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DESIGN OF STEEL STRUCTURES (VOLUME II) [S.I. UNITS]

By

Dr. Ram Chandra B.E., M.E. (Hons.), M.I.E., Ph.D. (Roorkee) , MIE Professor and Head Department of Structural Engineering M.B.M. Engineering College University of Jodhpur, Jodhpur (Rajasthan)

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Design of Steel Structures VOl–II

Published by:

RAJINDER KUMAR JAIN Standard Book House Unit of: Rajsons Publications Pvt. Ltd.

1705-A, Nai Sarak, Delhi - 110006 Post Box: 1074 Ph.: +91-(011)-23265506 Fax: +91-(011)-23250212 Showroom: 4262/3, First Lane, G-Floor, Gali Punjabian Ansari Road, Darya Ganj, New Delhi-110002 Ph.: +91-(011)-43551185 E-mail: sbhl0@ hotmail.com Web: www.standardbookhouse.in First Published Second Edition Third Edition Fourth Edition Fifth Edition Sixth Edition Seventh Edition Eight Edition Ninth Edition Tenth Edition Eleventh Edition Twelveth Edition Thirteenth Edition Fourteenth Edition Fifteenth Edition Sixteenth Edition Seventeenth Edition Eighteenth Edition Nineteenth Edition Twentieth Edition

: : : : : : : : : : : : : : : : : : : :

1971 1976 1981 1984 1987 1989 1991 1992 1994 1996 1998 2000 2003 2006 2008 2010 2012 2015 2016 2018

© Publishers All rights are reserved with the Publishers. This book or any part thereof, may not be reproduced, represent, photocopy or in any manner without the prior written permission of the Publishers.

` 495.00 ISBN: 978-81-89401-41-2 Typeset by: C.S.M.S. Computers, Delhi. Printed by: R.K. Print Media Company, New Delhi

FOREWORD

D EDICATED TO M Y TEACHERS

iii

Words from the Publishers Seventh edition of this book Design of Steel Structure Vol I and II is based on IS : 800–1984 as amended in 1997 and so also newly revised IS : 883–1994 for structural timber. New code of practice, IS : 800 is likely to be issued soon. In this it is likely to introduce ‘Limit State Design of Steel Structure’. For unsymmetrical bending and even for simple theory of bending, concept of shear-centre and axis of bending are essentially needed. These topics may be better appreciated, in case, the position and the location of shear centre for open thin-walled steel sections are nicely known. Illustrative examples have been given to explain the location of shear centre. Instead of IS : 883–1970, complete text for timber structures has also been revised as per revised and new code IS : 883–1994 Inspite of careful scrutiny of the manuscript, it is possible that some typograhical and computational errors are still left are noticed, publisher shall feel highly obliged to those, who bring these errors to their notice. Suggestions and comments from the readers for further improvement of the forthcoming editions of the shall be appreciated. 2015

Foreword Tables and clauses from the Indian Standard Specifications have been reproduced in the book with the kind permission of the Indian Standards Institution. It is desirable that for complete detail, reference be made to the latest versions of the Standards Institution, Manak Bhavan, 9, Bahadur Shah Zafar Marg, New Delhi-1, or from its branch offices at Mumbai, Kolkata, Kanpur and Chennai. SYSTEM INTERNATIONAL d’ UNITES (SI System of Units) In order to avoid the conversion of results obtained by engineers working with the Foot Pound Second (FPS) System (gravitational) of units in terms of centimetre-gram second absolute system of units used by the scientists, a need of common system of units was realised. The General Conference on Weights and Measures held at Paris in 1960 finalised the System International d’ Unites (SI). It is an absolute system of units. The mass is considered as fundamental unit and not the force. BIS has included a comment of transition in IS 3616– 1966. ‘Recommendation on the International System (SI) Units’ that this system has begun to replace older system of units in several branches of science and technology. The SI is a universal system of units and it has been adopted in France as a legal system and it is likely to become common in many countries. SI units have the following six basic units. Unit of Length (metre, m) The length equal to 1,650, 763.73 wavelengths, in vacuum, of the radiation corresponding to the transition between 2p19 and 5d5 levels of the krypton η atom of mass 86 is known as one metre. Unit of Mass (kilogram, kg) The mass of platinum-indium cylinder deposited at the International Bureau of Weights and Measures and declared as the international prototype of the kilogram by the First General Conference of Weights and Measures is called as one kilogram. Unit of Time (second, s) 1131, 566, 925, 974.7 of the length of the tropical year for 1900, the year commencing at 1200 hours universal time on the first day of January, 1900 is termed as one second. Unit of Electric Current (ampere, A) The constant current which flows in two parallel straight conductors of infinite length of negligible circular cross-section and placed at a distance of one metre from each other in vacuum producing a force of 2 × 10–7 New tons per metre length between the conductors is defined as an ampere.

FOREWORD

vii

Unit of Thermodynamic Temperature (degree Kelvin, K) The degree interval of the thermodynamic scale on which the temperature of triple point of water is 273.16 degrees, is known as one degree Kelvin. Units of Luminous Intensity (candela, cd) One sixtieth part of luminous intensity normally emitted by one hundred millimetre square of integral radiator (black body) at the temperature of solidification of platinum is called as one candela. The SI units make the use of multiples and sub-multiples 1000 times or 1/ 1000 times the unit quantity and in powers of 103 (kilo) or 10–3 (milli) in respect of still larger and smaller quantities respectively. The lengths are measured usually in kilometre (1 km = 1000 m), metre and millimetre (1 mm = 10–3 m).The symbols of units are not to be suffixed with V for plural. The force is a derived quantity and physical law connecting the quantity to the fundamental quantities or previously obtained derived quantities is force = mass × acceleration. It is defined as that force which produces unit acceleration i.e., 1 m per sec2 in a unit mass of 1 kg. Its unit is Newton (N). Though, the Newton is a small unit, a still larger unit kN may be used. The intensity of force (viz., stress) due to 1 Newton over a unit area of one metre square is known as one pascal. It is denoted by symbol, Pa. (1 Pa = 1 N/m2 and 106. Pa = 1 N/mm2, viz. 1 MPa = 1 N/mm2). SI system of units have many advantages. The units are very handy. The burden of non-decimal coefficients in foot-pound second system is avoided. It has relatively large main units in contrast to centimetre-gram-second system. At the same time, it is closely related to centimetre-gram-second system of units. In practice, it results in perfectly reasonable number when the value of g = 10 m/sec2 is used instead of 9.806 m/sec2. (Professor V.S. Mokashi, Visvesvaraya Regional College of Engineering, Nagpur in his paper titled as International System (SI) Units and their Application to Engineering published in Journal of Institution of Engineers, India, Vol. 19, March 1970 has highlighted the advantages and discussed SI units. A reference has been made to this paper). Structural Engineering is the science and art of planning, design, construction, operation, maintenance and rehabilitation of structures. The term “structures” includes bridges, buildings and all types of civil engineering structures (towers, shells, etc.) composed of any structural material. (Reference: Brochure, International Association for Bridge and Structural Engineering, Final Invitation to the International Conference, Structural Eurocodes, Daros, Switzerland September 14–16, 1992). Author

Preface to the First Edition In this book, the author with his long teaching experience in the subject has made an attempt to present the subject matter of design of steel structures in a way which lays emphasis on the fundamentals, keeping in view the difficulties experienced by the students. Every basic principle, method, equation or theory has been presented in simplified manner. Metric system of units has been used throughout the text. Indian Standards Specifications have been followed. The book is intended for the use of degree, diploma and A.M.I.E. students in various branches of engineering. The book deals with design of structural members and their connections. Each topic introduced is thoroughly described. A number of design problems including problems for examinations of the University of Jodhpur, and A.M.I.E. has been solved to illustrate the theory and practice. Slide-rule computation accuracy is adequate for the design and has been followed. The chapters have been so arranged that it facilitates self-understanding of the subject, during study. In spite of careful scrutiny of the manuscript, it is possible that some typographical and computational errors are still left. The author shall be highly obliged to any one who brings these errors to his notice. The author is thankful to Shri J.N. Srivastava and other colleagues who have very generously helped with their suggestions. The author is also thankful to the University of Jodhpur, Jodhpur and the Institution of Engineers, India, for following the use of their examinations, problems. Suggestions from the readers for the improvement of the book are welcome. 21 July, 1971

Ram Chandra

Preface to the Seventh Edition In the subsequent editions of this book, since first edition published in 1970 uptil now, the author enhanced the text by adding useful matter, fresh topics such as column formulae for axial stress in compression, design of built-up and perforated cover plate columns, modified and adjusted interaction formulae, equivalent axial load method of design of eccentrically loaded columns, approximate method of design of combined footing, graphical method of curtailment of flange plates, corrugated aluminium sheets used for roof covering and several examples. The author also added further text of design of high strength friction grip bolts. The twelveth edition of the book itself is a fourth edition in S.I. system of units (viz., system international d’ unites) and revised, rewritten and updated as per the latest code (viz., ‘Code of Practice for General Construction in Steel. IS : 800–1984) incorporating the revision of permissible stresses, effective length of the columns with idealised support conditions and columns in framed structures and Merchant Rankine formula for the allowable stresses. The concept of shear lag, design of semi-rigid connections, their behaviour (linear and nonlinear) and methods of analysis have also been included. The abbreviated symbols for Rolled Steel Sections as recommended in IS: 808–1989 have been used throughout the text of the book. Various definitions relating to the new and rational concept of Wind-Load as per IS: 875 (Part III)–1987 have been given in Chapter 2. Accordingly Chapter 9 (viz. Design of Roof Trusses) has been completely revised and determination of wind load has been thoroughly described and illustrated. Author expresses his sincere thanks to his colleagues, members of staff in various engineering colleges and students for appreciating the efforts made by them. The author also expresses his personal thanks for the Publishers Shri Rajinder Kumar Jain and Shri Sandeep Jain for getting the book prepared by latest technique and bringing out the book in such a nice getup. Author shall welcome the suggestions from the readers for the further improvement of the book in forthcoming editions. Jodhpur Dr. Ram Chandra August 21, 1991

Contents

PART 1 DESIGN OF STEEL BRIDGES CHAPTER 1

GENERAL

1.1 Introduction 1.2 Classification of Steel Bridges 1.3 Movable Bridges 1.3.1 Swing Bridge 1.3.2 Bascule Bridge 1.3.3 Rolling Bridge 1.3.4 Vertical Lift Bridge 1.3.5 Transporting Bridge 1.3.6 Floating Swing Bridge 1.4 Comparison of Fixed and Movable Bridges 1.5 Selection of Type of Bridges 1.5.1 Foundation Condition 1.5.2 Clearance Requirement 1.5.3 Length of the Bridge 1.5.4 Width of the Bridge 1.5.5 Live Load on the Bridge 1.5.6 Initial Cost 1.5.7 Operation and Maintenance Cost 1.5.8 Appearance 1.6 Erection Methods for Bridges 1.6.1 By Construction of Staging (Falsework) 1.6.2 By Rolling or Pushing from the Bank

3–32 3 4 10 10 11 12 12 13 13 13 14 14 14 14 15 15 15 15 15 15 16 17

XI

CONTENTS

1.6.3 By Cantilevering 1.6.4 By Using Cable Way 1.6.5 By Floating Spans into Position 1.7 Economical Span Length 1.8 Clearance 1.9 Width of Roadway and Footway 1.10 Dimension of Rolling Stock 1.11 Historical Development of Bridges 1.11.1 Beam Type (Timber) Bridges 1.11.2 Cantilever Type (Timber) Bridges 1.11.3 Arch Type (Stone and Brick Masonry) Bridges 1.11.4 Timber Bridges 1.11.5 Iron Bridge 1.11.6 Steel Bridges : (Arch and Truss Bridges) 1.11.6 Steel Bridges : (Cantilever Bridges) 1.11.7 Suspension Bridges 1.11.8 Cable Stayed Bridges 1.11.9 Reinforced Concrete Bridges 1.11.10 Prestress Concrete Bridges CHAPTER 2.

LOADS AND STRESSES

18 19 19 19 22 25 25 26 27 27 27 28 28 29 29 30 30 30 30 33–81

2.1 Introduction 2.2 Dead Load 2.3 Live Load 2.3.1 Railway Bridges 2.3.2 Highway Bridges 2.3.3 Foot Bridges and Foot-Paths (Attached to the Railway Bridges) 2.3.4 Footway (Attached to the Highway Bridges) 2.3.5 Combined Highway and Railway Bridges 2.3.6 Foot-Paths (Attached with the Combined Highway and Railway Bridges) 2.4 Impact Load 2.4.1 Railway Bridges 2.4.2 Highway Bridges 2.4.3 Foot Bridges 2.5 Wind Load 2.6 Lateral Load 2.7 Longitudinal Force 2.7.1 Railway Bridges 2.7.2 Highway Bridges 2.8 Centrifugal Force 2.8.1 Railway Bridge

33 33 34 34 45 52 53 54 55 55 55 57 58 58 59 59 59 64 65 65

XII

CONTENTS

2.8.2 Highway Bridge 2.9 Seismic Force 2.10 Erection Effects 2.11 Temperature Effects 2.12 Secondary Stresses 2.13 Relief Stresses 2.14 Combination of Loads 2.15 Allowable Stresses 2.15.1 Axial Stresses in Tension 2.15.2 Axial Stresses in Compression 2.15.3 Bending Stress 2.15.4 Shear Stress 2.15.5 Rivets, Bolts and Tension Rods 2.17 Allowable Stresses for Combination of Loads 2.17.1 Combined Stress 2.18 Fluctuation of Stresses 2.19 Endurance Limit 2.20 Design of Bridges CHAPTER 3.

DESIGN OF PLATE GIRDER BRIDGES

3.1 Introduction 3.2 Types of Floor Systems 3.2.1 Open Floor System 3.2.2 Solid Floor System 3.3 Deck Type Plate Girder Bridges 3.4 Through Type Plate Girder Bridges 3.5 Bracing of Deck Type Plate Girder Bridges 3.6 Bracing of Through Type Plate Girder Bridges 3.7 Self-weight of Plate Girders 3.8 Assumptions for the Design of Plate Girder Bridges 3.9 Design of Plate Girders for Deck Type Railway Bridges (Design of Maximum Section) 3.10 Design of Stringers, Cross-girders and Main Plate Girders for Through Type Railway Bridge 3.10.1 Design of Stringers 3.10.2 Design of Cross Girders 3.11 Curtailment of Flange Plates 3.12 Design of Connections of Flange Elements 3.13 Design of Stiffeners 3.13.1 Bearing Stiffeners 3.13.2 Intermediate Stiffeners 3.13.3 Vertical Stiffeners

65 66 66 66 66 67 67 67 67 68 69 74 74 75 75 76 80 81 82–162 82 82 83 83 84 85 85 86 87 88 88 96 97 99 105 105 106 107 107 107

XIII

CONTENTS

3.11.4 Horizontal Stiffeners 3.13.5 Connection of Intermediate Stiffeners to Web 3.14 Design of Stringers, Cross Girders and Plate Girders for Highway Bridges 3.14.1 Stringers 3.14.2 Cross-girders 3.14.3 Plate Girders 3.15 Wind Load on Plate Girder Bridges 3.15.1 For Unloaded Plate Girder Bridges 3.15.2 For Loaded Plate Girder Bridge 3.16 Wind Effect on Plate Girder Bridges 3.16.1 Overturning Effect 3.16.2 Horizontal Truss Effect 3.17 Design of Horizontal Truss Bracing 3.18 Design of Cross-frames 3.19 Design of Internal Gusset Plate Problems CHAPTER 4.

DESIGN OF TRUSS GIRDER BRIDGES

4.1 Introduction 4.2 Types of Truss Girder Bridges 4.2.1 Parallel Chord Truss Girder Bridges 4.2.2 Camel Back Truss Girder Bridges 4.2.3 Sub-divided Truss Girder Bridges 4.3 Deck Type and Through Types Truss Bridges 4.4 Component Parts of a Truss Bridge 4.5 Economic Proportions of Trusses 4.6 Self Weight of Truss Girders 4.6.1 Hudson’s Formula 4.6.2 Fuller’s Formula 4.7 Assumptions for the Design of Truss Bridges 4.8 Compression Members 4.9 Tension Members 4.10 Bracing of Deck Type Truss Girder Bridges 4.11 Bracing of Through Type Truss Girder Bridges 4.12 Wind Load on Truss Girder Bridges 4.13 Wind Effects on Truss Girder Bridges 4.13.1 Overturning Effect 4.14 Top Lateral Bracing 4.15 Bottom Lateral Bracing 4.16 Portal Bracing 4.17 Assumptions for the Analysis of Portal Frames

109 109 122 122 122 123 137 137 139 139 142 153 154 155 160

163–251 163 164 165 165 166 167 167 169 170 170 171 171 172 177 207 207 208 208 208 211 212 227 229

XIV

4.23 Sway Bracing Problems CHAPTER 5.

CONTENTS

249 249

DESIGN OF END BEARINGS FOR STEEL BRIDGES 252–290 5.1 Introduction 252 5.2 Functions of End Bearings 252 5.3 Types of End Bearing 253 5.3.1 Mechanical Bearings 254 5.3.2 Elastomeric Bearings 254 5.3.3 Combined Mechanical and Elastomeric Bearings 254 5.4 Selection of Type of end Bearings 255 5.4.1 Type of Super-Structure 255 5.4.2 Type of Supports 255 5.4.3 Length of Span 255 5.4.4 Loadings 255 5.4.5 Horizontal Movements 256 5.4.6 Rotational Movements 256 5.4.7 Plan Areas 256 5.4.8 Life and Maintenance 256 5.4.9 Inclination 256 5.4.10 Environment 256 5.4.11 Vibrations 257 5.4.12 Cost 257 5.5 Ferrous Mechanical Bearings 257 5.5.1 Plate Bearings 257 5.5.2 Rocker Bearings 259 5.5.3 Roller Bearings 261 5.5.4 Knuckle Pin Bearings 263 5.5.5 Bearing Adopted by Railway Board 264 5.5.6 Spherical Bearings 265 5.6 Allowable Stresses on Ferrous Bearings 265 5.6.1 Cylindrical Roller Bearing 265 5.6.2 Spherical Bearings 266 5.6.3 Sliding Bearings 267 5.7 Non-ferrous Mechanical Bearings 275 5.8 Elastomeric Bracings 276 5.9 Static Behaviour of Elastomer 278 5.9.1 Compression 278 5.10 Types of Elastomeric Bearings 283 5.11 Design of Elastomeric Bearings 284 5.12 Elastomeric Pot Bearings 287 Problems 288

CONTENTS

XV

PART 2 DESIGN OF STEEL AND OTHER STRUCTURES CHAPTER 6.

DESIGN OF STEEL AND OTHER STRUCTURES 291–334

6.1 Introduction 6.2 Types of Steel Chimneys 6.2.1 Self-Supporting Steel Chimneys 6.2.2 Guyed Steel Chimneys 6.3 Steel Plates for Chimney 6.4 Riveted Joints in Chimneys 6.5 Lining for Chimney 6.5.1 Fire Bricks 6.5.2 Insulating Refractory Bricks 6.5.3 Solid Grade Diatomaceous (moler earth) Bricks 6.5.4 Acid Resisting 6.5.5 Moler Concrete 6.5.6 Refractory Concrete 6.5.7 Sand and Cement Mixture 6.5.8 Guniting 6.6 Breech Opening 6.7 Ladder 6.8 Maintenance and Painting of Steel Chimney 6.8.1 Unlined Chimneys 6.8.2 Lined Chimneys 6.9 Lightening Conductor 6.10 Forces Acting on Steel Chimney 6.10.1 Self-Weight of the Chimney 6.10.2 Weight of Lining 6.10.3 Wind Pressure 6.10.4 Seismic Forces 6.11 Bending Moment on Self-Supporting Steel Chimney 6.12 Bending Stress on Steel Chimney due to Wind 6.13 Permissible Stresses 6.14 Design of Thickness of Steel Plates for Self-supporting Chimney 6.15 Design of Base-plate 6.16 Design of Anchor Bolts 6.17 Design of Foundation 6.18 Stability of Steel Chimney Problems

291 292 292 293 293 294 295 295 296 296 296 296 296 297 297 298 299 300 300 300 302 302 302 303 303 303 304 305 306 311 312 314 315 318 333

XVI

CONTENTS

Multiple Choice Questions Answers to Multiple Choice Questions CHAPTER 7. 7.1 7.2 7.3 7.4

DESIGN OF STEEL TANKS

Introcution Types of Steel Tanks Rectangular Steel Tanks Stand Pipes 7.4.1 Manhole 7.4.2 Pipe Connections 7.4.3 Outside Ladder 7.4.4 Roof Ladder 7.4.5 Painter’s Trolly Track 7.4.6 Roof Door 7.4.7 Overflow 7.5 Stresses in Stand Pipes 7.6 Design of Anchor Bolts for Stand Pipes 7.7 Elevated Circular Steel Tank 7.8 Riveted Joint In Elevated Circular Steel Tanks 7.9 Accessories of Elevated Circular Steel Tank 7.9.1 Tank Roof 7.9.2 Trap Door 7.9.3 Outside Ladder 7.9.4 Pipe Connections 7.9.5 Overflow 7.9.6 Painter’s Trolley Track 7.9.7 Stiffening Angles 7.9.8 Balcony 7.10 Forces Acting on the Circular Tank 7.10.1 Dead Load 7.10.2 Live Load 7.10.3 Seismic Forces 7.11 Stresses in Elevated Circular Steel Tanks 7.12 Stresses in Spherical Bottom 7.13 Stresses in Conical Bottom 7.14 Stresses in Connection between Side and Bottom Plates 7.15 Circular Girder 7.16 Staging for Circular Steel Tanks 7.17 Stresses in Columns 7.18 Wind Bracing 7.19 Pressed Steel Tanks 7.20 Capacity of Pressed Steel Tank

334 334 335–408 335 336 336 338 340 340 340 340 341 341 341 341 342 343 345 346 346 346 346 346 347 347 347 347 347 347 348 348 348 349 350 353 355 358 359 363 364 367

CONTENTS

7.21 7.22 7.23 7.24

Stays in Pressed Steel Tanks Accessories of Pressed Steel Tanks Staging for Pressed Steel Tanks Permissible Stresses Problems Multiple Choice Questions Answers to Multiple Choice Questions

CHAPTER 8. 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

DESIGN OF STEEL BUNKERS AND SILOS

Introduction Janssen’s Theory Airy’s Theory Design Parameters Design Criteria Analysis of Bins Hopper Bottom Design of Bins 8.8.1 Skin Plate Problems Multiple Choice Questions Answers to Multiple Choice Questions

CHAPTER 9.

DESIGN OF INDUSTRIAL BUILDINGS

9.1 Introduction 9·2 Structural Frame Work of Industrial Buildings 9.3 Floor Construction 9.4 Roof System 9.5 Wall Construction 9.6 Partition Walls 9.7 Staircases 9.8 Lighting 9.9 Heating and Ventilating 9.10 Crane Gantry Girders 9.11 Crane Columns 9.12 Bracing of Industrial Buildings 9.13 Bracing of Industrial Transverse Direction 9.14 Bracing of Industrial Buildings in Longitudinal Direction 9.14.1 Bents with Knee Braces 9.14.2 Bents without Knee Braces 9.15 Analysis of Industrial Building Bents 9.15.1 Columns Hinged at the Base

XVII

367 371 372 372 407 408 408 409–440 409 411 415 419 420 422 425 427 427 440 440 440 441–495 441 444 446 446 446 447 447 447 447 448 449 450 451 452 453 455 456 456

XVIII

CONTENTS

9.15.2 Columns Fixed at the Base 9.15.3 Columns Partially Fixed at the Base 9.15.4 Analysis of Stresses in the Members of Roof Truss 9.16 Design of Industrial Building 9.17 Unbraced Industrial Buildings (Gable Frames) Problems Multiple Choice Questions Answers to Multiple Choice Questions

457 459 460 461 493 494 494 495

CHAPTER 10. DESIGN OF MULTISTOREY BUILDINGS

496–544

10.1 10.2 10.3 10.4 10.5

Introduction Method of Analysis Number of Assumptions Required Bracing of Multi-storey Buildings Analysis of Multistorey Buildings with Moment Resistant Joints for Lateral Loads 10.6 Portal Method 10.7 Cantilever Method 10.8 Factor Method 10.9 Analysis of Multistorey Buildings with Moment-resistant Joints for Gravity Loads (Vertical Loads) 10.9.1 Uniformly Distributed Loads 10.10 High-rise Tubular Frames 10.11 Behaviour of Tubular Frames 10.12 Inital Preliminary Design Approach Problems Multiple Choice Questions Answers to Multiple Choice Questions CHAPTER 11. DESIGN OF LIGHT GAUGE STEEL MEMBERS 11.1 Introduction 11.2 Types of Cross-section 11.3 Definitions 11.3.1 Stiffened Elements 11.3.2 Multiple-Stiffened Elements 11.3.3 Unstiffened Elements 11.3.4 Flat-width Ratio 11.3.5 Effective Design Width 11.4 Local Buckling of Thin Elements 11.5 Post Buckling of Thin Elements 11.6 Light Gauge Steel Columns and Compression Members

496 498 498 499 502 504 506 508 513 513 536 540 542 542 544 544

545–598 545 546 548 548 548 548 548 549 550 552 554

XIX

CONTENTS

11.6.1 Crushing Failure of Columns 11.6.2 Local Buckling Failure of Columns 11.6.3 Primary Column Buckling of Columns 11.7 Form Factor For Columns and Compression Members 11.7.1 Members Composed Entirely of Stiffened Elements 11.7.2 Members Composed Entirely of Unstiffened Elements 11.7.3 Members Composed of Both Stiffened and Unstiffened Elements 11.8 Stiffened Compression Elements 11.9 Multiple Stiffened Compression Elements 11.10 Maximum Allowable Overall Flat-width Ratio 11.11 Unstiffened Compression Elements 11.12 Effective Length of Light Gauge Steel Compression Members 11.13 Basic Design Stress 11.14 Allowable Design Stress 11.15 Light Gauge Steel Tension Members 11.16 Light Gauge Steel Beams 11.17 Laterally Supported Light Gauge Steel Beams 11.18 Laterally Unsupported Light Gauge Steel Beams 11.19 Web Crippling of Beams 11.20 Deflection of Beams 11.21 Allowable Design Stresses in Beams 11.22 Beams Subjected to Combined Axial and Bendings Stresses 11.23 Connections 11.23.1 Bolted Connection 11.23.2 Welded Connection Problems Multiple Choice Questions Answers to Multiple Choice Questions CHAPTER 12. PLASTIC ANALYSIS AND DESIGN OF STEEL STRUCTURES 12.1 Introduction 12.2 Ductility of Steel 12.3 Applicability of Simple Plastic Theory of Bending for Different Materials 12.3.1 Aluminium Alloys 12.3.2 High Tensile Steel 12.3.3 Reinforced Cement Concrete 12.3.4 Brass 12.4 Perfectly Plastic Materials 12.5 Ultimate Load Carrying Capacity of Members Carrying Axial Tension

554 554 554 556 556 557 557 557 562 564 564 565 565 565 571 572 575 575 576 578 579 580 585 585 586 597 598 598

599–695 599 600 602 602 602 602 603 603 604

XX

CONTENTS

12.5.1 Single Member Carrying Axial Tension 12.5.2 More than one Member Carrying Axial Tension 12.6 Plastic Behaviour of a Simple Truss Structure 12.7 Plastic Bending of Beams 12.8 Fully Plastic Moment of a Section 12.9 Elastic-Plastic Bending of Rectangular Beam 12.10 Plastic Hinge 12.11 Shaper Factor 12.12 Moment Curvature Relationship 12.13 Load Factor 12.14 Marging of Safety 12.15 Fundamental Conditions for Plastic Analysis 12.15.1 Mechanism Condition 12.15.2 Equilibrium Condition 12.15.3 Plastic Moment Condition 12.16 Mechanism 12.16.1 Beam Mechanism 12.16.2 Panel or Sway Mechanism 12.16.3 Gable Mechanism 12.16.4 Joint Mechanism 12.17 Static Theorem or Lower Bound Theorem 12.18 Kinematic Theorem or Upper Bound Theorem 12.18.1 Principle of Virtual Work 12.19 Uniqueness Theorem or Combined Theorem 12.20 Plastic Collapse of Structure 12.20.1 Partial Collapse 12.20.2 Complete Collapse 12.20.3 Over Complete Collapse 12.21 Collapse Load for Standard Cases of Beams 12.21.1 Simply Supported Beam Subjected to Concentrated Load at the Centre 12.21.2 Simply Supported Beam Subjected to Eccentric Load 12.21.3 Simply Supported Beam Subjected to Uniformly Distributed Load 12.21.4 Propped Cantilever Beam Subjected to an Eccentric Concentrated Load 12.21.5 Propped Cantilever Beam Subjected to Uniformly Distributed Load 12.21.6 Fixed Beam Aubject to a Concentrated Load at the Centre 12.21.7 Fixed Beam Subjected to an Eccentric Load 12.21.8 Fixed Beam Subjected to Uniformly Distributed Load 12.22 Collapse Load for Continuous Beams

604 605 607 610 611 613 614 618 625 628 629 630 630 630 630 630 631 632 632 632 632 633 634 634 635 635 635 635 635 635 637 639 640 643 645 646 648 650

XXI

CONTENTS

12.22.1 Continuous Beam Subjected to Concentrated Loads 12.22.2 Continuous Beam Subjected to Uniformly Distributed Load 12.23 Combined Mechanism 12.24 Collapse Load for a Portal Frame 12.24.1 Beam Mechanism 12.24.2 Panel (Sway) Mechanism 12.24.3 Combined Mechanism 12.25 Design of Beams Problems Multiple Choice Questions Answers to Multiple Choice Questions

650 652 653 654 655 655 656 686 689 693 695

CHAPTER 13. DESIGN OF STEEL TOWERS, TRESTLES AND MASTS 696–732 13.1 Introduction 696 13.2 Loads on Towers 699 13.3 Shape, Sag (Dip) and Tension in Uniformly Loaded Conductors 703 13.4 Analysis of Towers 706 13.5 Masts 707 13.6 Trestles 708 13.7 Stresses in Trestles due to Vertical Loads 709 13.8 Stresses in Trestles due to Horizontal Loads 711 13.9 Design of Members in Towers 712 13.10 Design of Tower Foundation 715 Problems 731 CHAPTER 14.

DESIGN OF ALUMINIUM STRUCTURES

14.1 Introduction 14.2 Chemical Composition 14.3 Stress-strain Relationship for Aluminium 14.4 Properties of Aluminium Alloys 14.5 Factor of Safety 14.6 Advantages and Disadvantages 14.7 Permissible Stresses 14.8 Tension Members 14.9 Columns and Compression Members 14.9.1 Buckling or Bending Failure 14.9.2 Twisting Failure 14.9.3 Bending Combined with Twisting Failure 14.10 Permissible Stresses in Compression for Columns 14.11 Effective Length of Columns 14.12 Laced and Battened Columns

733–760 733 735 735 736 737 737 738 739 739 740 742 742 743 743 744

XXII

CONTENTS

14.13 Combined Compression and Bending 14.14 Beams 14.15 Local Buckling of Elements in Compression 14.16 Riveted Connections 14.16.1 Effective Diameter 14.16.2 Effective Bearing Area 14.16.3 Permissible Stresses 14.16.4 Reduction in Strength 14.16.5 Edge Distance of Rivet 14.16.6 Spacing of Rivet 14.16.7 Pitch of Rivets in Built up Compression Member Problems

744 745 746 750 750 751 751 751 752 752 752 758

PART 3 ANALYSIS OF STRUCTURES CHAPTER 15. INFLUENCE LINES FOR STRESSES IN FRAMES 761–836 15.1 Introduction 761 15.2 Influence Line Diagrams for Stresses in Members of a Pratt Truss with Parallel Chords 762 15.2.1 Influence Line Diagram for U3U4 (Top Chord Members)

762

15.2.2 Influence Line Diagram for L3 L4 (Bottom Chord Member)

764

15.2.3 Influence Line Diagram for U3U4 (Vertical Member)

765

15.2.4 Influence Line Diagram for U3L4 (Diagonal Member) 15.2.5 Influence Line Diagram for U1L1 15.3 Influence Line Diagram for Stresses in Members of a Pratt Truss with Curved Chord

766 766 767

15.3.1 Influence Line Diagram for U1U2

767

15.3.2 Influence Line Diagram for L1 L2

769

15.3.3 Influence Line Diagram for U1U2

770

15.3.4 Influence Line Diagram for Stress in U2U2

771

15.3.5 Influence Line Diagram for Stress in U1L1 15.4 Influence Line Diagram for Stresses in Members of a Warren Truss with Parallel Chords 15.4.1 Influence Line Diagram for U2 L3 (Top chord member)

772 772 772

CONTENTS

XXIII

15.4.2 Influence Line Diagram for Stress in Member L1L2 (Bottom Chord Member) 15.4.3 Influence Line Diagram for Stress in U2L2 (Diagonal Member) 15.4.4 Influence Line Diagram for Stress in U2L1 (Diogonal Member) 15.5 Influence Line Diagram for Stresses in Members of a Warren Truss with Curved Chord 15.5.1 Influence Line Diagram for U1U2 15.5.2 Influence Line Diagram for Stress in L1L2 15.5.3 Influence Line Diagram for Stress in U2 L1 15.5.4 Influence Line Diagram for Stress in U1L1 15.6 Influence Line Diagram for Stresses in Members of a Baltimore Truss with Sub-ties 15.6.1 Influence Line Diagram for Stress in Member U2U4 (Top Chord Member)

773 775 776 777 777 778 780 781 782 782

15.6.2 Influence Line Diagram for Stress in L2 L3 and L2 L4 (Bottom Chord Members) 15.6.3 Influence Line Diagram for Stress in U2 M3 (Diagonal Member) 15.6.4 Influence Line Diagram for Stress in U2 L2 (Vertical Member) 15.6.5 Influence Line Diagram for Stress in M3 L3 15.6.6 Influence Line Diagram for Stress in M3 M4(Sub -Tie) 15.6.7 Influence Line Diagram for Stress in M3 L4 15.6.8 Influence Line Diagram for Stress in U4 L4 15.7 Influence Line Diagrams for Stresses in Members of a Baltimore Truss with Substruts (Through Type) 15.7.1 Influence Line Diagram for Stress in U4U4 (Top Chord Member) 15.7.2 Influence Line Diagram for Stress in L2L3 and L3 L4 (Bottom Chord Members) 15.7.3 Influence Line Diagram for Stress in M3 L4 15.7.4 Influence Line Diagram for Stress in M3 L3 15.7.5 Influence Line Diagram for Stress in L2 M3 (Sub-Strut) 15.7.6 Influence Line Diagram for Stress in U2 M3 15.7.7 Influence Line Diagram for Stress in U2 L2 15.8 Influence Line Diagram for Stresses in Members of A Pettit of Pennsylvania Truss with Sub-ties 15.8.1 Influence Line Diagram for Stress in U4U6 15.8.2 Influence Line Diagram for Stress in L4 L5 (and L5 L6) 15.8.3 Influence Line Diagram for Stress in U4 M5 15.8.4 Influence Line Diagram for Stress in M5 L5 15.8.5 Influence Line Diagram for Stress in M5 U6 15.8.6 Influence Line Diagram for Stress in M5 L6

784 785 787 788 788 789 790 791 791 792 795 795 796 796 797 798 798 800 801 802 802 803

XXIV

CONTENTS

15.8.7 Influence Line Diagram for Stress in U4 L4 15.9 Influence Line Diagram for Stresses in Members of Pettit or Pennsylvania Truss with Sub-struts 15.9.1 Influence Line Diagram for Stress in L5 L6 and (L4 L5) 15.9.2 Influence Line Diagram for Stress in U4U6 15.9.3 Influence Line Diagram for Stress in M5 L6 15.9.4 Influence Line Diagram for Stress in M5 L5 15.9.5 Influence Line Diagram for Stress is L4 M5 15.9.6 Influence Line Diagram for Stress in U4 M5 15.9.7 Influence Line Diagram for stress U4L4 15.10 Influence Line Diagrams for Stresses in Members of a K-truss 15.10.1 Influence Line Diagram for S tress in U2U3 15.10.2 Influence Line Diagram for Stress in L2L3 15.10.3 Influence Line Diagram for Stress in M2U3 and M2L3 15.10.4 Influence Line Diagram for Stresses U3 M3 15.10.5 Influence Line Diagram for Stress in M3 L3 15.11 Influence Line Diagrams for Stresses in Members of a K-truss with Inclined Chord 15.11.1 Influence Line Diagram for Stress in U4U5 15.11.2 Influence Line Diagram for Stress in L4L5 15.11.3 Influence Line Diagram for Stress in U5U6 15.11.4 Influence Line Diagram for Stress in L5 L6 15.11.5 Influence Line Diagram for Stress in M4U5 15.11.6 Influence Line Diagram for Stress in M5 L6 15.11.7 Influence Line Diagram for Stress in U5 M5 15.11.8 Influence Line Diagram for Stress in M5 L5 15.12 Influence Line Diagrams for Stresses in Members of a Braced Cantilever and Suspended Span Girder 15.12.1 Influence Line Diagram for Reactions RA and RB 15.12.2 Influence Line Diagram for Stress in U3 U4 15.12.3 Influence Line Diagram for Stress in L3 L4 15.12.4 Influence Diagram for Stress in L3 L4 15.12.5 Influence Line Diagram for Stress U3 L3 Problems CHAPTER 16. 16.1 16.2 16.3 16.4 16.5

ANALYSIS OF SPACE FRAMES

Introduction Tension Coefficients Analysis of Plane Frames Procedure for the Application of Method Analysis of Space Frames Problems

804 806 806 807 809 810 811 811 812 814 814 814 816 817 818 819 819 821 822 822 823 824 825 826 828 828 830 831 832 833 834 837–855 837 837 838 840 844 853

CONTENTS

CHAPTER 17. 17.1 17.2 17.3 17.4 17.5 17.6 17.7

ANALYSIS OF SECONDARY STRESSES

Introduction Secondary Stresses due to Rigidity of Joints Determination of Changes of Angles in any Triangle Determination of Deflection Angles Secondary Moments Equilibrium of Joints Secondary Stresses due to Bending of Members Problems

CHAPTER 18.

SPECIAL STRUCTURES

18.1 Introduction 18.2 Orthotropic Steel Plate Panels and Bridges 18.3 Analysis of Orthotropic Panels 18.3.1 Panel with Open Stiffeners 18.3.2 Panels with Closed Stiffeners 18.4 Analysis of Orthotropic Bridges 18.5 Tensile Structures or Tension Loaded Structures 18.6 Cables 18.7 Cable Cross 18.8 Cable Nets 18.8.1 Orthogonal Cabic Nets 18.8.2 Axisymmetrical Cable Nets 18.9 Cable Systems and Nets Forming Vertical Surfaces 18.10 Cables Suspended Freely to form Surfaces 18.11 Cable Structures 18.12 Membranes 18.13 Difference between Cable Nets and Membrane Shapes 18.14 Prestressed Tension Loaded Structures 18.15 Tensile-stress Pneumatic Structures Index

XXV

856–869 856 857 858 860 861 862 863 869 870– 870 871 874 874 875 876 877 877 878 879 879 879 881 882 883 884 884 885 885 887

IS CODES Useful IS codes used in the Text of the book 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

226–1975 227 459 723 800–1984 806 808 812–1957 813–1961 816–1969

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

819–1957 875–1984 961–1975 1148–1973 1149–1982 1161 1173 1252 1261–1959 1323–1962

21. 1730 22. 1731 23. 1732 24. 25. 26. 27. 28. 29.

1911 1977–1975 2062–1984 2585 3139 3757–1972

30. 4000–1967 31. 456–1978 32. 6623–1972 33. 6639–1972 34. 6649–1972

Structural Steel (Standard quality) Galvanised (Plain and Corrugated) sheets Unreinforced Corrugated Asbestos Cement Sheets Mild Steel Wire Nails Code of Practice for General Construction in Steel Use of Steel Tubes in General Building Construction Rolled Steel Beams, Channels and Angle Sections Glossary of Terms Relating to Welding and Cutting Metals Scheme for Symbols of Welding Use of Metal Arc Welding for General Construction in Mild Steel Resistance Spot Welding for Light Assemblies in Mild Steel Loading Standard for Structural Safety of Buildings Structural Steel (High Tensile) Rivets Bars for Structural Purposes High Tensile Rivet Bars for Structural Purposes Steel Tubes for Structural Purposes Rolled Steel Sections, Tee-Bars Rolled Steel Sections, Bulb-Angles Seam Welding in Mild Steel Code of Practice for Oxy-Acetylene Welding for Structural Works in Mild Steel Dimensions for Steel Plates, Sheets and Strips for Structural and General Engineering Purposes Dimensions for Steel Flats for Structural General Engineering Purposes Dimensions for Round and Square Steel Bars for Structural and General Engineering Purposes Schedule of Unit Weight of Materials Structural Steel (Ordinary quality) Weldable Structural Steel Black Square Bolts and Nuts and Black Square Screws Dimensions for Screw Threads for Bolts and Nuts High Tensile Friction Grip Fasteners for Structural Engineering Purposes Assembly for Structural Joints using High Tensile Friction Grip Fasteners Code of Practice for Plain and Reinforced Concrete (Third Revision) High Tensile Friction Grip Nuts Hexagon Bolts for Steel Structures High Tensile Friction Grip Washers

1. General 2. Loads and Stresses 3. Design of Plate Girder Bridges 4. Design of Truss Girder Bridges 5. Design of End Bearings for Steel Bridges

CHAPTER

1

General

1.1

INTRODUCTION

The bridges are the structures, which provide means of communication (viz., passage) over a gap. In some gaps, the water flow for a part of the year or for whole of the year. Whereas, some other gaps remain dry throughout the year. The rivers, canyons and valleys form natural gaps. The railway and highway crossings, highway and canal crossings form artificial gaps. The bridges provide passage for the vehicular or other traffic over these gaps. The bridges constructed to carry highway traffic are known as highway bridges. The roads provide passage. The bridges built to carry railway traffic are known as railway bridges. The railway tracks provide passage. There are some bridges which carry the highway and railway traffic both, and these bridges are known as combined highway and railway bridges. There are some bridges for pedestrians only. Such bridges are known as foot bridges. There are some bridges which carry canals and pipe lines and these bridges are known as aqueduct bridges. The bridges are made of timber, stone masonry, brick masonry, reinforced cement concrete, prestressed cement concrete and steel. The timber bridges are used for short spans, light loads and for temporary bridges. The masonry bridges are also used for short spans. The economy of masonry bridges depends upon the availability of good building materials and the skilled labour. The masonry bridges are generally arch bridges. There are various types of reinforced cement concrete bridges, which are suitable for different spans and the different site conditions. The steel bridges have been discussed in this part of the book. The deck consisting of slab, girders (I-beams, plate girders, truss girders, etc.), bearings (which support girders) are the various components of the bridge. These components may be clubbed into first group. In addition to these, the abutments, piers, foundations, river training works, approaches, hand rails etc.,

4

DESIGN OF STEEL STRUCTURES–VOL. II

are also the components of the bridge. These components may be put into second group. The components included in first group and foundation are the structural systems. The components of second group forming the sub-structure are also considered as structural systems. The components placed below the level of bearings are grouped as sub-structure, whereas the components placed above the level of bearings are grouped as super structure. In human history, the bridges have figured prominently. The cities develop at a bridge head (e.g., London, Oxford, Cambridge, Innsbruck). The bridges abridge the distances and have rendered most for the civilization of our species. The bridges also have great importance in war also. The bridges help in advancement of the army. The bridges are destroyed in case, the army retreats. In military, special emphasis is given in training to build new bridges in short time and to destroy the existing bridges, if necessary.

1.2

CLASSIFICATION OF STEEL BRIDGES

The steel bridges may be classified according to the service they perform or according to their structural arrangements. The steel bridges are classified in various different manners. The following are different methods of classification of the steel bridges. 1. The steel bridges are classified according to make up of main load carrying element as follows: (i) I-beam bridges (ii) Plate girder bridges (iii) Truss girder bridges (iv) Suspension (cable) bridges (v) Box girder bridges (vi) Cable stayed bridges. In I-beam bridges, rolled steel wide flange I-beams are used as load carrying members for culverts. I-beam bridges as shown in Fig. 1.1 (a) are used upto 15 m span in railway bridges and upto 25 m span in highway bridges. The I-beam bridges are popular because of its simple design and construction. In the plate girder bridges as shown in Fig. 1.1 (b), the plate girders are used as load carrying members. The plate girder bridges are used upto 30 m span. The plate girder bridges are used both for highway bridges and railway bridges. The plate girders are used so long as the plate girders may be transported in one piece. The problems of fabrication, erection, and transportation usually limit the use of plate girders to depths of roughly 3 m to 4 m. When the depths of load carrying elements are more than this, then, the truss girder bridges are used. In the truss girder bridges, as shown in Fig. 1.1 (c), the truss girders are used as load carrying members. There are various types of the trusses which are suitable for different spans and different site conditions. The truss girder bridges are used from 30 m span to 250 m span. In the suspension (cable) bridges, as shown in Fig. 1.1 (d), high tensile steel wire cables are used. The suspension bridges are used for large spans. The suspension bridges are used for highway bridges for long spans above 250 m.

GENERAL

5

In box girder bridges, the box-shape sections are built from stiffened plates, angle sections, channel sections and/or beam sections. The box girder bridges are used for 15 m to 50 m spans. The cable stayed bridges are recently (viz., 1970) developed bridges. These bridges are used for 300 m to 600 m spans. The girders are supported by cables (which may be arranged in different arrangements). The cables are attached with tall towers built on two sides of the bridge spans. The tall towers are supported by heavy foundations. The cable stayed bridges have different profiles and the structural configurations.

(a ) B e am b rid g e

(b ) P late g ird er bridg e

(c) Tru ss g ird e r bridg e

(d ) S u sp e nsio n bridg e

Fig. 1.1

2. The steel bridges are classified according to the structural layout of the principal load carrying members as follows : (a) Simply supported span bridges (b) Continuous span bridges (c) Cantilever bridges (d) Arch bridges (e) Rigid frame bridges In the simply supported span bridges, the whole width to be bridged is divided into number of individual spans. In each span, the load carrying member is simply supported at both ends. Both, the plate girder and the truss girder load carrying elements are used in this type of bridges. A simply supported truss girder bridge is shown in Fig. 1.2. The analysis in this type of bridge is very

6

DESIGN OF STEEL STRUCTURES–VOL. II

simple. The simply supported span bridges are suitable where the uneven settlement of foundation may take place. The simple truss bridge J.J. Barry built in 1973 over Delware river with 250.546 m span and the simple girder bridge built in 1951 over Harlem river, New York city are the examples of their type with maximum spans.

Fig. 1.2 Simply supported span bridges

In the continuous span bridges the load carrying member of the bridge is continuous over more than two supports. The continuous span bridge is shown in Fig. 1.3. The continuous span bridges are statically indeterminate structures. The reactions at the supports depend upon the type of the structure and the

Fig. 1.3 Continuous span bridges

type of loadings. The reactions cannot be determined from the principles of static alone. The reactions at the supports are found by taking into account the dimensions and the material of the structure itself. The continuous span bridges are suitable where the uneven settlement of foundation does not occur. The stresses in each span are reduced due to the negative moments developed at the pier supports. After erection of one span in the continuous span bridges, the other subsequent spans may be erected by the cantilever method, and without the use of falsework. The continuous span bridges have several advantages over simple span bridges. There is an appreciable saving of material running as high as 10 to 20 per cent. The continuous span bridges need few supports as long spans are possible. The continuous span bridges can support higher loads than the simple span bridges. In case bending moment diagram is drawn for a continuous span bridge with three spans, loaded with uniformly distributed load, then, it will be observed that there are two points of contraflexures in the central span. If the continuous beam is cut at these points of contraflexures and the joints are made at these points which are capable of resisting shear but not bending, then, the resulting girder formed is known as cantilever bridge. The cantilever bridges have all the advantages of the original continuous girder, but the bending moment diagram will not now be affected by any settlement of the piers. Moreover, the bending moment diagram, and the shear force diagram may be drawn by applying simple principles of statics. The continuous truss bridge, Astoria built in 1966 over Columbus river, Oregon, with 375.514 m span (maximum in its own type) is the example of continuous truss bridge. The cantilever bridges consist of two simple spans, each having an overhanging or cantilever portion as shown in Fig. 1.4 and having simple span in between. The overhanging portions are continuous over piers. The cantilever bridge is

GENERAL

7

also defined as a bridge in which one or more of its trusses are extended beyond their supports thus forming cantilever arms. The extended arms or cantilever arms support other trusses at their ends. The central span is known as suspended span. This suspended span is supported by the cantilever portions.

Fig. 1.4 Cantilever bridges

The suspended span is supported at both ends on mechanical hinges. In general, each complete truss has only two supports. Therefore, reactions and the stresses are statically determinate. The stresses in each span are reduced due to the

(a ) S o lid rib arch b rid ge

(b ) B ra ced rib a rch b rd ge

(c) S p an drel bra ced a rch b rid ge

(d )

Tied arch b rid ge

(e ) Tied arch b rid ge (w ith pa rtially hu n g d e ckin g)

Fig. 1.5 Arch bridges

negative moments developed at the pier support. The chief advantage of the cantilever bridges is that the cantilever portions and the suspended span may be erected without the use of falsework or staging. The cantilever bridges are

8

DESIGN OF STEEL STRUCTURES–VOL. II

suitable where the uneven settlement of the foundation may occur. The cantilever bridges are used from 200 m span to 600 m span. The forth cantilever bridge built in 1889 at Scotland with 520 m is example of its type. Aesthetically, the arch bridges enhance the beauty of its surroundings more than the simple truss bridges. The various types of arch bridges are shown in Fig. 1.5. The arches are built by girder or trusses. The arches used for the arch bridges are fixed arches, two hinged arches, and the three hinged arches. The two hinged arches are most commonly used for arch bridges. The arch bridges are classified as solid ribbed, tied, braced rib, or spandrel braced. The solid ribbed steel arch bridges as shown in Fig. 1.5 (a) are used more for the highway bridges than for the railway bridges. These bridges are used for 150 m span to 200 m span. The braced rib arches as shown in Fig. 1.5 (b) are mainly used for highway bridges. These bridges are used for long spans from 500 m to 600 m. The spandrel braced arches as shown in Fig. 1.5 (c) are used for railway bridges. These bridges are used for spans upto 350 m. The horizontal thrust from this arch is resisted by supports (abutments and/or piers) or by horizontal ties. When the horizontal ties resist the horizontal thrust of the arch, the arch is known as tied arch as shown in Fig. 1.5 (d). The horizontal ties are in direct tension. These ties are suspended from the arch rib at intervals. These ties form side members of the deck construction. Figure 1.5 (e) shows tied arch bridge in which the deck is placed at a level above the apparent springing of the arch. The major portion of the length of deck is suspended by the suspenders from the arch rib. Towards the end, the deck is supported from top of the arch ring by columns or by cross-walls. The steel arch bridge built in 1976 over New River George at West Virginia with 518.160 m span is the example of its type. In the rigid frame bridges, the load carrying members are the rigid frames. The steel rigid frame bridge as shown in Fig. 1.6 (a) is quite oftenly used for single spans. This type of bridge is very satisfactory for over passages. This type of bridge is economical for spans from 10 m to 25 m. The verticals of rigid frame act as abutments and provide continuity. Figure 1.6 (b) shows a series of continuous span rigid frame bridge. The rigid frame bridges are suitable for rigid foundations.

(a )

(b )

Fig. 1.6 Rigid frame bridges

GENERAL

9

3. The steel bridges are classified according to the cross-section of the bridge, i.e., floor location as follows : (i) Deck type bridge (ii) Through type bridge (iii) Half through type bridge. In the deck type bridge, the floor rests on the top of main load carrying members. In the deck type plate girder bridge, the floor is placed on the top flanges. In the deck type truss girder bridge, the floor is placed on the top chords. In the deck type bridge, the bracing is not done over the top of the traffic. In the through type bridge, the floor rests on the bottom of main load carrying members. In the through type plate girder bridge, the floor is placed on the bottom flanges. In the through type truss girder bridge, the floor is placed on the bottom chords. In the through type bridge, the bracing is also done over the top of the traffic. In the half-through type bridge, the floor lies in between the top and the bottom of the main load carrying members. The half through type bridge is also termed a semi-through or a pony bridge. In the half-through type bridge, the over bracing is not done, and the load carrying members project above the floor level. There are also double deck type bridges. The decks are provided at two different levels. In the double deck type bridges, both the decks can be through type or one can be open deck type and the other can be the through type. The deck type and the through type plate girder bridges have been further discussed in Chapter 3. The deck type and the through type truss girder bridges have been further discussed in Chapter 4. The grade line of highway or railway track and the clearance required under the bridge decide the use of deck type or through type bridge. In case, the sufficient clearance is available under the bridge, the deck type bridges are the advantageous over the through type bridges. The deck type bridges are relatively economical. The load carrying members may be placed close together than in the through type bridges. This reduces the length of floor beams and the lateral moments in the floor system. The heights of piers and abutments are reduced in the deck type bridge. This reduces the masonry work. The deck type bridges have pleasing appearance. The deck type bridges are therefore more popular than through type bridges. 4. The steel bridges are classified according to the type of connections (or fasteners used) as follows: 1. Riveted bridges 2. Welded bridges 3. Pin-connected (and bolted) bridges. The majority of steel bridges are riveted bridges. The welded bridges are recently designed and constructed. The pin-connected bridges were constructed in the past. The riveted joints are rigid and have more secondary stresses. In order to reduce the secondary stresses, the pin-connections were introduced. It was found that the pin-connections are not as free to rotate as these were thought to be. The maintenance of the pin-joints is also difficult. The pin-connected bridges are not constructed nowadays.

10

DESIGN OF STEEL STRUCTURES–VOL. II

5. The steel bridges are classified according to the level of crossing of railway track and highway as follows : (i) Over bridges (ii) Under bridges. When the railway track crosses the highway, then, the railway track may be carried over or under the highway by means of a bridge or on the same level. When the highway bridge is carried over the railway track by means of a bridge, then, the bridge is known as over bridge. When the highway is carried under the railway track by means of a bridge the bridge is known as the under-bridge. 6. The steel bridges are classified according to the nature of movement of the bridge girders as follows : (i) Fixed (Permanent) bridges (ii) Movable bridges. The fixed bridges do not have any movement. These bridges remain in one position. The movable bridges can be opened either horizontally or vertically so as to allow rivet or channel traffic to pass. The movable bridges are provided over the navigable channels.

1.3

MOVABLE BRIDGES

When the bridges cross a navigable canal, channel or river near the water level, then, the clearance under the bridge girder is not sufficient for the passage of masted vessels or streamers. At such places, the fixed or permanent bridges cannot be provided. The movable bridges provide a satisfactory solution for crossing navigable streams. The movable bridges can be opened horizontally or vertically and leave the streams clear for navigation. When the movable bridges are closed, then, these provide passage for highway or railway or pedestrian traffic. The movement for movable bridges may be provided by (i) turning about a horizontal axis, or (ii) turning about a vertical axis, or (iii) rolling horizontally, or (iv) lifting vertically, or (v) floating in the stream. As regards the strength and the stiffness for the greatest load, which the bridge is to bear when closed, the movable bridges do not differ from the fixed bridges. The following are the various types of movable bridges. 1. Swing bridge 2. Bascule bridge 3. Rolling bridge 4. Vertical lift bridge 5. Transporting bridge 6. Floating swing bridge.

1.3.1

Swing Bridge

The swing bridge turns about a vertical axis in the horizontal direction as shown in Fig. 1.7. A pier is constructed in the centre of the stream. The pier supports a base plate. The diameter of base plate is kept nearly equal to the width of the bridge. The base plate has a pivot in the centre. There is a circular track round the circumference of a base plate as in a railway turn table. There is a roller frame with suitable conical rollers. The rollers rest on the track. The roller

GENERAL

11

frame turns about the central pivot. There is a circular revolving platform resting on the pivot and rollers. A toothed arc with suitable wheel work and fixed with

S w ing b ridge

Fig. 1.7 Swing bridge

the revolving platform gives motion to the platform. A set of parallel girders or cross girders resting on and fixed to the revolving platform support a continuous steel truss girder bridge. The truss girder may carry the floor at the top chord or at the bottom chord. The swing span bridge over Ft. Madison (a rail road bridge) at Mississippi river built in 1927 has 160.02 m span.

1.3.2

Bascule Bridge

The bascule bridge is also known as draw bridge. The bridge turns about a horizontal axis and in a vertical direction. The bridge is opened in the vertical position by means of a pinion driving a toothed sector. The cables are used for raising the bridge. The weight of bridge is balanced by the counter-weight. The

(a ) S ing le b ascu le b rid ge s

(b ) D o ub le ba scu le b rid ge s

Fig. 1.8 Bascule bridges

counter-weight facilitates the raising of bridge. The centre of gravity of the whole system of bridge lies on the horizontal axis about horizontal axis about which

12

DESIGN OF STEEL STRUCTURES–VOL. II

the bridge rotates. The strength of the bridge should be sufficient to support safely the overhanging weight of its structure, when lifted from the direct support. Figure 1.8 (a) shows single bascule bridge. In the single span bascule bridge, the bridge is turned up vertically at one end. Figure 1.8 (b) shows double bascule bridge. In the double bascule bridge the bridge is turned up vertically at both the ends. The bascule bridge is suitable for small spans. The bascule bridge can be opened more quickly than the swing bridge. The opening of the bridge becomes difficult in case heavy wind is blowing. The single-leaf bascule bridge 16th street (rail road bridge) at Chicago, Illinois built in 1919 and the bascule bridge over Sault Ste. Marie Michigan built in 1914 have 79.248 m and 102.413 m spans, respectively. The Pampan bridge at Rameshwaram (India) is the other example.

1.3.3

Rolling Bridge

The rolling bridge is shown in Fig. 1.9. The rolling bridge is also known as traversing bridge. The bridge rolls forward and backward horizontally. The bridge has sufficient overhanging portion to span the waterway. The rolling bridge has

Fig. 1.9 Rolling bridge

strong frame. The frame is supported by the wheels. The wheels roll on the rails. When the bridge is rolled forward and is in closed position, then, the rolling frame leaves a gap between one of the approach and the platform. A second rolling frame is placed in the gap. The rolling frame rolls sideways. When the rolling bridge is opened, then the second rolling frame is moved out of way. The centre of gravity of rolling bridge is so balanced that it remains over the rolling frame.

1.3.4

Vertical Lift Bridge

The vertical lift bridge is shown in Fig. 1.10. In the vertical lift bridge, the movable span is lifted vertically by means of chains or cables. The chains or cables pass

Fig. 1.10 Vertical lift bridge

GENERAL

13

over the pulleys. The pulleys are provided at the top of the towers erected at both the ends. At the other ends of chains or pulleys, the counter-weights are provided. The counter-weights facilitate the lift of the bridge. The vertical lift bridge is operated by the hydraulic lift. The vertical lift bridge Arthur Kill at Elizabeth, N.J. was built in 1959 with 170.078 m span.

1.3.5

Transporting Bridge

The transporting bridge is shown in Fig. 1.11. The towers are constructed at both the ends. These towers support a overhead bridge. A moving cradle is suspended

Fig. 1.11 Transporting bridge

from the overhead bridge. The persons or the goods are transported from one end to the other end by means of this moving cradle.

1.3.6

Floating Swing Bridge

The floating swing bridge is supported on the caisson or on the pontoons. The bridge is closed or opened by means of chains and windlasses. The recess is provided at the ends of a waterway. The bridge lies in the recess while it is in the open position. The pontoons are made of the sheet iron. The pontoons are designed to act as tubular girders, when the bridge is in the closed position.

1.4

COMPARISON OF FIXED AND MOVABLE BRIDGES

The movable bridges are constructed over the navigable streams when the clearance under the bridge is not sufficient for the passage of the naval traffic. The movable bridges have small initial cost. The movable bridges occupy less surrounding land area. The movable bridges are suitable for the narrow spans. The movable bridges require additional cost for persons, machines, and power for opening and closing of bridge. The fixed bridges are provided where the clearance available below the bridge is sufficient for the passage of the waterway traffic. The fixed bridges have high initial cost. The fixed bridges do not require additional cost for operation. The

14

DESIGN OF STEEL STRUCTURES–VOL. II

fixed bridges occupy more surrounding land area. The fixed bridges are suitable both for narrow and wide spans.

1.5

SELECTION OF TYPE OF BRIDGES

The various types of steel bridges have been discussed in Sec. 1.2. The selection of a particular type of a bridge, which suits for a particular case, depends upon the following factors : 1. Foundation condition 2. Clearance requirement 3. Length of the bridge 4. Width of the bridge 5. Live loads on the bridge 6. Initial cost 7. Operation and maintenance cost 8. Appearance

1.5.1

Foundation Condition

The soil under abutments and piers for the bridges may be poor. There may be considerable possibility of settlement to take place in the ground. The nature of the settlement may not be uniform. The type of bridge selected should be such that, the settlement of foundation does not affect the strength or permanency of the bridge structure. The possibility of settlement of the foundation rules out the choice of indeterminate structures e.g., fixed arch bridges, rigid frames, continuous span bridges etc. In such cases, the flexible type of bridges having maximum number of hinged joints e.g., hinged arches, simply supported span bridges, cantilever bridges etc., are selected.

1.5.2

Clearance Requirement

The difference between level of the soffit of a bridge and the height of flood water decides the clearance underneath the bridge structure. The clearance required underneath the bridge structure is also decided from the use of river or stream, whether it is used for the navigation purpose or not. In case the clearance available is large, and, there are number of spans, then, the deck type bridges may be selected. There is saving of masonry in the deck type bridges. In case, the clearance underneath the bridge structure is not sufficient for navigational purpose, and the waterway is to be used for navigation, then, the movable bridges are selected.

1.5.3

Length of the Bridge

The total length of a bridge depends upon the width of river to be spanned. In addition to the various factors, the site of bridge is selected, as far as possible, where the width of river is narrow. Besides the site of bridge, the total length of bridge also depends upon type of approaches. The comparison of costs of suspended approaches and the solid approaches decides the type of approaches for the bridge. The solid approaches are made by providing approach walls and filling material. In case the solid approaches are economical and suitable, these are provided for sufficient long distance. This reduces the total length of the

GENERAL

15

bridge. The number of spans depends on the total length of the bridge, area of waterway and the type of river bed. The areas of waterway existing before the construction of bridge, and after the construction of the bridge are kept equal. The regular profile with foundation bed at constant depth allows equal spans for the bridge. The short crossings with the highly irregular profile would compel to have unequal spans. The individual span decides the type of the bridge. The suitability of various types of bridges for different spans have been discussed in Sec. 1.2.

1.5.4

Width of the Bridge

The width of a bridge has slight influence on the type of bridge. When the width of bridge is small then, the parapet is provided by the bridge girders. This is economical construction. In such cases, it is likely that vehicles may collide against the bridge girders and damage them.

1.5.5

Live Load on the Bridge

When the bridges are designed for the heavy loads, then, the short spans are adopted for the bridge. When the bridges are constructed for light loads, then, the long spans are adopted. Depending upon span, a suitable type of bridge is selected.

1.5.6

Initial Cost

The initial cost of a bridge is an important factor to decide the type of bridge and it is given first consideration. In general, the economy of the bridge depends upon the number of spans and the type of construction. The number of spans in case of multiple span bridges is kept such that the cost of substructure (foundation) is equal to the cost of superstructure (the load carrying members). When the costs of substructure and the superstructure of a bridge are equal, then, the total cost of bridge would be minimum.

1.5.7

Operation and Maintenance Cost

The operation and maintenance cost is more for movable bridges as compared to the fixed bridges. For permanent type of various bridges the difference in maintenance cost and the actual cost of operation both are small.

1.5.8

Appearance

Aesthetically, the type of bridge selected should be such that the appearance of the bridge enhances the beauty of its surroundings. The purpose of bridge should be given greater importance in comparison to the beauty.

1.6

ERECTION METHODS FOR BRIDGES

An erection schedule is prepared carefully for the erection of a bridge by any method. This schedule is outlined actually before the final design is completed.

16

DESIGN OF STEEL STRUCTURES–VOL. II

The stresses developed in some members may be more than those due to the working loads. In some members, the character of stresses may be changed from tension to compression or vice versa. The large stresses are produced in some members as heavy cranes move on the trusses or during the forcing of parts to make them fit. The erection schedule shows the order of erecting various members. The erection methods for bridges depend upon the type of bridge (e.g., simply supported span bridges, cantilever bridges, suspension bridges, span of the bridges, the weight of the bridge girders, nature of the crossing (e.g., roads, railways, deep gorges, shallow bed, etc.), the maintenance cost of traffic during erection, time allowed for the erection, and risk to be taken. The type of bridge is an essential factor. Generally, for the erection of truss girders over a bridge, a derrick car is used. But in case a new bridge is to be constructed around an old bridge, and it is essential to maintain the traffic, then an overhead traveller is used. If there is limitation of time for erection, then, it is necessary to start erection simultaneously from both the ends of the bridge and to use more equipment. The risk taken is also an important factor. If there is a danger of flood at the site of erection, then, the method of cantilever erection is used instead of the falsework. The following are various methods of erection of bridges: 1. By construction of staging (falsework) 2. By rolling or pushing from the bank 3. By cantilevering 4. By using a cable way 5. By floating spans into position.

1.6.1

By Construction of Staging (Falsework)

The method of erection of bridge by constructing staging (falsework) is suitable for shallow bed and space underneath the span may be used. This method is not suitable for places where there is busy highway or railway lines or deep gorges. This method consists in erecting the staging (scaffolding) and the bridge may be built in position. The staging may be made of umber and/or steel. The strength of staging (scaffolding) is kept sufficient to support the weight of the bridge girder. The adequate cross-bracings are provided in the staging. The crossbracings provide lateral support to the staging. The staging is erected upto the level at which the bridge girder is to be built. The camber blocks are fixed on the top of staging to provide proper camber to the bottom members. The properly shaped members are first brought to the centre of the span. The members of the bridge girders are lifted by the crane. The assembly of various members is done over the staging and it proceeds from the centre towards the ends. The bottom chord members are placed first, and then, the floor system is fitted to the chords. It is followed by web members, top chord members, and bracing. The erection of the bridge by construction of staging (falsework) is shown in Fig. 1.12. If the traffic is not maintained during the erection, the falsework (staging) is

17

GENERAL

/ p ie r) Bed

L evel

(A bu tm en t

r) (A bu tm en t / pie

comparatively light. The falsework carries the weight of new span and erecting equipment. If the traffic is to be maintained during the erection, then, the

Fig. 1.12

falsework is comparatively heavy. The staging is subjected to heavy stresses due to moving loads. The moving loads should proceed with restricted and reduced speed. Instead of building falsework for each span, it is sometimes economical to use a light erection truss. The erection truss may be floated into position for each span.

1.6.2

By Rolling or Pushing from the Bank

The bridge girder is fabricated on the bank, and brought to the site. The bridge girder is placed over the rollers on the launching abutments. A launching nose is connected to the bridge girder at its forward end as shown in Fig. 1.13. A launching nose is a small truss. The length of launching nose is kept about onethird of the main span of bridge. A derrick pole is erected on the far abutment.

Fig. 1.13

A steel wire cable is connected to the launching nose. The cable passes over the derrick pole. The cable is controlled by a handling tackle. A cable is also attached at the rear end of main girder. This cable is controlled by the preventing cable. The main girder is moved forward by taking on the handling tackle and releasing the preventing tackle, till, the launching nose becomes simply supported, then, the main girder can be pushed forward and placed in correct position. The launching nose is detached from the girder. When the centre of gravity of the main girder enters the span, then, a counter-weight is placed on the main girder. The counter-weight prevents the overturning of the main girder.

18

DESIGN OF STEEL STRUCTURES–VOL. II

The connection between main bridge girder and the launching nose is made strong enough to transmit the developed stress. The stress developed would be maximum, when, the launching nose is about to touch the far abutment. The stresses in main girder, launching nose, and the cable should be checked for all positions.

1.6.3

By Cantilevering

The method of erection of bridges by cantilevering as shown in Fig. 1.14 is suitable for deep gorges, and places where, the method of construction of staging (falsework) cannot be adopted, due to deep gorges or deep and fast flowing water. This method is most suitable for arch bridges, and erection of subsequent spans of the continuous span bridges, and the erection of cantilever portions of the cantilever bridge. In this method, the construction may be started either from one end or from both the ends. The construction of bridge proceeds member by member and panel by panel towards the centre with the use of small crane

Fig. 1.14

travellers. In this method, the reversal of stresses take place in the members. The bottom chord of a simply supported bridge is subjected to tension due to dead load, live load and impact load and the top chord is subjected to compression. When the trusses are erected by the cantilever method, the bottom chord carries compression instead of tension, and the top chord carries tension instead of compression. The members should be checked for these stresses. This method is independent of crossing conditions. The erection of continuous span bridges by this method is shown in Fig. 1.15. The end span of continuous bridge may be erected by the method of construction of staging. The remaining spans may be completed by this method. The selfweight of finished span prevents the overtopping of over changing portions in the adjacent span.

Fig. 1.15

This is to note that this method has disadvantage of being more dangerous and expensive.

GENERAL

1.6.4

19

By Using Cable Way

The method of erection of bridges by using cable way is quite economical. This method is especially suitable for erecting the bridges over deep gorges in the hilly areas and the places where enough space is available at either end. In this method, the towers are erected first on the opposite banks as shown in Fig. 1.16. The steel wire cables are stretched over the towers. A saddle is placed on the cable. The saddle can travel along the span. The ends of cable are well anchored at the ends. The parts of members of the bridge are brought in the

Fig. 1.16

saddle and these are connected. These parts of members form the bridge girder. These connected parts are suspended from the cable. These parts are connected to the main girder by means of nuts and bolts. The bridge girder is completed panel by panel. The bolts and nuts are removed after completion of the bridge girder and the riveting is done. The bridge girder is then placed in position.

1.6.5

By Floating Spans into Position

The method of erection by floating spans into position is of particular advantages, when the structural member can be shipped easily to the sites for assembling. The depth of water should be deep. The condition of water should be such that this method is feasible. This method of erection saves false work. In this method, the trusses are assembled on the flat-bottomed boats. The trusses are riveted completely. The trusses are floated to the site. The trusses are anchored with the piers. The trusses are raised to the position on the piers in case, the bridge is high above the water. In case, the bridges are low above the water level, then, the trusses are raised higher than the piers by using cribs and the guy ropes. The trusses are then lowered in position.

1.7

ECONOMICAL SPAN LENGTH

The centre to centre distance between the end support of a bridge is termed as total span. The centre to centre distance between any two adjacent supports is termed as individual span or span only. When the total span of the bridge is large, then, the total span is divided into number of individual spans. In such case, the bridges are known as the multiple span bridges. In case of long crossings with regular profile and with the foundation bed at constant depth as shown in

20

DESIGN OF STEEL STRUCTURES–VOL. II

Fig. 1.17, the individual spans are made equal in length. The length of individual span is decided keeping in view the overall cost of the bridge. The total cost of the bridge is cost of superstructure (trusses and bracing) and the cost of substructure (abutments and piers). In case, the length of individual span is kept long, the cost of substructure decreases, but the cost of superstructure increases. In case, the length of individual span is kept short, the cost of superstructure decreases, but the cost of substructure increases. The economical span length of individual span is the length, which makes the total cost of superstructure and substructure minimum.

Fig. 1.17

It is difficult to derive mathematically an expression for the economical length, since, the total cost of bridge depends upon many factors. In addition to the cost of superstructure and the substructure, the cost of bridge also depends upon the operation and the maintenance cost, renewals, local conditions, and the conditions of the site. Dr. Wadell gave the following analysis for the economical span length after making certain assumptions : Let C = Total cost of bridge per unit length of the span F = Cost of flooring per unit length of the span S = Cost of substructure (abutments and piers) per unit length of the span T = Cost of the superstructure (trusses and bracings) per unit length of the span l = Length of individual span Then, the total cost of bridge is given by C = (T + S + F) ...(1.1) It is assumed that the cost of superstructure (trusses and bracings) per unit length of the span, T is directly proportional to the span. The bending moment in a simply supported span under the uniformly distributed load varies as the square of the span. The forces in the chord members of a truss girder are determined by dividing the moment by the height of the truss. The height of truss girder is expressed in terms of the span. As such, the forces in chord members are directly proportional to the span. The forces in the verticals and the diagonals are found from the shear force in the corresponding panel of the truss. The shear force is directly proportional to the span. The members are designed for the forces in the members, which are directly proportional to the span. Thus, the above assumption made is justified. Therefore, T ∝ l, T = (k 1 . l) …(a) Further it is assumed that the cost of substructure (abutments and piers) per unit length of the span is approximately inversely proportional to the span. In case the spans are long, the number of piers is less, In case the spans are short, the number of piers is more. The number of piers is inversely proportional to

GENERAL

21

the length of the individual span. Thus, the assumption is also justified. Therefore,

1 ⎛k ⎞ S = ⎜ 2⎟ ...(b) l´ ⎝ l ⎠ The cost of flooring per unit length of the span, F is independent of the span. The cost of flooring per unit length depends upon the cost of the material per unit length. The cost of flooring is not the function of the span. Therefore, this assumption is also justified. Hence, the cost of bridge per unit length of the span is given by S∝

k ⎛ ⎞ C = ⎜ k1 ⋅ l + 2 + F ⎟ l ⎝ ⎠ Differentiating the expression (c) with respect to l, k2 ⎞ ⎛ = ⎜ k1 − 2 ⎟ l ⎠ ⎝ For minimum or maximum value of C dC dl dC dl

= 0

...(c)

...(d)

...(e)

⎛ k2 ⎞ ...(f) ⎜ ⎟ ⎝ l ⎠ It is seen from expression (f), that the second derivative of C works out to be positive. Therefore, the expression (f) gives a condition for the minimum value of C. From expressions (a), (b), and (f), we obtain T = S ...(1.2) Thus, for economical span, in the multi-span bridges, the division of total span should be such that the cost of superstructure per unit length is equal to the cost of substructure per unit length. It is to note that this holds true for the multi-span simply supported truss girder bridges only. For economical span, or for the minimum cost of the bridge, the rule mentioned in the above paragraph can also be stated as the cost of pier is equal to one half the cost of trusses and bracing of the spans it supports. Therefore,

(k 1 . l) =

Fig. 1.18

The cost of pier per unit length varies inversely as the length of span. But in case of short crossing with highly irregular profile, as shown in Fig. 1.18, the cost of pier also depends upon its positions on the crossings in addition to the span. The above rule would give only rough guidance in such cases.

22

1.8

DESIGN OF STEEL STRUCTURES–VOL. II

CLEARANCE

When the bridges are through type or half through type, then the bridges are designed for adequate clearances, allowances being made for the displacement 6.80 06 m R a il 1 .37 16 m 1 .37 16 m 0 .60 96 m

(a)

1 .37 16 m

1 .37 16 m

C /L of tra ck C /L of tra ck

3

1 .06 68 m

2

3 .96 24 m cen tre to cen tre o f track 4 .4 19 6 m

4 .11 48 m 6 .80 06 m

1 .06 68 m

2 .13 36 m m in 3.50 52 m

1

1 .06 68 m

1 .06 68 m

2 .13 36 m m in

3 .50 52 m

23

Fig. 1.19

(b ) C lea ran ce d ia gra m fo r b roa d ga ug e

1 .90 5 m 1 .91 4 m 1 .90 5 m

1 .90 5 m

Top o f ra ils

C e ntre lin e of tra ck

0 .30 5 m

0 .76 2 m

2 .26 6 m

1 .06 7 m

2 .36 2 m

9 .48 8 m

2 .36 2 m 4 .72 4 m 2 .36 2 m

2 .13 4 m 0 .99 0 m

2 .13 4 m

0 .91 4 m 0 .91 4 m

C e ntre lin e of tra ck

1 .90 5 m

2 .36 2 m

2 .13 4 m 2 .13 4 m

0 .91 4 m

0 .91 4 m

GENERAL

of verticals due to any curvature and super elevation. The horizontal clearance is the clear width and the vertical clearance is the clear height available for the passage of live (moving) load. The clearance diagrams showing minimum clearance necessary for metre gauge and broad gauge railways as published by Railway Board are shown in Fig. 1.19 (a) and (b) respectively. The clearance diagram showing minimum clearance necessary for highway bridge as per IRC section I are showin in Fig. 1.20 (a) and (b).

2 .60 0 m

0 .82 5 m

4 .40 0 m m in.

(a ) S ing le la ne b rid ge

H a lf se ctio n sho w in g m ain fixed structu re b etw e en e nd p osts of arch rib

M ax. d im e nsio ns o f m ovin g ve hicles 3 .30 0 m x 4.5 00 m

3 .30 0 m

3 .80 0 m

0 .22 5 m

S a m e slo pe

0 .07 5 m

0.6 7 5 m

H o rizon ta l cle ara nce

S a m e slo pe

4 .50 0 m

H o rizon ta l cle ara nce

3 .8 00 m in .

C ro w n of roa d H a lf se ctio n sho w in g m ain fixe d structu re b e tw ee n e nd po sts on a rch b rid ge (b ) M ultip le la ne b rid ge

H a lf se ctio n sho w in g m ain fixe d structu re in the inte rm ed ia te po rtion s of a b rid ge

Fig. 1.20

0 .07 5 m

1 .30 0 m

N o t le ss tha n 6.8 0 0 for tw o traffic la ne s for ea ch a dd ition al tra ffic la ne increa se ro a d w idth b y 3.00 m

0 .22 5 m

1 .50 0 m

24 DESIGN OF STEEL STRUCTURES–VOL. II

25

GENERAL

1.9

WIDTH OF ROADWAY AND FOOTWAY

For high level bridges constructed for the use of road traffic only, the width of roadway shall not be less than 3.80 m for a single lane bridge, and shall be increased by a minimum of 3 m for every additional lane of traffic on multiple lane bridges. The road bridges shall be either one lane, two lanes, or four lanes. Three lane bridges are not constructed. In the case of four lanes, or multiple of two lanes, a central verge of at least 1.20 m shall be provided. For the bridges, constructed for the use of combined road and tramway traffic, the roadway width given above shall be increased by 4 m for a single track tramway and by 7.60 m for a double track tramway. For a bridge on a horizontal curve, the roadway width shall be increased. When a footway is provided, its width shall not be less than 1.50 m.

1.10 DIMENSION OF ROLLING STOCK In the railway trains, the passenger coaches and the goods wagons are called as rolling stock. The rolling stocks include coaching stock, goods wagons and the goods stocks. The coaching stock means all coaching vehicles including dining 6 10 m m

Fo r g uttering corn ice s side la m ps d estin ation b oa rd s an d a la rm sig na l d isc

2 49 0 m m 1 83 0 m m 9 30 m m

R a il le ve l

Fig. 1.21 Dimensions of coaching and goods stock for metre gauge

30 50 m m

25 6 5 m m

32 00 m m

R o lling sto ck fo r m etre g aug e

7 10 m m 30 5 mm

3 43 0 m m

Fo r n um ber p la te w in d ow b ars a n d re servatio n card h olde rs

26

DESIGN OF STEEL STRUCTURES–VOL. II

cars, mail vans, motor and carriage trucks, horse boxes, passenger brake vans, passenger road vans, refrigerator cars and dynamometer cars irrespective of contents whether attached to the passenger or goods train. The goods wagons intended for coaching traffic are marked as prescribed and used for coaching traffic only. They are treated as coaching stock for interchanging purpose. The goods stock means and includes all goods wagons i.e., all rolling stock other than coaching stock irrespective of contents and whether attached to passenger or goods train. The dimensions of coaching and goods stock for metre gauge are shown in Fig. 1.21. The dimensions of coaching and goods stock for broad gauge are shown in Figure 1.22. 6 10 m m

Fo r g uttering , co rnice s side la m ps d estin atio n b oa rd s a nd a la rm sign al d isc

3 35 5 m m

R o lling sto ck fo r b roa d g au g e

2 89 5 m m

411 5 m m 3 50 5 m m

Fo r n um b er p la te , w ind ow b ars a n d re se rvation card h olde rs

2 44 0 m m 102 m m

8 40 m m 30 5 m m

3 05 0 m m

1 60 0 m m R a il le ve l

Fig. 1.22 Dimensions of coaching and goods stock for broad gauge

These dimensions of rolling stocks are useful while considering the effect of wind on the moving load, for the loaded spans of the bridges.

1.11 HISTORICAL DEVELOPMENT OF BRIDGES The bridge construction technique developed with the development of human civilization. The various bridge structures may be grouped into four basic groups

GENERAL

27

from the view point of structural system (namely, beam bridges, cantilever bridges, suspension bridges and arch bridges).

1.11.1 Beam Type (Timber) Bridges The firstly formed bridges were the gifts of nature. In wind storms, accidently the trees fell across the water streams and dry gaps. These were the beam type bridges. Due to weathering action, and erosion, the natural rocks fissured and stone units (big and small rolled down). Accidently, these units formed arch bridges over the gaps. The creepers hanging from trees to trees acted like suspension bridges, which allowed the monkeys to cross the rivers from one bank to the other bank learning from the nature, the persons purposely, fell the trees intentionally across the water streams and crossed the rivers and gaps.

1.11.2 Cantilever Type (Timber) Bridges Around 4000 B.C., the persons started to live social life. In order to have permanent bridges, they gave more thought. In Switzerland, the lake dwellers constructed the timber trestles. Later on it led to build timber bridges. In India, at that time, the modern suspension bridges were developed. Two cables were stretched parallel between two banks and a levelled platform was suspended for the pedestrians. For the cantilever bridges, India also became birthplace. Wooden planks anchored with heavy stones at the two banks were corbeled out successively towards midstream. A simple plank was placed to span the gap.

1.11.3 Arch Type (Stone and Brick Masonry) Bridges During this time, the true arch bridges were developed by Mesopotamians. The bricks or stones were used to make the arch rib at ends. A pedestrian stone slab bridge across the Meles river in Smyrna Asia minor is the oldest bridge still standing. It was used by the ancient Greek epic poet Homer, it is at least 2500 years old. Many magnificent stone arch bridges were built by the Romans between 200 B.C. and 260 A.D. Massive piers were used with these semi-circular arches, so that in case one span were damaged in war, the other spans remain saved. The Pont du Gard aqueduct built at Nimes, France in 14 A.D. (a semi-circular three tiers arch) is the fine example of Roman bridge. The bridge at Alcantara in Span built by Caius Julius Lacer for emperor Trajan in 98 A.D. is another example of a Roman bridge. This bridge is nobel in its proportions and majestic in its simplicity, though it is a plain unadorned structure. There is inscription in Latin over the central tower, ‘I have left a bridge that shall remain for eternity.’ In ancient Rome, the bridge construction was treated so important that the Roman emperors adopted the title ‘Pontifex Maximus’ meaning that ‘chief bridge builders’. Since 250 B.C., the stone arch bridges were built by the Chinese. The most long-lived vehicular bridge today is the chao-chow bridge built around 600 A.D.

28

DESIGN OF STEEL STRUCTURES–VOL. II

It is situated about 350 km south of Peking. It is a single span stone arch bridge. This bridge has 37.4 m span, 7.23 m rise and 9 m roadway. The voussoirs of this bridge were correctly dressed to fit it and it was not necessary to put mortar in the joints. By virtue of this, this bridge has longevity. In Europe, in the middle ages after the fall of Rome, the bridge building activity was adopted by the religious orders. In 1345, in Italy, the Ponte Veechio Bridge was built at Florence using segmental curves first time. The decorative and defensive towers, chapels, statues, shops and dwellings were put as loads over the medieval bridges. The advances continued theory, technical skill and mechanical appliances with the dawn of Renaissance. The bridges were considered as civic works of art. The bridge builders were treated as leaders in progress and creator of monuments. Predominantly, the segmented arch bridges were built in stone masonry. In 1591, in Italy, the Rialto bridge built in Venice, Italy is a typical bridge of that period. The age of reason began in the eighteenth century. Professor Rubert Gautier, a French engineer published first treatise in bridge engineering in 1714. For the scientific advancement of construction of bridge a society (the Corps des Ingenieurs de Ponts et Chaussees) was founded. In 1747, the first engineering school (the Ecole de Ponts et Chaussees) was founded at Paris with Jean Perronet. He was called ‘father of modern bridge building’. He was the Director of this school. The masonry arches were perfected by Perronet. He built slender piers. The Pont de la Concorda at Paris built in 1791 was considered Perronet’s best work.

1.11.4 Timber Bridges In late eighteenth century, the covered timber bridges of each form became popular though the timber bridges had been built since early days. In order to protect the timber from weather, these wooden bridges were covered. Colossus bridge over the Schuylkill river at Fairmount, Pennsylvania was a notable timber bridge of that time. It had an arch span of 104 m. It was built in 1812. In 1838, this bridge was destroyed due to fire.

1.11.5 Iron Bridge In 1779, the first iron bridge was built at Coalbrookdale over the Severn river in England. It was built by Abraham Darby and John Wilkinson. Five semi-circular arch ribs in iron joined together side by side to have a single arch span of 30 m were used to build this bridge. A design of suspension bridge with wrought iron chain cables and level floor was patented by James Finley in Pennsylvania in 1808. In U.S.A., forty bridges of this design were built upto 1826. A suspension bridge with wrought iron chains (Menai Strait bridge) was built by Thomas Telford in Wales in 1826. This bridge had a record-breaking span of 177 m. In 1823, the first iron railway bridge was built by George Stephenson on the Stockton Darlington railway, Britania tubular bridge built by Robert Stephenson

GENERAL

29

in 1850 across the Menai strait is the most famous of the early iron railway bridges. Twin wrought iron tubes were used in this bridge. It is a continuous bridge. It has four spans of 70 m, 140 m, 140 m and 70 m. During the period 1840–1890, the cast iron was replaced by the wrought iron. During this period many truss bridges (e.g., Howe, Pratt, Whipple, Bollman, Fink and Warren) were built on railways. A number of wrought iron bridges failed. For example, in 1877, Howe truss bridge at Ashtabula, Ohio, and in 1879, the firth of Tay bridge in Scotland failed. As a result of which serious loss of life took place. A new era (an era of specialisation, research, careful detailing and thorough inspection) of bridge construction came. A more durable and stronger material (steel) was developed and used in building the bridges.

1.11.6

Steel Bridges : (Arch and Truss Bridges)

In 1874, the Eads Bridge at St. Louis, Missouri was built using extensively steel first time. It is a steel arch bridge. This bridge has three spans of 153 m, 158 m and 153 m. The cantilever method of erection was used to build this bridge. In 1878, all steel bridge was constructed at Glasgow, South Dakota. During 1869– 1883 period, steel cables were developed and Brooklyn bridge was built. In order to span more longer spans (120 m to 160 m), more efficient subdivided panel bridges (Baltimore, Parker and K-type) were built. For still larger spans (160 m to 200 m) sub divided panel and curved chord bridges (Pettit or Pennsylvania) were built. J.J. Barry bridge across Delawara river is the world’s longest simple truss bridge. It has a suspended span.

1.11.6

Steel Bridges : (Cantilever Bridges)

In 1867, Heinrich Gerber built the world’s first modern cantilever bridge across the river Main at Hassfurt, Germany. It has 129 m main span. The Firth of Forth bridge in Scotland is the world’s most famous cantilever bridge. In 1917, the world’s longest span cantilever bridge was built in Quebec, over the St. Lawrence river. It has a 549 m long main span. In 1943, Howrah bridge over the river Hoogly at Calcutta at present Kolkata (India) was built. It has a 457 m main span. It is the sixth longest bridge in the world. The bridge builders got inspiration from the success of the Eads bridge. They built many fine arch bridges. The Hell Gate bridge at New York with a span of 297 m built in 1917 and the Sydney Harbour bridge at Sydney, Australia with a span of 503 m built in 1932 are the notable arch bridges. The New river George bridge in West Virginia built in 1976 using weathering steel with a span of 159 m is the world’s longest arch bridge. The deck type arch span of this bridge is aesthetically the most pleasing. In 1936, the Henry Hudson bridge was built with a span of 244 m. In 1941, the Rainbow bridge at Niagara Falls was built with a span of 290 m. These bridges are the outstanding examples of beautiful steel arch bridges.

30

1.11.7

DESIGN OF STEEL STRUCTURES–VOL. II

Suspension Bridges

For very long spans, say 200 m and above, the suspension bridges were built. In 1883, the Brooklyn bridge was built with a main span of 486 m. It was longest bridge in the world at that time. Later on, other suspension bridges were built making the record span. In 1937, the Golden Gate bridge was built at San Francisco with a record span of 1280 m. In 1940, the Tacomo Narrows bridge was built and opened for traffic at Puget sound, Washington. Its span was 853 m with a stiffening truss girder 24 m deep. Due to 68 kilometre per hour gale and aerodynamic instability, this bridge failed. It forced to consider the aerodynamic effects on the suspension bridges and to make the theoretical studies. In 1964, the Verrazano Narrows bridge at New York with the longest Span in the world today was built with a main span of 1298 m. Professor D.B. Steinman looks a practical possibility to build a bridge with a span as large as 3000 m in future. At present a longest span bridge, the Humber bridge in England is being built with a main span of 1410 m.

1.11.8

Cable Stayed Bridges

For long spans say about 200 m, recently (1960 to 1990) cable strayed bridges have been developed. The Maracaibo lake bridge built at Venezuela in 1963 is the example of cable stayed bridges. Few more bridges of this type have been built in Europe.

1.11.9

Reinforced Concrete Bridges

In 1871, the first reinforced concrete bridge was built with 15 span over the Waveney at Homers field, England. In 1889, a concrete arch bridge of span 6 m was built at Golden Gate Park in San Francisco. In 1893, Professor Hennebique built a girder bridge as an approach to a will at Don, France. For architectural need, the concrete may be given any shape. The reinforced concrete has increased efficiency. Therefore, the reinforced concrete has been used at widespread for bridge building.

1.11.10

Prestress Concrete Bridges

After development of prestressing of concrete structural members, this technique opened new horizon since 1930. The Marne bridge built by Freyssinet in France is the one of the early prestressed concrete bridge. In 1954, the Palar bridge near Chingleput is the first prestressed concrete bridge built in India. It has 23 spans of 27 m each. In this country, since 1954, many prestressed concrete have been built successfully. The Gladesville bridge in Sydney, Australia with a span of 305 m is the longest span concrete arch bridge. Table 1.1 shows the progress in bridge building noted as successive record span lengths. Table 1.2 shows world’s longest spans for various types of bridges.

31

GENERAL

Table 1.1 Progress in bridge building (noted as successive record span lengths) Year

Type and bridges

Places

Main span (m)

1. Stone arch bridges 219 B.C.

Martorell

Spain

37.881

14 A.D.

* Nera river

Lucca, Italy

43.282

1377

* Trezzo

Italy

76.505

2. Timber arch bridges 104

* Trajan’s

Danube river

51.816

1758

* Wettingen

Switzerland

118.872

3. Cantilever bridges 1889

Forth

Scotland

518.160

1917

Quebec

Canada

548.640

1820

* Union (Tweed)

Berwick, England

136.855

1826

Menai Strait

Wales

176.784

1816

* Schuylkill falls

Philadelphia, Pa

124.358

1834

* Fribourg

Switzerland

265.176

1849

Wheeling

Ohio river

307.848

1851

* Lew is ton

Niagaria river

317.906

4. Chain bridges

5. Suspension bridges

1867

Cincinnati

Ohio river

322.174

1869

* Clifton

Niagara falls

386.486

1883

Brooklyn

New York City

486.461

1929

Ambassador

Detroit, Michigan

563.880

1931

George Washington

New York City

1937

Golden Gate

San Francisco, Calif.

1280.160

1964

Verrazano Narrows

New York City

1298. 448

1067.800

Note. These bridges are not standing Table 1.2 World’s longest spans (for various types of bridges) Year

Type

Bridge

Place

Main span (m)

1939

Concrete girder

Villeneuve

France

1919

Single-leaf bascule

* 16th street

Chicago, Illinois

1903

Masonry arch

Plauen

Germany

89.916

1951

Simple girder

Harlem river

New York

100.584

1914

Bascule

* Sautt STE. Marie Michigan

102.413

78.029 79.248

Contd.

32

DESIGN OF STEEL STRUCTURES–VOL. II

Contd. Table 1.2 Year

Type

Bridge

1758

Timber span

** Wettingen

Place

Main span (m)

Swtizerland

118.872

1850

Tubular girder

* Britannia

Menai strait

140.208

1927

Swing span

* Ft. Madison

Mississippi river

160.020

1937

Wichert truss

Homestread

Pittsburgh, Pa

162.763

1959

Vertical lift

Arthur Kill

Elizabeth, NJ.

170.078

1974

Prestressed concrete

Urato Bay

Japan

229.819

J.J. Barry

Delaware river,

250.546

Tiel

Waal river

girder 1973

Simple truss

1974

Cable stayed girder-concrete

Netherland

267.005

1974

Continuous girder

Niteroi

Rio de Janeiro, Brazil

299.923

1964

Concrete arch

Gladesville

Sydney Australia

304 .800

1926

Eyebar suspension

* Florianopolis

Brazil

339.547

1970

Cable stayed girder-steel Duisburg

Duisburg

350.520

1976

Steel arch

1917

Cantilever

1933

Transport bridge

1964

Cable suspension

New River Gorge West Virginia * Quebec

518.160

Canada

548.640

** Sky Ride

Chicago Illinos

563.880

Verrazano

New York

1298.448

Narrows Note. * Rail bridges. ** These bridges are not standing

CHAPTER

2

Loads and Stresses

2.1

INTRODUCTION

The design of a bridge structure has two aspects, one functional aspect, and the other structural aspect. The functional aspect takes into consideration the purpose for which the bridge is designed. In the structural aspect, it is ensured that the bridge is structurally safe, strong and durable. For the structural safety of the bridge, all loads should be taken into consideration, which the bridge should be required to bear. The following loads are taken into account for the design of bridge: 1. Dead load 2. Live load 3. Impact load 4. Wind load 5. Lateral load 6. Longitudinal force 7. Centrifugal force 8. Seismic force In addition to the above loads, the following effects are also considered for the design of bridge : 1. Erection effects 2. Temperature effects 3. Secondary stresses 4. Relief stresses The dead load, live load, impact load and the centrifugal force if any, are considered as normal loads. The wind load and lateral loads are considered as occasional loads. The seismic forces are considered as extraordinary loads.

2.2

DEAD LOAD

The dead load includes the self-weight of the bridge girder, and all other superimposed loads which are permanently attached to the structure. The dead load on a bridge may be divided into two parts : (i) weight of the floor and

34

DESIGN OF STEEL STRUCTURES–VOL. II

(ii) self-weight of the main structure. The weight of a bridge may be found by a preliminary design of the floor. The weight of floor may be used in calculating the loads on the main structure. The self-weight of the main structure itself should be estimated before the analysis and design of the structure. The importance of a correct estimate increases with the span length of the bridge. For a short span, the self-weight of the main structure may be a small portion of total live load and dead load. For a long span, the self-weight may be a large portion of the total live load and the dead load. The error in estimated dead load would affect the calculated stresses-appreciably. The self-weight of plate girders in case of the plate girder bridges has been discussed in Chapter 3. The selfweight of the truss girder in case of the truss girder bridges has been considered in Chapter 4. The self-weight of the bridge girder may be assumed depending on the experience and on the basis of similar existing bridges on the similar spans. The self-weight of the main structure may be found from previously established charts and formulae. These charts and formulae should be used with discretion because bridge specifications may differ with regard to live load and impact load, their method for load distribution and their permissible stresses. The superimposed dead load is determined from the various materials which are used, and their actual unit weights. If the actual unit weights of the materials used are not available, then the unit weights of the materials used are adopted from the Indian Standard Schedule of Unit Weights of Building Materials, IS : 1911–1961. On the completion of design, the total dead load is computed and compared with the dead load initially adopted for the design. In case, the actual dead load exceeds the dead load initially adopted by more than 2.5 per cent, the design is revised. If the effect due to the actual dead load and the adopted dead load varies to such an extent that the design is adversely affected, then also the design is revised.

2.3

LIVE LOAD

The live load in railway bridges and highway bridges consists of rolling or wheel loads. Theoretically, the bridges should be designed for the actual live loads, which the bridges are expected to carry. In practice, it is difficult to arrive at the actual wheel loads and therefore, the bridges are designed for the standard design loading. For the railway bridges, the Railway Board, Ministry of Railways, Government of India has specified the standard design loadings in Bridge Rules. For the highway bridges, the Indian Road Congress has specified the standard design loadings in IRC Section II. The standard design loading for the railway bridges, highway bridges, foot bridges, foot-path (foot way), and for the combined railway and highway bridges are given below.

2.3.1

Railway Bridges

The standard design loadings for the railway bridges has been specified in Bridge Rules for the broad gauge, metre gauge and narrow gauge tracks separately.

35

LOAD AND STRESSES

1 96 1 96

1 96

1 96

1 93

2 29

2 29

2 29

2 29

112

1 96

1 96

1 96

1 96

1 93

2 29

2 29

2 29

112

2 29

22 85 1 m m o ve r bu ffers

1 65 1

1 82 9

2 59 1

1 82 9

3 32 7

3 04 3

1 70 8

1 70 8

1 84 2

2 79 4

1 65 1 1 52 4

1 82 9

2 59 1

1 82 9

3 32 7

3 04 3

1 70 8

1 70 8

2 74 4 1 84 2

Train lo ad of 76 .7 k N /m

1 52 4

Axle s p ac in g in m m

Axle Loa d in kN

These standard design loading consists of number of wheel loads followed by a train of uniformly distributed load. The analysis of the bridge girder for the actual standard loadings is also difficult. In order to simplify the analysis, the actual standard loadings have been expressed in Bridge Rules as an equivalent uniformly distributed loads (EUDLL). The equivalent uniformly distributed loads are uniformly distributed loads which give the results equivalent to the wheel loads. The various standard design loading and their corresponding equivalent uniformly distributed loads for different tracks of the railways are as follows : For broad gauge 1676 mm (i) Standard main line (M.L.) loadings of 1926. The standard main line loading is adopted for all spans of the bridges on the main lines. This standard is considered to be adequate for all further requirements. The standard main line loading of 1926 for the broad gauge is shown in Fig. 2.1 (a). (ii) Standard branch line (B.L.) loadings 1926. The standard branch line loading is adopted for all spans of the bridges on branch lines. Such branch lines are never likely to be other than branch lines. In case a branch line is situated in heavy mineral area, then standard main line loading is used instead of the standard branch line loadings. The standard branch line loadings of 1926 for the broad gauge is shown in Fig. 2.1 (b). The equivalent uniformly distributed loads for the standard main line (M.L.) and the standard branch line (B.L.) for the purpose of determining bending moment and the shear force are given in Table 2.1 as per Bridge Rules.

23 85 1 m m o ve r bu ffers

14 3 14 3

14 9

14 9

16 3

17 3

17 3

17 3

17 3

91

14 9 14 9

14 9

14 9

16 3

17 3

17 3

17 3

17 3

91

22 96 2 m m o ve r bu ffers

22 96 2 m m o ve r bu ffers

(b) S ta nd ard bra nc h line lo ad in g

Fig. 2.1 Broad gauge standard loading of 1926

1 397

1 829

1 829

1 829

3 886

2 667

1 708

1 708

1 842

2 743

1 327 1 524

1 829

1 829 1 829

3 886

2 667

1 708

1 708

2 743 1 842

Train lo ad of 50 kN /m

1 524

A xle s pa cing in m m

A x le Lo ad in kN

(a) S ta nd ard m a in lin e loa ding

36

DESIGN OF STEEL STRUCTURES–VOL. II

Table 2.1 Equivalent uniformly distributed live loads (EUDLL) on each track and impact factors for broad gauge bridges L (metres)

Total load for B.M.

Total load for SF

Impact factor

(kN)

(kN)

⎛ 20 ⎞ ⎜ ⎟ ⎝ 14 + L ⎠

1.0 1.5 2.0

M.L. 458 458 458

B.L. 346 346 346

M.L. 458 458 524

B.L. 346 346 396

1.000 1.000 1.000

2.5 3.0 3.5

458 469 524

346 354 396

604 655 703

455 495 532

1.000 1.000 1.000

4.0 4.5 5.0

592 679 748

448 514 567

788 852 903

594 644 683

1.000 1.000 1.000

5.5 6.0 6.5

806 852 893

609 649 675

967 1049 1102

732 786 832

1.000 1.000 0.976

7.0 7.5 8.0

952 1007 1056

715 758 797

1152 1198 1239

871 906 935

0.952 0.931 0.909

8.5 9.0 9.5

1102 1140 1176

829 862 908

1286 1324 1367

972 1014 1050

0.889 0.870 0.851

10 11 12

1210 1336 1409

941 1025 1084

1406 1483 1557

1083 1140 1191

0.833 0.800 0.769

13 14 15

1472 1527 1606

1134 1172 1232

1637 1720 1806

1254 1316 1382

0.741 0.714 0.691

16 17 18

1688 1770 1859

1302 1371 1438

1881 1968 2050

1448 1516 1590

0.667 0.645 0.625

19 20 21

1939 2027 2111

1508 1580 1649

2141 2224 2305

1660 1726 1791

0.606 0.588 0.571

22 23 24

2187 2256 2329

1698 1758 1818

2387 2466 2548

1855 1916 1977

0.556 0.541 0.526 Contd.

37

LOAD AND STRESSES

Contd. Table 2.1 L (metres)

Total load for B.M.

Total load for SF

Impact factor

(kN)

(kN)

⎛ 20 ⎞ ⎜ ⎟ ⎝ 14 + L ⎠

25

M.L. 2410

B.L. 1880

M.L. 2627

B.L. 2037

0.513

26 27

2495 2560

1935 1988

2707 2789

2097 2158

0.500 0.488

28 29 30

2640 2710 2800

2052 2118 2165

2866 2944 3023

2230 2302 2375

0.476 0.465 0.455

32 34 36

2949 3095 3270

2300 2435 2577

3200 3375 3542

2518 2655 2780

0.435 0.417 0.400

38 40 42

3423 3590 3750

2705 2820 2950

3712 3877 4046

2912 3042 3186

0.385 0.370 0.357

44 46 48

3910 4080 4240

3085 3222 3345

4218 4384 4549

3316 3448 3576

0.345 0.333 0.323

50 55 60

4380 4775 5148

3470 3765 4052

4713 5122 5528

3702 4010 4310

0.313 0.790 0.270

65 70 75

5440 5918 6280

4335 4610 4865

5916 6322 6700

4600 4882 5154

0.253 0.238 0.225

80 85 90

6670 7035 7420

5130 5390 5680

7109 7504 7898

5438 5708 5978

0.213 0.202 0.192

95 100 105

7800 8200 8580

5913 6160 6425

8278 8686 9062

6246 6512 6772

0.183 0.175 0.168

110 115 120

8970 9350 9730

6685 6945 7195

9455 9848 10253

7042 7302 7564

0.161 0.155 0.149

125 130

10100 10485

7450 7700

10724 11133

7824 8804

0.144 0.139

Note. The intermediate values may be found by linear interpolation. The values of loads have been converted from metric tonnes to kilo-Newtons.

38

DESIGN OF STEEL STRUCTURES–VOL. II

For metre gauge 1000 mm (i) Standard main line (M.L.) loading of 1929. The standard main line loading is adopted for all spans of the bridge on the main lines. This standard loading is considered to be adequate for all future requirements. The standard main line loading of 1929 for metre gauge is shown in Fig. 2.2 (a). (ii) Standard branch line (B.L.) loading of 1929. The standard branch line loading is adopted for all spans of bridges on the branch lines. Such branch lines are never likely to be other than the branch lines. In case the branch line is situated in the heavy mineral area, then the standard main line loading is used instead of the standard branch line loading. The standard branch line loading of 1929 is shown in Fig. 2.2 (b). (iii) Standard C loading of 1929. The standard C loading of 1926 is shown in Fig. 2.2 (c). A xle S p acin g in m m 140 0

81 60

259 1 152 4 106 7

121 9 121 9 198 1

81 81 61 81 81 81 81 60

259 1 152 4 106 7 152 4

93 98 98

1 37 2 1 37 2 1 37 2 1 47 3 1 77 8 2 19 7

98 98 66 1 07 1 07

1 39 7 1 34 6 1 34 6

1 07 1 07

1 82 9

93

2 80 6 98 98

1 37 2 1 37 2 1 37 2 1 47 3

98 98

Fig. 2.2 Metre guage standard loading of 1929

1 39 7 1 34 6 1 34 6 2 13 3

A xle L oa d in kN 81 1 32 1 32 1 32 1 32 11 5

2 68 0 1 21

1 37 2 2 28 6 1 37 2 1 47 3 1 77 8 2 19 7

1 21 1 21 1 21 81 1 32 1 32

1 39 7 1 34 6

1 32 1 32

1 34 6 2 13 3

11 5 2 68 0 1 21 1 21

1 37 2 2 28 6 1 37 2 1 47 3

1 21 1 21 Train lo ad fo 3 8 .7 kn /m

Tra in lo ad fo 38 .7 kn/m

Train lo ad fo 3 8 .7 kn /m

140 0

81 81 81 81

1 07 1 07 1 07

1 82 9 2 80 6

1 77 8 2 19 7

66 1 07

1 93 80 m m o ve r b uffe rs

1 72 7 8 o ver bu ffers

127 0

81

18 28 8 m m ove r B u ffers

152 4 140 0 140 0 208 3

81

1 77 8 2 19 7 1 33 7 1 34 6 1 34 6

1 93 80 m m O ver bu ffers

198 1

81 81

A xle S p acin g in m m

A xle L oa d in kN

(a ) S tan da rd m a in lin e loa ding

121 9 121 9

81

(b ) S tan da rd bra n ch line loa ding

(c) S ta n da rd C Lo a din g

1 72 78 O ver B uffe rs

127 0

61 18 2 88 m m O ver B uffe rs

208 3

A xle S p acin g in m m

A xle L oa d in kN

39

LOAD AND STRESSES

The equivalent uniformaly distributed loads for the standard main line, the standard branch line and the standard C loadings are given in Table 2.2 as per Bridge Rules. Table 2.2 Equivalent uniformly distributed live loads (EUDLL) on each track and impact factors for broad gauge bridges L (metres)

Total load for B.M.

Total load for SF

Impact factor

(kN)

(kN)

⎛ 20 ⎞ ⎜ ⎟ ⎝ 14 + L ⎠

1.0 1.5 2.0

M.L. 264 264 264

B.L. 214 214 214

C 162 162 174

M.L. 264 291 350

B.L. 214 236 284

C 162 192 238

1.000 1.000 1.000

2.5 3.0 3.5

282 320 386

229 257 313

201 223 260

386 437 488

313 354 394

255 289 317

1.000 1.000 1.000

4.0 4.5 5.0

437 477 508

354 386 411

289 310 338

525 579 627

426 470 508

350 383 409

1.000 1.000 1.000

5.5 6.0 6.5

550 593 623

446 480 508

365 389 409

666 699 731

540 566 601

431 454 482

1.000 1.000 0.976

7.0 7.5 8.0

657 690 728

542 576 605

437 459 482

770 806 835

633 660 684

503 523 541

0.952 0.931 0.909

8.5 9.0 9.5

761 789 816

630 653 674

499 516 537

862 888 922

706 730 756

560 582 602

0.889 0.870 0.851

10 11 12

844 902 972

697 789 800

556 599 649

952 1020 1082

780 838 896

624 671 722

0.833 0.800 0.769

13 14 15

1035 1098 1160

864 921 980

698 748 796

1146 1206 1277

955 1018 1073

773 821 867

0.741 0.714 0.691

16 17 18

1226 1286 1342

1035 1085 1133

837 879 919

1338 1397 1454

1125 1176 1226

911 955 999

0.667 0.645 0.625

19 20 21 22

1395 1449 1495 1548

1182 1226 1268 1314

959 998 1041 1084

1509 1562 1614 1670

1275 1323 1368 1415

1041 1084 1126 1171

0.605 0.588 0.571 0.556 Contd.

40

DESIGN OF STEEL STRUCTURES–VOL. II

Table 2.2 Contd. L (metres)

23 24 25 26 27 28 29 30 32 34 36 38 40 42 44 46 48 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130

Total load for B.M.

Total load for SF

(kN)

(kN)

M.L. 1606 1658 1710 1778 1848 1906 1967 2033 2145 2265 2390 2515 2640 2745 2855 2960 3060 3160 3400 3630 3850 4050 4260 4480 4685 4881 5080 5280 5480 5675 5870 6070 6270 6460

B.L. 1364 1408 1455 1512 1568 1620 1669 1719 1817 1929 2040 2144 2242 2336 2430 2520 2612 2703 2921 3129 3332 3541 3737 3930 4139 4336 4528 4729 4919 5119 5305 5491 5681 5893

C 1127 1168 1213 1255 1301 1344 1384 1426 1520 1613 1699 1784 .1867 1950 2034 2116 2197 2276 2466 2667 2867 3069 3269 3644 3658 3855 4047 4232 4427 4614 4800 5005 5180 5389

M.L. 1734 1802 1869 1934 1999 2060 2122 2186 2312 2442 2568 2687 2802 2914 3023 3130 3235 3334 3579 3818 4047 4274 4501 4716 4927 5142 5433 5552 5765 5974 6177 6381 6583 6788

B.L. 1472 1528 1586 1640 1690 1744 1798 1853 1968 2075 2181 2284 2384 2481 2577 2671 2764 2886 3080 3299 3514 3727 3936 4143 4350 4550 4758 4961 5162 5363 5563 5763 5962 6161

Impact factor ⎛ 20 ⎞ ⎜ ⎟ ⎝ 14 + L ⎠

C 1217 1265 1309 1352 1396 1438 1498 1533 1631 1721 1810 1898 1985 2070 2156 2240 2324 2407 2613 2816 3020 3220 3422 3619 3819 4018 4214 4414 4622 4806 5002 5200 5396 5590

0.541 0.526 0.513 0.500 0.488 0.476 0.465 0.455 0.435 0.417 0 400 0.385 0.370 0.357 0.345 0.333 0.323 0.313 0.290 0.270 0.253 0.238 0.255 0.213 0.202 0.122 0.183 0.175 0.168 0.161 0.155 0.149 0.144 0 .139

Note. The intermediate values may be found by linear interpolation. The values of loads have been converted from metric tonnes to kilo Newtons.

89

89

89 97

16 76 97

25 9 1 3 04 8 91 4

97

89 97 97 91 4

97 84

97 18 29 91

91 91 4

3 0 48

91

16 76

91

91 4

91 4

91 4

O ver bu ffers 9 48 1

O ver bu ffers 9 44 9

Fig. 2.3 (a)

D ie sel electric 2 '–6 " G a ug e H class loa ding

O ver bu ffers 9 44 9

91

O ver bu ffers 8 56 6 O ver bu ffers 8 56 6 S tee l (Z F/1 ) d ie se l m e cha nica l 91

A xle lo ad (kN )

A xle sp acin g (m m )

A xle lo ad (kN )

89 17 27 91

91 4

A xle sp acin g (m m )

16 76

C – C o f C o –C o typ e 97 91

91 4

A xle lo ad (kN )

25 91 97 97 91 4 3 0 48

A xle sp acin g (m m )

A xle lo ad (kN )

A xle sp acin g (m m )

89 97

O ver bu ffers 9 48 1

91

89

97

16 76 3 04 8

O ver bu ffers 8 87 1 O ver bu ffers 8 87 1 B – B o f B o-B o typ e

91 4

76

89 97 97 91 4

97

17 27

91

18 29

97 76

97 97 84 91 91 91 4

Tra in lo a d 2 8.3 kN /m

Tra in lo a d 2 8.3 kN /m

Tra in lo a d 2 8.3 kN /m

Tra in lo a d 2 8.3 kN /m

LOAD AND STRESSES

41

For narrow gauge 762 mm The following are the standard loadings for narrow gauge (762 mm): 1. H (Heavy) class loading 2. A class loading 3. B class loading These standard loadings are shown in Fig. 2.3 (a), (b) and (c) respectively. The equivalent uniformly distributed loads are given in Table 2.3 as per Bridge Rules.

A x le s pac ing (m m )

A xle load (m m )

A x le sp acing (m m )

Fig. 2.3 (b) A x le loa d (m m )

A xle s pa cing (m m )

A x le load (m m )

46 914

1010

61

1753

61

A xle s pac in g (m m )

A x le load (m m )

51

965 965 965

61 61 61 61

208 3 61 127 0

61

61 1 676 61 2438 61 1 676 61

51 175 3

O ve r b uffe rs 88 39

O ve r b uffe rs 12 837

O ve r b uffe rs 99 76

61

965 965 965

61

61 O ve r b uffe rs 82 61

2'–6" G a uge B c las s loa ding

O ve r b uffe rs 88 39

S team e ngine (Ta nk Loc o)

61

61

B -B or B o-B o Type

O ve r b uffe rs 12 837

61 56

1 067

175 3

61 61

1010 2210 2337

61

10 10

61

10 10

61

22 10 56 23 37 10 67

O ve r b uffe rs 99 76

61

61 61

61

46

17 53

61 61

61 61

10 67

61

O ve r b uffe rs 82 61

914 243 8

914 914

1 067

61

61

9 14 9 14

61

243 8 9 14 9 14

61 61 61

208 3 61

1 676 61 243 8 61 1 676 61

127 0 61

61

Tra in load 28.3 kN /m

Tra in load 28.3 kN /m

Tra in load 28.3 kN /m

Tra in load 28.3 kN /m

42

DESIGN OF STEEL STRUCTURES–VOL. II

43

LOAD AND STRESSES

Axle spacing (m m ) 1415 1727

1295 1215 1295

81 81 81 81 71 71 71 71 71

1362 1451 1727

1295 1362

Axle load (kN )

81 81 81 81 71 71 71 71

76 76

76

2743

76 1676 76

76 914 76 914 76 914 76

76 2743 76

76 2743 76 914 76 914 76

914

76 1676

914

76 914 76 914 76 914 76

1676

76

76 76 76

71

76 1676 76

76 1676 76 2743 76 1676 76

Train load 28.3 kN /m

Train load 28.3 kN /m

Train load 28.3 kN /m

Train load 28.3 kN /m

Train load 28.3 kN /m

Fig. 2.3 (c)

2743

76

914

76

76

76

914

914

1676

Over buffers 8566

1295 1215

1676

Over buffers 9176

2375

Axle spacing (m m )

914

Over buffers 6872

2286

Axle load (kN )

76

76

61

Over buffers 9144

O ver buffers 15983

991 991 991

Axle spacing (m m )

O ver buffers 8566 B -B or - Bo-Bo type

2375

Axle load (kN )

O ver buffers 9176 C -C or - C o-C o type

2286

Axle spacing (m m )

O ver buffers 6872 Diesel m echanical or diesel electric

2'– 6" G auge a C lass Engine Loadidng

O ver buffers 15983

991 991

Axle load (kN )

61 O ver buffers 9144 D iesel m echanical or diesel electric (Articulated)

991

Axle Axle load spacing (kN ) (m m )

44

DESIGN OF STEEL STRUCTURES–VOL. II

Table 2.3 Equivalent uniformly distributed live load (EDULL) on each track and impact factor for 762 mm narrow gauge bridges L (metres)

Total load for B.M. (kN)

Total load for SF (kN)

Impact factor

⎛ 90 ⎞ ⎜ ⎟ ⎝ 90 + L ⎠ H class A class B class H class A class B class loading loading loading loading loading loading 1.0 1.5 2.0

193 193 245

163 163 210

120 122 160

242 270 315

193 220 255

152 170 200

0.989 0.984 0.978

2.5 3.0 3.5

310 345 380

265 285 305

190 220 240

370 405 430

300 330 370

230 255 280

0.973 0.968 0.963

4.0 4.5 5.0

400 425 450

350 380 405

255 270 295

460 495 515

400 430 460

310 330 350

0.957 0.952 0.947

5.5 6.0 6.5

475 495 520

432 450 465

340 380 405

535 555 580

485 510 530

370 390 410

0.942 0.938 0.933

7.0 7.5 8.0

555 580 600

480 495 515

425 445 455

610 640 670

550 570 590

430 450 470

0.928 0.925 0.918

8.5 9.0 9.5

615 630 640

535 560 590

465 475 480

695 725 755

615 630 650

485 505 525

0.914 0.910 0.905

10 11 12

655 695 765

620 645 670

485 500 520

785 850 915

670 720 765

545 585 625

0.900 0.891 0.882

13 14 15

880 950 990

690 725 800

550 600 700

970 1030 1090

810 855 900

665 710 750

0.874 0.865 0.857

16 17

1050 1100

860 900

770 800

1140 1200

960 1010

790 840

0.850 0.841

18 19

1150 1200

940 980

840 870

1270 1340

1060 1100

880 920

0.803 0.826

20 21

1250 1300

1020 1060

900 940

1410 1470

1150 1190

960 1000

0.818 0.811 Contd.

45

LOAD AND STRESSES

Table 2.3 Contd. L (metres)

Total load for B.M. (kN)

Total load for SF (kN)

Impact factor

⎛ 90 ⎞ ⎜ ⎟ ⎝ 90 + L ⎠

22 23

H class A class B class loading loading loading 1350 1100 970 1400 1150 1000

H class A class B class loading loading loading 1520 1230 1030 1560 1280 1060

0.804 0.796

24 25

1450 1500

1200 1250

1040 1070

1600 1640

1320 1370

1100 1130

0.790 0.783

26 27

1550 1560

1300 1350

1110 1140

1690 1730

1410 1450

1170 1200

0.776 0.770

28 29

1630 1670

1390 1440

1180 1210

1770 1820

1500 1540

1230 1260

0.763 0.756

30 32

1710 1790

1480 1570

1240 1310

1860 1940

1590 1670

1300 1360

0.750 0.738

34 36

1870 1940

1650 1730

1380 1450

2010 2090

1760 1840

1430 1490

0.726 0.714

38 40

2010 2080

1810 1890

1520 1580

2170 2240

1920 1990

1560 1620

0.703 0.692

42 44

2140 2210

1960 2020

1650 1720

2310 2380

2070 2140

1690 1750

0.662 0.672

46 48

2280 2340

2090 2150

1780 1850

2450 2510

2210 2280

1800 1860

0.662 0.652

50 55

2400 2560

2220 2390

1910 2060

2580 2740

2350 2510

1920 2070

0.643 0.621

60 65

2720 2830

2540 2690

2190 2320

2901 3080

2680 2840

2220 2360

0.600 0.581

70 75

2880 2910

2840 2990

2420 2490

3240 3400

3000 3150

2500 2630

0.563 0.545

Note. The intermediate values may be found by linear interpolation. (The values of loads have been converted from metric-tonnes to kilo Newtons.

2.3.2

Highway Bridges

The standard design loadings for the highway bridges have been specified in the Indian Road Congress, Section II, Loads and Stresses. The various standard loadings which are used for the highway bridges are as follows :

46

DESIGN OF STEEL STRUCTURES–VOL. II

IRC class AA loading This loading is adopted for the bridges on highways within certain Municipal limits, in certain existing or contemplated industrial areas and other specified areas. This loading is also adopted for the bridges on certain specified highways. The bridges designed for IRC class AA loading should be checked by IRC class A loading also, since, under certain conditions and in some cases, IRC class A loading may give heavier stresses than the IRC class AA loading. IRC class AA loading consisting of the tracked vehicle is shown in Fig. 2.4 (a) and IRC class AA loading consisting of the wheeled vehicle is shown in Fig. 2.4 (b). It is to note that the nose to tail distance between two successive IRC class C arria ge w ay w id th 0 .8 5 0 .8 5 m m

To ta l W eig ht 7 00 k N C 90 m m in

3 .8 m 7 .2 m

3 50 k N 2 .9 m

(a ) IR C class A A tra c k v e h ic le

(a) C a rria ge w a y w id th 2 .35 m m in 2 .5 m m in 1 .15 m m in 0 .85 m m in

3 7.5 kN

6 2.5 kN

3 00 m m 3 00 m m 3 00 m m

6 2.5 kN 7 00 m m

3 00 m m 3 00 m m 3 00 m m

D ire ctio n o f m otion

1 .00 m 1.2 m

6 2.5 kN

(b ) IR C C la ss A A w h e eled veh icle

Fig. 2.4 IRC class AA loading

3 50 k N

47

LOAD AND STRESSES

AA vehicles shall not be less than 90 m. The multilane bridges and culverts, one train of IRC class AA tracked or wheeled vehicle shall be considered for every two traffic lane width. No other live loads shall be considered on any part of the said two-lane width carriageway of the bridge, when the above said train of vehicle is crossing the bridge. The maximum load for the single axle wheeled vehicle is 200 kN. The maximum load for the bogie of two axles wheeled vehicle is 400 kN. The space between the two axles should not be more than 1.20 m centre to centre. IRC class AA tracked or wheeled vehicle may approach near to the kerb of road upto a minimum clear distance, C. The outer edge of the wheel or track shall be as per Table 2.4 as per IRC section II, given below. Table 2.4 Carriageway width

Minimum value of C

Single lane bridges : 3 8 m and above

0.3 m

Multilane bridges : Less than 5 5 m 5 5 m or above

0.6 m 1.2 m

IRC Class A Loading. This loading is to be normally adopted on all roads on which the permanent bridges and culverts are constructed. IRC class A loading is shown in Fig. 2.5 (a). IRC class A loading consists of one driving unit and two 1.2 m

27 27

114 114 68 68 (W heel loads in kN)

68

1.1 m

3m

3m

3m

4.3 m

1.2 m

3.2 m

20 m

18.4 m

20 m 68

27 27

(a) IRC Cla ss A train of vehicles B

B

B

B

W

W

W

a

1.8 m S ection on pp

Direction of m otion 1.1 m

3.2 m

1.2 m

B (b) Driving Vehicle

W

W

68

8.2 m

W

0.9 m

18.4 m

1.2 m

B

Fig. 2.5 IRC class A loading

a B

B

48

DESIGN OF STEEL STRUCTURES–VOL. II

trailers. The trailers attached to the driving unit are not detachable. The driving unit of IRC class A loading is shown in Fig. 2.5 (b) separately. The ground contact area of the wheels of the IRC class A loading shall be taken as per Table 2.5 as per IRC section II, as given below. Table 2.5 Ground contact area of wheels for IRC class A vehicles Ground contact area Axle load (kN)

B mm

114 68 27

W mm

250 200 150

500 330 200

The minimum clearance f, between outer edge of the wheel and the roadway face of the kerb, and, the minimum clearance g, between the outer edges of passing vehicles as shown in Fig. 2.5, as the multilane bridges shall be as per Table 2.5, as per IRC section II, as given below. C le a ran ce w idth

g

50 0 m m

50 0 m m

f

50 0 m m

50 0 m m

f

Fig. 2.6

Table 2.6 Minimum clearances, g and f for IRC class A vehicles Clear carriage way width 5.5 m to 7.5 m Above 7.5 m

g Uniformly increasing from 0.4 m to 1.2 m 1.2 m

f 150 mm for all carriageway widths

IRC Class B Loading. This loading is to be normally adopted for temporary bridges and for bridges in specified areas. IRC class B loading is shown in Fig. 2.7 (a). IRC class B loading also consists of one driving unit and two trailers. The trailers attached to the driving unit are not detachable. The driving unit of

49

LOAD AND STRESSES

IRC class B loading is shown in Fig. 2.7 (b) separately. The ground contact area of the wheels of the IRC class B loading shall be taken as per Table 2.7 as per IRC section II, as given below. 1.2 m

18 1 8 41 41 (Loads in kN )

16 16

41

1.1m

3m

3m

3m

4 .3 m

1.2 m

3.2 m

20 m

41

18.4 m

8.2 m

1.1 m

0.9 m

1 8.4 m

1.2 m

20 m 41

16 16

(a) IR C C lass B train of vehicles B

B

B

B

W

Direction of m otion 1.1 m

3.2 m

1.2 m

B

B

P

W

W

W

1.8 m S ection on pp

W

W

a

B

B

(b) Driving vehicle

Fig. 2.7 IRC class B loading

Table 2.7 Ground contact area of wheels for IRC class B vehicles Axle load (kN) 68 41 16

Ground contact area B (mm) 200 150 125

W (mm) 380 300 175

The minimum clearance f, between outer edge of the wheel and the roadway face of the kerb, and, the minimum clearance g, between the outer edges of passing vehicles as shown in Fig. 2.8, as the multilane bridges shall be as per Table 2.8, as per IRC section II, as given below.

50

DESIGN OF STEEL STRUCTURES–VOL. II

C le a ran ce w idth

g

300 m m

300 m m

f

300 m m

300 m m

f

Fig. 2.8

Table 2.8 Minimum clearances, g and f for IRC class B vehicles Clear carriage width 5.5 m to 7.5 m Above 7.5 m

g Uniformly increasing from 0.4 m to 1.2 m 1.2 m

f 150 mm for all carriageway widths

In case of IRC class A loading and IRC class B loading, the distance between successive trains shall not be less than 18.40 m. No other live load shall cover any part of the carriageway when a train of vehicles (or trains of vehicles in the multi-lane bridge) in crossing the bridge. Class 70 R loading. Class 70 R loading is used instead of IRC class AA loading for bridges over specified highways. The classification of vehicles for the highway bridges has been revised. According to the revised classification, various hypothetical vehicles have been considered. These vehicles have been

4 57 0 m m (G ro un d co ntact len gth) 7 92 0 m m (N o se to tail len gth ) C la ss 70 R track ed veh icle s

Fig. 2.9 Class 70 R tracked vehicles

given certain numbers along with letter ‘R’. The letter ‘R’ indicates the revised classification. The class 70 R loading is one of the various other hypothetical vehicles according to the revised classification. The class 70 R loading consisting of tracked vehicle is shown in Fig. 2.9. This vehicle is also having a weight of 700 kN. It is similar to that of class AA loading. It has a contact length of 4570 mm for the track. The length of vehicle measured from nose to tail is 7920 mm.

51

LOAD AND STRESSES

The width of track is 0.840 m. The width over track is 2.90 m similar to that of IRC class AA loading. Two successive tracked vehicles should maintain the minimum specified distance of 30 m. The class 70 R loading consisting of wheeled vehicle is shown in Fig. 2.10 (a). The total weight of wheeled vehicle is 1000 kN. This vehicle is 15220 mm long. There are seven axles in this wheeled vehicle. In addition to the effects of this

9 10 m m

8 0 kN

1 20 kN 1 20 kN 1 70 kN

3 96 0 m m

1 52 0 mm

M ax. sing le a xle loa d

2 13 0 mm

1 70 kN 1 70 kN 7 0 kN

1 37 0 mm

1 37 0 9 10 mm mm

M ax. B og ie a xle loa d

(a )

2 00 kN 1 22 0 mm

3 05 0 mm

4 00 0 kN

1 22 0 mm

1 83 0 mm

1 22 0 mm

1 83 0 mm

(b )

Fig. 2.10 Class 70 R wheeled vehicle

vehicle, the effects on the components due to a bogie loading of 400 kN are also to be considered. The single axle load and maximum bogie axle load are shown in Fig. 2.10 (b). The overall width of type and rim diameter of the type for both single axle and bogie axle loads are shown in Fig. 2.11. The tyre thread width may be taken as overall width minus 50 mm (since the tyre width is greater than 225 mm). The maximum tyre load on minimum tyre size is 800 kN. The actual maximum tyre load on 410 mm × 610 mm tyre size is 50 kN. The maximum tyre pressure is 0.5273 N/mm2. The contact areas of tyres on the deck (road surface) may be obtained from the corresponding tyre loads, maximum tyre pressure and width of the tyre threads. The minimum wheel spacings are also shown in Fig. 2.11. The spacing between successive vehicles would be 30 m. This spacing would be measured from the rear most point of the ground contact of the leading vehicle to the forward most point of the following vehicle in case of tracked vehicles; for wheeled vehicles, it is measured from the centre of the rear most axle of the leading vehicle to the centre of the first axle of the following vehicle. The minimum clearance between the road face of the kerb and the other edge of the wheel or track for any of the hypothetical vehicles (class 70 R) shall be same as for IRC class AA vehicles, when there is only one lane of traffic moving on a bridge. If a bridge is to be designed for the two lanes of any hypothetical vehicles, the clearance may be decided in each case depending upon the circumstances.

52

DESIGN OF STEEL STRUCTURES–VOL. II

2 79 0 m m n

8 60 m m (a )

S ing le axle B o gie A xle

8 60 m m 4 10 m m x 61 0 m m 4 10 m m x 61 0 m m 3 80 m m

2 79 0 m m (b ) S ing le axle B o gie a xle

4 10 m m × 6 10 m m 4 10 m m × 6 10 m m 2 79 0 m m

5 10 m m (c) S ing le axle b og ie axle

5 10 m m 2 30 m m × 5 10 m m 2 30 m m × 5 10 m m

Fig. 2.11 Minimum wheel spacing and tyre size of critical (heaviest) axles

IRC class AA, class B and class 70 R standard vehicles or train shall be assumed to travel parallel to the length of the bridge within the kerb to kerb width of the roadway. These vehicles shall occupy any position which would produce maximum stresses provided that the minimum clearances between a vehicle and the roadway face of kerb and between two passing or crossing vehicles are not encroached upon. For each standard vehicle or train, all the axles of a unit of vehicles shall be considered as acting simultaneously in a position causing the maximum stresses. The spaces on carriageway left uncovered by the standard train of vehicles shall not be assumed as subjected to any additional live load.

2.3.3

Foot Bridges and Foot-Paths (Attached to the Railway Bridges)

The live load due to pedestrian traffic shall be treated as uniformly distributed over the footway. For the design of foot-bridges or foot-paths attached to the railway bridges, the live load including impact shall be taken as 4.90 kN/m2 of

LOAD AND STRESSES

53

the foot-path area. The live load on foot-path for the purpose of designing the main girders shall be taken as follows : (a) For effective spans of 7.5 m or less, the live load is taken as 4.15 kN/m2. (b) For effective spans over 7.5 m, but not exceeding 30 m, an intensity of load reducing uniformly from 4.5 kN/m2 for a span of 7.5 m o 2.95 kN/m2 for a span of 30 m. (c) For effective spans over 30 m, the live load is adopted according to the formula P =

1 ⎛ 400 ⎞ ⎛ 17 − W ⎞ ⎜ 13.3 + ⎟⎜ ⎟ kN/m2 100 ⎝ L ⎠ ⎝ 1.4 ⎠

...(2.1)

P = Live load in kN/m2 L = Effective span of the bridge in metres W = Width of foot-way in metres. The kerbs 600 mm or more in width shall be designed for the loads mentioned above in addition to the lateral loading of 7.50 kN/m run of the kerb applied horizontally at the top of the kerb. If the kerb width is less than 600 mm, no live load shall be applied in addition to the lateral load specified above. These loads are not taken for the design of the supporting structure. The above values of live loads for the foot bridges and the foot-paths attached to the railway bridges are as per Bridges Rules.

where

2.3.4

Footway (Attached to the Highway Bridges)

For all parts of the bridge floors, accessible only to pedestrians, and animals and for all footways, the loading shall be taken as 4.00 kN/m2. Where crowd loads are likely to occur, such as on bridges located near towns, which are either centres of pilgrimage or where large congregetional fairs are held seasonally, the intensity of footway loading shall be increased from 4.00 kN/m2 to 5.00 kN/m2. In calculating stresses in the members of the structures with the cantilevered footway, the footway shall be considered as loaded on one side or on both the sides or unloaded, whichever condition gives the maximum stresses. In the bridges designed for any of the standard deign loadings described under the highway bridges, the main girders, trusses, arches or other member supporting the footway shall be designed for the following live loads per square metre of footway area, the loaded length of footway taken in each case being such as to produce the worst effect in the members under consideration: (a) For effective spans of 7.5 m or less, the live load is taken as 4.00 kN/m2 or 5.00 kN/m2 as mentioned above. (b) For effective spans over 7.5 m but not exceeding 30 m, the intensity of loads shall be determined according to the equation P = P´ −

1 ⎛ 40L − 300 ⎞ 2 ⎜ ⎟ kN/m 100 ⎝ 9 ⎠

...(2.2)

54

DESIGN OF STEEL STRUCTURES–VOL. II

(c) For effective spans over 30 m, the intensity of load shall be P =

1 ⎛ 4800 ⎞ ⎛ 16.5 − W ⎞ ⎜ 100 P´−260 + ⎟⎜ ⎟ 100 ⎝ 15 ⎠ L ⎠⎝

...(2.3)

P´ = 4.00 kN/m2 or 5.00 kN/m2 as mentioned above P = Live load in kN/m2 L = Effective span of the main girder truss, or arch in metres W = Width of the footway in metres. Each part of the footway shall be capable of carrying a wheel load of 40 kN which shall be deemed to include impact, distributed over a contact area 300 mm in diameter, the working stress shall be increased by 25 per cent to meet this provision. This provision need not be made where the vehicles cannot mount the footway as in the case of a footway separated from the roadway by means of an insurmountable obstacle, such as truss or a main girder. The kerbs 600 mm or more in width shall be designed for the loads mentioned above in addition to the lateral loading of 7.50 kN/m run of the kerb applied horizontally at the top of the kerb. If the kerb width is less than 600 mm, no live load shall be applied in addition to the lateral load specified above. These loads are not token for the design of supporting structure. It is to note that the footway kerb shall be considered mountable by the vehicles. The above values of the live loads for the footways are as per IRC section II. Further it is to note that the above live loads are not applicable to the foot bridges.

where,

2.3.5

Combined Highway and Railway Bridges

Where the railway and highway decks are not common, that is, they are at different levels orside by side, the main girders shall be designed for the worst combination of the live loads with full impact on the train-loads only. No impact shall be allowed for the highway loading. Where the railway and highway decks are common, the effect of the highway and the foot-path loads on the main girders shall be provided for by an allowance of 1.95 kN/m2 as a minimum over the whole area of the highways and the footpaths not occupied by the train load. The railway floor members shall be designed for the full effect of the maximum live load including impact which may occur on the roadway. The floor members which carry or may carry the roadway and railway loads simultaneously shall be designed by the maximum effect including impact, which may be imposed by either class of load separately or together. In cases where the roadway are railway are on the same alignment, the floor members and their connections shall be designed for the maximum effect of either class of load. The roadway floor system of combined bridges carrying two traffic lanes for the loads for IRC class AA loading shall be designed on the assumption that two IRC class AA vehicles may be placed opposite to each other of the centre lines of each traffic lane at any position in a panel. Under this circumstance of loading, the over stresses specified for the occasional loads shall apply.

55

LOAD AND STRESSES

2.3.6

Foot-Paths (Attached with the Combined Highway and Railway Bridges)

Where the foot-paths are provided on a combined highway and railway bridge, the load on foot-path for purpose of designing the main girder shall be taken as 1.95 kN/m2. In case of a foot-path on a combined highway and railway bridges, where the failure of a foot-path due to highway vehicle mounting the kerb is likely to endanger the railway traffic, the foot-path may be designed for a heavier standard of loading.

2.4

IMPACT LOAD

The dynamic effect of moving loads is known as impact. The impact is caused due to unbalanced weight of the driving wheels of the locomotive, rough and uneven track, flat or irregular wheels, eccentric wheels, rapid application of loads, deflection of floor beams and rail bearers, and the wheel to another. The unbalanced or eccentric weights are provided on the driving wheels of the locomotives. These weights balance the reciprocating and the rotating parts of the locomotive mechanism. These weights result in unbalanced forces and hammer blow effect. The upward and downward forces in the hammer blow effect are periodic and tend to set the bridge structure into vibrations. In case the period of vibration of the bridge structure and the period of rotation of the moving weights synchronise, then, the resonance takes place, and large deflection and large additional or impact loads are caused. The effect of impact depends upon the nature of moving load, speed of the moving load, type of the bridge structure, the material of the bridge structure and the loaded length of the bridge structure. The provisions for the effect of impact is made by impact allowance or impact load. The impact load is determined as a product of impact factor and live (moving) load. The impact factor is also termed as impact coefficient. The expression for impact factor is given in terms of the loaded length of structure. The various other parameters are given due consideration by the constants in the expression of the impact factor. The impact factor is also expressed as percent or fraction. The impact factor is specified by different authorities for different types of bridges, and for different types of moving loads separately. The impact factor for railway bridges, highway bridges and foot bridges are as follows:

2.4.1

Railway Bridges

For broad gauge and metre gauge railway bridges of steel and iron, the impact factor is taken as under: (a) For single track Impact factor,

⎛ 20 ⎞ > 1.00 i = ⎜ ⎟| ⎝ 14 + L ⎠

...(2.4)

56

DESIGN OF STEEL STRUCTURES–VOL. II

where,

L = Loaded length of span in metre for the position of the train giving the maximum stresses in the member under consideration. (b) For the main girder of double track with two girders Impact factor,

⎛ 20 ⎞ > 0.72 i = 0.72 × ⎜ ⎟| ⎝ 14 + L ⎠

...(2.5)

(c) For the intermediate main girder of multiple track spans Impact factor,

⎛ 20 ⎞ i = 0.60 × ⎜ > 0.60 ⎟| ⎝ 14 + L ⎠

...(2.6)

(d) For the outside main girders of the multiple track spans with intermediate girders, the impact allowance shall be that specified in (a) or (b) whichever applies. (e) For cross girders carrying two or more tracks Impact factor,

⎛ 20 ⎞ > 0.72 i = 0.72 × ⎜ ⎟| ⎝ 14 + L ⎠

...(2.7)

where,

L = Loaded length giving maximum load on the cross-girder i.e., two panel lengths in the case of intermediate cross girders. (f) For rails with ordinary fish plate joints and supported directly on transverse steel troughing or sleepers. For broad gauge, impact factor, ⎛ 7.32 ⎞ i = ⎜ ⎟ ⎝ B + 5.49 ⎠

...(2.8)

For metre gauge, impact factor, 9.5 ⎛ ⎞ i = ⎜ ⎟ ⎝ 91.5 + 4.27 ⎠

...(2.9)

where, B = Spacing of main girders in metres. This reduced impact factor should be used for calculating the stresses in the main girders upto 7.5 m effective span, stringer with span upto 7.5 m, and also for the stresses in the chords of triangulated girders supporting the troughs or sleepers. For narrow gauges (762 mm and 610 mm) railway bridges of steel and iron, the impact factor is taken as under: Impact factor, where,

⎛ 9.5 ⎞ i = ⎜ ⎟ ⎝ 91.5 + L ⎠

...(2.10)

L = Loaded length of span in metres for the position of the train giving the maximum stresses in the member under consideration.

57

LOAD AND STRESSES

2.4.2

Highway Bridges

For IRC class AA loading and class 70 R loading, the value of the impact percentage shall be taken as follows. (a) For spans less than 9 m (i) For tracked vehicle

(ii) For wheeled vehicle (b) For spans of 9 m or more (i) For tracked vehicles (ii) Wheeled vehicle

25 percent for spans upto 5 m, and linearly reducing to 10 percent for spans of 9 m 25 percent 10 percent for all spans 25 percent for spans upto 23 m and in accordance with the curve indicated in Fig. 2.12 for spans in excess of 23 m

For IRC class A loading and IRC class B loading, the impact percentage shall be determined from the curve indicated in Fig. 2.12. This impact percentage can also be determined from the following expression which is applicable for spans between 3 m and 45 m. Impact factor,

9 ⎛ ⎞ i = ⎜ ⎟ 13.5 + L ⎝ ⎠

...(2.11)

where, L = Loaded length in metres. In any bridge structure where there is filling of a not less than 0.6 m including the road crust, the impact percentage to be allowed shall be assumed to be onehalf of those specified above. For calculating the pressure on the bearings and on the top surface of the bed blocks, full values of the appropriate impact percentage shall be allowed, but for the design of piers, abutments, and structures, generally below the level of the top of bed block, the appropriate-impact percentages shall be multiplied by the factor given below : (a) For calculating the pressure at 0.5 the bottom surface of the bed block (b) For calculating the pressure on the top 3 m of the structure below the

0.5 decreasing uniformly to zero

bed block (c) For calculating the pressure on the

Zero

portion of the structure more than 3 m below the bed block

58

DESIGN OF STEEL STRUCTURES–VOL. II

53 30 45 40 35 30 25 20 15 10 5 0

0

3

6

9

12

15

18

21

24

27

30

33

36

39

42

45

S p an in m e tre s

Fig. 2.12 Impact percentage curve for highway bridges for IRC class A and IRC class B loading

2.4.3

Foot Bridges

No impact allowance is made for foot bridges and footways attached to the bridges.

2.5

WIND LOAD

The wind load or wind force is assumed as a horizontal force acting in such a direction that the resultant stresses in the members under consideration are the maximum. The wind pressures are expressed in terms of basic wind pressure, p. The basic wind pressure is static pressure in the windward direction. For the purpose of the design, these basic wind pressures in kN per square metre at various heights (in metres) are adopted from the maps and tables given in IS : 875–1984. The wind pressure is determined from the appropriate wind pressure adopted and the exposed area of bridge-girder. The exposed area and wind pressures on moving loads have been further discussed in Chapter 3 for the plate girder bridges and in Chapter 4 for the truss girder bridges. The railway and foot bridges shall not carry any live load, when the wind pressure at deck level exceeds the following limits. Broad gauge bridges l.50 kN/m2

LOAD AND STRESSES

59

Metre gauge bridges 1.00 kN/m2 Foot bridges 0.75 kN/m2 The highway bridges shall not be considered to be carrying any live load when the wind velocity at deck level exceeds 130 km per hour. A wind pressure of 2.40 kN/m2 is adopted for the unloaded span of the railway, highway and foot bridges.

2.6

LATERAL LOAD

The wind load described in Sec. 2.5 is also a lateral load. In addition to the wind in the railway bridges, nosing effect is present due to the lateral movement of the railway coaches (rolling stocks). This produces a lateral force, which is known as racking force. The lateral bracings of loaded deck of the railway bridges are designed in addition to the windload and the centrifugal loads, a racking force of 6.00 kN/m treated as a moving load. This lateral load need not be taken into account when calculating stresses in the chords of flanges of the main girders. In the case of effective spans upto 20 m, it is not necessary to calculate wind stresses but in the railway bridges, the lateral bracing are designed for a lateral load due to wind and racking forces of 9.00 kN/m treated as a moving load in addition to the centrifugal force, if any. The horizontal forces on parapet are also included in the lateral loads. The railings or parapets shall have a minimum height above the adjacent roadway or footway surface 100 m less one-half the horizontal width of the top rail or top of the parapet. The parapets are designed to resist lateral horizontal force and a vertical force each of 1.50 kN/m applied simultaneously at the top of the railing or parapet.

2.7

LONGITUDINAL FORCE

The longitudinal forces are caused due to one or more of the following : (a) Tractive effect caused through couple or driving wheels. (b) Braking effect from the application of the brakes to the wheels. (c) Frictional resistance offered to the movement of free bearing due to change of temperature and other causes. The longitudinal forces are not increased on account of impact. The longitudinal force acts along the bridge. The longitudinal forces for the railway bridges and the highway bridges are as follows.

2.7.1

Railway Bridges

These loads are considered as acting horizontal through the knuckle pins or through the girder seats, where the girders have sliding bearing. For the spans supported on sliding bearing, the horizontal load shall be considered as being divided equally between the two ends of the bridges. For spans which have roller bearing at one end, the whole of the horizontal load shall be considered to act through fixed end. The values of longitudinal loads due to either the tractive

60

DESIGN OF STEEL STRUCTURES–VOL. II

effort or the braking force for the loaded length shall be adopted from Table 2.9 for broad gauge and Table 2.10 for metre gauge. These values are as per Bridge Rules. Table 2.9 Longitudinal loads (without deduction for dispersion) for broad gauge 1676 mm L (metres)

Tractive efforts (kN)

Braking force (kN)

M.L.

B.L.

M.L.

B.L.

1.0 1.5 2.0

157 154 151

118 116 115

113 112 111

86 85 84

2.5 3.0 3.5

149 150 165

112 113 125

110 112 124

83 84 93

4.0 4.5 5.0

184 207 224

139 157 170

139 157 172

105 119 130

5.5 6.0 6.5

238 248 256

180 187 194

184 193 200

139 146 151

7.0 7.5 8.0

269 282 291

202 211 219

212 222 232

159 167 175

8.5 9.0 9.5

298 304 309

225 230 239

239 245 250

180 185 193

10 11 12

315 338 348

245 259 268

257 279 289

200 214 221

13 14 15

355 359 368

273 275 282

298 304 316

229 233 243

16 17 18

378 388 398

292 300 308

327 339 352

253 262 272

19 20 21

405 416 424

315 324 331

360 374 384

280 291 300 Contd.

61

LOAD AND STRESSES

Table 2.9 Contd. L (metres)

Tractive efforts (kN)

Braking force (kN)

22 23 24

M.L. 431 435 440

B.L. 335 339 344

M.L. 393 399 408

B.L. 304 311 318

25 26 27

448 455 458

350 352 356

416 427 432

325 331 336

28 29 30

465 469 476

361 366 368

442 447 457

343 349 353

32 34 36

476 476 476

368 368 368

472 482 500

368 382 394

38 40 42

476 476 476

368 368 368

514 531 544

405 417 428

44 46 48

476 476 476

368 368 368

555 571 585

438 451 462

52 55 60

476 476 476

368 368 368

591 621 644

468 489 507

65 70 75

476 476 476

368 368 368

670 692 712

525 539 550

80 85 90

476 476 476

368 368 368

734 753 772

564 577 591

95 100 105

476 476 476

368 368 368

788 810 832

598 610 633

110 115 120

476 476 476

368 368 368

852 870 885

635 648 655

125 130

476 476

368 368

899 912

663 570

Note. Intermediate values may be found by linear interpolation. The values of loads have been converted from metric-tonnes to kilo-Newton.

62

DESIGN OF STEEL STRUCTURES–VOL. II

Table 2.10 Longitudinal loads (without deduction for dispersion) for metre gauge 1m L (metres) M.L.

Tractive efforts (kN) B.L. C

M.L.

Braking force (kN) B.L. C

1.0 1.5 2.0

91 88 86

74 72 70

56 54 57

58 57 56

47 46 46

35 35 37

2.5 3.0 3.5

89 99 116

73 79 94

65 69 78

60 67 80

49 54 65

43 44 54

4.0 4.5 5.0

128 137 142

104 111 115

85 89 95

90 97 103

73 79 83

60 63 69

5.5 6.0 6.5

151 159 165

122 129 133

100 104 107

111 118 124

90 96 100

73 78 81

7.0 7.5 8.0

169 174 179

139 145 149

112 115 119

129 134 140

106 112 117

86 89 93

8.5 9.0 9.5

184 187 190

152 155 157

121 122 125

146 150 154

121 124 127

96 98 101

10 11 12

192 198 206

152 162 170

127 132 138

158 166 176

130 136 145

104 110 118

13 14 15

212 218 223

177 183 188

143 148 153

185 193 201

154 162 170

125 132 138

16 17 18

228 233 236

193 196 199

156 159 161

210 217 223

177 183 189

143 141 153

19 20 21

238 240 242

202 203 205

164 166 168

229 234 239

194 198 202

157 161 166

22 23 24

244 246 248

207 209 211

171 173 175

244 250 255

207 212 216

171 175 180 Contd.

63

LOAD AND STRESSES

Table 2.10 Contd. L (metres)

Tractive efforts (kN) B.L. C 211 175 211 175 211 175

M.L. 260 266 273

Braking force (kN) B.L. C 221 184 227 188 232 193

25 26 27

M.L. 248 248 248

28 29 30

248 248 248

211 211 211

175 175 175

279 284 290

237 241 245

197 200 204

32 34 36

248 248 248

211 211 211

175 175 175

300 309 320

254 263 273

212 220 227

38 40 42

248 248 248

21,1 211 211

175 175 175

329 339 345

281 287 294

234 239 245

44 46 48

248 248 248

211 211 211

175 175 175

352 358 363

299 305 310

251 256 261

50 55 60

248 248 248

211 211 211

175 175 175

368 379 388

315 322 335

265 275 285

65 70 75

248 248 248

211 211 211

175 175 175

395 400 406

342 350 356

294 303 312

80 85 90

248 248 248

211 211 211

175 175 175

412 417 421

362 368 374

319 326 332

95 100 105

248 248 248

211 211 211

175 175 175

424 428 432

378 383 388

338 343 349

110 115 120

248 248 248

211 211 211

175 175 175

435 438 442

392 396 400

354 359 364

125 130

248 248

211 211

175 175

445 448

403 408

368 374

Note. Intermediate values may be found by linear interpolation. The values of loads have been converted from metric-tonnes to kilo-Newton.

64

DESIGN OF STEEL STRUCTURES–VOL. II

The loaded length L shall be taken as equal to (a) the length of one span when considering the effect of the longitudinal loads (i) on the girder (ii) on the stability of abutments (iii) on the stability of pier under the condition of one span loaded or (iv) on the stability of piers carrying one fixed end and one roller bearing (b) the length of two spans when considering under the conditions of both spans loaded, the stability of piers carrying fixed or sliding bearings. In this case, the total longitudinal load shall be divided between the two spans in proportion to their lengths. For determining the value of tractive effort, the loaded length, L shall not by taken to exceed the following : (a) For broad gauge : 30 m for main line and branch line loadings. (b) For metre gauge : 24 m for all loadings. Where the structure carries more than one track longitudinal loads shall be considered to act simultaneously on all tracks.

2.7.2

Highway Bridges

In all highway bridges, provisions are made for longitudinal forces. The braking effect on a simply supported span or a continuous unit of spans or on any other type of bridge unit shall be assumed to have the following values. (a) In the case of a single lane bridge or two lane bridge. 20 percent of the first train load plus 10 percent of the loads of the succeeding trains or part thereof, the train loads in one lane only being considered for this purpose. Where the entire first train is not on the full spans, the braking force shall be taken as equal to 20 percent of the loads actually on the span. (b) In the case of bridges having more than two lanes. As in (a) above for the first two lanes plus 5 percent of the loads on the lines excess of two lanes. The forces due to braking effect shall be assumed to act along a line parallel to the highway and 1.20 m above it. While transferring the force to the bearings, the change in the vertical reaction at the bearing should be taken into account. The longitudinal force at any free bearing shall be limited to the sum of dead and live load reactions of the bearings multiplied by the co-efficient of friction. The coefficient of friction of the bearing shall be assumed to have the following values : For roller bearing ... 0.03 For sliding bearing of hard copper alloy ... 0.15 For sliding bearing of steel on cast iron or steel on steel ... 0.25 For sliding bearing of steel on ferro-asbestos ... 0.20 The longitudinal force at the fixed bearing shall be taken as the algebraic sum of the longitudinal force at the free bearing in the bridge unit under consideration and force due to the braking on the wheels as mentioned above.

LOAD AND STRESSES

65

The effects of braking force on the bridge structure without bearings such as arches, rigid frames etc., shall be calculated in accordance with the approved methods of analysis of the indeterminate structures. The effects of longitudinal forces and all other horizontal forces should be calculated upto a level where the resultant passive earth resistance of the soil below the deepest scour level (floor level in case of bridge having pucca floor) balances these forces.

2.8

CENTRIFUGAL FORCE

While the traffic lane or track over a bridge is situated on a curve, the allowance for the centrifugal force of the moving load is made in designing the members of the bridge. All the lanes and tracks are considered as occupied by the moving load. The values of centrifugal force for the railway bridges and highway bridges are determined from the following expressions. No addition for impact need be considered for centrifugal force.

2.8.1

Railway Bridge

The horizontal load due to centrifugal force for the railway bridge ⎛ WV 2 ⎞ C = ⎜ ⎟ ⎝ 127 R ⎠

...(2.12)

where,

C = Horizontal force in kN per metre run of the span W = Equivalent distributed live load in kN per metre run V = Maximum speed in km per hour R = Radius of the curve in metres The line of action of the centrifugal force is assumed at a height of 1.83 m above the rail level for broad gauge and 1.45 m above the rail level for metre gauge.

2.8.2

Highway Bridge

The centrifugal force for the highway bridge is determined from the following formula

where,

⎛ W V2 ⎞ C1 = ⎜ 1 1 ⎟ kN …(2.13) ⎝ 127 R1 ⎠ C1 = Centrifugal force in kN acting normally to the traffic (1) at the point of action of the wheel loads or (2) uniformly distributed over every length on which a uniformly distributed load acts. W1 = Live load (1) in kN in case of wheel loads, each wheel load being considered as acting over the ground contact length specified for the live load for the highway bridges in Sec. 2.3 and (2) in kN per linear metre run in case of uniformly distributed live load. V1 = Design speed of the vehicles using the bridge in km per hour.

66

DESIGN OF STEEL STRUCTURES–VOL. II

R1 = Radius of the curve in metres. The line of action of the centrifugal force is assumed at a height of 1.20 m above the level of the carriageway.

2.9

SEISMIC FORCE

If a bridge is situated in a region subjected to the earthquake, allowance is made in the design for the seismic force and the earthquake resistant and features are embodied in the structural details of the design. The seismic force is taken as a horizontal force, but shall not exceed 0.12 of the gravity. For the bridges situated in the epicentral tracts, where the large devastations have occurred in the past due to the earthquakes, the percentage is fixed depending upon local conditions regarding the intensity of the earthquakes generally experienced in those regions. The horizontal forces due to seismic force shall be taken to act through the centre of gravity of all the loads under consideration. The direction of this force should be such that the resultant stresses in the members under considerations are not maximum. IS: 1983–62, Indian Standards Recommendations for Earthquake Resultant Design of structure may be referred to for actual design. It should be noted that the seismic force and the wind load should not be considered to act simultaneously.

2.10

ERECTION EFFECTS

The erection effects include the weight of all permanents and temporary material of a bridge together with all other forces and effects which may operate on any part of the bridge during transportation and erection. The erection effects also take into account the placing or storage of construction material and the erection equipments. The proper provisions (e.g., temporary bracings) shall be made to take care of all the stresses caused during the erection. The erection stresses may be opposite in nature from those which the member will be subjected to during actual working.

2.11

TEMPERATURE EFFECTS

The variation in temperature results in expansion and contraction of the structural material. The range of the variation in the temperature varies from locality to locality, season to season and day to day. Where any portion of the structure is not free to expand or contract under the variation of temperature, the allowance is made for the stresses resulting from this condition. The coefficient of thermal expansion or contraction for each degree (Centigrade) in variation or temperature above or below the normal temperature is adopted as 0.0000117, for steel.

2.12

SECONDARY STRESSES

The secondary stresses are additional stresses brought into play in the structural members of the bridge. The secondary stresses fall into two groups.

LOAD AND STRESSES

67

(a) The stresses, which are the result of eccentricity of connections generally, and off panel point loading (e.g., loads rolling direct on chords, self-weight of members, and wind load on the members. It is to note that the secondary stresses due to unknown eccentricities arising out of inaccuracies in fabrication are already allowed for in the factor of safety. (b) The stresses which are the result of the elastic deformation of the bridge structure combined with the rigidity of the joint. All the bridges shall be designed, fabricated and erected in a manner as to minimize, as far as possible, the secondary stresses. These secondary stresses are properly accounted for in the design. The analysis of secondary stresses have been discussed in Chapter 17.

2.13

RELIEF STRESSES

In the determining the maximum stress in any member of a bridge, any relief afforded to the member by the adjoining parts may be taken into account. In determining the amount of relief, the secondary stress, if any in the member shall be determined and considered with the other co-existent stresses.

2.14

COMBINATION OF LOADS

The following combination of loads described above are considered which give the most severe conditions of loads and the stresses. All the members of the bridge are designed for the most severe combination of loads and stresses. (i) Dead load + Live load + Impact load + Centrifugal force, if any. (ii) (Dead load + Live load + Impact load + Centrifugal force, if any) + Wind load + Other lateral loads + Longitudinal forces + Temperature stresses. (iii) The stresses due to loads during erection or stresses due to lifting during maintenance. (iv) The stresses due to second combination of loads plus the seismic forces excluding the wind.

2.15

ALLOWABLE STRESSES

The following allowable stresses are used for the design of the steel bridges. The actual values may be noted from respective codes.

2.15.1

Axial Stresses in Tension

The allowable stresses in axial tension or direct stresses in tension on the net effective sectional area of the section are adopted as 140 N/mm2 as per IS : 1915–1961. The allowable stresses in axial tension on the effective sectional area as per code of practice published by Railway Board for mild steel conforming to IS : 226 with yield stress 236 N/mm2 is 141 N/mm2.

68

DESIGN OF STEEL STRUCTURES–VOL. II

The allowable stresses in axial tension, σat on the effective cross-sectional area as per IS : 800–1984 is adopted as 0.6 fy (where fy is the yield stress for the steel used).

2.15.2

Axial Stresses in Compression

The allowable axial or direct stresses in compression on the gross sectional area of axially loaded compression members are adopted as per Table 2.11. IS : 1915–1961 specifies the values of allowable stresses for the design of the steel bridges. These values converted from metric system of units to S.I. units have been described in the subsequent sub-articles. These values are yet to be revised by ISI. As such the values of permissible stresses recently revised in IS : 883–1994 are used in the design of bridges. The permissible stresses for axial compression as per IS : 1915–1961 for steel conforming to IS : 226 have been listed in column (i) and those as per code of practice published by Railway Board have been listed in column (ii). Table 2.11 Slenderness ratio 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

Permissible stress in N/mm2 (i) (ii) 140 139 137 134 130 125 119 111 101 91 81 72 64 56 50 44 40 36

140.0 136.0 130.0 118.0 101.0 80.5 63.0 49.4 39.0

Note. The intermediate values may be found by linear interpolation. The values have been converted from metric units to S.I. units.

The permissible stresses for axial compression as per IS : 883–1994, σac is 0.6 fcr (which fcr is the buckling stress and it is function of slenderness ratio of the member.

69

LOAD AND STRESSES

2.15.3

Bending Stress

The allowable stress in bending in tension and compression for laterally supported structural members are adopted as per Table 2.12. The permissible bending stresses as per IS : 883–1994 are as follows : 1. Bending in tension, σ bt 0.66 fy (i) for flats, rounds etc. 0.66 fy (ii) for rolled girders (single or multiple) 0.63 fy (iii) for plate girder 0.60 fy 2. Bending in compression, σ bc (i) Compression flange restrained As in step (i) above. against lateral movement. (ii) Compression flange restrained only As per expression at some points (based on later buckling). derived below. Table 2.12 Permissible bending stress as per IS : 1915–1961 for steel conforming to IS : 226–1958 Description

Permissible bending stress (N/mm2)

Parts in tension or compression on the effective section for the extreme fibre stress : (i) For plates, flats, tubes, rounds, squares and other similar section (ii) For plate girder with single or multiple webs and for rolled sections

158

d |> 85 t

(iii) For plate girders with single or multiple webs

150 d > 85 t

142

The expression for determining critical stress as given in Indian Railway Standard Code of Practice for the design of steel or wrought iron bridges is different from that given in IS : 800. The bending stress in compression for sectional shapes with Iyy smaller than Ixx, (where Iyy is moment of inertia of the whole section about the axis lying in the plane of bending yy-axis, and is moment of inertia of the whole section about the axis normal to the plane of bending xx-axis is reduced for lateral buckling in proportion to critical stress, Cs found as follows for sections of single web. (a) Where the flanges have equal moment of inertia about yy-axis, is subjected to couple M at the ends, the critical value of bending moment M is given by 1/2

Mcr

⎛ ⎞⎤ π⎡ π2 Ely ⋅ h2 ⎟ ⎥ ⎢ Ely ⋅ GK ⎜1 + 2 ⋅ = l⎣ l 4GK ⎝ ⎠⎦

...(i)

70

DESIGN OF STEEL STRUCTURES–VOL. II

⎛ π2 Ely h2 ⎞ By rearranging the terms ⎜ ⎟ , the expression (i) may be written as ⎝ l2 ⋅ 4GK ⎠ follows 1/2

Mcr

1 4GK l2 ⎞ ⎤ π2 Ely ⋅ h ⎡⎛ ⋅ ⋅ ⎢⎜⎜1 + 2 ⋅ ⎟⎥ = 2 Ely h2 ⎟⎠ ⎥⎦ 2 π l ⎣⎢⎝

...(ii)

Substituting G = 04E, K Ω 0.9bft f3 and ly = t fbf3/12 where t f and bf are the thickness and the width of the flange 1/2

Mcr

3 2 ⎡⎛ 1 4 × 0.4 E × 0.9bf tf l ⎞ ⎤ π Ely ⋅ h ⎢⎜1 + ⎟⎥ × 2 3 2 ⋅ = ⎢⎜ ⎟⎥ E t b h 2 π × × × 2 f f 2 l ⎢⎜ ⎟⎥ 12 ⎠⎦ ⎣⎝

Mcr

3 2 π2 Ely ⋅ h ⎡⎛ l2 ⎞ ⎤ 8.64 t f l ⎢ ⎜ ⎟⎥ 1 + × × = 2 2 2 2 ⎟ ⎢⎜⎝ 2l b h π f ⎠ ⎥⎦ ⎣

2

1/2

...(iii)

For an I-section and I-section of wide flanges, the approximate radii of gyration are given by ry = 0.22b (viz., b = 4.545 ry) and ry = 0.25b (viz., b = 4.000 ry) respectively. As an approximation bf= 4.2 ry, then 1/2

Mcr

2 π2 ⋅ Ely ⋅ h ⎡⎛ 8.64t f l2 ⎞ ⎤ ⎢ ⎜ ⎟⎥ 1 + ⋅ = 2 2 2 ⎟ ⎢⎜⎝ 2l2 r h 17.65 π × y ⎠ ⎥⎦ ⎣

...(iv)

The elastic critical stress for bending for beams with ly smaller than Ix is given by (3 Mcf = fcb . Zxx) Mcr

π2 ⋅ Ely ⋅ h ⎡ ⎛ l ⋅ tf ⎞ ⎢1 + ⎜ = ⎟ 2 ⋅ Z x l2 ⎢⎣ ⎜⎝ ry ⋅ h ⎟⎠

2 ⎤1 / 2

⎥ ⎥⎦

...(v)

The value of modulus of elasticity E is taken as 2 × 105 N/mm2 (MPa) and Iy = A.ry2 . Substituting these values in the expression (v) fcb

The value of

9.8596 × 105 A ⋅ h ⎡ ⎛ l ⋅ tf ⎞ ⎢1 + 1 ⎜ = ⎟ 2 ⎢⎣ 20 ⎜⎝ ry ⋅ h ⎟⎠ ⎛ l ⎞ ⎜ r ⎟ Z xx ⎝ y⎠

2 ⎤1 / 2

⎥ ⎥⎦

A ⋅h for I-section may be taken as 2.688 approximately. Z xx

...(vi)

71

LOAD AND STRESSES

Then 1/2

fcb

2 26.5 × 105 ⎡ 1 ⎛ l ⋅ tf ⎞ ⎤⎥ ⎢ = + 1 ⎜ ⎟ 2 ⎢⎣ 20 ⎜⎝ ry ⋅ h ⎟⎠ ⎥⎦ ⎛ l ⎞ ⎜r ⎟ ⎝ y⎠

N/mm2

...(2.14)

The value of critical stress, Cs is increased by 20 percent for rolled beams, channels and plate girders provided ⎛ tc ⎞ > 2 ⎜t ⎟ | ⎝ ω⎠ ⎛ d1 ⎞ > 85 ⎜t ⎟| ⎝ ω⎠ where l = Effective length of compression flange. ry = Radius of gyration about yy-axis of the gross-section of the whole girder at the point of maximum bending moment. h = Overall depth of the girder, at the point of maximum bending moment. tf = Effective thickness of the compression flange. = k1 × mean thickness of the horizontal portion of the compression flanges at the point of maximum bending moment. k 1 makes the allowance for reduction in thickness and breadth of flanges between points of effective lateral restraint and depends on y the ratio of the total area of both flanges at the point of least bending moment to the corresponding point of greater bending moment between such points of restraints. d1 = clear depth of web tw = thickness of web. The values of k1 have been given in Table 2.13. Table 2.13 ψ k1

1.0 1.0

0.9 1.0

0.8 1.0

0.7 0.9

0.6 0.8

0.5 0.7

0.4 0.6

0.3 0.5

0.2 0.4

0.1 0.3

0.0 0.2

Note. Flanges shall not be reduced in width to give a value of ψ lower than 0.25. Where the value of ψ calculated for compression flange alone is smaller than that when both the flanges are combined, this smaller value of ψ shall be used.

(b) Where the moment of inertia of the compression flange about yy-axis exteeds that of tension flange. A paper titled as ‘the basis for design of beams and plate girder in the revised BS : 153 (British Standards)’ structural paper 48, proceedings of the Institution of Civil Engineers, London, August 1956 was published by Kerensky, O.A., Flint, A.R.; and Brown, W.C. Professor, Kerensky gave the expression for critical bending moment for beams and plate girders as under:

72

DESIGN OF STEEL STRUCTURES–VOL. II

Mcr =

1/2 1/2 ⎤ 2 π 1/2 ⎡ ElyGK ) ⎢⎛⎜1 + π Ely h2 ⎞⎟ + (2λ − 1 ) πh ⎛⎜ EI y ⎞⎟ ⎥ ( l 2 ⋅ l ⎝ GK ⎠ ⎦⎥ l2 4GK ⎠ ⎣⎢⎝

...(i)

where, λ is the ratio of moment of inertia of compression flange about yy-axis. If to the moment of inertia of the whole section about yy-axis, Iy.(2λ – 1) may be replaced by k2. For the values of λ = 1 or zero k 2 is equal to + 1.0 to –1.0. Various approximations used as in Eq. 2.14, introduces inaccuracies and results in a wide scatter. These approximations are allowed for by reducing k3 by fifty percent. Above expression (i) may be written as below as expression (iv) derived in Eq. 2.14. 1/ 2

2 π2 Ely ⋅ h ⎡ 1 ⎛ l ⋅ tf ⎞ ⎥⎤ ⎢ Mcr = 1 + ⎜ ⎟ 2 ⋅ l2 ⎢⎣ 20 ⎜⎝ ry ⋅ h ⎟⎠ ⎥⎦

π2 El´y ⋅ h ⎡ ⎛ l ⋅t ⎞ ⎢1 + 1 ⎜ f ⎟ Mcr = 2 ⎜ ⎟ 2⋅l ⎣⎢ 20 ⎝ ry⋅h ⎠

2 ⎤1 / 2

⎥ ⎥⎦

⎛ π2 EI h ⎞ + k2 ⋅ ⎜ ⋅ y ⎟ l2 ⎠ ⎝ 2

...(ii)

⎛ π2 EI h ⎞ + k2 ⋅ ⎜ ⋅ y ⎟ l2 ⎠ ⎝ 2

...(iii)

where Iy´ is the modified moment of inertia. It is equal to k1 . Iy, (Iy is the moment of inertia of the whole section about yy-axis at the section of maximum bending moment). The elastic critical stress for bending is given by (Mcr = fcb . Zxx) 1/ 2

2 π2 Ely ⋅ h ⎡ 1 ⎛ l ⋅ tf ⎞ ⎤⎥ ⎢ fcb = 1 + ⎜ ⎟ 2 ⋅ Z xx ⋅ l2 ⎣⎢ 20 ⎝⎜ ry ⋅ h ⎠⎟ ⎦⎥

⎛ π2 EI y′ ⋅ h ⎞ + k2 ⋅ ⎜ ⎟ 2 ⎟ ⎜ ⎝ 2 ⋅ Z xx ⋅ l ⎠

...(iv)

This expression (iv) may be written as follows (as Eq. 2.14 has been derived) : 1/2

fcb

2 26.5 × 105 k1 ⎡ 1 ⎛ l ⋅ t f ⎞ ⎥⎤ ⎢ ⋅ = 1 + ⎜ ⎟ 2 ⎢⎣ 20 ⎜⎝ ry ⋅ h ⎟⎠ ⎥⎦ ⎛ l ⎞ ⎜r ⎟ ⎝ y⎠

+ k2 ⋅

26.5 × 105 k1 ⎛ l ⎞ ⎜r ⎟ ⎝ y⎠

fcb = k 1 . (X .+ k 2 .Y) 1/2

where

2 ⎡ 1 ⎛ l ⋅ tf ⎞ ⎤ X = Y . ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ 20 ⎝ r2 ⋅ h ⎠ ⎥⎦

Y =

26.5 × 105 N/mm2 ⎛ l ⎞ ⎜r ⎟ ⎝ y⎠

N/mm2

2

...(2.15)

...(2.16)

73

LOAD AND STRESSES

where

tf = Effective thickness of flange = k 1 × mean thickness of the horizontal portion of the flange of greater moment of inertia about yy-axis of the girder at the point of maximum bending moment. k2 = A coefficient to allow for inequality of tension and compression flanges, and depends on ω, the ratio of moment of inertia of compression flange alone to the sum of moments of inertia of the compression and tension flanges each calculated about its own axis parallel to the yy-axis of the girder at the point of maximum bending moment. Value of k 2 is adopted from Table 2.14.

Table 2.14

ω k2

1.0 0.5

0.9 0.4

0.8 0.3

0.7 0.2

0.6 0.1

0.5 0.0

0.4 –0.2

0.3 0.2 –0.4 –0.6

0.1 –0.8

0 –1.0

(c) Where the moment of inertia of the tension flanges about yy-axis exceeds that of compression flange ⎡ ⎤ ⎢ ⎥ 2 1 ⎛ l ⋅ ts ⎞ 26.5 × 105 ⎥ yc ⎢ 26.5 × 105 2 ⋅ 1+ fcb = ⎢ ⎜ r ⋅ D ⎟ + k2 ⋅ ⎥ × y N/mm ...(2.17) 2 2 20 y l l t ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥ ⎜r ⎟ ⎢ ⎜ ry ⎟ ⎥ y ⎝ ⎠ ⎣ ⎝ ⎠ ⎦

= k 1 . (X + k 2 . Y) .

yc N/mm2 (MPa) yt

...(2.18)

where

yc = Distance from the neutral axis of the girder to extreme fibre in compression flange yt = Distance from neutral axis of the girder to extreme fibre in tension flange Permissible stress in bending in compression is adopted as per Table 2.15 for steel as per IS : 226. Table 2.15 Cc (N/mm2) 30 40 50 60 70 80 90 100

Pbc (N/mm2) 15 20 25 30 35 38 42 46 Contd.

74

DESIGN OF STEEL STRUCTURES–VOL. II

Table 2.15 Contd. Cc (N/mm2)

Pbc (N/mm2)

120 140 160 180 200 220 240 260 280 300

53 60 67 72 76 88 84 88 92 96

350 400 450 500 550 600 650 700 750 800 900 1000 1250 1500 2000 2150

105 112 119 124 129 133 136 139 141 144 149 153 158 158 158 158

Note. The values have been converted from metric units to S.I. units.

2.15.4

Shear Stress

The average allowable shear stress on the gross- sectional area of the unstiffened web is adopted as per IS : 1915–1961 as 87 N/mm2, which is very low. The average allowable shear stresses for the stiffened web have also been described in Sec. 8.10, Vol. 1, are adopted. The value of average shear stress as per IS : 883–1994 is 0.36 fy.

2.15.5

Rivets, Bolts and Tension Rods

The allowable stresses in rivets, holts, and tension rods shall not exceed the values given in Table 2.16. These values are also as per IS : 883–1994.

75

LOAD AND STRESSES

Table 2.16 Permissible stresses in rivets, bolts and tension rods as per IS : 803– 1984 for steel conforming to IS : 226–1975 Description In tension Axial stress on gross area of rivets and on net area of bolts and tension rods Power driven rivets Hand driven rivets Close tolerance and turned bolts Bolts in clearance holes In shear Shear stress on gross area of rivets and bolts Bolts in clearance holes Power driven rivets Hand driven rivets Close tolerance and turned bolts Bolts in clearance holes In bearings Bearings stress on gross diameter of rivets and bolts Power driven rivets Hand driven rivets Close tolerance and turned bolts Bolts in clearance holes

2.17

Permissible stress N/mm 2

100 80 120 120

120 100 80 100 80

300 250 300 250

ALLOWABLE STRESSES FOR COMBINATION OF LOADS

The allowable stresses for the combination of normal loads and the occasional loads are adopted as 1.25 times the corresponding values as discussed in Sec. 2.15. The allowable stresses during erection or lifting during maintenance are adopted as 1.30 times the corresponding values described in Sec. 2.15. Allowable stresses due to extraordinary loads are adopted as per the discretion of the appropriate authority but not exceeding 90 percent of yield stress. The end cross girders or other members which are used for lifting the span shall be so proportioned that the maximum stress while lifting span during maintenance including the stress due to dead load or any other co-existing load shall not exceed the permissible stress by more than 25 percent.

2.17.1

Combined Stress

When the structural members of a bridge are subjected to the different system of combined stress, then, these members are checked as follows :

76

DESIGN OF STEEL STRUCTURES–VOL. II

When the members are subjected to both axial stresses and bending stresses (compressive or tensile), then, the members are so proportioned that the quantity ⎛ σac.cal σbc.cal ⎞ + deos not exceed unity ⎜ σ σbc ⎟⎠ ⎝ ac

where, σac.cal = Calculated axial stress σac = Appropriate allowable stress in axially loaded members σbc.cal = Calculated maximum bending (compressive or tensile) stress about both principal axes, including secondary stresses, if any σbc = Appropriate allowable stress in bending. When the members are subjected to bending and shear stresses, then, irrespective of any permissible increases of allowable stresses, the equivalent stress, σe due to this combination should not exceed 220 N/mm2 for the structural steel conforming to IS : 226. The equivalent stress, σe , is obtained from the following formula σee.cal = [σbc2.cal + 3τvm.cal]1/2 ...(2.12) 2 1/2 σe.cal = [σbt .cal + 3τvm.cal] ...(2.20) where, τvm.cal = Calculated co-existing shear stress σbt.cal = Calculated bending tensile stress σbc.cal = Calculated bending compressive stress. Where a bearing stress is combined with tensile bending or compressive bending and shear stresses approaching the maximum allowable value under the most unfavourable conditions of loading the equivalent stress σe.cal shall not exceed 0.9fy for structural steel conforming to IS : 226. The equivalent stress, σe.cal is obtained from the following formula σe.cal = [σbt2.cal + τp2. cal + σbc.cal σy.cal + 3τvm2.cal ]1/2 ...(2.21) 2 2 2 1/2 σe.cal = [σbc .cal – τp . cal + σbc.cal σp.cal + 3τvm .cal ] ...(2.22) where, σe.cal = σbc.cal and τvm.cal are as specified above σp.cal = Calculated co-existing bending stress.

2.18

FLUCTUATION OF STRESSES

The ultimate tensile strength of steel is the stress determined by dividing the maximum load on the specimen by the initial cross sectional area. The load is applied gradually. The specimen fails at ultimate load in its first application. In case the specimen is loaded and unloaded large number of times, then the specimen fails at smaller load. The magnitude of load to produce failure decreases as number of cycles of loading increases. This also depends on range of stress. The range of stress is defined as the algebraic difference of maximum and minimum values of the repeated stress. The phenomenon of decrease of

LOAD AND STRESSES

77

resistance of material due to repeated stresses is known as fatigue. In case of cantilever test, specimen for fatigue, the maximum compressive and tensile stresses occur alternately due to reversed bending, the tensile strength decreases to a one-third of its original value. The variation of stresses, and, or reversal of stresses, i.e., the fluctuation of stresses in the structural members of a bridge results in the fatigue failures of members or connections or both. The fatigue failures may take place at lower stresses than those at which the member would fail under the static loads, if the members are subjected to the number of cycles of fluctuation of the stresses. The number of repetitions of the stress cycle decreases as the range between the maximum and minimum stresses in the member increases. The fatigue failures are primarily due to the connections of the stresses in members. The concentration of stresses in the members occur due to constructional details, e.g., sharp corners, sudden changes in cross-section, rivets, weld, etc. To allow for the effect of fatigue in case of tension elements, the allowable working stress is determined by multiplying the appropriate allowable stress by a factor, k, the values of which are given in Table 2.17 as per IS : 1915–1961. The factor k depends on the ratio of minimum stress (fmin) to the maximum stress (fmax), the number of repetitions of the stress cycle, the quality of steel, the method of fabrication, and the type of connections. In determining the ratio ⎛ fmin ⎞ ⎜f ⎟ , gross areas of members are used. ⎝ max ⎠ In case of railway bridges, generally, for plate girders, floors and web system of the truss girders, the member of repetition of stress cycles may be taken as 21 × 105 cycles, and that for the chord members of truss girders as 6 × 105 cycles. No allowance for fatigue need be made in the case of highway bridges and foot overbridges. In case the reversal of the stresses takes place in the web members of a truss girder, then counter-brace members are provided. The actual distribution of the stress in the counter-brace members is statically indeterminate. The exact value of compressive force in the counter-brace members can be determined, if desired by considering the buckling and shortening of such struts. In order to simplify, it is generally assumed that one member remains active at a time and takes tension only. No allowance for fatigue shall be made in the design of riveted and the bolted connections but all rivets and bolts subjected to reversal of stress during the passage of the live loads shall be designed for the sum of the maximum and the reversed loads on them. In case of wind bracing the connections shall be designed to resist the greater stresses. No allowances for fatigue shall be made for compression members or for shear stresses in webs except that when open holes larger than those normally used for rivets or fastenings exist. The fatigue factors shall be applied to the resulting tensile stress near the hole. The allowance for fatigue shall be made for normal loading only. The stresses due to wind, temperature and secondary stresses shall be ignored in considering fatigue.

(1) 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 –0.1 –0.7 –0.3 –0.4

⎛ f min ⎞ ⎜f ⎟ ⎝ max ⎠

1×105 (2) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

6×105 (3) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 0.95 0.91 0.87 0.83

20×105 (4) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 0.98 0.93 0.88 0.83 0.79 0.76 0.72 0.69

Number of cycles

k for steel members conforming to IS : 226–1958, fabricated with or connected by rivets, bolts or butt welds, and for those fabricated with continuous longitudinal fillet welds

Table 2.17 Factor K for fluctuating stress

1×105 (5) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 0.91 0.83 0.77 0.71

6×105 (6) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 0.97 0.86 0.78 0.71 0.65 0.60 0.56

20×105 (7) 1.0 1.0 1.0 1.0 1.0 1.0 0.93 0.79 0.69 0.62 0.56 0.50 0.46 0.43 0.40

Number of cycles 1×105 (8) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

6×105 (9) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 0.94 0.88 0.83 0.79

Number of cycles 20×105 (10) 1.0 1.0 1.0 1.0 1.0 1.0 0.91 0.83 0.77 0.71 0.67 0.62 0.59 0.56 0.53 Contd.

k for steel members conforming to k for high tensile steel members IS : 226–1958, fabricated with conforming to IS : 961–1957 intermittent or transverse fabricated or connected fillet welds, or connected by by rivets or bolts transverse or longitudinal fillet welds

78 DESIGN OF STEEL STRUCTURES–VOL. II

5

6×10 (3) 0.80 0.77 0.74 0.71 0.69 0.67 5

20×10 (4) 0.67 0.64 0.62 0.60 0.58 0.56

Number of cycles 5

1×10 (5) 0.67 0.62 0.59 0.56 0.53 0.50 5

6×10 (6) 0.52 0.49 0.46 0.43 0.41 0.39 5

20×10 (7) 0.37 0.35 0.33 0.31 0.29 0.28

Number of cycles 5

1×10 (8) 1.0 1.0 0.96 0.91 0.88 0.84 5

6×105 (9) 0.75 0.71 0.68 0.65 0.62 0.60

Number of cycles

2. The intermediate value of k may be found by linear interpolation.

Note : 1. The ratio (fmin /fmax) positive or negative respectively if the maximum stresses are of like or unlike sign.

(1) –0.5 –0.6 –0.7 –0.8 –0.9 –1.0

1×10 (2) 1.0 0.96 0.93 0.89 0.86 0.83

Table 2.17 Contd.

20×105 (10) 0.50 0.48 0.46 0.44 0.42 0.40

LOAD AND STRESSES

79

80

2.19

DESIGN OF STEEL STRUCTURES–VOL. II

ENDURANCE LIMIT

The bridge structure are subjected to repeated loading and unloading or variation and reversal of stresses i.e., the fluctuation of stresses as discussed in Sec. 2.18. As a result of this the strength of material of bridge elements for the repeated stresses decreases. This phenomenon of decrease in strength to repeated stresses is known as fatigue. If the members are subjected to number of cycles of fluctuation of the stresses, the fatigue failure may take place at lower stresses than those at which the members would fail under the static loads. The members of bridge structures are subjected to number of cycles of repetitions of the stress. The members of the bridge structures should be such that maximum stress does not exceed the endurance limit and the failure of the members does not take place. The endurance limit is defined as the maximum stress that can be applied to a material for an infinite number of cycles of repeated stress without causing failure. The endurance limit is also called as fatigue limit.

S -N o rm inal stress N /m m 2

7 70 7 00 6 30 5 60 4 90 4 20 3 50 1 02

1 04 1 05 N -N u m be r o f cycles

1 06

1 07

Fig. 2.13

The endurance limit is found by conducting series of fatigue tests (endurance test) on number of specimens of the material at different stresses. The stress endured by each specimen is plotted against the number of cycles to failure. The diagram plotted is known as the stress-cycle diagram, or S–N diagram. Smaller and smaller values of stresses are taken, and the endurance tests are conducted. At the beginning, the values of stress reduce rapidly as the number of repititions of stress cycles increases. A value of stress is obtained at which the specimen does not fail irrespective of the number of stress cycles applied. In the stress cycle diagram, the curve corresponding to the value of stress becomes asymptotic to the horizontal line. This stress is known as endurance limit. Instead of plotting the stress against the number of cycles, it is convenient to plot either the stress against the logarithm of number of cycles or both, the stress and number of cycles on logarithmic scale. A typical stress-cycle diagram is shown in Fig. 2.13 in which the number of cycles of stress has been plotted on

LOAD AND STRESSES

81

logarithms scale. Near the endurance limit both the diagrams show a sharp bend. The endurance limit for most of the steels may be established with sufficient accuracy on the basis of 2 × 106 and 10 × 106 cycles. In order to simplify, certain national codes have given an expression to find the fatigue limit for cyclic loading or range of stress. IS 807–1976 for design of cranes specifies an expression as under F =

T ⎡ ⎛ fmin ⎞ ⎤ ⎢2 − ⎜ f ⎟⎥ ⎝ max ⎠ ⎦ ⎣

where, T is the tensile strength and F is the fatigue limit. This may also be adopted for bridges also. Most of non-ferrous metals do not show clearly defined endurance limit. The values of stresses of such metals continuously decrease even upto several million cycles.

2.20

DESIGN OF BRIDGES

Ministry of Railways, Government of India has laid down the specifications for the design of railway bridges in the code ‘Indian Railway Standard Code of Practice for the design of steel and wrought iron bridges carrying rail, road or pedestrian traffic’. All the railway bridges are to be designed conforming to these specifications. The various loads specified in Bridge Rules—rules specifying the loads for designing the super and sub-structures of bridges and for assessing the strength of existing bridges are to be considered. Indian Road Congress has drafted the specifications for the design of highway bridges in IRC section I, IRC section II, IRC section III, IRC section IV, and IRC section V. In addition to the above, the specifications given by ‘Code of Practice for design of steel bridges’ IS : 1915 are also followed for the design of bridges. The design of plate girder bridges have been described in chapter 3 and that of truss girder bridges have been covered in chapter 4. The design of bearings for the various types of bridges have been discussed in Chapter 5.

CHAPTER

3

Design of Plate Girder Bridges

3.1

INTRODUCTION

The plate girders are used as the main load carrying members in the plate girder bridges. The plate girder bridges are also known as solid web girder bridges. The plate girders are economically used for the railway bridges for span from 15 m to 30 m and for the highway bridges for spans from 20 to 40 m. The plate girders are comparatively free from the secondary stresses. The minimum height of plate girders in case of railway bridges are one-twelfth of the span, and in case of highway bridges one-twenty fifth of the span. The effective height (depth) of plate girder is measured from gravity axis of top flange upto the gravity axis of the bottom flange. The span is taken as the distance between centres of bearings in the case of simply supported bridges. In addition to the main (stock) rails, the guard rails are provided on all the bridges in order to avoid the derailment of trains and serious disasters. The design of plate girder and its various components have been discussed in detail in author’s Steel Structures Vol. I. The various types of sections of plate girders are given in Vol. I. The use of deck type or through type bridges depends upon the clearance available and the depth of the plate girder.

3.2

TYPES OF FLOOR SYSTEMS

The floor systems may be classified in accordance with the type of traffic carried, such as railway or highway floors ; in accordance with the principal material used, such as steel, timber, concrete or masonry floor; or in accordance with the structural action of the floor, as one way or two way slabs, composite concrete floors. The selection of the appropriate floor system for a particular bridge is governed by the following considerations ; quality of road surface, proper

DESIGN OF PLATE GIRDER BRIDGES

83

drainage, weight of the floor system, required construction time, and overall cost, including that of maintenance. There are two types of floor systems, which are used in the railway bridges. 1. Open floor system 2. Solid floor system.

3.2.1

Open Floor System

The open floor system is also termed as open deck type floor. This is most common type of floor system used both in the deck type and through type plate girder bridges. In the deck type plate girder bridges, the wooden sleepers, which are also known as timber ties are carried directly over the main girders. The open floor system in deck type plate girder bridge is shown in Fig. 3.3. The timber ties are placed with 0.10 m to 0.20 m intermediate openings. The river bed or other crossing remains clearly visible through the openings (spacings). In case of the through type bridges, and the half through type bridges, the timber ties are placed over the stringers (longitudinal girders or rail bearers). The open floor system in case of the through type plate girder bridge is shown in Fig. 3.4.

3.2.2

Solid Floor System

The solid floor system is also known as ballasted deck type floor or the continuous floor. The solid floor system consists of steel or concrete or timber troughs. The troughs are supported over stringers and floor beams (cross girders). The ballast is placed in these troughs. The sleepers-supporting railway track are placed in the ballast. The common types of steel troughs used in the solid floor are shown in Fig. 3.1.

Fig. 3.1 Type of steel troughs

By varying the thickness and depth of troughs, the solid floor can be used for any span and the load. The solid floor system used in case of a through type bridge is shown in Fig. 3.2.

84

DESIGN OF STEEL STRUCTURES–VOL. II

M ain girder Rails

Sleepers

Fig. 3.2 Solid floor system

The floor system in case of highway bridges generally consists of reinforced concrete slab. The reinforced concrete slab is supported on steel stringers, which are further supported by the floor beams. The reinforced concrete slab provides its own traffic surface. Many times, in addition to this, the bituminous or carpet surface is furnished. This acts as a wearing surface.

3.3

DECK TYPE PLATE GIRDER BRIDGE

The deck type plate girder bridge is shown in Fig. 3.3. R ails

Sleepers (Tim ber ties)

M ain girder

Fig. 3.3 Deck type plate girder bridge

The deck type plate girder bridge consists of two plate girders. The width between centres of main girders is kept sufficient to resist overturning due to specified wind pressure and load conditions. The width between the main girders should not be less than one-twentieth of the effective span. The timber ties (wooden sleepers) are directly placed over the plate girders. The plate girders are suitably braced for lateral forces. A deck type plate girder bridge is shown in Plate 3.1. Sixteen spans out of twenty spans of the bridge over river Luni, Samdari (Rajasthan) are that of deck type plate girder bridge.

DESIGN OF PLATE GIRDER BRIDGES

3.4

85

THROUGH TYPE PLATE GIRDER BRIDGE

The through type plate girder bridge is shown in Fig. 3.4. When the clearance between the high water level and the underside of bridge is not sufficient to allow the use of deck type bridge, then through type girder bridge is used. The floor system in case of through type plate girder is supported M ain girde r S lee pe rs

R a ils

S trin ge rs

Floo r b ea m s

Fig. 3.4 Through type plate girder bridge

over beams which are placed parallel to the main girders. These beams are known as stringers or longitudinal girders or rail bearers. These stringers have span from 3 m to 5 m. These stringers are supported and framed into the crossbeams. The cross-beams are termed as cross-girders or transverse girders or floor beams. The floor beams are connected near the bottom of plate girders as possible. These points of connections are known as panel points. The plate girders are suitably braced for the lateral loads. A through type plate girder bridge is shown in Plate 3.2. Four spans out of twenty spans of the bridge over river Luni, Samdri (Rajasthan) are that of through type plate girder bridge. Earlier these spans were also deck type, which were washed-off in heavy rain and re-built as through type.

3.5

BRACING OF DECK TYPE PLATE GIRDER BRIDGES

In addition to the dead load, live load and impact load, the bridge is also subjected to lateral and longitudinal forces. In order to provide lateral stability and the torsional rigidity, the bridge is suitably braced. The bracing consists of horizontal bracing and the transverse bracing. The horizontal bracing is also known as horizontal truss bracing. The main horizontal truss bracing is provided in between the loaded flanges. The loaded flanges in case of deck type plate girder are top flanges (compression flanges). Formerly in addition to the’ horizontal bracing in the loaded flanges, the horizontal bracing also used to be provided with unloaded flanges. The transverse bracing for deck type plate girder bridges consists of cross-frames. The cross frames are provided in the parallel vertical

86

DESIGN OF STEEL STRUCTURES–VOL. II

G ua rd ra ils M ain rails Tim b er tie s

H o rizon ta l truss b racin g M ain girde rs C ro ss fram e (vertical b racin g)

(A b utclea r / P ie r)

Fig. 3.5 Bracing of deck type plate girder bridges

planes. The horizontal truss bracing and the cross-frames in the deck type girder bridge are shown in Fig. 3.5. The horizontal truss bracing is also shown in Fig. 3.3.

3.6

BRACING OF THROUGH TYPE PLATE GIRDER BRIDGES

The bracing of through type plate girder bridge consists of the horizontal bracing and the transverse bracing. The horizontal truss bracing is provided in the loaded flange. The loaded flange in case of the through type bridge is the bottom flange (tension flange). The upper flanges of through type plate girder are braced by

In te rna l g usse t pla te M ain girde r

G ua rd ra il S tock rail

S lee pe r Floo r b ea m

S trin ge r H o rizon ta l truss b racin g (A b utclea r / Pie r)

Fig. 3.6 Bracing of through type plate girder bridges

means of triangular gusset plates. These gusset plates are connected on the internal sides with the floor beams and the main girders. These are called

DESIGN OF PLATE GIRDER BRIDGES

87

internal gusset plates. The horizontal bracing and the internal gusset plates are shown in Fig. 3.6. The horizontal truss bracing of the through type plate girder bridge is also shown in Fig. 3.4. The effective length of compression flange for half through type bridge has been described in Sec. 4.8.

3.7

SELF-WEIGHT OF PLATE GIRDERS

The dead load acting on plate girder bridges also include the self-weight of plate girders of the bridges. The self-weight of plate girders of the bridges are either assumed depending upon experience and by comparing with the existing plate girder bridges on the similar spans or determined by the following formula: ⎛ WL ⎞ W1 = ⎜ ⎟ ⎝ 300 ⎠

...(3.1)

where,

W = Total superimposed load on both the plate girders in kN L = Span of the plate girders in metres W1 = Total weight of both the plate girders in kN The self-weight of both the plate girder can also be determined from Fuller’s formula for 10 m to 30 m span w =

(a.L + b)

...(3.2) 100 kN/m where, w = Weight of both the plate girders in kN per metre a, b = Constants Constants a and b depend upon the type of the bridge, the load for which the bridge is designed and the unit stress used. The term aL represents principally the weight of those portions whose weight per linear length increases approximately in proportions to span. The term b represents mainly weight of floor. The weight of floor per linear length depends of panel length and width of the bridge. It is approximately constant. The values of a and b constants are adopted as 20 and 100 respectively. By substituting the values of constants, Eq. 3.2 may be written as w =

1 ⋅ (20L + 100 ) kN/m 100

...(3.3)

The weight of steel plate girder railway bridges carrying single track may be found from the following formula also. This formula has been given by using Dr. Wadell’s extensive date: w = k . L (W)1/2 ...(i) where, w = Weight of two plate girders and bracings in kN per metre length of bridge k = Constant. Its value is about 0.0522 for deck type bridges L = Effective span of the bridge in metres

88

DESIGN OF STEEL STRUCTURES–VOL. II

W = Heaviest axle load of engines in kN In the broad gauge standard loading the heaviest axle loads for the main line and branch line loadings are 229 kN and 173 kN respectively. Therefore, ...(ii) w1 = 0.790L kN/m (Main line) w2 = 0.686 L kN/m (Branch line) ...(iii) In the metre gauge standard loading the heaviest axle loads for the main line, branch line and C-class loadings 132 kN, 107 kN and 81 kN respectively. Therefore, w3 = 0.600 L kN/m (Main line) ...(iv) ...(v) w4 = 0.5401 kN/m (Branch line) ...(vi) w5 = 0.470 L kN/m (C-class)

3.8 ASSUMPTIONS FOR THE DESIGN OF PLATE GIRDER BRIDGES Following assumptions are made in the design of plate girder bridges : 1. The web plates of the plate girders resist the shear force and the shear stress is uniformly distributed over the entire cross-sectional area of the web. 2. The flanges of plate girders resist the bending moment.

3.9 DESIGN OF PLATE GIRDERS FOR DECK TYPE RAILWAY BRIDGES (DESIGN OF MAXIMUM SECTION) The plate girders in deck type railway bridges are spaced with sufficient width to develop lateral strength and rigidity. As per code of practice for the design of steel bridges the width between centres of plate girders should be sufficient to resist the overturning with specified wind pressure and load conditions. The plate girders are placed about 150 to 200 mm outside the main or stock rails, so that the effect of impact is less. The plate girders carry the weight of stock rails, guard rails, fastenings, weight of sleepers (timber ties) and the self-weight. It is quick to determine the selfweight of plate girders by Fuller’s formula. In addition to the dead load, the plate girders carry live load and the impact load. The live loads for the railway bridges for broad gauge, metre gauge, and narrow gauge railway tracks have been described in Sec. 2.3 as per Bridge Rules. The impact load has been described in Sec 2.4. The calculations are done for dead load, live load and impact load per track for whole of the span. The total of dead load, live load and impact load is then found for one girder. The total load acts as uniformly distributed load. The maximum bending moment, M, due to this loading occurs at the centre. The economical depth of web plate may be found from the following expression :

⎛M ⎞ × tw ⎟ d = 1.1 ⎜ ⎝ fb ⎠ where,

M = Maximum bending moment

...(3.4)

DESIGN OF PLATE GIRDER BRIDGES

89

fb = Allowable bending stress tw = Thickness of web The thickness of web, t w is assumed suitably. The minimum thickness of web plate shall not be less than the following as per IS : 883–1994. The thickness of stiffened web plate shall be not less than the following (i)

(ii)

τw.min = d1

(t .cal ) ⋅ wa

τw.min = d1

(f ) ⋅

1 2

816

y

11 2

1344

whichever is more, but

d1 85 where, d1 is the depth of the web, and τva· cal is the calculated average shear stress in the web. The thickness of the vertically stiffened web plate shall be not less than

τw.min = 1500 kN (Support reaction). Hence, safe.

119

DESIGN OF PLATE GIRDER BRIDGES

Step 4. Connection of bearing stiffener with the web plate Use 22 mm diameter power driven rivets. Strength of rivet in double shear = 2×

π (23.5 )2 × 100 × = 86.70 kN 4 1000

Strength of rivet in bearing = Rivet value, Number of rivets required

23.5 × 8 × 30 = 96.40 kN 1000

R = 56.40 kN

⎛ 1500 ⎞ = ⎜ ⎟ = 26.59 ⎝ 56.4 ⎠ The filler plates (packings are properly fitted with the bearing stiffeners. These filler plates are subjected to direct compression only. Provide 30 rivets in 2 rows. Design of Intermediate stiffeners Step 1. Clear depth between flanges angles on the plate girder d1 = (2500 – 2 × 150) = 2200 mm Thickness of web, tw = 8 mm. In case, the web plate is to be unstiffened, the minimum thickness of web necessary is as follows :

⎛ 1500 × 1000 ⎞ 2 t va· cal = ⎜ ⎟ = 75 N/mm ⎝ 2500 × 8 ⎠ (i)

1/2 ⎛ τw· min = ⎜ d1 ⋅ tva.cal 816 ⎝

⎞ ⎛ 2200 × 751 / 2 ⎟=⎜ 816 ⎠ ⎝

⎞ ⎟ mm ⎠

= 23.34 mm (ii)

(iii)

1/2 ⎞ 1/2 ⎞ ⎛ ⎛ τw· min = ⎜ d1 ⋅ fv ⎟ = ⎜ 2200 × 250 ⎟ mm 1344 ⎠ ⎝ 1344 ⎠ ⎝

τw· min =

d1 ⎛ 2200 ⎞ =⎜ ⎟ = 25.88 mm 85 ⎝ 85 ⎠

The actual thickness of web 8 mm is less than the above values of t w· min, as such the vertical stiffeners are provided. Step 2. Now, if the horizontal stiffeners are not to be used, then, the thickness of web necessary is as below :

120

DESIGN OF STEEL STRUCTURES–VOL. II

d2 = (i)

(ii)

1 × 2200 = 1100 mm. 2

⎛ d ⋅ f 1 / 2 ⎞ ⎛ 1100 × 2501 / 2 ⎞ τw . min = ⎜ 2 y ⎟ = ⎜ ⎟ = 5.435 mm 3200 ⎠ ⎝ 3200 ⎠ ⎝

τw . min =

d2 ⎛ 1100 ⎞ =⎜ ⎟ = 5.5 200 ⎝ 200 ⎠

The actual thickness of web 8 mm is more than t w.min, then, the horizontal stiffeners are not necessary. Step 3. Design of vertical stiffeners At support, shear force = 1500 kN Actual average shear stress in the web plate

⎛ 1500 × 1000 ⎞ 2 t va.cal = ⎜ ⎟ = 75 N/mm ⎝ 2500 × 8 ⎠ Ratio,

d tw

⎛ 220 ⎞ = ⎜ ⎟ = 275 ⎝ 8 ⎠

The smaller clear panel dimension for the actual thickness of web = 180 × 8 = 1440 mm The greater clear panel dimension for the actual thickness of web = 270 × 8 = 2160 mm The vetical stiffeners may be provided at a spacing smaller than 1440 mm. Let the spacing of vertical stiffeners be = 0.6 × d = 0.6 × 2200 = 1320 mm From IS 800–1984, Table 6.6 (A), the permissible average shear stress, tva in the stiffened web plate of steel with fy = 250 N/mm2 and 0.6 d spacing and

d ratio tw

tw = 81 N/mm2 > (t va . cal = 75 N/mm2) Length of outstanding leg of the vertical stiffener

⎛ 1 ⎞ = ⎜ × clear depth of girder + 50 ⎟ mm ⎝ 30 ⎠ ⎛ 2200 ⎞ = ⎜ + 50 ⎟ = 123.33 mm ⎝ 30 ⎠

DESIGN OF PLATE GIRDER BRIDGES

121

IS A 1 25 m m x 95 m m x 8 m m W eb 8 m m th ick

1 26 0 m m IS A 1 25 m m x 95 m m x 8 m m 1 32 0 m m

Fig. 3.16

Provide ISA 125 mm × 95 mm × 8 mm. The length of outstanding leg of the angle section is 90 mm. Clear distance between vertical stiffeners c = (1320 – 60) = 1260 mm Depth of plate girder = 2500 mm Minimum required thickness of web

⎛ 1500 × 1000 ⎞ tw1 = ⎜ ⎟ = 7.41 mm ⎝ 2500 × 81 ⎠ Required moment of inertia I =

1.5d2 ⋅ tw31 C

2

⎛ 1.50 × 22002 × 7.413 ⎞ =⎜ ⎟ mm4 12602 ⎝ ⎠

= 409 × l04 mm4 Moment of inertial about the face of web plate provided = (266 + 3.802 × l6.98) ×.l04 mm4 = 511 × 104 mm4. Step 4. Connection of vertical stiffeners to web plate Shear force

2 ⎞ ⎛ = ⎜ 125 t ⎟ kN/m ⎝ h ⎠ 2 ⎞ ⎛ = ⎜ 125 × 8 ⎟ = 88.89 kN/m ⎝ 90 ⎠

Use 22 mm diameter power driven field rivets Strength of rivet in single shear 2 ⎛ ⎞ = ⎜ π × 23.5 × 100 ⎟ = 43.35 kN 1000 ⎝4 ⎠

122

DESIGN OF STEEL STRUCTURES–VOL. II

Strength of rivet in bearing

⎛ 23.5 × 8 × 300 ⎞ = ⎜ ⎟ = 56.4 kN 1000 ⎝ ⎠ Rivet value, Pitch of rivets

R = 43.35 kN

⎛ 43.35 ⎞ = ⎜ ⎟ = 0.487 m ⎝ 88.89 ⎠

Provide rivets at a pitch = 200 mm Provide ISA 125 mm × 95 mm × 8 mm and 22 mm rivets to connect the stiffners with the web at 200 mm pitch. The vertical stiffeners are shown in Fig. 3.16.

3.14 DESIGN OF STRINGERS, CROSS GIRDERS AND PLATE GIRDERS FOR HIGHWAY BRIDGES The floor system in case of highway bridges generally consists of reinforced concrete slab. In case of deck type plate girder highway bridges, the slab is supported directly by the plate girders. In case of through type highway bridges, the reinforced concrete slab is supported on stringers, and cross-girders, or by the cross-girders alone. Many times, the reinforced concrete slab provides its own traffic surface. In addition to this, the bituminous, asphalt or carpet surface is also furnished. This acts as a wearing surface. The design of reinforced concrete slab has not been discussed in the text.

3.14.1

Stringers

The stringers support the reinforced concrete slab in case of through type highway bridges. The stringers are supported by the cross-girders. The stringers may be supported on the top of cross-girders or may be framed into the cross-girders by the use of suitable connections. When the reinforced concrete slab is used, then either the stringers should be supported on the top of the cross-girders or in case the stringers are framed into the cross-girders, then the top of stringers should be on the same level as the cross-girders. The stringers carry the dead load, which consists of the weight of wearing coat, the weight of reinforced concrete slab and the self-weight. In addition to this, the stringers also support the live load and the impact load due to highway standard vehicle or trains. The design of stringers has been illustrated in the worked out examples.

3.14.2

Cross-girders

The load from floor system is carried to the cross-girders by means of the stringers or the loads may be carried to the cross-girders directly by the reinforced concrete slab. The cross-girders carry dead load, which consists of the weight of wearing coat, the weight of reinforced concrete slab, the reaction from the stringers and the self-weight. In the addition to this, the cross-girders carry live load and

DESIGN OF PLATE GIRDER BRIDGES

123

impact load due to highway standard vehicles or trains. The design of cross-girders has been illustrated in the worked out example.

3.14.3

Plate Girders

In the deck type highway bridges, the spacing between plate girders is kept sufficient to develop lateral strength and rigidity, and to resist the overturning with the specified wind pressure and the load conditions. In the through type highway bridges the spacing between plate girders is kept sufficient to suit the clearance requirement. The spacing of plate girders required for the clearance requirement is sufficient to resist the overturning with the specified wind pressure and load conditions, and to develolp lateral strength and rigidity. In the dick type highway bridges, two plate girders are used for single lane carriageway width and three or four plate girders depending upon the design are used for two lane carriageway highway bridge. The reinfoced concrete slabs inclusive of wearing coat is suppoted directly by the plate girders. The plate girders carry dead load. The dead load consists of the wearing coat, the weight of reinforced concrete slab and the self-weight of plate girders. The self-weight of plate girders may be found by Fuller’s formula. In addition to this, the plate girders carry the live load and the impact load due to the highway standard vehicles or trains. In case of through type highway bridges, the plate girders carry the dead load. The dead load consists of the weight of wearing coat, the weight of reinforced concrete slab, the weight of stringers, the weight of cross-girders, and the selfweight. The self-weight of the plate girders may be found by Fuller’s formula. In addition to this, the plate girders carry the live load and the impact load due the highway standard vehicles or trains. Within the kerb to kerb width of the roadway, the standard vehicles or trains are assumed to travel parallel to the length of bridge, and to occupy any positions, which produce the maximum stresses provided that the minimum clearances between a vehicle and the roadway face of a kerb and between two passing or crossing vehicles as specified in Tables 2.4, 2.6 and 2.8, are not encroached upon. For each standard vehicle or train, all the axles of a unit of vehicles are considered as acting simultaneously in position causing maximum stresses. The vehicles in adjacent lanes are taken as headed in the direction producing maximum stresses. The space on carriageway left uncovered by the standard train of vehicles shall not be assumed as subjected to any additional live load. The maximum reaction due to number of highway standard train of vehicles shall not be assumed as subjected to any additional live load. The maximum reaction due to number of highway standard vehicles or trains passing simultaneously over the carriageway depending upon the number of lanes, is found by the transverse location of loading. Further the maximum bending moment and maximum shear force on the plate girders are found by longitudinal location of loading. After determining the maximum bending moment and the maximum shear force, the section, of plate girder is either selected from ISI Handbook No. 1 or

124

DESIGN OF STEEL STRUCTURES–VOL. II

the section may be designed as discussed in Sec. 3.9. The maximum section is further checked for lateral loads, and the suitable bracing is provided. The designs of deck type and through type highway plate girder bridges have been illustrated in worked out examples. Example 3.7 The effective span of a through type plate girder two lane bridge is 30 m. The reinforced concrete slab is 250 mm thick inclusive of the wearing coat. The foot-paths are provided on their sides of carriageway. The cross girders are provided at 3 m centres. The stringers are spaced at 2.45 m centre to centre. The spacing between the main girders is 9.80 m. Determine the maximum sections for the stringers and the cross-girders, if the bridge is to carry IRC class A standard loading. Solution (A) Design of stringers : Step 1. From IRC section I Width of roadway for single lane = 3.80 m For second lane = 3.80 m Width of foot-paths 2 × 1.50 = 3.00 m Total width = 9.80 m Stringers Effecive span = 3 m. M ain girde r 2 .45 m

2 .45 m

S trin ge rs

9 .80 m

C ro ss g ird ers

S trin ge rs

2 .45 m

2 .45 m 3m M ain girde r

S trin ge rs

3m

3m

3m

Fig. 3.17

Step 2. Dead load The stringers support weight of slab inclusive of wearing coat and the selfweight. Weight of slab + wearing coat per stringer

⎛ 2.45 × 3.0 × 250 × 24 ⎞ = ⎜ ⎟ = 44.l kN 1000 ⎝ ⎠

125

DESIGN OF PLATE GIRDER BRIDGES

Self-weight of stringers (1 kN/m assumed) = l × 4.0 = 3.0 kN Total dead load per stringer = 47.1 kN Step 3. Bending moment due to dead load The maximum bending moment M1 due to dead load occurs at the centre

⎛ 47.1 × 3.0 ⎞ M1 = ⎜ ⎟ 17.66 kN-m 8 ⎝ ⎠ Step 4. Shear force due to dead load The maximum shear force F1, due to dead load occurs at the supports 47.1 = 23.55 kN 2 Step 5. Bending moment and shear force due to live load and impact load Road width between kerbs = 6.80 m. Therefore two trains of IRC class A loading pass over the bridge simultaneously. Step 6. Transverse location of loading For the design of central stringer, the loads are so placed that the reaction on the stringer is maximum. From IRC section II Destance between centre of two wheels of one train = 1.80 m The minimum distance between adjacent edges of wheels of two trains

F1 =

1.2 − 0.4 ⎡ ⎤ × (6.8 − 5.5 )⎥ = 0.92 m ⎢⎣0.40 + ⎦ 2 The distance between centre to centre of adjacent wheels of two trains = (0.92+ 0.52)= 1.42 m The minimum clearance between outer edge of the wheel and the roadway face of the kerb f = 0.150 m (minimum). Transverse location of wheels of trains is as shown in Fig. 3.18.

2 .45 m

P

1.80 m

P

1.42 m

M in f = 0 .15 m

P

1.80 m

0 .4 0

1 .50 m

g =

P

0 .92 m

2 .45 m

9 .80 m

Fig. 3.18

2 .45 m

2 .45 m

126

DESIGN OF STEEL STRUCTURES–VOL. II

Reaction on the central stringer

⎡ (2.45 − 1.20) (2.45 − 0.22 (2.45 − 0.22 − 1.80) ⎤ =P ⎢ + + ⎥⎦ = 1.595 P 2.45 2.45 2.45 ⎣ From IRC section II, impact factor for steel bridge

9 9 ⎛ ⎞ ⎛ ⎞ i = ⎜ ⎟=⎜ ⎟ = 0.545 ⎝ 13.5 + L ⎠ ⎝ 13.5 + 30 ⎠ Reaction on the central girder including impact = 1.595 P ×1.545 = 2.455 P Maximum one wheel load of IRC class A standard loading 114 = 57.0 kN 2 Step 7. Longitudinal location of loading The maximum bending moment occurs when the two heaviest wheels of one train are as shown in Fig. 3.19. The distance between wheels is l.20 m.

P =

2 .45 5 P 0 .9 m

2 .45 5 P 0 .9 m

1 .20 m

3m

Fig. 3.19

Maximum bending moment due to both the trains and including impact M2 = 2.455 × (57.0) × 0.90 = 125.94 kN-m The dispersion of wheel load takes place at 45° inclination. Therefore, length of dispersion of wheel load = (0.25 +B + 0.25) = (0.25 + 0.25 + 0.25) = 0.75 m The maximum shear force occurs in the stringer when the heaviest wheel load is as shown in Fig. 3.20. 2 .45 5 P

2 .45 5 P 2 .0 m

4 5°

4 5° 0 .75 m

0 .25 m

0 .25 m

3m

Fig. 3.20

Maximum shear force due to both the trains and including impact

127

DESIGN OF PLATE GIRDER BRIDGES

0.75 ⎞ ⎛ 0.75 ⎞ ⎤ ⎛ 1 ⎞ ⎡⎛ F2 = (2.455)(57) ⎜ ⎟ ⎢⎜ 3 − ⎟ + ⎜ 3 − 1.20 − ⎟ kN 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎝ 3 ⎠ ⎣⎝

= 188.91 kN Total bending moment due to dead load, live load and impact load M = M1+ M2 = (17.66 + 125.94) = 143.60 kN-m Total shear force due to dead load, live load and impact load F = F1 + F2 = (23.55 + 188.91) = 212.46 kN Section modulus required (allowable stress in bending for unstiffened section, σb = 165 N/mm2) 6 ⎞ ⎛ Z = ⎜ 143.60 × 10 ⎟ = 870 × l03 mm3 165 ⎝ ⎠

From ISI Handbook No. 1 Select ISMB 400, @ 0.616 N/m Zxx = 1022.9 × 103 mm3, t w = 89 mm, h = 400 mm. Step 8. Check for shear 212.46 × 1000 = 59.68 N/mm2 400 × 8.9 < (040 × 250 = 100 N/mm2). Hence, safe.

τva.cal = (B) Design of Cross-girder Step 1. Effective span

= 0.98 m

9 9 ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟=⎜ ⎟ = 0.386 ⎝ 13.5 + L ⎠ ⎝ 13.5 + 9.80 ⎠

Impact factor

The maximum reaction on cross-girder occurs for the positions of loads as shown in Fig. 3.21. 5 7 kN

5 7 kN 3 .20 m

3 .20 m

3m

4 .30 m

3m

Fig. 3.21

Step 2. Maximum reaction on central cross-girder

1.80 ⎞ ⎛ = ⎜ 57.0 + 57.80 × ⎟ = 91.2 kN 3.00 ⎠ ⎝

128

DESIGN OF STEEL STRUCTURES–VOL. II

Step 3. Maximum bending moment The trains of IRC class A loading pass simultaneously on the carriageway. Therefore, there are four wheel loads, on the cross-girder. The maximum bending moment is caused for the positon of loads as shown in Fig. 3.22. 9 1.2 kN 2 .39 m

9 1.2 kN

1 .80 m

9 1.2 kN

1 .80 m

9 1.2 kN

1 .80 m

2 .39 m

9 .80 m 2 x 92 kN

2 x 91 .2 kN

Fig. 3.22

Maximum bending moment occurs at the centre 9.80 ⎡ 1.42 1.42 ⎞ ⎤ ⎛ − 91.2 × ⎜1.80 + − 91.2 × = ⎢2 × 91.2 × ⎟ = 600 kN-m 2 ⎝ 2 2 ⎠ ⎥⎦ ⎣

Maximum bending moment due to the live load and impact load = 1.386 × 600 = 831.6 kN-m Step 4. Maximum shear force Maximum shear force occurs when the position of loads is as shown in Fig. 3.23. 9 1.2 kN 1 .50 m

0 .4

9 1.2 kN

1 .80 m

9 1.2 kN

1 .42 m

9 1.2 kN

1 .80 m

9 .80 m

Fig. 3.23

1 [(9.80 – 1.90) + (9.80 – 3.70) 9.00 + (9.80 – 5.12) + (9.80 – 6.92)] = 200 kN Maximum shear force due to live lod and impact load = 1.386 × 200 = 277.2 kN. = 91.2 ×

DESIGN OF PLATE GIRDER BRIDGES

129

Step 5. Dead load on cross-girder Weight of slab over cross-girder

⎛ 3.0 × 9.80 × 250 × 24 ⎞ = ⎜ ⎟ 1000 ⎝ ⎠ = 176.4 kN The cross-girder supports five stringers. The stringers transmit the loads as concentrated loads at 2.45 m spacing. However, the loads are treated as uniformly distributed load. Weight of stringers

⎛ 5 × 616 × 3.0 ⎞ = ⎜ ⎟ = 9.34 kN 1000 ⎝ ⎠

Self-weight of stringers (assumed @ 3 kN/m) = 9.80 × 3 = 29.30 kN Total dead load = 215.04 kN Step 6. Maximum bending moment due to dead load

⎛ 215.04 × 9.80 ⎞ = ⎜ ⎟ = 263.424 kN-m 8 ⎝ ⎠ Step 7. Maximum shear force due to dead load

⎛ 215.04 ⎞ = ⎜ ⎟ = 107.52 kN ⎝ 8 ⎠ Total bending moment due to dead load, live load and impact load =

(263.424 + 831.6) = 1095.024 kN-m

Maximum shear force due to dead load, live load and impact load = (107.52 + 277.2) = 384.72 kN From, ISI Handbook No. 1 A built-up plate girder section as shown in Fig. 3.24 is provided for the crossgirder. Maximum allowable moment = 1112 kN-m >(Maximum BM) Maximum allowable shear = 907 kN > (Maximum shear force) Weight per metre = 2.097 kN/m < Assumed weight. Hence, safe. If desired, the plate girder section may be designed as discussed in Sec. 3.9.

130

DESIGN OF STEEL STRUCTURES–VOL. II

4 12 m m

2 IS A 2 00 m m x 1 00 m m x 15 m m 12 m m x

x 8 00 m m

2 IS A 2 00 m m x 1 00 m m x 15 m m

Fig. 3.24

Example 3.8 The effecive span of a through type plate girder two lane highway bridge is 30 m. The reinforced concrete slab is 250 mm thick inclusive of the wearing coat. The foot paths are provided on both sides of the carriageway. The cross-girders are provided at 3 m centres. The stringers are spaced at 2.45 m centre to centre. The spacing between main girders is 9.80 m. Design the maximkum section of plate girder, if the bridge if to carry IRC class A standard loading. Solution Design : Step 1. From IRC section I Width of roadway for single lane = 3.80 m For second lane = 3.00 m Width of foot-paths 2 × 15.0 = 3.00 m Total width = 9.80 m Cross-girder’s are used at 3 m spacing Number of panels = 10 There are five stringers in one panel Total number of stringers = 5 × 10 = 50 Number of cross-girders = 11 Step 2. Dead load Weight of reinforced concrete slab inclusive of wearing coat

⎛ 30 × 9.80 × 250 × 24 ⎞ = ⎜ ⎟ = 1764 kN 1000 ⎝ ⎠ Weight of stringers (assumed @ 1 kN/m) = 50 × 3 × 1 = 150 kN

DESIGN OF PLATE GIRDER BRIDGES

131

Weight of cross-girders (assumed @ kN/m) = 11 × 9.80 × 3 = 323.4 kN Self-weight of the plate girders by Fuller’s formula

⎛ 20L + 100 ⎞ ⎛ 20 × 30 + 100 ⎞ = ⎜ ⎟=⎜ ⎟ = 7 kN/m 100 ⎝ 100 ⎠ ⎝ ⎠ Total weight of plate girders Total dead load of the bridge

=

7 × 30 = 210 kN

= (1764 + 150 + 323.4 + 210) = 2447.4 kN Total dead load per girder = 1223.7 kN Step 3. Bending moment due to dead load The maximum bending moment M1, due to dead load occurs at the centre

⎛ 1223.7 × 30 ⎞ M1 = ⎜ ⎟ = 4588.88 kN-m 8 ⎝ ⎠ Step 4. Shear force due to dead load The maximum shear force F1, due to dead load occurs at the support 1 × 1223.7 = 611.85 kN 2 Step 5. Bending moment and shear force due to live load and impact load Road width between kerbs = 6.80 m Therefore, two trains of IRC class A loading which pass over the bridges simultaneously. Step 6. Transverse location of loading For the design of maximum section of the plate girders, the loads are so placed that the reaction on one plate girder is maximum. From IRC Section II Distance between centre to centre of two wheels of one train = 1.80 m The minimum distance between adjacent edges of wheels of two trains

F1 =

g =

91.2 − 0.4) ⎡ ⎤ × (6.8 – 5.5) ⎥ = 0.92 m. ⎢⎣0.40 + 2 ⎦

The distance between centre to centre of adjacent wheels of two trains = (0.92 + 0.60) = l.42 m The minimum clearance between outer edge of the wheel and the roadway face to kerb, for IRC class A vehicle f = 0.150 m The transverse location of wheels of trains are as shown in Fig. 3.25. Reaction on the plate girder A.

132

DESIGN OF STEEL STRUCTURES–VOL. II

P [(9.80– 1.80) + ( 7.90 – 1.80) + (6.10 – 1.42) + (4.68 – 1.80)] = 2.2 P 9.80 From IRC section II, impact fator for steel bridge

=

9 9 ⎛ ⎞ ⎛ ⎞ i = ⎜ ⎟=⎜ ⎟ = 0.207 ⎝ 1.35 + L ⎠ ⎝ 1.35 + 30 ⎠ Reaction on the plate girder A, including impact = 2.2 × 1.207P = 2.655 P where, P represents the one wheel load of IRC class A train of vehicle.

M in f = 0 .15 m

P

P

0.92 m 1 .42 m

1.5 0m

P

P

9 .80 m

Fig. 3.25

In order to account for the effect of both the trains of vehicles, and the impact effect, either the wheel loads of IRC class A vehicles may be increased or bending moment and shear force may be increased by multiplying by factor 2.655. Step 7. Longitudinal location of loading From IRC section II One train of wheel loads of IRC class A train of vehicles is as shown in Fig. 3.26. 9 .7 2 m

5 7 kN

3 4 kN

1 3 .5 kN 1 3 .5 kN 1 .1 m

5 7 kN 1 .2 3 .2 m m

3 4 kN 4 .3 m

3m

3 4 kN 3m

3 4 kN 3m

3 .4 9

O 15 m

15 m

R A = 1 2 1 .8 7 kN

R A = 1 5 5 .1 3 kN

Fig. 3.26

The centre of gravity of train of loading, c.g., is at a distance y from the front wheel and it is found by making moment about the same,

DESIGN OF PLATE GIRDER BRIDGES

133

⎡ 3.4(0 + 3 + 6 + 9) + 5.7(13.3 + 14.5) + 1.35(17.7 + 18.8) ⎤ y = ⎢ ⎥⎦ = 9.72 m. 4 × 3.4 + 2 × 5.7 + 2 × 1.35 ⎣ It is to note that the absolute maximum bending moment occurs under a wheel load, such that the centre of span, O is midway between the c.g. of load system and the wheel load under consideration. Thus, the absolute maximum bending moment M2, occurs under the fifth wheel load from the front, at X, as shwon in Fig. 3.26. Reaction RA =

1 – [34(3.49 + 6.49 + 9.49 + 12.49) 30

+ 57 (16.79 + 17.99) + 1.35 (21.19 + 22.29)] kN = 121.87 kN RB = 155.13 kN M2 = [121.87 × (15 – 1.79) – 57 × 12 – 135 × 4.4 – 13.5 × 5.5] kN-m = 1407.85 kN-m The absolute bending moment including effect of both the trains and impact = 2.655 × 1407.85 = 3737.85 kN The absolute maximum shear force occurs for the position of loading as shown in Fig. 3.27. 0 .37 5

5 7 kN 5 7 kN 1 .2 4 .3 m m

5 4 kN

5 4 kN 5 4 kN 5 4 kN 3 m

3 m

3 m

30 m

Fig. 3.27

Reaction,

RB =

1 [57(0.375 + 1.575) + 34(5.875 + 8.875 30

+ 11.875 + 14.875)] = 48.998 kN RA = (250 – 48.998) = 201.002 kN Maximum shear force including the effect of both the trains and impact = 2.655 × 201.002 = 533.66 kN Total bending moment due to dead load, live load and impact load M = M1 + M2 = (4588.88 + 3737.85) = 3826.73 kN-m Total shear force due to dead load, live load, and impact load F = F1 + F2 = (611.85+ 533.66) From ISI Handbook No. 1 Select a built-up plate girder as shown in Fig. 3.28.

134

DESIGN OF STEEL STRUCTURES–VOL. II

5 50 m m 20 m m 20 m m

2 IS A 20 0 m m x 1 50 m m x 18 m m

1 60 0 m m

16 m m

2 IS A 20 0 m m x 1 50 m m x 18 m m

20 m m 20 m m

Fig. 3.28

P

9 .80 m 6 .80 m P

1 .58 m

1.4 2 m

1 .50 m

1 .42 m

0 .4

P

1.8 0 m

= 874.3 kN-m > (Maximum moment) Maximum allowable shear = 2117 kN > (Maximum shear). Hence, safe. The maximum section of plate girder may be designed in detail if desired, as discussed in Sec. 3.9. Example 3.9 The effective span of a deck type plate girder two lane highway bridge is 30 m. The reinforced concrete slab is 250 mm thick inclusive of the wearing coat. The foot-paths are provided on either side of the carriageway. Design the maximum section of plate girder, if the bridge is to carry IRC class A loading. 1.8 0 m

Maximum allowable moment

P

1 .8 m

1 .2 m 3m

Fig. 3.29

1 .50 m 1 .50 m

0 .4

DESIGN OF PLATE GIRDER BRIDGES

135

Solution Design : Step 1. From IRC Section I Width of roadway for single lane = 3.80 For second lane = 3.00 m Width of foot-paths 2 × 1.50 = 3.00 m Total width = 9.80 m Provide four plate girders as shown in Fig. 3.29. The intermediate plater girder B or C is subjected to maximum load. Therefore, the section of intermediate plate girder C is designed and the same section is provided for all the four plate girders. Step 2. Bending moment and shear force due to live load and impact Road width between kerbs = 6.80 m Therefore, two trains of IRC class A loading pass over the bridge simultaneously. Step 3. Transverse location of loading For the design of maximum section of the intermediate plate girder the loads are so placed that the reaction on one plate girder is maximum. From IRC Section II Distance between centre to centre of two wheels of one train = 1.80 m The minimum distance between adjacent edges of wheels of two trains g =

(1.2 − 0.4) ⎡ ⎤ × (6.8 − 5.5) ⎥ = 0.92 m ⎢⎣0.40 + 2 ⎦

The distance between centre to centre of adjacent whels of two trains = (0.92 + 0.50) = 1.42 m The minimum clearance between outer edge of the wheel and the roadway face of kerb, for IRC class A vehicle, f = 0.150 m The transverse location of wheels of trains is as shown in Fig. 3.29. Reaction on plate girder C,

1.2 1.42 ⎤ ⎡ = 1.873P = P ⎢1.0 + + 2 3 ⎥⎦ ⎣ From IRC Section II, impact factor for steel bridges

9 9 ⎛ ⎞ ⎛ ⎞ i = ⎜ ⎟=⎜ ⎟ = 0.207 ⎝ 13.5 + L ⎠ ⎝ 13.5 + 30 ⎠ Reaction on plate girder C, including impact = 1.873 × 1.207 P = 2.26 P where, P represents the one wheel load of IRC class A train of vehicle.

136

DESIGN OF STEEL STRUCTURES–VOL. II

In order to account for the effect of both the trains of vehicles, and the impact effect, either the wheel loads of IRC class A vehicles may be increased or bending moment and shear force may be increased by multiplying them by factor 2.26. Step 4. Longitudinal location of loading Effective spans in Example 38 and Example 39 are equal. Therefore, the longitudinal locations for maximum bending moment and for maximum shear force for both the examples are same and these are shown in Figs. 3.26 and 3.27. From Example 3.8 Maximum bending moment = 1407.85 kN-m Maximum shear force = 200.002 kN The maximum bending moment including effect of both the trains and impact = 2.26 × 1407.85 = 3181 kN-m The maximum shear force including effect of both the trains and impact = 2.26 × 200.002 = 452.004 kN Step 5. Dead load bending moment and shear force Weight of reinfoced slab inclusive of wearing coat

⎛ 30 × 3 × 25 × 24 ⎞ = ⎜ ⎟ = 540 kN 1000 ⎝ ⎠ Self-weight of the plate girders by Fuller’s formula

⎛ 20L + 100 ⎞ ⎛ 20 × 30 + 100 ⎞ = ⎜ ⎟=⎜ ⎟ = 7 kN/m 100 ⎝ 100 ⎠ ⎝ ⎠ Weight of one plate girder for complete span =

1 × 7 × 30 = 105 kN 2 (540 + 105) = 645 kN

Total dead load

=

Bending moment

⎛ 645 × 30 ⎞ = ⎜ ⎟ = 2418.75 kN-m 8 ⎝ ⎠

1 × 645 = 322.5 kN 2 Total bending moment due to dead load, live load and impact load M = (3181.74+ 2418.75) = 5600.49 kN-m Total shear force due to dead load, live load and impact load F = (452.004 + 322.5) = 774.504 kN From ISI Handbook No. l Shear force

=

137

DESIGN OF PLATE GIRDER BRIDGES

Select a built-up plate girder as shown in Fig. 3.30. Maximum allowable moment = 5747 kN-m > (Maximum moment) Maximum allowable shear = 1890 kN > (Maximum shear) Hence, safe. The maximum section may be designed in detail if desired as discussed in Sec. 3.9.

3.15

WIND LOAD ON PLATE GIRDER BRIDGES

The wind load on plate girder bridges is determined as the product of appropriate basic wind pressure and exposed area of the bridge. The appropriate basic wind pressure for highway and railway, loaded and unloaded bridges have been discussed in Sec. 2.5. The exposed area of unloaded and loaded plate girder bridges are as follows :

3.15.1

For Unloaded Plate Girder Bridges

The exposed area of unloaded span of the bridge consists of area of windward girder and a part of leeward girder depending upon the height (depth) and spacing of the main girders. The exposed part area of leeward girder is expressed as fraction of area of windward girder and added to the area of windward girder. The factor k, to account for the part of exposed area of leeward girder as per bridge rule is mentioned below : S. No.

Conditions

I> |

1. 1 2

1 2

D

Factor, k k = 0.00

D I ≥ 2 D

k = 1.00

2.

where,

s = spacing between the main girder h = height (depth) of the main girder. Exposed area of uncloaded plate girder bridge = (1 + k) × Area of windward girder Area of windward girder is the area in elevation i.e., the product of height of girder and length of girder.

3.15.2

For Loaded Plate Girder Bridge

The exposed area of loaded span of bridge consists of area of windward girder plus the part of leeward girder not covered by the moving load multiplied by the factors as shown above plus the area of moving load.

138

DESIGN OF STEEL STRUCTURES–VOL. II

In the case of railway bridges the area of moving load is taken as from 600 mm abvoe the rail level to the top of the highest stock for which the bridge is designed. The dimensions of rolling stocks for broad gauge and metre gauge have been shown in Fig. 1.21 and Fig. 1.22 respectively in Sec. 1.10. The height of rolling stock above 600 mm from the rail level upto the top is 3500 mm in case of broad gauge and 2830 mm in case of metre gauge. The entire span of the bridge is assumed to be occupied by the moving load. Thus, the area of moving load is 3.50 m × span of bridge in metres in case of broad gauge and 2.830 m × span of bridge in metres in case of metre gauge. The wind load acts at the centre of gravity of area of moving load. 5 00 m m 20 m m 20 m m

2 IS A 20 0 m m x 1 00 m m x 15 m m

16 m m

1 25 0 m m

2 IS A 20 0 m m x 1 50 m m x 18 m m

20 m m 20 m m

Fig. 3.30

In the case of highway bridges the clear distance between the trailer of a train of vehicle shall not be omitted. The lateral wind load against any exposed moving load is assumed to have a value of 3 kN per linerar metre for the ordinary highway bridges and 4.50 kN per linear metre for the highway bridges carrying tramway. This load is assumed to act at 1.50 m above the road surface. The total assumed wind load on highway bridges shall, however, not be less than 4.50 kN per linear metre in the plane of the loaded flange in case of deck type bridges and 2.25 kN per linear metre in case of through type bridges. In the case of foot bridges, the height of moving load is to be taken as 2 m throughout the length of span.

139

DESIGN OF PLATE GIRDER BRIDGES

3.16

WIND EFFECT ON PLATE GIRDER BRIDGES

The wind effects on plate girder bridges are considered on the unloaded span, and on the loaded spans of the bridges, and the worst effects are taken into consideration and accounted for properly. The wind load has following two distinct effects on the plate girder bridges: 1. Overturning effect 2. Horizontal truss effect.

3.16.1

Overturning Effect

The wind load P, on plate girders is determined as discussed in Sec. 3 15. The wind load P, acting on unloaded deck type plate girder bridge is shown in Fig. 3.31. The wind load is assumed to be acting at half the height of the bridge structure (i.e. height of the girder plus sleepers plus rails). The wind load has the effect of overturning the bridges about the lower flange of the leeward girder.

P

h h 2 2R

S = (S pa cing )

2R

Fig. 3.31

As a result of overturning effect, an additional thrust 2R is caused over the leeward girder, and an uplift, 2R is caused on the windward girder. These equal forces form a resisting couple. The additional thrust 2R, causes reaction of 2R (R, R at each end of the girder), as shown in Fig. 3.31. The additional thrust, 2R due to overturning has the tendency to increase the bending compressive stress already existing in the upper flange and the bending tensile stress already existing in the lower flange of the’leeward girder. Whereas, the uplift 2R, due to overturning in the windward girder causes the opposite kind of stresses in the flanges and, therefore, relieves the upper flange and lower flange from the already existing stresses. The value of reaction R, may be found by taking the moment of wind load about the bottom flanges.

140

DESIGN OF STEEL STRUCTURES–VOL. II

2R × s =

P.

h 2

⎛1 P ⋅h ⎞ R = ⎜ × ⎟ s ⎠ ⎝4

...(3.11)

The value of reaction, R is expressed in term of percentage of dead load reaction. Since the tendency of additional thrust over the leeward girder is similar to the action of dead load, the percentage increases of bending compressive

R o lling sto ck

(3 .50 m for B .G .) or (2 .83 0 m fo r M .G .)

(Fo r M .G . 1.41 5 m ) or (Fo r B .G . 1 .7 5 m )

0 .60 m

h1

P

h

h2

2 R1

S = (S p acing )

2 R1

Fig. 3.32

stress in the upper flange and bending tensile stress in the lower flange of leeward girder are found directly from the stresses due to dead load only. The nature of increase of stresses due to overturning is additive in the already existing stresses in the upper and lower flanges of leeward girder.

141

DESIGN OF PLATE GIRDER BRIDGES

The wind load P´, acting on the loaded deck type plate girder bridge (broad gauge/metre gauge) and wind load P1´, acting on the moving load are shown in Fig. 3.32. The wind loads acing on plate girders and moving loads are determined as discussed in Sec. 3.15. The wind load acting on the plate girders is assumed to be acing at half the height of girder. The wind load on moving load acts at the centre of moving stock as shown in Fig. 3.32. Both these wind loads P´ and P1´ have the effect of overturning the bridge about the lower flange of the leeward girder. As a result of the overturning effect, and additional thrust 2R1 is caused over the leeward girder and an uplift 2R1 is caused on the windward girder. These equal and opposite forces furnish a resulting couple. The additional thrust 2R, causes reaction 2R1 (R1,R1 at each end of the girder), as shown in Fig. 3.32. The additional thrust 2R, due to overturning also has the tendency to increase the bending compressive stress already existing in the upper flange and the bending tensile stress already existing in the lower flange of the leeward girder. Whereas the uplift 2R1 due to overturning in the windward, girder causes the opposite nature of stress in the flanges, and, therefore, relieves the upper and lower flanges from the already existing stresses. The value of reaction R1, may be found taking the moment of wind load P´ and P1´ about the bottom flanges

⎛ h ⎞ 2R1 × s = ⎜ P´ + P1´ h1 ⎟ ⎝ 2 ⎠ ´ ⎛ ⎞ R1 = ⎜ 1 P ´h + P1 h1 ⎟ 2s ⎠ ⎝4 s

...(3.12)

The value of reaction R1 is mentioned in terms of percentage of dead load, live load and impact load reaction determined from the loads found for calculating bending moment. Since, the tendency of additional thrust over the leeward girder

h h 2

2R

S = (S p acing )

Fig. 3.33

2R

142

DESIGN OF STEEL STRUCTURES–VOL. II

is similar to the action of dead load, live load and impact load, the percentage increase of bending compressive stress in the upper flange and bending tensile stress in the lower flange of leeward girder are determined directly from the stresses due to normal loads (i.e. dead load, live load and impact load) only. The nature of increase of increase of stresses due to over turning is additive in the already existing stresses in the upper and lower flanges of leeward girder. The wind loads acting on unloaded and loaded through type plate girder bridges are shown in Fig. 3.33 and Fig. 3.34.

P1

R o lling sto ck

0 .60 m

h1 P

h

h2

Fig. 3.34

The overturning effect due to wind over the through type plate girder bridge is same as over the deck type plate girder bridge (as discussed above).

3.16.2

Horizontal Truss Effect

The horizontal truss bracing is provided in the loaded flange of the plate girder bridge. The horizontal bracing consists of chords, laterals (diagonals), and floor beams. In case of deck type bridge, the top (loaded) flanges of the plate girder act as chords and separate floor beams and diagonals are used so as to form approximate square panels. In case of through type bridge, the bottom (loaded) flange of the plate girders act as chords. The laterals (diagonals) are connected with the floor beams in case of the through type bridge. These laterals (diagonals in the horizontal truss bracing may be single or double or K-type as shown in

DESIGN OF PLATE GIRDER BRIDGES

143

Fig. 3.35 (a), (b) and (d) respectively. The cross-diagonal or double lateral bracing are most commonly used. The double diagonal bracing of a through type plate girder bridges is shown in Fig. 3.3. p = In te nsity o f w in d loa d pe r u nit len gth S (a ) p = In te nsity o f w in d loa d pe r u nit len gth S (b ) p = In te nsity o f w in d loa d pe r u nit len gth S

Fig. 3.35

The horizontal truss bracing resists the lateral loads. It is to note that the wind load is only considered as lateral load, and the racking force is not taken into acccount while calculating increase of stresses in the flanges of the plate girders. The wind load acts as uniformly distributed load in the horizontal plane of the loaded flange of the girder. The maximum wind load from loaded and unloaded bridge is considered. The horizontal truss bracing acts as a simply supported girder. It is subjected to the maximum bending moment at the centre.

⎛ P ⋅l ⎞ M = ⎜ ⎟ ⎝ 8 ⎠ where,

...(i)

P = Total wind load l = Span in metre Let F, F be the two equal and opposite forces acting in the flanges of the main girders due to the horizontal truss effect. From the steel theory, these form a resisting couple, M1. M1 = F × s ...(ii) where, s = Spacing between the main girders The resisting couple M1, and the maximum bending moment M, are equal. Therefore,

⎛ P ⋅l ⎞ F×S = ⎜ ⎟ ⎝ 8 ⎠

144

DESIGN OF STEEL STRUCTURES–VOL. II

⎛ P ⋅l ⎞ F = ⎜ ⎟ ⎝ 8⋅s ⎠

...(3.13)

The intensity of stress in a flange due to horizontal truss effect f =

F A

... (3.14)

where, A = Appropriate cross-sectional area of one flange of the main girder. The nature of stress due to hroizontal truss effect is compressive in the windward girder and tensile in the leeward girder. In case of the deck type plate girder bridges, the stress due to horizontal truss effect in the top flange of the leeward girder is tensile and therefore, relieves the flange from existing compressive bending stresses. In case of the through type plate girder bridges, the stress due to horizontal truss effect in the bottom flange of leeward girder in tensile and therefore, additive. In through type plate girder bridges, the overturning effect and the horizontal truss effect, both are additive for the bottom flange of the leeward girder. The overall stresses in flange of the plate girder due to dead load, live load, impact load and wind load should be less than 1.25 times and allowable bending stresses in the flanges. In case the overall stresses exceed, then either the spacing between main girders is increased or the section of main girder is redesigned. It is to note in case the effective spans are upto 20 m, it is not necessary to determine the wind stresses. Example 3.10 The effecive span of a deck type plate girder railway bridge for single metre gauge track is 24 m. The depth of plate girder is 1864 mm. The spacing between the plate girder is 1.30 m. The rail level is 400 mm above the top of the plate girders. The design reaction is 750 kN. The net area of tension flange is 19652 mm2 and the gross area of compression flange is 23304 mm2. The moment of inertia of the plate girder section about xx-axis is 3749172 × 104 mm4. Determine the increase of stresses in the flanges of leeward girder in the following cases : (a) Overturning effect due to wind, when the bridge is unloaded. (b) Horizontal truss effect due to wind, when the bridge is unloaded. (c) Overturing effect due to wind, when the bridge is loaded. (d) Horizontal truss effect due to wind when the bridge is loaded. Solution The deck type plate girde bridge, when the bridge is unloaded is shown in Fig. 3.36. The height of deck type plate girder bridge structure, including sleepers and rails = (1864 + 400) = 2264 mm.

145

DESIGN OF PLATE GIRDER BRIDGES

1 .0 m

M etre ga ug e track 0 .40 0 m

2 .26 4 m P 1 .86 4 m 1 .13 2 m

Fig. 3.36

Half the height of bridge structure 1 × 2264 = 1332 mm 2 Step 1. Bridge unloaded (overturning effect) Intensity of wind pressure = 2.40 kN/m2 Wind force on windward plate girder

=

⎛ 2.40 × 2264 × 24 ⎞ = ⎜ ⎟ = 130. 41 kN 1000 ⎝ ⎠ Spacing between main girders, s = 1.30 m Depth of the main girder = 1864 mm The spacing between main girders is greater than half the depth, and less than the full depth. Therefore, factor to account for the wind effect on leeward girder = 0.25. Wind force on leeward plate girder ⎛ 2.40 × 2264 × 24 ⎞ = 0.25 × ⎜ ⎟ = 32.60 kN 1000 ⎝ ⎠ Total wind force, P = (130.41 + 30.60) = 163.01 kN The wind pressure is assumed to act as half the depth of bridge structure, i.e., at 1.132 m as shown in Fig. 3.36. Let the reaction at each end of the leeward girder be R. The total reaction on the leeward girder due to overturning is 2R.

146

DESIGN OF STEEL STRUCTURES–VOL. II

Taking moment about the bottom of girders 2R × 1.30 = 163.01 × 1.132 R = 70.97 kN Due to overturning effect, a uniformly distributed load 2 × 70.97 = 141.94 acts downward on the leeward girder. The maximum bending moment, M occurs at centre

⎛ 141.94 × 24 ⎞ M1 = ⎜ ⎟ = 425.82 kN-m 8 ⎝ ⎠ Distance to the extreme fibre of plate girder from the neutral axis y =

1 × 1864.0 = 932.0 mm 2

Increase of stress (above the stress in flanges due to dead load only in the compression flange, of the leeward girder)

⎛ 425.82 × 106 × 932.0 ⎞ ⎟ = 10.58 N/mm2 σbc1 = ⎜ ⎝ 3749172 × 104 ⎠ Increase of stress in the tension flange of the leeward girder

23304 ⎞ ⎛ 2 σbt = ⎜10.58 × ⎟ = 12.55 N/mm . 19652 ⎠ ⎝ Step 2. Bridge unloaded (Horizontal truss effect) Total wind pressure acting on the bridge structure P = 163.01 kN The horizontal truss bracing is provided in between the compression (loaded) flanges. The total wind force is assumed to act in the plane of horizontal truss. The horizontal truss acts as horizontal girder. The maximum bending moment =

P ⋅ L ⎛ 163.01 × 24 ⎞ =⎜ ⎟ = 489.03 kN-m 8 8 ⎝ ⎠

The bending moment is resisted by two equal and opposite forces F in the compression flanges of both the girders F × 1.30 = 489.03, F = 376.177 kN The force F is tensile in the compression flange of the leeward girder. It causes tensile stress only in the compression flange of the leeward girder. Therefore, the decrease of stress in the compression flange of leeward girder 3 ⎛ σ = ⎜ 376.177 × 10 23304 ⎝

⎞ 2 ⎟ = 16.14 N/mm ⎠

DESIGN OF PLATE GIRDER BRIDGES

147

P F 1 .30 m F 1 6 p an e ls @ 1.5 m = 2 4 m

Fig. 3.37

Step 3. Bridge loaded (overturning effect) Maximum intensity of wind in case of metre gauge = 1.00 kN/m2 Wind load on moving train, P1 = 1.00 × 24 × 2.830 = 67.92 kN Height of line of action of wind load P1, above the bottom of plate girder = (2.264 + 0.60 + 1.415) = 4.279 m Wind load on the deck type bridge structure (windward girder and leeward girder both) P2 = 1.00 × 24 × 2.264 × 1.25 = 67.92 kN Height of line of action of wind load P2, above the bottom of the plate girders = 1.132 m. Taking moment about the bottom of the plate girders. 2R × 1.30 = 67.92 × 4.279 + 67.92 × 1.132 R = 141.35 kN Due to overturning effect, of uniformly distributed load 2 × 141.35 = 282.70 kN acts downward on the leeward girder, The maximum bending moment M2 occurs at the centre

⎛ 282.70 × 24 ⎞ M2 = ⎜ ⎟ = 848.1 kN-m 8 ⎝ ⎠ Increase of stress (above the existing stress in the flange due to dead load, live load and impact laod) in compression flange of the leeward girder

⎛ 848.1 × 106 × 932.0 ⎞ σbc2 = ⎜ ⎟ = 21.108 N/mm2 ⎝ 3749172 × 104 ⎠ Increase of stress in the tension flange in the leeward girder

23304 ⎞ ⎛ 2 σbc2 = ⎜ 21.02 × ⎟ = 25 N/mm . 19652 ⎝ ⎠

148

DESIGN OF STEEL STRUCTURES–VOL. II

(M etre ga ug e) R o lling sto ck

P1

2 .83 0 m

1 .41 5 m

0 .60 m 4 .27 9 m

P2 2 .26 4 m 1 .86 4 m 1 .13 2 m

1 .30 m

Fig. 3.38

Step 4. Bridge loaded (Horizontal truss effect) The wind loads acting on the moving train and the deck type bridge structure, both are assumed to act in the horizontal plane of truss bracing. Total wind load = (P1 + P2) = (67.92 + 67.92) = 135.84 kN P1 + P2

F 1 .30 m F R

R

Fig. 3.39

For unloaded bridge, the wind load = 163.01 kN Tensile force in the compression flange of leeward girder

149

DESIGN OF PLATE GIRDER BRIDGES

F = 376.177 kN Therefore, for loaded bridge, the tensile force in the compression flange of the leeward girder

135.84 ⎞ ⎛ F = ⎜ 376.177 × ⎟ = 313.477 kN 163.01 ⎠ ⎝ It causes tensile stress only in the compression flange of the leeward girder. Therefore, the decrease of stress in the compression flange of the leeward girder ⎛ 313.477 × 103 ⎞ σb = ⎜ ⎟ = 13.45 N/mm2. 23304 ⎝ ⎠ Example 3.11 The effective span of a through plate girder railway bridge for a single broad gauge track is 30 m. The depth of plate girder is 2664 m. The spacing between the plate girders is 5.0 m. The rail level is 800 mm above the bottom of the girders. The net area of flange is 27986 mm2 and the gross-area of compression flange is 31983 mm2. The gross-moment ofinteria of the plate girder section about xx-axis is 10761879.6 104 mm4. Determine the increase of stresses in the flanges of the leeward girder in the following cases : (a) Overturning effect due to wind, when the bridge is unloaded. (b) Horizontal truss effect due to wind, when the bridge is unloaded. (c) Overturning effect due to wind when the bridge is loaded. (d) Horizontal truss effect due to wind, when the bridge is loaded. Also determine the percentage stress increment in case (c), if the dead load, live load and impact load reaction is 1210 kN. Solution The through type plate girder bridge, when the bridge is unloaded is shown in Fig. 3.40. Height of the plate girders = 2664 mm Half the height of plate girders = 1332 mm

B ro ad g au ge track

P

2 .66 4 m

1 .33 2 m

2R

0 .8 m

5 .0 m

Fig. 3.40

2R

150

DESIGN OF STEEL STRUCTURES–VOL. II

Step 1. Bridge is unloaded (overturning effect) Intensity of wind pressure = 2 .40 kN/m2 Wind pressure on windward plate girder

⎛ 2.40 × 2664 × 30 ⎞ = ⎜ ⎟ = 191.81 kN. 1000 ⎝ ⎠ Spacing between the main girders, s = 5.00 m Depth of the plate girders = 2.664 m The spacing between main girders is greater than one and a half times the depth of plate girder and less than twice its depth. Therefore, factor to account for the wind effect on the leeward girder, k = 1.00 Wind pressure on the leeward girder

⎛ 2.40 × 2664 × 30 ⎞ = 1.00 × ⎜ ⎟ = 191.81 kN 1000 ⎝ ⎠ Total wind pressure, P = (191.81 + 191.81) = 383.62 kN The wind pressure is assumed to act at half the depth of plate girder, i.e., at 1332 mm as shown in Fig. 3.40. Let the reaction at each end of the leeward girder be R. The total reaction on the leeward girder due to overturning is 2R. Taking moment about the bottom girders 2R × 5.00 = 383.62 × 1332, R = 51.1 kN Due to overturning effect, a uniformly distributed load 2 × 51.1 = 102.2 kN acts downward on the leeward girder. The maximum bending moment M1 occurs at the centre

⎛ 102.2 × 30 ⎞ M1 = ⎜ ⎟ = 383.25 kN-m 8 ⎝ ⎠ Distance to the extreme fibre of the plate girder from the neutral axis =

1 × 2664 = 1332 mm 2

Increase of stress (above the stress in flanges due to dead load only) in the compression flange of the leeward girder

⎛ 383.25 × 106 × 1332 ⎞ ⎟ = 4.74 N/mm2 σbc1 = ⎜ ⎝ 10761879.7 × 104 ⎠ Increase of stress in tension flange in leeward girder

DESIGN OF PLATE GIRDER BRIDGES

151

31983 ⎞ ⎛ σbc1 = ⎜ 4.74 × ⎟ = 5.42 N/mm2. 27986 ⎠ ⎝ Step 2. Bridge unloaded (Horizontal truss effect) The horizontal truss bracing is provided in between the tension flanges (loaded flanges). The total wind pressure is assumed to act in the plane of horizontal truss. The horizontal truss acts as a horizontal girder. The maximum bending moment

⎛ 383.62 × 30 ⎞ M = ⎜ ⎟ = 1438.575 kN-m 8 ⎝ ⎠ P F 5 .0 m F 6 pa ne ls @ 5 m = 3 0 m

Fig. 3.41

This bending moment is resisted by two equal and opposite forces, F in the tension flanges of both the girders due to moment, M. F × 5.0 = 1438.575, F = 287.715 kN This force is tensile in the tension flange of the leeward girder. It causes tensile stress, only in the tension flange of leeward girder. Therefore, the increase of stress ⎛ 287.715 × 103 ⎞ σ = ⎜ ⎟ = 10.28 N/mm2 27986 ⎝ ⎠

Step 3. Bridge loaded (overturning effect) Maximum intensity of wind in case of broad gauge = l.50 kN/m2 Wind load on moving train P1 = 150 × 30 × 350 = 157.50 kN Height of line of action of wind load P1 above the bottom of plate girders = (0.80 + 0.60 + l.75) = 3.15 m Wind load on windward and leeward girder both, P2 = 150 × 30 × 2.664 × 2.0 = 239.76 kN Height of line of action of wind load P2 above the bottom of plate girders = 1.332 m

152

DESIGN OF STEEL STRUCTURES–VOL. II

B ro ad g au ge ro lling sto ck

P1

3 .50 m

1 .75 0 m

0 .60 m P2

2 .66 4 m

1 .33 2 m

5 .0 m

2R

2R

Fig. 3.42

Taking the moment above the bottom of plate girders 2R × 5.0 = 157.50 × 3.15 + 239.76 × l.332 R = 81.55 kN Due to overturning effect, a uniformly distributed load 2 × 81.55 =163.10 kN acts downward on the leeward girder. The maximum bending moment M2, occurs at the centre

⎛ 163.10 × 30 ⎞ M2 = ⎜ ⎟ = 611.625 kN-m 8 ⎝ ⎠ Increase of stress (above the existing stress in the flange due to dead load, live load and impact lod) in compression flange in the leeward girder σbc2

⎛ 611.625 × 106 × 1332 ⎞ ⎟ = 7.57 N/mm2 = ⎜ ⎝ 10761879.6 × 104 ⎠

Increase of stress in the tension flange of the leeward girder

31983 ⎞ ⎛ 2 σbc2 = ⎜ 7.57 × ⎟ = 8.745 N/mm 27686 ⎠ ⎝

DESIGN OF PLATE GIRDER BRIDGES

153

Step 4. Bridge loaded (Horizontal truss effect) The wind loads acting on the moving train and the through type bridge structure both are assumed to act in the horizontal plane of truss bracing. Total wind load = (P1 + P2) = (157.50 + 239.76) = 397.26 kN For unloaded bridge, the wind load = 383.62 kN Tensile force in the tension flange = 287.715 kN Therefore, for the loaded flange, the tensile force in the tension flange of the leeward girder F = 287.715 ×

397.26 = 297.945 kN 383.62

P1 + P2 F 5 .0 m F 6 pa ne ls @ 5 m = 3 0 m

Fig. 3.43

It causes the tensile stress only in the tension flange of the leeward girder. Therefore, the increase of stress in the tension flange of leeward girder ⎛ 297.945 × 103 ⎞ F = ⎜ ⎟ = 10.646 N/mm2. 27986 ⎝ ⎠ It is to note that due to overturning effect and horizontal truss effect both, the stress in tension flange of a leeward girder of through type bridge increases. Step 5. Reaction due to overturning effect in case the bridge is loaded at the end of leeward girder = 81.55 kN Dead load, live load and impact load reaction = 1210 kN The increase of stress due to overturning may be expressed as percentage increase of stress due to dead load, live load and wind load. The percentage incremental stress is directly proportional to the reactions in these two cases. Therefore, pecentage incremental stress

⎛ 81.55 ⎞ = ⎜ × 100 ⎟ = 6.74 percent. ⎝ 1210 ⎠

3.17

DESIGN OF HORIZONTAL TRUSS BRACING

The cross or double diagonal horizontal truss bracing as shown in Fig. 3.44 is most commonly provided in the horizontal plane of the loaded flange of the main girders. The horizontal truss bracing is designed to resist the lateral loads (wind load, racking force, and centrifugal force, if any). The wind load whichever is maximum in case of loaded or unloaded spans of the bridge, is taken into consideration.

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DESIGN OF STEEL STRUCTURES–VOL. II

L ate ra l loa d p e r un it le ng th

R1

R1

Fig. 3.44

In case of deck type bridge, the laterals are connected so as to form approximate squares and uniform panels throughout the span. In case of through type bridge, the diagonals are connected with the floor beams. For the analysis, it is assumed that the diagonal members subjected to tension remain active and others remain dummy for one direction of wind. In case, the direction of wind is reversed, the active diagonals become dummy, and the dummy diagonals become active. The end struts (or floor beams) are subjected to maximum compression and the end diagonals are subjected to maximum tension. The forces intermediate struts (floor beams) and intermediate diagonals are comparatively less. The angle sections are provided for the struts and the diagonals. The forces in end members are small and require nominal size of sections only. The same size of angle sections are provided for the intermediate members.

3.18

DESIGN OF CROSS-FRAMES

The cross-frames are provided as the transverse bracing for the deck type bridges at all panel points in the vertical plane. Figure 3.45 shows an end cross-frame. The end cross-frame is subjected to maximum lateral load. The cross-frame is designed for lateral load for the loaded

P1

h

θ S

Fig. 3.45

or unloaded bridge whichever is maximum less the corresponding force at the end panel of horizontal truss bracing. The diagonal member of cross-frame subjected to tension is assumed to be effective member. The force in diagonal

155

DESIGN OF PLATE GIRDER BRIDGES

member is R1 sec θ, where R1 is the reaction at the end of horizontal truss. This force depends on the type of truss bracing used and assumption made. The diagonal member is designed as the tension member. The angle section is used for the diagonal members.

3.19

DESIGN OF INTERNAL GUSSET PLATE

The internal gusset plates are provided as the transverse bracing for the through type plate girder bridges at all the panel points in the vertical plane. The internal gusset plates transmit lateral forces acing on the top (compression) flanges of the plate girders. The internal gusset plates used in the through type bridge are shown in Fig. 3.46. The internal guset plates are connected with the beams and the main girders. P

In te rna l g u sset p late

h

Fig. 3.46

Example 3.12 Design the horizontal truss bracing and cross-frames for the deck type plae girder railway bridge for a single metre gauge track as in Example 3.10. Solution (A) Design of Horizontal truss bracing Step 1. The horizontal truss bracing with cross-diagonals as shown in Fig. 3.47 is provided between the loaded flanges (top or compression flanges) only. The diagonals which carry tension, remain active. Tota l la teral Lo ad = P H

1 30 m 1 .50 m

1 .50 m 1 6 p an e ls @ 1 .5 m = 2 4 m

Fig. 3.47

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DESIGN OF STEEL STRUCTURES–VOL. II

From Example 3.10 When the bridge is unloaded, the wind load acting on the bridge structure (inclusive of windward and leeward girder) is 163.01 kN. Step 2. When the bridge is loaded Wind load on moving train = 67.92 kN Wing load on the bridge structure = 67.92 kN Racking force, 6.00 × 24 = 144.0 kN Total lateral load, PH = 279.84 kN The lateral load acing in the plane of horizontal truss bracing is maximum in case the bridge is loaded. Therefore, the horizontal truss bracing is designed for the lateral load = 279.84 kN Reaction at the end = 139.92 kN Spacing between the main girder = l.30 m Span = 24 m 16 panels @ 15 m are provided in the horizontal truss bracing so that each panel is approximaely square. Lateral load at each intermediate panel

⎛ 279.84 ⎞ P = ⎜ ⎟ = 17.49 kN ⎝ 16 ⎠ Lateral load at end panel = 8.745 kN End strut carries maximum compression = 139.92 kN Shear force in the end panel F = (139.92 – 8.745)= 131.175 kN tan θ =

1.30 = 0.8626, ∴ θ = 40° 47' 1.50

sin θ = 0.653, cosec θ = 1.532 Force in the end diagonal = (131.175 × 1.532) = 200.96 kN End strut Effective length of end strut = 1.30 m Compresson in the end strut = 139.92 kN The slenderness ratio for the double angle compression member and the value of yield stress for steel may be assumed as 100 and 260 N/mm2. The permissible stress in axial compression σac = 82 N/mm2 Cross-sectinal area required =

⎛ 139.92 × 1000 ⎞ 2 ⎜ ⎟ = 1706.34 mm 82 ⎝ ⎠

Provide 2 ISA 100 mm × 65 mm × 6 mm with long legs connected.

DESIGN OF PLATE GIRDER BRIDGES

157

Cross-sectional area provided = 2 × 955 = 1910 mm2 From IS 833–94, the permissible stress in axial compression = 2 × 955 = 1910 mm2 rxx = 31.8 mm, ryy = 25.5 mm, ∴ rmin = 25.5 mm

⎛ 1.30 × 1000 ⎞ = ⎜ ⎟ = 50.98 25.5 ⎝ ⎠ 2 σac = 135 N/mm The safe load carrying capcity is more. Hence, safe. End diagonal Force in end diagonal = 200.96 kN As per IS : 833–94, allowable stress in axial tension = 0.60 × 260 = 156 N/mm2 Net cross-sectional area required

Slenderness ratio

⎛ 200.96 × 1000 ⎞ 2 = ⎜ ⎟ = 1288.21 mm 156 ⎝ ⎠ Allowing area for rivet holes = 460 mm2 Gross area required = 1748.21 mm2 Provide 2 ISA 100 mm × 65 mm × 6 mm for the diagonal membes with the short leg connected together. Gross area provided = 1910 mm2 Deduction for rivet hole (assuming that 22 mm diameter rivets are used) = 2 ×6 × 23.5 = 282 mm2 Net area provided = 1628 mm2 > Net are required. Hence, safe. 2 ISA 100 mm × 65 mm × 6 mm are provided for all struts and all diagonal members. (B) Design of Cross-frames The cross-frames are provided between the windward girder and the leeward girder in the vertical plane and at all the plane points of horizontal truss bracing. Figure 3.48 shows an end cross-frame. When the bridge is loaded, then the lateral load acting in plane of the horizontal truss is maximum. End reaction in the horizontal truss bracing = 139.92 kN It is assumed that the diagonal member carrying tension remains active member. Length of the diagonal member = (1.802 + 1.302)1/2 = 2.22 m ⎛ 1.30 ⎞ cos θ = ⎜ ⎟ = 0.585 ⎝ 2.22 ⎠ ⎛ 139.92 ⎞ Force in the diagonal member = ⎜ ⎟ = 239.18 kN ⎝ 0.585 ⎠

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DESIGN OF STEEL STRUCTURES–VOL. II

F

0 .32 m θ

θ

1 .80 m

0 .32 m 1 .30 m

Fig. 3.48

Allowable stress in axial tension =156 N/mm2 Net cross-sectional area required

⎛ 23.18 × 1000 ⎞ 2 = ⎜ ⎟ = 1533.20 mm 156 ⎝ ⎠ Allowing area for rivet hole = 300 mm2 Gross-area required = 1833.20 mm2 Provide 2 ISA 100 mm × 65 mm × 8 mm for the diagonal members with the short legs connected together. Gross cross-sectional area provided = 2514 mm2 Reduction for rivet hole =2 × 8 × 23.5 = 376.0 mm2 Net cross-sectional area provided = 2514–376 = 2138 mm2 > Net area required. Hence, safe. 2 ISA 100 mm × 65 mm × 8 mm are provided for all the cross-diagonal members of all the cross-frames. Example 3.13 Design the horizontal truss bracing and the internal gusset plates for the through type plate girder bridge for a single broad gauge track as in Example 3.11. Tota l la te ral lo ad = P H

5 .0 m

6 pa ne ls @ 5.0 m = 30 m

Fig. 3.49

DESIGN OF PLATE GIRDER BRIDGES

159

Solution (A) Design of Horizontal truss bracing Step 1. The horizontal truss bracing with cross-diagonls as shown in Fig. 3.49, is provided btween the loaded flanges (bottom or tension flanges) of plate girders in the throught type bridge. The diagonals which carry tension remain active. From Example 3.11 When the bridge is unloaded, the total wind load on the bridge = 383.62 kN Step 2. When the bridge is loaded Wind load on moving train = 157.50 kN Wind load on windward and leeward girder both = 239.76 kN Racking force, 6.00 × 30 = 180 kN Total lateral load, PH = 577.26 kN The lateral load acting in the plane of horizontal truss is maximum in case the bridge is loaded. Therefore, the horizontal truss bracing is designed for lateral load = 577.36 kN Reaction at the end = 288.63 kN Spacing between the main girders = 5 m 6 panels @ 5.0 m are provided in the horizontal truss, so that each panel is a square. Lateral load at each intermediate panel point

⎛ 577.26 ⎞ = ⎜ ⎟ = 96.21 kN ⎝ 6 ⎠ Lateral load at the end panel point = 48.105 kN End strut carries maximum compression = 288.63 kN The forces in struts are carried by the floor beams Shear force in the end panel = (288.63 – 48.105) kN = 240.525 kN θ = 45° Force in end diagonal = 240.525 × sec θ = 24.06 × End diagonal Force in end diagonal As per IS : 833–94, allowable stress in axial tension = 156 m/mm2 Net area provided = 2730 mm2. Hence, safe.

2 = 340.15 kN

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DESIGN OF STEEL STRUCTURES–VOL. II

Pi

2 IS A 7 5 m m x 5 0 m m x 6 m m

10 m m th ick

10 m m th ick

Fig. 3.50

Design of Internal gusset plates (see Fig. 3.50) Wind load when the bridge is unloaded = 383.62 kN Wind load when the bridge is loaded = 397.26 kN The wind load is maximum when the bridge is loaded. Wind load per panel

P1 =

3976.26 = 66.21 kN. 6

Problems 3.1 A plate girder single track main line broad gauge is of 24 m span between bearings. Design a suitable section for the plate girder and calculate the curtailment of flanges. 3.2 A deck type plate girder bridge is provided for a single broad gauge track and standard main line loading. The total span of main girder from centre to centre of bearing is 28 m. The main girders are provided at a spacing of 2 m between their centre lines, 0.60 kN per metre stock rails and 0.40 kN per metre guard rails are provided. The weight of fastenings may be taken a 0.20 kN per metre. The sleepers are spaced at 0.50 m from centre to centre and are of 2.8 m × 250 mm × 250 mm size. The unit weight of timber may be assumed as 7.50 kN per cubic metre. The floor is open deck type. Design the maximum section of the main girder. 3.3 A through type plate girder is provided for a single metre gauge track. The cross-girders are spaced 2.80 m apart, the total span of the main girder from centre to centre of bearings being 28 m. The stringers are

DESIGN OF PLATE GIRDER BRIDGES

3.4

3.5

3.6

3.7

3.8

3.9

161

spaced at 120 m between centre lines. 0.60 kN per metre stock rails and 0.40 kN per metre guard rails are used. The sleepers are spaced at 0.50 m from centre to centre and are of size 2 m × 250 mm × 250 mm. The weight of timber may be assumed as 750 kN per cubic metre. Design the maximum section of the plate girder. The main girders are provided at a spacing of 4.8 m between their centre line. Adopt loading from Bridge Rule. In Problem 3.2, determine the increase or decrease of stresses in the flanges of the leeward girder in the following cases : (a) Overturning effect due to wind, when the bridge is unloaded. (b) Overturning effect due to wind, when the bridge is loaded. (c) Horizontal truss effect due to wind, when the bridge is unloaded. (d) Horizontal truss effect due to wind, when the bridge is loaded. In Problem 3.3 determine the increase of stresses in the leeward girder in the following cases : (a) Overturing effect due to wind, when the bridge is unloaded. (b) Overturing effect due to wind, when the bridge is loaded. (c) Horizontal truss effect due to wind, when the bridge is unloaded. (d) Horizontal truss effect due to wind, when the bridge is loaded. Design the horizontal truss bracings and the cross-frames for the deck type plate girder railway bridge for a single broad gauge track as in Problem 3.4. Design the horizontal truss bracings and the internal gusset plates for the through type plate girder bridge for a single metre gauge track as in Problem 3.5. In a plate girder through bridge carrying a single broad gauge track, the cross-girders are spaced at 4 m centres. The stringers are spaced at 2 m centres. Design the stringers if the spacing between main girders is 4 m centre to centre. Determine the increase or decrease of stresses in the flanges of leeward girder of a deck type plate girder bridge for the broad gauge single track due to overturning effect and the horizontal truss effect due to wind for loaded and unloaded span. The particulars of the bridge are as follows: Effect span of bridge = 30 m Spacing between main girder = 2m Depth of plate girder = 2.16 m Intensity of wind for unloaded span = 2.40 kN/m2 Intensity of wind for loaded span = 1.50 kN/m2 Height of rolling stock above 600 mm from rail level = 3500 mm Height of rails above plate girder = 400 mm Net area of tension flange = 26000 mm2

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DESIGN OF STEEL STRUCTURES–VOL. II

Gross-area of compression flange = 30000 mm2 Gross moment of inertia of the plate girder section about xx-axis = 12 × 1010 mm4 3.10 Design the horizontal truss bracing and the cross-frames for the deck type single track broad gauge plate girder railway bridge for the following particulars: Effective span = 26 m Spacing between main girders = 2.00 m Depth of plate girder = 2.00 m Intensity of wind for unloaded span = 2.40 kN/m2 Intensity of wind for loaded span = 1.50 kN/m2 Height of coach above 600 mm from rain level = 3500 mm Height of rails above the plate girder including sleepers = 400 mm Racking force = 6 00 kN/m.

CHAPTER

4

Design of Truss Girder Bridges

4.1

INTRODUCTION

The truss girders are used as the main load carrying members in the truss girder bridges. The truss girder bridges are also known as the open web girder bridges. The truss girder bridges are also termed as the lattice girder bridges. The truss girder bridges have opened web work in contradistinction to the plate girders with the continuous web. The truss girders are the triangulated framed structures. The arrangement of the members and the connections at their ends are such that the external loads are applied to the joints. The centroidal axes of members meet at a point. The members are subjected to direct tension and compression. The members of the truss girders are classified as chord members and the web members. The uppermost members constitute the top chord or top boom. The lowermost members form the bottom chord or bottom boom. The vertical members (or verticals) and the diagonal members (or diagonals, which are internally sloping members) are the web members. Those members which are absolutely necessary for the stability of the truss bridge girders are known as the main members. The main members are stressed by loads anywhere on the bridge girder. There are sub-members in some truss girder bridges. In case the sub-members are removed, even then the trusses would be stable. The sub-members act only when loads act directly upon them or on certain portions of the span. The sub-members are also known as secondary members. In most of the truss girder bridges, the chord members carry bending moment in the form of direct tension or compression and the vertical and/or diagonal members carry the shear force in the form of direct tension or compression. The members of truss girder bridges are joined through the gusset plates by riveting, bolting or welding. The thickness of gusset plate in light truss bridges is kept as 10 mm or 12 mm. The thickness of gusset plate in the heavy truss bridges is kept

164

DESIGN OF STEEL STRUCTURES–VOL. II

as 20 mm to 22 mm. The truss girder bridges are of spans greater than those economical for the plate girder bridges. The truss girders of bridges may be either single plane type or the double plane type. In the single plane type truss, the gusset plates are used in one plane only. There is only one gusset plate at each joint of the truss. The single plane type truss is suitable for light loads, small members, and light bracing. In the double plane type trusses, the gusset plates are used in two parallel planes.

4.2

TYPES OF TRUSS GIRDER BRIDGES

The various types of truss girder bridges are as shown in Fig. 4.1.

(a ) H o w e truss

(b ) W h ip p le tru ss

(c) W a rre n truss

(d ) W a rre n truss (w ith ve rticals)

(e ) D o ub le wa rren truss

(f) W a rren tru ss (w ith ve rtica ls a nd curved cho rd )

(g ) W a rre n truss (w ith sub -divide d pa ne ls)

(h ) W a rre n truss (curve d w ith sub -divide d pa ne ls)

(j) P ra tt truss

(k) P ra tt tru ss (cu rve d ch o rd)

(l) P ra tt truss w ith su b-divide d p an els (B altim o re tru ss)

(m ) P ratt truss w ith curve d cho rd an d sub-d ivid e d p an e ls (P e tit o r P e nn sylva nia tru ss)

(n ) P ra tt tru ss (su b-d ivid ed p an els) (B a ltim o re )

(p ) P ra tt tru ss (cu rve d ch ord sub -divide d pa ne ls) (P etit or P e nn sylva nia tru ss)

(r) K -tru ss

(t) P a rab olic tru ss

(s) Z -tru ss

(w ) Viere nd ee l tru ss

Fig. 4.1 Various types of truss girders for the bridges

DESIGN OF TRUSS GIRDER BRIDGES

165

Various truss bridges shown in Fig. 4.1 are grouped as parallel chord truss girder bridges, camel back or curved (inclined) chord truss girder bridges and sub-divided truss girder bridges. The truss bridges are described as under.

4.2.1

Parallel Chord Truss Girder Bridges

Figure 4.1 (a) shows a Howe truss bridge. The diagonal members are in compression and the vertical members are in tension. Howe truss is seldom made entirely in steel. The top chord is constructed in timber and the bottom chord, and the vertical ties are made in steel. Figure 4.1 (b) shows Whipple truss bridge. It is a double intersection Pratt truss bridge. In long span bridges, with considerable depth, this truss was designed to give short panels. It is a statically intermediate structure. Whipple truss bridge has become obsolete. Figure 4.1 (c), (d) and (e) show single Warren truss, Warren truss with verticals and double Warren truss bridges respectively. It is considered that Warren trusses have more attractive appearance as compared to Pratt truss. Warren trusses consist of combination of isosceles triangles. The inclination of diagonals varies from 45° to 60°. The diagonal members carrying compression and tension. The vertical members are occasionally introduced in Warren trusses to support intermediate loads. The verticals stiffen the top chord. In case of double Warren truss, it is assumed for analysis, that the shear force in each panel is shared equally by both the diagonals. Warren trusses with verticals have relatively high secondary stresses. Figure 4.1 (j) shows Pratt truss bridge. The vertical members carry direct compression and the diagonal members carry direct tension. The vertical members are comparatively short members. The reversal of stresses may take place in some of the middle diagonal members of Pratt truss bridges. Then the counter brace members are used as shown by dotted lines in Figure 4.1 (j). In addition to this, the secondary stresses are also comparatively less in these bridges. Therefore, it is desirable to use Pratt truss. Pratt trusses are preferred next to Warren truss bridges. The parallel chord trusses are used for 30 m to 60 m span.

4.2.2

Camel Back Truss Girder Bridges

The camel back truss girder bridges are also known as inclined or curved chord truss bridges. Figure 4.1 (f) shows Warren truss bridges with curved chord. Figure 4.1 (k) shows a Pratt truss bridge with curved chord. The camel back truss girder bridges are used for longer spans than the parallel chord truss girder bridges. In camel back trusses, the height of the trusses are larger where the moments are large. The bending moments in simply supported span under the uniformly distributed loads varies as the square of the span. The forces in chord members of a truss girder are found by dividing the bending moment by the height of truss girder. The bending moment is maximum in the centre. The heights of inclined chord truss is also kept maximum at the centre. The bending moment decreases towards the ends of span. The height of inclined chord truss also decreases towards the ends. The forces in the chord members are kept approximately equal. The height of vertical members also becomes short near the supports. The values

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DESIGN OF STEEL STRUCTURES–VOL. II

of compressive stresses in the vertical member are large, it is economical to design short compression members. Further the inclined chord members share the shear force of their corresponding panels. The vertical components of stresses in these inclined chord members relieve the web members. Thus, it is seen that camel back trusses become economical in weight. The camel back trusses are lighter in weight than the parallel chord trusses. The cost of fabrication is slightly more. The camel back trusses are suitable for 60 m to 120 m span.

4.2.3

Sub-divided Truss Girder Bridges

From the economic considerations, it is seen that inclinations of diagonals may be kept approximately 45° with the horizontal and the height of truss to the span ratio may vary from one-fifth to one-eighth. If the trusses are designed for larger spans in the light of these recommendations, the panel lengths and the unsupported lengths of the compression members would be excessive. Because of the large panel lengths the floor system between the panel points becomes unreasonably heavy and expensive. In order to shorten the panel length (the span of stringers), the panels of trusses are sub-divided. Figures 4.1 (g), and (h) show sub-divided Warren trusses. Figure 4.1 (l), (m), (n) and (p) show sub-divided Pratt trusses. The sub-divided trusses shown in Figure 4.1 (l) and (n) are known as Baltimore trusses. Baltimore trusses are Pratt trusses with parallel chord and sub-divided panels. In sub-divided trusses, the additional sub-members or secondary members are also introduced. The additional vertical members are known as sub-verticals. The stresses from sub-verticals are transmitted to the lower panel points by substruts as in Fig. 4.1 (l) or to the upper panel points by sub-ties as in Figure 4.1 (n). The sub-members are stressed by the loads for certain limited position on the span. Further, it is more economical to use inclined chord sub-divided trusses for large spans. Figures 4.1 (h) shows Warren truss with inclined chord and subdivided panels. Figures 4.1 (m) and (p) show the inclined chord and sub-divided Pratt trusses. These trusses are known as Pettit or Pennsylvania trusses. In this truss, the advantages of Pratt truss, sub-divided truss and inclined chord truss have been used. Pennsylvania truss bridges are statically determinate bridge structure for all positions of the loads. These are economical in construction and satisfactory in service. These trusses (Baltimore and Pennsylvania trusses) have almost entirely replaced Whipple trusses for long spans. The sub-divided trusses also have their own disadvantages. The secondary trusses in these trusses are very high. There are large number of members in these type of trusses. These trusses do not appear too attractive. Figures 4.1 (r), (s) and (t) show K-truss, Z-truss and parabolic truss bridges. Ktrusses have less secondary stresses and have the advantage of sub-divided trusses. K-trusses keep desirable inclination for the diagonals. It has short panel lengths. These are economical in material mainly on account of shortness of compression members. The simple span truss girder bridge can be used upto 200 m span. Figure 4.1 (w) shows Vierendeel truss girder bridge. Actually it is not a truss by usual definition. The joints in Vierendeel truss bridge are moment resisting. Vierendeel trusses support the load by means of bending resistance of its short

Contd.

DESIGN OF TRUSS GIRDER BRIDGES

167

heavy members. The analysis of Vierendeel truss is very difficult. These are efficient structures.

4.3

DECK TYPE AND THROUGH TYPES TRUSS BRIDGES

In the deck type truss bridges, the floor system rests on the top chord. Figures 4.2 (a) (i) and (ii) show deck type Warren truss bridges with the verticals. Figures 4.2 (b) (i) and (ii) show deck type Pratt truss bridges. In the through type truss bridges, the floor system rests on the bottom chord. Figure 4.2 (a) (iii) shows through type Warren truss bridge with verticals. Figure 4.2 (b) (iii) shows through

(i) D eck typ e

(ii) D e ck typ e

(iii) Th ro ug h typ e (a ) W a rren trusses w ith ve rtica ls

(i) D eck typ e

(ii) D e ck typ e

(iii) Th ro ug h typ e (b ) P ra tt tru sses

Fig. 4.2

type Pratt truss bridge. The various types of through type truss bridges have been shown in Figure 4.1. Although, the parallel chord trusses are used for the through type truss bridges, but the lines of various members of parallel chord truss bridges are more in harmony with deck type truss bridges. The camel back or inclined chord trusses as shown in Figure 4.1 are more graceful for the through type bridges.

4.4

COMPONENT PARTS OF A TRUSS BRIDGE

Figure 4.3 shows the various component parts of a through type railway bridges. The various component parts consists of (i) the main vertical trusses, (ii) the

168

DESIGN OF STEEL STRUCTURES–VOL. II

Top la teral b racin g

Top C h ord S w a y bra cin g E n d flo or b ea m P o rta l b racin g

B o tto m la te ral bracin g In te rm edia te flo or b ea m s B o tto m cho rd

End p osts D iag on als Ve rtica ls S trin ge rs E n d flo or be am

Fig. 4.3 Diagrammatic sketch of a through type Pratt truss railway bridge

floor system, (iii) the bottom lateral bracing, (iv) the top lateral bracing, (v) the portal bracings and (vi) the sway bracings. These component parts are separately shown in Fig. 4.4. P o rta l b racin g Top la te ral b racin g

Sway b racin g

M ain ve rtica l trusses

B o tto m la te ral b racin g

Fig. 4.4 Component parts of a truss bridge

The floor system transfers the applied load to the main trusses. The floor system consists of floor, longitudinal girders (stringers or rail bearers) and the transverse girders (floor beams or cross-girders). The floor is supported by the longitudinal girders. The longitudinal girders run parallel to the length of bridge. The

DESIGN OF TRUSS GIRDER BRIDGES

169

longitudinal girders span between two adjacent panel points of the main truss girders. The longitudinal girders are supported by the transverse girders. The transverse girders span between the main girders. These transverse girders transmit the applied load to the main truss girders. The bridge consists of two main truss girders. The truss girders are the main load carrying members. The truss girders transmit the applied load to the abutments or piers through the bearings. The top lateral bracings, bottom lateral bracings, sway bracings and portal bracings are provided for the lateral stability and the torsional rigidity of the bridge, and to resist the lateral loads and the longitudinal forces. The design of various components of a truss girder bridge is done in the order in which the applied load is transmitted from one component to the other.

4.5

ECONOMIC PROPORTIONS OF TRUSSES

The cost of a bridge includes the cost of superstructure and the cost of substructure. The cost of superstructure includes the cost of materials, the cost of fabrication, the cost of floor system, the cost of construction and the cost of maintenance. Thus, it is seen that the economy of a bridge depends on many factors. It is difficult to determine a mathematical expression for the most economy of the truss bridge. The cost of fabrication depends on number and types of joints. It may not be least when the weight of truss is minimum. A two panel truss with diagonals inclined at 45° gives the least weight theoretically for given span and loading. In this, the permissible stresses in tension and compression are assumed constant. The permissible stress in compression is not constant. These stresses depend on the slenderness ratios for the members. When the number of panels in a truss is small, then the lengths of members are large. The slenderness ratios for the members are also large. As a result of this, the permissible stresses in compression have small values. The section of compression members adopted become heavy. The weight of truss is increased. The weight of floor system also increases for large panel length. The number of panels in a truss should be 6 to 8 for overall economy. The ratio of height of truss to the span of a bridge producing the greatest economy of material is that which makes the weight of chord members equal to the weight of the web members of truss. The economic height to span ratio of truss bridge is one-sixth to one-eighth. It varies with the type of truss bridges. The smaller ratios are used for long span bridges. The larger ratios are adopted for the relatively short span bridges. The minimum height (depth) of highway and railway bridges as per IS: 1915–1961 is one-tenth of the span. The effective height (depth) of riveted trusses is the distance between the gravity axes of chord. The span for the purpose of determining the minimum height is taken as the distance between the centres of bearings in the case of simply supported span. The effective height (depth) between gravity axes should not be greater than three times the width between centres of main girders as per code of practice for the design of steel bridges published by the Railway Board. The total length of all the diagonals in a truss inclined at an angle α with the horizontal is equal to L sec α, where L is the span of truss. For given span and loading the shear force in a panel F may be assumed to be constant. The force in a

170

DESIGN OF STEEL STRUCTURES–VOL. II

diagonal is equal to F. cosec α as an average. The cross-sectional area of a diagonal is proportional to F. cosec α. The volume of diagonal is proportional to F cosec α (L sec α). When the inclination a is 45°, then, the volume of diagonals is minimum. Therefore, for economic consideration for truss bridges, the inclination of diagonals should be kept approximately 45° with the horizontal. In general, this inclination may vary from 45° to 60°. The panel length may vary from 6 m to 9 m. The distance or width between centres of trusses should be sufficient to develop lateral strength and rigidity. As per code of practice for the design of steel bridges, the width between centres of main girders should be sufficient to resist overturning with the specified wind pressure and load conditions, otherwise, the provisions should be made to prevent the overturning. In no case this width should be less than one-twentieth of the effective span. In the through type truss bridges, the clearance requirements usually provide this width more than what it is necessary.

4.6

SELF WEIGHT OF TRUSS GIRDERS

The dead load acting on the truss girder bridges also include the self-weight of truss girders of the bridge. The self-weight of truss girder of the bridge is either assumed depending upon the experience and by comparing with the existing truss girder bridges on similar spans or determined by using the following empirical formulae.

4.6.1

Hudson’s Formula

Mr. Clarence W. Hudson gave an empirical formula for the determination of the self-weight of truss girders and bracings for the bridges. The self-weight determined from this formula, is sufficiently accurate. The self-weight of trusses and the bracings in kN per metre w, was expressed in terms of the net area of maximum tension chord, in square mm, A. The total areas of two girders are as follows : Area of the bottom chord = 2 × 100 A Area of the top chord = 2 × 1.25 A Area of the web system = 2 × 1.25 A Area of the details = 2 × 1.00 A Area of the bracing = 2 × 0.50 A Total area of two trusses and bracings = 10.00 A The unit weight of steel is 78.50 kN/m2. Therefore,

⎛ 78.50 × 10.00 A ⎞ w = ⎜ ⎟ kN/m ⎝ 1000 × 1000 ⎠ w = 0.785 ×10–3A kN/m ...(4.1) In order to determine the value of net area of maximum tension chord A, the self-weight of trusses and bracing is first assumed. Then, maximum force in tension chord is determined due to dead load including weight of floor system and selfweight, live load and impact load. The net area of maximum tension chord A, is

DESIGN OF TRUSS GIRDER BRIDGES

171

found by dividing the maximum force by the allowable stresses in axial tension (140 N/mm2). The self-weight of truss girders including bracing w, in kN/m is determined from Eq. 4.1, and compared with the assumed self-weight. The value of assumed self-weight is modified, and the calculations are repeated, till the assumed self-weight and the self-weight from Eq. 4.1 come in close agreement. This method is sometimes convenient and reasonably accurate. But this method can lead to quite erroneous results in certain instances. When the depth/span ratio for a truss is high, then the weight of the web members is more. Similarly a shallow truss girder will have more weight in the flanges. Hence, the above weight proportions may not apply. However, the above proportionate weight of different parts help in the work of estimating.

4.6.2

Fuller’s Formula

The self-weight of truss girders and bracing in kN per metre as per Fuller formula is given by w = (a . l + b) ...(4.2) where, l = Span (length) of the bridge in metres a, b = Constants The dead load exercises a greater influence on the weight. The term representing weight of truss girder increases faster than the span length. The exponent of l in the term al becomes greater than one. The exponent increases with increase in span length. Constants a and b depend upon the type of bridge. For the truss girder bridges, the weight of trusses and bracings in kN per metre may be obtained from the formula given below upto 100 m span for the bridges.

⎛ 15 l + 550 ⎞ w = ⎜ ⎟ kN/m ⎝ 100 ⎠

4.7

...(4.3)

ASSUMPTIONS FOR THE DESIGN OF TRUSS BRIDGES

In the design of truss girder bridges, the axial stresses (forces) in the various members are determined on the following assumptions : (a) All the members of the truss girder bridges are straight and free to rotate at the panel points. (b) All the panel points (joints) of the truss bridges lie at the intersection of the gravity axes of the members. (c) All the loads including the weight of the members are applied to the panel points.

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DESIGN OF STEEL STRUCTURES–VOL. II

All the axial stresses are termed as primary stresses. The forces in the various members due to dead load may be determined by the method of resolution of forces about the joint or by the method of section. The dead load acting on the truss girder bridges is uniformly distributed load. The forces in various members of the truss girders may be found by drawing the influence line diagrams. The forces in various members for the self-weight are the algebraic sum of the areas of influence line diagram multiplied by the intensity of the load. The forces in various members due to live load or moving load are determined by drawing the influence line diagrams for the force for the respective members. The influence line diagrams for the various members of the various types of trusses have been discussed in chapter 15. The live loads for the members are noted from Bridge Rules for bending moment in case the influence line diagrams are drawn from criterion of bending moment, and from shear force, in case the influence line diagrams are drawn from criterion of shear force.

4.8

COMPRESSION MEMBERS

The properties of cross-section of compression members are found from the effective sectional area. The effective sectional area of compression member is the gross area less the deductions for excessive width of plates and the maximum deductions for open holes, including holes for pins and black bolts occurring in a section perpendicular to the axis of member. The unsupported width of a plate measured between adjacent lines of rivets connecting the plate to other parts of the sections shall be 50 t for steel conforming to IS : 226, where t is the thickness of single plate or the aggregate thickness of two or more plates provided these plates are adequately tacked together. Any excess over this width shall not be included in the effective sectional area in computing the direct compressive stress. The unsupported width of a plate forming any part of a compression member measured between adjacent lines of rivets connecting the plate to other parts of the section, unless effectively stiffened, shall not exceed 20 t for steel conforming to IS : 226. The open sides of built up compression members of U or I sections shall be connected by lacing, battening or perforated plates where the length of the outstand towards the open sides exceeds 16 times the mean thickness. The effective length of compression members in riveted trusses shall be taken as in Table 4.1 except for battened struts for which all values given in Table 4.1 shall be increased by 10 percent. The ratio of effective length to the least radius of gyration shall not exceed 120 far main members and 140 for wind bracing and subsidiary members.

DESIGN OF TRUSS GIRDER BRIDGES

173

Table 4.1 (As Per IS : 1915–1961)

Member

For including in the plane of truss

Effective Length T of members For buckling normal to the plane of truss Compression Compression chord chord or or compression member compression effectively braced member unbraced by lateral bracing

(a) Chord

0.85 × distance between centres of intersection with the web member

0.85 × distance between centres of intersection with lateral bracing members or rigidly connected cross girder

(b) Web (1) Single triangulated system

0.70 × distance between 0.85 × distance centres of intersection between centre of with the main chord intersection

(2) Multiple intersection system where adequate connections are provided at all points of intersections

0.85 × greatest distance between centres of any two adjacent intersection

0.70 × distance between centres of intersections with the main chord

as per IS : 1915–1961

Distance between centre of intersections 0.85 × distance between centres of intersections with the main chords

Note. The intersections referred to are those of the centroidal axes of members.

Floor beams alongwith the vertical members of the main truss girders (or floor beams along the stiffeners of the main plate girders) form U-frames in half-through type bridges. These U-frames act like elastic springs. The U-frames provide lateral support to compression chord (compression flange). The effective length of the compression chord (compression flange) is determined as follows : (a) When δ is not greater than a3/40 EI Effective length, l = a. (b) When δ is greater than a3/40 EI Effective length, l = 25 (E.I.a.δ)1/4

| a and but l < l| > effective span

174

DESIGN OF STEEL STRUCTURES–VOL. II

where,

a = Distance between U-frames I = Maximum moment of inertia of compression chord (compression flange) about yy-axis of the girder δ = Virtual lateral displacement at the centroid of compression chord (compression flange) at the frame nearest to the mid-span of the girder as shown in Fig. 4.5 (i). δ is taken as horizontal deflection at the gravity axis of the frame (or at the point of intersection of stiffeners with the centroid of the compression flange). The horizontal deflection is found under the action of unit horizontal force applied at this point in the plane of the frame only. δ

δ Top cho rd R e straine d le g (C olum n )

Ic

R o ad w a y

do d c

C ro ss-g ird er

a = spa cing o f truss g ird ers I b U -F am e (P o ny fra m e)

Fig. 4.5 (i)

In cases of symmetrical U-frames where cross-girders and stiffeners are each of constant moment of inertia throughout their own length ⎡ (d )3 (d )2 b ⎤ c c ⎥ + δ = ⎢ EI EI 3 ⎢⎣ c g ⎥ ⎦

where,

dc = Distance of the centroid of the compression flange from the top of the cross-girder. du = Distance of the centroid of the compression flange from neutral axis of the cross-girder. b = Half the distance between centres of the main girders. Ic = Moment of inertia of a pair of stiffeners about the centre of the web or a single stiffeners about the face of the web in its plane of bending. Ig = Moment of inertia of the cross-member (cross-girder) in its plane of bending. These U-frames shall have rigid connections in order to provide effective lateral support. These are designed to resist in addition to the effect of wind and other

175

DESIGN OF TRUSS GIRDER BRIDGES

applied forces, a horizontal force, F acting normal to the compression chord (or flange) of the girder at the level of the centroid.

B =

⎡ ⎤ ⎢ ⎥ −3 ⎢1.4 × 10 × l ⎥ ⎢ ⎛C ⎞⎥ ⎢ δ ⋅ ⎜ s − 1.7⎟ ⎥ ⎠ ⎦⎥ ⎣⎢ ⎝ fcc

where,

Cs = Critical stress l = 2.5 (E.I.A.δ)1/4 fcc = Calculated working stress in the chord (or flange) δ = As mentioned above. The various forms of sections used for top chord members are shown in Fig. 4.5 (ii).

(a )

(b )

(d )

(c)

(e )

Fig. 4.5 (ii)

The top chord members are most highly stressed compression members of the truss. In the top chord members, much care is given in proportioning and detailing to obtain the desired strength with economy and to provide satisfactory relations with other connecting members. In general, it is preferable to provide as few parts as possible. The thickness of webs is kept as thick as economic fabrication allows. It gives proper distribution of stress. In the top chord members, the cover plates are necessarily used on the top of sections. These plates connect the web plates and make whole of the section to act together effectively as one unit. These plates prevent rain water entering the section, and protect the parts of the members and the joints from corrosion. The bottom of these top chord members are connected only by lacing or by tie or stay plates, which are not continuous elements. The centre of gravity of section is raised above the centre line of web because of cover plate. This is not a desirable feature. Therefore, only one cover plate is used and

176

DESIGN OF STEEL STRUCTURES–VOL. II

the thickness of this cover plate is kept as thin as possible. This thickness of cover plate is kept sufficient to prevent it from buckling. A uniform section is provided throughout the top chord of the truss girder. The minimum section may be designed for the end panel. The strength or gross area of member may be increased by providing plates, in the sides of members. In order to provide most economical section, the relation between depth to width of the top chord compression members may be such that the radii of gyration about xx-axis and yy-axis are equal. However, the depth of top chord section may be adopted as 1/10 th of the distance between centroidal axes of top and bottom chords. The clear width of top chord section may be adopted as 1/10th of the distance between centroidal axes of top and bottom chords plus twice the thickness of gusset plates. The thickness of each gusset plate is kept 22 mm. The approximate radii of gyrations are noted (Refer Sec. 4.8, in author’s Steel Structures Vol. I) for the trial section. The effective length of top chord member is adopted as per Table 4.1. The maximum slenderness ratio is then found for the trial section. The allowable stress in axial compression is found corresponding to the maximum slenderness ratio. The area of top chord section is found for the force. The area of top chord section is provided accordingly. The top chord members are also subjected to bending due to self-weight. The self-weight of members is computed. The members are checked for ⎛ σ ac.cal σbc.cal ⎞ + ⎜⎝ σ σbc ⎟⎠ ac

< 1.00

where, σac.cal = Actual calculated axial stress in compression in the member σac = Allowable axial stress in compression in the member σbc.cal = Actual calculated bending stress in the extreme fibre of member σbc = Allowable bending compressive stress in the member and, it is also checked for the equivalent stress. For the end posts, the sections similar to those of top chords are provided. The forces in end posts are approximately equal to those in the end panels. In addition to the top chord, some of the web members are also subjected to compression. In a Pratt truss, the vertical members carry compression, while in Warren truss each alternate diagonal carries compression. In case, the counterbracing is not provided and the reversal of stress takes place, then, such members are also designed as compression members. The various forms of compression members used as web members are as shown in Fig. 4.6. The form of such compression members should be such that these should be convenient for attachment to floor beams, and these may enter the top chord conveniently. Usually, a built-up I-section is adopted for the web compression members. The depth of such member is adopted as 1/10th the distance between centroidal axes of the top and bottom chords. The member is designed as described above.

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DESIGN OF TRUSS GIRDER BRIDGES

(a )

(d )

(b )

(c)

(e )

(f)

Fig. 4.6

4.9

TENSION MEMBERS

The tension members shall preferably be of rigid cross-section. The properties of the cross-section shall be computed from the effective sectional area. When plates are provided solely for the purpose of lacing or battening, they shall be ignored in computing the radius of gyration of the section. The open sides of built-up tension members of U- or I-section shall be connected by lacing, battening or perforated plates, when the length of the outstand towards the open side, exceeds 16 times the mean thickness of the out stand. The net sectional area of a bolt or screwed tension rod shall be taken as the area at the root of thread. The various forms of sections used for bottom chord members are shown in Figure 4.7. The bottom chord members are most highly stressed tension members. In the bottom chord, more sectional area is concentrated mainly in the two web plates and plates connected to them. The angle sections are mainly used for connecting the lacing. The angle sections and the flanges of rolled steel sections are turned inward. Therefore, the connections of floor beams and bottom lateral bracing become convenient. The material required for lacing is also reduced. The distance between back to back of outside plates is made equal to the overall depth of the web members. In this case, the gusset plates are fitted on the outer sides of the members at the bottom chord joints. There is no advantage as to the strength of the members in turning the angle, and the flanges outward. The connection of floor beams and the bottom lateral bracing also becomes difficult. When the flanges are turned outward, then the gusset plates are placed inside the members. A uniform section is provided throughout the bottom chord of the truss girder. The minimum section may be designed for end panels. The strength of section or net sectional area is increased from member to member by providing additional plates in the sides. However the distance between back to back of angles is kept constant. Therefore, the distance between gusset plates and the length of floor beams remain same for whole of the span.

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DESIGN OF STEEL STRUCTURES–VOL. II

(a )

(b )

(c)

(d )

(e )

(f)

(g )

(h )

(i)

Fig. 4.7

The net-sectional area is found for the force in the bottom chord tension members. The gross area of the member required is determined by adding suitable allowance for the estimated rivet holes for connection. The dimensions of cross-section of the bottom chord member are kept same as that of the top chord member except for top plates. The plates are not used either at the top or at the bottom of the section, so that the rain water does not collect and the corrosion of elements is not caused. The gross area is provided accordingly. From the gross area, the net area is again computed. The net area provided should be greater than the net area required. The bottom chord members are also subjected to bending due to self-weight. Therefore, these are checked for ⎛ σat.cal σbc.cal ⎞ + ⎜ ⎟ σbc ⎠ ⎝ σat

< 1.00

where, σat.cal = Actual calculated axial stress in tension σat = Allowable axial stress in tension and, it is also checked for the equivalent stress. In addition to the bottom chord, some of the web members are also subjected to

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DESIGN OF TRUSS GIRDER BRIDGES

tension. In a Pratt truss, all the internal diagonal members are tension members, while in a Warren truss each alternate diagonal carries tension. The various forms of tension members used as web members are shown in Fig. 4.8.

(a )

(b )

(d )

(c)

(e )

Fig. 4.8

For the web tension members the ratio of unsupported length to the least radius of gyration shall not exceed 250 for railway bridges and 300 for highway bridges. The form of such tension members should be such that, the gusset plates attached to these members may enter the top chord conveniently. Usually, a built-up Isection is adopted for the web tension members. The depth of such members is adopted as 1/10th the distance between centroidal axes of the top chord and the bottom chord. The member is designed as described above. Example 4.1 A Pratt truss girder through bridge is provided for single broad gauge track. The effective span of bridge is 50 m. The cross-girders are spaced 5 m apart. The stringers are spaced 2 m between centre lines. 0.60 kN per metre stock rails and 0.40 kN per metre check rails are provided. Sleepers are spaced at 0.45 m from centre to centre and are of size 2.8 m × 250 m × 250 mm. Weight of timber may be assumed as 7.50 kN per cubic metre. The main girders are provided at a spacing of 7 m between their centre lines. Design the central top chord member and bottom chord member and the vertical and diagonal of central panel. Also, design the joint, where the central top chord, vertical and diagonal members meet. The bridge is to carry standard main line loading. Solution Design : Preliminary Effective span or truss girder bridge = 50 m Panel length of bridge girder = 5 m Number of panels = 10 Spacing between main girders = 7 m.

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DESIGN OF STEEL STRUCTURES–VOL. II

Height of truss girder For economical consideration 1 × span = 7

1 × 50 = 7.14 m 7

As per IS 1915–1961 (Minimum) 1 1 × span = × 50 = 5 m 10 10

Provide 6 m height between centre of gravity of top chord members to centre of gravity of bottom chord members. A Pratt truss girder bridge as shown in Fig. 4.9 is used. U1

U2

U3

U4

U5

U6

U7

U8

U9 6m

L0

L1

L2

L3

L4 L5 L6 L7 1 0 P an els a 5 m = 5 0 m (E leva tio n)

L8

L9

L 10

2m 7m 5 m

5 m

5 m

5 m

5 5 m m (P lan )

5 m

5 m

5 m

5 m

Fig. 4.9

Step 1. Dead load The dead loads acting on truss girder are as follows : Weight of stock rails per track per metre = 2 × 0.60 = 1.20 kN/m Weight of check rails per track per metre = 2 × 0.40 = 0.80 kN/m Weight of fastenings (assumed) = 0.20 kN/m Weight of stringers per track per metre (assumed)

10 ⎛ 2.8 × 250 × 250 × 7.50 ⎞ ×⎜ ⎟ = 2.92 kN/m 0.45 ⎝ 1000 × 1000 ⎠ Weight of stringers per track per metre (assumed) = 3.00 kN/m Weight of cross-girders per track per metre (assumed) = 5.00 kN/m Self-weight of both the truss girders by Fuller’s formula, as per Eq. 4.3.

DESIGN OF TRUSS GIRDER BRIDGES

181

⎛ 15l + 550 ⎞ ⎛ 15 × 50 + 550 ⎞ ⎜ ⎟=⎜ ⎟ = 13.00 kN/m 100 ⎝ 100 ⎠ ⎝ ⎠ Total dead load per track per metre is 26.12 kN/m Total dead load per girder per metre is 13.06 kN/m Ω 13.2 kN/m. Step 2. Influence line diagrams for forces in the members The influence line diagrams for forces in U4L5, L4L5, L4 L4, L4L5 and U3U4 are shown in Fig. 4.10. (i) I.L.D. for force in U4U5 The influence line diagram for force in U4U5 is shown in Fig. 4.10 (b). The maximum ordinate of the triangle M L5 6

=

1 ⎛1 ⎞ × ⎜ × 2.5 ⎟ = 2.08 units (compression) 6 ⎝2 ⎠

(ii) I.L.D. for force in U3U4 The influence line diagram for force in U3U4 is as shown in Fig. 4.10 (c). The maximum ordinate of the triangle M L4

=

1 ⎛ 20 × 30 ⎞ ×⎜ ⎟ = 200 units (compression) 6 ⎝ 50 ⎠

=

1 ⎛ 20 × 30 ⎞ ×⎜ ⎟ 2.00 units (tension) 6 ⎝ 50 ⎠

6 (iii) I.L.D. for force in L4L5 The influence line diagram for force in L4L5 is an shown in Fig. 4.10 (d). The maximum ordinate of the triangle M U4 6

(iv) I.L.D. for force in U4U4 The influence line diagram for force in U4L4 is as shown in Fig. 4.10 (e). The ordinates are as below : y1 =

m 4 = = 0.4 units n 10

⎛ n − 1 − m ⎞ ⎛ 10 − 1 − 4 ⎞ y2 = ⎜ ⎟=⎜ ⎟ = 0.5 units 10 n ⎝ ⎠ ⎝ ⎠ (v) I.L.D. for force in U4 L5 The influence line diagram for force in U4L5 is as shown in Fig. 4.10 (f). The ordinates are as below : y3 =

4 m × 1.3 = 0.52 units cosec θ = n 10

⎛ 10 − 1 − 4 ⎞ ⎛ n − 1 − m⎞ cosec θ = ⎜ y4 = ⎜ ⎟ ×1.3 ⎟ ⎝ ⎠ 10 n ⎝ ⎠ = 0.65 units

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DESIGN OF STEEL STRUCTURES–VOL. II

The point of intersection of influence line curve in Fig. 4.10 (e) base is at a distance AO from left hand support. U1

U2

U3

U4

U5

U6

U7

U8

U9

6m L0

L1

L2

L3

L4 L5 L6 L7 1 0 P an els @ 5 m = 5 0 m (a ) (– )

(b )

L8

L9

L10

2 .0

1 .L.D fo r U 4 U 5

(– ) 2 .00

(c)

(+ )

1 .L.D fo r U 3 U 4

2 .00

(d )

1 .L.D fo r L 4 L 5 2 7.7 8 m

(+ ) y 1 = 0 .4

y 2 = 0 .5 (–)

2 2.2 2 m

1 .L.D fo r U 4 L 4

(e ) 2 2.2 2 m

y 4 = 0 .65 (+ )

(– ) y 3 = 0 .52 2 7.7 8 m (f)

1 .L.D fo r U 4 L 5

Fig 4.10

AO = (20 + 222) = 22.22 m Step 3. Forces in the members due to dead load Dead load = 133 kN/m Force in member U4U5 =

1 L × 2.08 × 50 × 13.2 = – 636.4 kN (i.e., compression) 2

DESIGN OF TRUSS GIRDER BRIDGES

183

Force in member U3U4 1 –2.00 × 50 × 13.2 = – 660 kN (i.e., compression) 2 Force in member L4L5

=–

1 × 2 × 50 × 13.2 = + 660 kN (i.e., compression) 2 Force in member U4L4

=+

1 1 × 27.78 × 0.5 – × 22.22 × 0.4] ×13.2 2 2 = – 33.0 kN (i.e., compression)

=

[

Force in member U4U5 1 1 –× 27.78 × 0.65 – × 22.22 × 0.52] × 13.2 2 2 = + 42.92 kN (i.e., tension) Step 4. Forces in the members due to live load and impact load (i) Member U4U5 Loaded length = 50 m

= [

Impact factor

⎛ 20 ⎞ ⎛ 20 ⎞ = ⎜ ⎟=⎜ ⎟ = 0.313 ⎝ 14 + L ⎠ ⎝ 14 + 50 ⎠

From Bridge Rules, for broad gauge, 50 m loaded length Live load + Impact load per girder 1 × 4380) = 2875.47 kN 2 Force in the member due to live load and impact load

= 1.313 × (

=

⎛ 2875.47 ⎞ 1 × 2.08 × 50 × ⎜ ⎟ = – 2990.49 kN (i.e., compression) 2 ⎝ 50 ⎠

(ii) Member U3U4 Loaded length = 50 m Impact factor = 0.313 Live load + Impact load per girder

⎛1 ⎞ = 1.313 × ⎜ × 4380 ⎟ = 2875.47 kN 2 ⎝ ⎠ Force in the member due to live load and impact load =–

⎛ 2875.47 ⎞ 1 × 2 × 50 × ⎜ ⎟ = – 2875.47 kN (i.e., compression) 2 ⎝ 50 ⎠

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DESIGN OF STEEL STRUCTURES–VOL. II

(iii) Member L4L5 Loaded length = 50 m Impact factor = 0.313 Live load + Impact load per girder = 2875.47 kN Force in the member due to live load and impact load =+

⎛ 2875.47 ⎞ 1 × 2 × 50 × ⎜ ⎟ = + 2875.47 kN (i.e., tension) 2 ⎝ 50 ⎠

(iv) Member U4U4 (For tension) Loaded length = 22.22 m

20 ⎛ ⎞ Impact factor = ⎜ ⎟ = 0.552 14 22.22 + ⎝ ⎠ Live load + Impact load per girder

⎛1 ⎞ = 1.552 × ⎜ × 2404.4 ⎟ = 1865.81 kN ⎝2 ⎠ Force in member due to live load + Impact load =

⎛ 1865.81 ⎞ 1 × 0.4 × 22.22 × ⎜ ⎟ = + 373.16 kN (i.e., tension) 2 ⎝ 22.22 ⎠

(For compression) Load length = 27.78 m Impact factor = 0.479 Live load + Impact load per girder =

1.479 × (

1 × 2849.06) = 2106.88 kN 2

Force in member due to live load + Impact load =

1 ⎛ 2106.88 ⎞ × 0.5 × 27.78 × ⎜ ⎟ = 526.72 kN (compression) 2 ⎝ 27.78 ⎠

(v) Member U4U5 The force in member U4L5 may be found from force in member U4L5 by multiplying by cosec θ = 1.3. Compression = 1.3 × 373.16 = 485.11 kN Tension = 1.3 × 526.72 = 684.74 kN

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DESIGN OF TRUSS GIRDER BRIDGES

Design forces inthe members are as follows : Forces in the member D.L. Member

U4U5 U3U4 L4 L 5 U 4L 5 U 4L 5

Compression (kN) 686.4 660 — 33.0 —

Tension (kN) — — 660 — 42.92

Design forces D.L. + L.L. + I.L.

(L.L. + L.L.) Compression (kN) 2999.49 2875.47 — 536.72 485.11

Tension

Compression (kN) 3676.89 3535.47 — 559.72 442.19

(kN) — — 2875.47 373.16 684.74

Step 5. Design of top chord member Forces in the member = 3676.89 kN (compression) Allowance is not made for fatigue for the compression members. Depth of truss girder = 6 m = 6000 mm Depth of top chord member =

1 × 6000 = 600 mm 10

8 50 m m 7 56 m m 16 m y = 18 0. 80 m m

X 12 m m

6 44 m m

X

6 00 m m

12 m m

4 35 .2 m m

4 IS A 8 0 m m x 5 0 m m x 6 m m

Fig. 4.11

Width of top chord member

⎛1 ⎞ = ⎜ × 6000 + 2 × 22 ⎟ = 644 mm 10 ⎝ ⎠ Try the section as shown in Figure 4.11. Approximate radius of gyration, rx = 0.39 h = 0.39 × 600 = 234 mm ry = 0.55 b = 0.55 × 654 = 354 mm

Tension (kN) — — 3535.47 340.16 727.66

186

DESIGN OF STEEL STRUCTURES–VOL. II

rmin = 234 mm Length of member from centre to centre of intersection is 5 m. Effective length of member is 0.85 × 5000 = 4250 mm Maximum slenderness ratio is

4250 = 18.16 234

From IS : 833–94 allowable stress in axial compression, for the steel having value of yield stress as 260 N/mm2 σac = 154.37 N/mm2

⎛ 3676.89 × 1000 ⎞ Area required = ⎜ ⎟ = 23818.86 mm2 154.37 ⎝ ⎠ Width of top cover plate of the section between centre to centre of rivet line b = 762 mm

⎛ 762 ⎞ ⎛b⎞ ⎜ t ⎟ = ⎜ 16 ⎟ = 47.625 < 50 ⎝ ⎠ ⎝ ⎠ Therefore, the whole width of top cover plate is effective in compression. The area of section provided is as follows : Top cover plate = 850 ×16 = 13600 mm2 Web = 2 × 600 × 12 = 14400 mm2 4 ISA 80 mm × 50 mm × 6 mm = 4 × 746 = 2984 mm2 Total = 30984 mm2 More area is provided in order to adjust increase of force due to wind effect. The c.g. of section from top is at a distance O Ratio

⎛ 850 × 16 × 8 + 2 × 746 × (11.6 + 16) + 2 × 736 ⎞ ⎜ × (600 − 11.6 + 16) + 2 × 600 × 12 × 316) ⎟ ⎟ = 180.80 mm y = ⎜ ⎜ 850 × 16 + 2 × 746 + 2 × 746 + 2 × 600 × 12 ⎟ ⎜ ⎟ ⎝ ⎠

1 ⎡ ⎤ 3 2 ⎢2 × 12 × 1.2 × 60 + 2 × 60 × 1.2 (30 − 18.08) + 85 × 1.6 ⎥ ⎢ ⎥ Ixx = ⎢ × (18.08 − 0.8)2 + 2 × 7.46 × 18.08 – 2.76)2 + 2 × 7.46 ⎥ × 104 mm ⎢ × (43.52 − 1.16)2 + 4 × 14.4 ⎥ ⎢ ⎥ ⎣⎢ ⎦⎥ = 134601.18 × l04 mm4

⎡1 ⎤ 2 2 Iyy = ⎢ × 1.6 × 85 + 2 × 60 × 1.2 × (32.1 + 0.6) + 4 × 48 ⎥ × 104 mm4 ⎣18 ⎦ = 236053.09 × 104 mm4

187

DESIGN OF TRUSS GIRDER BRIDGES

rmin

⎛ 134601.18 × 104 = ⎜⎜ 30984 ⎝

1/2

⎞ ⎟⎟ ⎠

= 208.43 mm

4250 = 20.39 208.43 From IS : 833–94 allowable stress in axial compression for the steel having yield stress as 260 N/mm2 σac = 150.883 N/mm2 Force carrying capacity of the member

Slenderness ratio =

⎛ 150.883 × 30984 ⎞ = ⎜ ⎟ 1000 ⎝ ⎠ = 4674.96 kN. Hence safe. The member may be checked for combined stresses and equivalent stress. Step 6. Design of bottom chord member Force in member = 3535.47 kN To allow for the effect of fatigue fmin = 660 kN, fmax = + 3535.47 kN ⎛ fmin ⎞ ⎜ ⎟ = ⎝ fmax ⎠ From IS : 1915–1961, K = 0.92 ∴ Allowable stress in axial tension =

⎛ 660 ⎞ ⎜ ⎟ = 0.186 ⎝ 3535.47 ⎠

0.92 ×0.6 × 260 = 143.52 N/mm2

⎛ 3535.47 × 1000 ⎞ = ⎜ ⎟ = 24633.99 mm2 143.52 ⎝ ⎠ The overall dimension of the section is kept same as that for top chord section, i.e., as shown in Figure 4.12.

X

X 4 IS A 2 00 m m x15 0 m m x 10 m m

6 00 m m

Fig. 4.12

6 00 m m

Net area required

188

DESIGN OF STEEL STRUCTURES–VOL. II

The area provided is as follows : 2 web plates 600 mm × 12 mm = 2 × 600 × 12 = 14400 mm2 2 additional plates with web = 2 × 600 × 8 = 9600 mm2 4 ISA 200 mm ×150 mm × 10 mm = 4 × 3400 = 13600 mm2 Total gross area provided = 37600 mm2 Use 22 mm diameter rivets. Assuming that 16 rivets would be necessary for connecting main members, and for lacing. Area of holes = [(4 × 23.5 × 10) + 4 × 23.5 (10 + 12 + 8) + 12 × 235 (8 + 12)] mm2 = 9400 mm2 Net area provided (37600 – 9400) = 28200 mm2 > 24633.99 mm2 The member may be checked for combined and equivalent stress. Step 7. Design of vertical member Compressive force = 559.72 kN Tensile force = 340.16 kN

4 IS A 12 5 m m x 75 m m x 8 m m

6 00 m m

L acin g

Fig. 4.13

The overall depth of vertical member is kept equal to the internal width of top cord members less twice the thickness of gusset plate i.e., as shown in Figure 4.13. Depth of members = (644 – 2 × 22) = 600 mm

189

DESIGN OF TRUSS GIRDER BRIDGES

Length between centre to centre of intersection = 6000 mm Effective length of member = 0.7 × 6000 = 4200 mm Assuming allowable stress in axial compression, for the steel having yield stress as 260 N/mm2 and slenderness ratio as 70. σac = 115 N/mm2

⎛ 559.72 × 1000 ⎞ 2 Area required = ⎜ ⎟ = 4867.13 mm 115 ⎝ ⎠ Provide 4 ISA 125 mm × 75 mm × 8 mm as shown in Fig. 4.13. Area provided = 4 × 1538 = 6152 mm2 Ixx = [5 × 67.2 + 4 × 15.38 × (30 –1.68)2] × 104 mm4 = 49676.42 × 104 mm4 Iyy = [4 × 245.5 + 4 × 15.38 × (4.15 + 0.5)2] × 104 mm4 = 2312.22 ×104 mm4 rmin

⎛ 2312.22 × 104 ⎞ = ⎜ ⎟ 6152 ⎝ ⎠

1/2

= 61.31 mm

⎛ 4200 ⎞ ⎜ ⎟ = 68.51 ⎝ 61.31 ⎠ From IS : 833–1994 for steel having yield stress as 260 N/mm2 σac = 116.64 N/mm2 Force carrying capacity of the member Slenderness ratio

⎛ 116.64 × 6152 ⎞ = ⎜ ⎟ = 717.56 kN. Hence safe. 1000 ⎝ ⎠ To check the member for tension fmin = + 340.16 kN fmax = – 559.72 kN ⎛ fmin ⎞ ⎜ ⎟ ⎝ fmax ⎠

= (– 0.0608)

From IS : 1915–1961, K = 0.638 Allowable stress in axial tension, allowing the effect of fatigue = 0.638 × (0.6 × 260) = 99.528 N/mm2 Net area required

Gross area provided

⎛ 340.16 × 1000 ⎞ = ⎜ ⎟ = 3416.73 mm2 99.528 ⎝ ⎠ = 6150 mm2

190

DESIGN OF STEEL STRUCTURES–VOL. II

Area of rivet holes = 4 × 23.5 × 8 = 752 mm2 = (6150 – 752) = 5398 mm2 > 3417. 73. Hence safe. Step 8. Design of diagonal member Tensile force = 727.66 kN Compressive force = 442.19 kN fmin = 442.19 kN fmax = + 727.66 kN ⎛ fmin ⎞ ⎛ 442.19 ⎞ ⎜ ⎟ = ⎜ = – 0.607 f 727.66 ⎠⎟ ⎝ ⎝ max ⎠ From IS : 1915–1961, K is 0.6372 Allowable stress in axial tension, allowing the effect of fatigue = 0.6372 × 0.6 × 260 = 99.40 N/mm2 Net area required

= Area of rivet holes

727.66 × 1000 = 7320.52 mm2 99.40

= 4 × 23.5 ×10 = 940 mm2

Gross area needed

6 00 m m

= (7320 + 940) = 8260.52 mm2 Provide 4 ISA 125 mm × 95 mm × 12 mm Gross area provided = 4 × 2498 = 9992 mm2. Hence, safe. The section for U4L5 is shown in Figure 4.14.

4 IS A 1 25 m m x 95 m m x 12 m m

Fig. 4.14

DESIGN OF TRUSS GIRDER BRIDGES

191

Step 9. Design of joint U4 Use 22 mm diameter power-driven rivets. Strength of rivet in single shear =

π (23.5)2 × 100 × = 43.35 kN 4 1000

=

23.5 × 10 × 300 = 70.5 kN 1000

Strength of rivet in bearing

Rivet value, R = 43.35 kN Force in U3U4 = 3535.47 kN U4U5 = 3676.89 kN The top chord member is a continuous member. The rivets are provided for the difference of forces. Number of rivets required

⎛ 3637.89 – 3535.47 ⎞ = ⎜ ⎟ = 3.26 43.35 ⎝ ⎠ Provide 16 rivets Force in U4L5(max.) = 559.72 kN Number of rivets required

⎛ 559.72 ⎞ = ⎜ ⎟ = 12.91 ⎝ 43.35 ⎠ Force in U4L5 (max.) = 727.66 kN Number of rivets required

⎛ 727.66 ⎞ = ⎜ ⎟ = 16.78 (say 20) ⎝ 43.35 ⎠ Step 10. The joint U4 is as shown in Figure 4.15. The gusset plates are kept 22 mm thick. Example 4.2. The effective span of a through type truss girder highway through two lane bridge is 64 m. The reinforced concrete slab is 250 mm thick inclusive of the wearing coal. The foot paths are provided on either side of the carriageway. The spacing between centre to centre of truss girder is 13 m. Suggest a suitable truss girder for the bridge. Design the central top chord member, the central bottom chord member, the vertical and diagonal member of the central panel. Design the joint, where the central bottom chord, vertical and diagonal members. The highway bridge is to carry IRC class A standard loading.

192

DESIGN OF STEEL STRUCTURES–VOL. II

M em b er U 3 U 4

Joint U 4

U 4U 5 R ive ts 2 2 m m in d ia m e te r

M em b er U 4L 4

Fig. 4.15

Solution Design : A Pratt type truss girders for 64 m span with 16 panels as shown in Fig. 4.16 are provided for highway bridge. U1

U2

U3

U4

L3

L4

U5

U6

U7

U8

U9

U10

U 11

U 12

U 13 U 14

U 15

L 11

L 12

L 13

L 15 L 16

8m L0 L1

L2

L5

L6 L7 L8 L9 L10 1 6 = P a ne ls @ 4 m = 6 4 m

L14

Fig. 4.16

From IRC section I Width of roadway for single lane = 3.80 m For second lane = 3.00 m Width of foot-paths 2 × 1.50 m = 3.00 m Total width = 9.80 m The cross-girders are provided at all the panel points. ∴Numbers of cross-girders = 17 Handrails would be provided on both sides. Height of truss girder

=

1 1 × span = × 64 = 8 m 8 8

(The minimum height of truss girders for highway bridge is of reinforced concrete slab inclusive of wearing coat

1 ×span ). Weight 10

DESIGN OF TRUSS GIRDER BRIDGES

193

⎛ 64 × 9.80 × 250 × 24 ⎞ = ⎜ ⎟ = 3763.2 kN 1000 ⎝ ⎠

Step 1. Dead load Weight of cross-girders (assumed @ 5.00 kN/m) = 17 ×13 × 5 = 1105 kN Weight of handrails for both the sides (assumed @ 1.50 kN/m = 64 × l.50 = 96 kN Self-weight of both the truss girders, from Fuller’s formula, as per Eq. 4.3

⎛ 15l + 550 ⎞ ⎛ 15 × 64 + 550 ⎞ = ⎜ ⎟=⎜ ⎟ × 64 100 ⎝ 100 ⎠ ⎝ ⎠ = 966.4 kN

Total dead load is 5930.6 kN Dead load per truss girder is 2965.3 kN Dead load per metre per girder is 46.233 kN/m Transverse location of live load For the design of members of the truss girder, the live loads are so placed that the reaction on one truss girder is maximum. For IRC section II. Distance between centre to centre of two wheels of IRC class A train of loading = 1.80 m Width of carriageway is 6.80 m The minimum distance between adjacent edge of wheels of two trains

(1.2 − 0.4) ⎡ ⎤ × (6.8 − 5.5) ⎥ = 0.92 m ⎢0.40 + 2 ⎣ ⎦ The distance between centre to centre of adjacent wheels of two trains The minimum clearance between outer edge of the wheel and the roadway face of kerb, for IRC class A loading f = 0.150 mm The transverse location of wheels of trains is as shown in Fig. 4.17. g =

8 .52 m 6 .72 m 3 .50 m

P

P

P

13 m

Fig. 4.17

P

194

DESIGN OF STEEL STRUCTURES–VOL. II

Reaction on the truss girder, A

⎡P ⎤ = ⎢ [(13 − 3.50) + (13 − 5.30) + (13 − 6.72) + (13 − 8.52) ⎥ = 2.15 P ⎣13 ⎦ For IRC section II, impact factor for steel bridges

9 9 ⎛ ⎞ ⎛ ⎞ =⎜ i = ⎜ ⎟ ⎟ = 0.116 ⎝ 13.5 + L ⎠ ⎝ 13.5 + 64 ⎠ Reaction on the truss girder A, including impact = 2.15 × 1.116 P = 2.5 P where, P represents the one wheel load of IRC class A train of loading. In order to account for the effect of both the trains and the impact effect, either the wheel loads of IRC class A loading may be increased or the forces in the members may be increased by multiplying them by factor 2.5 P. Step 2. Influence line diagrams for forces in members (i) Member U7U8 The influence line diagram force U7U8 is as shown in Fig. 4.18 (b). The maximum ordinate of the triangle =

M L8 8

=

1 ⎛ 32 × 32 ⎞ × 8 ⎝⎜ 64 ⎠⎟

= 2.00 units (Compression) (ii) Member L7 L8 The influence line diagram for force in L7L8 is as shown in Fig. 4.18 (c). The maximum ordinate of the triangle =

1 1 ⎛ 24 × 36 ⎞ × MU7 = × ⎜ 8 8 ⎝ 64 ⎠⎟

= 1.97 units (Tension) (iii) Member L6 L7 The influence line diagram of force in L6L7 is as shown in Fig. 4.18 (d). The maximum ordinate of the triangle =

1 1 ⎛ 24 × 40 ⎞ × MU8 = × ⎜ 8 8 ⎝ 64 ⎠⎟

= 1.875 units (Tension) (iv) Memeber U7U7 The influence line diagram for force in U7L7 is as shown in Fig. 4.18 (e). The ordinates are as follows : y1 =

m 7 = = 0.437 units n 16

⎛ n − m − 1 ⎞ ⎛ 16 − 7 − 1 ⎞ y2 = ⎜ ⎟ = ⎜ 16 ⎟ = 0.5 units n ⎝ ⎠ ⎝ ⎠

195

DESIGN OF TRUSS GIRDER BRIDGES

The influence line diagram for force in U7L7 is as shown in Figure 4.18 (f). The ordinates are as follows : cosec θ = y3 =

(8

2

+ 42 8

)

1/2

=

(80 )1 / 2 8

= 1.118

7 m × 1.118 = 0.488 units cosec θ = n 16

8 ⎛ n − m − 1⎞ y4 = ⎜ ⎟⎠ cosec θ = 16 × 1.118 ⎝ n

= 0.559 units The influence line diagram for force in U6L7 is as shown in Fig. 4.18 (g). The ordinates are as follows : y5 =

m 6 cosec θ = × 1.118 = 0.418 units n 16

9 ⎛ n − m − 1⎞ × 1.118 y6 = ⎜ ⎟⎠ cosec θ = ⎝ 16 n = 0.628 units Step 3. Forces in the members due to dead load Dead load = 46.33 kN/m (i) Force in member U7U8 1 × 2 × 64 × 46.33 = – 2965.12 kN (i.e., compression) 2 (ii) Force in member L7L8

=–

1 × 1.97 × 64 × 46.33 = + 2920.64 kN (i.e., compression) 2 (iii) Force in member L6L7 1 = + × 1.875 × 64 × 46.33 = + 2779.8 kN (i.e., tension) 2

=+

(iv) Force in member U7L7 1 1 × 34.13 × 0.5 – × 29.87 × 0.437 × 46.33] 2 2 = –92.26 kN (i.e., compression) (v) Force in member U7L8

= –[

1 1 × 34.13 × 0.559 – × 29.87 × 0.437 × 46.33] 2 2 = + 102.93 kN (i.e., tension)

=–[

196

DESIGN OF STEEL STRUCTURES–VOL. II

(vi) Force in member U5L7 1 1 × 38.4 × 0.628 – × 25.6 × 0.418 × 46.33] = 303.16 kN 2 2 Step 4. Forces in members due to live load and impact load (i) Force in member U7L8

=[

1 × ML8 8 The force in member U7U8 is maximum when the bending moment ML8 is maximum. The bending moment ML8 about panel point L8 is maximum, when the shear force changes sign (in other words, when the average loading on the portion left of L8 and the average loading on the portion right on L8 change sign from heavy to light or from light to heavy). Consider from right hand support, when first four loads are on portion L8B. Average loading on L8B

Force in member U7U8 =

1 (75 + 13.5 + 13.5 + 3 × 34) = 5.81 kN/m (Light) 32 Average loading on AL8

=

1 (57 + 4 × 34) = 6.03 kN/m (Heavy) 32 Consider from right support, when first five loads are on portion L8B. Average loading on L8B

=

1 (2 × 57 + 2 × 135 + 3 × 34 32 = 7.59 kN/m (Heavy)

= Average loading on AL8

1 (4 × 34) = 4.25 kN/m (Light) 32 Thus, it is seen that the force in member U7U8 is maximum for the longitudial position of loading as shown in Figure 4.18 (b). (i) Force in member U7U8

=

2 [34 (18.7 + 21.7 + 24.7 + 27.7) + 57 (23 + 30.8) 32 + 135 (27.6 + 26.5) + 34 (6.5 + 3.5 + 0.5)] = 488.88 kN Force in member U7U8 including the effect of adjacent passing vehicle and effect of impact = 2.5 × 488.88 = 122.22 kN (Compression) (ii) Force in member L7L8 The force in member L7L8 is maximum for the longitudinal position of loading as shown in Fig. 4.18 (c). This position of loading has been fixed as discussed for force in member U7U8.

=

197

DESIGN OF TRUSS GIRDER BRIDGES

3 m

3 m

4 .3 m

U 11

U 12

U 13 U14

U 15

L 11

L12

L13

L15

U7 U8

I.L.D . fo r

3 .2 m

L14

20 m

3 m 3 m

(b )

0 .5 m

U7 U8

I.L.D . fo r

2 .00

B L 16

3 .4 t

U 10

3 .4 t

U9

L6 L7 L8 L9 L 10 1 6 = P a ne ls @ 4 m . 64 m

3 m

(– )

U8

L5

3 .4 t

(a )

U7

3 .4 t

L4

U6

1 .1 m 1 .3 5 1 .3 5

L3

U5

5 .7 t 1 .2 m 5 .7 t

L2

U4

3 .4 t

L1

U3

3 .4 t

U2

3 .4 t

U1

34 (c)

(+ )

3 m

3 m

135 1 35

1 .97 34

5 75 7 34 34 3 4 .3 3 .2 m m m

(+ )

3 m

34

3 m

34 57 57 57 4 .3 m

y 1 = 0 .43 7 (e )

I.L.D . Fo r L 6 L 7 3 4.1 4 m

(+ ) 34

34

34 34

34 34 34 34 3 3 3 m m m

20 m

1.35 1.35 57

3 m

1 .3 5 1 35

(d )

1.2 3.2 m 1 35 1.1 1 35

34

L7 L 8

I.L.D . fo r

1 .87 5 34

34 3 34 34 34 3 3 1 .5 m m m m

20 m

34

34 34

34 (-)

57 57

2 9.8 6 m

(f)

34

34 34

34

57 57

(g )

34

34 34

y 5 = 0 .41 8

34

(+ ) 57 57

2 5.6 m

34 34

57 57 34 34

1 .3 5 1.35

34

(+ ) 34

I.L.D . fo r U 7 U 8 3 3.4 m y 6 = 0 .62 8

1 35

2 9 . 8 6m

U7 U7

y 4 = 0 .55 9

57 57 y 3 = 0 .48 8

I.L.D . fo r

1 .3 5

1 .3 5

1.35

y 2 = 0 .5

34 34

I.L.D . fo r

Fig. 4.18

U6 U7

4 .3 m

57 57 1 .2 m

198

DESIGN OF STEEL STRUCTURES–VOL. II

Force in member L7L8 =

1.97 1.97 [34(14.7+17.7 + 20.7 + 23.7) + 57 × 28] + [ 57 × 34.8 + 13.5 28 36

(31.6 + 30.5) + 34 (10.5 + 7.5 + 4.5 + 1.5)] = 495.08 kN Force in member L7L8 including the effect of adjacent passing vehicle and effect of impact = 2.5 × 495.08 = 1237.70 kN (Tension). (iii) Force in member L6L7 The force in member L6L7 is maximum for the longitudinal position of loading as shown in Fig. 4.18 (d). This position of loading has been fixed as discussed for member U7U8. Force in member L6L7 1.875 1.875 [34(10.7 + 13.7+ 16.7 + 19.7) + 57 × 24] + [57 × 38.8 +13.5 40 24 + (35.6 + 34.5) + 34(14.5 + 11.5 + 8.5 + 5.5) + 57(1.2 + 0)] = 483.36 kN Force in member L6L7 including the effect of impact and adjacent passing vehicles = 2.5 × 483.36 = 1208.4 kN (Tension). (iv) Force in member U7L7 Force in U7L7 = S.F. in panel L7L8 The force in U7L7 is maximum, when shear force in panel L7L8 is maximum. The maximum shear force in this panel occurs when the load in that panel is equal to the total load divided by number of panels. Total load = (2 × 13.5 + 2 × 57 + 4 × 34) = 277 kN Number of panels = 16 Load in the panel L7L8 for maximum shear force

=

227 = 17.31 kN 16 For IRC class A train of vehicles, when wheel load of 13.5 kN is on the panel, then maximum shear force occurs in the panel L7L8. (A) Compression in member U7L7 The compression in U7L7 is maximum, when train of loading passes from right to left. The position of wheel loads is shown in Fig. 4.18 (e) by firm lines. Compression in U7L7

=

=

0.5 [57 (32 + 30.8) + 34 (26.5 + 23.5 + 20.5 + 17.5)] 32

0.437 0.437 × 27.7 + 13.5 × 1.06] = 93.48 kN 28 1.85 Force in member U7L7 including the effect of adjacent passing vehicle and the impact = 93.48 × 25 = 233.7 kN.

– [l3.5 ×

199

DESIGN OF TRUSS GIRDER BRIDGES

(B) Tension in member U7L7 The tension in U7L7 is maximum when train passes from left to right. The longitudinal position of wheel loads is shown in Fig. 4.18 (e) by dotted lines. Tension in U7L7 =

0.437 0.5 [57(28 + 26.8) + 34(22.5 + 19.5 + 16.5 + 13.5)] – [13.5 × 28 0.24

× 1.34 + 13.5 ×

0.5 × 13.7] = 76.23 kN 32

Tension in U7L7 including the effect of adjacent passing vehicle and the impact = 25 × 76.23 kN = 190.575 kN (Tension) (v) Force in U7L8 The position of loading is fixed as discussed for member U7L7. (A) Tension in U7L8 The tension in U7L8 is maximum, when train passes from right to left. The longitudinal position of wheel loads is shown in Fig. 4.18 (f) by firm lines. Tension in U7L8 =

0.599 [57(32 + 30.8) + 34 (26.5 + 23.5 + 20.5 + 17.5)] 32

0.488 0.488 × 27.7 + 13.5 × × 1.06] = 112.74 kN 28 1.86 The tension in U7L8 including the effect of adjacent passing vehicle and the impact = 112.74 × 2.5 = 281.85 kN (Tension). (B) Compression in U7L8 The compression in U7L8 is maximum when train passes from left to right. The longitudinal position of wheel loads is shown in Fig. 4.18 (f) by dotted lines. Compression in U7L8

–[13.5 ×

=

0.488 [57(28 + 26.8) + 34 (22.5 + 19.5 + 16.5 + 13.5)] 28

0.559 0.559 × 1.34 + 13.5 × × 31.7] = 85.90 kN 32 2.14 Compression in member U7L8 including the effect of adjacent passing vehicles and impact = 2.5 × 85.90 = 214.75 kN (Compression). Force in member U6L7 The position of loading is fixed as discussed for member U7L7 . (vi) Tension in U6L7 The tension in U7L7 is maximum when the train of loading passes from right to

– [1.35 ×

200

DESIGN OF STEEL STRUCTURES–VOL. II

left. The longitudinal position of wheel loads is shown in Fig. 4.18 (g) by firm lines. (A) Tension in U6L7 =

0.628 [57(36 + 34.8) + 34(30.5 + 27.5 + 24.5 + 21.5+ 1.5)] 36

0.418 0.418 × 23.7 + 13.5 × × 0.8] 124.8 kN. 25 1.6 Tension in the member U6L7 including the effect of adjacent passing vehicle and impact = 124.8 × 2.5 = 312 kN (Tension). (B) Compression in U6L7 The compression U6L7 is maximum when the train of loading passes from left to right. The position of wheel loads is shown in Fig. 4.18 (g) by dotted lines.

–[ 13.5 ×

=

0.418 [57(24+ 22.8) + 34 (18.5 + 15.5 + 12.5 + 9.5)] 24

0.628 13.5 × 35.7 + 13.5 × × 1.6] = 61.36 kN 2.4 24 Compression in the member U6L7 including the effect of adjacent passing vehicle and the impact = 61.36 × 2.5 = 153.4 kN (Compression). Design forces in the members are as follows :

– [0.628 ×

Forces in the member (Dead Load)

(L.L. + L.L.)

Member

Compression (kN)

Tension

U7U8 L7L8 L6L7 U7L7 U7L8 U6L7

2965.12 — — 92.26 — —

D.L. + L.L. + I.L. Tension

(kN)

Compression (kN)

Tension

(kN)

Compression (kN)

— 2920.64 2779.80 — 102.93 303.16

1222.2 — — 233.7 214.75 153.4

— 1237.70 1208.40 190.575 281.85 312.00

4187.32 — — 325.96 111.82 —

— 4158.34 3988.20 98.315 384.78 149.76 615.16

(kN)

Step 5. Design of chord member Force in the member = 4187.32 kN (Compression). Allowance is not made for the effect of fatigue for the compression members. Depth of the truss girder = 8000 mm Depth of top chord member

201

DESIGN OF TRUSS GIRDER BRIDGES

⎛1 ⎞ = ⎜ × 8000⎟ = 800 mm ⎝ 10 ⎠ Width of top chord member

⎛1 ⎞ = ⎜ × 8000 + 2 × 22⎟ = 844 mm ⎝ 10 ⎠ Try the section as shown in Fig. 4.19. Approximate radius of gyration rx = 0.39 h = 0.39 × 800 = 312 mm ry = 0.55 b = 0.55 × 844 = 465 mm rmin = 312 mm Length of the member from centre to centre of intersection = 4000 mm Effective length of member = 0.85 × 4000 = 3400 mm Maximum slenderness ratio =

3400 = 10.9 312 Y 1 05 0 m m 9 70 m m 12 m m

y = 3 94 .86 m m

16 m m

8 44 m m

16 m m

8 00 mm

4 17 .14 m m

Y

Fig. 4.19

From IS: 833–1994, allowable stress in axial compression for the slenderness ratio 10.9 and the steel having yield stress as 260 N/mm2 σac = 155.82 N/mm2

202

DESIGN OF STEEL STRUCTURES–VOL. II

Area required

⎛ 4187.32 × 1000 ⎞ 2 A = ⎜ ⎟⎠ = 26872.80 mm ⎝ 155.82 Width of top cover plate of the section between centre to centre of rivet line h = 970 mm Ratio ⎜⎛ b ⎟⎞ = ⎛⎜ 970 ⎞⎟ ⎝ t ⎠ ⎝ 12 ⎠

=

80.8 > 50.

Hence, length of plate (50 × 12) mm is only considered as effective. The area of section provided is as follows : Top cover plate = (50 × 12) × 12 = 7200 mm2 Web plates = 2 × 800 × 16 = 25600 mm2 Angles 4 ISA 80 mm × 50 mm × 6 mm = 4 × 746 = 2984 mm2 Total = 35784 mm2 The c.g. of section from top is at a distance y

⎡105 × 12 × 6 + 2 × 746 × (11.6 + 12) + 2 × 746 ⎤ y = ⎢ × (800 − 11.6 + 12) + 2 × 800 × 16 × 412 ⎥⎥ ⎢ ⎢⎣ 105 × 12 × 2 × 746 + 2 × 746 + 2 × 800 × 16 ⎥⎦ = 394.86 mm 1 ×1.6×803 + 2 × 80 × 2.6 12 × (40 – 39.486)2 + 105 × 1.2 × (39.486 – 0.6)2 + 2 × 7.46 (39.486 – 2.76)2 + 2 × 7.46 × (41.714 – 1.16) + 4 ×12.3] × 104 mm4 = 371839.33 × 104 mm4

Ixx = [2 ×

1 × 1.2 × 1053 + 2 × 80 × 1.6 × 432 12 + 4 × 48] × 104 mm4 = 589298.5 × 104 mm4

Iyy = [

4⎞ ⎛ rmin = ⎜ 371839.30 × 10 ⎟ ⎝ ⎠ 35784

1/2

= 322.35 mm

340 = 10.55 322.35 From IS : 833–94, allowable stress in axial compression, for the steel having yield stress as 260 N/mm2 σac = 155.89 N/mm2

Slenderness ratio =

203

DESIGN OF TRUSS GIRDER BRIDGES

Force carrying capacity of the member

⎛ 155.89 × 35784 ⎞ = ⎜ ⎟⎠ = 5578.36 kN ⎝ 1000 The member may be checked for combined and equivalent stresses. Step 6. Design of bottom chord member L7L8 Force in member = 4158.34 kN To allow for the effect of fatigue fmin = + 2920.64 kN, fmax = + 4158.34 kN ⎛ fmin ⎞ ⎜⎝ f ⎟ max ⎠

⎛ 2920.64 ⎞ = 0.702 = ⎜ ⎝ 4158.34 ⎟⎠

From IS : 1915–1961, the value of K is 1.00 Allowable stress in axial tension = 0.6 × 260 =156 N/mm2

⎛ 4158.34 × 1000 ⎞ Net area required = ⎜⎝ ⎟⎠ = 26656.03 mm2 156 The overall dimension of the section is kept same as that for the top chord section i.e., as shown in Figure 4.20. The area provided is as follows: 2 web plates 800 mm × 10 mm = 2 × 800 × 10 mm2 = 16000 mm2 2 additional plates 800 mm × 8 mm = 2 × 800 × 8 mm2 = 12800 mm2

8 00 x mm

x 4 IS A 8 0 m m x 50 m m x 6 mm

8 00 m m

Fig. 4.20

4 ISA 80 mm × 50 mm × 6 mm = 4 × 746 = 2984 mm2 Gross area provided = 31784 mm2

204

DESIGN OF STEEL STRUCTURES–VOL. II

Use 22 mm diameter rivets. Assume that 4 rivets would be necessary for connecting main members and 4 rivets for lacing. Area of rivet holes = 4 × (10 + 8 + 6) × 23.5 + 4 × 6 × 2.35 = 2820 mm2 Net area provided = (31784 – 2820) = 28964 mm2 > 26656.03 mm2. Hence, safe. The member may be checked for combined and equivalent stresses. Step 7. Design of vertical member U7L7 Compressive force = 325.96 kN Tensile force = 983.15 kN The overall depth of vertical member is kept equal to the internal width of top and bottom chord members less twice the thickness of gusset plate i.e., as shown in Figure 4.21. Depth of member = (844 – 2 × 22) = 800 mm Length between centre to centre of intersections = 8000 mm Effective length of member = 0.7 × 800 = 5600 mm Assuming allowable stress in axial compression, for the slenderness ratio 90 and the steel having yield stress as 260 N/mm2 σac = 92 N/mm2 Area required,

⎛ 325.96 × 1000 ⎞ ⎟⎠ A = ⎜⎝ 91 = 3543.04 mm2 1 25 m m

1 25 m m

L acin g

4 IS A 1 25 m m x 75 m m x 8 m m

1 25 m m

1 25 m m

Fig. 4.21

8 00 mm

205

DESIGN OF TRUSS GIRDER BRIDGES

Provide 4 ISA 125 mm × 75 mm × 8 mm as shown in Fig. 4.21. Area provided = 4 × 1538 = 6152 mm2 Ixx = [4 × 67.2 + 4 × 15.38 × (40 – 1.68)2] × 104 mm4 = 90606.15 × 104 mm4 Iyy = [5 × 245.5 + 4 × l5.38 ×(415 + 0.5)4] ×104 mm4 = 2312.22 × 104 mm4 4⎞ ⎛ rmin = ⎜ 2312.22 × 10 ⎟ ⎝ ⎠ 6152

Slenderness ratio =

1/2

= 61.31 mm

5600 = 91.34 61.31

From IS: 833–1994, allowable stress in axial compression for steel having yield stress as 260 N/mm2 = 90.66 N/mm2 Force carrying capacity of the member

⎛ 90.66 × 6152 ⎞ ⎜⎝ ⎟⎠ 100

= 557.74 kN. Hence, safe.

Step 8. Design of diagonal member U7L8 Tensile force = + 284.78 kN Compressive force = – 111.82 kN To allow for the effect of fatigue fmin = – 111.82 kN fmax = + 384.78 kN ⎛ fmin ⎞ ⎜⎝ f ⎟ max ⎠

⎛ −111.82 ⎞ ⎟ = – 0.29 = ⎜⎝ 384.78 ⎠

From IS: 1915–1961, K = 0.724 Allowable stress in axial tension = 0.724 × 0.6 × 260 = 112.94 N/mm2

⎛ 384.78 × 1000 ⎞ Net area required = ⎜⎝ ⎟⎠ = 3406.82 112.94 Area of rivet holes = 4 × 23.5 × 10 = 940 mm2 Gross area required = 4346.82 mm2 Provide 4 ISA 90 mm × 60 mm × 10 mm. Gross area provided = 1401 × 4 = 5604 mm2. Hence, safe. The section for U7L8 is shown in Fig. 4.22.

206

DESIGN OF STEEL STRUCTURES–VOL. II

90 m m

90 m m

4 IS A 9 0 m m x 60 m m x 10 m m

90 m m

8 00 mm

90 m m

Fig. 4.22

Step 9. Design of joint L7 Use 22 mm diameter power driven rivets. Strength of rivet in single shear

π (23.5)2 × 100 × 4 1000 Strength of rivet in bearing

= 43.35 kN

23.5 × 10 × 300 ⎞ = ⎛⎜ ⎟⎠ kN ⎝ 1000 = 70.5 kN Rivet value, R = 43.35 kN Force in L7L8 = 4158.34 kN L6L7 = 3988.2 kN The bottom chord member is a continuous member. The rivets are provided for the difference of the forces. Number of rivets

⎛ 4158.34 − 3988.2 ⎞ ⎟⎠ = 3.92 = ⎜⎝ 43.35 Provide 8 rivets. Number of rivets

Force in U7L7 = 325.96 kN

⎛ 395.96 ⎞ ⎟ = ⎜⎝ 43.35 ⎠ = 7.52. Provide 8 rivets.

DESIGN OF TRUSS GIRDER BRIDGES

Force in U6L7

= 615.16 kN

Number of rivets

⎛ 615.16 ⎞ = ⎜ = 14.19. Provide 16 rivets. ⎝ 43.35 ⎟⎠

207

The joint L7 is shown in Fig. 4.23. The gusset plates are kept 22 mm thick.

M em b er U 7L 7 M em b er U 6 L7

M em b er L 6 L 7

M em b er L 7 L 8

Fig. 4.23

4.10

BRACING OF DECK TYPE TRUSS GIRDER BRIDGES

In addition to the dead load, live load and impact load, the truss girder bridge is also subjected to lateral and longitudinal forces. In order to provide lateral stability and torsional rigidity, the truss girder bridges are suitably braced. The bracing consists of horizontal bracing and transverse bracing. The horizontal bracings are also termed as horizontal truss bracings. The main horizontal truss bracing is provided in the loaded chord of the girder. In case of the deck type truss girder bridge, the top chord is the loaded chord, and the bottom chord is the unloaded chord. The horizontal truss bracing is also provided in the unloaded chord of this type of truss girder-bridge. The various type of horizontal truss bracings used are shown in Sec. 3.16. The transverse bracings in case of the deck type truss girder bridges consist of cross-frames. The cross-frames are provided in the vertical planes at all the panel points of bridge.

4.11

BRACING OF THROUGH TYPE TRUSS GIRDER BRIDGES

The bracings of through type truss girder bridge consist of top lateral (horizontal) truss bracing, a bottom lateral horizontal truss bracing, portal bracings and sway bracings. The top lateral (horizontal) truss bracing is provided in the horizontal plane of the top chords of the truss girders. The bottom lateral (horizontal) truss

208

DESIGN OF STEEL STRUCTURES–VOL. II

bracing is provided in the horizontal plane of the bottom chords of the truss girders. The portal bracings are provided in the inclined plane of end posts of the truss girder. The sway bracings are provided in the vertical planes at the panel points along with the verticals of the truss girders.

4.12

WIND LOAD ON TRUSS GIRDER BRIDGES

The wind load on truss girder bridge is determined as the product of appropriate basic wind pressure and exposed area of the bridge. The appropriate basic wind pressure for the railway bridges and the highway bridges have been given in Sec. 2.5. The exposed areas of unloaded and loaded truss girder bridges are as follows:

4.12.1

For Unloaded Truss Girder Bridges

The exposed area of unloaded span of the bridge as per Bridge Rules for the railway bridges, and as per IRC section II for the highway bridges are considered as one and half times the area of elevation of truss girders. However, this exposed area is taken as 1.75 times or twice the projected area of the span, which is on safer side.

4.12.2

For Loaded Truss Girder Bridges

The exposed area of loaded span of the bridge as per Bridge Rules for the railway bridges and as per IRC section II for the highway bridges, is considered as one and half times the area of elevation of truss girders plus the area of moving load. However, this exposed area is taken as 1.75 timers or twice the area of elevation of truss girder plus the area of moving load, which is on safer side. The areas of moving loads for the railway bridges, highway bridges, and foot bridges have been discussed in Sec. 3.15.

4.13

WIND EFFECTS ON TRUSS GIRDER BRIDGES

The wind effects on truss girder bridges are considered on the unloaded spans and the loaded spans of the bridges, and the worst effects are taken into consideration and properly accounted for. The wind load has the following four distinct effects on the truss girder bridges: 1. Overturning effect of wind on the span. 2. Lateral effect on the top chords and wind bracing considered as horizontal girder. 3. Lateral effect on the bottom chords and wind bracing considered as horizontal girder. 4. Bending and direct stresses in the members transmitting the wind load from the top to the bottom chords.

4.13.1

Overturning Effect

The wind loads P1 and P2 acting on the unloaded through type truss girder bridge are shown in Fig. 4.24. The wind loads on truss girder are found as discussed in Sec. 4.12.

209

DESIGN OF TRUSS GIRDER BRIDGES

P1

H e ig ht of tru ss g irde r h

P2 0 .8 m

S (S pa cing ) 2R

L evel of b ea rin gs

2R

Fig. 4.24

The wind load acting over unloaded truss girder is determined by multiplying the basic wind pressure and 1.75 times or twice the area of truss girders in elevation, which one is on safer side. A part of this wind load, P1 is assumed to be acting on the top chord. This part of wind load P1 can also be determined by multiplying the basic wind pressure and twice the area of top chord, area of verticals, area of diagonals and end posts and twice the area of top panel points (top gusset plates). The remaining part of the wind load P2, is assumed to be acting on the bottom chord. This part P1 of wind load can be found by the product of basic wind pressure and twice the area of bottom chord, twice the area of bottom panel points (bottom gusset. plates), area of verticals, area of diagonals and area of the end posts. These wind loads P1 and P2 are assumed to act at the centre of gravities of top chord and bottom chord respectively. The wind loads P1 and P2 result in an overturning couple. The overturning couple has a tendency to overturn the bridge about the level of bearings of truss girders. The overturning moment creates two equal and opposite reactions 2R as shown in Fig. 4.24. These reactions furnish a resisting moment to balance the overturning couple. Due to the overturning couple, an additional thrust 2R is caused on the leeward girder. The overturning couple gives an uplift 2R to the windward girder. The additional thrust is similar in action to the dead load. The additional thrust assumed to be acting on the bottom chord. This part P2 of wind provides reaction 2R (R, R at each end of the leeward girder) as shown in Fig. 4.24. The additional thrust has a tendency to increase the compressive forces already existing in the top chord and the tensile forces already existing in the bottom chord of leeward girder. Whereas the uplift 2R due to overturning couple in the windward girder causes opposite kind of forces in the top and bottom chords and therefore, relieves the top chord and bottom chord from the already existing force.

210

DESIGN OF STEEL STRUCTURES–VOL. II

The value of reaction R may be found by taking the moment of wind loads about one of bearings 2R × s = P1 (h + 0.80) + P2 × 0.8 1 [P1h + 0.80) + P2 × 0.80]. 2s The value of reaction R is expressed in terms of percentage of dead load reaction. The increase of compressive force in top chord and tensile force in bottom chord of the leeward girder are determined from those due to dead load only. The nature of forces is additive in the already existing forces in the top and bottom chords of the leeward truss girder. The wind loads acting on loaded through type truss girder in railway bridges are shown in Fig. 4.25 and in highway bridges are shown in Fig. 4.26 respectively. These forces cause overturning couple about the level of bearings. The overturning couple has the tendency to overturn the bridge. The overturning couple causes additional thrust 2R1 in the leeward girder and uplift 2R1in the windward truss girder. The nature of additional thrust over the leeward girder is similar to that due to dead load, live load and impact load. The additional thrust causes additional compressive force in top chord and additional tensile force in the bottom chord of the leeward girder. The uplift causes opposite nature of forces in the chords of windward girder and, therefore, relieves them from already existing forces. The values of reactions can be found by taking the moment about one of the bearings. The wind load on area of moving load, in railway bridges acts at the centre of gravity of the rolling stock as shown in Fig. 4.25. The wind load on area

R =

W ind o n m o ving loa d

1 .50 m R o ad surface

S

1 .45 m

Fig. 4.25

211

DESIGN OF TRUSS GIRDER BRIDGES

of moving load in the highway bridges acts at 1.50 m above the road surface as shown in Fig. 4.26. The road surface is provided 1.45 m above the level of bearings. The increase in compressive forces in the top chord and in tensile forces in the bottom chord of leeward truss girder are determined from those due to normal loads (dead load, live load and impact load) only. P1

P3 (1 .41 5 m , M .G .) (1 .75 m , B .G .)

h 0 .60 m

h3

P2 S L evel of b ea rin gs 2R1

h 2 = 0 .80 m

2R 1

Fig. 4.26

The lateral effect on top chord, the lateral effect on bottom chord and the bending stresses in the members transmitting the wind load from top chord to the bottom chord have been discussed in Secs. 4.14, 4.15, 4.16 and 4.23 respectively.

4.14

TOP LATERAL BRACING

In case of the through type truss girder bridges, the top lateral (horizontal truss) bracings are provided in the horizontal plane between the top chords of bridges. W ind loa d pe r u nit le n gth Top cho rd of w in d-w ard g ird er

D ia g on als

S truts Top cho rd of le ew a rd girde r

Fig. 4.27

212

DESIGN OF STEEL STRUCTURES–VOL. II

The top lateral bracing provides rigidity to the bridge structure. The top chords of through type bridge are the compression chords. The top lateral bracing connects both the top (compression) chords. The top lateral bracing also acts as lacing between the top (compression) chords. This provides stability to these chords. The top chords of main truss girders also act as chords of top horizontal lateral bracing as shown in Fig. 4.27. The cross or double diagonal horizontal bracing is most commonly used. This type of diagonal bracing gives better appearance. The diagonals are connected at the points of their intersections. The struts are used to connect opposite points and to complete the horizontal truss. The top lateral (horizontal) truss bracing resists wind load, acting at the centre of gravity of top chord. The maximum wind load on the top chord is in case of unloaded span and the same is taken into consideration. In addition to the wind, the top lateral bracing is also designed to resist transverse shear force at any section. The transverse shear force is assumed equal to 2.5 percent of the sum of compressive force in both the top chords at the section under consideration. For the analysis, it is assumed that the diagonal member which is in tension, remains effective. The other diagonal member remains dummy. When the direction of wind changes, the former effective diagonal members become dummy members and the former dummy diagonal members become effective tension members. The end strut and the end diagonal member carry maximum forces. The strut and diagonal are designed for these forces. The values of forces in strut and diagonal are small. These require nominal sizes. The same sizes are adopted for intermediate struts and intermediate diagonals. The single or double angle sections are adopted for the struts and diagonals. As regards lateral effect on the top chords, the maximum wind load act as uniformly distributed load on the top chords as shown in Fig. 4.17. The top lateral bracing acts as a simply supported horizontal girder. The top lateral bracing is subjected to maximum moment in the centre. It is assumed from steel theory that this moment is resisted by two equal and opposite forces in the chords acting at a distance equal to the spacing of truss girders. The values of forces are determined for each panel by equating the resisting moment to the corresponding moment. This force is compressive in the top chord of the windward girder and tensile in the top chord of the leeward girder. The nature of this force is additive for the top chord of windward girder and subtractive for the top chord of the leeward girder.

4.15

BOTTOM LATERAL BRACING

In case of the through type bridge, the bottom lateral (horizontal truss) bracing is provided in the horizontal plane between the bottom chords. The bottom chords of main truss girders also act as chords of bottom horizontal (lateral) bracing. The cross or double diagonal bracing as shown in Fig. 4.28 is most commonly used.

213

DESIGN OF TRUSS GIRDER BRIDGES

The cross or double diagonal bracing gives better appearance. The diagonals are connected the floor beams and also at the point of their intersections. The bottom lateral bracing resists wind load acting at the centre of gravity of the bottom chord. In addition to the wind load, the bottom horizontal truss bracing also resists racking force and the longitudinal force. The racking forces has been discussed in Sec. 2.6. The longitudinal force has been discussed in Sec. 2.7. The maximum lateral load (wind load in case of unloaded span and wind load plus racking force in case of loaded span) is taken into consideration and properly accounted for. The longitudinal forces have a tendency to develop bending stresses in the flanges of the floor beams. In order to avoid this, the longitudinal forces are transmitted to L0 c d

L1

L2

L3

L4

L5

b

b

d

L0

L6

d

d L1

L2

L3

L4

c

L5

L6

L5

L6

(a ) S ing le tra ck L0 e f g h L0

L1

L2

L3

L4

d

d

c

c b

b

a

a L1

L2

L3

L4

L5

e f g h L6

(b ) D o ub le tra ck

Fig. 4.28

the main truss girders by means of traction frames. In single track bridge, the transverse struts are connected with the bottom flanges of the stringers (rail bearers) at the points of their intersection with the diagonals of the bottom horizontal (lateral) bracing as shown in Fig. 4.28 (a). Such transverse strut, ab with the diagonals makes a Lo ab Lo c d Lo´ queen post truss. This queen post truss (traction frame) transmits the longitudinal forces to the main truss. Similar traction frame is also provided at the other end. In double-track bridges, the traction frame Lo’abcd Loefgh Lo, is made as shown in Fig. 4.28 (b). Such traction frames are provided in the end panels. It is not necessary to provide such traction frames in each panel. It is desirable to provide such traction frames at intervals of 30 m. The longitudinal forces are transmitted to these frames by the stringers. The magnitude of longitudinal force depends upon the number of panels over which tractive or braking forces are applied. These traction frames may also be provided at the top flange of the stringers.

214

DESIGN OF STEEL STRUCTURES–VOL. II

For the purpose of analysis, it is assumed that the diagonal members which are in tension, remain effective. The other diagonal members remain dummy. When the direction of wind changes, the former effective diagonal members become dummy and the former dummy members become effective tension members. The diagonal members are designed for the maximum lateral load of loaded and unloaded spans. As regards the lateral effect on the bottom chord, the bottom lateral bracing act as a horizontal truss girder. It is to note that for this purpose, only wind load is considered and racking force is not taken into consideration. The bottom lateral bracing is subjected to bending moment due to uniformly distributed wind load acting on the bottom chord. It is assumed from steel theory, that this moment is resisted by two equal and opposite forces in the bottom chords, acting at a distance equal to the spacing of main truss girders. The value of force is found by equating the moment and resisting moment. This force is compressive in the bottom chord of the windward girder and tensile in the bottom chord of the leeward girder. The nature of this force is additive for the bottom chord of the leeward girder and subtractive for the bottom chord of windward girder. Example 4.3 The effective span of a through type Pratt truss girder railway bridge for a single broad gauge track is 50 m. Pratt truss girder consists of 10 panels @ 5 m. The height of girder between c.g. to c.g. of chords is 6 m. The spacing between main truss girders is 7 m. The rail level is 800 mm above the c.g. of bottom chord. The chord members are 600 mm deep × 644 mm wide. The inner web members are 600 mm deep × 260 wide. The end posts are 600 mm deep × 644 mm wide. Determine the increase or decrease of forces in the central chord member of the leeward truss girder in the following cases : (a) Overturning effect due to wind, when the bridge is unloaded. (b) Lateral effects of top chord and bottom chord bracings, when the bridge is unloaded. (c) Overturning effect due to wind, when the bridge is loaded. (d) Lateral effects of top chord and bottom chord bracings, when the bridge is loaded. Solution A Pratt truss girder is as shown in Fig. 4.29.

L0

U1

U2

U3

L1

L2

L3

U4

U5

L4

L5

Fig. 4.29

Step 1. (Case I) Bridge is unloaded Wind pressure = 2.40 kN/m2

U6

L6

U7

L7

U8

L8

U9

L9

L10

215

DESIGN OF TRUSS GIRDER BRIDGES

The details of exposed area of truss girder area as follows : S.No. 1. 2. 3. 4. 5. 6.

Details of exposed area Top chord Bottom chord End posts Verticals Diagonals Gusset plate area @ 0 5 m2 for each gussets 9 × 0 5 m2 at top and 11 × 0.5 m2 at bottom

Depth 0.60 0.60 0.60 0.60 0.60

m m m m m

Width 0.644 0.644 0.644 0.26 0.26

m m m m m

Face 0.60 0.60 0.60 0.26 0.26

m m m m m

Total length/area 40 m 50 m 2 ×7.8 m 9 × 6.0 m 8 × 7.8 m

—

—

—

4.5 m2

—

—

—

5.5 m2

Wind load on windward girder is as follows : 1. Wind load on top chord = (40 × 0.60 × 2.40) = 57.60 kN 2. Wind load on bottom chord = (50 × 0.60 × 2.40) = 72 kN 3. Wind load on verticals = (9 × 6 × 0.26 × 2.40) = 33.70 kN 4. Wind load on diagonals = (8 × 7.8 × 0.26 × 2.40) = 38.94 kN 5. Wind load on end posts = (2 × 7.8 × 0.60 × 2.40) = 22.46 kN 6. Wind load on top gussets = (4.5 × 2.40) = 10.8 kN 7. Wind load on bottom gussets = (5.5 × 2.40) = 13.2 kN Spacing between main girders = 7 m. Wind load is assumed 75 percent of wind load on windward girder for leeward girder. Wind load on leeward girder is as follows : 1. Wind load on top chord = 0.75 × 57.60 = 43.20 kN 2. Wind load on bottom chord = 0.75 × 72.0 = 54 kN 3. Wind load on verticals = 0.75 × 33.70 = 25.27 kN 4. Wind load on diagonals = 0.75 × 38.94 = 29.21 kN 5. Wind load on end posts = 0.75 × 24.4 = 18.30 kN 6. Wind load on top gussets = 0.75 × 10.8 = 8.1 kN 7. Wind load on bottom gussets = 0.75 × 13.2 = 9.9 kN Wind load acting on top chord 1 1 wind load on verticals + wind load on 2 2 diagonals and end posts + wind load on top gussets

P1 = Wind load on top chord +

P1 = [(57.60 + 43.20) +

1 1 1 (33.70 + 25.27) + (38.94 + 29.21) + (22.46 + 18.30) 2 2 2 + (108 + 8.1)] = 203.64 kN

216

DESIGN OF STEEL STRUCTURES–VOL. II

Wind load acting on bottom chord 1 1 wind load on vertical + wind load on 2 2 diagonals and end posts + wind load on bottom gussets

P2 = Wind load on bottom chord +

1 1 1 (33.70 + 25.27) + (38.94 + 20.21) + (22.46 + 18.30) 2 2 2 + (13.2 + 9.9)] = 233.04 kN Wind loads acting on top and bottom chords, when the bridge is unloaded are shown in Fig. 4.30.

P2 = [(72 + 54) +

P1

6m

P2 0 .80 m

7m 2R

L evel o f b ea rin gs

2R

Fig. 4.30

Step 2. Overturning effect due to wind, when bridge is unloaded. Take the moment about level of bearings 2R × 7 = (P1 × 6.80 + P2 × 0.80) = (203.64 × 6.80 + 233.04 × 0.8) R = 112.23 kN, 2R = 224.45 kN Due to overturning effect, a thrust of 224.45 kN acts downward on leeward girder. Increase of stress in central top chord member U4U5 of the leeward girder

1 ⎛ 1 25 × 25 ⎞ ⎛ 224.45 ⎞ × ⋅ (50) ⋅ ⎜ ⎜ ⎟ ⎟ 2⎝6 50 ⎠ ⎝ 50 ⎠ = 233.80 kN (compression). Increase of stress in central bottom chord L4L5 of the leeward girder =

1 ⎛ 1 25 × 30 ⎞ ⎛ 224.45 ⎞ × × ⋅ (50) ⋅ ⎜ ⎟ 2 ⎝⎜ 6 50 ⎠⎟ ⎝ 50 ⎠ = 224.45 (tension) =

DESIGN OF TRUSS GIRDER BRIDGES

217

Step 3. Lateral effect of top chord bracing when the bridge is unloaded. P1

7m 8 P a ne ls @ 5 m = 40 m

Fig. 4.31

Wind pressure acting on top lateral bracing is shown in Fig. 4.31. The top chord members of the leeward girder are subjected to tension due to lateral effect of the top lateral bracing. Therefore, the forces in these members decrease. Decrease in force in central top chord member, U4U5 due to top lateral bracing.

⎛ 203.64 × 40 1 ⎞ × ⎟ = 145.46 kN = ⎜ 8 7⎠ ⎝ Step 4. Lateral effect of bottom bracing when the bridge is unloaded. Wind pressure acting on bottom lateral bracing is shown in Fig. 4.32. P2 7m 1 0 P an els @ 5 m = 5 0 m

Fig. 4.32

The bottom chord members of leeward girder are subjected to tension. Therefore, the forces in these members increase. Increase of force in central bottom chord member due to bottom lateral bracing

⎛ 233.04 × 50 ⎞ 1 = ⎜ ⎟ × = 208.07 kN (Tension) 8 ⎝ ⎠ 7 Step 2. (Case II) Bridge is loaded Wind pressure = 1.50 kN/m2 Wind load on top chord

⎛ 203.64 × 1.50 ⎞ = ⎜ ⎟ = 127.275 kN 2.40 ⎝ ⎠

Wind load on bottom chord

⎛ 233.04 × 1.50 ⎞ = ⎜ ⎟ = 145.65 kN 2.40 ⎝ ⎠

Wind load on moving train = 50 × 3.50 × l.50 = 262.50 kN Wind loads acting on the loaded bridge are shown in Fig. 4.33.

218

DESIGN OF STEEL STRUCTURES–VOL. II

B ro ad g au ge ro lling sto ck

0 .8 0 m

0 .6

1.7 5

P3

3 .5 m

P1

P2

0 .80 m

7m 2R

L evel o f b ea rin gs

2R

Fig. 4.33

Step 6. Overturning effect due to wind, when bridge is loaded Take moment about one of bearings 2R × 7 = [127.275 × 6.80 + 145.65 × 0.80 + 262.50 (0.75 + 0.60 + 0.80 + 0.80)] R = 125.45 kN, 2R = 250.91 kN Due to overturning a thrust of 250.91 kN acts downward on leeward girder. Increase of stress in central top chord U4U5 member of leeward girder =

1 1 ⎛ 25 × 25 ⎞ ⎛ 250.91 ⎞ × × ⋅ (50 ) ⋅ ⎜ ⎟ 2 6 ⎜⎝ 50 ⎟⎠ ⎝ 50 ⎠

= 261.36 kN (Compression) Increases of stress in central bottom chord member L4L5 =

1 ⎛ 1 ⎞ ⎛ 20 × 30 ⎞ ⎛ 250.91 ⎞ × × ⋅ (50 ) ⋅ ⎜ ⎟ 2 ⎝⎜ 6 ⎠⎟ ⎝⎜ 50 ⎠⎟ ⎝ 50 ⎠

= 250.91 kN (Tension). Step 7. Lateral effect or top chord bracing when the bridge is loaded Wind load on top lateral bracing, when the bridge is unloaded = 203.64 kN Wind load on top lateral bracing, when the bridge is loaded = 127.275 kN Decreases in force in central top chord member due to lateral bracing when bridge is unloaded = 145.46 kN

DESIGN OF TRUSS GIRDER BRIDGES

219

Decrease in force in central top chord member due to top lateral bracing when bridge is loaded

⎛ 145.46 × 127.275 ⎞ = ⎜ ⎟ = 90.91 kN 203.64 ⎝ ⎠ Step 8. Lateral effect of bottom chord bracing when bridge is loaded Wind pressure acting on bottom lateral bracing when the bridge is unloaded = 233.04 kN Increase of force in central bottom chord member due to bottom lateral bracing = 208.07 kN (Tension) Wind pressure acting on bottom lateral bracing when the bridge is loaded = (P2 + P3) = (145.65 + 262.50) = 408.15 kN Increase of force in central bottom chord members due to bottom lateral bracing ⎛ 408.15 ⎞ = 208.07 × ⎜ ⎟ = 363.23 kN (Tension). ⎝ 233.80 ⎠ Example 4.4 The effective span of a through type Pratt truss girder two lane highway bridge is 64 m. The highway bridge carries IRC class A loading. Pratt truss girder consists of 16 panels @ 4 m. The depth of girder between c.g. to c.g. of chord is 8 m. The spacing between main girders is 13 m. The chord members are 800 mm deep × 844 mm wide. The vertical members are 800 mm deep × 260 mm wide and the diagonal members are 800 mm deep × 190 mm wide. The end posts are 800 mm deep × 844 mm wide. Determine the increase or decreases in the central chord members of the leeward truss girder in the following cases : (a) Overturning effect due to wind, when the bridge is unloaded (b) Lateral effects of top chord and bottom chord bracing, when the bridge is unloaded (c) Overturning effects due to wind, when the bridge is loaded (d) Lateral effects of top chord and bottom chord bracings, when the bridge is loaded. Solution A Pratt truss girder is shown in Fig. 4.34. Step 1. (Case I) Bridge is unloaded Wind pressure = 240 kN/m2 Length of diagonals = (82 + 42)1/2 = 8.94 m The details of exposed area of truss girder are as follows : U1

U2 U3

U4 U5

U6 U7

U8 U9

U 1 0 U 11 U 1 2 U 1 3 U 1 4 U 1 5

L1

L2

L4 L5

L6

L8 L9

L 1 0 L 11 L 1 2 L 1 3 L 1 4 L 1 5 L 1 6

L3

L7

1 6 P an els @ 4 m = 6 4 m

Fig. 4.34

220 S.No. 1. 2. 3. 4. 5. 6.

DESIGN OF STEEL STRUCTURES–VOL. II

Details of exposed area Top chord Bottom chord End posts Verticals Diagonals Gussets for top chord @ 0.5 m2 15 × 0.5 m2 For bottom chord @ 0.5 m2 17 × 0.5 m2

Depth 0.80 0.80 0.80 0.80 0.80

m m m m m

Width 0.844 0.844 0.844 0.26 0.19

m m m m m

Face 0.80 0.80 0.80 0.26 0.19

Total length/area m 56 m m 64 m m 8 × 8.94 m m 15 × 8.0 m m 14 × 8.94 m 7.5 m2 8.5 m2

Wind load on windward and leeward girders (assumed 100 percent of wind load on windward girder) both is as follows : 1. Wind load on top chord = 56 × 0.80 × 2.40 × 2 = 215.04 kN 2. Wind load on bottom chord = 64 × 0.80 × 2.40 × 2 = 245.76 kN 3. Wind load on verticals = 15 × 8 × 2.40 × 0.26 × 2 = 149.76 kN 4. Wind load on diagonals = 14 × 8.94 × 0.19 × 2.40 × 2 = 114.15 kN 5. Wind load on end posts = 2 × 8.94 × 0.80 × 2.40 × 2 = 68.66 kN 6. Wind load on top gussets = 7.4 × 2.40 × 2 = 36 kN 7. Wind load on bottom gussets = 8.5 × 2.40 × 2 = 40.8 kN Wind load acting on top chord 1 1 wind load on verticals + wind load on 2 2 diagonals and end posts + wind load on top gussets

P1 = Wind load on top chord +

1 1 1 × 149.76 + × 114.14 + × 68.65 + 36] = 417.28 kN 2 2 2 Wind load acting on bottom chord

P1 = [215.04 +

1 1 wind load on verticals + wind load on 2 2 diagonals and end posts + wind load on bottom gussets

P2 = Wind load on bottom chord +

P2 = [245.76 +

1 1 1 × 149.76 + 114.15 + × 68.65 + 40.8] = 452.83 kN 2 2 2

221

DESIGN OF TRUSS GIRDER BRIDGES

Wind loads acting on top chord and on bottom chord, when the bridge is unloaded are shown in Fig. 4.35. P1

6m

P2 0 .80 m

13 m 2R

2R

Fig. 4.35

Step 2. Overturning effect due to wind when bridge is unloaded Take the moment about the level of bearings 2R × 13 = P1 × 8.80 + P2 × 0.80 = 417.28 × 8.80 + 452.84 × 0.80 R = 155.17 kN, 2R = 310.33 kN Due to overturning effect, a thrust of 310.33 kN acts downward on leeward girder. Increase of stress in central top chord member U7U8 of the leeward girder =

1 ⎛ 1 ⎞ ⎛ 32 × 32 ⎞ ⎛ 310.33 ⎞ ×⎜ ⎟×⎜ ⋅ (64 ) ⋅ ⎜ ⎟ ⎟ 2 ⎝ 8 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠

= 310.33 kN (Compression) Increase of stress in central bottom chord member L7L8 of the leeward girder =

1 ⎛ 1 ⎞ ⎛ 28 × 36 ⎞ ⎛ 310.33 ⎞ × × ⋅ (64 ) ⋅ ⎜ ⎟ 2 ⎝⎜ 8 ⎠⎟ ⎝⎜ 64 ⎠⎟ ⎝ 64 ⎠

= 305.48 kN (Tension). Step 3. Lateral effect of top chord bracing when the bridge is unloaded Wind pressure acting on top lateral bracing is shown in Fig. 4.36.

222

DESIGN OF STEEL STRUCTURES–VOL. II

P1 13 m 1 4 P an els @ 4 m = 5 6 m

Fig. 4.36

The top chord members of leeward girder are subjected to tension due to lateral effect of top lateral bracing. Therefore, the forces in these members decrease. Decrease in force in central top chord member due to top lateral bracing P1 × 56 1 × 8 13

⎛ 417.28 × 56 1 ⎞ × ⎟ kN = ⎜ 8 13 ⎠ ⎝

= 224.69 kN (Tension) Step 3. Lateral effect of bottom chord bracing when the bridge is unloaded Wind load acting on bottom chord bracing is shown in Fig. 4.37. P2 13 m 1 6 P an els @ 4 m = 6 4 m

Fig. 4.37

The bottom chord members of leeward girder are subjected to tension. Therefore, the forces in these members increase. Increase of forces in central bottom chord member due to bottom lateral bracing P2 × 64 1 × 8 13

⎛ 452.84 × 64 1 ⎞ × ⎟ kN = ⎜ 8 13 ⎠ ⎝ = 278.67 kN (Tension) Step 4. (Case II) Bridge is loaded. Wind pressure = 1.50 kN/m2 Wind load on top chord ⎛ 417.28 × 1.50 ⎞ P1 = ⎜ ⎟ = 260.8 kN 2.40 ⎝ ⎠ Wind load on bottom chord

⎛ 452.54 × 1.50 ⎞ P3 = ⎜ ⎟ = 283.03 kN 2.40 ⎝ ⎠ From IRC section II Length of IRC class A loading

223

DESIGN OF TRUSS GIRDER BRIDGES

= (8.2 + 1.2 + 4.9 + 1.2 + 4.9) = 20.4 m Intensity of wind on ordinary highway bridge = 3.00 kN/linear metre Number of trains = 2 Wind load on moving trains P3 = 20.4 × 3.00 × 2 = 122.4 kN The wind load on moving train acts 1.5 m above road surface. The road surface is 1.45 m above the level of bearings. The wind loads on loaded bridge are shown in Fig. 4.38. Step 5. Overturning effect due to wind, when the bridge is loaded. Take moment about one of the bearings 2R × 13 = P1 × 8.80 + P2 × 0.80 + P3 × 2.95 2R × 13 = 260.8 × 8.80 + 283.03 × 0.80 + 122.4 × 2.95 R = 110.87 kN, 2R = 221.73 kN Due to overturning a thrust of 221.73 kN acts downward on leeward girder. Increase of stress in central top chord member U7U8 of the leeward girder

1 ⎛ 1 ⎞ ⎛ 32 × 32 ⎞ ⎛ 221.73 ⎞ ×⎜ ⎟×⎜ ⋅ (64) ⋅ ⎜ ⎟ ⎟ 2 ⎝ 8 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠ = 221.73 kN (Compression) Increase of stress in central bottom chord member L7L8 =

1 ⎛ 1 ⎞ ⎛ 28 × 36 ⎞ ⎛ 221.63 ⎞ ×⎜ ⎟×⎜ × (64) × ⎜ ⎟ ⎟ 2 ⎝ 8 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠ = 218.26 kN (Tension). =

P1

W ind load on m ovin g tra in P3

8m

1 .50 m

P2

1 .45 m 0 .80 m

13 m 2R

2R

Fig. 4.38

224

DESIGN OF STEEL STRUCTURES–VOL. II

Step 6. Lateral effect of top chord bracing when the bridge is loaded. Wind load on top lateral bracing when the bridge is unloaded P1 = 417.28 kN Wind load on top lateral bracing when the bridge is loaded P1´ = 260.8 kN Decrease in force in central top chord member due to top lateral bracing when the bridge is unloaded = 224.69 kN Decrease in force in central top chord member due to top lateral bracing when the bridge is loaded = 224.69 ×

260.8 = 140.43 kN 417.28

Step 7. Lateral effect of bottom chord bracing when the bridge is loaded Wind load acting on bottom lateral bracing when the bridge is unloaded P1 = 452.84 kN Increase in force in central bottom chord member due to bottom lateral bracing = 278.67 kN Wind load acting on bottom lateral bracing when the bridge is loaded = (P2 + P3) = (283.03 + 122.4) = 405.43 kN Increase in force in central bottom chord member due to bottom lateral bracing =

278.67 ×

405.43 = 249.49 kN 452.84

Example 4.5 Design the top lateral bracing and bottom lateral bracing for the through type truss girder railway bridge for a single broad gauge track as in Example 4.3. Solution Design : Step 1. To lateral bracing The horizontal truss bracing with cross-diagonals as shown in Fig. 4.39 is provided between the top chords of windward and leeward truss girders. The diagonals which carry tension remain active. P1

7m 8 P a ne ls @ 5 m = 40 m

Fig. 4.39

DESIGN OF TRUSS GIRDER BRIDGES

225

From Example 4.3 When the bridge is unloaded, the wind load acting on the bridge structure on windward and leeward girders P1 = 203.64 kN. This is also maximum. Reaction at the end 1 × 203.64 = 101.82 kN 2 8 panels @ 5 m are provided in the top lateral bracing. Lateral load at each intermediate panel point

=

=

⎛ 203.64 ⎞ ⎜ ⎟ = 25.45 kN ⎝ 8 ⎠

Lateral load at end panel = 12.73 kN End strut carries maximum compression = 101.82 kN Shear force in end panel = (101.82 – 12.73) = 89.09 kN 5 = 1.2 , θ = 50° 12´ 7 sin θ = 0.768, cosec θ = 1.305

tan θ = Force in the end diagonal

= 89.09 × 1.305 =116.26 kN Step 2. End strut Effective length = 0.85 × 7000 = 5950 mm Compression in end strut = 101.82 kN Assume allowable stress in axial compression for the slenderness ratio 120 and the steel having yield stress as 260 n/mm2 τac = 64 N/mm2 Cross-sectional area required

⎛ 101.82 × 1000 ⎞ 4 = ⎜ ⎟ = 1590.94 mm 64 ⎝ ⎠ Maximum allowable slenderness ratio for compression members of wind bracing = 140 Minimum radius of gyration ⎛ 5950 ⎞ rmin = ⎜ ⎟ = 42.5 mm ⎝ 140 ⎠ From ISI Handbook No. 1. Provide 2 ISA 150 mm × 115 mm × 8 mm with 10 mm gusset plate rx = 47.6 mm, ry = 46.2 mm Area, A = 4116 mm2 Slenderness ratio

⎛ 5960 ⎞ = ⎜ ⎟ = 126.86 < 140 ⎝ 46.2 ⎠

226

DESIGN OF STEEL STRUCTURES–VOL. II

From IS: 833–1994, allowable stress in axial compression for the steel having yield stress as 260 N/mm2 = 59.24 N/mm2 Force carrying capacity of member

⎛ 59.24 × 4116 ⎞ ⎜⎝ ⎟⎠ 1000

= 243.83 kN. Hence safe.

It is to note that the extra force carrying capacity of end strut also takes into account 2½percent of axial compression of the member in the panel under consideration for bracing. Step 3. Diagonal member Force in diagonal = 116.26 kN

⎛ 116.26 × 1000 ⎞ 2 Net area required = ⎜ ⎟ = 745.26 mm ⎝ 0.6 × 260 ⎠ Area of rivet hole = 235 mm2 Gross area required = 980.26 mm2 Provide ISA 90 mm × 60 mm × 10 mm Gross area provided = 1401 mm2. Hence safe. Step 4. Bottom lateral bracing The bottom lateral bracing with cross-diagonals as shown in Fig. 4.40 provided between the bottom chords of windward and leeward truss girders. The diagonals which carry tension remain active. P2 + P3 + P4

1 0 P an els @ 5 m = 5 0 m

Fig. 4.40

From Example 4.3 Wind load on bottom chord when bridge is unloaded P2 = 233.04 kN Wind load on bottom chord when bridge is loaded P2´ = 145.65 kN Wind load on moving train P3´ = 262.50 kN Racking force @ 6.00 kN/m P4´ = 6.00 × 50 = 300 kN Total lateral load= P2´ + P3´+ P4´ = 708.15 kN The lateral load action in the plane of bottom lateral bracing is maximum in case the bridge is loaded. Therefore, the bottom lateral bracing is designed for lateral force = 708.15 kN.

DESIGN OF TRUSS GIRDER BRIDGES

227

End reaction = 354.075 kN Number of panels = 10 Wind load at each intermediate panel point = 70.815 kN Wind load at end panel =

1 × 70.815 = 35.41 kN 2

Shear force in end panel = (354.075 – 35.41) = 318.65 kN cosec θ = 1.305 Force in diagonal member = 318.655 × 1.305 = 415.86 kN Allowable stress in axial tension = 0.6 × 260 = 156 N/mm2

⎛ 415.86 × 1000 ⎞ 2 Net area required = ⎜ ⎟ = 2665.76 mm 156 ⎝ ⎠ Assuming area for rivet hole = 640 mm2 Gross area required = 3305.76 mm2 Provide 2 ISA 150 mm × 115 mm × 10 mm. Gross area provided = 2 × 2552 = 5104 mm2. Hence safe.

4.16

PORTAL BRACING

The top lateral bracing and bottom lateral bracing stabilize the bridge structure in the horizontal plane. In order to prevent the distortion of rectangular section of a through type bridge, the end posts are tied together to form a rigid frame. This rigid frame in combination with the end posts of truss girders provides transverse bracing. Since, this rigid frame is provided at the entrance of bridge, it is known as portal bracing. The portal bracing may be provided in a single plane or in two planes. In case the portal bracing is provided in single plane, it is located preferably in the central transverse plane of the end posts, with diaphragms between the webs of the end posts to provide for a distribution of the portal stresses. The portal bracing should be as deep as clearance (head-room) allow. (The clearance diagrams have been given in Sec. 1.8). The deep portal frames minimize the bending stresses in the end posts and provide maximum rigidity. The portal bracing is designed to resist the full end reaction due to wind load on the top lateral bracing. The maximum wind load of unloaded span is taken into consideration. The end posts transfer this reaction to the truss bearings. In addition to the wind, the portal bracing is designed for an assumed cross1 shear at any section equal to 1 percent of the sum force in both the end posts or 4 both the top chords of the end panel whichever is more. The various types of portal bracing which are used as shown in Fig. 4.41.

228

DESIGN OF STEEL STRUCTURES–VOL. II

Fre e L en gth (a )

(b )

(c)

(d )

(e )

(f)

Fre e L en gth

(g )

Fig. 4.41

The portal bracing shown in Fig. 4.41 (a) consists of a plate girder bracing. This is most effective where the head-room available is small. The portal bracing shown in Figure 4.41 (b) is known as A-frame. This type of portal bracing is suitable where greater head-room is available. This type is common for single track railway bridges. The portal bracing shown in Figs. 4.41 (c), (d), (e) and (f) are used for shallow head-room. These are common in double track railway bridges. The portal bracing shown in Figs. 4.41 (g) and (h) are used where deep head rooms are available. The deep head-rooms occur in heavy trusses. These rigid portal bracings are necessary to assure sufficient rigidity. As regards the portal effect in the bottom chords of main truss girders, it would be noted that the inclined end posts are subjected to direct stresses. The windward end post would be subjected to tension and the leeward end post would be subjected to compression. The direct stresses in the end post produce horizontal reactions at their bottom ends. One end of the truss girder is supported on rollers. Therefore,

DESIGN OF TRUSS GIRDER BRIDGES

229

the horizontal reactions at the bottom ends of the end posts are provided by the bottom chords if the truss girders. The magnitude of this horizontal reaction is equal to the horizontal component of the direct stress in the end post of the portal bracing. As a result of the portal effect, the bottom chord of windward truss girder is subjected to compression and the bottom chord of leeward truss girder is subjected to tension. The compression and tension due to portal effect in the bottom chords of the truss girders are uniform throughout their length. The vertical component of direct stresses in the and posts are transmitted to the bearings.

4.17

ASSUMPTIONS FOR THE ANALYSIS OF PORTAL FRAMES

The end posts of a through type truss girder bridge are assumed to have fixed supports. Therefore, the portal bracings act as rigid frames. Such portal bracings are statically indeterminate to third degree. In order to have the analysis of the portal bracings by the approximate method, three assumptions are necessary. These assumptions are as follows : 1. The horizontal shear on the portal bracing is shared equally by both the end posts. 2. A point of contraflexure occurs in the middle of free length of one end post. 3. A point of contraflexure also occurs in the middle of free length of other end port. First assumption is reasonable in case the portal bracing is symmetrical and the bending stiffnesses of the end posts are equal. Second and third assumptions are also reasonable, since the points of reversal of curvature would occur near the centre of free length of end posts. Second and third assumptions are equivalent to assume hinges at the middle of free lengths of the end posts. Therefore, the portal bracing with fixed bases may be analysed by considering end posts hinged at the points of contraflexures.

4.18

ANALYSIS OF PORTAL BRACING

Figure 4.42 shows a portal bracing, drawn parallel to the plane of end posts. This type of portal bracing is used where head-room is small. The member CC may be a beam or a plate girder. The member A´ C´ and AC are the end posts. The bottom ends of end posts are assumed to have fixed supports. The points of contraflexure are assumed to occur at the centre of AD and A´D´ i.e., at A1 and A1´. The horizontal shears in end posts are equal and are shown in Fig. 4.42. The vertical reactions are found by taking the moment of all the forces acting on the portal bracing above A1 and A1´ about A1 and A1´ . ∴

h⎞ ⎛ V.s = P ⎜ a + ⎟ 2⎠ ⎝

P⎛ h⎞ a+ ⎟ ⎜ 2⎠ s⎝ The bending moment at the bottom ends of end posts are maximum and are as follows: V =

230

DESIGN OF STEEL STRUCTURES–VOL. II

⎛ P h ⎞ ⎛ P ⋅h ⎞ MA = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ ⎛ P h ⎞ ⎛ P ⋅h ⎞ MA = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ The bending moment at D and D´ on the end posts are also maximum and are as under: X

P

F1

C

C'

d

S D

D'

F2

X

h 2 P 2 P 2

A '1 V'

A1 V

h 2

A'

A

Fig. 4.42

⎛ P h ⎞ ⎛ P ⋅h ⎞ MD = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ ⎛ P h ⎞ ⎛ P ⋅h ⎞ MD = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ The shear force in each end post is equal to posts are equal to vertical reactions,

P . The axial stresses in the end 2

⎡ h ⎤ P ⎛⎜ a + ⎞⎟ ⎥ ⎢ ⎝ 2⎠ ⎢i.e., V = ⎥ s ⎣ ⎦ The axial stress in A´C´ is tensile and that in AC is compressive.

231

DESIGN OF TRUSS GIRDER BRIDGES

In order to find the forces in the beam or plate girder, consider a section XX at a distance x from C. Consider the equilibrium of right hand side portion of the portal bracing. F1 and F2 represent force in the flanges and S represents shear force at the section XX. It is assumed that the bending moment is resisted completely by the flanges. Taking the moment at the centre of upper flange. F2· a =

P⎛ h⎞ a + ⎟ –V ⋅x ⎜ 2⎝ 2⎠

F2 =

1 ⎡P ⎛ h⎞ h⎞x⎤ ⎛ a + ⎟ − P ⎜a + ⎟ ⎥ a ⎢⎣ 2 ⎜⎝ 2⎠ 2⎠ s⎦ ⎝

F2 =

P⎛ h ⎞ ⎡1 x ⎤ a + ⎟⎢ − ⎥ ⎜ 2 ⎠ ⎣2 s ⎦ a⎝

when

x = 0, F2 =

when

x =

when

x

P ⎛ h⎞ a+ ⎟ ⎜ 2a ⎝ 2⎠

s , F2 = 0 2 P ⎛ h⎞ = s, F2 = − ⎜ a + ⎟ 2a ⎝ 2⎠

P ⎛ h⎞ a + ⎟ at the ⎜ 2a ⎝ 2⎠ ends, compression at the leeward side and tension at the windward side. Taking the moment at the centre of bottom flange Thus, the force in bottom flange is zero at the centre and

F1· a = F1 =

P h ⋅ −V ⋅x 2 2 1 ⎡ Ph P ⎛ h⎞ ⎤ − ⎜a + ⎟x⎥ ⎢ a⎣ 4 s⎝ 2⎠ ⎦

when

x = 0, F1 = –

when

x =

P ⋅h 4

s Ph , F1 = – 2 2

1 ⎡P ⋅h ⎤ ⎡P ⋅h ⎤ + P ⋅ a⎥ = − ⎢ + P⎥ 4 a ⎢⎣ 4 a ⎦ ⎣ ⎦ Thus, the force in top flange is maximum tensile at right end, and maximum

when

x = s, F1 = −

compressive at the left end. The force in top flange at the centre is The shear force at all points is equal to S = V=

P⎛ h⎞ a+ ⎟ ⎜ 2⎠ s⎝

P compressive. 2

232

4.19

DESIGN OF STEEL STRUCTURES–VOL. II

ANALYSIS OF PORTAL BRACING

Figure 4.43 shows a portal bracing drawn parallel to the plane of the end posts. This type of portal bracing is used where head-room is small. The member CC´ is a continuous beam. It resists moment, shear and direct stress. FD and F´D´are two knee braces. These members carry direct stresses only. The end posts are also continuous. The knee braces are hingned to the end posts. In this type of portal bracing a section cannot be passed through the end posts or the beam except at the ends of these members without taking into account the moment, shear and direct stress in the unknowns. The ends of the beam C and C´ are assumed as hinged. The bottom ends of end posts AC and A´C´ are assumed to have fixed supports as shown in Fig. 4.43 (a). The points of contraflexures are assumed at the centre of AD and A´D´, i.e., at A1 and A1´. The horizontal shears in end posts are equal and are shown in Fig. 4.43. The vertical reaction on this portal bracings are equal to the axial forces in the posts. The vertical reactions may be found by taking moments of all the forces acting on the portal bracing above A1 and A1´ about A1 and A1´. ∴

h⎞ ⎛ V. s = P ⎜ a + ⎟ 2⎠ ⎝ V =

P⎛ h⎞ a+ ⎟ 2⎠ s ⎝⎜

P . The 2 horizontal shears below A1 and A1´ are as shown in Fig. 4.43 (a); and are equal P to

The horizontal shears above A1 and A1´ in A1C and A1´ C are equal to

P . The moment MA at bottom end of post AC i.e., at A is equal to the horizontal 2 shear at the point of contraflexure in the end post multiplied by the distance from the bottom to the point of contraflexure. The moments are maximum and are as follows :

⎛ P h ⎞ ⎛ P ⋅h ⎞ MA = ⎜ × ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ Moment at A´,

⎛ P h ⎞ ⎛ P ⋅h ⎞ MA´ = ⎜ × ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ Moment at

⎛ P h ⎞ ⎛ P ⋅h ⎞ D = ⎜ × ⎟=⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠

Moment at

⎛ P h ⎞ ⎛ P ⋅h ⎞ D´ = ⎜ × ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠

Moment at

C = 0

233

DESIGN OF TRUSS GIRDER BRIDGES

P

d

F'

F

Ph

X

d

4

C

C´

a X D'

D

h 2

A1

P 2 P 2

P 2 P V 2

B M D for A C / A 'C '

A1 V h 2

A

A

S A

Ph 4 B

Fig. 4.43

The bending moment diagram for the end post is shown in Fig. 4.43 (b). The forces in the various members are as follows: Forces in FD and F´D´ In order to find force in member FD, pass a section XX through the point C (assumed as hinged). Resolve the force in FD in its horizontal and vertical components at point D. Take moment about C. FD sin θ, a

=

P⎛ h⎞ a+ ⎟ ⎜ 2⎝ 2⎠

FD

=

P⎛ h⎞ 1 a+ ⎟ ⎜ 2⎝ 2 ⎠ a ⋅ sin θ

=

P⎛ h ⎞ ( a 2 + d 2 )1 / 2 (Compression) a+ ⎟ ⎜ 2⎝ 2⎠ ad

Similarly force in F´D´, FD =

P⎛ h ⎞ ( a 2 + d 2 )1 / 2 (Tension) + a 2 ⎝⎜ 2 ⎠⎟ ad

234

DESIGN OF STEEL STRUCTURES–VOL. II

Force in CD and C´D´ Force in CD = Vertical component of forces in FD – Direct force in A1D. ∴

CD

=

FD cos θ –V

=

P⎛ h ⎞ ( a2 + d2 )1 / 2 a a + ⋅ 2 ⎜ ⎟ 2⎝ 2⎠ ad ( a + d2 )1 / 2 h⎞ ⎛ P ⎜a + ⎟ 2⎠ − ⎝ s

=

P ⎛ h⎞ P ⎛ h⎞ a + ⎟ − ⎜a + ⎟ ⎜ 2d ⎝ 2⎠ s ⎝ 2⎠

=

P⎛ h ⎞ ⎡1 2⎤ a + ⎟ ⎢ − ⎥ (Tension) ⎜ 2⎝ 2 ⎠ ⎣d s ⎦

=

P⎛ h ⎞ ⎡1 2⎤ a + ⎟ ⎢ − ⎥ (Compression). ⎜ 2⎝ 2 ⎠ ⎣d s ⎦

Similarly force in C´D´, C´ D´

Bending moment at F By taking moment from right hand side MF = V ⋅ d −

P⎛ h⎞ a+ ⎟ ⎜ 2⎝ 2⎠

h⎞d P ⎛ h⎞ ⎛ = P ⎜a + ⎟ − ⎜a + ⎟ 2⎠ s 2 ⎝ 2⎠ ⎝ =

P⎛ h ⎞ ⎡ 2d ⎤ a + ⎟⎢ − 1⎥ 2 ⎝⎜ 2⎠⎣ s ⎦

Bending moment at F´ By taking moment from left hand side MF´ = V´d –

P⎛ h⎞ a+ ⎟ 2 ⎝⎜ 2⎠

h⎞d P ⎛ h⎞ ⎛ P = ⎜a + ⎟ − ⎜a + ⎟ 2⎠ s 2 ⎝ 2⎠ ⎝ =

P⎛ h ⎞ ⎡ 2d ⎤ a + ⎟⎢ − 1⎥ ⎜ 2⎝ 2⎠⎣ s ⎦

235

DESIGN OF TRUSS GIRDER BRIDGES

The bending moment at F´ is of opposite sign of that at F. The bending moment at the centre of portion FF´ is zero. The values of bending moments at C and C´ are also zero. The distribution of bending moment for the beam CC´ is shown in Fig. 4.43 (a). Force in EE´ The force in EE´ is found by considering equilibrium from right half of the portal bracing EE = P –

P P = (Compression). 2 2

Force in FC The force in FC is found by considering equilibrium of joint F FC

=

FD sin θ – EF

=

P⎛ h ⎞ ( a 2 + d 2 )1 / 2 a+ ⎟ ⎜ 2⎝ 2⎠ ad

=

P⎛ h ⎞ 1 P P ⋅h (Tension). a+ ⎟ − = ⎜ 2⎝ 2⎠a 2 4a

d 2

a +d

2

−

P 2

Force in C´F´ The force in C´F´ is found by considering equilibrium of joint F C´ F ´

=

F ´C´ sin θ + FF´

=

d P P⎛ h ⎞ ( a 2 + d 2 )1 / 2 + a+ ⎟ ⎜ 2⎝ 2⎠ ad ( a 2 + d2 )1 / 2 2

P⎛ h⎞1 P Ph a+ ⎟ + =P+ (Compression). ⎜ 2⎝ 2⎠a 2 4a Shear force in C´F´ = V =

h⎞ 1 ⎛ = P ⎜a + ⎟ ⎝ 2⎠ s Shear force in FC and F´C´ Shear force in FC is equal to shear force in FF´ less the vertical component of force in FD = V – FD cos θ =

P⎛ h ⎞1 P ⎛ h⎞ a + ⎟ − ⎜a + ⎟ 2 ⎝⎜ 2⎠s 2 ⎝ 2⎠ ×

( a 2 + d2 )1 / 2 a × 2 ad ( a + d2 )1 / 2

h ⎞⎛1 1 ⎞ ⎛ = P ⎜a + ⎟⎜ − 2 ⎠ ⎝ s 2d ⎠⎟ ⎝

236

DESIGN OF STEEL STRUCTURES–VOL. II

Shear force in F´C´ is also equal to shear force in FC. Shear force in CD Shear force in CD is equal to shear in A1D less horizontal component of force in FD

4.20

=

P − FD sin θ 2

=

a P P ⎛ h ⎞ ( a 2 + d 2 )1 / 2 × 2 − ×⎜a + ⎟ × 2 2 ⎝ 2⎠ ( a + d2 )1 / 2 ad

=

P P⎛ h⎞1 − ⎜a + ⎟ 2 2⎝ 2⎠a

.

ANALYSIS OF PORTAL BRACING

Figure 4.44 shows a portal bracing drawn parallel to the plane of the end posts.

P

C´

X

S 2

o

F

A '1

S 2

X

G

a θ

X2

X'

X'

P 2 P V' 2

Ph 4

C o

o E

D'

X1

P 2 P 2 V

D

BM D fo r A C / A 'C '

A1 h 2

h 2

A'

A

S (a )

Fig. 4.44

Ph 4 (b )

DESIGN OF TRUSS GIRDER BRIDGES

237

The portal bracing shown in Fig. 4 44 is known as A-frame. This type of portal bracing is used where greater head-room is available and it is common for single track railway bridges. This type of portal bracing is similar to that shown in Fig. 4.43 except that the braces are made to meet at the centre as shown in Fig. 4.44. This type of portal bracing is very simple. This is effective form of portal bracing. The moment at F is zero (this has also been shown in Sec. 4.19). The member CC´ is not subjected to bending moment or shear. Generally, the members shown by dotted lines are introduced in this type of portal bracing. This provides better appearance. The members DF and D´F are also stiffened by the introduction of C´E´, E’E and CE, (the stresses in these members are zero) as the effective lengths of the members DF and D´F become DE and DE´. This is also shown in the analysis. The assumptions discussed in Sec. 4.17 are made for its analysis. Vertical reactions The vertical reactions are found by taking moment of all the above A1 and A1´. ∴

h⎞ ⎛ V . s = P ⎜a + ⎟ 2⎠ ⎝ h⎞ 1 ⎛ V = P ⎜a + ⎟ ⋅ 2⎠ s ⎝

Also

h⎞ 1 ⎛ V´ = P ⎜ a + ⎟ ⋅ 2⎠ s ⎝

The direct or axial stresses ion end posts are equal to the vertical reactions. Axial force in the end posts AA1

h⎞ 1 ⎛ = P ⎜ a + ⎟ ⋅ (Compression) 2⎠ s ⎝ Axial force in the end posts A´A1´

h⎞ 1 ⎛ = P ⎜ a + ⎟ ⋅ (Tension). 2⎠ s ⎝ Moment at the fixed supports The moments at the fixed supports are maximum, and are found as follows:

⎛ P h ⎞ ⎛ P ⋅h ⎞ MA = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ ⎛ P h ⎞ ⎛ P ⋅h ⎞ MA´ = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ The moments are also maximum at D and D´. The values of moments at these points are as follows :

⎛ P h ⎞ ⎛ P ⋅h ⎞ MD = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠

238

DESIGN OF STEEL STRUCTURES–VOL. II

⎛ P h ⎞ ⎛ P ⋅h ⎞ MD´ = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ 2 2⎠ ⎝ 4 ⎠ But, it is to note that the moments at these points are of opposite sign than those at supports. The moments at the top of end posts are zero. These points C and C´ are assumed as hinged. The forces in the various members are as follows : Forces in ED and E´D´ In order to find force in member ED, pass the section XX as shown in Fig. 4.44 (a). Resolve the force in ED in its horizontal and vertical components at point D. Take the moment about C. ED sin θ ⋅ a

=

P 2

h⎞ ⎛ ⋅ ⎜a + ⎟ 2⎠ ⎝

ED

=

P 2

h⎞ 1 ⎛ ⋅⎜a + ⎟ 2 ⎠ a sin θ ⎝

=

⎡ 2 ⎛ s⎞2⎤ ⎢a + ⎜ ⎟ ⎥ ⎝ 2⎠ ⎥ P ⎛ h ⎞ ⎢⎣ ⎦ ⋅⎜a + ⎟ s 2 ⎝ 2⎠ a⋅ 2

1/2

1/ 2

⎡ 2 ⎛ s⎞2 ⎤ ⎢a + ⎜ ⎟ ⎥ ⎝ 2⎠ ⎥ h ⎞ ⎣⎢ ⎛ ⎦ = P ⋅⎜a + ⎟ ⎝ 2⎠ a⋅s

(Compression)

Similarly, force in E´D´ 1/ 2

E´ D´

⎡ 2 ⎛ s ⎞2 ⎤ ⎢a + ⎜ ⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ h ⎞ ⎢⎣ ⎛ = P ⎜a + ⎟ a ⋅s 2⎠ ⎝

(Tension).

Forces in CE and C´E´ The force in CE may be found by considering the section XX and taking the moment F. Moment about point F ⎛ ⎞ MF C´E´ = ⎜ ⎟= F CE Perpendicular distance from to ⎝ ⎠ xCE

239

DESIGN OF TRUSS GIRDER BRIDGES

Consider the equilibrium of right hand portion MF =

= ∴

CE´

P⎛ h⎞ a + ⎟ −V 2 ⎝⎜ 2⎠

⎛s⎞ ⋅ ⎜ ⎟. ⎝2⎠

h⎞ ⎛ P ⎜a + ⎟ P⎛ h⎞ 2⎠ s ⋅ =0 a+ ⎟− ⎝ s 2 ⎝⎜ 2⎠ 2

∴ C´E´ = 0.

= 0, Also

Forces in FC and FC´ The force in FC may be found by considering the section XX and taking moment about D FC ⋅ a FC

⎛P h⎞ = ⎜ ⋅ ⎟ ⎝ 2 2⎠ ⎛ P ⋅h ⎞ = ⎜ ⎟ (Tension) ⎝ 4a ⎠

Consider a section X´X´. Take moment about D´ P ⋅a +

P h ⋅ + FC a = 0 2 2

∴

FC´

=

⎡ Ph ⎤ + P ⎥ (Compression). –⎢ ⎣ 4a ⎦

Force in EF´ The force in EF´ is found by considering section X1X1 and taking moment about F

EE´

=

MF x ´ EE

As shown above, moment MF is zero. ∴ EE´ = 0. Forces in EF and E´F´ The force in EF is found by considering the equilibrium of joint E. The forces in EF´ and CE are zero. 1/2

⎡ 2 ⎛ s ⎞2 ⎤ ⎢a + ⎜ ⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ h ⎞ ⎢⎣ ⎛ = P ⎜a + ⎟ 2⎠ ⎝ s⋅a

Similarly, E´F´ = E´ D´

(Compression)

240

DESIGN OF STEEL STRUCTURES–VOL. II

1/2

⎡ 2 ⎛ s ⎞2 ⎤ ⎢a + ⎜ ⎟ ⎥ ⎝ 2 ⎠ ⎦⎥ h ⎞ ⎣⎢ ⎛ = P⎜a + ⎟ s⋅a 2⎠ ⎝

(Tension)

The direct or axial stress in the portion CD of end post AC is equal to the vertical reaction less the vertical component of the member ED. Similarly, the axial stress in the portion C´D´ of end post A´C´ is also equal to the vertical reaction less the vertical component of the force in member E´D´. The forces in the various members of the portal bracings as shown in Fig. 4.41 (c) and (d) are found in the same manner as that shown in Fig. 4.44. It is to note all the members of the portal bracings shown in Figs. 4.41 (c) and (d) carry forces. The analysis of portal bracing as shown in Fig. 4.41 (e) is also done in the same manner as that shown in Fig. 4.44.

4.21

ANALYSIS OF PORTAL BRACING

Figure 4.45 shows a portal with simple diagonal bracing. The portal bracing is shown in the plane parallel to the end posts. This type of portal bracing is used where deep head room is available. The assumptions discussed in Sec. 4.17 are made in the analysis of this portal bracing. The forces in the posts are same as discussed in the previous portal bracings, except the shear above D and D´. The diagonal members CD´ and CD are made to take tension. The members CC´ and DD´ are then compression members. When the wind acts in the direction as shown in Fig. 4.45, then the tension member CD acts as dummy member. In order to determine the forces in the members, consider a section XX as shown in Fig. 4.45. Force DD´ Take moment about point C

DD´ . a =

P⎛ h⎞ P ⎛ h⎞ a+ ⎟= a+ ⎟ 2 ⎝⎜ 2 ⎠ 2a ⎝⎜ 2⎠

Force in CD´ Force in member CD´ is equal to shear in panel DD´ multiplied by cosec θ CD´ h ⎞ 1 ( a 2 + s2 )1 / 2 ⎛ P ⎜a + ⎟ ⋅ 2⎠s a ⎝

= V . cosec θ =

h ⎞ 1 ( a 2 + s2 )1 / 2 ⎛ P ⎜a + ⎟ ⋅ 2⎠s a ⋅s ⎝

Force in CC´ Take moment about D´ P ⋅ a + CC´ ⋅ a −

P h ⋅ 2 2

= 0

241

DESIGN OF TRUSS GIRDER BRIDGES

CC´

=

1 ⎡ Ph ⎤ P ⋅h + Pa ⎥ = +P ⎢ 4a a⎣ 4 ⎦

The diagonal members C´D and CD may also be made as compression members, then the members CC´ and DD´ are made as tension members. Analysis of the portal bracing as shown in Fig. 4.41 (f) is done by assuming that the forces are taken by system of bracing in which diagonal members are in tension. The maximum moments, shears and the axial stresses in the end posts are same as discussed for the previous portal bracings. P

x

C'

C θ a

D'

D x

h 2

A '1

P 2 P V' 2

P 2 P 2 V

A1

h 2

A1

S

A

Fig. 4.45

4.22

ANALYSIS OF PORTAL BRACING

Figure 4.46 shown a portal bracing, drawn parallel to the posts. The assumptions discussed in Sec. 4.17 are made for the analysis of this portal bracing. The forces in the end posts are same as in the portal bracing discussed in Sec. 4.18. It is assumed that the bending moment is completely resisted by the flanges CC´ and DD´ and the shear is resisted by the web members. The shear force is constant throughout the lattice girder (CC´D´DC). Therefore, the forces in the web members are also constant. Consider any section XX at a distance x, as shown in Fig. 4.46. The shear force S at any section XX is assumed as taken equally by the web

242

DESIGN OF STEEL STRUCTURES–VOL. II

members cut by the section XX. Let n be the number of members cut by the section XX. Then, vertical components of force in any web member is

P ⎛ h⎞ ⎛V ⎞ = ⎜ ⎟= ⎜a + 2 ⎟ n s n ⋅ ⎝ ⎠ ⎝ ⎠ Half of the web members carry tension and the other half of web members carry compression. In case the direction of wind changes, and acts in the opposite direction, then the nature of forces in the web members also changes. In order to determine the forces in the flange members, the section XX is drawn such that it passes through the points of intersection of the web members. The moment of forces may be taken about G. The forces in all the web members are numerically equal. Half of the web members carry tension and half of the web members carry compression. Therefore, the sum of moments due to the forces in the web members about any point in the vertical line GH is zero. Let F1 and F2 represent forces in the top flange and in the bottom flange respectively. x

x C F1

C'

P a

G

C

S

D'

D C2 H

x x

x

D

x

h 2

A '1

P 2 P V' 2

P 2 P 2

P 2 A1 P 2 V

A1 V

h 2

A'

A

S (a )

(b )

Fig. 4.46

Taking moment about the point G, F2 . a =

A

P⎛ h⎞ a + ⎟ –V ⋅x ⎜ 2⎝ 2⎠

DESIGN OF TRUSS GIRDER BRIDGES

=

243

1 ⎡P ⎛ h⎞ h⎞x⎤ ⎛ a + ⎟ − P ⎜a + ⎟ ⎥ ⎜ ⎢ a⎣2 ⎝ 2⎠ 2⎠ s⎦ ⎝

P⎛ h ⎞⎛1 x ⎞ a + ⎟⎜ − ⎟ 2 ⎠⎝ 2 s ⎠ a ⎝⎜ x = 0 =

when

F2 = when when

x = F2 x F2 x

= = = =

P ⎛ h⎞ a+ ⎟ ⎜ 2a ⎝ 2⎠ s 2 0 s 0 s

P ⎛ h⎞ a+ ⎟. ⎜ 2a ⎝ 2⎠ Thus, the maximum tension in the bottom flange occurs at D´ and the maximum compression occurs at D. Taking moment about point H as shown in Fig. 4.46 (b) F2 =

F1. a = = when

–

P h ⋅ –V ⋅x 2 2 1 ⎡ Ph P ⎛ h⎞ ⎤ − ⎜a + ⎟x⎥ ⎢ a⎣ 4 s⎝ 2⎠ ⎦

x = 0

⎛ P ⋅h ⎞ F1 = ⎜ ⎟ ⎝ 4a ⎠ when

x =

s 2

F1 = – when

P 2

x = s

1 ⎡P ⋅h ⎤ + Pa ⎥ ⎢ a⎣ 4 ⎦ Thus, the maximum tension in the top flange occurs at C, and the maximum compression occurs at C´. Example 4.6 Figure 4.47 shows a portal frame in a through type lattice girder bridge subjected to a lateral load P = 97.5 kN. Determine the forces in the various members of the portal frame. Also find the portal effect in the end post and in the F1 = –

244

DESIGN OF STEEL STRUCTURES–VOL. II

bottom chords of the lattice girders. The inclination of end post with the horizontal is 50 ° 12'. Solution Figure 4.47 shows the portal bracing drawn parallel to the plane of end posts. A and A´ are assumed as fixed supports. A1 and A1´ are assumed as points of contraflexures. The horizontal shears in the end posts are assumed as equal. X' P = 9 7.5 kN C '

S = 3.5 m 2

F

X1 X S = 3.5 m 2

o

a=3m

o E'

D'

C

o

X'

E X1

D

X

A 1' V'

P 2 P 2

P 2 P 2

7.8 m

h = 2.4 m 2

A1 V

h = 2.4 m 2

A'

S=7m

A

Fig. 4.47

Step 1. Reaction Horizontal reactions in the end posts at the point of contraflexures P 1 = × 97.5 = 48.75 kN 2 2 The vertical reaction, V is found by taking moment about A1´ V × 7 = 97.50 × (3 + 2.4) ∴ V = 75.27 kN Step 2. Portal effect in the end post Axial force in the end post, AC = 75.2 kN (Compression) Axial force in the end post, A´B´ = 75.2 kN (Tension)

DESIGN OF TRUSS GIRDER BRIDGES

245

Moment at the fixed support

⎛P h⎞ ⎜ + ⎟ = 48.75 × 2.4 = 117 kN-m ⎝ 2 2⎠ Moment at the point D ⎛P h⎞ = ⎜ + ⎟ = 48.75 × 2.4 = 117 kN-m ⎝ 2 2⎠ Moment at the top i.e., at C = 0 Step 3. Forces in the various members (i) Forces in ED and E´D´ Consider a section XX as shown in Fig. 4.48. Resolve the force in ED in its horizontal and vertical components at point D. Take moment about C.

ED sin θ × 3 = sin θ =

P (3 + 2.4) 2 3.5 2

(3.5 + 32 )1 / 2

= 0.76

ED = 115.46 kN (Compression) Similarly, force in E´D´ = 115.46 kN (Tension). (ii) Forces in CE and C´E´ Consider the section XX. Take moment about E,

∴

⎛ MF ⎞ ⎟ = ⎜ ⎝ xCE ⎠ Consider equilibrium of right hand portion of the bracing CE

P (3 + 2.4) – 75.2 × 3.5 = 48.75 2 × 5.4 – 75.2 × 3.5 = 0 = 0

MF = ∴

CE

Similarly,

C´E´

= 0

(iii) Forces in FC and FC´ Consider the section XX. Take moment about D´ FC´×3

=

P × 2.4 2

⎛ 97.5 2.4 ⎞ × = ⎜ = 39 kN (Tension) 3 ⎠⎟ ⎝ 2 Consider section X´X´. Take moment about D´ FC

⎛ 97.5 ⎞ 97.5 × 3 + ⎜ × 2.4 ⎟ + FC´×3 = 0 ⎝ 2 ⎠ FC´ = –136.50 kN (i.e., Compression)

246

DESIGN OF STEEL STRUCTURES–VOL. II

(iv) Force in EE´ Consider the section X1X1. Take moment about F

EF ´

⎛ MF ⎞ = ⎜ x´ ⎟ ⎝ EE ⎠

MF = 0

3

∴ EE´ = 0 (v) Force in EF and E´F´ Consider equilibrium of joint E

EF = ED = 115.46 kN (Compression) ∴ Also E´F´ = 115.2 kN (Tension) Step 4. Portal effect in the bottom chord of the lattice girder Additional tension in the bottom chord of the leeward girder = V × cos θ (cos 50° 12´ = 0.64) = 75.2 × 0.64 = 48.128 kN Compression in the bottom chord of the windward girder = 48.128 kN. Example 4.7 Figure 4.48 shows a portal bracing in a through type truss girder bridge, subjected to lateral force P = 130 kN. Determine the force in the various members of the portal bracing. Also determine the portal effect in the bottom chords of the truss girders. The inclination of end post with the horizontal is 63 °26'. p = 1 30 kN

C'

2

4 .33 m

E D'

4 .33 m G'

2

4 .33 m

G

4 F

4 .33 m 3 1 C

4 .33 m 4 3

a=3m

E 1

D

h = 2 .97 m 2

A '1

P 2

P 2

P 2

P 2

A1 V h = 2 .97 m 2

A

S = 13 m

Fig. 4.48

A

DESIGN OF TRUSS GIRDER BRIDGES

247

Solution Figure 4.48 shows the portal bracing drawn parallel to the plane of end posts. A and A´ are assumed as fixed supports. A1 and A1´ are assumed as points of contraflexures. The horizontal shear in the end posts are assumed as equal. Step 1. Reaction Horizontal reactions in the end posts at the points of contraflexures P 2

=

130 = 65 kN. 2

The vertical reaction V, is found by taking moment about A1 ∴ V × 13 = 130 × (3 + 2.97) V = 59.7 kN ∴ V = 59.7 kN Step 2. Portal effect in the end post Axial force in the end post AC = 59.7 kN (Compression) Axial force in the end post A´C´ = 59.7 kN (Tension) Moment at the fixed support

⎛P h⎞ = ⎜ × ⎟ = 65 × 2.97 = 193 kN-m ⎝ 2 2⎠ Moment at the point D

⎛P h⎞ = ⎜ × ⎟ = 65 × 2.97 = 193 kN-m ⎝ 2 2⎠ Moment at the top i.e., at C = 0 Step 3. Forces in the various members (i) Forces in ED and E´D´ Consider section 1–1 as shown in Fig. 4.48. Resolve the force in ED in its horizontal and vertical components at point D. Take moment about point C sin θ =

4.33 (4.332 + 32 )1 / 2

ED sin θ × 3 =

Similarly,

= 0.822

P (3 + 2.97) 2

E´ D´

⎛ 65 × 5.97 ⎞ = ⎜ ⎟ = 157.36 kN (Compression) ⎝ 3 × 0.822 ⎠

E´ D´

= 157.36 kN (Tension)

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DESIGN OF STEEL STRUCTURES–VOL. II

(ii) Forces in CE and CE´ Consider section 1–1. Resolve the force in CE in its vertical and horizontal components at C. Take moment about point F. P (3 + 2.97) − V × 4.33 + CE cos θ × 4.33 = 0 (cos θ = 0.5714) 2 CE

⎛ 5.97 × 4.33 − 65 × 5.97 ⎞ = ⎜ ⎟ kN 4.33 × 0.5414 ⎝ ⎠

= – 52.36 kN (i.e., Compression) Similarly, C´E´ = + 52.36 kN (Tension) (iii) Force in FC and E´C´ Consider section 1–1. Resolve force in CE in its horizontal and vertical components. Take moment about D P 2

= 2.97 + 52.5 sin θ × 3 – FC × 3 = 0

FC = 106.8 kN (Tension) Consider section 2–2. Resolve the force C´E´ in its horizontal and vertical components. Take moment about D´

13 × 3 + 52.5 sin θ × 3 + 65 × 2.97 + F ´C´ × 3 = 0 F´C´ = – 236.8 kN (i.e., Compression) (iv) Forces in GE and G´E´ Consider section 3–3. Take moment about F – 59.7 × 4.33 + 65 × (3 + 2.97) + GE ×

GE

=

3 =0 2

−65 × 5.97 + 59.7 × 4.33 1.5

= – 211 kN (i.e., Compression) Similarly, G´E´ = 21.1 kN (Tension) (v) Forces in GF and GF´ Consider section 3–3. The vertical component of force in GF is equal to shear force at section 4–4. GF cos θ = 59.7 GE

Similarly,

=

59.7 = 104.5 kN (Tension) 0.5714

GF´ = 104.5 kN (Compression)

DESIGN OF TRUSS GIRDER BRIDGES

249

(vi) Force in FF´ Consider section 4–4. Take moment about point G from right side of the section 65 × (2.97 + 1.5) – 59.7 × 6.5 + FF ´ × 1.5 = 0

FF ´

=

+66 × 59.7 − 66 × 4.47 1.5

= 65 kN (Compression) Step 4. Portal effect on the bottom chords of the truss girder Force in end post AC = 59.7 kN (Compression) ∴ Additional tension in the bottom chord of the leeward girder = 59.7 cos 63° 26´ = 59.5 × 0.4472 = 26.7 kN. Force in end post A´C = 59.7 kN (Tension) ∴ Compression in the bottom chord of windward girder = 59.7 × cos 63° 26´ = 59.7× 0.4472 = 26.7 kN.

4.23

SWAY BRACING

The sway bracings are provided to stabilize and to add rigidity to the bridge structure in transverse vertical planes. The sway bracings are used to maintain rectangular cross-section of the bridge. The sway bracings are provided at all the intermediate panel points of the through type truss girder bridges. The sway bracings do not provide any relief to the lateral system unless the stresses are calculated for the complete space frame. The sway bracings are also made as deep as head-room allows. As per IS : 1915–1961 and also Code of practice for design of steel bridges published by Railway Board, the sway bracings are proportioned to transmit to the lower chords through the web members at least 50 percent of the top panel lateral load and the vertical web members are designed to resist the resulting bending moment. When the top lateral bracing is stiff, them, theoretically, there are no stresses in the sway bracings. The above recommendation is made only for guidance to arrive at suitable size of the members. The heavy bracings tend to equalize the loads in double track and highway bridges, in case one track or one side is only loaded. In general, the various forms of the sway bracings and their analysis are similar to those of portal bracings.

Problems 4.1 A top boom member of a highway bridge is subjected to a compressive stress of 2500 kN. Its length from node to node is 4.00 m and the full length of top boom is 20.00 m and it is not supported transversely by any sway bracing for its entire length. It is formed of

250

DESIGN OF STEEL STRUCTURES–VOL. II

(i) two channels ISMC 350 × 100 @ 0.421 kN/m spaced at 275 mm between backs of channels, (ii) two web plates 225 mm × 19 mm and between roots attached centrally to the channel web, (iii) a top cover plate 500 mm × 22 mm, and (iv) suitable lacing and batten plates are provided at the bottom of the section. Give all calculations to show that the section is adequate to resist the stress imposed on the section. Use any formula you know for estimating the permissible stress as applied to such structures. 4.2 A through type highway steel bridge 48 m span, is supported on two Ngirders each consisting of 10 bays of 48 m each, the height of the N-girder being 48 m. The dead load of the bridge including self-weight of the two N-girders is 90 kN/m and the rolling load on the bridge, to be carried by the two girders is equivalent to 100 kN/m. Design the top and bottom chords at the fifth panel of the bridge and the diagonal member in the third bay from left. 4.3 A Pratt truss girder through bridge is provided for a single metre gauge track. The effective span of the bridge is 50 m. The cross-girders are 4 m apart. The stringers are spaced at 1.20 m between centre lines. 0’60 kN per metre stock rails and 040 kN per metre guard rails are provided. The sleepers are spaced at 0.50 m from centre to centre and are of size 2 m × 250 mm × 250 mm. Weight of timber may be assumed as 7.50 kN per cubic metre. The main girders are provided at spacing of 5 m between their centre lines. Design the central top chord member, the central bottom chord members and the vertical and diagonal members of central panel. Design the joint, where the central top chord, vertical and diagonal members meet. The bridge is to carry standard main line loading. 4.4 Determine the increase or decrease of forces in the central chord members of the leeward truss girder in Problem 4.3, in the following cases : (a) Overturning effect due to wind, when the bridge is unloaded. (b) Lateral effects of top chord and bottom chord bracings, when the bridge is unloaded. (c) Overturning effect due to wind, when the bridge is loaded. (d) Lateral effects of top chord and bottom chord bracings, when the bridge is loaded. 4.5 Design the top lateral bracing and bottom lateral bracing for the through type truss girder railway bridge for a single metre gauge track as in Problem 4.3. 4.6 An A-type portal bracing of a truss girder is subjected to a horizontal reaction of 100 kN at the top, from left to right. The total length of portal bracing is 6.60 m. The knee braces have been connected at 2. 80 m from the top. Determine the forces in the various members of the portal frame.

251

DESIGN OF TRUSS GIRDER BRIDGES

Also determine the portal effect in the end post and in the bottom chords of the truss girder. The inclination of the end post with the horizontal is 45°. 4.7 Analyse completely the portal bracing shown in Fig. P.4.7 carrying a lateral load of P = 100 kN. Also draw the B.M., S.F., and axial force diagrams for AD and BC. 1 00 kN D

4m

4m

C

E 3m

H F

J G

6m

A

B

Fig. P.4.7

CHAPTER

5

Design of End Bearings for Steel Bridges

5.1

INTRODUCTION

The use of plate girder and truss girder steel bridges over the large spans became possible with the development of steel as a structural material. The rapid response of steel bridges to changes in temperature was seen and need to accommodate expansion and contraction was seriously considered. The sliding bearings (one steel plate sliding over the other) was attached with the one end of the bridge structure was kept fixed. The structural components of the bridge structure become light as the quality of the material (steel) was improved. The appreciable deflections along the span and rotations at the supports of the bridge structures under the loads were noticed. The rocker bearings were used at the fixed ends of the bridge structures to allow the rotation and rocker and roller bearings were used at the free ends to permit horizontal movement and rotation. Recently, the bearings made of materials like plastic and elastomers evolved by some special companies have been used under modern steel and concrete bridges. In addition to provide longitudinal movements and rotations, the elastomeric bearings have a useful capacity of damping down the vibration of the bridge structures. For all spans in excess of 9 m, the provisions are made for change in length due to temperature and stress variation. The provisions for expansion and contraction should be such as to permit movement of the free bearings to the extent of 25 mm for every 30 m of length. For spans greater than 15 m on rigid pier or abutment, the bearings, which permit angular deflection of the girder ends, are provided, and at one end, there shall be a roller, rocker or other effective type of expansion bearing. For wide bridges and skew spans, consideration shall be given to the lateral expansion and contraction.

5.2

FUNCTIONS OF END BEARINGS

The longitudinal movements and rotation occur in the steel bridges due to

DESIGN OF END BEARINGS FOR STEEL BRIDGES

253

superimposed loads and temperature changes. The bearings are provided at both the ends of a bridge girder. One end of the bridge girder is kept fixed in position and it is referred as fixed end. While the other end of the bridge girder is kept free for the horizontal movement and it is referred as free end. The bearings are provided for the following functions : 1. The bearings are provided to transmit the end reaction to the abutments and/or piers and to distribute it uniformly, so that the bearing stress does not exceed the allowable bearing stress of the material. 2. The bearings are provided to allow free movement in the longitudinal direction (expansion and contraction) due to change in temperature and stresses. 3. The bearings are provided to allow rotation at the ends, when the bridge girders are loaded and deflections take place. The accurate function of bearings vary between different bridge girders and even from point to point within the same bridge girder. An individual bearing may transmit to a limited extent, some combination of forces and longitudinal movements and rotations about three references axes. Separate bearings may be used to reduce the effect of impact due to live loads, to damp off structural vibrations, or to limit the transmission of sound waves. The precise functions of bearings are carefully considered in the beginning of the complete design. The finally selected bearings should closely match in performance, and the initial assumptions. The real behaviour of a bridge girder will depend upon its end bearings provided. The bearings influence the performance of the bridge girder. The span of a bridge girder is determined from the positions of bearings. The frictional resistance of bearings to horizontal movements and rotations may set up forces and bending moments in the surrounding structural components. For accurate analysis of the bridge girder, the characteristics of the bearings should be assessed in the beginning of the complete design. The horizontal movements and rotations in a bridge girder may be reversible or irreversible. The temperature changes and temporary superimposed loads cause reversible effects. While the permanent loads and settlement of supports cause irreversible effects. For the approximate initial assessment, the maximum range of longitudinal movement, due to all causes of steel bridges may be assumed 0.09 percent (the proportion of expansion length). It is more difficult to estimate the rotations. However, the end rotation of beam of uniform section may be assumed as four times the allowable central deflection divided by the span. In the design of bearings, provisions shall be made for the transmission of longitudinal and lateral forces to the bearings and the supporting structures. Provisions shall be made against any uplift to which the bearing may be subjected. All bearings are designed to permit inspection and maintenance.

5.3

TYPES OF END BEARING

The end bearings may be broadly classified in the following three categories:

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DESIGN OF STEEL STRUCTURES–VOL. II

1. Mechanical bearings 2. Elastomeric bearings 3. Combined mechanical and elastomeric bearings.

5.3.1

Mechanical Bearings

The mechanical bearings permit longitudinal movements and rotations by sliding, rocking or rolling actions, generally on metal parts. The traditional types of ferrous end bearings allow horizontal movements by steel sliding upon steel, cast iron or hard copper alloy and rotational movements on rockers, rollers or pins. Non-ferrous mechanical bearings were developed and manufactured from aluminium alloy castings. The pure aluminium has very good resistance to corrosion. Most of aluminium casting alloys in this characteristic. The aluminiumsilicon and aluminium-magnesium alloys have been seen to possess satisfactory resistance to corrosion. The aluminium alloy bearings should never be used in contact with copper or its alloys, since alloys of aluminium are very sensitive to galvanic action. The aluminium alloys having oxidised surface may safely be used in contact with steel in rural areas and not in industrial or marine environments.

5.3.2

Elastomeric Bearings

The elastomeric bearings provide horizontal movements and rotations by compressing and shearing actions of layers of rubber-like materials. Several rubberlike layers interleaved with and bonded to thin sheets of steel are used in the elastomeric bearings. These bearings are manufactured from long strips, plain pads or laminated layers. The elastomeric bearings are not suitable when the bridge girders are subjected to repeated uplift forces, since the material of these bearings are placed in tension. The elastomers are essentially compression material. The elastomeric bearings may provide horizontal movements of about 70 mm. The elastomeric bearings may provide more than approximately 0.02 radian rotations about horizontal axis.

5.3.3

Combined Mechanical and Elastomeric Bearings

The combined mechanical and elastomeric bearings are used whenever the horizontal movements exceed the practical limit for an elastomeric bearing. All the horizontal movements are permitted by the mechanical bearings while the elastomeric bearings allow rotations. The bearings may also be classified in the following two categories. (i) Fixed bearings, and (ii) Expansion bearings.

5.3.3.1

Fixed Bearings

The fixed bearing retain their positions and permit the rotational movements only.

DESIGN OF END BEARINGS FOR STEEL BRIDGES

5.3.3.2

255

Expansion Bearings

The expansion bearing allow horizontal movements. A fixed bearing is provided at one end of a simply supported bridge girder and an expansion bearing is used at it is other end. A continuous bridge girder needs only one fixed bearing and expansion bearings at all supports except one.

5.4

SELECTION OF TYPE OF END BEARINGS

The initial selection of a particular type of an end bearing which suits for a particular bridge depends upon the following factors : 1. Type of super-structure 2. Type of supports 3. Length of span 4. Loadings 5. Horizontal movements 6. Rotational movements 7. Plan area 8. Life and maintenance 9. Inclination 10. Environment 11. Vibrations 12. Cost

5.4.1

Type of Super-Structure

High level slab bridges need no special bearing. The submersible slab bridges require bearings to resist the uplift forces. The girder bridges are provided with fixed and expansion bearings.

5.4.2

Type of Supports

In case the piers support the bridge girders, then the bearings having resistance to horizontal movement should not be used. Such bearings influence the entire design of the piers.

5.4.3

Length of Span

The bridge structures over spans below 10 m require no formal bearings. The traditional steel bearings are used for small to medium span steel bridges. It is usual to adopt proprietary bearings in all other cases. Some special companies manufacture the standard mechanical bearings to suit the particular requirements. Such bearings are referred as proprietary mechanical bearings. These manufacturers publish catalogues. These catalogues are revised and improved regularly. The positions of the bearings determine the span for the bridge structure. The type of bearing influences the performance of the bridge structure. The complete analysis of a bridge structure depends upon the type of bearings used at the ends.

5.4.4

Loadings

The standard mechanical bearings are suitable to transmit the vertical loads over 30,000 kN and horizontal loads over 3000 kN while the elastormeric bearings are

256

DESIGN OF STEEL STRUCTURES–VOL. II

capable of resisting vertical loads upto 3000 kN and horizontal forces upto 20 kN only. The elastomeric bearings are not suitable to resist repeated uplift forces. The properly designed mechanical bearings may be used for such cases.

5.4.5

Horizontal Movements

The mechanical bearings have practically unlimited capacity to accommodate horizontal movements. The elastomeric bearings may be provided for limited horizontal movements upto 70 mm only. Such bearings offer smallest resistance in case the horizontal movement is upto 20 mm only. Its resistance increases nearly linearly with the horizontal movement.

5.4.6

Rotational Movements

The mechanical bearings may permit rotation of 0.08 radian about horizontal axis whereas it is difficult to design the elastomeric bearings with appreciable loadings to allow rotations more than 0.02 radian. Correctly selected mechanical bearings may permit greater (virtually unlimited) rotations in plan than the elastomeric bearings.

5.4.7

Plan Areas

Theoretically, the elastomeric bearings may be designed and manufactured to any plan shape whereas the shape of proprietary bearing is fixed.

5.4.8

Life and Maintenance

The elastomeric bearings deteriorate steadily with time. Oil, oxygen and ozone attack such bearings. The life of elastomeric bearings is considerable (45 to 80 years) if these are enclosed. However, these bearings may be replaced after 50 years. These bearings require no maintenance. The properly designed mechanical bearings last upto the life of bridge structures (100–120 years or more). These bearings need regular inspection and maintenance, painting and greasing. These bearings should be accessible for maintenance.

5.4.9

Inclination

The mechanical bearings are provided at right angles to the longitudinal axis of the skew bridges where the skew angle or inclination is small (less than 20 degrees). The elastomeric bearings may be placed parallel with the supported member. The sliding bearings are provided at both the supports if the skew angle is more than 20 degrees and the span length along the longitudinal axis is less than 10 m. The fixed bearings are used when this span exceeds 10 m.

5.4.10

Environment

The aluminium alloy mechanical bearings are used when the bridge structures are constructed in polluted industrial atmosphere.

DESIGN OF END BEARINGS FOR STEEL BRIDGES

5.4.11

257

Vibrations

The elastomeric bearings are excellent to resist the vibrations, impact and sound waves. These bearings have industrial atmosphere.

5.4.12

Cost

The initial cost of elastomeric bearings is low and these bearings prove to be cheap. The initial cost of mechanical bearings is high. However, the selection is seldom governed by economics. It would not be possible to achieve all the desirable characteristics simultaneously. However, efforts are made to select the bearings which suit the particular type of bridge structure.

5.5

FERROUS MECHANICAL BEARINGS

The ferrous mechanical bearings are attached with the steel girder bridges. These bearings provide horizontal movements by metal-to-metal sliding on rollers. The rotational movements are provided by rockers, rollers or pins. The different types of ferrous bearings used for the steel bridges (depending upon the magnitude of end reaction and the span of bridge) are as follows: 1. Plate bearings 2. Rocker bearings 3. Roller bearings 4. Knuckle pin bearings 5. Bearing adopted by Railway Board 6. Spherical bearings.

5.5.1

Plate Bearings

The plate bearings are simplest type of bearings. The plate bearings are used for small spans upto 20 m and small end reaction of the bridge. Figure 5.1 shows a plate

B ridg e girde r

S o le plate (S h oe p la te) B e d pla te (W a ll plate) C ircu la r h ole for fixed b ea rin g

E llip tica l slotted ho le fo r e xp an sio n b ea rin g

Fig. 5.1

258

DESIGN OF STEEL STRUCTURES–VOL. II

bearing. The plate bearing consists of two plates. A sole plate or shoe plate is attached to the bridge girder. The sole plate rests on bed plate or wall plate. The wall plate is anchored to the masonry. Two anchor bolts fixed in masonry pass through the wall plate and the sole plate.The size of wall plate is found by the end reaction and the allowable bearing pressure on the masonry. The plates are made rigid to distribute the end reaction as uniformly as possible over the required area of the masonry. When the anchor bolts pass through the circular holes in the sole plate, then the plate bearings act as fixed bearings. One end of the bridge girder is fixed or anchored to the masonry through the fixed bearings. The fixed bearings are designed for the end reaction (vertical load) and the longitudinal forces. The magnitude of end reactions used are large. Therefore, the fixed bearings designed for end reactions (vertical loads) only are strong enough to take the longitudinal forces. In order to allow the longitudinal movement, the slotted holes are provided in the sole. In order to reduce the friction, the surfaces of sole plate and wall plate in contact are well machined and smoothly finished. The sole plate can slide upon the

B ridg e girde r

Fig. 5.2

wall plate. The plate bearings act as expansion bearings of sliding type. In the expansion bearing, the longitudinal movement (expansion or contraction) takes place with change of temperature and loads. As per the recommendations of IRC section II, the longitudinal force at any free bearing shall be limited to the sum of dead load and live load reactions at the bearings multiplied by the coefficient of friction. The coefficients of friction for the different surfaces in contact are given in

DESIGN OF END BEARINGS FOR STEEL BRIDGES

259

Sec. 2.7. The magnitudes of such longitudinal forces are high in long span bridges and the dead load and live load reactions are also high. The plate bearings have three disadvantages. The edge of plate nearest to the end of span has a tendency to lift along with the deflection of bridge girder. Therefore, the end reaction is not distributed uniformly. Secondly, in order to have longitudinal movement, the sliding friction is to be overcome. Thirdly, the plane bearings have a tendency to force the deck beyond the normal travel during the hot summer. In cold climates, these bearings tend to freeze (i.e., the horizontal movement is restricted). The abutments are pulled away from the approach fill due to enough frictional resistance. Therefore, for the large span bridges, the more efficient devices are necessary.

B ridg e girde r

C ircu lar h ole for fixe d bearing

S lotted h ole for e xp an sion b ea rin g

Fig. 5.3

The end reaction is distributed uniformly by providing a deep cast steel bed block as shown in Fig. 5.2. Such bed blocks have adequate rigidity. The sole plates of plate bearings are many times made curved as shown in Fig. 5.3. The curved sole plate allows rotation. For large spans, the plate bearings are not suitable. The rocker bearings and roller bearings are used in such cases. The coefficient of friction for machine contact surface for steel sliding upon steel or cast iron is 0.25 and that for steel upon copper alloy is 0.15. After some corrosion has taken place, these values both become nearly 0.33 for steel sliding upon both cast iron and copper alloy. Under the service condition, the coefficient of friction for steel sliding upon steel is 0.50.

5.5.2

Rocker Bearings

The fixed end rocker bearings are of various types. It includes cylindrical rockers rotating on flat plates, cylindrical or spherical rockers rotating inside the concave

260

DESIGN OF STEEL STRUCTURES–VOL. II

seatings and knuckle pin bearings. A fixed rocker bearing consists of a cylindrical rocker rotating on a flat surface is shown in Fig. 5.4. Such types of bearings are suitable for small loads. The dowels are used to restrain the horizontal sliding. These dowels pass from one to another. The load carrying capacity of the rocker depends upon the diameter of the rocker’s surface. The load carrying capacity of such rollers may be increased by hardening the rocker and plate. In such rocker bearings, instead of dowels, nibs projecting from the sides of the plates are used.

Fig. 5.4

Figure 5.5 shows a typical rocker bearing. The cast steel shoe and cast steel bed block are used in these types of bearings. A cylindrical pin is inserted in between the cast steel shoe and the cast steel bed block. The pin allows rotation at the ends of bridge girder. The rocker bearing acts as hinged bearing. The end reaction of a bridge girder is transmitted to the pin by direct bearing through the shoe attached with the girder. The vertical plates are used to transmit the end reaction. The

B ridg e girde r

Fig. 5.5

DESIGN OF END BEARINGS FOR STEEL BRIDGES

261

number of plates (two or three) depends upon the magnitude of end reaction. The end reaction is further transmitted to the cast steel bed block and then to the supporting structure. Two outer vertical plates completely encircle the pin. In case the bearing is subjected to an uplift, then the uplift is resisted by these plates. The middle plates provide only bearing with the cylindrical surface of the pin. The cylindrical pin is subjected to bending, shearing and bearing. The required bearing area is provided by the product of total thickness of plates and the diameter of pin. The thickness of all the plates is kept equal. Therefore, the end reaction is transmitted equally by these plates. The value of bending movement is found by multiplying force transmitted by outer plate of the shoe to the outer plate of bed block and centre to centre distance between these plates. The size of base plate is found as the allowable bearing stress in the masonry and the end reaction. The rocker bearings are also subjected to lateral and longitudinal forces in addition to the end reaction (vertical loads). The increase of end reaction due to lateral and longitudinal forces are also taken into consideration. The lateral forces and the longitudinal forces are assumed to act at the level of cylindrical pin of the rocker bearing. The base plate is subjected to movement along both the directions. The total bearing stress in the masonry should not exceed the allowable bearing stress. The rocker bearings are designed for the end reaction and then checked for lateral forces and longitudinal forces. Figure 5.5 shows the rocker bearing for the fixed end. In the rocker bearing for the free end of the bridge girder the underside of shoe is curved, which rotates on the horizontal bearing plate and allows longitudinal movement. This acts as rocker type expansion bearing.

5.5.3

Roller Bearings

The roller bearings as shown in Fig. 5.6 are also used for the long span bridges. Figure 5.6 (a) shows a single roller used in the bearing. The rollers provide the rotation as well as the longitudinal movement. Figure 5.6 (b) shows number of rollers used in the bearing. The roller bearings act as roller type expansion bearings. The rollers are kept in position by means of dowels, lugs or keys as shown in Figure 5.6 (a). The roller bearings for spans above 35 m should preferably be protected from dirt by oil or grease boxes. So long as the size of rollers is small the complete circular rollers are provided. When the size of rollers becomes large, then the sides of rollers are cut in order to reduce the length of shoe, and to make the bearings more compact. These rollers with cut sides are known as segmental rollers. In order to avoid overturning or displacement of these rollers, these are geared with upper and lower plates. The spacing between segmental rollers and the width of rollers may be found as shown below :

262

DESIGN OF STEEL STRUCTURES–VOL. II

B ridg e girde r

S e gm e nta l rolle r

(a )

B ridg e girde r

S e gm e nta l ro lle rs

(b )

Fig. 5.6

It is assumed that the rollers do not slip but only roll during rolling. When, the roller rolls to the maximum position, as shown in the Fig. 5.7 (a), then, the vertical axis of roller turns through an angle θ, and the centre of roller travels through a forward motion, B. Then, where,

⎛d⎞ sin θ = ⎜ ⎟ ⎝D⎠

...(i)

d = Width of segmental roller D = Diameter of segmental roller B = Horizontal travel

⎛ 2B ⎞ ⎛ 114.6B ⎞ tan θ = θ = ⎜ radians = ⎜ ⎟ ⎟ ⎝ D ⎠ ⎝ D ⎠

...(ii)

DESIGN OF END BEARINGS FOR STEEL BRIDGES

⎛ 114.6B ⎞ d = D sin ⎜ ⎟ ⎝ D ⎠

Therefore,

263

...(5.1)

(d + a ) d

d

D

b

B

D D 2

θ

θ

θ

θ

a

(a )

(b )

Fig. 5.7

The distance between adjacent segmental rollers a (i.e., the spacing between the segmental rollers) should be such that the rollers do not come in contact during the forward motion as shown in Fig. 5.7 (b). Then, (a + d) = (d + b) sec θ ...(iii) a = [b sec θ + d (sec θ – 1)] ...(5.2) where, b = Least allowable perpendicular distance between the faces of adjacent rollers, after their revolved positions. The spacing between adjacent segmental rollers a, is found, knowing b, d and θ. A roller is required to move between guides. The flanges attached on the sides of contact plates serve as guide usually. Any possibility of lateral movement or skewing of a roller is prevented. The pinions are attached to the ends of a roller. The pinions move in racks connected to the sides of the contact plates. This ensures the correct rolling movement. The possibility of displacement of a roller by relative rotation between the two contact plates is also eliminated. The multiple rollers are carefully arranged. The load is distributed uniformly over all rollers. Any single roller must not be overloaded. The use of multiple rollers is preferred as compared to that of single roller. The considerations of serviceability decide the load carrying capacity of the roller bearing. It does not depend on the collapse load. The overloading of roller bearing may cause either indentation of the contact plates or development of flats on the roller surface, which will increase the resistance to movement. The rolling motion is likely to be ceased completely. In practice, the contact plates are slightly hardened to avoid the indentation of the plates.

5.5.4 Knuckle Pin Bearings A knuckle pin is inserted between top casting (attached at the end of a bridge girder) and bottom casting. The knuckle pin bearings allow rotation.

264

DESIGN OF STEEL STRUCTURES–VOL. II

The roller bearings are also used to support the cast steel shoe with pin bearings as shown in Fig. 5.8. In such cases the roller also acts as hinged bearing.

Fig. 5.8

The following points are kept in mind while designing a shoe and a pedestal for the roller bearing. 1. The shoe transmits the end reaction to the pin. The end reaction must be distributed from the pin to the various rollers uniformly. 2. The size and number of rollers provided should be adequate to have proper stress and free movement. 3. The rollers should be so arranged that these can be readily cleaned of accumulated dirt and dust. 4. The detail should be so arranged that the rollers do not become displaced from their positions. 5. The end reaction should be distributed from rollers or from the shoe uniformly to the masonry.

5.5.5

Bearing Adopted by Railway Board

The design of bearing adopted by Railway Board, as shown in Fig. 5.9 has top casting and bottom casting, which makes a total height about 1.0 m. If the rollers are used alone, the overall height would be much less. The whole bearing is enclosed in a sheet metal box. The box is completely filled with lubricating oil. This design does not require practically any maintenance. In this bearing, the continuous support (bearing) is provided to the pin. The pin is subjected to bearing pressure only.

DESIGN OF END BEARINGS FOR STEEL BRIDGES

265

The design of knuckle pin in such bearings is based simply upon the allowable bearing stress of the material used. The full projected area of the knuckle pin (the

B ridg e g ird er

Top casting P in o ver con tin u ou s b ea rin g B o tto m casting

Fig. 5.9

product of pin diameter and the length of seating) is determined for the vertical loading. The lateral load is taken by the flanges on the pin in the similar manner as in the roller bearings.

5.5.6

Spherical Bearings

In spherical bearing, a spherical convex rocker (attached with distributed plate connected with the bottom of bridge girder) rotates in a spherical concave seating (attached to a base plate). The spherical bearings allow rotations nearly 0.01 radian in any direction.

5.6

ALLOWABLE STRESSES ON FERROUS BEARINGS

The allowable stresses on different types of bearings are as follows. These stresses are as per IS : 1915–1961 (but converted in S.I. units).

5.6.1

Cylindrical Roller Bearing

(i) Cylindrical rollers on curved surfaces The permissible load per unit length shall be as follows : (a) Single or double rollers :

266

DESIGN OF STEEL STRUCTURES–VOL. II

⎡ ⎤ ⎢ ⎥ 1 1 ⎥ kN/mm ⋅ K3 ⎢ 100 ⎢ 1 − 1 ⎥ ⎢⎣ Ds Ds2 ⎥⎦ 1

...(i)

(b) Three or more rollers : ⎡ ⎤ ⎢ ⎥ 1 1 ⎥ kN/mm ⋅ K4 ⎢ 100 ⎢ 1 − 1 ⎥ ⎢⎣ Ds Ds2 ⎥⎦ 1

...(ii)

where, for mild steel conforming to IS : 226. K3 = 0.80, K4 = 0.50 DSI and Ds2 are the diameters in mm of the convex and concave surfaces, respectively. These expressions (i) and (ii) are used for a convex rocker rotating in a concave seating even though the rocker may be only a segment of a roller. The vertical load carrying capacity with a concave seat is more than with a flat surface, as the contact surface is increased. (ii) Cylindrical roller on flat surfaces For rollers of diameter not less than 100 mm, the permissible load shall be as follows: (a) Single or double rollers 1 ⋅ K 3 Dr kN/mm 100 (b) Three or more rollers 1 ⋅ K 4 Dr kN/mm 100

...(iii)

...(iv)

where K3 and K4 are as mentioned above. Dr = Diameter of roller in mm. The expressions (i) and (ii) reduce to the expressions (iii) and (IV), since the seating is over a flat surface, where Ds2 is infinite.

5.6.2

Spherical Bearings

The permissible load on the spherical bearings of mild steel as per IS : 226 are as follows : 2

1 1 ⎡ 1 ⎤ kN ⋅ 1 ⎥ 100 125 ⎢ 1 − ⎢ ⎥ ⎣⎢ Ds1 Ds2 ⎥⎦

...(v)

DESIGN OF END BEARINGS FOR STEEL BRIDGES

267

For high tensile structural steel 2

1 1 ⎡ 1 ⎤ kN ⋅ 1 ⎥ 100 75 ⎢ 1 − ⎢ ⎥ ⎢⎣ Ds1 Ds2 ⎥⎦

...(vi)

where Ds1 and Ds2 are as mentioned above. B.S. 153 specifies the load carrying capacity of a spherical bearing of diameter Ds1in a spherical concave seating of diameter Ds2 as below : (a) For mild steel 2

1 1 ⎛ Ds1 ⋅ Ds2 ⎞ ⋅ ⎜ ⎟ kN 100 130 ⎝ Ds2 − Ds1 ⎠

(vii)

(b) For high yield stress steel 2

1 1 ⎛ Ds1 ⋅ Ds2 ⎞ ⋅ ⎜ ⎟ kN 100 78 ⎝ Ds2 − Ds1 ⎠

(viii)

(c) High load steel for rocker and seating 2

1 1 ⎛ Ds1 ⋅ Ds2 ⎞ ⋅ ⎜ ⎟ kN 100 15 ⎝ Ds2 − Ds1 ⎠

...(ix)

When the steels of intermediate strength are used the interpolation may be done.

5.6.3

Sliding Bearings

The permissible pressure for steel sliding on steel, hard copper alloys or on cast iron, shall not exceed 32 N/mm2. The permissible stresses in pins of mild steel as per IS : 226, as per IS : 833–1994 shall be as follows : In shear 100 N/mm2 In bearing 300 N/mm2 In bending 600 σy For turned and fitted knuckle pins and sphere in bearing, the permissible stress on the projected area shall not exceed 120 N/mm2. The allowable working pressure under bearings for bed plates as per Code of practice for the design of steel bridges published by Railway Board are as follows : The area of bearings or bed plates shall be so proportioned that with the eccentricity of loads due to combination of dead load with live load, impact load and forces due to curvature, the maximum pressure on material forming the bed shall not exceed the following limits : Granite 3.6 N/mm2 Sand stone 2.95 N/mm2

268

DESIGN OF STEEL STRUCTURES–VOL. II

(As per Indian Railway Standard, concrete bridge code) Cement concrete M 10 2 N/mm2 M 15 3 N/mm2 M 20 4 N/mm2 M 25 5 N/mm2 Reinforced concrete M 10 2.5 N/mm2 M 15 3.75 N/mm2 M 20 5 N/mm2 M 25 6.25 N/mm2 The above mentioned limits may exceeded by 33 13 percent when wind pressure effects and forces and forces effects due to earthquake are also taken into consideration. The centre of pressure under flat bearing plates attached to the girders shall be assumed to be at one-third of the length from the front edge. Slab bases for bearings. The effective area for distributing the load to the foundation shall be taken as the contact area of the member communicating the load to the slab plus the area given by a projection of twice the thickness of the slab around the contact area of the member. Example 5.1 The effective span of a plate girder deck type bridge for a single metre gauge track is 24 m. The dead load, live load and impact load reaction is 750 kN. The vertical reaction due to overturning effect of wind at each of the girder is 115 kN. Design a suitable bearing. Solution Design : Step 1. Dead load, live load and impact load reaction = 750 kN Step 2. Longitudinal force per girder Tractive effort =

1 × 248 = 124.0 kN 2

Braking force =

1 × 255 = 127.5 kN 2

Allowable pressure in cement concrete = 4 N/mm2 Area of bed plate required 750 × 1000 = 1875 × 100 mm2 4 Provide 550 mm × 400 mm bed plates, and 550 mm × 250 mm rocker plates.

=

DESIGN OF END BEARINGS FOR STEEL BRIDGES

269

Step 3. Design of Rocker Plate Maximum bending moment

⎛ 750 × 1252 ⎞ = ⎜⎜ 250 × 2 × 1000 ⎟⎟ = 23.4 kN-m ⎝ ⎠ Let t 1 be the thickness of rocker plate then, 1 ×550 × t 12 × 185 = 23.4 × 106 6

t1

⎛ 6 × 23.4 × 106 = ⎜⎜ ⎝ 550 × 185

1/2

⎞ ⎟⎟ ⎠

= 37.15 mm

Provide 36 mm thick rocker plate. Step 4. Design of Anchor Bolts It is assumed that longitudinal force is taken equally by bearings at both ends. Longitudinal force for one bearing 1 × 127.5 = 63.75 kN 2 Provide two anchor bolts in one bearing. Allowable shear = 100 N/mm2

=

Area of each bolt =

63.75 × 1000 = 318. 75 mm2 2 × 100

Provide 25 mm diameter bolts. Step 5. Design of Thickness of base plate

⎛ 720 × 1000 ⎞ 2 ⎜ ⎟ = 3.41 mm ⎝ 550 × 400 ⎠ Projection = 75 mm Let t 2 be the thickness of bed plate and consider 1 mm wide strip. Maximum bending moment Bearing pressure =

⎛ 752 ⎞ = ⎜⎜ 3.41 × ⎟ = 9590.625 N-mm 2 ⎟⎠ ⎝ ∴

1 × 1 × t22 × 185 6

= 9590.625 1/ 2

⎛ 6 × 9590.625 ⎞ t2 = ⎜ ⎟ 185 ⎝ ⎠ = 17.64 mm Provide 18 mm thick bed plate.

270

DESIGN OF STEEL STRUCTURES–VOL. II

The sole plate is also kept 18 mm thick. The bearing designed is shown in Figure 5.10.

7 50 kN

S o le plate R o cker plate

38 m m

B e d pla te

75 m m

4 00 m m

75 m m

Fig. 5.10

Example 5.2 The effective span of a plate girder through type bridge for a single broad gauge track is 30 m. The dead load, live load and impact load reaction is 1200 kN. The vertical reaction due to overturning effect of wind at each end of the girder is 80 kN. The lateral load due to wind at each bearing is 34 kN. Design the rocker bearing. Solution Design : Step 1 Dead load, live load and impact load reaction = 1200 kN Vertical reaction due to wind = 80 kN Longitudinal load from Bridge Rule, for 30 m span, for broad gauge per track. Tractive effort = 476 kN Braking force = 457 kN Longitudinal load per girder =

1 × 476 = 238.0 kN 6

Allowable stress in cement concrete = 4 N/mm2 Area of bed plate required =

1200 × 1000 = 3000 × 100 mm2 4

Provide 700 mm × 700 mm bed plate as shown in Fig. 5.11. A large bed plate is provided, since, it will also be subjected to lateral load and longitudinal force.

80 0 m

DESIGN OF END BEARINGS FOR STEEL BRIDGES

3 mm

7 00 m m

271

3 mm

1 00 m m 75 m m 25 m m 2 37 .5 4 00 kN 4 00 kN 4 00 kN mm 25 mm 2 37 .5 mm 25 28 m m 28 m m mm 75 m m

Fig. 5.11

Step 2. Design of Rocker bearing As per IS : 800–1984 Allowable stress in bearing on projected area = 0.75 fy Allowable stress on mild steel pins In shear = 100 N/mm2 In bearing = 300 N/mm2 In bending = 0.66 fy Provide 150 mm diameter pin Thickness of rib plates required

⎛ 1200 × 1000 ⎞ ⎟ = 42.66 mm = ⎜⎝ 0.750 × 250 × 150 ⎠ Provide 25 mm thick, 3 rib plates Total thickness = 75.0 mm Load shared by each plate 1 × 1200 = 400 kN 3 Let the gap between two outer plates of top casting and bottom casting be 3 mm. Distance between centre to centre of two outer plates = (25 + 3) = 28 mm

=

272

DESIGN OF STEEL STRUCTURES–VOL. II

Maximum bending moment in pin 28 = 11.2 kN-m 103 Maximum bending stress in the extreme fibre of pin,

= 400.0 ×

⎛ 11.2 × 106 ⎞ σb = ⎜ ⎟ = 33.82 N/mm2. Hence safe. π 3 ⎜ × 150 ⎟ ⎝ 32 ⎠ Maximum shear stress, τv.cal =

4 ⎛ 400.0 × 1000 ⎞ × 3 ⎝⎜ π / 4 × 1502 ⎠⎟

= 30.2 N/mm2. Hence safe. Vertical load due to dead load, live load, impact load and wind load = (1200 + 80) = 1280 kN Let the height of pin above base = 300 mm Lateral load due to wind at each bearing = 34 kN Moment due to wind at the base = 34.0 × 0.30 = 10.2 kN-m Moment due to longitudinal force = 238 × 0.30 = 71.4 kN-m Maximum pressure below the base 10.2 × 106 ⎛ 1280 × 1000 ⎞ ⎛ = ⎜ ⎟ + ⎜⎜ ⎝ 700 × 700 ⎠ ⎝ 1 / 6 × 700 × (700 )2

⎞ ⎛ 71.43 × 106 ⎟⎟ + ⎜⎜ 2 ⎠ ⎝ 1 / 6 × 700 × (700 )

⎞ ⎟⎟ ⎠

= 4.04 N/mm2 < 1.25 × 4 N/mm2. Hence, safe. Allowable stress for occasional loads is 1.25 × allowable stress for normal loads. Step 3. Design of base plate Therefore, design pressure for base 4.04 = 3.23 N/mm2 1.25 Let the projection of the base = 75 mm Maximum bending moment per mm strip

=

=

3.23 × (75)2 = 9084.38 mm 2

Let t be the thickness of base plate. Then,

1 × 1× t 2 × 185 = 9084.38 6 t = 17.16 mm

273

DESIGN OF END BEARINGS FOR STEEL BRIDGES

Provided 20 mm thick base plate. The base plate is tested at section XX as shown in Fig. 5.12. 25 m m

25 m m

25 m m

1 59 .12 m m 2 10 m m X 7 0.8 8 m m

X 3 .23 N /m m 2

20 mm 7 00 m m

Fig. 5.12

Maximum bending moment =

⎡ 3.23 × 700 × (350 )2 ⎤ ⎢ ⎥ = 138.486 kN-m 2 × 1000 ⎣ ⎦

Let the distance of XX-section from the base be y , ⎡ ⎛ 210 ⎞⎤ ⎢ 700 × 20 × 10 + 3 × 210 × 25 × ⎜⎝ 2 + 20⎟⎠ ⎥ ⎥ y = ⎢ 700 × 20 + 3 × 210 × 25 ⎣ ⎦

= 70.88 mm The moment of inertia of base block

⎡1 3 × 2.5 × (21)3 ⎤ 3 2 70 2 70 2 × × + × × + ( ) 6.088 ⎢ ⎥ 4 4 I = 12 12 ⎢ ⎥ × 10 mm 2 ⎢ +3 × 2.5 × 21 × (5.412 ) ⎥⎦ ⎣ = 15636.85 ×104 mm4 Bending stress,

⎡138.486 × 106 × (230 − 70.88) ⎤ ⎢ ⎥ 15636.85 × 104 ⎣ ⎦ 2 = 140.92 N/mm . Hence, safe. Example 5.3. The dead load, live load and impact load reaction at the end of a bridge girder as 1300 kN. The vertical reaction of each end of the girder due to overturning effect of wind is 60 kN. Design the roller bearing. The least allowable perpendicular distance between the faces of adjacent rollers after their revolved position may be taken as 6 mm. The centre of roller travels 25 mm. σb =

274

DESIGN OF STEEL STRUCTURES–VOL. II

Solution Design : The maximum end reaction at the support due to dead load, live load, impact load, and vertical reaction at each end of the girder due to overturning = (1300 + 60)= 1360 kN. Step 1. Design of Pin The pin bearing is provided with continuous seating Minimum size of pin = 100 mm Assume length of pin = 400 mm Bearing area provided 400 × 100 = 4 × 104 mm2 Bearing stress on the pin 1360 × 103

= 34 N/mm2 4 × 104 < 1.33 × 300 N/mm 2 (Allowable bearing stress) Step 2. Design of Rollers on flat surface Provide 100 mm diameter, three rollers. Permissible load for 3 or more rollers =

= K 4 . Dr

1 kN/mm 100

⎛ 0.50 × 100 ⎞ ⎟⎠ = 0.50 kN/mm = ⎜⎝ 100 Total length of rollers required

⎛ 1360 ⎞ ⎜⎝ ⎟ 0.50 ⎠

= 2720 mm

⎛ 2720 ⎞ ⎟ = 906.66 mm Length of each roller = ⎜⎝ 3 ⎠ Provide 100 mm diameter, 3 rollers each 0.950 m long. Least allowable perpendicular distance between the faces or adjacent rollers after their revolved position b = 6 mm Horizontal travel B = 25 mm Diameter of rollers D = 100 mm From Eq. 5.1, the width of segmental rollers d =

°

⎛ 114.6 × B ⎞ D sin ⎜ ⎟ D ⎝ ⎠

DESIGN OF END BEARINGS FOR STEEL BRIDGES

275

°

⎛ 114.6 × 2.5 ⎞ = 100sin ⎜ ⎟ mm 10 ⎝ ⎠ = 100 × 0.48 = 48 mm Provide 50 mm width for the rollers From Eq. 5.2, the spacing between adjacent rollers a = b sec θ + d (sec θ –1) °

⎛ 114.6B ⎞ θ = ⎜ ⎟ = 28.65° ⎝ D ⎠ ∴

a = 6 × sec (28.66)° + 50 (sec 26.65° – 1) = 13.7 mm The spacing between adjacent rollers is kept equal to 150 mm. The bearing designed is shown in Figure 5.13. Top casting

P in 1 0 0 m m d ia m e te r 4 00 m m lo ng

B o tto m casting

3 rolle rs e ach 0.9 50 m lo ng

1 00 m m

65 m m

65 m m

Fig. 5.13

5.7

NON-FERROUS MECHANICAL BEARINGS

Non-ferrous mechanical bearings based upon aluminium alloy castings were developed along with the increase in the concrete bridges, which do not need painting. The aluminium alloy castings have satisfactory resistance against

276

DESIGN OF STEEL STRUCTURES–VOL. II

corrosion. Non-ferrous mechanical bearings are manufactured by some companies having special rights. As such, these bearings are referred as proprietary mechanical bearing. These bearings are similar in concept to the traditional designs of ferrous bearings. Adequate trade literature is published by the manufacturers of such bearings. The manufacturer’s catalogues are revised and improved with time. These bearings are selected after going through the latest editions of these catalogues. The mechanical bearings based upon the principle of ball and socket are also manufactured from non-ferrous metal. Such bearings are referred as ball and socket type bearings. These bearings offer freedom of rotation about any horizontal axis, normally upto 0.09 radians. These bearings have ability to accommodate inclination of the superstructure. These bearings also permit the rotations in plan about the vertical axis. These ball and socket type of bearings have convex rocker in the form of segment of a sphere and a concave seating. The concave seating has a lining of polytetra fluorethylene. The complete assemblies are restrained in elastomeric discs. The vertical load is transmitted nearly radially. The seating behaves like a spherical shell.

5.8

ELASTOMERIC BRACINGS

The natural rubber and synthetic materials having many rubber-like characteristics are referred as elastomers. 1. Natural rubber is cheap and it is adequate for most of the civil engineering applications. It has moderate weathering resistance. It is inflammable material. It burns readily. It is attacked by oxygen, but this remains confined to the surface. It is also readily attacked by ozone. The extensive deep cracks are produced in the material in the direction perpendicular the tensile strain. The rubber compounds specially prepared for bearings are protected by incorporating anti-oxidants and anti-ozonants inside the material. Natural rubber absorbs all the derivatives of petroleum and aromatic hydrocarbons (in case such materials remain in direct contact), swells and looses its mechanical properties. The hardness of a small specimen of natural rubber by about 4 IRHD (International rubber hardness). It is equivalent to about 20 per cent change in the elastic modulii. The elongation at break for this material may fall to about 85 percent of its original value. The natural rubber bridge bearings should have a useful life of at least fifty years. At present, the rubber bearings are in use for about twenty years. The natural rubber has a number of shortcomings. As such, the synthetic materials have been developed and used as elastomers. 2. Many elastomers are produced with a wide range of physical properties, depending upon the composition and the manufacturing process. The elastomers are generally characterized by measuring its hardness. The hardness of elastomer is measured by reversible elastic indentation under a steel point. It is measured in degrees on International Rubber scale (IRHD). The IRHD scale ranges from 0 for very soft material to 100 for very hard material. The hardness of erasers and car tyres are about 30 and 60 IRHD respectively. The approximate properties of some elastomers are given in Table 5.1.

277

DESIGN OF END BEARINGS FOR STEEL BRIDGES

Table 5.1 Approximate properties of elastomers* K

εU

Hardness

E

G

IRHD

N/mm 2

N/m 2

50

2.3

0.6

0.75

> 450

60

3.7

1.0

0.60

400

70

6.2

1.4

0.56

300

Percent

* Long, J.E. ‘Bearings in Structural Engineer published by Newnes-Butterworths, London, 1974 p. 12). E = Elastic modulus G = Shear modulus K = Empirically determined constant

εU = Ultimate tensile strain. The elastic modulus of elastomer increases with increase in its hardness. The elastic modulii in tension and compression are roughly equal. The elastic modulus of an elastomer is always small as compared with most of the engineering materials, whereas the bulk modulus is high. The bulk modulus of an elastomer is compared with that of water. The actual value of bulk modulus varies with hardness and other factors. Its value is normally in excess of 1000 N/mm2. For many applications, an elastomer may be treated as incompressible. The tensile breaking stress based on original cross-sectional area of most elastomers exceeds 10 N/mm2 at normal temperature. 3. Neoprene (chloroprene rubber) is best known synthetic rubber. It was introduced about 40 years ago. Chloroprene synthetic rubber was given a trade name as ‘Neoprene’. As regards resistance to ageing, weathering, flames and attack by oils and fuels, neoprene is superior to natural rubber. Neoprene may be produced with hardness ranging from 30 to 90 IRHD. Neoprene elongates less than the natural rubber of same hardness. A neoprene bearing tends to be larger than the equivalent natural rubber bearing. Neoprene has good resistance to flame ; tearing and abrasion. Neoprene bonds readily with metals. It weathers well. The useful life of Neoprene bearings is at least sixty years. Provision for the removal and replacement of Neoprene bearings may be kept as a precautionary measure during the life of structure, if this exceeds about 50 years. In India, Neoprene bearings were first used over the north abutment of a prestressed concrete across river Coleroon in Tamil Nadu state. The total thickness of each bearing is 30 mm. Five Neoprene layers 300 mm × 500 mm reinforced with galvanised iron mesh of 1 mm wire size and 4.6 mm mesh spacing. Since then, Neoprene bearings have been provided in many bridges. Butyl rubber, silicone rubber, nitrile rubber, Butadienceslyrene rubber etc. are the numerous synthetic rubbers. Some of these have been used in bearings.

278

5.9

DESIGN OF STEEL STRUCTURES–VOL. II

STATIC BEHAVIOUR OF ELASTOMER

The behaviour of elastomer subjected to static loading in (i) compression, (ii) shear and (iii) torsion in plan may be described as under :

5.9.1

Compression

A rectangular block of an elastomer subjected to vertical load P is shown in Fig. 5. 14. The block is under compression. The loaded faces of the block are prevented from slipping by friction or physical bonding. The vertical stiffness of such a block depends upon its freedom to bulge at the sides as shown in Fig. 5.14 (a) and it is

dc t B

L

(a ) E levatio n

(b ) S ide vie w

Fig. 5 .14

expressed in terms of shape factors, S. The shape-factor, S is defined as the ratio of loaded plan area to the force-free surface area. The shape-factor, S for different shape of an elastomer block is as follows : (i) A rectangular block L.B ⎡ ⎤ ⎢2( ⎥ ) ⋅ t ⎣ L+B ⎦

...(5.3)

S =

L 4t

...(5.4)

S =

1 1 D πD 2 × = 4 πD ⋅ t 4t

...(5.5)

S = (ii) For square block (L = B)

(iii) For circular block

(iv) For very long rectangular block A very long rectangular block in which the length, L is greater than five times the width, B, the bulging at the ends is neglected. The shape-factor of such a strip is found by considering unit-length B ...(5.6) 2⋅t The vertical stiffness of an elastomer block increases rapidly with the increase in shape-factor. From Eqs. (5.5 to 5.6) it is seen that the shape-factor shall be large

S =

DESIGN OF END BEARINGS FOR STEEL BRIDGES

279

when the thickness of block, t is small and the vertical stiffness shall also be large. In case, the thickness of two blocks is the same, and the plan area of one is larger than the other, the block of larger plan area shall be more stiffer than the other. For the vertically stiff-block, the compression in bulk cannot be neglected. The vertical compressive strain is determined by Lindley, P.B. for a rubber block carrying uniform compressive stress σc is given by σc ...(5.7) E (1 + 2 K ⋅ S 2 ) where K is an empirically found constant. For a completely incompressible material, the value of K may be taken as unity. The value of K decreases with the increase in the hardness of an elastomer. The term E (1 + 2K . S2) in the denominator of Eq. 5.7 may be called as the apparant modulus Ea. Therefore,

ec =

ec =

σc Ea

...(5.8)

The total vertical strain, e v in the bulk of an elastomer is equal to sum of the vertical strain e c and the bulk strain, e k . ev = e c + ek σ ⎞ ⎛σ ev = ⎜ c + c ⎟ ⎝ Ea Ek ⎠

or

or

ev = Ea ⋅ Ek

( Ea + Ek ) Therefore,

σc ⎡ ⎤ ⎢ E ⋅E ⎥ a k ⎢ ⎥ ⎣⎢ ( Ea + Ek ) ⎦⎥

...(5.9)

...(5.10)

may be termed as ‘modified modulus’, Em. ⎛σ ⎞ ev = ⎜ c ⎟ ⎝ Em ⎠

...(5.11)

The product of total vertical strain and the thickness of the elastomer block gives the total compression or vertical deflection. ⎛ σc ⋅ t ⎞ dc = e v . t = ⎜⎝ E ⎟⎠ m

...(5.12)

The average compressive stress, σc is obtained by dividing the vertical load by plan area, A. P ⋅t ⎛P⎞ σc = ⎜ ⎟ = ⎝ A ⎠ A ⋅ Em

280 or

DESIGN OF STEEL STRUCTURES–VOL. II

⎛ P ⋅t⎞ dc . A = ⎜ ⎝ Em ⎟⎠

or

P dc

or

⎛ Em ⋅ A ⎞ ⎟ Kc = ⎜⎝ t ⎠

...(5.13)

⎛ E ⋅ A⎞ = ⎜ m ⎝ t ⎟⎠ ...(5.14)

where Kc is termed as compressive rating of block. It is defined as the vertical load required to produce unit vertical deflection. The apparent modulus for a long strip as found by Lindley, P.B., it is given by 4 E (1+K .S2) ...(5.15) 3 Therefore, the vertical compressive strain in the elastomer strip under the uniform compressive stress is given by

Ea =

ec =

3σ c σc = Ea 4 E (1 + K ⋅ S 2 )

...(5.16)

It is to note that the compressive strain in an elastomer strip is independent of length L, for uniform loading. Therefore, it is referred per unit length. The vertical deflection may be determined as per Eq. 5.12. ⎛ σc ⋅ t ⎞ dc = ⎜ ⎝ Em ⎟⎠

...(i)

Multiplying both the sides of the above equation by width, B, ⎛ B × σc × t ⎞ A × dc = ⎜ ⎟⎠ Em ⎝

or or

⎛ P ⋅t ⎞ B . dc = ⎜⎝ L ⋅ E ⎟⎠ m ⎛ P 1⎞ ⎜⎝ d × L ⎟⎠ 0

⎛ B ⋅ Em ⎞ = ⎜ ⎝ t ⎟⎠

(3 P = σc × B × L) ...(5.17)

⎛ B ⋅ Km ⎞ B⎞ ⎛ ⎟ = 2S . Em ⎜3 S = ⎟ Kc = ⎜⎝ ...(5.18) ⎝ t⋅ ⎠ 2t ⎠ where, Kc is termed as compressive rating per unit length. It is defined as the uniformly distributed load per unit length to produce unit vertical deflection. The force-free surface of both circular and very long rectangular blocks bulge in the form of a parabola due to symmetry. Lindley, P.B. ‘Engineering with natural rubber’, NR Technical Report, The Natural Rubber Producers Research Association, or

281

DESIGN OF END BEARINGS FOR STEEL BRIDGES

London (1986). Lindley, P.B. derived the following expression for the maximum central bulge, b, from the initial unstrained position, due to a vertical deflection d. (i) For circular block

⎛ 3 D ⋅ dc ⎞ ⎛ 3 ⎞ ⎟⎠ = ⎜⎝ S ⋅ dc ⎟⎠ b = ⎜⎝ 2 2 t where,

...(5.19)

D = Diameter of the block

⎛ 3 B ⋅ dc ⎞ ⎛ 3 ⎞ ⎟⎠ = ⎜⎝ S ⋅ dc ⎟⎠ b = ⎜⎝ ⋅ 2 t 4

...(5.20)

where, B = Width A short rectangular block also bulges parabolically in plan. Eq. 5.20 may be used to determine the bulge approximately. In direct compression, the strength of most of the elastomers is generally 150 N/mm2 or more. In case, the slip occurs in the loaded forces freely in the lateral direction, the vertical stiffness of a block decreases. The compressive rating of block, Kc is approximately obtained as below : E⋅A t For small deflection, the bulge is uniform as shown in Fig. 5.15. (i) For circular block

Kc =

...(5.21)

0.5

b =

⎫ D ⎧⎛ t ⎞ − 1⎬ ⎨⎜ ⎟ 2 ⎩⎝ t − dc ⎠ ⎪⎭

...(5.22)

B ⋅ dc 2(t + dc )

...(5.23)

(ii) For long rectangular block b =

dc t

b

B

b

Fig. 5.15

Shear. The load/deflection relationship of an elastomer block in shear nearly linear in case the loaded faces do not slip. This relationship is also independent of shape-factor. In case, the shear stress in the block is τ, then, the shear strain is found as

282

DESIGN OF STEEL STRUCTURES–VOL. II

⎛τ⎞ es = ⎜ ⎟ ... (5.24) ⎝G ⎠ The tangent of the angle of deformation, tan φ as shown in Fig. 5.16 gives the shear strain. es = tan φ

⎛H ⎞ τ = ⎜ ⎟ ⎝A⎠

Since,

⎛ H ⎞ es = tan φ ⎜ ⎟ ⎝A G⎠

...(5.25)

where, H = Total horizontal force. φ

t

sap sp

H

Fig. 5.16

The relatively horizontal displacement, ds, between top and bottom faces of the block is approximately given by

⎛ t⋅H ⎞ ...(5.26) ds = t . tan φ = ⎜ ⎟ ⎝ A ⋅G ⎠ The shear rating of the block, Ks is defined as the shear load necessary to produce unit relative displacement. It is given by Ks =

H ⎛ A ⋅G ⎞ = ds ⎜⎝ t ⎟⎠

...(5.27)

or For a strip, the shear rating is expressed per unit length ∴

⎛ H 1 ⎞ ⎛ A ⋅G ⎞ Ks = ⎜ ⋅ ⎟ = ⎜ ⎟ ⎝ ds L ⎠ ⎝ t ⋅ L ⎠

...(5.28)

⎛ B ⋅G ⎞ ⎛ A ⎞ 3 = B⎟ Ks = ⎜ = 2S . G ⎟ ⎜ ⎝ t ⎠ ⎝ L ⎠ where B is the width of the strip. It is to note that the shear stiffness of a block or a strip remains same in all the horizontal directions. In case the angle of deformation, φ due to shear is large, then the relative horizontal displacement is to be obtained as below :

or

ds =

⎡ tan φ ⎤ ⋅ t⎥ ⎢ 2 ⎣ (1 + tan φ) ⎦

...(5.29)

DESIGN OF END BEARINGS FOR STEEL BRIDGES

283

The vertical displacement, dvs due to shear is determined as under: ...(5.30) dvs = ds . tan φ Torsion in plan The structural bearings twist in plan occasionally only. The parallel force twists in relation to its other face held. In the context of structural bearing, this is a secondary problem. It needs only approximate treatment. The relative torsional stiffness is defined as torque required to cause unit angle of twist. Kτs =

T θ

...(5.31)

(i) For circular elastomeric disc 4 ⎛ T ⎞ ⎛ πG ⋅ D Kt = ⎜ θ ⎟ = ⎜⎜ 32t ⎝ ⎠ ⎝

⎞ ⎟⎟ ⎠

...(5.32)

where, the diameter of D is such greater than the thickness of the block. (ii) For a circular disc with a central hole of diameter, d Kt =

T πG 4 D − d4 = θ 32t

(

)

The angular distortion, θ causes the shear strain in both the cases

⎛ 16T eφ = ⎜ ⎝ πGD3

5.10

⎞ ⎛ D⋅θ⎞ ⎟ = ⎜ 2t ⎟ ⎠ ⎝ ⎠

...(5.33)

TYPES OF ELASTOMERIC BEARINGS

Unconfined pad bearings and confined pot bearings are two types of elastomeric bearings. Unconfined pad bearings are further sub-divided into two categories : ‘unreinforced’ and ‘reinforced’. Elastomeric bearing of rectangular shape consisting of one single layer of synthetic material without any metal is known as unreinforced bearing. Slab bridges with spans upto 10 m and transferring uniformly distributed loads are supported on unreinforced bearings. Elastomeric bearings of rectangular shape consisting of several rubber-like layers strengthened with metal (steel) plates is known as reinforced bearing. Girder bridges transmitting heavy concentrated loads are supported on reinforced bearings. Elastomeric layers are bonded to thin metal plates during the manufacturing process. The metal plates may remain visible at the sides or may be covered with elastomeric layer. Laminated reinforced bearings and moulded reinforced bearings are further two classes of reinforced bearings. In a laminated reinforced bearing, elastomeric layer is bonded with suitable adhesives on both the sides of one metal plate. Such metal plates with elastomeric layers on their both the sides are further bonded in the next operation. In a moulded reinforced bearing, the several elastomeric layers including the alround cover with metal plates are moulded together in one operation. Moulded

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DESIGN OF STEEL STRUCTURES–VOL. II

reinforced bearings are preferred. The plan dimensions of the overall bearing may be selected from the preferred numbers recommended in IS : 1076–1967.

5.11

DESIGN OF ELASTOMERIC BEARINGS

In Britain, the elastomeric bearings for static loading are designed as per specifications issued by the department of the Environment in its memorandum No. 802 (Provisional rules for the use of the rubber bearings in highway bridges, H.M.S.O., London, 1962 revised in 1972). The bearings for highway bridges are designed by following the specifications as laid down in memorandum No. 802. In addition to this, the specifications drafted by International Union of Railway Pairs, (January 1969) are also available. These specifications have been referred in U.I.C. Code 772 R ‘Code for the use of rubber bearings for rail bridges’. Indian Road Congress has also drafted specifications (Draft specifications for elastomeric bearings) in 1976, but it is still unpublished. The specifications in memorandum No. 802 are based on the concept of limiting strains. The strains are limited to the specified values. Such limitations guard against the failure of the elastomer in fatigue. The unreinforced pad bearings are designed as follows : (i) The area of elastomeric bearing in plan is determined for the total vertical load, P ⎛P ⎞ A´ = ⎜ ⎟ ⎝ σc ⎠

...(5.34)

The average vertical compressive stress is limited to the allowable limits (σc< 2 SG). When a horizontal movement, ds occurs in the bearing, the vertical load acts on the reduced effective area. In case this horizontal movement, ds takes place in the direction of length L of a rectangular block as shown in Fig. 5.17, the effective area shall be reduced to (L – ds). Due to this fact, the shape factor of the bearing is not changed.

ds

ds

( L –d s ) L

Fig. 5.17

Total compression i.e., the vertical deflection as found from Eq. 5.12 permitting for the effects of compression in bulk shall not be greater than 15 percent of the total thickness, t of the elastomer in the bearing. By this restriction, the linear elastic design theory is not violated.

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DESIGN OF END BEARINGS FOR STEEL BRIDGES

(ii) In order to avoid the tension at the rear edge of the elastomer, Fig. 5.18, the rotation about the horizontal axis is limited. The life of the elastomer is likely to be reduced due to direct tensile strain. ⎛ 2d 2P 0.2 t ⎞ = α < ⎜ c = ⎟ LK c L ⎠ ⎝ L

...(5.35)

(iii) The thickness of elastomer is limited to ensure the stability against overturning. t