 1.3Fnt −

Step 4:

Fnt ⎛ 90 ⎞ Fnv′ = 1.3(90) − ⎜ ⎟ (33.9) = 60.5 ≤ 90 Fnv ⎝ 54 ⎠

Check the design tensile stress versus the required tensile stress. φFnt′ = 0.75 ( 60.5 ) = 45.4 > 14.6 ksi Thus, by LRFD, the bolts are adequate.

EXAMPLE 11.9b Bolts in Combined Shear and Tension in Bearing Type Connections by ASD

Goal:

Determine the strength of a connection using bolts in combined shear and tension and compare it to the applied load.

Given:

An inclined hanger that supports a dead load of 10 kips and a live load of 50 kips is shown in Figure 11.18. The connection uses four 1 in. A325-N bolts.

SOLUTION

Step 1:

Determine the required strength for the appropriate load combination. Ra = 10.0 + 50.0 = 60.0 kips

Step 2:

Determine the force assigned to each bolt in tension and shear. ⎛ 60.0 ⎞ Bolt tension = sin ( 30 ) ⎜ ⎟ = 7.50 kips ⎝ 4 ⎠ ⎛ 60.0 ⎞ Bolt shear = cos ( 30 ) ⎜ ⎟ = 13.0 kips ⎝ 4 ⎠ 7.50 = 9.55 ksi Bolt tension stress = f rt = 0.785 13.0 = 16.6 ksi Bolt shear stress = f rv = 0.785

Step 3:

Determine the reduced nominal tensile stress. The nominal shear and tensile stress values are, from Specification Table J3.2, Fnv = 54 ksi Fnt = 90 ksi and the nominal shear stress including the effects of tension-shear interaction is Fnv′ = Ωf rv = 2.00 (16.6) = 33.2 ksi Thus, from Equation 11.7 F ⎛ 90 ⎞ Fnt′ = 1.3Fnt − nt Fnv′ = 1.3(90) − ⎜ ⎟ (33.2) = 61.7 ≤ 90 Fnv ⎝ 54 ⎠

Step 4:

Check the allowable tensile stress versus the required tensile stress. Fnt′ Ω = 61.7 2.00 = 30.9 > 9.55 ksi Thus, by ASD, the bolts are adequate.

Chhapter 11

Siimple Connecctions 533

When W high-sttrength bolts are installedd with an inittial pre-tensioon, they act aas a clamp, holding the t two conneected elementts together. F Figure 11.19 sshows a typiccal tension haanger where the bolts are expected d to carry thee applied tenssion load. Anny pre-tension from the bbolt actually causes a compressivee force to dev velop betweenn the conneccted parts. Appplication of the applied load redu uces the con ntact force bu ut has little eeffect on thee bolt tensionn, as long ass contact is maintained between the plates. Once O the plattes are separrated, the iniitial conditionns have no influencee and the bolt force, 2B, mu ust equal the applied load,, 2T.

If I the attacheed element, in this case the flange oof the tee, iss permitted to deform, as show wn in Figure 11.20a, 1 additiional forces ddevelop at thee tips of the flange. Thesee additional forces, q, q are shown in Figure 11 1.20b and aree the result oof prying actiion. They aree called the prying fo orces. There is a relationsship between the thickness of the flangge and the prrying force. When the thickness iss large, the pllate does not bend and noo prying actioon takes placee. When the thicknesss is small, beending of thee plate may bbe extensive and the pryiing force maay be large. Prying action may bee completely eliminated inn a design byy selecting a sufficiently thick plate, although this may nott be a practicaal solution. Itt may also bee avoided if w washers are uused to keep the flang ge from com ming into con ntact with thhe support; hhowever, this, too, is noormally not desirablee.

Figure 11.19 1 Hangeer Connection n with Bolts inn Tension

(a)

Figure 11.20 Tee Deformation with Prying Actio on Part (a) Photo o Courtesy of Patrick P McMan nus

(b)

5334 Chapter 11 1

Simple Connections C

Figure 11.21 Force Equ uilibrium Con nsidering Pryying Action The details for deesign of this type t of conneection includiing prying action are givenn in Part 9 of the Manual. It suggests that the minimum m plaate thickness needed to eliiminate pryinng action be determin ned. If this is a reasonable thickness, noo further action is required. If this thicckness is not reasonab ble for the deetails of the design, a desiggn that takes into account prying actionn should be undertak ken with the goal of havin ng a reasonabble combinattion of strenggth and stiffnness that results in an n economical connection. c Figu ure 11.21 sho ows a WT secction used as a hanger attacched to a supporting membber with bolts. The dimensions d given are used to determinee the relationsship betweenn the flexural strength of the flange and the app plied load. The applied looad is 2T, so the load perr bolt is T. It is not a simple matteer to determin ne the actual moment in thhe flange. Thhe design apprroach recognizes that the bolt heaad provides some s stiffenin ng to the flannge so that tthe moment aarm, b′, takenn as the distance from m the stem to o the edge off the bolt, willl provide a goood representtation of the required moment M r = Tb ′ . It hass also been fo ound through tests that thee flexural strenngth is best ppredicted by use of Fu rather than Fy. So, using g a tributary w width of platee associated w with each bollt, p, the nominal moment strength h is pt 2 (11.9) M n = Fu Z = Fu min 4

φ = 0.9 9 (LRFD)

Ω = 1.67 (ASD)

minimum thickkness to Settting the requiired strength equal to the available strrength, the m eliminate prrying action can be deterrmined. The Manual provvides these equations in tterms of LRFD and ASD A as 4Tu b′ Ω4Ta b ′ tmin = (LRFD) tmmin = (ASD) (Manuual 9-18) φpFu pFu

EX XAMPLE 11.10a 1 Ha Hanger Coonnection byy LR RFD

Goa al:

Deterrmine whetheer the WT haanger connecttion is adequuate to be dessigned witho out considerin ng prying acti on.

Giv ven:

A WT T9×48.5 sectiion, A992 steeel, is used as shown in Figgure 11.21 to carry a deaad load of 15 5 kips and a live load of 45 kips. Fouur 7/8 in. diaameter A325 5 bolts are useed in a 9 in. loong fitting.

Chapter 11

SOLUTION

Step 1:

Simple Connections 535

Determine the moment arm, b′, from the web to the edge of the bolt. Based on the properties of the section. tf = 0.870 in., tw = 0.535 in., bf = 11.1 in., gage = 5-1/2 in., p = 9.0/2 = 4.5 in. gage − t w (5.5 − 0.535) = = 2.48 in. 2 2 d 78 b′ = b − b = 2.48 − = 2.04 in. 2 2 b=

Step 2:

Determine the force per bolt. Tu = 1.2 (15.0 ) + 1.6 ( 45.0 ) = 90.0 kips Tu =

Step 3:

Determine the minimum flange thickness to ignore prying action using Manual Equation 9-20.

tmin = Step 4:

90.0 = 22.5 kips/bolt 4

4 ( 22.5 )( 2.04 ) 4Tu b′ = = 0.835 in. 0.9 ( 4.5 )( 65) φpFu

Compare the available thickness with the required thickness. tmin = 0.835 < t f = 0.870 in. Because the actual flange thickness is greater than the minimum, the WT9×48.5 does not require consideration of prying action.

EXAMPLE 11.10b Goal: Hanger Connection by Given: ASD

Determine whether the WT hanger connection is adequate to be designed without considering prying action.

Step 1:

Determine the moment arm, b′, from the web to the edge of the bolt. Based on the properties of the section.

SOLUTION

A WT9×48.5 section, A992 steel, is used as shown in Figure 11.21 to carry a dead load of 15 kips and a live load of 45 kips. Four 7/8 in. diameter A325 bolts are used in a 9 in. long fitting.

tf = 0.870 in., tw = 0.535 in., bf = 11.1 in., gage = 5-1/2 in., p = 9.0/2 = 4.5 in. gage − t w (5.5 − 0.535) = = 2.48 in. 2 2 d 78 b′ = b − b = 2.48 − = 2.04 in. 2 2 b=

536 Chapter 11

Simple Connections Step 2:

Determine the force per bolt. Ta = 15.0 + 45.0 = 60.0 kips

Ta = Step 3:

Determine the minimum flange thickness to ignore prying action using Manual Equation 9-20.

tmin = Step 4:

60.0 = 15.0 kips / bolt 4

1.67 ( 4 )(15.0 )( 2.04 ) Ω4Ta b′ = = 0.836in. 4.5 ( 65 ) pFu

Compare the available thickness with the required thickness. tmin = 0.836 < t f = 0.870 in. Because the actual flange thickness is greater than the minimum, the WT9×48.5 does not require consideration of prying action.

11.11 BEAM BEARING PLATES AND COLUMN BASE PLATES The connections discussed throughout this chapter transfer force through a series of connecting elements to a supporting member. Two other types of simple connections deserve mention here, the beam bearing plate and the column base plate, both addressed in Part 14 of the Manual. These plates transfer a force through direct bearing from one member to another member or directly to a support. Although these plates are used in two very different applications, their actual behavior is quite similar. Some aspects of beam bearing were discussed in Section 6.14, and column base plates were briefly discussed in Section 5.11. For design of the plate, three properties must be determined: the width and breadth, which results in an appropriate area, and the thickness. The area of the plate is determined by assessing the limit states of the supporting member or material and those of the member applying the force to the plate. The thickness of the plate is determined through the limit state of flexural yielding of the plate. To determine the required plate thickness for either type of plate, two primary assumptions are made: (1) the plate exerts a uniform pressure on the supporting material, and (2) the plate is treated as a cantilevered strip that is 1.0 in. wide. For a bending cross section 1.0 in. wide with a thickness, tp, the nominal flexural strength for the limit state of yielding is ⎛ 1.0t 2p ⎞ (11.10) M n = Fy Z = Fy ⎜ ⎟ ⎝ 4 ⎠ For a uniform contact pressure between the plate and the supporting material, fp, and a cantilever length, l, the required moment strength for the cantilever is f pl 2 (11.11) Mr = 2 For LRFD, the required plate thickness can be obtained by setting the design moment equal to the required moment, where the required moment is obtained using fp = fu and φ = 0.9; thus, φM n = M r φFy t 2p f l2 = u 4 2

Chhapter 11

Siimple Connecctions 537

which yields

t p = 1.49l

fu Fy

D) ( LRFD

(11.12a)

Similarly, S for ASD, using fp = fa and Ω = 1.67, Mn Ω = Mr Fy t 2p f a l 2 = 4Ω 2

which yields

t p = 1.83l

fa Fy

D) ( ASD

(11.12b)

The determination of the cantileverr distance, l, tto be used in the case of a beam bearinng plate or a column base b plate is addressed in n Manual Parrt 14. For a bbeam bearingg plate, shown in Figure 11.22, th he cantilever distance d is meeasured from the point idenntified as the k distance froom the web centerline to the edge of the plate, noted n in the ffigure as n. Manual Tablle 1-1 provid des two valuues for the seection properrty k, kdes annd kdet. The smaller value, v kdes, caalled k-design n, should be cchosen so thaat, conservativvely, a longeer cantilever distance is used in thee design. For column basee plates, the ccantilever dim mensions are bbased on an assumptiion for the po oint under thee column secttion where caantilever actioon begins. Thhis is shown in Figuree 11.23, where the cantileever dimensioon, n, is meaasured from a line 0.1bf iin from the flange tip ps and the caantilever dimension, m, iss measured frrom a line 0.0025d in from m the flange face. A third cantilever dimensio on is also givven to accounnt for those caases where the cantilever from the web, in the area between n the flanges,, is critical. T The Manual ggives this as λn′ , where n′ =

bf d 4

and λ can be conservatiively taken ass 1.0.

The T design beearing streng gth, φc Pp , andd the allowabble bearing sttrength, Pp Ω c , under a beam bearing plate orr a column base b plate are each a functtion of the m material that ssupports the or a concrete support, Speccification Secction J8 provvides that whhen the plate is the same plate. Fo size as th he concrete su upport,

Figure 11.22 1 Beam Bearing B Plate

5338 Chapter 11 1

Simple Connections C

Figure 11.23 Column Baase Plate

Pp = 0.855 f c′A1

and when th he plate is lesss than the fulll area of the suupport, A Pp = 0.85 f c′A1 2 ≤ 1.7 f c′A1 A1

(AIS SC J8-1)

(AIS SC J8-2)

where A1 = area of the pllate; A2 = max ximum area oof the supportt that is geom metrically the same as the plate; fc′ = compressiive strength of o the concrette; φc = 0.65 ffor LRFD; and Ωc = 2.31 ffor ASD. The limit on n Equation J8--2 means thatt A2 A1 ≤ 2 .0 . Thus, A2 may be up too 4 times A1. C Concrete bearing stren ngth was also o discussed in n Section 5.11.

EX XAMPLE 11.11a 1 Beeam Bearing g Pllate by LRFD D

Goa al:

Deterrmine the requ uired thicknesss of a beam bbearing plate.

Giv ven:

A W18×40 A992 beam is suppported by a 7×4 in. A366 bearing plaate as shown n in Figure 11.24. The bbeam end reaaction is Ru = 50.0 kips. From Manu ual Table 1-1,, kdes = 0.927 iin.

SO OLUTION

Step p 1:

Deterrmine the presssure under thhe plate. 50.0 fu = = 1.779 ksi 77.0 ( 4.0 )

Step p 2:

Deterrmine the canttilever dimennsion.

Step p 3:

Deterrmine the min nimum plate tthickness usinng Equation 111.12a

l = n = 33.5 − 0.927 = 2.57 in.

t p = 1.49l Step p 4:

fu 1.79 = 1.49(2.57) = 0.854 in. Fy 36

Select the required d plate thickneess. Usse a 7/8 in. plaate

Chhapter 11

Siimple Connecctions 539

Figure 11.24 1 Beam Bearing B Plate for f Example 11.11 G EXAMPLE E 11.11b Goal: Beam Bearring Given: G Plate by AS SD

SOLUTION N

Deetermine the required r thickkness of a beaam bearing pllate. A W18×40 A9 992 beam is supported byy a 7×4 in. A36 bearing plate as sh hown in Figu ure 11.24. Thhe beam end reaction is Ra = 34.0 kipps. From Manual M Table 1-1, 1 kdes = 0.9927 in.

Step S 1:

Deetermine the pressure p undeer the plate. 34.0 fa = = 1.21 ksi 7.0 ( 4.0 )

Step S 2:

Deetermine the cantilever c dim mension.

Step S 3:

Deetermine the minimum m plaate thickness uusing Equatioon 11.12b.

l = n = 3.5 − 0.9277 = 2.57 in.

t p = 1.83l Step S 4:

fa 1.21 = 0.8862 in. = 1.83(2.577) Fy 36

Seelect the requiired plate thicckness. Use a 7/8 inn. plate

Bearing B on th he beam web must also be checked for the limit stattes of web loccal yielding were discusseed in Sectionn 6.15 and illlustrated in and web b local cripplling. These liimit states w Examplee 6.21.

5440 Chapter 11 1

Simple Connections C

Figure 11.25 Column Baase Plate for Example E 11.1 2

EX XAMPLE 11.12a 1 Coolumn Base Plate byy LRFD

Goa al:

Deterrmine the requ uired thicknesss of a colum mn base plate.

Giv ven:

A W14×74 A992 column is ssupported byy a 13×17 inn. A36 base plate, which h itself is sup pported on a 19×23 in. cooncrete pier aas shown in F Figure 11.25 5. The column n carries a dessign load Pu = 620 kips annd the concrete has a com mpressive strength of fc′ = 5 ksi .

SO OLUTION

Step p 1:

Deterrmine the desiign strength oof the concrette under the pplate. Area of the base pllate

A1 = 13(17) = 2211 in.2

Area of the concreete support

A2 = 19( 23) = 4377 in.2

2 = 1.41 < 2..0 , the nominnal bearing strrength using Thus,, since 437 221 Equattion J8-2 is A 437 Pn = Pp = 0.85 f c′A1 2 = 0.885 ( 5)( 221) = 939 (1..41) = 1320 kiips A1 221

Step p 2:

Deterrmine if the deesign bearingg strength is suufficient to suupport the loaad. φPp = 0.65 (13320 ) = 858 kipps > 620 kipss Thus,, the concrete strength is addequate.

Chapter 11

Simple Connections 541

Step 3:

Determine the pressure under the plate. 620 fu = = 2.81 ksi 13(17 )

Step 4:

Determine the cantilever dimension. B − 0.8b f 13.0 − 0.8(10.1) n= = = 2.46 in. 2 2 N − 0.95d 17.0 − 0.95(14.2) m= = = 1.76 in. 2 2 and n′ =

bf d 4

=

10.1(14.2) = 2.99 4

Thus, the maximum cantilever dimension is l = 2.99 in.

Step 5:

Determine the minimum plate thickness using Equation 11.12a. f 2.81 t p = 1.49l u = 1.49(2.99) = 1.24in. Fy 36

Step 6:

Select the required plate thickness. Use a 1-1/4 in. plate Note: A more efficient plate dimension could be selected if the B and N dimensions were adjusted so that m and n were closer to being equal and the conservative assumption of λ = 1.0 was not made. Manual Part 14 includes a complete discussion of the use of λ in determining base plate required thickness.

EXAMPLE 11.12b Goal: Column Base Plate Given: by ASD SOLUTION

Step 1:

Determine the required thickness of a column base plate. A W14×74 A992 column is supported by a 13×17 in. A36 bearing plate as shown in Figure 11.25. The column carries an allowble load Pa = 410 kips. Determine the allowable strength of the concrete under the plate. Area of the base plate

A1 = 13(17) = 221 in.2

Area of the concrete support

A2 = 19( 23) = 437 in.2

Thus, since 437 221 = 1.41 < 2.0 , the nominal bearing strength using Equation J8-2 is

542 Chapter 11

Simple Connections

Pn = Pp = 0.85 f c′A1 Step 2:

A2 437 = 0.85 ( 5 )( 221) = 939 (1.41) = 1320 kips A1 221

Determine if the allowable bearing strength is sufficient to support the load. Pp Ω = 1320 2.31 = 571 kips > 410 kips Thus, the concrete strength is adequate.

Step 3:

Determine the pressure under the plate. 410 fa = = 1.86 ksi 13(17 )

Step 4:

Determine the cantilever dimension. B − 0.8b f 13.0 − 0.8(10.1) n= = = 2.46 in. 2 2 N − 0.95d 17.0 − 0.95(14.2) m= = = 1.76 in. 2 2 and λn′ =

bf d

4

= (1.0 )

10.1(14.2) = 2.99 4

Thus, the maximum cantilever dimension is l = 2.99 in.

Step 5:

Determine the minimum plate thickness using Equation 11.12b. f 1.86 t p = 1.83l a = 1.83(2.99) = 1.24 in. Fy 36

Step 6:

Select the required plate thickness. Use a 1-1/4 in. plate Note: A more efficient plate dimension could be selected if the B and N dimensions were adjusted so that m and n were closer to being equal and the conservative assumption of λ=1.0 was not made. Manual Part 14 includes a complete discussion of the use of λ in determining base plate required thickness

11.12

PROBLEMS

1. Design an all-bolted double-angle connection for a W18×65 A992 beam to carry a dead load reaction of 15 kips and a live load reaction of 45 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The uncoped beam is connected to the flange of a W14×109 A992 column. Design by (a) LRFD and (b) ASD.

2. Design an all-bolted double-angle connection for a W27×114 A992 beam to carry a dead load reaction of 30 kips and a live load reaction of 90 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The uncoped beam is connected to the web of a W36×135 A992 girder. Design by (a) LRFD and (b) ASD.

Chapter 11 3. Design an all-bolted double-angle connection for a W24×146 A992 beam to carry a dead load reaction of 25 kips and a live load reaction of 75 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The uncoped beam is connected to the flange of a W14×132 A992 column. Design by (a) LRFD and (b) ASD. 4. Design an all-bolted double-angle connection for a W16×77 A992 beam to carry a dead load reaction of 20 kips and a live load reaction of 60 kips. Use 3/8 in. A36 angles and 3/4 in. A490-N bolts in standard holes. The beam is uncoped and the supporting member is not critical. Design by (a) LRFD and (b) ASD. 5. Design an all-bolted double-angle connection for a W18×143 A992 beam to carry a dead load reaction of 25 kips and a live load reaction of 75 kips. Use 3/8 in. A36 angles and 3/4 in. A490-N bolts in standard holes. The beam is uncoped and the supporting member is not critical. Design by (a) LRFD and (b) ASD. 6. Design an all-bolted double-angle connection for a W8×40 A992 beam to carry a dead load reaction of 8 kips and a live load reaction of 24 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is uncoped and the supporting member is not critical. Design by (a) LRFD and (b) ASD. 7. Design an all-bolted double-angle connection for a W30×191 A992 beam spanning 40 ft and carrying a total uniformly distributed dead load of 60 kips and live load of 180 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is uncoped and the supporting member is not critical. Design by (a) LRFD and (b) ASD.

Simple Connections 543

edge distance is 1-1/4 in. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 11. Design an all-bolted double-angle connection for a W18×76 A992 beam to support a dead load reaction of 16 kips and a live load reaction of 48 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 12. Design an all-bolted double-angle connection for a W21×73, A992 beam spanning 20 ft and carrying a total uniformly distributed dead load of 28 kips and live load of 84 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 13. Design an all-bolted double-angle connection for a W24×84, A992 beam to support a dead load reaction of 25 kips and a live load reaction of 75 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 14. Design an all-bolted double-angle connection for a W12×87 A992 beam to support a dead load reaction of 14 kips and a live load reaction of 42 kips. Use 3/8 in. A36 angles and 3/4 in. A490-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. The supporting member is not critical. Design by (a) LRFD and (b) ASD.

8. Design an all-bolted double-angle connection for a W18×76 A992 beam to support a dead load reaction of 16 kips and a live load reaction of 48 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is uncoped and the supporting member is not critical. Design by (a) LRFD and (b) ASD.

15. Design an all-bolted double-angle connection for a W16×77 A992 beam spanning 20 ft and carrying a total uniformly distributed dead load of 23 kips and live load of 69 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. The supporting member is not critical. Design by (a) LRFD and (b) ASD.

9. Design an all-bolted double-angle connection for a W21×73 A992 beam spanning 20 ft and carrying a total uniformly distributed dead load of 28 kips and live load of 84 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is uncoped and the supporting member is not critical. Design by (a) LRFD and (b) ASD.

16. Design an all-bolted double-angle connection for a W16×77 A992 beam to carry a dead load reaction of 20 kips and a live load reaction of 60 kips. Use 3/8 in. A36 angles and 3/4 in. A490-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. and the supporting member is not critical. Design by (a) LRFD and (b) ASD.

10. Design an all-bolted double-angle connection for a W30×191 A992 beam spanning 40 ft and carrying a total uniformly distributed dead load of 60 kips and live load of 180 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is coped so that the

17. Design an all-bolted double-angle connection for a W18×143 A992 beam to carry a dead load reaction of 25 kips and a live load reaction of 75 kips. Use 3/8 in. A36 angles and 3/4 in. A490-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. and

544 Chapter 11

Simple Connections

the supporting member is not critical. Design by (a) LRFD and (b) ASD. 18. Design an all-bolted double-angle connection for a W8×40 A992 beam to carry a dead load reaction of 8 kips and a live load reaction of 24 kips. Use 5/16 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is coped so that the edge distance is 1-1/4 in. and the supporting member is not critical. Design by (a) LRFD and (b) ASD. 19. Design a welded-bolted double-angle connection for an uncoped W18×65 A992 beam to carry a dead load reaction of 15 kips and a live load reaction of 45 kips. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam and 3/4 in. A325-N bolts in standard holes to connect to the supporting member. The beam is connected to the flange of a W14×109 A992 column. Design by (a) LRFD and (b) ASD. 20. Design a welded-bolted double-angle connection for an uncoped W27×114 A992 beam to carry a dead load reaction of 30 kips and a live load reaction of 90 kips. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam and 3/4 in. A325-N bolts in standard holes to connect to the supporting member. The beam is connected to the web of a W36×135 A992 girder. Design by (a) LRFD and (b) ASD. 21. Design a welded-bolted double-angle connection for an uncoped W24×146 A992 beam to carry a dead load reaction of 25 kips and a live load reaction of 75 kips. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam and 3/4 in. A325-N bolts in standard holes to connect to the supporting member. The beam is connected to the flange of a W14×132 A992 column. Design by (a) LRFD and (b) ASD. 22. Design a welded-bolted double-angle connection for an uncoped W33×241 A992 beam to carry a dead load reaction of 50 kips and a live load reaction of 150 kips. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam and 7/8 in. A490-N bolts in standard holes to connect to the supporting member. The beam is connected to the flange of a W14×370 A992 column. Design by (a) LRFD and (b) ASD. 23. Design a welded-bolted double-angle connection for a coped W18×76 A992 beam to support a dead load reaction of 16 kips and a live load reaction of 48 kips. Assume the beam is coped so that the edge distance is 11/4 in. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam

and 3/4 in. A325-N bolts in standard holes to connect to the supporting member. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 24. Design a welded-bolted double-angle connection for a coped W12×87 A992 beam to support a dead load reaction of 14 kips and a live load reaction of 42 kips. Assume the beam is coped so that the edge distance is 11/4 in. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam and 3/4 in. A325-N bolts in standard holes to connect to the supporting member. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 25. Design a welded-bolted double-angle connection for a coped W16×77 A992 beam spanning 20 ft and carrying a total uniformly distributed dead load of 23 kips and live load of 69 kips. Assume the beam is coped so that the edge distance is 1-1/4 in. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam and 3/4 in. A325-N bolts in standard holes to connect to the supporting member. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 26. Design a welded-bolted double-angle connection for a coped W21×73 A992 beam spanning 25 ft and carrying a total uniformly distributed dead load of 30 kips and live load of 90 kips. Assume the beam is coped so that the edge distance is 1-1/4 in. Use 5/16 in. A36 angles and 70 ksi welding electrodes to connect to the web of the supported beam and 3/4 in. A325-N bolts in standard holes to connect to the supporting member. The supporting member is not critical. Design by (a) LRFD and (b) ASD. 27. Design a bolted-bolted single-angle connection for an uncoped W18×65 A992 beam to carry a dead load reaction of 8 kips and a live load reaction of 24 kips using 3/8 in. A36 angles and 3/4 in. A325-N bolts in standard holes. The beam is connected to the web of a W36×150 A992 girder. Design by (a) LRFD and (b) ASD. 28. Design a bolted-welded single-angle connection for a coped W12×87 A992 beam to support a dead load reaction of 7 kips and a live load reaction of 21 kips. Assume the beam is coped so that the edge distance is 11/4 in. Use 3/8 in. A36 angles, E70 electrodes, and 3/4 in. A325-N bolts in standard holes. Determine the minimum thickness for the web of the supporting A992 girder. Design by (a) LRFD and (b) ASD. 29. Design a welded-welded single-angle connection for a coped W16×77 A992 beam spanning 20 ft and carrying a total uniformly distributed dead load of 12 kips and live load of 36 kips. Assume the beam is coped so that the

Chapter 11 edge distance is 1-1/4 in. Use 3/8 in. A36 angles and E70 electrodes. Design by (a) LRFD and (b) ASD. 30. Design a shear tab connection for an uncoped W18×65 A992 beam to carry a dead load reaction of 10 kips and live load reaction of 30 kips. Use a 3/8 in. thick A36 shear tab and 3/4 in. A325-N bolts. Assume that the supporting member is not critical. Design by (a) LRFD and (b) ASD. 31. Design a shear tab connection for an uncoped W27×114, A992 beam to carry a dead load reaction of 15 kips and a live load reaction of 45 kips. Use a 5/16 in. thick A36 shear tab and 3/4 in. A490-N bolts. Assume that the supporting member is not critical. Design by (a) LRFD and (b) ASD. 32. Design a shear tab connection for a coped W21×73 A992 beam spanning 20 ft and carrying a total uniformly distributed dead load of 23 kips and live load of 70 kips. Assume that the edge distance at the cope is 1-1/4 in. Use a 5/16 in. thick A36 shear tab and 3/4 in. A490-N bolts. Assume that the supporting member is not critical. Design by (a) LRFD and (b) ASD. 33. Design a shear tab connection for a coped W18×76 A992 beam to carry a dead load reaction of 10 kips and a live load reaction of 30 kips. Assume an edge distance of 1-1/4 in. Use a 3/8 in. thick A36 shear tab and 3/4 in. A325-N bolts. Assume that the supporting member is not critical. Design by (a) LRFD and (b) ASD. 34. Design a welded seated connection for a W16×26 A992 beam framing into the web of a W14×99 A992 column. The seat must carry a dead load reaction of 6 kips and a live load reaction of 18 kips. Use an equal-leg A36 angle and E70 electrode. Design by (a) LRFD and (b) ASD. 35. Design a welded seated connection for a W18×40 A992 beam framing into the web of a W14×109 A992 column. The seat must carry a dead load reaction of 8 kips and a live load reaction of 24 kips. Use an equal-leg A36 angle and E70 electrode. Design by (a) LRFD and (b) ASD. 36. Design the connection for an A36 double-angle tension member connected to a uniform-width A36 gusset plate. The angles are L4×4×1/2 and carry a dead load of 10 kips and a live load of 30 kips. The angles are connected to the gusset plate by a single line of 3/4 in. A325-N bolts. The gusset plate is welded perpendicular to the axis of the member with welds from E70 electrodes. Design by (a) LRFD and (b) ASD. 37. An inclined WT hanger is used to support a tension member carrying a dead load of 8 kips and a live load of

Simple Connections 545

24 kips. The force is applied at an angle of 45 degrees from the horizontal and is transferred by four 3/4 in. A325-N bolts. Determine whether the bolts have sufficient strength to carry the applied load by (a) LRFD and (b) ASD. 38. A 9 in. long WT7×24, A992 steel, is used as a tension hanger with four 3/4 in. A325-N bolts in the flanges similar to the setup shown in Figure 11.19. The hanger must resist a dead load of 11 kips and a live load of 33 kips. Determine whether prying action must be included to determine the connection strength by (a) LRFD and (b) ASD. 39. Determine the required thickness of an A36 beam bearing plate to support an A992 W16×77 beam carrying a dead load reaction of 10 kips and a live load reaction of 30 kips. The plate is 12.0 in. wide and 6.0 in. along the beam web. Design by (a) LRFD and (b) ASD. 40. Determine the required thickness of an A36 beam bearing plate to support an A992 W18×65 beam carrying a dead load reaction of 15 kips and a live load reaction of 45 kips. The plate is 7.5 in. wide and 10.0 in. along the beam web. Design by (a) LRFD and (b) ASD. 41. Determine the required thickness of a 24 in.×24 in. A36 column base plate that supports an A992 W14×193 column and carries a dead load of 100 kips and a live load of 300 kips. Design by (a) LRFD and (b) ASD. 42. Determine the required thickness of a 17 in.×15 in. A36 column base plate that supports an A992 W12×79 column and carries a dead load of 130 kips and a live load of 390 kips. Design by (a) LRFD and (b) ASD. 43. Integrated Design Project. Design double-angle shear connections for the gravity beams and girders designed in Chapter 6. Some of these connections will be to girder webs, some to column webs, and some to column flanges. Reconsider the types of shear connections, and design using single-plate connections.

C Chapterr 12

M Momen nt Co onnecttions

Four World T Trade Center, N New York

122.1

TYPES S OF MOM MENT CONN NECTIONS S Although th hey are called d moment co onnections, thhese connections are expeected to transsfer both shear and moment m betw ween the co onnected mem mbers. Mom ment connecttions are deffined in Specification n Section B3 3.4b as fully restrained, r Tyype FR, or ppartially restraained, Type P PR. The impact of th hese connectio on types on th he behavior oof a steel fram me was discusssed in Chapteer 8. For fully restrain ned connectio ons, the mom ment is transfeerred while thhe relative rotaation of the m members remains at zero. For paartially restraained connecttions, the mooment is trannsferred while some predictable relative r rotatiion is permitted. For the siimple shear cconnection as discussed in Chapter 11, no mom ment is expecteed to be transferred and thee connection is assumed too rotate freelyy. Figure 8.21, redraw wn here as Figure F 12.1, shows s three moment-rotaation curves ffor connectioons with distinctly diifferent behav vior. The rig gid and simpple connectionn behavior shhown in Figuure 12.1 illustrates that t real con nnection beh havior does not exactly follow the ideal behavvior, as demonstrateed by the verrtical axis forr a rigid connnection and tthe horizontaal axis for thee simple connection. Fivee moment con nnections thatt are commonnly used for cconnecting beeams framing into the strong axis of columns are a illustrated d in Figure 122.2. For the ffirst four exam mples, whichh include welded flange (Figure 12.2aa), the weldedd flange platee (Figure 12.22b), the bolteed flange the direct-w plate (Figurre 12.2c), an nd the bolted d tee (Figuree 12.2d), sheear is transfeerred throughh a web connection similar s to tho ose discussed in Chapter 111, whereas tthe moment iis transferred through the various flange conneections. In th he extended eend plate connnection, Figgure 12.2e, shhear and moment are combined an nd transferred through the cconnecting pllate and bolts. Because shear is resisted prim marily by the w web of a widee-flange beam m, it is logicall that the shear force is transferreed through th he web connnection to thhe supportingg member. Siimilarly, because the moment is resisted r primaarily throughh the flange oof a wide-flannge beam, thhe flange connections transfer the moment to the t supportingg member. B Because of thhe moment reesistance provided by the flanges, there t is no neeed to consideer eccentricityy in the designn of the web sshear

5446

Chappter 12

Mooment Connecctions 547

Figure 112.1 Connection Behavioor connectio on. Thus, th he web platee or angles are sized too resist only shear, simpplifying the connectio on design. Th he flange con nnection is deesigned to ressist the full m moment, even though the flanges do d not actually y carry this fu ull moment. T Through strainn hardening, and in combiination with some unccalculated mo oment strengtth of the web connection, the flange coonnections aree capable of developin ng the full mo oment. The T welded flange f connecction, Figure 12.2a, is the most direct moment connnection and requires the fewest paarts. The flan nges are field--welded to thhe supportingg member witth complete joint pen netration groov ve welds. Thee web connecction is usuallly a single plaate welded to the column and bolteed to the beaam. In this arrrangement, thhe flange forrce, Pf, is deteermined by ddividing the moment by the distancce between fllange midpoinnts. Thus, Mr (12.1) Pf = (d − t f ) The T flange plated p connecctions, Figuree 12.2b and c, transfer tthe flange foorces to the correspon nding plates through t eitherr bolt shear o r weld shear. The plate forrce is then traansferred to

Figure 12.2 1 Momen nt Connection ns

548 Chapter 12

Moment Connections

Table 12.1 Sections of Specification and Parts of Manual Covered in This Chapter Specification B4.1 B4.3 D3 F13 J2 J3 J4 J10

Classification of Sections for Local Buckling Gross and Net Area Determination Effective Net Area Proportions of Beams and Girders Welds Bolts and Threaded Parts Affected Elements of Members and Connecting Elements Flanges and Webs with Concentrated Forces Manual

Part 7 Part 8 Part 9 Part 10 Part 15

Design Considerations for Bolts Design Considerations for Welds Design of Connecting Elements Design of Simple Shear Connections Design of Hanger Connections, Bracket Plates, and Crane-Rail Connections

the supporting member through welds. The flange plate connectors, bolts or welds, are sized to resist the force developed at the plate-flange interface. Thus, M (12.2) Pf = r d The bolted tee connection, Figure 12.2d, is needed when the connection to the supporting member must be bolted. Although this connection is not as clean and simple as the flange plate connections, it provides a solution when there is a compelling reason to require an all-bolted connection. The connection to the beam flange is treated as with the flange plate connections, and the connection to the support is treated similarly as the tension connection discussed in Chapter 11. The extended end plate connection, shown in Figure 12.2e, may take a variety of forms. The end plate is fully welded to the end of the beam and then bolted to the support. The end plate must extend beyond the beam flange on the tension side so that a minimum of four bolts can be symmetrically spaced, with the flange located at the bolt group centroid. If an extended end plate connection is called upon to resist a moment that is always acting in the same direction, it may be extended on only one side. However, if the moment is expected to reverse, the plate must be extended beyond both the top and bottom flanges. Table 12.1 lists the sections of the Specification and parts of the Manual discussed in this chapter.

12.2 LIMIT STATES The limit states that control the strength of these connections are the same as those already considered for the shear connections. Their specific application depends on the complete connection geometry and the forces that the elements are expected to carry. These limit states include 1. Bolts a. Shear rupture b. Tension c. Shear-tension interaction d. Bearing and tearout

Chapter 12 2. 3.

4.

Welds a. b. Plates a. b. c. d. e. f. Beam a. b. c.

Moment Connections 549

Tension rupture Shear rupture Compression buckling Tension yielding Tension rupture Shear yielding Shear rupture Block shear Flexure of reduced section Shear yield Shear rupture

In addition to these limit states, which are all associated with the beam side of the connection, the designer must consider the impact of the connection on the column to which it is attached. Here limit states include 5. Column a. Flange local bending b. Web local yielding c. Web local crippling d. Web compression buckling e. Web panel zone shear

12.3 MOMENT CONNECTION DESIGN Design of moment connections is presented in two parts. First, examples are given for a directwelded beam-to-column connection, a welded flange plate connection, and a bolted flange plate connection. These examples treat the beam side of the connection without considering the column to which the connection is attached. This is followed by a discussion of the limit states associated with the column, and examples are given to illustrate that design process. 12.3.1

Direct-Welded Flange Connection The direct-welded flange moment connection provides an FR connection with very few connecting elements. As the name implies, the flanges are directly welded to the supporting member, usually the flange of a column. These welds are either complete joint penetration groove welds or a pair of fillet welds on each side of the beam flanges. The groove weld provides a weld that can be made in the downward position for both flanges, whereas the fillet welds require overhead welding on the bottom of each flange. The only limit state to consider for the flange connection is tension or shear rupture of the weld. The web connection is usually made with a single plate, the same as for the shear tab simple connection discussed in Chapter 11. However, unlike the shear tab connection, the web plate in this FR connection does not need to take into account any eccentricity because the flanges are designed to carry all of the moment. The limit states for the web connection are those previously discussed for the shear tab. Although the Specification specifically identifies fully restrained moment connections as Type FR moment connections, it is normal engineering practice to simply refer to them as moment connections. Thus, unless specified as Type PR connections, all moment connections are assumed to be Type FR. Since that is normal practice, that is the approach taken in this book.

5550 Chapter 12 1

Momentt Connectionss

Figure 12.3 Connection n for Example 12.1

EX XAMPLE 12.1a 1 Diirect-Welded d M Moment Coonnection byy LR RFD

Goa al:

Desig gn a direct-weelded beam-too-column mom ment connecttion.

Giv ven:

A dirrect-welded beam-to-colum b mn moment connection iss shown in F Figure 12.2aa. The beam iss a W24×76 aand the colum mn is a W14× ×109. Bolts aare 3/4 in. A3 325-N and th he electrodes are E70. Thee shapes are A A992 steel annd the plate is A36. The required r strenngth is Mu = 5500 ft-kips and Vu = 60.0 kkips.

SO OLUTION

Step p 1:

Obtaiin the beam an nd column prroperties from m Manual Tabble 1-1. m: W24×76 Beam d = 23.9 in., bf = 8.99 in., tw = 0.440 in.,, tf = 0.680 inn., Z = 200 in.3 Colum mn: W14×109 9 d = 14..3 in., bf = 14..6 in., tw = 0.5525 in., tf = 0..860 in.

Step p 2:

Check k the flexurall strength of thhe beam. d have been m made during design of thee beam. The beam This check should on is not redu uced becausee of bolt holees in the flannge, so Mp ccan be sectio determ mined using the gross secction plastic ssection moduulus and the ddesign streng gth is 0.99(50)(200) = 750 ft-kips > 500 ft-kips φM n = φM p = φFy Z = 12 Thus,, the flexural strength is addequate.

Step p 3:

Desig gn the flange-to-column weeld. f umn weld c an be eitherr a completee joint penettration The flange-to-colu (CJP)) groove weld d or fillet wellds. CJP weldds are used in this examplee. E70 electrrodes match th he strength o f the plates, aas indicated bby the User N Note in

Chapter 12

Moment Connections 551

Section J2.6. Therefore, the CJP welds will develop the full strength of the beam flanges without tension rupture of the weld and no further calculations are needed. Step 4:

Design the web plate. First consider the shear rupture strength of the bolts to determine the minimum number of bolts required. For a 3/4 in. A325-N bolt, φrn = 17.9 kips; therefore, the minimum number of bolts based on bolt shear is 60 n= = 3.35 17.9 Thus, try a four-bolt connection with bolt spacing of 3.0 in. and end distances of 1.5 in. Thus, l = 12.0 in., which is greater than T/2 = 10.0 in. Try a plate with t = 3/8 in. See Figure 12.3.

Step 5:

Determine the bolt bearing strength. Since the beam web is thicker and has a higher yield strength than the plate, bearing and tearout of the plate will be more critical than the beam web. For the last bolt, determine the clear distance. 1 1 lc = lev − d h = 1.5 − (3/4 + 1/16) = 1.09 < 2(3/4) = 1.5 in. 2 2 Thus, tearout controls over bearing and the bolt nominal strength is Rn = 1.2lc tFu = 1.2 (1.09 )( 0.375 )( 58 ) = 28.4 kips The design strength for tearout is φRn = 0.75 ( 28.4 ) =21.3 kips Since this is greater than the design shear rupture strength, shear rupture will control for this bolt. For the other bolts lc = s − d h = 3.0 − ( 3 4 + 1 16 ) = 2.19 > 2 ( 3 4 ) = 1.5in. Therefore, bearing will control over tearout, and the bolt nominal strength is Rn = 2.4dtFu = 2.4 ( 3 4 )( 0.375 )( 58 ) = 39.2 kips The design strength for bearing is φRn = 0.75 ( 39.2 ) =29.4 kips Since this is greater than the design shear rupture strength, shear rupture will control for these bolts also.

552 Chapter 12

Moment Connections Thus, for the four-bolt connection, the design strength is φRn = 4 (17.9 ) = 71.6 > 60.0 kips as expected from step 4. Step 6:

Check the plate for shear yield. The gross shear area is Agv = tl = 0.375 (12.0 ) = 4.50 in.2 and the design shear yield strength is φVn = φ0.6 Fy Agv = 1.0 ( 0.6 ( 36 )( 4.50 ) ) = 97.2 > 60.0 kips

Step 7:

Check the plate for shear rupture. The net shear area is Anv = ( l − n ( d h + 1 16 ) ) t = (12.0 − 4 ( 3 4 + 1 8) ) ( 0.375) = 3.19 in.2 And the design shear rupture strength is φVn = φ0.6 Fu Anv = 0.75 ( 0.6 ( 58 )( 3.19 ) ) = 83.3 > 60.0 kips

Step 8:

Check block shear of the plate. First calculate the areas. 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜1.5 − ( 3 4 + 1 8 ) ⎟ ( 0.375 ) = 0.398 in.2 2 ⎝ ⎠ Agv = ltw = 10.5 ( 0.375 ) = 3.94 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.5 − 3.5 ( 3 4 + 1 8 ) ) ( 0.375 ) = 2.79 in.2

Determine the tension rupture strength Fu Ant = 58 ( 0.398) = 23.1 kips Consider shear yield and shear rupture and select the one with least strength; thus, 0.6 Fy Agv = 0.6 ( 36 )( 3.94 ) = 85.1 kips 0.6 Fu Anv = 0.6 ( 58 )( 2.79 ) = 97.1 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection block shear strength, with Ubs = 1.0, of φRn = 0.75 ( 85.1 + 23.1) = 81.2 kips > 60.0 kips Step 9:

Select the plate-to-column weld.

Chapter 12

Moment Connections 553

Based on a fillet weld on each side of the plate, with a weld design strength of 1.392 kips per 1/16 in. of weld per in. of length 60.0 D= = 1.80 sixteenths of in. (2(1.392)(12.0)) Therefore, use a 3/16 in. weld, the minimum weld for the 3/8 in. plate, as given in Specification Table J2.4. Step 10:

Develop the final design. Figure 12.3 shows the final design using Four 3/4 in. A325-N bolts in a 3/8×3-1/2×1’-0 plate The impact of the connection on the column flange and web must be checked. This will be addressed in Section 12.4.

EXAMPLE 12.1b Direct-Welded Moment Connection by ASD

Goal:

Design a direct-welded beam-to-column moment connection.

Given:

A direct-welded beam-to-column moment connection is shown in Figure 12.2a. The beam is a W24×76 and the column is a W14×109. Bolts are 3/4 in. A325-N and the electrodes are E70. The shapes are A992 steel and the plate is A36. The required strength is Ma = 333 ft-kips and Va = 40.0 kips.

SOLUTION

Step 1:

Obtain the beam and column properties from Manual Table 1-1. Beam: W24×76 d = 23.9 in., bf = 8.99 in., tw = 0.440 in., tf = 0.680 in., Z = 200 in.3 Column: W14×109 d = 14.3 in., bf = 14.6 in., tw = 0.525 in., tf = 0.860 in.

Step 2:

Check the flexural strength of the beam. This check should have been made during design of the beam. The beam section is not reduced because of bolt holes in the flange, so Mp can be determined using the gross section plastic section modulus and the allowable strength is (50)(200) 1.67 M n Ω = M p Ω = Fy Z Ω = = 499 ft-kips > 333 ft-kips 12

Thus, the flexural strength is adequate. Step 3:

Design the flange-to-column weld. The flange-to-column weld can be either a complete joint penetration (CJP) groove weld or fillet welds. CJP welds are used in this example. E70 electrodes match the strength of the plates, as indicated by the User Note in Section J2.6. Therefore CJP welds will develop the full strength of the beam flanges and no further calculations are needed.

554 Chapter 12

Moment Connections Step 4:

Design the web plate. First consider the shear rupture strength of the bolts to determine the minimum number of bolts required. For a 3/4 in. A325-N bolt, rn Ω = 11.9 kips ; therefore, the minimum number of bolts based on bolt shear is 40 n= = 3. 36 11.9 Thus, try a four-bolt connection with bolt spacing of 3.0 in. and end distances of 1.5 in. Thus, l = 12.0 in., which is greater than T/2 = 10.0 in. Try a plate with t = 3/8 in. See Figure 12.3.

Step 5:

Determine the bolt bearing strength. Since the beam web is thicker and has a higher yield strength than the plate, bearing and tearout of the plate will be more critical than the beam web. For the last bolt, determine the clear distance. 1 1 lc = lev − d h = 1.5 − (3/4 + 1/16) = 1.09 < 2(3/4) = 1.5 in. 2 2 Thus, tearout controls over bearing and the bolt nominal strength is Rn = 1.2lc tFu = 1.2 (1.09 )( 0.375 )( 58 ) = 28.4 kips The allowable strength for tearout is Rn Ω = 28.4 2.00 = 14.2 kips Since this is greater than the allowable shear rupture strength, shear rupture will control for this bolt. For the other bolts lc = s − d h = 3.0 − ( 3 4 + 1 16 ) = 2.19 > 2 ( 3 4 ) = 1.5in. Therefore, bearing will control over tearout, and the bolt nominal strength is Rn = 2.4dtFu = 2.4 ( 3 4 )( 0.375 )( 58 ) = 39.2 kips The allowable strength for bearing is Rn Ω = 39.2 2.00 = 19.6 kips Since this is greater than the allowable shear rupture strength, shear rupture will control for these bolts also.. Thus, for the four-bolt connection, the allowable strength is Rn Ω = 4 (11.9 ) = 47.6 > 40.0 kips

Chapter 12

Moment Connections 555

as expected from step 4. Step 6:

Check the plate for shear yield. The gross shear area is Agv = tl = 0.375 (12.0 ) = 4.50 in.2 and the allowable shear yield strength is Vn Ω = 0.6 Fy Agv Ω = ( 0.6 ( 36 )( 4.50 ) ) 1.50 = 64.8 > 40.0 kips

Step 7:

Check the plate for shear rupture. The net shear are is Anv = ( l − n ( d h + 1 16 ) ) t = (12.0 − 4 ( 3 4 + 1 8) ) ( 0.375) = 3.19 in.2 and the allowable shear rupture strength is Vn Ω = 0.6 Fu Anv Ω = ( 0.6 ( 58 )( 3.19 ) ) 2.00 = 55.5 > 40.0 kips

Step 8:

Check block shear of the plate. First calculate the required areas. 1 ⎛ ⎞ Ant = ⎜ leh − ( d h + 1 16 ) ⎟ tw 2 ⎝ ⎠ 1 ⎛ ⎞ = ⎜1.5 − ( 3 4 + 1 8 ) ⎟ ( 0.375 ) = 0.398 in.2 2 ⎝ ⎠ Agv = ltw = 10.5 ( 0.375 ) = 3.94 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.5 − 3.5 ( 3 4 + 1 8 ) ) ( 0.375 ) = 2.79 in.2

Determine the tension rupture strength Fu Ant = 58 ( 0.398) = 23.1 kips Consider shear yield and shear rupture and select the one with least strength; thus, 0.6 Fy Agv = 0.6 ( 36 )( 3.94 ) = 85.1 kips 0.6 Fu Anv = 0.6 ( 58 )( 2.79 ) = 97.1 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection block shear strength, with Ubs = 1.0, of Rn Ω = ( 85.1 + 23.1) 2.00 = 54.1 kips > 40.0 kips

5556 Chapter 12 1

Momentt Connectionss

Step p 9:

Select the plate-to--column weldd. Based d on a fillet weld w on each sside of the plaate, with a weeld design strrength of 0.9 928 kips per 1/16 1 in. of weeld per in. of llength 40.0 D= = 1.80 sixteenths of in. (2(0.928)(112.0)) Thereefore, use a 3/16 3 in. weld,, the minimum m weld for thhe 3/8 in. plaate, as given n in Specificattion Table J2..4.

Step p 10:

Devellop the final design. d Figurre 12.3 shows the final desiign using Four 3/4 3 in. A325-N N bolts in a 3/8×3-1/2×1’-0 plate The impact i of th he connectionn on the coluumn flange and web muust be check ked. This will be addressedd in Section 12.4.

122.3.2

Weldeed Flange Pla ate Connectio on The welded flange plate connection c reeplaces the dirrect-welded fflanges with pplates that aree welded orted beam fllange and to the supportinng column. T The web connnection is usuually the to the suppo typical single-plate shearr connection.. To accomm modate the pllate-to-beam flange weld, the top flange plate must be keptt to a width att least 1.0 in. less than the beam flange w width and thee bottom l 1.0 in. greater g in widdth than the bbeam flange. Manual Figuure 8-13 flange plate must be at least nimum shelf dimensions for fo specific filllet weld sizess. provides min The limit states associated a wiith the tensioon flange platte are yieldingg, rupture, annd block shear, wherreas those asssociated witth the comprression flangge plate are yielding, loccal plate buckling, and a compresssion bucklin ng. These lim mit states aare evaluatedd in Examplle 12.2.

Figu ure 12.4

EX XAMPLE 12.2a 1 W Welded Flang ge Pllate Momentt Coonnection byy LR RFD

Co onnection for Example 12. 2

Goa al:

Desig gn a welded fllange plate beeam-to-colum mn moment coonnection.

Giv ven:

A weelded flange plate beam-tto-column mooment conneection is show wn in Figurre 12.2b. The beam is a W W18×50 and tthe column iss a W14×90. Bolts are 7//8 in. A325-N N and the eleectrodes are E E70. The shappes are A9922 steel and th he plates are A36. A The LR RFD required strength is Mu = 250 ft-kipps and Vu = 45 4 kips. Assu ume the mom ment will causee the top flannge to be in teension and th he bottom flan nge to be in ccompression.

Chapter 12

SOLUTION

Step 1:

Moment Connections 557

Obtain the beam and column properties from Manual Table 1-1. Beam: W18×50 d = 18.0 in., bf = 7.50 in., tw = 0.355 in., tf = 0.570 in. Column: W14×90 d = 14.0 in., bf = 14.5 in., tw = 0.440 in., tf = 0.710 in.

Step 2:

Determine the force to be carried in each flange plate. Since the plate thickness is not yet known, conservatively assume the moment arm is the depth of the beam. Thus, using Equation 12.2 gives M 250(12) Pu = u = = 167 kips d 18.0

Step 3:

Determine the minimum plate area based on the limit state of yielding for the tension flange plate. P 167 Ap = u = = 5.15 in.2 φFy 0.9(36) The top flange plate should be narrower than the beam flange by at least 1.0 in. to facilitate welding in the down position. Therefore, try a 7/8×6.5 plate. Ap = 5.69 in.2.

Step 4:

Check the plate for tension rupture. Figure 12.4a illustrates the expected weld pattern and the top flange plate. The shear lag factor, U, for a welded joint is given in Specification Table D3.1 case 4 as 3l 2 ⎛ x ⎞ U= 2 ⎜1 − ⎟ 3l + w2 ⎝ l ⎠ In order to establish a starting point for checking plate rupture, assume that the weld length will be at least equal to the plate width. That will give 3l 2 ⎛ x ⎞ ⎛ 7 16 ⎞ U= 2 1 − ⎟ = 0.75 ⎜1 − ⎟ = 0.75 ( 0.933) = 0.70 2 ⎜ 3l + w ⎝ l⎠ 6.5 ⎠ ⎝ This will likely be a lower bound on the actual U for this connection. Thus, with An = Ap for a welded plate using Equation D3-1 Ae = AnU = 5.69 ( 0.70 ) = 3.98 in.2

and using Equation J4-2 Pn = Fu Ae = 58 ( 3.98) = 231 kips Thus,

φPn = 0.75 ( 231) = 173 > 167 kips

The tension rupture limit state will not control as long as the final weld

558 Chapter 12

Moment Connections length is at least equal to the plate width. Step 5:

Select the fillet weld size based on weld rupture. The minimum size fillet weld for a 7/8 in. plate attached to a 0.570 in. beam flange, based on Specification Table J2.4, is 1/4 in. Therefore, determine the required length of a pair of 1/4 in. fillet welds on the sides of the flange plate. 167 l= = 15.0 in. 2(1.392(4)) This length appears to be reasonable for this connection and it exceeds the 6.5 in. plate width. Thus, the top plate is 7/8×6.5×1’-5, as shown in Figure 12.4c.

Step 6:

Check the rupture strength of the base material at the welds for the tension flange. For the plate tmin =

3.09 D 3.09 ( 4 ) = = 0.213 in. < t p = 7 8 in. Fu 58

For the beam flange 3.09 D 3.09 ( 4 ) tmin = = = 0.190 in. < t f = 0.570 in. Fu 65 Step 7:

Consider the block shear rupture of the top flange of the beam. Because the plate is welded to the flange, the critical shear limit state is shear yielding. For the two blocks on each side of the web, as shown in Figure 12.4b, the areas are Agv = 16.5(0.570) = 9.41 in.2 Ant = 0.5(0.570) = 0.285 in.2

The nominal strength for tension rupture is Fu Ant = 65 ( 0.285) = 18.5 kips and the nominal strength for shear yield is 0.6 Fy Agv = 0.6 ( 50 )( 9.41) = 282 kips Thus, the design strength for block shear of the beam flange is φRn = 2 ( 0.75 ( 282 + 18.5 ) ) = 451 kips > 167 kips

Chapter 12 Step 8:

Moment Connections 559

Determine the required compression flange plate. This plate must be checked for local buckling and overall buckling. As a starting point, assume that yielding will be the controlling limit state. Thus, the same area will be required as for the tension plate. However, this plate should be 1.0 in. wider than the beam flange so that the welds can again be placed in the downward position. Assume a plate width of 8.5 in. Thus, 5.15 tp = = 0.606 in. 8.5 Select a 5/8 in. plate for further consideration.

Step 9:

Check the compression plate for local buckling. Local buckling of the compression plate is checked against the width-tothickness limits from Specification Table B4.1a. The width of plate between welds is treated as a stiffened plate and the width that projects beyond the weld to the flange tip is treated as an unstiffened plate. For the stiffened plate, case 8, b 7.5 = = 12.0 < 1.49 E Fy = 1.49 29,000 36 = 42.3 t 0.625 For the unstiffened plate, case 1, b 0.5 = = 0.80 < 0.56 E Fy = 0.56 29,000 36 = 15.9 t 0.625 So the plate strength is not limited by local buckling.

Step 10:

Determine the compressive strength of the plate. Since the force to be transferred between the beam and the compression plate is the same as that transferred for the tension plate, use the same 1/4 in. welds 15.0 in. long on each side. The plate has a length for compression buckling of 2.0 in. from the column flange to the end of the weld, as shown in Figure 12.4a. The effective length factor is taken as 0.65, the value recommended in the Commentary for a fixed-fixed column. Determine the slenderness ratio for this plate.

r=

I h2 = = A 12

(0.625) 2 = 0.180 12

lc KL 0.65(2.0) = = = 7.22 0.180 r r For compression elements that are part of connections, Specification Section J4.4 indicates that, when the slenderness ratio is less than 25, Fcr = Fy. Thus, the selection of this plate for yielding, as was originally done, is correct and the 5/8×8-1/2 plate is acceptable for the compression limit states.

560 Chapter 12

Moment Connections

Step 11:

Check the rupture strength of the base material at the welds for the compression flange. For the plate tmin =

3.09 D 3.09 ( 4 ) = = 0.213 in. < t p = 5 8 in. Fu 58

For the beam flange the calculation is the same as it was at the tension flange 3.09 D 3.09 ( 4 ) tmin = = = 0.190 in. < t f = 0.570 in. Fu 65 Step 12:

Determine the welds required to connect the flange plates to the column flange. The force to be transferred is the same for both plates. It can be determined by dividing the design moment by the beam depth plus the average of the tension and compression plate thicknesses. Thus, 250 (12 ) Mu Pu = = = 160 kips ( d + t pt 2 + t pc 2 ) (18.0 + 0.875 2 + 0.625 2 ) A comparison of the plate width with the column flange width shows that both plates are compatible because each plate is narrower than the column flange width, bf = 14.5 in. The force is perpendicular to the weld, so by Equation J2-5 the weld strength can be increased by 1.5. Thus, for fillet welds on both the top and bottom of the plate 160 38.3 D= = bp 1.5(1.392)(2bp ) For the top flange plate D=

38.3 = 5.89 sixteenths of in. 6.5

Therefore use a pair of 3/8 in. welds. For the bottom flange plate 38.3 D= = 4.51 sixteenths of in. 8.5 Therefore use a pair of 5/16 in. welds. The minimum thickness of the column flange to match the rupture strength of the weld is given by Manual Equation 9-2 as 3.09 D 3.09 ( 5.89 ) tmin = = = 0.280 < t f = 0.710 in. Fu 65

Chapter 12

Moment Connections 561

So the column flange has sufficient thickness to transfer the force.

Step 13:

Develop the final design. The web connection design that was demonstrated in Example 12.1a must also be carried out here. In addition, the impact of the connection on the column flange and web must be checked. This will be addressed in Section 12.4. The final geometry for the welded flange plate connection is shown in Figure 12.4c.

EXAMPLE 12.2b Welded Flange Plate Moment Connection by ASD

Goal:

Design a welded flange plate beam-to-column moment connection.

Given:

A welded flange plate beam-to-column moment connection is shown in Figure 12.2b. The beam is a W18×50 and the column is a W14×90. Bolts are 7/8 in. A325-N and the electrodes are E70. The shapes are A992 steel and the plates are A36. The ASD required strength is Ma = 167 ft-kips and Va = 30 kips. Assume the moment will cause the top flange to be in tension and the bottom flange to be in compression.

SOLUTION

Step 1:

Obtain the beam and column properties from Manual Table 1-1. Beam: W18×50 d = 18.0 in., bf = 7.50 in., tw = 0.355 in., tf = 0.570 in. Column: W14×90 d = 14.0 in., bf = 14.5 in., tw = 0.440 in., tf = 0.710 in.

Step 2:

Determine the force to be carried in each flange plate. Since the plate thickness is not yet known, conservatively assume the moment arm is the depth of the beam. Thus, using Equation 12.2 gives M 167(12) Pa = a = = 111 kips d 18.0

Step 3:

Determine the minimum plate area based on the limit state of yielding for the tension flange plate. P 111 Ap = a = = 5.15 in.2 Fy Ω 36 1.67 The top flange plate should be narrower than the beam flange by at least 1.0 in. to facilitate welding in the down position. Therefore, try a 7/8×6.5 plate. Ap = 5.69 in.2.

Step 4:

Check the plate for tension rupture.

562 Chapter 12

Moment Connections Figure 12.4a illustrates the expected weld pattern and the top flange plate. The shear lag factor, U, for a welded joint is given in Specification Table D3.1 case 4 as 3l 2 ⎛ x ⎞ U= 2 ⎜1 − ⎟ 3l + w2 ⎝ l ⎠ In order to establish a starting point for checking plate rupture, assume that the weld length will be at least equal to the plate width. That will give 3l 2 ⎛ x ⎞ ⎛ 7 16 ⎞ U= 2 1 − ⎟ = 0.75 ⎜1 − ⎟ = 0.75 ( 0.933) = 0.70 2 ⎜ 3l + w ⎝ l⎠ 6.5 ⎠ ⎝ This will likely be a lower bound on the actual U for this connection. Thus, with An = Ap for a welded plate using Equation D3-1 Ae = AnU = 5.69 ( 0.70 ) = 3.98 in.2

and using Equation J4-2 Pn = Fu Ae = 58 ( 3.98) = 231 kips Thus, Pn Ω = 231 2.00 = 116 > 111 kips

The tension rupture limit state will not control as long as the final weld length is at least equal to the plate width. Step 5:

Select the fillet weld size based on weld rupture. The minimum size fillet weld for a 7/8 in. plate attached to a 0.570 in. beam flange, based on Specification Table J2.4, is 1/4 in. Therefore, determine the required length of a pair of 1/4 in. fillet welds on the sides of the flange plate. 111 l= = 15.0 in. 2(0.928(4)) This length appears to be reasonable for this connection and it exceeds the 6.5 in. plate width. Thus, the top plate is 7/8×6.5×1’-5, as shown in Figure 12.4c.

Step 6:

Check the rupture strength of the base material at the welds for the tension flange. For the plate tmin =

3.09 D 3.09 ( 4 ) = = 0.213 in. < t p = 7 8 in. 58 Fu

For the beam flange 3.09 D 3.09 ( 4 ) tmin = = = 0.190 in. < t f = 0.570 in. Fu 65

Chapter 12 Step 7:

Moment Connections 563

Consider the block shear rupture of the top flange of the beam. Because the plate is welded to the flange, the critical shear limit state is shear yielding. For the two blocks on each side of the web, as shown in Figure 12.4b, the areas are Agv = 16.5(0.570) = 9.41 in.2 Ant = 0.5(0.570) = 0.285 in.2

The nominal strength for tension rupture is Fu Ant = 65 ( 0.285) = 18.5 kips and the nominal strength for shear yield is 0.6 Fy Agv = 0.6 ( 50 )( 9.41) = 282 kips Thus, the allowable strength for block shear of the beam flange is Rn Ω = 2 ( ( 282 + 18.5 ) 2.00 ) = 301 kips > 111 kips Step 8:

Determine the required compression flange plate. This plate must be checked for local buckling and overall buckling. As a starting point, assume that yielding will be the controlling limit state. Thus, the same area will be required as for the tension plate. However, this plate should be 1.0 in. wider than the beam flange so that the welds can again be placed in the downward position. Assume a plate width of 8.5 in. Thus, 5.15 tp = = 0.606 in. 8.5 Select a 5/8 in. plate for further consideration.

Step 9:

Check the compression plate for local buckling. Local buckling of the compression plate is checked against the width-tothickness limits from Specification Table B4.1a. The width of plate between welds is treated as a stiffened plate and the width that projects beyond the weld to the flange tip is treated as an unstiffened plate. For the stiffened plate, case 8, b 7.5 = = 12.0 < 1.49 E Fy = 1.49 29,000 36 = 42.3 t 0.625 For the unstiffened plate, case 1, b 0.5 = = 0.80 < 0.56 E Fy = 0.56 29,000 36 = 15.9 t 0.625 So the plate strength is not limited by local buckling.

564 Chapter 12

Moment Connections Step 10:

Determine the compressive strength of the plate. Since the force to be transferred between the beam and the compression plate is the same as that transferred for the tension plate, use the same 1/4 in. welds 15.0 in. long on each side. The plate has a length for compression buckling of 2.0 in. from the column flange to the end of the weld, as shown in Figure 12.4a. The effective length factor is taken as 0.65, the value recommended in the Commentary for a fixed-fixed column. Determine the slenderness ratio for this plate.

r=

I h2 = = A 12

(0.625) 2 = 0.180 12

lc KL 0.65(2.0) = = = 7.22 0.180 r r For compression elements that are part of connections, Specification Section J4.4 indicates that, when the slenderness ratio is less than 25, Fcr = Fy. Thus, the selection of this plate for yielding, as was originally done, is correct and the 5/8×8-1/2 plate is acceptable for the compression limit states. Step 11:

Check the rupture strength of the base material at the welds for the compression flange. For the plate tmin =

3.09 D 3.09 ( 4 ) = = 0.213 in. < t p = 5 8 in. Fu 58

For the beam flange the calculation is the same as it was at the tension flange 3.09 D 3.09 ( 4 ) tmin = = = 0.190 in. < t f = 0.570 in. Fu 65 Step 12:

Determine the welds required to connect the flange plates to the column flange. The force to be transferred is the same for both plates. It can be determined by dividing the allowable moment by the beam depth plus the average of the tension and compression plate thicknesses. Thus, 167 (12 ) Ma Pa = = = 107 kips ( d + t pt 2 + t pc 2 ) (18.0 + 0.875 2 + 0.625 2 ) A comparison of the plate width with the column flange width shows that both plates are compatible because each plate is narrower than the column flange width, bf = 14.5 in. The force is perpendicular to the weld, so by Equation J2-5 the weld strength can be increased by 1.5. Thus, for fillet welds on both the top and bottom of the plate

Chapter 12 D=

Moment Connections 565

107 38.4 = 1.5(0.928)(2bp ) bp

For the top flange plate D=

38.4 = 5.91 sixteenths of in. 6.5

Therefore use a pair of 3/8 in. welds. For the bottom flange plate 38.4 D= = 4.52 sixteenths of in. 8.5 Therefore use a pair of 5/16 in. welds. The minimum thickness of the column flange to match the rupture strength of the weld is given by Manual Equation 9-2 as 3.09 D 3.09 ( 5.91) tmin = = = 0.281 < t f = 0.710 in. Fu 65 So the column flange has sufficient thickness to transfer the force. Step 13:

Develop the final design. The web connection design that was demonstrated in Example 12.1a must also be carried out here. In addition, the impact of the connection on the column flange and web must be checked. This will be addressed in Section 12.4. The final geometry for the welded flange plate connection is shown in Figure 12.4c.

12.3.3

Bolted Flange Plate Connection

The bolted flange plate connection is similar to the welded flange plate connection except that the attachment of the plate to the beam flange is through bolts. The addition of bolts to the beam tension flange means that a new limit state, the flexural strength of the beam based on rupture of the tension flange, must be assessed. The other limit states that result from the use of bolts have been described several times and are applicable again here. The bolted flange plate connection is an effective connection from the erection standpoint. The plates can be shop-welded to the column flange, with the beam inserted between the plates and bolted in the field. To accommodate this field erection process, the top plate is usually set a bit high and a filler plate used once the beam is in place. The following example demonstrates the limit state checks associated with the transfer of the flange force, as was done for Example 12.2. The web connection will not be designed because no new limit states are to be considered.

5666 Chapter 12 1

Momentt Connectionss

(a)

(c)

((b)

(d)

n for Example 12.3 Figure 12.5 Connection

EX XAMPLE 12.3a 1 Boolted Flangee Pllate Momentt Coonnection byy LR RFD

Goa al:

Desig gn a bolted flaange plate beaam-to-columnn moment connnection.

Giv ven:

A bo olted flange plate p beam-too-column mooment connection is show wn in Figurre 12.2c. Thiss connection iis to be desiggned for the ssame conditioons as those in Examplee 12.2a. The beam is a W W18×50 andd the columnn is a W14× ×90. Bolts aree 7/8 in. A3225-N and the electrodes arre E70. The sshapes are A992 A steel and d the plate is A36. The reequired strenggth is Mu = 2250 ftkips and a Vu = 45.0 0 kips. Assum me the momennt will cause tthe top flangee to be in ten nsion and the bottom b flangee to be in com mpression.

SO OLUTION

Step p 1:

Deterrmine the beam m and column mn properties. m dimeensions are thee same as thoose given for Example 12.2a. In The member additiion, for the W18×50, W from m Manual Tabble 1-1, Sx = 88.9 in.3.

Step p 2:

Check k the net beam m section for flexure. ough the conn nection has nnot yet been designed, it iis known thaat at a Altho sectio on through th he connectionn there will bbe two bolt hooles in the teension flangee. This may reduce the strength of the beam below the reqquired

Chapter 12

Moment Connections 567

strength. If that is the case, there will be no reason to continue with this connection design. Thus, the provisions of Specification Section F13.1 must be applied for the limit state of rupture of the tension flange. Determine the gross and net areas of the tension flange. A fg = b f t f = 7.5 ( 0.570 ) = 4.28 in.2

Afn = ( b f − n ( d h + 1 16 ) ) t f = ( 7.5 − 2 ( 7 8 + 1 8) ) ( 0.570 ) = 3.14 in.2

Check the yield stress to tensile strength ratio to determine a value for Yt. Fy 50 = = 0.76 < 0.8 Fu 65 Therefore, for all A992 shapes Yt = 1.0, and for this W18×50 beam with a pair of holes for 7/8 in. bolts, Yt Fy Afg = 1.0 ( 50 )( 4.28 ) = 214 kips Fu Afn = 65 ( 3.14 ) = 204 kips

Because FuAfn < YtFyAfg, the nominal moment strength is limited by Specification Equation F13-1 to Fu A fn 204 ⎛ 1⎞ (88.9) ⎜ ⎟ = 353 ft-kips Mn = Sx = 4.28 A fg ⎝ 12 ⎠ and

φM n = 0.9 ( 353) = 318 > 250 ft-kips

So the beam flexural strength is adequate. Step 3:

Check the flange plate for tension yield. Use the force to be carried in each plate determined in step 2 of Example 12.2a, Pu = 167 kips The flange plate will likely be similar to the one used in Example 12.2a. Try a 3/4×7-1/4 plate. The gross area is Ag = 7.25 ( 0.75 ) = 5.44 in.2 and the design yield strength is φRn = φFy Ag = 0.9 ( 36 )( 5.44 ) = 176 > 167 kips

Step 4:

Check the plate for tension rupture. The net area is

An = ( 7.25 − 2 ( 7 8 + 1 8 ) ) ( 0.750 ) = 3.94 in.2

and the design rupture strength is φRn = φFu An = 0.75 ( 58 )( 3.94 ) = 171 > 167 kips So the plate size is adequate based on tension.

568 Chapter 12

Moment Connections Step 5:

Determine the number of bolts required based on the bolt shear rupture strength. First consider the shear rupture strength of the bolts to determine the minimum number of bolts required. For a 7/8 in. bolt, from Manual Table 7-1, φrn = 24.3 kips; therefore, the minimum number of bolts is 167 n= = 6.87 24.3 Thus, try an eight-bolt connection with bolt spacing of 3.0 in. and end distances of at least twice the bolt diameter so that the full bolt bearing strength can be used.

Step 6:

Determine the bolt bearing strength on the plate. Rn = 2.4dtFu = 2.4 ( 7 8)( 0.750 )( 58 ) = 91.4 kips Thus, the design strength per bolt for bearing on the flange plate is φRn = 0.75 ( 91.4 ) = 68.6 > 24.3 kips which is greater than the bolt design shear strength so bolt shear controls.

Step 7:

Determine the bolt bearing strength on the beam flange, again assuming that all bolts have sufficient clear distance to be controlled by bearing. Rn = 2.4dtFu = 2.4 ( 7 8 )( 0.570 )( 65 ) = 77.8 kips Thus, the design strength per bolt for bearing on the flange is φRn = 0.75 ( 77.8) = 58.4 > 24.3 kips which is greater than the bolt design shear strength so bolt shear controls. The assumption of an end distance of 2db is not required. If the end distance in the beam flange, the element with the lower bearing strength, was 1-1/8 in. which is the minimum for the 7/8 in. diameter bolts as given in Table J3.4, the nominal tearout strength with 1 lc = 1.125 − ( 7 8 + 1 16 ) = 0.656 in. 2 is Rn = 1.2lc tFu = 1.2 ( 0.656 )( 0.570 )( 65 ) = 29.2 kips and the design tearout strength is φRn = 0.75 ( 29.2 ) = 21.9 < 24.3 kips Since this is less than the bolt shear strength, bolt shear strength will still control for six bolts and tearout will control for two bolts. Thus, the strength of the 8 bolts for bearing and tearout is φRn = 6 ( 24.3) + 2 ( 21.9 ) = 190 > 167 kips

Chapter 12 Step 8:

Moment Connections 569

Check the plate for block shear rupture. Check the plate for block shear using the geometry shown in Figure 12.5. There are three possible block shear failure patterns shown, one with the center portion failing in tension, Figure 12.5a, one with the two outside portions failing in tension, Figure 12.5b, and one with only one shear failure plane and the tension failure over the middle portion with one outside portion, Figure 12.5c. The worst case must be identified. For the first two possibilities, the critical tension area for block shear will be the one associated with the least tension width. In this case it will be for the middle 3-1/2 in. section as shown in Figure 12.5a, and the critical net tension area is Ant = ( b − ( d h + 1 16 ) ) t p = ( 3.5 − ( 7 8 + 1 8 ) ) (0.750) = 1.88 in.2

and the shear areas for the two shear planes are Agv = 2ltw = 2 (10.75(0.750) ) = 16.1 in.2 Anv = 2 ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = 2 (10.75 − 3.5 ( 7 8 + 1 8 ) ) (0.750) = 10.9 in.2

Determine the tension rupture strength Fu Ant = 58 (1.88) = 109 kips Consider the shear yield and shear rupture and select the one with least strength; thus, 0.6 Fy Agv = 0.6 ( 36 )(16.1) = 348 kips 0.6 Fu Anv = 0.6 ( 58 )(10.9 ) = 379 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection design block shear strength, with Ubs = 1.0, of φRn = 0.75 ( 348 + 1.0 (109 ) ) = 0.75 ( 457 ) = 343 > 167 kips

Since the shear contribution is so much greater than the tension contribution, the pattern shown in Figure 12.5c should be checked. For a tension width equal to 5.38 in., the net tension area is Ant = ( b − 1.5 ( d h + 1 16 ) ) t p = ( 5.38 − 1.5 ( 7 8 + 1 8 ) ) (0.750) = 3.29 in.2

and the shear areas for a single shear plane are

570 Chapter 12

Moment Connections Agv = ltw = 10.75(0.750) = 8.06 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.75 − 3.5 ( 7 8 + 1 8 ) ) (0.750) = 5.44 in.2

Determine the tension rupture strength Fu Ant = 58 ( 3.29 ) = 191 kips Consider shear yield and shear rupture and select the one with the least strength; thus 0.6 Fy Agv = 0.6 ( 36 )( 8.06 ) = 174 kips 0.6 Fu Anv = 0.6 ( 58 ) 5.44 = 189 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection design block shear strength, with Ubs = 1.0, of φRn = 0.75 (174 + 1.0 (191) ) = 0.75 ( 365 ) = 274 > 167 kips Of the three block shear patterns illustrated, this is clearly the most critical. However, it should be noted from step 4 that tension rupture of the plate is more critical than any of the block shear failure patterns. When the shear strength of the block shear patterns is larger than all the potential tension pattern strengths, tension rupture of the plate will likely control over any of the potential block shear modes. Thus, any time that the pattern illustrated in Figure 12.5c would be the controlling block shear pattern, tension rupture of the plate will be more critical. Step 9:

Check the beam flange for block shear. In this case, the beam web prevents a block shear failure in a pattern similar to Figure 12.5a. Thus, check a pattern similar to that shown in Figure 12.5b where the flange width of 7.5 in. yields a tension width of 2.0 in.. The net tension area is 1 ⎛ ⎞ Ant = 2 ⎜ b − ( d h + 1 16 ) ⎟ t p 2 ⎝ ⎠ 1 ⎛ ⎞ = 2 ⎜ 2.00 − ( 7 8 + 1 8 ) ⎟ (0.570) = 1.71 in.2 2 ⎝ ⎠ and the shear areas for the two shear planes are Agv = 2ltw = 2 (10.75(0.570) ) = 12.3 in.2 Anv = 2 ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = 2 (10.75 − 3.5 ( 7 8 + 1 8 ) ) (0.570) = 8.27 in.2

Determine the tension rupture strength

Chapter 12

Moment Connections 571

Fu Ant = 65 (1.71) = 111 kips Consider the shear yield and shear rupture and select the one with least strength; thus, 0.6 Fy Agv = 0.6 ( 50 )(12.3) = 369 kips 0.6 Fu Anv = 0.6 ( 65 )( 8.27 ) = 323 kips

Selecting the shear rupture term and combining it with the tension rupture term gives a connection design block shear strength, with Ubs = 1.0, of φRn = 0.75 ( 323 + 1.0 (111) ) = 0.75 ( 434 ) = 326 > 167 kips Step 10:

Check the compression plate for local buckling. Try the same plate as was used for the tension plate using the geometry given in Figure 12.5a. Check the plate for local buckling in a similar fashion as for the welded plate. In this case, the stiffened plate width is the distance between the bolt lines, and the unstiffened width is from the bolt line to the free edge. Thus, for the stiffened plate, case 8 in Table B4.1a, b 3.5 = = 4.67 < 1.49 E Fy = 1.49 29,000 36 = 42.3 t 0.750 For the unstiffened plate, case 1 in Table B4.1a, b 1.875 = = 2.50 < 0.56 E Fy = 0.56 29,000 36 = 15.9 t 0.750 So the plate strength is not limited by local buckling.

Step 11:

Check the compression plate for buckling over its length. The distance from the column flange to the first bolt is taken as the buckling length of the plate; thus, L = 1.75 + 0.50 = 2.25 in. Assuming the effective length factor of a fixed-fixed column, k = 0.65, is appropriate,

r=

h2 = 12

( 0.750 ) 12

2

= 0.217 in.

and Lc KL 0.65(2.25) = = = 6.74 < 25 r r 0.217 Thus, according to Section J4.4, Fcr = Fy and the strength is the same as for the tension yield limit state. Step 12:

Determine the welds required to connect the flange plates to the column flange. The force to be transferred is the same for both plates. A comparison of the plate width with the column flange width shows that they are compatible because the plates are narrower than the column flange width, bf = 14.5 in. In addition, the force is perpendicular to the weld, so the weld strength can

572 Chapter 12

Moment Connections be increased by 1.5 according to Equation J2-5. Thus, for fillet welds on both the top and bottom of the plate, 167 40.0 D= = 1.5(1.392)(2bp ) bp For both flange plates D=

40.0 = 5.52 sixteenths of in. 7.25

Therefore, use a pair of 3/8 in. welds, which exceeds the minimum for this plate thickness. Step 13:

Consider the web connection. The same web shear connection as would be used in the welded flange plate connection could be used in this connection.

Step 14:

Develop the final design. The flange plates of this connection are PL3/4×7-1/4×1’-1 with eight 7/8 in. A325-N bolts and a pair of 3/8 fillet welds, as shown in Figure 12.5d. As was the case for the previous connection, the impact of the connection on the column flange and web must be checked. This will be addressed in Section 12.4.

EXAMPLE 12.3b Bolted Flange Plate Moment Connection by ASD

Goal:

Design a bolted flange plate beam-to-column moment connection.

Given:

A bolted flange plate beam-to-column moment connection is shown in Figure 12.2c. This connection is to be designed for the same conditions as those in Example 12.2b. The beam is a W18×50 and the column is a W14×90. Bolts are 7/8 in. A325-N and the electrodes are E70. The shapes are A992 steel and the plate is A36. The required strength is Ma = 167 ftkips and Va = 30.0 kips. Assume the moment will cause the top flange to be in tension and the bottom flange to be in compression.

SOLUTION

Step 1:

Determine the beam and column properties. The member dimensions are the same as those given for Example 12.2b. In addition, for the W18×50, from Manual Table 1-1, Sx = 88.9 in.3.

Step 2:

Check the net beam section for flexure. Although the connection has not yet been designed, it is known that at a section through the connection there will be two bolt holes in the tension flange. This may reduce the strength of the beam below the required strength. If that is the case, there will be no reason to continue with this connection design. Thus, the provisions of Specification Section F13.1 must be applied for the limit state of rupture of the tension flange. Determine the gross and net areas of the tension flange.

Chapter 12

Moment Connections 573

A fg = b f t f = 7.5 ( 0.570 ) = 4.28 in.2

Afn = ( b f − n ( d h + 1 16 ) ) t f = ( 7.5 − 2 ( 7 8 + 1 8) ) ( 0.570 ) = 3.14 in.2 Check the yield stress to tensile strength ratio to determine a value for Yt. Fy 50 = = 0.76 < 0.8 Fu 65 Therefore, for all A992 shapes Yt = 1.0, and for this W18×50 beam with a pair of holes for 7/8 in. bolts, Yt Fy Afg = 1.0 ( 50 )( 4.28 ) = 214 kips Fu Afn = 65 ( 3.14 ) = 204 kips

Because FuAfn < YtFyAfg, the nominal moment strength is limited by Specification Equation F13-1 to Fu A fn 204 ⎛1⎞ (88.9) ⎜ ⎟ = 353 ft-kips Mn = Sx = 4.28 Afg ⎝ 12 ⎠ and M n Ω = 353 1.67 = 211 > 167 ft-kips

So the flexural strength is adequate. Step 3:

Check the flange plate for tension yield. Use the force to be carried in each plate determined in step 2 of Example 12.2b, Pa = 111 kips The flange plate will likely be similar to the one used in Example 12.2b. Try a 3/4×7-1/4 plate. The gross area is Ag = 7.25 ( 0.75 ) = 5.44 in.2 and the allowable yield strength is Rn Ω = Fy Ag Ω = ( 36 )( 5.44 ) 1.67 = 117 > 111 kips

Step 4:

Check the plate for tension rupture. The net area is

An = ( 7.25 − 2 ( 7 8 + 1 8 ) ) ( 0.750 ) = 3.94 in.2

and the allowable rupture strength is Rn Ω = Fu An Ω = ( 58)( 3.94 ) 2.00 = 114 > 111 kips So the plate size is adequate based on tension. Step 5:

Determine the number of bolts required based on the bolt shear rupture strength. First consider the shear rupture strength of the bolts to determine the

574 Chapter 12

Moment Connections minimum number of bolts required. For a 7/8 in. bolt, from Manual Table 7-1, rn Ω = 16.2 kips ; therefore, the minimum number of bolts is 111 n= = 6.85 16.2 Thus, try an eight-bolt connection with bolt spacing of 3.0 in. and end distances of at least twice the bolt diameter so that the full bolt bearing strength can be used. Step 6:

Determine the bolt bearing strength on the plate. Rn = 2.4dtFu = 2.4 ( 7 8)( 0.750 )( 58 ) = 91.4 kips Thus, the allowable strength per bolt for bearing on the flange plate is Rn Ω = 91.4 2.00 = 45.7 > 16.2 kips which is greater than the bolt allowable shear strength so bolt shear controls.

Step 7:

Determine the bolt bearing strength on the beam flange, again assuming that all bolts have sufficient clear distance to be controlled by bearing. Rn = 2.4dtFu = 2.4 ( 7 8 )( 0.570 )( 65 ) = 77.8 kips Thus, the allowable strength per bolt for bearing on the flange is Rn Ω = 77.8 2.00 = 38.9 > 16.2 kips which is greater than the bolt allowable shear strength so bolt shear controls. The assumption of an end distance of 2db is not required. If the end distance in the beam flange, the element with the lower bearing strength, was 1-1/8 in. which is the minimum for the 7/8 in. diameter bolts as given in Table J3.4, the nominal tearout strength with 1 lc = 1.125 − ( 7 8 + 1 16 ) = 0.656 in. 2 is Rn = 1.2lc tFu = 1.2 ( 0.656 )( 0.570 )( 65 ) = 29.2 kips and the allowable tearout strength is Rn Ω = 29.2 2.00 = 14.6 < 16.2 kips Since this is less than the bolt shear strength, bolt shear strength will still control for six bolts and tearout will control for two bolts. Thus, the strength of the 8 bolts in bearing and tearout is Rn Ω = 6 (16.2 ) + 2 (14.6 ) = 126 > 111 kips

Step 8:

Check the plate for block shear rupture.

Chapter 12

Moment Connections 575

Check the plate for block shear using the geometry shown in Figure 12.5. There are three possible block shear failure patterns shown, one with the center portion failing in tension, Figure 12.5a, one with the two outside portions failing in tension, Figure 12.5b, and one with only one shear failure plane and the tension failure over the middle portion with one outside portion, Figure 12.5c. The worst case must be identified. For the first two possibilities, the critical tension area for block shear will be the one associated with the least tension width. In this case it will be for the middle 3½ in. section as shown in Figure 12.5a, and the critical net tension area is Ant = ( b − ( d h + 1 16 ) ) t p = ( 3.5 − ( 7 8 + 1 8 ) ) (0.750) = 1.88 in.2

and the shear areas for the two shear planes are Agv = 2ltw = 2 (10.75(0.750) ) = 16.1 in.2 Anv = 2 ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = 2 (10.75 − 3.5 ( 7 8 + 1 8 ) ) (0.750) = 10.9 in.2

Determine the tension rupture strength Fu Ant = 58 (1.88) = 109 kips Consider the shear yield and shear rupture and select the one with least strength; thus, 0.6 Fy Agv = 0.6 ( 36 )(16.1) = 348 kips 0.6 Fu Anv = 0.6 ( 58 )(10.9 ) = 379 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection allowable block shear strength, with Ubs = 1.0, of Rn Ω = ( 348 + 1.0 (109 ) ) 2.00 = 457 2.00 = 229 > 111 kips Since the shear contribution is so much greater than the tension contribution, the pattern shown in Figure 12.5c should be checked. For a tension width equal to 5.38 in., the net tension area is Ant = ( b − ( d h + 1 16 ) ) t p = ( 5.38 − ( 7 8 + 1 8 ) ) (0.750) = 3.29 in.2

and the shear areas for a single shear plane are Agv = ltw = 10.75(0.750) = 8.06 in.2 Anv = ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = (10.75 − 3.5 ( 7 8 + 1 8 ) ) (0.750) = 5.44 in.2

576 Chapter 12

Moment Connections Determine the tension rupture strength Fu Ant = 58 ( 3.29 ) = 191 kips Consider shear yield and shear rupture and select the one with the least strength; thus 0.6 Fy Agv = 0.6 ( 36 )( 8.06 ) = 174 kips 0.6 Fu Anv = 0.6 ( 58 ) 5.44 = 189 kips

Selecting the shear yield term and combining it with the tension rupture term gives a connection allowable block shear strength, with Ubs = 1.0, of Rn Ω = (174 + 1.0 (191) ) 2.00 = 365 2.00 = 183 > 111 kips Of the three block shear patterns illustrated, this is clearly the most critical. However, it should be noted from step 4 that tension rupture of the plate is more critical than any of the block shear failure patterns. When the shear strength of the block shear patterns is larger than all the potential tension pattern strengths, tension rupture of the plate will likely control over any of the potential block shear modes. Thus, any time that the pattern illustrated in Figure 12.5c would be the controlling block shear pattern, tension rupture of the plate will be more critical. Step 9:

Check the beam flange for block shear. In this case, the beam web prevents a block shear failure in a pattern similar to Figure 12.5a. Thus, check a pattern similar to that shown in Figure 12.5b where the flange width of 7.5 in. yields a tension width of 2.0 in. The net tension area is 1 ⎛ ⎞ Ant = 2 ⎜ b − ( d h + 1 16 ) ⎟ t p 2 ⎝ ⎠ 1 ⎛ ⎞ = 2 ⎜ 2.00 − ( 7 8 + 1 8 ) ⎟ (0.570) = 1.71 in.2 2 ⎝ ⎠ and the shear areas for the two shear planes are Agv = 2ltw = 2 (10.75(0.570) ) = 12.3 in.2 Anv = 2 ( l − ( n − 0.5 )( d h + 1 16 ) ) tw = 2 (10.75 − 3.5 ( 7 8 + 1 8 ) ) (0.570) = 8.27 in.2

Determine the tension rupture strength Fu Ant = 65 (1.71) = 111 kips Consider the shear yield and shear rupture and select the one with least strength; thus, 0.6 Fy Agv = 0.6 ( 50 )(12.3) = 369 kips 0.6 Fu Anv = 0.6 ( 65 )( 8.27 ) = 323 kips

Chapter 12

Moment Connections 577

Selecting the shear rupture term and combining it with the tension rupture term gives a connection allowable block shear strength, with Ubs = 1.0, of Rn Ω = ( 323 + 1.0 (111) ) 2.00 = 434 2.00 = 217 > 111 kips Step 10

Check the compression plate for local buckling. Try the same plate as was used for the tension plate using the geometry given in Figure 12.5a. Check the plate for local buckling in a similar fashion as for the welded plate. In this case, the stiffened plate width is the distance between the bolt lines, and the unstiffened width is from the bolt line to the free edge. Thus, for the stiffened plate, case 8 in Table B4.1a, b 3.5 = = 4.67 < 1.49 E Fy = 1.49 29,000 36 = 42.3 t 0.750 For the unstiffened plate, case 1 in Table B4.1a, b 1.875 = = 2.50 < 0.56 E Fy = 0.56 29,000 36 = 15.9 t 0.750 So the plate strength is not limited by local buckling.

Step 11:

Check the compression plate for buckling over its length. The distance from the column flange to the first bolt is taken as the buckling length of the plate; thus, L = 1.75 + 0.50 = 2.25 in. Assuming the effective length factor of a fixed-fixed column, k = 0.65, is appropriate,

r=

h2 = 12

( 0.750 ) 12

2

= 0.217 in.

and Lc KL 0.65(2.25) = = = 6.74 < 25 r r 0.217 Thus, according to Section J4.4, Fcr = Fy and the strength is the same as for the tension yield limit state. Step 12:

Determine the welds required to connect the flange plates to the column flange. The force to be transferred is the same for both plates. A comparison of the plate width with the column flange width shows that they are compatible because the plates are narrower than the column flange width, bf = 14.5 in. In addition, the force is perpendicular to the weld, so the weld strength can be increased by 1.5 according to Equation J2-5. Thus, for fillet welds on both the top and bottom of the plate, 111 39.9 D= = bp 1.5(0.928)(2bp ) For both flange plates

578 Chapter 12

Moment Connections D=

39.9 = 5.50 sixteenths of in. 7.25

Therefore, use a pair of 3/8 in. welds, which exceeds the minimum for this plate thickness. Step 13:

Consider the web connection. The same web shear connection as would be used in the welded flange plate connection could be used in this connection.

Step 14:

Develop the final design. The flange plates of this connection are PL3/4×7-1/4×1’-1 with eight 7/8 in. A325-N bolts and a pair of 3/8 fillet welds, as shown in Figure 12.5d. As was the case for the previous connection, the impact of the connection on the column flange and web must be checked. This will be addressed in Section 12.4.

12.4 COLUMN STIFFENING The connection designs illustrated in the previous examples treated the beam side of the connection. That is, they looked at only the connecting elements and their influence on the beam to which they were attached. They did not consider the influence of the connection and transfer of forces to the supporting element. Normally, a fully restrained moment connection is made to the flange of a column. This is the most efficient use of the column because the strong axis is resisting the transferred moment. As with each connecting element, the application of force to a supporting element requires a check of all applicable limit states. The typical moment connection, such as those illustrated in Figure 12.2a through c, results in the transfer of a concentrated force to the column flange. The limit states for flanges and webs with concentrated forces that are applicable to the beam-column connection are defined in Specification Section J10 as 1. Flange local bending 2. Web local yielding 3. Web local crippling 4. Web compression buckling 5. Web panel zone shear Application of these limit states varies depending on whether the applied force is tension or compression and on whether the connection is on one side or both sides of the column. If the limit states are exceeded, either the column section should be changed or stiffeners and web doubler plates employed. 12.4.1

Flange Local Bending

Flange local bending is addressed in Section J10.1 and is illustrated in Figure 12.6. This limit state is applicable only where a tensile force is applied to the column flange. The primary concern addressed through this limit state is the stress distribution in the weld if the column flange

Chappter 12

Mooment Connecctions 579

deformattion is excesssive. Thus, the limit onn the appliedd force is seet to preventt excessive deformattion. The nom minal strength h for flange loocal bending iis (A AISC J10-1) Rn = 66.25 Fyf t 2f and φ = 0.9 (LRFD) Ω = 1.67 ((ASD) If I the force is applied oveer a small ceentral portionn of the colum mn flange, leess than 15 percent of o the flange width, this limit state dooes not needd to be checkked because tthe force is applied close c to the co olumn web an nd very little fflange deform mation occurs. If I the force iss applied closse to the end oof the columnn, the distribuution of the fforce within the flang ge is limited by the proximity of the end of the ccolumn, and the resultingg deflection increasess. Thus, if thee force is appllied closer thaan a distance of 10tf to thee end of the m member, the nominal strength must be reduced by 50 percennt. When this limit state is exceeded, a ppair of halfded. depth traansverse stiffeeners are need

12.4.2

Weeb Local Yiellding Web loca al yielding is addressed in Section J10.22. It is the sam me limit statee that was connsidered for bearing of o the web in a seated connection, as diiscussed in booth Sections 6.14 and 11.99. Although previouslly discussed for compresssive forces, thhis limit statee is also applicable to tennsile forces. Figure 12 2.7 illustratess the application of a conceentrated forcee to the web oof the column. This T force co ould be transfferred througgh a direct-weelded beam fflange or a bbeam flange plate. Th he bearing len ngth, lb, is tak ken as the thicckness of the plate applyinng the force, or the plate thicknesss plus the welld width, zw, when w attachedd with fillet w welds. The foorce is distribuuted in both direction ns, provided th he connection n is at a distannce equal to aat least the deepth of the coolumn away from thee column end d. This distrib bution is on a slope of 22.5:1 over thee depth givenn by kdes in Manual Table T 1-1. In this case Rn = Fyywtw ( 5kdes + lb ) (A AISC J10-2) and φ = 1.0 (LRFD) Ω = 1.50 ((ASD) If I the spacing g from the con nnection to thhe end of thee column is leess than the ddepth of the column, d, the distrib bution of the force can takke place in oonly one direcction and thee factor 5 is replaced by 2.5. This is the relatio onship that w was used for thhe seated connnection on tthe end of a beam and d is given by Rn = Fywtw ( 2.5kdes + lb ) (A AISC J10-3) When W this lim mit state is excceeded, a paiir of half-deptth stiffeners oor a web doubbler plate is needed.

Figure 12.6

Flange Locall Bending

5880 Chapter 12 1

Momentt Connectionss

Figure 12.7 Distribbution of Conncentrated Forces on Column n Web

122.4.3

Web lo ocal cripplin ng ddressed in Seection J10.3 aand applies onnly to compreessive forces. It is the Web local crrippling is ad limit state th hat predicts the t crumpling g of the webb beneath a ccompressive fforce. It is siimilar to local web yiielding but occcurs in relativ vely slender w webs compareed with local web yieldingg. Web b local cripp pling strength h depends onn how close to the colum mn end the force is applied. Forr illustration here, it is assumed that thhe force is appplied at a ppoint greater tthan d/2 from the collumn end. When W this limiit state was coonsidered forr the seated cconnection in Section 11.9, the force was assu umed to be ap pplied at a ppoint less thann d/2 from thhe end. The nominal strength forr that case iss also a function of the bearing lenggth to depthh ratio. For tthe case considered here, h where th he force is asssumed to be applied at a distance greaater than d/2 ffrom the end of the member, m the Specification Sp gives g 1.5 ⎡ Fyywt f ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EF 2 ⎢ Qf Rn = 0.80tw 1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ (AISC C J10-4) tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ and φ = 0.75 (LRFD) Ω = 2.00 (AS SD) bers. If the weeb local cripppling limit staate is exceeded, a pair where Qf = 1.0 for wide flange memb hree quarter-ddepth doublerr plate is requuired. of three quarrter-depth stifffeners or a th

122.4.4

Web Compression C Buckling Web compression buckliing is addresssed in Sectioon J10.5 andd applies onlyy when com mpressive a on opposite o sidess of the coluumn, putting the web intto compressiion. The forces are applied strength is given g by 24tw3 EFyw Rn = Qf (AISC C J10-8) h and φ = 0.90 (LRFD) Ω = 1.67 (AS SD) 1 for wide flange f membeers. If the forcces are less thhan d/2 from tthe end of thee where Qf = 1.0 column, the strength is reeduced by 50 percent.

Chapter 12

Moment Connections 581

If the web compression buckling limit state is exceeded, a single full-depth stiffener, a pair of full-depth stiffeners, or a full-depth doubler plate are required. 12.4.5

Web Panel Zone Shear and Stiffeners

Web panel zone shear, addressed in Section J10.6, within the boundaries of a fully rigid connection may be significant. The strength of the panel zone is based on shear yielding of the web unless a significant axial force also exists. In this case, shear-axial interaction is considered. When the effects of panel zone deformation on frame stability are not considered in the structural analysis, the panel zone is expected to behave elastically. If the behavior of the panel zone is included in the structural analysis, the nonlinear behavior of the panel zone can be included and its strength increased accordingly. When panel zone deformations are not included in the analysis, panel zone strength as given by Specification Section J10.6 is For α Pr ≤ 0.4 Py (AISC J10-9) Rn = 0.6 Fy d c t w For α Pr > 0.4 Py ⎛ αP ⎞ Rn = 0.60 Fy d c tw ⎜ 1.4 − r ⎟ Py ⎠ ⎝

(AISC J10-10)

and φ = 0.90 (LRFD)

Ω = 1.67 (ASD)

where α = 1.0 for LRFD and 1.6 for ASD, Pr is the required column axial strength and Py is the yield strength. The panel zone strength must be sufficient to resist the total shear in the panel zone, including the story shear carried by the column web. When this limit state is exceeded, a full-depth doubler plate or a pair of diagonal stiffeners is required. In the discussion of each of these limit states, the concluding statement indicated that if the limit state is exceeded, a stiffener or doubler plate is required. Thus, this is a “go/no-go” decision. It is possible that a stiffener may be required by a very small margin of exceedance of only one of these limit states. Unfortunately, stiffeners are an expensive element to add to a connection, especially if they must be fitted between the column flanges as for a full-depth stiffener. In many cases, it is much more economical to select a column section that may be larger than required for the axial load but avoids the requirement of stiffeners. Thus, stiffener requirements should not be left for the detailing stage, but addressed early in the design process when member sizes can still be revised. If stiffeners cannot be avoided, they are designed to resist a force calculated as the applied force, either tension or compression, minus the resisting force as defined for each limit state. This net force is resisted by the cross section of the stiffeners, which are sized based on the provisions for tension or compression connecting elements in Specification Section J4. The Specification provides additional criteria for stiffeners and doubler plates in Sections J10.8 and J10.9. The dimensional requirements for stiffeners are as follows: 1. The width of each stiffener plus half the column web thickness must be greater than or equal to one-third of the width of the attached plate delivering the force: bs ≥ (bp/3 – tw/2). 2. The thickness of the stiffener must be at least half the thickness of the attached plate and at least equal to the plate width divided by 16: ts ≥ tp/2 and ts ≥ bp/16. 3. Transverse stiffeners must extend at least one-half the depth of the column. The strength requirements are also found in these sections. For stiffeners that resist tension forces, the provisions of Specification Section J4.1, which refers to Chapter D, must be

582 Chapter 12

Moment Connections

satisfied. The weld between the loaded flange and stiffener must be sized to transfer the load carried by the stiffener, and the weld to the web must transfer the difference between the forces on the ends of the stiffener. The strength of a compression stiffener must satisfy the requirements for compression connecting elements found in Specification Section J4.4. The stiffener may be designed to bear on the loaded flange or welded to transfer the force that the stiffener is required to resist. The weld to the web is designed to transfer the difference between the forces on the ends of the stiffener. Doubler plates, when needed, must be designed for the forces they are required to resist according to the provisions for those forces. These include, for compression, the provisions of Chapter E; for tension, the provisions of Chapter D; and for shear, the provisions of Chapter G. Additional limitations are as follows: 1. The plate thickness and size must provide sufficient additional material to equal or exceed the strength requirements. 2. The welds attaching the doubler plate to the column web must develop the force transmitted to the doubler plate.

EXAMPLE 12.4a Column Side Limit States by LRFD

Goal:

Check the column side limit states for a moment connection, and design any needed stiffeners and doubler plates.

Given:

Consider the bolted flange plate connection of Example 12.3a. The flange plates are 3/4×7-1/4 and resist a required force of Pu = 167 kips.

SOLUTION

Step 1:

Determine the column flange strength based on flange local bending as addressed in Section J10.1. From Manual Table 1-1 tf = 0.710 in. This limit state is applicable only for a tension force, and the nominal strength is given by Equation J10-1 as

Rn = 6.25Fyf t 2f = 6.25 ( 50 )( 0.710 ) = 158 kips 2

and the design strength for flange local buckling is φRn = 0.9 (158) = 142 < 167 kips Since the available strength is less than the required strength, a pair of halfdepth stiffeners are required. Step 2:

Determine the column web strength based on web local yielding as addressed in Section J10.2. Assume the connection is not near the column end. This limit state applies to both tension and compression forces applied to the column web. The bearing length, lb, is the sum of the plate thickness and the 3/8 in. fillet weld on each side of the plate; thus, lb = 3 4 + 2 ( 3 8) = 1.50 in. and from Manual Table 1-1, kdes = 1.31, so the web nominal strength is Rn = Fyf tw ( 5kdes + lb ) = 50 ( 0.440 ) ( 5 (1.31) + 1.50 ) = 177 kips and the design strength for web local yielding is φRn = 1.00 (177 ) = 177 > 167 kips

Chapter 12

Moment Connections 583

Therefore, this limit state is not exceeded and does not call for stiffeners. Step 3:

Determine the column web strength based on web local crippling as addressed in Section J10.3. This limit state is applicable only for a compressive force applied to the column. The column nominal web strength when the force is applied at a distance greater than d/2 from the column end is, using Equation J10-4, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFyf t f 2 Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ 1.5 ⎡ ⎛ 1.50 ⎞⎛ 0.440 ⎞ ⎤ 29,000(50)(0.710) Rn = 0.80(0.440) 2 ⎢1 + 3 ⎜ (1.0 ) ⎟⎜ ⎟ ⎥ 0.440 ⎝ 14.0 ⎠⎝ 0.710 ⎠ ⎦⎥ ⎣⎢

= 274 kips and the design strength for web local crippling is φRn = 0.75(274) = 206 > 167 kips Therefore, no stiffeners are required for this limit state. Step 4:

Determine the column web strength for web compression buckling as addressed in Section J10.5. This limit state does not need to be checked unless there are opposing compressive forces on opposite sides of the column. The connection described for this example did not mention any connection on the other side of the column. This limit state can be checked to establish any limits on future connections to this column. The value for h is not given explicitly in the Manual; however, h/tw = 25.9 is given. Thus, h = 25.9(0.440) = 11.4 in. The column web nominal strength, based on Equation J10-8, is then 24tw3 EFyw Rn = Qf h 24(0.440)3 29,000(50) = (1.0 ) = 216 kips 11.4 and the design strength for web compression buckling is φRn = 0.9(216) = 194 kips Thus, this column web does not experience compression buckling if opposing forces less than 194 kips are applied on opposite sides of the column.

Step 5:

Determine the strength of the web for panel zone shear as addressed in Section J10.6.

584 Chapter 12

Moment Connections Based on yielding of the panel zone, without the interaction of any axial force in the column, the nominal panel zone shear strength as given by Equation J10-9 is Rn = 0.6 Fy d c tw = 0.6 ( 50 )(14.0 )( 0.440 ) = 185 kips and the design strength for web panel zone shear is φRn = 0.9 (185 ) =167 kips Because this is equal to the force applied by the connection, the panel zone cannot accommodate any additive story shear. For a typical exterior column connection, the story shear and the shear from the connection forces are not additive, so this panel is not likely to have a panel zone shear problem unless the column required axial strength, αPr, is greater than 0.4Py, in which case the panel zone shear strength will be reduced. Step 6:

Determine the force to be transferred by stiffeners. The only column web limit state that calls for a stiffener in this example is that of flange local bending, which is an issue for the tension flange only. The force to be transferred through the stiffener is the difference between the applied force and that available through the web; thus, from step 1, Ru = 167 − 142 = 25 kips This is clearly a small force to be transferred. Careful review of the limit state of flange local bending shows that if the column flange were 0.770 in. thick instead of 0.710 in. thick, no stiffener plates would be required. In this case, a W14×99 would have eliminated the stiffener problem.

Step 7:

Determine the required stiffener size. Based on the dimensional requirements for a stiffener, the minimum width of each stiffener is ⎛ bp t ⎞ ⎛ 7.25 0.440 ⎞ bs ≥ ⎜ − w ⎟ = ⎜ − ⎟ = 2.20 in. 2 ⎠ ⎝ 3 2⎠ ⎝ 3 The thickness of the stiffener must be at least t p 0.750 bp 7.25 ts ≥ = = 0.375 in. and t s ≥ = = 0.453 in. 2 2 16 16 Transverse stiffeners must also extend at least one-half the depth of the column. Therefore, try a 2.25×1/2 in. stiffener with a 3/4 in. corner cutoff, as shown in Figure 12.8. For the tension stiffener, the gross area of the stiffener is Ag = ( 2.25 − 0.750 )( 0.50 ) = 0.750 in.2 and the design tensile strength of one stiffener is φRn = φFy Ag = 0.9 ( 36 )( 0.750 ) = 24.3 kips Therefore, the pair of stiffeners provide 2(24.3) = 48.6 kips, which is

Chappter 12

Mooment Connecctions 585

greater than thee required streength of 25 kiips.

Step S 8:

Deetermine the required r weldd size. Th he weld between the loadeed flange andd stiffener muust be sized too transfer thee 12.5 kips carried by eacch stiffener. F Fillet welds w will be used on the top an nd bottom of the t stiffener. T Thus, 122.5 D= = 2.00 sixteeenths of in. 1.5(1.392) (22) (2.25-0.7500) an nd a minimum m 3/16 in. welld is required by Specificattion Table J2..4. he weld to thee web must trransfer the diifference betw ween the forces on the Th en nds of the stiff ffener. Becausse this is a haalf-depth stiffe fener, the total force in thee stiffener mu ust be transferrred to the coolumn web; thhus, 122.5 D= = 0.781 sixteeenths of in. (1.392)(2)(66.50 − 0.750) an nd a minimum m 3/16 in. welld is required by Specificattion Table J2..4.

Step S 9:

Co onclusion. or the conditiions as checkked in this exxample, excluuding those inn steps 4 Fo an nd 5, the PL1//2×2-1/4×0’-66-1/2 stiffeneers shown in Figure 12.8 w with 3/16 in. fillet welds will w be adequuate.

Figu ure 12.8

EXAMPLE E 12.4b Column Siide Limit States by ASD A

Co olumn Stiffen ner for Examp ple 12.4

Goal: G

Ch heck the colu umn side lim mit states for a moment coonnection, annd design an ny needed stifffeners and dooubler plates.

Given: G

Co onsider the bo olted flange pplate connecttion of Exam mple 12.3b. Thhe flange plaates are 3/4×7 7-1/4 and resiist a required force of Pa = 111 kips.

586 Chapter 12 SOLUTION

Moment Connections Step 1:

Determine the column flange strength based on flange local bending as addressed in Section J10.1. From Manual Table 1-1 tf = 0.710 in. This limit state is applicable only for a tension force, and the nominal strength is given by Equation J10-1 as

Rn = 6.25Fyf t 2f = 6.25 ( 50 )( 0.710 ) = 158 kips 2

and the allowable strength for flange local buckling is Rn Ω = 158 1.67 = 94.6 < 111 kips Since the available strength is less than the required strength, a pair of halfdepth stiffeners are required. Step 2:

Determine the column web strength based on web local yielding as addressed in Section J10.2. Assume the connection is not near the column end. This limit state applies to both tension and compression forces applied to the column web. The bearing length, lb, is the sum of the plate thickness and the 3/8 in. fillet weld on each side of the plate; thus, lb = 3 4 + 2 ( 3 8) = 1.50 in. and from Manual Table 1-1, kdes = 1.31, so the web nominal strength is Rn = Fyf tw ( 5kdes + lb ) = 50 ( 0.440 ) ( 5 (1.31) + 1.50 ) = 177 kips and the allowable strength for web local yielding is Rn Ω = 177 1.50 = 118 > 111 kips Therefore, this limit state is not exceeded and does not call for stiffeners.

Step 3:

Determine the column web strength based on web local crippling as addressed in Section J10.3. This limit state is applicable only for a compressive force applied to the column. The column nominal web strength when the force is applied at a distance greater than d/2 from the column end is, using Equation J10-4, 1.5 ⎡ ⎛ lb ⎞ ⎛ tw ⎞ ⎤ EFyf t f 2 Rn = 0.80tw ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ Qf tw ⎝ d ⎠ ⎝ t f ⎠ ⎥⎦ ⎢⎣ 1.5 ⎡ ⎛ 1.50 ⎞⎛ 0.440 ⎞ ⎤ 29,000(50)(0.710) Rn = 0.80(0.440) 2 ⎢1 + 3 ⎜ (1.0 ) ⎟⎜ ⎟ ⎥ 0.440 ⎝ 14.0 ⎠⎝ 0.710 ⎠ ⎥⎦ ⎢⎣ = 274 kips and the allowable strength for web local crippling is Rn Ω = 274 2.00 = 137 > 111 kips

Chapter 12

Moment Connections 587

Therefore, no stiffeners are required for this limit state. Step 4:

Determine the column web strength for web compression buckling as addressed in Section J10.5. This limit state does not need to be checked unless there are opposing compressive forces on opposite sides of the column. The connection described for this example did not mention any connection on the other side of the column. This limit state can be checked to establish any limits on future connections to this column. The value for h is not given explicitly in the Manual; however, h/tw = 25.9, is given. Thus, h = 25.9(0.440) = 11.4 in. The column web nominal strength, based on Equation J10-8, is then 24tw3 EFyw Rn = Qf h 24(0.440)3 29,000(50) = (1.0 ) = 216 kips 11.4 and the allowable strength for web compression buckling is Rn Ω = 216 1.67 = 129 kips Thus, this column web does not experience compression buckling if opposing forces less than 129 kips are applied on opposite sides of the column.

Step 5:

Determine the strength of the web for panel zone shear as addressed in Section J10.6. Based on yielding of the panel zone, without the interaction of any axial force in the column, the nominal panel zone shear strength as given by Equation J10-9 is Rn = 0.6 Fy d c tw = 0.6 ( 50 )(14.0 )( 0.440 ) = 185 kips and the allowable strength for web panel zone shear is Rn Ω = 185 1.67 =111 kips Because this is equal to the force applied by the connection, the panel zone cannot accommodate any additive story shear. For a typical exterior column connection, the story shear and the shear from the connection forces are not additive, so this panel is not likely to have a panel zone shear problem unless the column required axial strength, αPr, is greater than 0.4Py, in which case the panel zone shear strength will be reduced.

588 Chapter 12

Moment Connections Step 6:

Determine the force to be transferred by stiffeners. The only column web limit state that calls for a stiffener in this example is that of flange local bending, which is an issue for the tension flange only. The force to be transferred through the stiffener is the difference between the applied force and that available through the web; thus, from step 1, Ra = 111 − 94.6 = 16.4 kips This is clearly a small force to be transferred. Careful review of the limit state of flange local bending shows that if the column flange were 0.770 in. thick instead of 0.710 in. thick, no stiffener plates would be required. In this case, a W14×99 would have eliminated the stiffener problem.

Step 7:

Determine the required stiffener size. Based on the dimensional requirements for a stiffener, the minimum width of each stiffener is ⎛ bp t ⎞ ⎛ 7.25 0.440 ⎞ bs ≥ ⎜ − w ⎟ = ⎜ − ⎟ = 2.20 in. 2 ⎠ ⎝ 3 2⎠ ⎝ 3 The thickness of the stiffener must be at least t p 0.750 bp 7.25 ts ≥ = = 0.375 in. and t s ≥ = = 0.453 in. 2 2 16 16 Transverse stiffeners must also extend at least one-half the depth of the column. Therefore, try a 2.25×1/2 in. stiffener with a 3/4 in. corner cutoff, as shown in Figure 12.8. For the tension stiffener, the gross area of the stiffener is Ag = ( 2.25 − 0.750 )( 0.50 ) = 0.750 in.2 and the allowable tensile strength of one stiffener is Rn Ω = Fy Ag Ω = ( 36 )( 0.750 ) 1.67 = 16.2 kips Therefore, the pair of stiffeners provide 2(16.2) = 32.4 kips, which is greater than the required strength of 16.4 kips.

Step 8:

Determine the required weld size. The weld between the loaded flange and stiffener must be sized to transfer the 8.20 kips carried by each stiffener. Fillet welds will be used on the top and bottom of the stiffener. Thus, 8.20 D= = 1.96 sixteenths of in. 1.5(0.928) (2) (2.25-0.750) and a minimum 3/16 in. weld is required by Specification Table J2.4. The weld to the web must transfer the difference between the forces on the ends of the stiffener. Because this is a half-depth stiffener, the total force in the stiffener must be transferred to the column web; thus,

Chapter 12 D=

Moment Connections 589

8.20 = 0.768 sixteenths of in. (0.928)(2)(6.50 − 0.750)

and a minimum 3/16 in. weld is required by Specification Table J2.4. Step 9:

Conclusion. For the conditions as checked in this example, excluding those in steps 4 and 5, the PL1/2×2-1/4×0’- 6-1/2 stiffeners shown in Figure 12.8 with 3/16 in. fillet welds will be adequate.

12.5

PROBLEMS

For all moment connections, design a bolted single-plate shear connection.

required, design the stiffeners and doublers by (a) LRFD and (b) ASD.

1. Design a bolted flange plate connection to connect a W21×44 beam to the flange of a W14×99 column. The connection must transfer a dead load moment of 28 ftkips and a live load moment of 84 ft-kips, and a dead load shear of 7.2 kips and a live load shear of 22 kips. The plates are A36 steel and welded to the column with E70 electrodes. Use 3/4 in. A325-N bolts. The beam and column are A992 steel. Design by (a) LRFD and (b) ASD.

7. Design a welded flange plate connection to connect a W24×68 beam to the flange of a W14×132 column. The connection must transfer a dead load moment of 35 ftkips and a live load moment of 105 ft-kips, and a dead load shear of 10 kips and a live load shear of 30 kips. The beam and column are A992 steel and the plates are A36. Use E70 electrodes and 3/4 in. A325-N bolts. Design by (a) LRFD and (b) ASD.

2. For the design from Problem 1, determine the column stiffening requirements. If stiffeners or doubler plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD.

8. For the design from Problem 7, determine the column stiffening requirements. If stiffeners or doubler plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD.

3. Design a welded flange plate connection to connect a W30×99 beam to the flange of a W14×176 column. The connection must transfer a dead load moment of 167 ftkips and a live load moment of 500 ft-kips, and a dead load shear of 12 kips and a live load shear of 35 kips. The beam and column are A992 steel and the plates are A36. Use E70 electrodes and 3/4 in. A325-N bolts. Design by (a) LRFD and (b) ASD.

9. Design a bolted flange plate connection to connect a W21×83 beam to the flange of a W14×120 column. The connection must transfer a dead load moment of 55 ftkips and a live load moment of 165 ft-kips, and a dead load shear of 12 kips and a live load shear of 36 kips. The plates are A36 steel and welded to the column with E70 electrodes. Use 3/4 in. A325-N bolts. The beam and column are A992 steel. Design by (a) LRFD and (b) ASD.

4. For the design from Problem 3, determine the column stiffening requirements. If stiffeners or doubler plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD.

10. For the design from Problem 9, determine the column stiffening requirements. If stiffeners or doubler plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD.

5. Design a bolted flange plate connection to connect a W18×60 beam to the flange of a W14×109 column. The connection must transfer a dead load moment of 40 ftkips and a live load moment of 120 ft-kips, and a dead load shear of 8 kips and a live load shear of 24 kips. The plates are A36 steel and welded to the column with E70 electrodes. Use 7/8 in. A325-N bolts. The beam and column are A992 steel. Design by (a) LRFD and (b) ASD.

11. Design a welded flange plate connection to connect a W21×132 beam to the flange of a W14×176 column. The connection must transfer a dead load moment of 150 ft-kips and a live load moment of 450 ft-kips, and a dead load shear of 10 kips and a live load shear of 30 kips. The beam and column are A992 steel and the plates are A36. Use E70 electrodes and 3/4 in. A325-N bolts. Design by (a) LRFD and (b) ASD.

6. For the design from Problem 5, determine the column stiffening requirements. If stiffeners or doubler plates are

12. For the design from Problem 11, determine the column stiffening requirements. If stiffeners or doubler

590 Chapter 12

Moment Connections

plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD. 13. Design a direct-welded flange moment connection to connect a W21×83 beam to the flange of a W14×90 column. The connection must transfer a dead load moment of 80 ft-kips and a live load moment of 240 ftkips, and a dead load shear of 15 kips and a live load shear of 45 kips. The beam and column are A992 steel and the web plate is A36. Use E70 electrodes and 3/4 in. A325-X bolts. Design by (a) LRFD and (b) ASD. 14. For the design from Problem 13, determine the column stiffening requirements. If stiffeners or doubler plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD. 15. Design a direct-welded flange moment connection for the conditions of Problem 1. Design by (a) LRFD and (b) ASD. 16. For the design from Problem 15, determine the column stiffening requirements. If stiffeners or doubler plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD. 17. Design a direct-welded flange moment connection for the conditions of Problem 3. Design by (a) LRFD and (b) ASD. 18. For the design from Problem 17, determine the column stiffening requirements. If stiffeners or doubler plates are required, design the stiffeners and doublers by (a) LRFD and (b) ASD. 19. Integrated Design Project. Design bolted flange plate beam-to-column moment connections for the frame designed in Problem 37 or Problem 38 of Chapter 8. Reconsider the type of moment connections and design using welded flange plate connections.

Chapter

13

Steel System ms for Seeismicc Resistance

California Pacificc Medical Centter, Van Ness & Geary Camppus Hospital. Photogrraph by Brett D Drury; Courtessy of Degenkollb Engineers.

13.1

INT TRODUCTIION For wind d and gravity loads, structtural analysis and design aare normally performed by assuming that the structural ressponse remain ns elastic. In seismic desiign this assum mption is too restrictive, particularly for appliications that involve siggnificant grouund motion. That is, thee structural response in a strong earthquake e iss naturally ineelastic, and aan elastic anaalysis may unnnecessarily overestim mate the resullting forces an nd incorrectlyy underestimaate the deform mations. This T chapter provides p an in ntroduction too the design oof steel building structures for seismic resistance. The requ uirements of the Specificcation are suupplemented by the AIS SC Seismic Provision ns for Structu ural Steel Buiildings, ANSII/AISC 341-116, to providee appropriate gguidance in designing g for seismic loads. In thiss chapter, this standard is reeferred to as tthe Seismic P Provisions. To T account fo or the inelastiicity that is exxpected in the response off a structure tto a seismic event, th he approach used u in the Seeismic Provissions, the Naational Earthqquake Hazardd Reduction Program (NEHRP) Provisions, P ASCE A 7, andd the Internaational Buildiing Code inccorporate a seismic response r mod dification coeefficient, R, a deflection am mplification factor, Cd, annd a system overstren ngth factor, Ωo. Although more m advanceed analyses ccan be used, thhis approach permits the use of an a elastic anaalysis to pred dict an inelaastic responsee. These facttors are incorporated as follows: ● R is used u as a divissor in determiining the seissmic force forr which the sttructure will be deesigned. High her R values represent higgher ductilityy levels in thhe structural system m, which ressults in seism mic forces bbeing reduced in proporttion to this ductillity. ● Cd iss used as a multiplier inn determiningg the story drift. Lowerr Cd values repressent higher leevels of structtural stiffnesss and thereforee lower story drift. ● Ωo is used as a multiplier m in seeismic load ccombinationss. It increasess the design loads to account for the leveel of overstrrength presennt in the seiismic force resistiing system so o that the anaalysis results in a more acccurate predicction of the inelasstic behavior.

591

5992 Chapter 13 1

Steel Sy ystems for Seiismic Resistan nce

To determine d thee values of R, Cd, and Ωo that are approopriate for a design, builddings are categorized based on occcupancy and use. u In the NE EHRP Provisiions, buildinggs are assigneed to one of three seissmic use grou ups and then to a seismic ddesign categorry based on thhe seismic use group, the expected d acceleration n, soil charactteristics, and tthe period off the building.. Provisions in ASCE 7 and the Intternational Bu uilding Code vary slightlyy but are simillar. Seissmic design categories c A, B, and C gennerally corresspond to a claassification oof low to moderate seismicity. For category A, there t are no sspecial seismiic design requuirements, andd design may proceed d without refference to thee Seismic Proovisions. In seeismic designn categories B and C, the engineerr can choose to use a basic steel structuure with no sspecial detailiing, for whichh R = 3, Cd = 3, and Ωo = 3, if peermitted by th he building ccode; such buuilding designns can be bassed upon the Specificcation and do o not require consideratioon of requirem ments in thee Seismic Proovisions. Alternatively y, the engineer can choosee to use a systtem defined iin the Seismicc Provisions aand take advantage of a higher R factor. f Seissmic design categories c D, E, and F geenerally correespond to a cclassification of high seismicity. In such casees, the engin neer must usee a structuraal system deffined in the Seismic Provisions. The remaindeer of this chap pter discussess the structuraal systems proovided in the Seismic Provisions for f resisting seismic forcess—those in wh which R is takeen greater thaan 3.

133.2

EXPEC CTED BEH HAVIOR For gravity loads, wind d loads, and seismic loadds associatedd with smalleer earthquakes, it is expected th hat the structtural response will be ellastic. Howevver, for largeer earthquakes, it is recognized that t it may be impracticall or impossib le to preventt some inelasttic behavior. For this reason, and because there is no guaraantee that an actual earthqquake will be less severe tthan that defined for design purposes in the building code, tthe Seismic P Provisions aree based on a capacity design meth hodology. Acccordingly, thee provisions iincluded for eeach system aare intended to result in a structurre in which co ontrolled inellastic deform mations can occcur during a strong earthqquake to dissipate thee energy impaarted to the building b by thhe ground motion. These innelastic deforrmations are forced to t occur in a predictable manner andd to be restriicted to speccific elementss and/or locations in n the structurral system. The T remaindder of the sttructure remaains elastic aas these deformation ns occur, proteected in much h the same w way that a fusee protects thee wiring in ann electric circuit from overload. Given this basic premise of th he capacity ddesign methoddology, the ““fuse” elemennts often establish thee design requ uirements for the memberss and connecttions that surrround them. T This has varying imp plications for different d typees of systems..

Frame System ms Figure 13.1 Moment-F

Chapteer 13

Steel S Systems for S Seismic Resisstance 593

Figure 13.2 1 Braced-Frame Systeems ●

●

As illlustrated in Figure F 13.1, inn a moment fframe, the fusse element iss typically a plastic hinge thaat forms in the girders just outsidee the girderr-to-column conneection. Accorrdingly, the giirder-to-colum mn connectioons, column ppanel zones, and columns c mustt all be designned to develoop the flexuraal strength off the girders conneected to them. As illlustrated in Fiigure 13.2, inn a concentrically braced frrame, the fusee element is usuallly a compreession buckliing–tension yyielding mecchanism form med in the bracin ng member itself. i Accorddingly, the bbrace-to-gusseet connectionns, gussets, gusseet connections, beams, annd columns m must all be ddesigned to ddevelop the tensio on yield strength and coompression bbuckling strenngth of the braces that conneect to them.

Regardless R of the system m chosen, fuuse elements must deform m in a predictable and controlleed manner and d must provid de a ductility that exceeds the level of deformation anticipated. Thus, the systems arre configured d so that lim mit states withh higher ducctility, such aas yielding, control over o limit statees with lesserr ductility, succh as rupture.. The T actual material properrties, such ass steel yield sstrength and strain hardenning effects, can influ uence the behaavior of the system. s As disscussed throuughout this boook, steel is sspecified by ASTM designation, d which identtifies the sppecified miniimum yield strength, am mong other characterristics. The acctual yield strrength, howevver, is most llikely higher tthan the speccified value. Also, oncce strain hard dening beginss to take plac e, the effects of load reversals will tennd to further elevate th he apparent yield y strength.. The differennce between tthe actual yielld strength annd specified minimum m yield stren ngth and straain hardeninng effects aree important in the capaccity design methodology becausee they increaase the strenggth required in the remainnder of the sstructure to he fuse elements. promote yielding in th These T effects are treated directly with m multipliers in the Seismic P Provisions. A multiplier, Ry, is giv ven for each grade g of steel.. When applieed to the speccified minimuum yield strenngth, Fy, the resulting quantity is the expected d yield strenggth, RyFy. Second, an alloowance is made for the effects of strain hardeening, generaally with a faactor of 1.1. Thus, the Seeismic Provisiions use an elevated yield strengtth, generally equal to 1.1R RyFy, in calcuulating the exxpected strenngth of fuse elementss in order to determine thee design forcces for conneections and m members surroounding the fuse elem ments. Table 13.1 1 lists selected examplees of ASTM ssteel with the correspondinng Ry.

5994 Chapter 13 1

Steel Sy ystems for Seiismic Resistan nce Ta able 13.1 Sam mple Expected d Yield Strenggth Multiplierrs Ho ot-rolled shap pes

Ry

A36 A A1043 A Gr. 36 6 A992 A A913 A

1.5 1.3 1.1 1.1

HS SS A500 A Gr. C A53 A

133.3

1.3 1.6

MOME ENT-FRAM ME SYSTEM MS The momen nt-frame systeems, addressed in Chapteer E of the SSeismic Provvisions, generrally use flexural fuse elements, usually plasttic hinges thhat form in tthe girders juust outside tthe fully restrained (F FR) girder-to--column mom ment connectiions. Three tyypes of momeent-frame systtems are addressed in n the Seismic Provisions: special momeent frames (S SMF), in Secttion E3; interrmediate moment fram mes (IMF), in n Section E2;; and ordinaryy moment fraames (OMF), in Section E1. Three other system ms are addreessed in Chaapter E: speccial truss mooment framees (STMF), oordinary cantilever co olumn system ms (OCCS), and a special ccantilever collumn systemss (SCCS). ST TMF are discussed fu urther in Sectiion 13.5. SMF F and IMF usse connectionns that have ddemonstrated inelastic rotation capaacities of at leeast 0.03 radiians and 0.01 radians, respectively, in teesting. OM MF use a presccriptive conneection that prrovides for sm mall inelasticc demands. Assuming that the elasstic drift of a moment fraame is 0.01 radians and the inelastic drift is equaal to the SMF, IMF, aand OMF proovide for inteerstory drifts of 0.04, inelastic rotation at the connections, c 0.02, and 0.01 radians, respectively. Some typicaal seismic m moment conneections are shhown in Figures 13.3 3 and 13.4.

Figuree 13.3 Typiccal Seismic R Reduced Beam m Sectionn (RBS) Mom ment Connecttion

Chapteer 13

Steel S Systems for S Seismic Resisstance 595

F Figure 13.4 T Typical Seism mic End-Platee M Moment Connnection

The T values off R, Cd, and Ωo provided in the ASCE E 7 for the thrree systems aare given in Table 13.2. Table 13.2 1 Seismic Coefficients C ffor Moment F Frames System SMF IMF OMF

R

Ωo

Cd

8 4-1//2 3-1//2

3 3 3

55-1/2 4 3

The use of SMF is no ot limited in any seismic ddesign categoories, whereaas IMF or OM MF usage is restricted d according to o seismic desiign category, building heigght, and structural configurration. 13.3.1

Speecial Momen nt Frames (SM MF) SMF aree configured to t form fusess through pla stic hinging iin the beams, usually adjaacent to the beam-to--column conn nection, to accommodate a e significant inelastic deeformation duuring large seismic events. e Theree may also be some inelasttic deformatioon in the coluumn panel zoone. Several requirem ments are inclu uded in the Seeismic Provissions to prom mote this behaavior, as described in the ensuing sections. s Fuse Strrength With the plastic hingees forming in the beams, thhe fuse flexurral strength iss 1.1RyMp. Thhe girder-tocolumn connections, c column c panell zones, and ccolumns musst all be desiggned to allow w the fuse to develop this flexurall strength. Alternative A appproaches reccognized in the Seismic Provisions include moment-fram m me systems wiith partially rrestrained (PR R) connectionns and weak panel-zone systems, wherein the fuses would d form througgh connectionn deformationns and panel--zone shear deformattions, respectiively.

5996 Chapter 13

Steel Sy ystems for Seiismic Resistan nce

Figure 13.5 Reduced Beeam Section (RBS) ( Momennt Connectionn olumn Connections Beam-to-Co The momen nt connections used in SM MF must havee supporting tests demonsstrating confoormance with the du uctility requirrements, such h as throughh the use of a connectionn listed in thhe AISC Prequalified d Connection ns for Specia al and Interrmediate Steel Moment F Frames for Seismic Applicationss (AISC 358--16). Alternatively, the usse of connecttions qualifieed by prior teesting or project-speccific testing is acceptable. The FR connnections shoown in Figurees 13.3 and 13.4 are prequalified d connections according to o AISC 358-116. Figure 133.5 shows a rreduced beam m section (RBS) conneection in a sp pecial momentt frame. Panel Zone Requiremen nts umn panel zonne is permitteed, and is in m many cases beeneficial Some inelasstic deformation in the colu to the system m performan nce. A panel zone z consisteent with testeed assembliess is required, and the requirementts in the Seism mic Provision ns generally reesult in stiff ppanel zones w with limited yyielding. It may also be necessary to reinforce the t column w with a web dooubler plate foor shear, Figuure 13.4, ners, also callled continuityy plates, in tthe column aat the beam flanges, and/or transsverse stiffen Figure 13.3,, to transfer th he beam flang ge forces into the column. C Com mpactness Beam and Column The compacctness criteriaa in the Speciffication are b ased on a reqquired ductilitty level of 3, whereas the expected d member ducctility demands for beamss and columnss in SMF cann be on the orrder of 6 or 7. Accord dingly, beamss and column ns in SMF muust meet the more stringennt width-to-thhickness limits of hig ghly ductile members m set fo orth in the Seiismic Provisioons. Prevention of Story Mechanisms In SMF, a strong-colum mn, weak-beeam relationsship must be satisfied iin proportionning the columns. Th his requiremen nt is formulatted as a checkk of the mom ment ratio betw ween the beam m(s) and column(s) att each momen nt-connected joint in the sttructure. How wever, this cheeck is not inteended to eliminate alll column yiellding. Rather,, it is a simpllified approacch that resultss in a framingg system with column ns strong eno ough to force flexural yiellding in beam ms at multiplee levels of thee frame. This reducees the likelihood of develloping a storry mechanism m, as shown in Figure 133.6, and achieves a higher h level of o energy disssipation. Som me exceptions are permittedd, as in the caase for a one-story bu uilding, wheree it would not increase eneergy dissipatiion if the beam ms yielded innstead of the columnss.

Chapteer 13

Steel S Systems for S Seismic Resisstance 597

Figure 13.6 1 Momen nt Frames Witthout Strong Columns/Weeak Beam Rellationship y Bracing Requirements Stability Special stability s braccing requirem ments apply iin SMF becaause the bracing must be suitable to maintain the position n of the braced elementss well into tthe inelastic range. For beams, the permitted d unbraced leength is generally reducedd, and bracingg is required near the locaation where the plastiic hinge is ex xpected to forrm. The Seism mic Provisionns provide sevveral options for bracing the beam ms at the beaam-to-column n connection and the coluumn. Often, tthe configuraation of the gravity framing f and interconnectio i on of the flooor slab and thhe beam can be used to satisfy these requirem ments. Protecteed Zones The fusee regions in SMF—the plastic p hinge regions in thhe beams—aare expected to undergo significan nt inelastic deformations. d . Accordinglyy, attachmennts and other potential nootch effect– inducing conditions arre prohibited in these areass. 13.3.2

Inteermediate Moment M Fram mes (IMF) an nd Ordinary Moment Fraames (OMF) IMF and d OMF system ms are similarr in configuraation to SMF but do not prrovide as highh a capacity to accom mmodate inellastic deform mation duringg large seism mic events. T The Seismic Provisions emphasizze that IMF and a OMF aree subject to leess stringent special requiirements com mpared with SMF. In n fact, OMF are often subject s to noo additional requirementts beyond thhose in the Specifica ation. IMF I are based on the use of a tested coonnection dessign with a qqualifying inteerstory drift angle of 0.02 radianss. That is, IM MF are subjecct to the samee connection testing requirements as SMF, bu F are based oon a prescripptive design ut with a lesser required in nterstory drifft angle. OMF procedurre and an ex xpected interrstory drift aangle of 0.001 radians, w which correspponds to a nominallly elastic resp ponse.

13.4

BRA ACED-FRA AME SYSTE EMS The bracced-frame sy ystems coverred in Chap ter F of thee Seismic Prrovisions falll into two categoriees: concentricc and eccenttric. Concenttrically braceed frames generally use axial fuse elementss—usually thee braces them mselves, whicch yield in teension and/orr buckle in coompression. Eccentriccally braced frames gen nerally use sshear and/or flexural fusse elements— —usually a segment,, called a link k, in the beams themselves adjacent to thhe braces.

5998 Chapter 13 1

Steel Sy ystems for Seiismic Resistan nce

Threee types of braced-fram me systems aare addressedd in Chapteer F of the Seismic Provisions: special conceentrically braaced frames ( SCBF), in Seection F2; ordinary concentrically braced fram mes (OCBF), in n Section F1;; and eccentriically braced frames (EBF)), in Section F F3. Two other system ms are addresssed in Chapteer F: bucklingg restrained bbraced framess (BRBF) andd special plate shear walls w (SPSW W). These are discussed furrther in Sectioon 13.5. The values of R, Cd, and Ωo provided d in ASCE 7 for f the first th hree systems aare given in T Table 13.3. Table 13.3 Seismic Coeffficients for B Braced Framees System

R

Ωo

Cd

SCBF OCBF EBF

6 3-1/4 8

2 2 2

5 3-1//4 4

All braced-framee systems hav ve building heeight restrictioons that vary based on thee seismic design categ gory. 133.4.1

Specia al Concentriccally Braced Frames (SCB BF) SCBF are configured c to form fuses through t tensiion yielding and compresssion buckling of the braces betw ween the end connections,, to accommoodate significcant inelasticc deformationn during major seism mic events. Sev veral requirem ments are inclluded in the SSeismic Proviisions to prom mote this behavior, as described in the ensuing sections. s gth Fuse Streng The fuse axial strength in n tension is equal e to RyFyA g, a value thhat is usuallyy higher than the fuse axial strengtth in compresssion. In com mpression the fuse axial sttrength is the lesser of RyFyAg and (1.0 0.877 ) Fcre Ag where Fcre is the expected crritical bucklinng strength. The brace-to-gusset connections, gussets, gu usset-to-beam m and gusset--to-column connections, bbeams, and ccolumns must all be designed d to permit the bracce to developp this full axiaal strength in ttension.

ure 13.7 Tyypical Seismicc Bracing Connnection Figu

Chapteer 13

Steel S Systems for S Seismic Resisstance 599

Gusset Requirement R ts Most braaces and gusseets are detaileed such that oout-of-plane bbrace bucklinng occurs befoore in-plane buckling. When this is i the case, weak-axis w bennding is inducced in the gussset by the ennd rotations and the gusset must be detailed to t accommoddate these rottations. One method is too use a free length eq qual to twice the plate thicckness providded between th the end of thee brace and thhe bend line in the gu usset plate, as illustrated in Figure 13.7. The bend linne in the gusseet is a line peerpendicular to the braace axis that passes p throug gh the point oon the gusset edge connecttion that is neearest to the brace end d. Alternativeely, the bracin ng connectionn can be detaailed to force the deformation into the bracing member, m with h either in-plaane or out-of-pplane bucklinng occurring. Brace Sllenderness The slenderness ratio, Lc/r, of the brace affectss post-bucklinng cyclic performance of the system. Accordin ngly, a maxim mum brace sleenderness ratiio of 200 is peermitted. Brace Net Section Liimitations The effective net areaa of the brace must not bbe less than thhe brace grosss area to inssure tension yielding will control over tension n rupture. M Most end connnections incclude a net ssection that requires reinforcemen r nt to satisfy th his requiremennt. ution of Braciing Distribu Braces must m be used d in a manneer such that tthe lateral foorces in all sstories are resisted by a combinattion of tensio on yielding an nd compressioon buckling oof the brace m members. Althhough a 5050 distrib bution is conssidered ideal, the provisionns allow up to 70 percent of the lateral force to be resisted by b tension orr compression braces, unlless it can bee shown thatt the system response is essentiallly elastic. Th he mixing off tension andd compressionn braces impproves the buuckling and post-buck kling strength h of the systeem and helps prevent the aaccumulation of inelastic ddrifts in one direction n. Beam, Column, C and Brace Comp pactness The com mpactness criteeria in the Specification arre based on a required ducctility level off 3, whereas the expeccted member ductility dem mands for beaams and colum mns in SMF can be on thee order of 6 or 7. Acccordingly, beaams and colu umns in SCBF F must meet tthe more strinngent width-tto-thickness ratio limiits of highly ductile d memb bers in the Seiismic Provisiions. These m more stringentt limitations also imprrove the fractture resistancee and post-buuckling cyclicc performancee of the bracess.

Figure 13.8 Bracing Co onfigurations of Concentriically Braced Frames

6000 Chapter 13

Steel Sy ystems for Seiismic Resistan nce

Figure 13.9 Two-S Stories of a Sppecial Concenntrically Braceed Frame (SC CBF) Photo coourtesy Degenkkolb Engineerss

Bracing Co onfigurationss A variety of bracing configuration ns are avaiilable. Somee configuratiions require special consideratio ons; others—ssuch as K-braacing, where a column is braced betw ween two connsecutive floors—are not permitteed. Figure 13 3.8 illustratess several braacing configuurations. Figuure 13.9 shows two stories s of a special concenttrically bracedd frame. In V-braced V and inverted-V-b braced framess, the expecteed yielding annd buckling bbehavior of the bracess creates an unbalanced u veertical force bbecause the teension brace rremains effecttive as it yields but th he compressiion brace is ineffective a fter bucklingg. This unbalanced force must be resisted by the t intersectin ng beam, as well w as by its connections and supportinng members. That is, the beam must m be design ned for the corresponding c g load redistrribution in adddition to thee gravity loads. Alterrnatively, thee bracing con nfiguration ccan be altereed to eliminaate the potenntial for unbalanced loading. For example, e the V and inverteed-V configuurations can be alternated too form a two-story X configuratio on. Another approach a invvolves the adddition of a zzipper columnn. These bracing conffigurations arre illustrated in i Figure 13.110. The use of K-braacing (and kneee bracing) iss prohibited inn SCBF becaause the configuration results in an n unbalanced force f in the co olumns. X-bracing is allowed. Howeveer, the comm on tension-onnly design appproach used ffor wind forces is nott permitted. Both B diagonalss of the X-braacing must bee designed to resist the tension and compression n forces that result r from cy yclic load reveersals.

Figure 13.10 Two-Story X andd Zipper Configuurations

Chapteer 13

Steel S Systems for S Seismic Resisstance 601

Single-diagon S nal braced fraames, where all braces are oriented inn the same direction, are not perm mitted becausee all braces in n this configuuration are caalled upon too resist the saame type of force, either tension or compression, at the ssame time. T This configurration would violate the requirem ment that, for one o direction of loading, aat least 30 butt no more thaan 70 percentt of the load be resistted by bracess in tension. Thus, the diiagonal braceed frame shoould be implemented as previouslly shown in Figure F 13.2. y Bracing Requirements Stability Special stability s bracing requirem ments apply inn SCBF becaause the braccing must be suitable to the position maintain of the braced d elements weell into the ineelastic range. Often, the coonfiguration of the grravity framing g and interco onnection of the floor slabb to the beam m can be used to satisfy these req quirements. Protecteed Zones The fuse regions in SC CBF—the braaces and gusssets—are exppected to undeergo significaant inelastic deformattion. Accordingly, attachm ments and othher potential notch effect––inducing connditions are prohibiteed in these areeas. 13.4.2

Ord dinary Concentrically Brraced Frames (OCBF) OCBF are a similar in i configurattion to SCB BF, but do nnot provide as high a ccapacity to accommo odate inelasttic deformatiion during large seismiic events. T The Seismic Provisions emphasizze that OCBF F are subject to t less stringeent requiremennts than SCBF.

13.4.3

Ecccentrically Braced Framees (EBF) EBF are configured to o form fuses through sheaar yielding, fl flexural yieldiing, or a com mbination of the two in i the EBF lin nk, in order to t accommoddate significannt inelastic deeformation duuring major seismic events. e Unlik ke the behavior of SCBF F and OCBF,, the braces in EBF are intended to remain nominally n elaastic and aree not expectted to bucklee. The requiirements in tthe Seismic Provision ns to promotee this behavio or are describeed in the ensuuing sections. Fuse Strrength The fusee strength in shear s for I-sh haped links is 1.25RyVn, whhere Vn is thee nominal sheear strength of the lin nk, which is th he lesser of th he nominal pplastic shear sstrength and tthe shear assoociated with the plastiic moment caapacity of the link. The straain hardeningg multiplier foor EBF links oof 1.25 is

Figure 13.11 1 Examp ple Configuraations of Ecceentrically Braaced Frames

6002 Chapter 13

Steel Sy ystems for Seiismic Resistan nce

Figure 1 3.12

Eccenntrically Bracced Frame (EB BF)

Photo couurtesy Degenkoolb Engineers

higher than the correspon nding multipllier for other ssystems becaause EBF linkks are associaated with higher deforrmation demaands. The beam segmentss outside the link, gussets, gusset-to-beeam and gusset-to-co olumn connecctions, and co olumns mustt all be desiggned to deveelop the sheaar and/or flexural yiellding mechanism in the lin nks. Link Locatiion EBF links are a usually loccated as segm ments within tthe length off the beams, eeither betweenn braces or between a brace and a beam-to-co olumn conneection. Alternnatively, linkss can be provvided as vertical elem ments between beams and V or inverte d-V bracing. EBF configuurations are shown in Figure 13.11. Figure 13.12 shows an eccentricallyy braced fram me. In the foreeground the E EBF link is in the upp per beam and in the backgrround the linkk is in the low wer beam. Wheen links are located as seegments withiin the lengthh of the beam ms, it is prefeerable to locate the lin nks between the t ends of th he braces. Whhen the links are located bbetween a brace and a beam-to-collumn connecction, the beeam-to-colum mn connectionn requires sspecial consiideration because thee rotational demands aree substantiallly higher thhan those att a beam-to-column connection in an SMF. In application ns involving a significantt flexural dem mand, a preqqualified d by testing m must be usedd. When the links used aare short connection or a connecttion qualified enough that shear yieldin ng dominates,, the need forr qualificationn testing is eliiminated if thhe beamto-column connection c is reinforced with w haunchees or other suuitable reinfoorcement desiigned to preclude ineelastic action in i the reinforced zone adjaacent to the coolumn. Link Rotatiions Plastic rotatiion of link beeams in EBF systems is lim mited to 0.02 radians for fl flexural-yieldiing links and 0.08 rad dians for sheear-yielding liinks. For linkks that deform m in combinned shear andd flexure yielding, thee rotation limiit is determined by linear iinterpolation bbetween these limits. Link Stiffen ning and Bra acing To ensure that t the required rotationss can be achhieved and thhat yielding occurs withoout local buckling weell into the inelastic rang ge, links are stiffened as prescribed inn Seismic Prrovisions Section F3.5 5(b). The use of web doub bler plates to stiffen links is not permittted because tthis type of reinforcem ment does no ot deform con nsistently withh the web defformations annd would incrrease the strength of the t fuse. Add ditionally, beam web peneetrations withhin the link aare not permiitted and the link musst be braced against a out-off-plane displaccement and tw wist at the endds of the link.

Chapteer 13

Steel S Systems for S Seismic Resisstance 603

Braces and a Beam Segments Outsside of Linkss Because the inelastic action in EBF F is intended to occur prim marily within the links, thee braces and beam seg gments outsid de of the link ks must be ddesigned to rremain nominnally elastic aas the links deform. Limited L yield ding outside of o the links is allowed, as loong as the beam segments outside the links and d braces have sufficient strrength to deveelop the fullyy yielded and strain-hardenned strength of the lin nks. The bracces and beam segments ouutside the linkks are normallly designed aas members subject to o the combineed effects of axial a force annd flexure.

13.5

OTH HER FRAM MING SYST TEMS Several S other systems are provided for in the Seism mic Provisionss, including sspecial truss moment frames (STM MF), in Sectio on E4; bucklling-restrainedd braced fram mes (BRBF),, in Section F4; and special s plate shear s walls (S SPSW), in Seection F5. The values of R R, Cd, and Ωo provided in ASCE 7 for each of th hese three sysstems are giveen in Table 133.4. Table 13.4 4 Seismic Coeefficients for Other Framinng Systems System

R

Ωo

Cd

STMF BRBF SPSW

7 8 7

2 2-1//2 2

55-1/2 5 6

Each of these system ms has buildiing height reestrictions thaat vary basedd on the seissmic design category. Composite C steeel and reinfo orced concretee systems aree also provideed for in Chappters G and H of the Seismic Provvisions.

Figure 13.1 3 Special T Truss Momentt Frame Conffiguration

6004 Chapter 13 1 133.5.1

Steel Sy ystems for Seiismic Resistan nce

Specia al Truss Mom ment Frames (STMF) STMF are configured to form fusess through yi elding in a special segm ment of the ttruss, to accommodaate significantt inelastic defformation durring large seiismic events. The special segment can be eitheer a truss paneel with diagon nals or a Vierrendeel truss panel. The reemainder of tthe truss and framing g in the systtem is design ned to remaiin elastic as the special segment defoorms. A schematic STMF is illusttrated in Figurre 13.13. Wheen diagonals are used, th he yielding off the special segment occcurs by axiall tension yielding and d compression n buckling off the diagonaals. Diagonal web memberrs used in thee special segments off STMF systtems are limiited to flat b ars only, andd must meet a limiting w width-tothickness rattio of 2.5. Wh hen a Vierend deel panel—aa panel withoout diagonals— —is used, yieelding of the special segment s occurrs by flexurall yielding of tthe chord mem mbers. The size of the truss and the size and loocation of thhe special seggment are lim mited to correspond with w the research on whicch the system m is based. It is required thhat the STMF F special segment be located withiin the middle half of the trruss span beccause shear duue to gravity loads is generally low wer in that reegion. Other than t the norm mal gravity loaads carried byy the frame, nno major structural loading is perm mitted in the sp pecial segmennt.

133.5.2

Buckliing-Restrained Braced Frrames (BRBF) BRBF repreesent a type of o concentricaally braced frrame system tthat has speccial bracing ellements, as shown in n Figure 13.14. These brracing elemeents provide essentially thhe same respponse in compression n as they do in n tension. The bracing elem ments are com mposed of a lload-bearing ccore and a surroundin ng sleeve eleement that resstrains the gllobal buckling of the coree, forcing yieelding in compression n rather than buckling. BR RBF are conffigured to forrm fuses throough tension yielding and compresssion yielding g of these speecial bracing elements, to accommodatee significant inelastic deformation n during large seismic even nts. A three-sttory bucklingg restrained brraced frame iis shown in Figure 13 3.15. The bracing elem ments must bee qualified byy testing to ennsure that thee braces used provide the necessary strength and a deformaation capacityy. The requirred deformattion capacityy during testing is am mplified beyon nd what is req quired by an aanalysis in recognition thaat actual deforrmations can be larger than those predicted p by analysis. a

B Frame Bracing Elem ment Figure 13.14 Buckling Restrained Braced

Chapteer 13

Steel S Systems for S Seismic Resisstance 605

Fiigure 13.15 Applicationn of a Bucklinng R Restrained Braace in a Conceentrically Braaced Frrame (BRBF)) Phhoto courtesy S Star Seismic

The T steel corre has a yield ding segmentt that is desiggned with a cross-sectionnal area and length based on strength, stiffnesss, and strainn demands. Because eacch bracing ellement is a manufacttured item, th he designer caan specify ann array of braaces that prom mote distributted yielding througho out the frame. The T steel corre projections beyond thee yielding seegment are ddesigned to pprovide the transition n from the co ore and conneection to the rremainder off the framing system. The projections are desig gned so that they remain nominally eelastic like thhe rest of thee frame as thhe yielding segment deforms. 13.5.3

Speecial Plate Sh hear Walls (S SPSW) SPSW have slender, unstiffened plate p element s surroundedd by and connnected to horrizontal and vertical boundary elements that are rigidly interconnectted. A schem matic SPSW system is illustrated in Figure 13 3.16 and an example e buildding applicatioon is shown iin Figure 13.117 SPSW S are co onfigured to form f fuses thhrough web pplate yieldingg and buckling (tension field actiion in the weeb element), along with pplastic-hinge formation inn the ends off horizontal boundary y elements to t accommod date significaant inelastic deformationn during majjor seismic events. The T tension-ffield action in n SPSW is annalogous to thhat in a plate girder, but thhe behavior and stren ngth of SPSW W differs from m that of platee girders. Acccordingly, thee design requuirements in the Seism mic Provision ns for SPSW differ d from thhose in the Sppecification ffor plate girdeers. A more detailed discussion d off SPSW system ms is found inn AISC Desiggn Guide 20, Steel Plate Shhear Walls.

6006 Chapter 13 1

Steel Sy ystems for Seiismic Resistan nce

Fig gure 13.16 133.5.4

Special Platee Shear Wall S System

Composite Systems A variety off composite structural s sysstems are provvided for in Chapters G aand H of the Seismic Provisions. These system ms include, in n Chapter G, ccomposite mooment framess: composite ppartially restrained moment m framees (C-PRMF), composite special mom ment frames (C-SMF), coomposite intermediatee moment frrames (C-IM MF), and com mposite ordinnary momennt frames (C C-OMF). Chapter H in ncludes comp posite ordinarry braced fram mes (C-OBF)), composite sspecial concenntrically braced fram mes (C-SCBF)), composite eccentricallyy braced fram mes (C-EBF), composite oordinary shear walls (C-OSW), composite c sp pecial shear w walls (C-SSW W), compositte plate sheaar wallsconcrete enccased (C-PSW W/CE), and Composite platte shear wallss-concrete fillled (C-PSW/C CF).

133.6 133.6.1

OTHER GENERA AL REQUIR REMENTS S Bolted d and Welded d Connection ns Connectionss in the seism mic force resiisting system m must be connfigured suchh that a ducttile limit state in the fuse controlls—that is, th he deformatioons occur in the fuse eleements beforee failure occurs in th he connection ns. This generrally means tthat connections in the seeismic force rresisting system are much larger than they would w be if deesigned for ggravity, windd, and low-seeismicity pecial requireements for thhe use of bolts and weldds in the applications. There are additional sp Seismic Provisions. Boltted joints in shear are dessigned as pree-tensioned beearing joints with faying surfaces prepared as for Class A or o better slip--critical conneections. These are not slipp-critical connnections; w some slip p resistance. B Because slip cannot and nneed not be prrevented they are beaaring joints with during largee ground mo otions, the inttent is to conntrol slip duuring lesser gground motioons. Pretensioned bo olts are used because largee ground mottions can causse full reversal of design lload and thus have a negative n effecct on snug-tig ght bolt perforrmance. Hole type is resstricted to staandard holes and short-sllotted holes perpendiculaar to the loading direection, unless another holee type is show wn to be acceeptable by teesting. One exxception provided is that t oversized d holes are peermitted in braace diagonalss within certain limits. For design purposes, bolt beearing checkss are requiredd to be madee at the defoormation considered level to prevent excessiv ve deformatiions of bolteed joints duee to bearingg on the connected material, m primaarily in order to minimize damage in leesser ground m motions.

Chapteer 13

1 Figure 13.17

Steel S Systems for S Seismic Resisstance 607

Speciial Plate Sheaar Wall for a H High-Rise Buuilding

Photo cou urtesy Michael Dickter/Magn nusson Klemenncic Associatess

In I welded con nnections, filller metal withh a minimum m specified Chharpy V-notchh toughness of 20 ft-llbs at 0ºF is required r in alll welds invo lved in the seeismic load ppath, except ffor demandcritical welded w joints,, which have more m stringennt notch toughhness requirem ments. Sharing S of a common fo orce between bolts and w welds is prohhibited becauuse seismic deformattion demandss generally exceed e the ddeformation ccompatibility required forr such load sharing. If this load sh haring were used, u the weldd may fail beefore engaging the full streength of the bolt. 13.6.2

Pro otected Zones n the variou us systems coovered in thhe Seismic P Provisions maay undergo The fusee elements in significan nt inelastic deformationss when subj bjected to laarge ground motions. A Accordingly, constructtion operation ns that mightt cause disconntinuities muust be barred from these aareas. Thus, the Seism mic Provision ns designate protected p zonnes in each syystem that muust be kept frree of sharp transition ns, penetrations, notches, and a so forth. As mentioneed, an example of a proteccted zone is the expeccted hinge reegion in beam ms of SMF, w where web oppenings or stuuds on the topp flange are not allow wed. Disconttinuities that are inadverttently createdd in these zoones must generally be repaired or elements replaced. r

13.6.3

Loccal Buckling quires a mem mber ductilityy of 6 or 7, which is moore than the The yielding of fuse elements req d of 3 used in the developmentt of the comppactness criteeria in the Sppecification. normal ductility Thus, in the Seismic Provisions, more m stringennt seismic coompactness crriteria for higghly ductile and moderately ductille members arre provided inn Table D1.1..

13.6.4

Collumn Requirrements Special requirements r for columns and column splices in thhe seismic foorce resisting system are stipulated d in Section D1.4 of the Seismic Proovisions. Miniimum designn forces are sspecified to

608 Chapter 13

Steel Systems for Seismic Resistance

preclude column and column splice failure in compression or tension. This approach does not necessarily preclude yielding of the column, and some guidance is provided in the Commentary for cases in which yielding of the column might be of concern. Column splices must be located away from the beam-to-column connections—generally within the middle third of the story height in which the splice occurs—to reduce the effects of flexure. Additionally, if partial-joint-penetration groove welds are used to make column splices, a 100 percent increase in required strength is specified and the use of notch-tough filler metal is required. There are also requirements for columns that are not a part of the seismic load resisting system, because these columns must still distribute a fraction of the seismic shear between the floors. 13.6.5

Column Bases To increase frame stiffness, column bases are normally treated similarly to beam-to-column moment connections, accounting for the inherent differences, such as the increased flexibility due to deformations in the base plate, longer anchor rods, compressibility of the grout and concrete, and foundation rocking effects.

13.7

CONCLUSIONS This introduction to the design of steel structures for seismic force resistance is intended to provide a starting point for further study. The detailed provisions are found in the Seismic Provisions for Steel Buildings, ANSI/AISC 341-16. Additional guidance is found in the AISC Seismic Design Manual and AISC Facts for Steel Buildings 3, Earthquake and Seismic Design. The interested student is encouraged to study these two documents for a more in-depth treatment of seismic design of steel structures.

13.8

PROBLEMS

1. What is the major difference between the analysis and design of a structure for wind and gravity loads and the analysis and design of a structure for seismic loads? 2. Explain the use of the R, Cd, and Ωo factors. What do these factors account for? 3. How are the R, Cd, and Ωo factors determined for a particular analysis? 4. What is the purpose of fuse elements in seismic design? Provide some examples of structural fuse elements. 5. How is the R factor used to influence the seismic forces a structure is designed to resist? 6. What are the seismic design categories specified by the building codes and how do they relate to the level of expected seismicity?

7. Explain the capacity design method in relation to material yield strength. 8. What types of fuse elements are typically used in moment frame systems? 9. What are the three primary types of moment frames considered in the Seismic Provisions? What are the respective values for R, Cd, and Ωo for each of these systems? 10. For special moment frames, what type of relationship should exist between the column and beams to reduce the likelihood of a story mechanism? 11. Which type of moment frame has a ductility requirement for connections of an interstory drift angle of 0.02 radians? 12. Name the two categories of braced frames provided for in the Seismic Provisions. What type of fuse element is used by each of these?

Chapter 13 13. List the three primary types of braced frame systems addressed in the Seismic Provisions and their corresponding values for R, Cd, and Ωo. 14. How do SCBF and OCBF allow for inelastic deformations in structures? 15.

List some examples of CBF configurations.

16. How do EBF differ from CBF in their performance during large seismic events? 17. Where are the fuse eccentrically braced frames?

elements

located

for

18. How do BRBF resist seismic forces and what is the fuse element in this system? 19. Describe how SPSW resist seismic forces. 20. Compare the expected behavior of STMF with other moment frame systems. 21. How does the size of connections in seismic force resisting systems differ from that of connections designed for gravity and wind systems? Why? 22. Is it permissible to share a force between bolts and welds in seismic design? Why or why not? 23. For seismic design, where should column splices be located and why? 24. Integrated Design Project. The same lateral load resisting system is used for resisting wind forces as is used for seismic forces. The lateral load resistance in the north-south direction is provided by a concentrically braced frame with a chevron configuration as depicted in Figure 1.24. Before the forces in these bracing members can be determined, the specified seismic load must be determined. For this system, R = 3 may be selected and special detailing according to the AISC Seismic Provisions is not required. At this stage in the design, a simplified approach to seismic load calculation might yield the following loads at each level in the north-south direction. Roof Fourth floor Third floor Second floor Total seismic load

56.0 kips 162.0 kips 108.0 kips 54.0 kips 380.0 kips

Lateral load resistance in the east-west direction is provided by two perimeter moment frames as shown in Figure 1.24. For these frames, R = 3 will also be used.

Steel Systems for Seismic Resistance 609

Thus, no special detailing will be required. A simplified approach to seismic load calculation might yield the following loads. Roof Fourth floor Third floor Second floor Total seismic load

54.0 kips 152.0 kips 98.0 kips 46.0 kips 350.0 kips

Using the member sizes determined in earlier work, carry out an analysis of both lateral load resisting systems and check the members to be sure that they are adequate to resist the seismic forces. Reconsider the analysis for the use of special concentrically braced frames and special moment frames. Determine the appropriate loading according to ASCE 716 using an approximate building weight as given below, and compare the required member strengths to the previous analysis. Roof Fourth floor Third floor Second floor Total building weight

550 kips 2150 kips 2150 kips 2150 kips 7000 kips

INDEX A ACI 318, 181, 378, 392, 412, 414 Advanced Analysis, 3, 227 AISC Code of Standard Practice, 58-59, 136, 326 AISC Specification, 1-6 AISC Steel Construction Manual, 1, 4-5 AISC Web Resources, 5 Alignment Charts, 148-153, 315, 320, 325, 353-358 Allowable Strength Design (ASD), 1, 18-19 American Welding Society (AWS), 77-78, 451-461 Amplification Factor, 306-349 ANSI/AISC 341, 41, 59, 148-149, 591 ANSI/AISC 360, 1 ANSI/AWS D l.l, 77-78 Aon Center, 15-16 Approximate Effective Length, 148-149 Area, Effective Net, 85-87, 91, 96 Gross, 87 Influence, 35-37, 44-50 Net, 87-89, 91-95, 462 Tributary, 35-36 ASCE 7, 27, 44-51 ASTM, A6, 63 A36, 70 A53, 70 A108, 72 A242, 75 A283, 75 A307, 75 A370, 61-62 A500, 70 A50l, 72 A514, 75 A529, 72 A572, 72 A588, 73 A618, 73 A709, 72 A847, 73-75 A913, 73 A992, 68, 73 A1043, 72 A1065, 73 A1066, 75 A1085, 72 F3125, 77

B Base Plates, 4, 181-182, 535-542, 608 Base Shear, 40-41 Beam-and-Column Construction, 12-14

610

Beam Line, 350-352 Beam-Columns, 8-10, 124, 297-366 Composite, 421-425 Design Tables, 336-337 Effective Axial Load, 333-336 Interaction, 297-298, 300-304 Selection, 333-336 Truss Members, 297 Beams, See Bending Members Bearing Connection, 517-523 Bearing Plates, 4, 252, 471, 535-538 Bearing Stiffeners, 250 Bearing Strength Concrete, 537-538 Pin-connected members, 113 Bearing Wall, 11-12 Bending Members, Beams, 8-10, 12, 188-251 Composite, See Composite Beams Double Angle, 243-246 Flange Local Buckling, 216-219, 234 Girders, 188 Lateral-Torsional Buckling, 201-216, 239-241 Leg Local Buckling, 238-239 Open Web Steel Joists, 256-258 Single Angle, 237-241 Tee, 232-235 Web Local Buckling, 219, 266-272 Yielding, 189-192 Biaxial Bending, 247 Blast, 33-34, 412 Block Shear, 104-105, 463-466, 474-524 Bolts, A307, 75, 434, 440 A325, 77, 435-436, 439-442 A490, 77, 435-436, 439-442 Bearing, 438-442 Combined Forces, 450-451 Common, See A307 F1852, 77, 435-436 F2280, 77, 435-436 F3125, 77 Group A, 435, 439 Group B, 435, 439 High Strength, 435-436 Holes, 436-437 Limit States, 438-442 Shear, 438-440 Slip-Critical, 438, 448-451 Tension, 429-430 Tension Control, 435-436 Tearout, 438, 440-442

Index Braced Frame, 13-14, 117, 147, 305-311, 593, 597-605 Buckling-Restrained, BRB, 604-605 Eccentric, EBF, 598, 601-603 Ordinary Concentric, OCBF, 598, 601 Special Concentric, SCBF, 598-601 Seismic, 597-606 Bracing, Beam, 206-208, 360-361 Column, 124, 131-132, 359-360 Design, 359-361 Frame, 361 Panel, 359-361 Point, 359-362 Stability, 2-3, 358-361, 597, 601 Tension, 82 Bracing Member, 80, 82, 84, 117, 358-363 Buckling Load, 127-131 Building Codes, 6, 19-22, 27, 591-592 Built-up Girder, See Plate Girder Burnham, Daniel, 56

C Calibration, 51-52 California Pacific Medical Center, 591 Cambering, 408 Carnegie-Phipps Steel Company, 56 Chemical Components, 68-70 Carbon, 68 Chromium, 70 Columbium, 69 Copper, 69 Manganese, 69 Molybdenum, 70 Nickel, 69 Phosphorus, 69 Silicon, 69 Sulfur, 69 Vanadium, 69 Coefficient of Variation, 23-25 Collapse Load, 20 Collapse Mechanism, 20, 229-231 Column, 7-9, See also Compression Member Column Stiffening, 578-589 Doubler Plates, 578-582 Flange Local Bending, 578-579 Stiffeners, 578-589 Web Compression Buckling, 580-581 Web Local Crippling, 580 Web Local Yielding, 579 Web Panel Zone Shear, 581-582 Combined Force Member, 8-9, See also Beam-Columns Compact Beams, 188-216 Composite Beams, 372-411 Advantages, 375 Design, 386-394 Tables, 386-391 Preliminary, 403-404 Disadvantages, 375 Effective Flange Width, 376 Encased, 403

611

Flexural Strength, 377-391, 403 Fully Composite, 378-380 Lower Bound Moment of Inertia, 409-412 Metal Deck, 394-403 Negative Moment Strength, 391 Partially Composite, 383-386 Plastic Neutral Axis, 377 Serviceability, 408-411 Shored, 375-376 Steel Headed Stud Anchors, 391-394 Unshored, 375-376 Composite Columns, 411-425 Beam-Columns, 421-425 Encased Shapes, 412-416 Filled HSS, 416-421 Composite Construction, 375 Composite Systems, Seismic, 606 Compression Member, 7-8, 124-181 Base Plate, 181-182 Behavior, 126-137 Boundary Conditions, 130-132 Bracing, 124, 131-132, 359-360 Built-up, 177-178 Design Tables, 163-168 Effective Length, 130-134, 138, 146-155 End Conditions, 130-132 Euler Column, 126-130 Flexural-Torsional Buckling, 169-170, 174-176 Length Effects, 146-157 Real Column, 134-135 Shapes, 124-126 Single Angle, 174-176 Slender Elements, 157-163 Strength, 126-145 Torsional Buckling, 169-170 Concentrated Forces, 250-252, 286-292, 578-589 Connecting Elements, 461-464 Compression, 462 Flexure, 462 Shear, 462 Tension, 461-462 Connections, 6-7, 10-11, 429- 589 Beam-to-Column, 430-432 Bolted Flange Plate, 565-578 Bolted Tee, 546, 548 Bracing, 524-528 Bracket, 430-431 Clip Angle, 434 Direct-Welded Flange, 546, 549 Double-Angle, 471-498 Bolted-Bolted, 473-491 Bolted-Welded, 496-498 Welded-Bolted, 492-496 Welded-Welded, 498 Fixed, 431-432 Fully Restrained, 432-433 Moment, Fully Restrained, 432-433, 546-578 Limit States, 548-549

612 Index Partially Restrained, 349-353, 432-433 Moment-Rotation Curves, 349-353 Partially Restrained, 349-353, 432-433, 546-547 Seated, 518-523 Seismic, 606-608 Shear, 471-523 Shear Tab, See Single-Plate Simple, 431-434, 471-542 Single-Angle, 498-509 Single-Plate, 509-517 Tension, Butt, 446-448 Hanger, 429, 530-535 Lap, 442-446 Prying Action, 533-535 Type 2 with Wind, 352 Welded Flange Plate, 556-565 Whitmore Section, 524-527 Construction Types, 11-17 Continuous Beams, 227-231 Critical Buckling, 128-129

D Dead Load, 23-26, 30-31, 34-35, 52 Deflection, 26, 17-18,247-249, 408-412 Deflection Amplification Factor, 591, 595, 598, 603 Direct Analysis Method, 324-325, 330-333 Drift, 26, 248

E Eads Bridge, 56-57 Effective Area, Compression, 159-162 Effective Area, Welds, 454-455 Effective Flange Width, 376 Effective Length, 130-134, 138, 146-155 Elastic, 146-149 Inelastic, 152-155 Method, 130-181, 325-326, 328-329 Elastic Buckling, 126-132 Encased Beam, 403 Euler Buckling, 126-130 Euler Column, 126-130 Eyebars, 81-83, 115-116 Exploration Tower, 56

F Filled Column, 412-414, 416-421 Fillet Welds, 453-457 First-Order Effects, 298-300 Flange Local Bending, 250-251, 266-273 Flexible Moment Connections, 252-258, 433-434 Flexural Buckling, 127-132 Flexural-Torsional Buckling, 169-170, 174-176 Four World Trade Center, 546 Framing Systems, 11-17

G Gable-Frame, 16 Geometric Imperfections, 325-327 Girders, 8-9, 12, 14 Girts, 7-9 Grades of Steel, See ASTM

Gravity Only Columns, 314-324, 337-349 Ground Snow Load, 32, 37-38

H High-Rise Construction, 15-16 High Strength Bolts, 434-436 Holes, Oversize, 88, 91 Patterns, 91-92 Placement, 91 Size, 88-90 Slotted, 88, 91 Standard, 88-89 Home Insurance Building, 56-57

I Impact, 33 Inelastic Buckling, 138-159 Inelastic Design, 18, 20 Influence Area, 35-37 Interaction, 247, 297-298, 300-304 Interaction Diagram, 301-304 Interaction Equation, 301-304 International Building Code, 27, 591-592

J Jenney, William LeBaron, 56 John Hancock Center, 15-16 Joists, 7-9, 12, 256-258

K K-factors, 146-155 KFC Yum! Center, 297

L Lateral Bracing, 201-216, 360-361 Lateral-Torsional Buckling, 201-216, 239-241 Leaning Columns, 314, 337, See also Gravity Only Columns Limit States, 18-19, 26-27 Lintel, 8-9 Live Load, 36-37, 42-43, 52 Arbitrary Point in Time, 30-31 Reduction, 36-37 Load and Resistance Factor Design (LRFD), 18-25 Load Combinations, 42-43, 51-52 Load Effect, 7-11, 17-18, 21-24 Load Factor, 42, 52 Loads, 30-41 Local Buckling, 157-163, 216-219, 234-235, 238, 244246, 266-270, 272 Long-Span Construction, 14 Lower Bound Moment of Inertia, 409-412

M Margin of Safety, 17-18, 20-21 Marlins Park, 429 Massachusetts College of Art and Design, 164 Mean Load Effect, 23-24 Mean Resistance, 23 Member Effects, 299-300, 314, 329-330 Metal Deck, 394-403 Minor Axis Bending, 222-223 Modular Ratio, 409

Index Modulus of Elasticity, 61-62 Moment Connection, 546-589 Moment Frame, 147-149, 312-324 IMF, 594, 597 OMF, 594-595, 597 Seismic, 594-597 SMF, 594-597 Moment Gradient, 206-214 Moment Redistribution, 227 Moment-Rotation Curves, 351-353

N National Earthquake Hazard Reduction Program (NEHRP), 591-592 Nelson Stud Company, 373 Net Area, 87-89, 91-95, 462 New York University Kimmel Center, 264 Nominal Strength, 17-19, 23, 26 Nomograph, See Alignment Chart Noncompact Beams, 216-219, 266-270 Notional Load, 325-330

O One Eleven South Main Street, 80 One World Trade Center, 1 Open Web steel Joist, 12, 256-258

P Partially Restrained Frames, 349-352 Pentagon, 34 Perfect Column, 127-128 Pin Connected Member, 112-115 Plastic Analysis, 227-231 Plastic Design, 227-231 Plastic Hinges, 227-231 Plastic Moment, 20, 194-201 Plastic Neutral Axis (PNA), 191 Plastic Region, 61-62 Plastic Section Modulus, 191, 194 Plate Girder, 264-295 Bending Strength Reduction Factor, 271-272 Compact Web, 279-280 Homogeneous, 266-267 Hybrid, 264 Noncompact Web, 267-271 Nontension Field Action, 282-284 Shear, 281-286 Slender Web, 271-274 Tension Field Action, 281-282, 284-286 Transverse Stiffeners, 281-282, 286-288 Pretensioned Bolts, 438 Probability, 21-22 Proportional Limit, 61-62 Protected Zones, 597, 601, 607 Prying Action, 533-535 Pure Column, 127 Purlin, 7-9 P-δ Effects, See Member Effect P-Δ Effects, See Structure Effect

R Rand-McNally Building, 56-57

613

Reduced Beam Section, 594, 596-597 Reliability index, 22-26 Required Strength, 17-19, 23 Residual Stress, 134-136 Resistance Factor, 19-26 Response Modification Factor, 591, 595, 598, 603 Root, John, 56 Rutgers University School of Business, 124

S Safety, 6, 17-19 Safety Factor, 18-26 Sears Tower, See Willis Tower Second-Order Effects, 15, 298-300, 312-315, 324-331 Seismic Behavior, 59-594 Seismic Design, 591-608 Capacity Design, 592-593 Categories, 592 Fuse Elements, 593 Protected Zones, 597, 601, 607 Seismic Load, 30, 33, 40-41, 43 Semirigid Connections, See Partially Restrained Frames September 11 Memorial Pavilion, 30 Seven Bryant Park, 372 Serviceability, 18, 26-27 Shake Down, 352-353 Shapes, Bars, 66-67, 74-75 Built-up, 67-68 C, 63 HP, 63-64 HSS, 65-66 L, 65 M, 64 MC, 65 MT, 65 S, 64 ST, 65 W, 63-64 WT, 65 Pipe, 66 Plates, 66-67 Shear, Beams, 224-225 Center, 237-238 Plate Girders, 281-286 Postbuckling Strength, 281-286 Tension Field Action, 284-286 Shear Lag, 95-97 Shear Stud, 78, 372-374, 391-394 Metal Deck, 395-403 Placement, 392-393 Strength, 391-392 Strength Reduction Factors, 392 Shored Construction, 375-376 Sidesway, Inhibited Frame, 147-149 Permitted Frame, 147-149 Prevented Frame, 147-149 Uninhibited Frame, 147-149

614 Index Web Buckling, 251, 290-291 Single Angle Compression Member, 174-176 Single-Story Frame, 16 Slender Elements, 157-163, See also Plate Girders Slenderness Parameter, 139 Snow Load, 32, 37-38, 42-43 Spandrel, 8-9 Special Plate Shear Walls, 598, 603, 605-607 Special Truss Moment Frame, 594, 603-604 St. Louis Gateway Arch, 56 Stability Analysis, 324-333 Direct Analysis Method, 324-325, 330-333 Effective Length Method, 325-326, 328-331 First-Order Analysis Method, 326, 329-330 Stanford University Bing Concert Hall, 471 Stem Local Buckling, 334-335 Stiffeners, 250-251 Bearing, 286, 288-291 Design, 291 Intermediate, 288-291 Stochastic Analysis, 30, 52 Strain Hardening, 62 Structural Stability Research Council (SSRC), 137 Structural Steel, Definitions of, 6-17 Design, 1-6 Structure Effect, 298-300, 312-315 System Overstrength Factor, 591, 595, 598, 603

T Tangent Modulus of Elasticity, 135-136, 152 Tensile Strength, 62 Tension Field Action, 284-286 Limitations, 283-284 Tension Member, 7-8, 80-119 Behavior, 84-87 Bracing, 80, 82, 117 Built-up, 83-84, 116 Eyebar, 81-83, 115-116 Fracture, See Tension Member, Rupture Hanger, 7-8, 80-81 Pin-connected, 112-113 Rod, 82-83, 85, 115-116 Rupture, 85-87 Sag Rod, 7, 80-81, 116 Shapes, 82-84 Slenderness, 84, 101 Truss, 82-84, 116-117 Yielding, 85 Terrorism, 34 Thermal Effects, 32-34 Torsional Buckling, 169-170 Transient Live Load, 32 Tributary Area, 35-36, 39

U Unbraced Frame, 147-148, 327-333 Unshored Construction, 375-375

V Vibration, 247-248, 409

W Weak Axis Bending, 222-223 Web, Local Buckling, 219-220,266-274 Local Crippling, 251-252, 288-290 Local Yielding, 250-252, 288-290 Plastification Factor, 268 Sidesway Buckling, 290-291, 288, 290-291 Welds, Fillet Weld, Strength, 455-457 Groove Weld, Strength, 461 Limit States, 454-455 Positions, 454 Sizes, 454 Types, 453-454 Welding Process, FCAW, 453 SAW, 452 GMAW, 453 SMAW, 452 Whitmore Section, 524-528 Willis Tower, 15-16 Wind Load, 30-33, 38-40, 42-43 World Trade Center, 1, 15-16, 546

Y Yield, Moment, 190-192 Point, 61-62 Stress, 61-62