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COD

APPLICATUO

EXAMPLES

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Table of Con tents

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CopyrightlPub lisher/Editor/Disclaimer Preface Acknowledgments Suggestions for Improvement / Errata Notifi cation Introduction How to Use This Document Notation Definitions

ii ix xi I 2 3 18

EXAMPLE

ASCEISEI 7-05

VlJ

DESCRIPTION

Example i

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Classification/Importance Factors Seismic Design Category Example I Earthquake Load Combinations: Strength Design Example 2 Comb inations of Loads Example 3 Design Spectral Respon se Accelerations Introduction to Vertical Irregularities Example 4 Vertical Irregul arity Type l a and Type Ib Example 5 Vertical Irregul arity Type 2 Example 6 Vertical Irregularity Type 3 Example 7 Vertical Irregularity Type 4 Example 8 Vertical Irregular ity Type 5a Example 9 Vertical Irregularity Type 5a Introduction to Horizontal Irregularities Example 10 Horizontal Irregularity Typ e Ia and Type Ib Example I I Horizontal Irregularity Type 2 Example 12 Horizontal Irregularity Typ e 3 Example 13 Horizontal Irregularity Type 4 Example 14 Horizontal Irregularity Type 5 Example 15 Redu ndancy Factor p Example 16 P-delta Effects Example 17 Seismic Base Shear

..

PAGE

§11.5-1 §11.6

25 26

§ 12.4.2.3 §2.4 § 11.4 §12.3.2.2 §12.3.2.2 §12.3.2.2 §12.3.2.2 §12.3.2.2 §12.3.2.2 §12.3.3.1 §12.3.2.1 §12.3.2.1.. §12.3.2.1 §12.3.2.1 §12.3.2.1 §12.3.2.1 §12.3.4 §12.8.7 §12.8. 1...

27 32 36 41 42 46 48 50 52 54 58 59 63 65 67 68 69 74 78

2006 IBC Stru ctural/S eismic Design Manual, Vol. I

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EXAMPLE

DESCRIPTION

ASCE/SEI 7-05

Example 18 Example 19 Example 20 Example 2 1

§12.8.2.I §I 2.14 §I 2.2.3.I

80 83 86

§I2.2.2

90

Example 23 Example 24 Example 25 Example 26 Example 27 Example 28

Appro xima te Fundamental Period Simplified Alternative Structural Design Procedure Combination of Structural Systems: Vertical Comb ination of Framin g Systems: in Different Directions Combin ation of Structural Systems : Along the Same Axis Vertical Distribution of Seismic Force Horizontal Distribution of Shear Amplification of Accident al Torsion Elements Supporting Discontinuous Systems Elements Supporting Disconti nuous Walls or Frames Soil Pressure at Foundati ons

Example 29 Example 30 Exampl e 31 Example 32 Example 33 Example 34 Example 35

Drift Story Drift Lim itations Vertical Seismic Load Effect. Design Response Spectrum Dual Systems Lateral Forces for One-Story Wall Panels Out-of-Plane Seismic Forces for Two-Story Wall Panel

Example 22

Example 36 Example 37 Example 38 Example 39 Example 40 Example 4 1 Exampl e 42 Example 43 Example 44 Example 45 Example 46 Example 47 Example 48 Example 49 iv

Rigid Equipment.. Flexible Equipment Relative Motion of Equipment Attachments Deformation Compatibility for Seismic Design Categories D, E, and F .· Adjoining Rigid Elements Exterior Elements: Wall Panel Exterior Nonstructural Wall Elements: Precast Panel. Beam Horizontal Tie Force Collector Elements Out-of-Plane Wall Anchorage of Concrete or Masonry Walls to Flexible Diaphragms Wall Anchorage to Flexible Diaphragms Determination of Diaphragm Force Fpx : Lowrise Determination of Diaphragm Force Fpx : Highrise Building Separations

2006 IBC Structural/Seismic Design Manual, Vol. I

§I 2.2.3.2 §12.8.3 §12.8.4 §I2.8.4.3 §I 2.3.3.3 §12.3.3.3 §2.4 §I2.I3.4 §12.8.6 § 12.12 §12.4.2.2 §11.4.5 §12.2.5. I §12.11 §12.11. I §I2.11.2 §I 3.3.1 §13.3.1 §I3.3.2

PAGE

92 93 97 102 106 I 10 I I3 I 16 I 19 121 124 126 129 133 137 140 143

§ I2.12.4 §12.7.4 §I3.5 .3 §13.5.3 §12.1.3 §12.10.2 §12.11.2 §12.11.2.1 § 12.11.2.1

145 148 150 153 160 162

§12.10 .1.1

170

§12.10.1 §12.12.3

174 176

165 167

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Table of Conten ts

1

EXAM PLE

DESCRIPTION

ASCE /SEI 7-05

Example 50 Example 5 1 Example 52 Example 53 Example 54 Example 55 Example 56 Example 57

Flexible Nonbuilding Structure Lateral Force on Nonb uilding Structure Rigid No nbuilding Structure Tank With Supported Bottom Pile Interconnections Simplified Wind Loads on 2-Story Buildings Simplified Wind Loads on Low-Rise Buildings Wind Loads - Ana lytica l Procedure

PAGE

§15.5 §15.0 § 15.4.2 §15.7.6 IBC § 1808.2.23. 1 §6.4 §6.4 §6.5

179 182 186 188 190 193 200 205

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2006 IBC Structural/Seismic Design Manual, Vol. I

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Preface

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This document is the initial volume in the three-volume 20061BC Structural/Seismic Design Manual, It has been developed by the Structural Engineers Association of California (SEAOC) with funding provided by SEAOC. Its purpose is to provide guidance on the interpretation and use of the seismic requirements in the 2006 l llfem ational Building Code (IBC), published by the International Code Council , Inc., and SEAOC's 2005 Recommended Lateral Force Requirements and Commentary (also called the Blue Book). The 2006 lBC Structural/Seismic Design Manual was developed to fill a void that exists between the commentary of the Blue Book, which explains the basis for the code provisions, and everyday structural engineering design practice . The 2006 lBC Structural/Seismic Design Manual illustrates how the provisions of the code are used. Volume 1: Code Application Examples, provides step-by-step examples for using individual code provisions, such as computing base shear or building period . Volumes 1I and lIl: Building Design Examples, furnish examples of seismic design of common types of buildings. In Volumes" and III, important aspects of whole buildings are designed to show, calculation-bycalculation, how the various seismic requirements of the code are implemented in a realistic design.

I I

The examples in the 2006 lBC Structural/Seismic Design Manual do not necessarily illustrate the only appropriate methods of design and analysis. Proper engineering judgment should always be exercised when applying these examples to real projects. The 20061BC Structural/Seismic Design Manual is not meant to establish a minimum standard of care but; instead, presents reasonable approaches to solving problems typically encountered in structural /seismic design .

1

The example problem numbers used in the prior Seismic Design Manual - 2000 IEC Volume I code application problems have been retained herein to provide easy reference to compare revised code requirements.

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SEAOC, NCSEA and ICC intend to update the 2006 lBC Structural/Seismic Design Manual with each edition of the building code. Jon P. Kiland and Rafael Sabelli Project Managers

2006 IBC Structural/Seismic Design Manual, Vol. I

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2006 IBC Structural/Seismic Design Manual, Vol. I

Acknowledgments

The 2006 IBC Structural/Seismic Design Manual Volume J was written by a group of highly qualified structural engineers. They were selected by a steering committee set up by the SEAOC Board of Directors and were chosen for their knowledge and experience with structural engineering practice and seismic design. The consultants for Volumes I, II, and III are: Jon P. Kiland, Co-Project Manager Rafael Sabell i, Co-Project Manager Douglas S. Thompson Dan Werdowatz Matt Eatherton

John W. Lawson Joe Maffei Kevin Moore Stephen Kerr

A number of SEAOC members and other structural engineers helped check the examples in this volume. During its development, drafts of the examples were sent to these individuals. Their help was sought in review of code interpretations as well as 'detailed checking of the numerical computations.

I'

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'l,.

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Close collaboration with the SEAOC Seismology Committee was maintained during the development of the document. The 2004-2005 and 2005-2006 committees reviewed the document and provided many helpful comments and suggestions. Their assistance is gratefully acknowledged.

ICC

2006 IBC Structural/Seismic Design Manual, Vol. I

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Sugges tion s for Impro vem ent

I In keep ing with SEAOC's and NCSEA's Mission Statemen ts: "to adva nce the structural engineering profession" and "to provide structural engineers with the most current informa tion and tools to improve their practice," SEAOC and NCSEA plan to upd ate this document as structural/seismic requirements change and new research and better understand ing of building performa nce in earthqu akes becomes ava ilable. Comm ents and suggestions for improvements are welcome and shou ld be sent to the following: Structural Engi neers Association of Cal ifornia (SEAOC) A ttention : Executive Director 14 14 K Street, Suite 260 Sacramento, California 95814 Telephone: (9 16) 447-1198 ; Fax : (916) 932-2209 E-ma il: leeiWseaoc.org; Web address: www .seaoc .org

SEAOC and NCSEA have made a substantial effort to "ensure that the information in this document is accurate. In the event that corrections or clarifi cations are needed, these will be posted on the SEAOC web site at h/lP://11 1111'.seaoc.org or on the ICC website at http:// wll1l..iccsaf e.org. SEAOC. ati ts sole discretion, mayor may not issue written errata

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2006 IBC Structural/Seismic DesIgn Manual, Vol. I

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Introduction

Volume I of the 2006 lBC Structural/Seismic D esign Manual: Code App lication Examples deals with interpretation and use of the structural/seismic provisions of the 2006 lntem ational Building Code'" (!BC). The 2006 lBC Structural/Seismic Design Manual is intended to help the reader understand and correctly use the mc structural/seismic provisions and to provide clear, concise, and graphic guidance on the application of specific provisions of the code. It primarily addresses the major structural/seismic provisions of the !BC, with interpretation of specific provisions and examples highl ighting their proper application.

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The 2006 !BC has had structural provisions removed from its text and has referenced several national standards documents for structural design provisions. The primary referenced document is ASCE/SEI 7-05, which contains the "Minimum Design Loads for Buildings and Other Structures." ASCE/SEI 7-05 is referenced for load and deformation design demands on structural elements, National Material design standards (such as ACI, AISC, MSJC and NOS) are then referenced to take the structural load demands from ASCE/SEI 7-05 and perform specific materia l designs. Volume I presents 58 examples that illustrate the application of specific structural/seismic provisions of the !Be. Each example is a separate problem , or group of problems, and deals primarily with a single code provision. Each example begins with a description of the problem to be solved and a statement of given information. The problem is solved through the normal sequence of steps, each of which is illustrated in full. Appropriate code references for each step are identified in the right-hand margin of the page. The complete 2006 lBC Structural/Selsmic Design Ma nual will have three volumes. Volumes II and III will provide a series of structural/seismic design examples for buildings illustrat ing the seismic design of key parts of common building types such as a large threestory wood frame building, a tilt-up warehouse, a braced steel frame building, and a concrete shear wall building. While the 2006 lBC Structural/Seismic Design Manual is based on the 2006 !BC, there are some provision ofSEAOC 's 2005 Recommended Lateral Force Provisions and Commentary (Blue Book) that are app licable. When differences between the !BC and Blue Book are significant they are brought to the attention of the reader. The 2006 lBC Structural/Seism ic Design Manual is intended for use by practicing structural engineers and structural designers, building departments, other plan review agencies , and structural engineering students.

I 2006 IBC Structural/Seismic Design Manual , Vol. I

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Ho w to Use This Do cum ent

] • :JC . .'>;

The various code application examp les of Volume I are organized by topic consistent with previous editions. To find an example for a particular provision of the code, look at the upper, outer come r of each page, or in the table of contents. Generally, the ASCE/SEI 7-05 notation is used throughout. Som e other notation is defined in the followi ng pages, or in the examples. Reference to ASCE/SEI 7-05 sections and formulas is abbreviated. For example, "ASCE/SE I 7-05 §6.4.2" is given as §6.4.2 with ASCE/SEI 7-05 being understood. " Equation (12 .8-3)" is designated (Eq 12.8-3) in the right-hand margins. Similarly, the phrase "T 12.3-1" is understood to be ASCE/SEI 7-05 Table 12.3-1, and "F 22-15 " is understood to be Figure 2215. Throughout the document, reference to specific code provisions and equations is given in the right-hand margin under the category Code Reference. Generally, the examples are presented in the following format. First, there is a statement of the example to be solved, including given information, diagram s, and sketches. This is followed by the "Calculations and Discussion" section, which provides the solution to the example and appropriate discussion to assist the reader. Finally, many of the examples have a third section designated "Commentary." In this section, comm ents and discussion on the example and related material are made. Commentary is intended to provide a better understanding of the exampl e and/or to offer guidance to the reader on use of the information generated in the example. In general, the Volume I examples focus entirely on use of speci fic provisions of the code. No building design is illustrated . Building design examples are given in Volumes II and III. The 2006 lEe Structural/Seismic Design Manual is based on the 2006 IBC, and the referenced Standard ASCE/SEI 7-05 unless otherwise indicated. Occasionally, reference is made to othe r codes and standards (e.g., 2005 AISC Steel Construction Manual 13th Edition, ACI 318-05, or 2005 NOS). When this is done, these documents are clearly ident ified.

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2006 lac Structural/Seismic Design Manual, Vol. I

Notation

The following notations are used in this document. These are generally consistent with those used in ASCE/SEI 7-05 and other Standards such as ACI and AISC. Some new notations have also been added . The reader is cautio ned that the same notation may be used more than once and may carry entire ly different mean ings in different situations, For example, E can mean the tabulated elastic modulus under the AISC definition (steel) or it can mean the earthquake load under § 12.4.2 of ASCE /SEI 7-05 . When the same notation is used in two or more definitions, each definition is prefaced with a brief descript ion in parentheses (e.g., steel or loads) before the definition is given. A ABM

area of floor or roof supported by a member =

cross-sectional area of the base material area of anc hor, in square inches the combined effective area, in square feet, of the shear walls in the first story of the structure

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A ch

=

cross-sectional area of a struct ural member measured out-to-out of transverse reinforcement net area of concrete section bounded by web thickness and length of section in the directio n of shear force considered

Ae

the minimum cross-sectional area in any horizontal plane in the first story, in square feet of a shea r wall

AI

flange area

Ag

=

gross area of section

Ag

=

the gross area of that wall in which ADis identified

Ai

=

the floor area in square feet of the diaphragm level immediately abo ve the story under consideration

=

area of the load-carrying foun dation

=

the effective area of the projection of an assumed concrete failure surface upon the surface from wh ich the anchor protrudes , in square inches

=

area of non-prestressed tension rein forcement

2006 lac Stru ctural/S eism ic Design Manual, Vol. I

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Nota tion

4

Ash

=

total cross-sectional area of transverse reinforcement (including supplementary crossties) having a spacing s" and crossing a section with a core dimension of he

Ask

=

area of skin reinforceme nt per unit height in one side face

A Slmin

=

area having minimum amount of flexural reinforcement

As,

=

area of link stiffener

AT

=

tributary area

Av

=

area of shear reinforcement within a distance s, or area of shear reinforcement perpendicular to flexura l tension reinforcement within a distance s for deep flexural members

A,'J

=

required area of leg reinforcement in each group of diagonal bars in a diagonally reinforced coupling beam

Avr

=

area of shear-friction reinforcement

All'

=

(web) link web area

Aw

=

(weld) effective cross-sectional area of the weld

A.,

=

the torsional amp lification factor at Level x - § 12.8.4.3

a

=

(concrete) depth of equivalent rectangu lar stress block

a

=

(concrete spandrel) shear span, distance between concentrated load and face of supports

ae

=

coefficient defin ing the relative contribution of concrete strength to wall strength

ad

=

incrementa l factor relating to the P-delt a' effects as determined in §12.8.7

a,

=

the acceleration at Level i obtained from a modal analysis (§ 13.3. 1)

ap

=

amplification factor related.to the response of a system or component as affected by the type of seismic attachment determined in §13.3.1

b

=

(concrete) width of compression face of member

br

=

flange width

bu.

=

web width

2006 IBC Structural/S eism ic Design Manual, Vol. I

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Notation

member width-thickness ratio

bit

Cd

=

deflection amplification factor as given in Tables 12.2-1 or 15.4-1 or 15.4-2

Ce

=

snow exposure factor

em

coefficient defined in §Hl of AISC/ASD, 9th Edition

C,

=

the seismic response coefficient determined in § 12.8.1.1 and §19.3.1

Cr

=

building period coefficient - § 12.8.2.1

C, C1'X

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snow thermal factor =

vertical distribution factor - §12.8.3

c

distance from extreme compression fiber to neutral axis of a flexural member

D

dead load, the effect of dead load

De

=

gross weight of helicopter

Dh Dp

the length, in feet, of a shear wall in the first story in the direction parallel to the applied forces

=

d

relative seismic displacement that a component must be designed to accommodate - §13.3.2 effective depth of section (distance from extreme compression fiber to centroid of tension reinforcement)

db

=

(anchor bolt) anchor shank diameter

db

=

(concrete) bar diameter

d,

=

column panel zone depth

E

=

(steel) modulus of elasticity combined effect of horizontal and vertical earthquake-induced forces (§12.4)

E

Em

=

seismic load effect including overstrength factors (§§ 12.4.3.2 and 12.14.2.2.2)

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N otation

£1

=

flexural stiffness of compression member

1

s,

=

modules of elasticity of concrete , in psi

E.,

=

(concrete) modu lus of elastic ity of reinforcement

I

e

=

EBF link length

F

=

load due to fluids

Fa

=

site coefficient defined in §11 .4.3

Fa

=

axial comp ressive stress that would be permitted if axial force alone existed

Fa

=

flood load

Fb

=

bending stress that would be permitted if bending moment alone existed

FaM

=

nominal strength of the base material to be welded

Fexx

=

classification number of weld metal (minimum specified streng th)

Fi,Fu,P, =

6

portion of seismic base shear, V, induced at Level i.n, or x as determined in §12.8.3.

Fp

=

seism ic force, induced by the parts being connected, centered at the component's center of gravity and distributed relative to the compon ent' s mass distribution, as determined in §12.8.3

Fp ,

=

the diap hragm desig n force

F"

=

specified minimum tensile strength, ksi

F"

=

through -thickness weld stresses at the beam-co lumn interface

Fill

=

minimum specified tensile strength of the anchor

F,.

=

long period site coefficient (at 1.0 second period) see § 11.4.3

F,

=

the design lateral force applied at Level x

r,

=

the lateral force induced at any Level i - § 12.8.3

Fw

=

(stee l LRFD) nomina l strength of the weld electrode material

F II'

=

(steel ASD) allowable weld stress

2006 1BC S tructura l/Seis mic Design Manual, Vo/./

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notation

=

assumed web depth for stability

h;, hn,h,

=

height in feet above the base to Level i, 11 or x, respectively

h,

=

height in feet of the roof above the base

hsx

=

the story height below Level x

hll'

=

height of entire wall or of the segment of wall considered

I

=

the importan ce factor determined in accord ance with §11.5.1

I

=

moment of inertia of section resisting externally applied factored loads

Ia

=

moment of inertia of cracked section transformed to concrete

Ig

=

(concrete, neglecting reinforcement) moment of inertia of gross concrete section about centroidal axis

t;

=

moment of inertia of reinforcement about centroidal axis of member cross section .

I,

=

moment of inertia of structural steel shape, pipe or tubing about centroidal axis of composite member cross section .

Ig

=

(concrete, neglecting reinforcement) moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement.

t,

=

component importance factor that is either 1.00 or 1.5, as determined in §13.3.1

K

=

(steel) effecti ve length factor for prismatic member

k

=

a distribution exponent - §12.8.3

L

=

live load, except rooflive load, including any permitted live load reduction (i.e, reduced design live load). Live load related internal moments or forces. Concentrated impact loads

Lo

=

unreduced design live load

Lb

=

(steel) unbraced beam length for determining allowable bending stress

Lp

=

limiting laterally unbraced length for full plastic flexural strength, uniform moment case

L,

=

roof live load including any permitted live load reduction

he

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Notation

1

Ie

(steel RBS) length of radius cut in beam flange for red uced beam section (RBS) design

t:

length of a compression member in a frame, measured from center to center of the joints in the frame

/;,

=

distance from column centerl ine to centerline of hinge for reduced bending strength (RBS) connection design

I"

=

clear span measured face-to-face of supports unsupported length of compression member

I"

Ill'

=

Level i

length of entire wall, or of segment of wall considered, in direction of shear force level of the structure referred to by the subscript i. " i = I" designates the first level above the base

Leveln

=

that level that is upperm ost in the main portion of the structure

Level,r

=

that level that is under design consideration. "x = I" designates the first level above the base

=

(steel) maximum factored moment

M

factored moment to be used for design of compression member

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moment at centerline of colum n

Mer

=

moment at which flexural cracking occurs in response to externally applied loads limiting laterally unbraced length for full plastic flexural strength, uniform moment case

"1;-

=

moment at face of column (concrete ) modified moment (steel) maximum moment that can be resisted by the member in the absence of axial load (steel) nominal moment strength at section (concrete) required plastic moment strength of shearhead cross section

2006 IBC Stru ctural/Se ismic Des ign Manual, Vol. J

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Notation

PDL,

10

u,

=

(steel) nominal plastic flexural strength, FyZ

Mpa

=

nominal plastic flexural strength modified by axial load

u;

=

nominal plastic flexural strength using expected yield strength of steel

Mpr

=

(concrete) probable moment strength determined using a tensile strength in the longitudinal bars of at least 1.25;;. and a strength reduction factor cjJ of 1.0

Mpr

=

(steel RBS) probable plastic moment at the reduced beam section (RBS)

M,

=

(concrete) moment due to loads causing appreciable sway

M,

=

torsional moment

M,a

=

accidental torsional moment

M"

=

(concrete) factored moment at section

M"

=

(steel) required flexural strength on a member or joint

M,•.

=

moment corresponding to onset of yielding at the extreme fiber from an elastic stress distribution

MI

=

smaller factored end moment on a compression member, positive if member is bent in single curvature, negative if bent in double curvature

M)

=

larger factored end moment on compression member, always positive

N

=

number of stories

P

=

ponding load

P

=

(steel) factored axial load

P

=

(wind) design wind pressure

i» , r.; =

unfactored axial load in frame member

Pb

=

nominal axial load strength at balanced strain conditions

Pbl

=

connection force for design of column continuity plates

r,

=

(concrete) critical load

2006 IBC Structural/S eismic Design Manual, Vol. I

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Notation

Pc

=

(concrete anchorage) design tensile strength

P"

=

nominal axial load strength at given eccentricity, or nominal axial strength of a column

Po

=

nominal axial load strength at zero eccentricity

P si

=

FyA (concrete) factored axial load, or factored axial load at given eccentricity

p" p"

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=

(steel) nominal axial strength ofa column, or required axial strength on a column or a link

p"

(concrete anchorage) required tensile strength from loads

r,

nominal axial yield strength ofa member, which is equal to F),A g

p.,

total unfactored vertical design load at and above Level x

PE

=

axial load on member due to earthquake

Pu

=

axial live load

QE

=

the effect of horizontal seismic forces

R

=

rain load

R

The response modification factor from Table 12.2-1

R"

nominal strength

Rp

=

required strength

R" R)'

R, Rl R2

component response modification factor that varies from J.00 to 3.50 as set forth in Table J 3.5- J or Table J 3.6-1

=

ratio of expected yield strength F)'c to the minimum specified yield strength Fy

=

live load reduction in percent -

!Be §§ 1607.9.2/1607. J 1.2

r

rate of reduction equal to 0.08 percent for floors

r

(steel) radius of gyration of cross section of a compression member

ry

=

radius of gyration about y axis 2006 IBC Structural/Seismic Design Manual, Vol. I

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No tation

12

S

=

snow load

So

= = =

design spectral response acceleration 0.6 (SosITo) T + 0.4 (Sos), for T less than or equal to To (SOl ) 1 T, for T greater than T,

Sos

=

5% damped, design, spectral response acceleration parameter at short period (i.e., 0.2 seconds) = (2/3) S,«.. - § 11.4.4

S,

=

Mapped, MCE, 5% damp ed, spectral acceleration parameter at short periods (i.e., 0.2 seconds) as determin ed by § 11.4.1

SOl

=

5% damped , design, spectral response acceleration parameter at l -second perio d = (2/3) S sn

SI

=

Mapped, MCE, 5% damped, spectral acceleration parameter for a l-s econd period as determined in § 11.4.1

s'IIS

=

MCE, 5% damped, spectral response acce leration parameter for short periods (i.e., 0.2 seconds) = FoS,. adj usted for site class effects

S,I/I

=

MCE, 5% damped , spectral response acceleration parameter for l-second peri od = F•. SJ, adjusted for site class effects

SRBS

=

sectio n modu lus at the reduced beam section (RBS)

S

=

spacing of shear or torsion reinforcement in direction parallel to longitudinal reinforcemen t, or spacing of transverse reinforcement measured along the longitud inal axis

T

=

self-straining force arising from contracti on or expansion resultin g from temperature change, shrinkage, moisture change, creep in comp onent materials, movement due to differential settlement or combinations thereof

T

=

elastic fundamental period of vibration, in seconds, of the structure in the direction under consideration, see § 11.4.5 for limitations

To

=

approximate fundamental period as determined in accordance with § 12.8.2.1

To

=

0.2 (SOl 1Sos)

T,

=

SOl 1Sos

If

=

thickness of flange

tw

=

thickness of web

2006 IBC Structural/Seismic D esign Manual, Vol. I

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Notation

ratio of expected yield strength F ,.

D

Beam A-B and Column C-D are elements of the special moment-resisting fram e. Structural ana lysis has provi ded the follow ing beam moments at A, and the column axial loads and moments at C due to dead load, office building live load, and left- to-right ( ~) and right-to-left (-) directio ns oflateral seismic loading. Dead Load D

Live Load L

Left-to-Right Seism ic Load

Right-to-Left Seismic Load

(--+QI;;)

Beam Mome nt at A Column C-D Axial Load Column Moment at C

- 100 kip -ft +90 kips +40 kip-ft

-50 kip-ft +40 kips +20 kip-ft

+120 kip- ft +110 kips +160 kip-ft

(-- QI;;)

-120 kip-ft - 110 kips -) 60 kip-It

Sign Convention: Positive moment induces flexural tension on the bottom side of a beam and at the right side of a column. Positi ve axial load induces compression . Note that for the pa rticular location of Column C-D, the seism ic Axial Load and Moment at C are both positive fo r the left-to-right ( ~) load ing and are both negative for the right-to -Ieft (-) loading. This is not necessarily true for the other elements of the structure. Find the following.

ILJ [!J [!J

Strength design seismic load combinations (Comb .) Strength design moments at beam end A for seismic load combinations Strength design interaction pairs of axial load and moment for the design of column section at C for seismic load combinations 2006 IBC Structural/Seismic Design Manual, Vol . I

27

§12.4.2.3

Example 1 • Earthquake Load Combinations: Strength Design

I [LJ

Governing strength design seismic load combinations

1

1.2D + I.OE + 0.5L ... (Note 0.2S =0)

(Comb . 5)

0.9D + 1.0E

(Comb. 7)

where for a given type of load action such as moment M or axial load P

E=E,,+E,.

(Eq 12.4-1)

E; = PQE E,. =0.2S DSD

(Eq 12.4-3) (Eq 12.4-4)

Combined, these yield

E

=PQE+ 0.2SDSD

(Eq 12.4-3)

when the algebraic sign, ±, of QE is taken as the same as that for D, and

E

=PQE -

0.2SDSD

when the algebraic sign, ±, of QE is taken as opposite to that for D. For the given values of: p

= 1.3, SDS= 1.10, the load combinations are

1.2D + 1.3QE+ (0.2)( I.I)D + 0.5L

= 1.42D + 1.3QE+ 0.5L

(Comb. 5)

when the signs of QEand D are the same, and

1.2D + 1.3QE - (0.2)(1.1)D + 0.5L

=0.98D + 1.3QE+ 0.5L

(Comb. 5)

when the signs of QEand D are opposite .

0.9D + 1.3QE + (0.2)(1.1)D

= 1.I2D + 1.3QE

when the signs of QE and D are the same, and

28

2006 IBC Structural/Seismic Design Manual, Vol. I

(Comb. 7)

1 I I I I I I I I I I I I I

I I

I 1

Example 1 •

Earthqua ke Lo ad Combinations: Stren g th Design

0.9 D + 1.3QE- (0 .2)( 1.l)D =0. 68D + 1.3Q£

§1 2. 4.2.3

(Comb . 7)

when the signs of Q£ and D are oppos ite.

]

By inspection, the governing seismic load combinations are

when the signs of Q£ and D are the same, 0.68D + 1.3QE

when the signs of QE and D are opposite.

Streng th design moments at beam end A for seismic load combinations

~ For the governin g load combin ation when the signs of Q£ and D are the same 1.42D + 1.3QE+ 0.5L

I

I I I I I I I

MA

= 1.42 (- 100) + 1.3 (-120) + 0.5(-50) =- 323 kip-ft

~ For the governing load combination when the signs of Q£ and D are opposite 0.68 D + 1.3QE

with

D = M D = - 100 and QE = 120 MA = 0.68(-100) + 1.3(120) = 88 kip-ft

: . Beam section at A must be designed for M A = - 323 kip-ft and + 88 kip-ft

2006 IBC Structural/Se ism ic Design Manual, Vol. I

29

§12.4.2.3

[!J

Example 1 •

Earthquake Load Combinations: Strength Design

Strength design interaction pairs of axial load and moment for the design of column section at C for seismic load combinations

The seismic load combinations using the definitions of E given by Equations 12.4-1 through 12.4-4 can be used for the design requirement of a single action such as the moment at beam end A, but they cannot be used for interactive pairs of actions such as the axial load and moment at the column section C. These pairs must occur simultaneously because of a common load combination. For example , both the axial load and the moment must be due to a common direction of the lateral seismic loading and a common sense of the vertical seismic acceleration effect represented by 0.2 SDsD. There can be cases where the axial load algebraic signs are the same for QE and D, while the moment algebraic signs are different. This condition would prohibit the use of the same load combination for both axial load and moment. To include the algebraic signs of the individual actions, the directional property of the lateral seismic load effect QE, and the independent reversible property of the vertical seismic load effect 0.2 SDsD, it is proposed to use

E = p(-->QE) ± 0.2 SDSD, and p( -Qd ± 0.2 SDsD. The resulting set of combinations is

1.2D + p( -->QE) - 0.2 SDsD + L

0.9D + p( -->Qd - 0.2 SDsD 0.9D + p( -Qd + 0.2 SDsD 0.9D + p(-Qd - 0.2 SDsD (Note : a factor of 0.5 applies to L if L :0: 100 psf [except at garages and public assembIy areas]) For the specific values of p = 1.3 and SDS = l.l 0, the load combinations provide the following values for MA , and the interaction pair Pc and Me. Note that the interaction pair Pc and Me must occur simultaneously at a specific load combination of gravity load, and lateral and vertical seismic load effects. The interaction design of the column section must satisfy all of the eight pairs of P e and Me from the seismic load 30

2006 IBC Structural/Seismic Design Manual, Vol. I

I I I I I I I I I I I I I I I I

I I

Example 1 a Earthq uake L oad Co m b i n ations : Strength Des i g n

§1 2.4 .2 .3

combinations along with the pairs from the gravity load combinations and wind load combinati ons.

Combination 1.420 + 1.3 (Q i;) + O.5L O.98D + 1.3 (Q E) + O.5L 1.42D - 1.3 (Qd + 0.5L 0.98D - 1.3 (Qd + 0.5L 1.12D + 1.3 (Qd 0.68D + 1.3 (Q E) 1. 12D - 1.3 (Qd 0.68D - 1.3 (QE)

MA kip-ft

Pc kips

and

Me kip-ft

-35 -9 -299 -255 + 20 +64 -244 -200

+268.8 +26.8 +229.2 - 12,8 +22 1.8 + 182.2 -2 0.2 -59 .8

and and and and · and and and and

+242. 8 +225.2 - 109.2 -126. 8 +220 .8 +203 .2 -131.2 - 148.8

The govern ing va lues are und erlined fo r MA [sam e as det erm ined in Part (2) ] and for the interaction pairs of Pc and Me required for the design of the column section at C.

... The eight seismic load combinations resulting from the proposed definition of E pro vid e an automatic method of considering the individual algebraic signs of the load actions, the direction of the lateral seismic load, and the independent ± action of 0.2 SDCD. There is no nee d to use the "same sign" and "opposite sign" limitations of Equations 12 .4-2 and 12.4-3 since all possible com binations are represented . Thi s is imp ortant for interactive pa irs of actions that must be evaluated for a common load combination. When the Modal Respo nse Spectrum Analysis' procedure of § 12.9 is used, the algebraic signs of seismic load actions are lost because of the process of combining the individual modal responses. The signs to be used for an interaction pair of actions due to a given direction of lateral loading can be obtained from the primary mode response where the prim ary mode is the mode having the largest participation fac tor for the given direction of lateral seism ic loading. Or, alternatively, the signs can be obtained from the equivalent lateral force procedure of § 12.8.

2006 IBC Structural/Seismic D esign Manual, Vol. I

31

§2.4

Example 2 •

Combinations o f Loads

-:

The code permits the use of allowable stress design for the design of wood members and their fastenings (ASCE/SEI 7-05 §2.4 and § 12.4.2.3). Section 2.4 defines the basic load combinations for allowable stress design. This example illustrates the application ofthis method for the plywood shear wall shown below. The wall is a bearing and shear wall in a light wood framed build ing. Gravity loads

The following information is given. Seismic Design Catego ry B

ITnTTTTm Plywood shear wall

J = 1.0 P = 1.0

5 DS = 0.3 E = Ell = bQ£ = 4 kips (seismic force due to the base shear determined from § 12.4.2) Grav ity loads Dead lVD = 0.3 kif (tributary dead load, including weight of wall Live lIIL = (roof load supported by other elements)

Shear Wall Elevati on

Moment arm from center of post to center of hold-down bo lt L = 10 ft - (3.5 + 2.0 +3.5/2) = 10 ft -7.25 in = 9.4 ft Determine the required design loads for shear capacity q and hold-down capacity T for the following load combinations.

[!J

Basic allowable stress design

32

2006 IBC Structural/Seismic Design Manual, Vol. I

1 I

Example 2 • Combinations o f Loads

r-

[IJ

§2.4

..

.

,.

.,.,.~

Basic allowable stress design

§12.4.2.3

The governing load combinations for basic allowable stress design are Basic ASD Combinations 5, 6, and 8, as modified in § 12.4.2.3. These are used without the usual onethird stress increase.

§ 12.4.2 defines the seismic load effect E for use in load combinations as (Eq 12.4-1) (Eq 12.4-3)

I

1

= QE+ 0.06D when D and QE are in the same sense

and

(Eq 12.4-4)

E = PQE - 0.2SosD

=QE- 0.06D

when D and QE have oppos ite sense

For ASD Basic Combination 5 the load comb ination is:

I I I I

D +0.7E

(Comb . 5)

= D(I .O) + 0.7 (0.6D + QE) =

and

( 1.042)D + 0.7QE for D and QE with the same sense

D(1.0)+0 .7(-0.6D-Qd =

0.958D - O.7QE for D and QE with opposite sense

For ASD Basic Combination 6 the load combination is:

\

D + 0.75(0.7E) + 0.75 (L + Lr)

I

=

(Comb. 6)

D(1.0 + (0.75)(0 .7)(0.06)) + (0.75)(0.70)(1.0)QE+ 0.75 L,.

= J.032D + 0.75L,. + 0.525 QE for D and QEwith the same sense

I

I I

=

D(J.O- 0.968) + 0.75 L,. - 0.525 QE for D and QE with the opposite sense

For ASD Basic Combination 8 the load combination is:

0.6D + 0.7E

(Comb. 8) 2006 IBC Struct ural/S eismic Design Manual, Vo l. I

33

§2.4

Example 2 •

Combinations of Loads

=D(0.06) + 0.7(1.0) QE+ 0.7(0.06)D =

(0.6 + 0.042)D + 0.7QE

= 0.642D + O. TQE for D and QE in the same sense =

(0.6 - 0.042)D - 0.7QE

= 0.558D - O. TQE for D and QE in the opposite sense

For the determination of design shear capacity, dead load and live load are not involved, and all load combinations reduce to

For the design hold-down tension capacity the governing load combination is

0.558D - 0.7QE For the wall boundary element compression capacity, the governing load combination would be

1.042D + 0.7QE

~ Required unit shear capacity q Base shear and the resulting element seismic forces QE determined under §12.8.1 are on a strength design basis. For allowable stress design, QE must be factored by 0.7 as indicated. For design shear capacity the seismic load effect is

QE

=

4000 Ib

For the governing load combination ofO.7QE, the design unit shear is

q

= 0.7QE L

=

0.7(4000) = 280 If 10ft P

This unit shear is used to determine the plywood thickness and nailing requirements from lBe Table 2306.4.1, which gives allowable shear values for short-time duration loads due to wind or earthquake. For example, select 15/32 structural I sheeting (plywood) with 10d common nails having a minimum penetration of 1-1/2 inches

34

2006 IBC Structural/Seismic Des;gn ·Manual, Vol. I

1

I I

Example 2 • Comb/nations of Loads

§ 2.4

into 2x members with 6-inch spacing of fasteners at panel edges; allowable shear of 340 plf.

Required hold-down tensile capacity T Taking moments about point 0 at center of post at right side of wall with = oQ£ = 4000 Ib, the value of the hold-down tension force T due to horizontal seismic forces is computed

E"

0.558(300 pit) I0 ft(5 ft -

~ ) - o. 7(4000 Ib)(9 ft) + T(9.4 ft) = 2(12)

0

Thus : 8125.88 Ib ft - 25,200 Ib ft + 9.4 ft(T) = 0 T = 1816.39 Ib tension

Similarly the boundary element comp ression capacity is computed 1.042(300 pit) lOft (5 ft -

~ ) + 0.7 (4000 Ib)(9 ft) 2(12)

C(9.4 ft) = 0

Thu s:

I

15,1741b ft + 25,200 Ib ft-9.4 ft C=O

c = 4295 Ib compression

1

I I I I I I

The tension value is used for the selection of the pre-manufactured hold-down elements. Manufacturer's catalogs commonly list hold-down sizes with their "1.33 x allowable" capacity values. Here the 1.33 value represents the allowed Load Duration factor for resisting seismic loads. This is not considered a stress increase (although it has the same effect). Therefore, the catalog "1.33 x allowable" capacity values may be used to select the appropriate hold-down element.

Equations 12.4-1 and 12.4-2 for E create algebraic sign problems in the load combinations. It would be preferable to use

E = pQ£ + 0.2 SDsD and use ± E in the load combinations. 2006 IBC Structural/Seismic Design Manual, Vol. I

I

35

§11.4

Des ign Spectral Response Acc el eraOons

For a given building site, the maximum conside red earthquake spectral response accelerations S, at short periods, and S) at I-second period are given by the acceleration contour maps in §22. This example illustrates the general procedure for determin ing the design spectral response parameters Sos and SDl from the mapped values of Ss and 8). The parameters Sos and So, are used to calculate the design base shear in §12.8 and the Design Response Spectrum in § 11.4.5. Note that by far the most accurate, easiest, and most effic ient way to obtain the spectral design values is to use the USGS website iwww.eqhazmaps.usgs.govr. Given the longitude and latitude of the site, the website provides va lues of Ss and S). The site longitude and latitude can be obtained from an internet site such as u\I~"H'. geocode.com " by simply inputting the address. From u\I'lI'\I'. geocode.com " it is determ ined that a building site near Sacramento, California is located at Latitud e 38. 123° North and Longitude - 121.123 (or 121.123 west). The soil profile is Site Class D. D

D

Determin e the following.

[!.J

!TI [!J

Maximum considered earthquake spectral response accelerations and Sl

Site coefficients and adjusted maximum considered earthquake spectral response acceleration parameters SMS and SMl Design spectral response acceleration parameters Sos and

~ Plot the general procedure response spectrum ~ Calculation of seismic response coefficient c, Given: so il site class D, R = 6, T = 0.60 sec, and I = 1.0

36

s,

2006 IBC Structural/Seismic D esign Manual, Vol. I

I

I I

I I

I I I

I I

SOl

I I I

Design Spec tral Res p onse Accelerations

. ""

,..

[!J

§1 1.4

Code.8eference· "

Maximum considered earthquake spectral response accelerations

§11.4.1

For the given position (Near Sonora - NW of Sacramento, California) of 38° North (Latit ude = 38.123°) and 121.123° West (Longitu de =- 121.123'), USGS provides the values of

5s = 46.2%g = 0.462g

5,

=

20.3%g = O.203g

~ Site coefficients and adjusted maximum considered earthquake spectral §1 1.4.3

response accelerations

From the USGS for the given site class D, and Ss = 0.462g, 5\ = O.203g, the site coefficients are as follows

I I I I I I

Fa = 1.58

TII.4-1

= 1.99

T 11.4-2

F,.

The adjusted maximum conside red earthqu ake spectral response accelerations (based §11.4.3) are also given on the CD ROM as follows

011

SMS =FaS, = 1.58(0.462g) =O.730g

(EqI1.4-1)

S'/I =F,S,

(Eq 11.4-2)

= 1.99(O.203g) =0.404g

I 2006 lac Stru ctural/S eismic De sign Manu al, Vol. I

37

§ 1 1. 4

D es ign Sp ectral Respon se Acc el erations

~ Design spectral response acceleration parameters S DS

=-2

3

SMS

=-2 (0.73g) =0.49g 3

(Eq 11.4-3)

1

?

'J

SOl

§1l.4.4

= ~ S.I{I = ~ (0.404g) =0.27g

(Eq 11.4-4)

General procedure response spectrum

§1l .4.5

For periods less than or equal to To, the design spectra l response shall be given by

So

=

SDS

0.6-

r:

T + 0.4 Sos

I 1

(Eq 11 .4-5)

For periods greater than or equal to To and less than or equal to T" the design spectral response acceleration So shall be taken equal to Sos For periods greater than T:" and less than TL , the design spectral response acceleration Sa shall be given by

Sa = (SOI) / T

I I I

(Eq 11.4-6)

I I I I

Where : To = 0.20 (SOl / Sos)

T,

Tt.

38

=

0.2 (0.27 /0.49)

=

O. I I sec

=

SOl / 50s

=

0.27 / 0.49

=

0.55 sec

=

8 sec

2006 IBC Structural/Seismic Design Manu al, Vol. I

(F 22- 15)

I I I I I I I

Des ign Spectral R esponse A cc ele ra ti ons

§11 A

Thus: T

=Period

ScJg

0.00 0.11 0.55 0.80 1.00 1.20 1.40 1.60 2.00

Computation for Sa 0.4 (0.49) 0.49 0.27 /0.55 0.27 / 0.8 0.27 / 1.000.27 / 1.2 0.27 / 1.4 0.27 / 1.6 0.27 /2.0

0.18 0.49 0.49 0.34 0.27 0.23 0.19 0.17 0.135

S. in g 's 0.5 S DS

= 0.4 9g 0.4

0.3

0.2 0.18

... .....

0. 1

o

o

0.2

To =0.11 sec

0.4

0.6

0.8

1.0

1.2

1.4

... ...

...

-... - ... -

1.6

1.8

2.0

T, =0.55 sec

General Procedure Response Spectrum

2006 IBC Structural/Seismic Design Manual, Vol. I

39

§11.4

1

Design Spectral Response Accelerations

I Calculation of seismic response coefficient Cs (Recall Soil Site Class D, f = 1.0 and T= 0.60) §12.8.1 The seismic response coefficient shall be determined by C,

=

SDS I (RIl)

=

0.49 I (6.0/1.0)

=

0.08 2 ... Governs

(Eq 12.8-2)

The value of C, need not exceed C,.

=

SDI I (RIlE) T

(Eq 12.8-3)

= 0.27 I (6.0/1.0) (0 .6) = 0.D75

But shall not be taken less than C,

=

0.01

where SI 2: 0.6g C, shall not be less than C,

40

= 0.5S 1 I (RIl)

2006 IBC Structural/Seismic Design Manual, Vol. I

(Eq 12.8-5)

1 I I

I I I I I I I I I I I I I I

Intro duction to Vertica l Irregul arities

§12.3.2.2

-,':

'.

1

§12.3.2.2

Tab le 12.3-2 defines vertical structural irregularities and assigns analysis and design procedures to each type and seismic design category. These irregularities can be divided into two categories. The first, dyn amic force-distribution irreg ularities, which are Types Ia, Ib, 2, and 3. The second, irregularities in load path or force tran sfer, which are Types 4 and 5. The vertical irregularities are Ia. Stiffness Soft Story Irregu larity Ib. Stiffness Extreme Soft Story Irregularity 2. Weight (mass) irregularity 3. Vertical geometric irregu larity 4. In-plane discontinuity in vertical latera l-force-resisting element Sa. Discontinu ity in Lateral Stength - Weak Story Irregularity 5b. Discontinui ty in Lateral Strength - Extreme Weak Story Irregularity Structures in Seismic Design Categories D, E, and F possessing dynamic force distribut ion irregularities shall be analyzed using the dynamic analysis procedure (or moda l analysis procedure) prescribed in §12.7. (Refer to Tab le 12.6.1) Structure Description 3. The vert ical force distribution provided by § 12.8.3 may be assumed to be adequate for structures lacking vertical irregulari ty Types Ia, Ib, 2, and 3. However, stiffness and mass discontinuities may significantly affect the vertical distribution of forces and, for this reason the modal analysis procedure, which can account for these discontinuities, is necessary. Although designers may opt to use the dynamic analysis procedure and bypass checks for irregularity Types Ia, Ib, 2, and 3, the reference sections listed in Tab le 12.3-2 should still be checked for limitations and design requirements. Note that § 12.3.3.1 prohibits structures with vertical irregularity Types Ib, Sa, or 5b for Seismic Design Categories E and F. Regular structures are assumed to have a reasonably uniform distribution of inelastic behavior in elemen ts throughout the lateral- force-resisting system. When vertical irregu larity Types 4 and 5 exist. there is the possibility of having localized concentrat ions of excessive inelastic deformations due to the irregular load path or weak story. In this case, the code prescribes addi tiona l strengthening to correct the deficienc ies for structures in cert ain seismic design categories (SDCs) . In the case of vertical irregularity Type 5b, limits are placed on the building height for all SDCs except S DC A.

2006

lac

Structural/S eismic D esign Manu al, Vol. I

41

Ex ample 4 •

§12.3.2.2

~

1

Vertic al1rregula rlty Typ e 1

,

&le4 .Vertical Irr egularity Type 1a a nd Type 1b

§ 12.3.2 .2

A Seismic Design Category D five-story concrete special moment-resisting frame is shown below. The code-prescribed lateral forces F, from Equation 12.8- 11 have been applied and the corresponding floor level displacem ents O.re at the floors' centers-of-mass have been determined as shown below. Ft + Fs 10'

J

~ [';'==::;-1';::=::::::;11-

F, - . .

i:

.• DD /· DD ./ , DD .I,'

F, - . .

0 " . 2.02

I I

/ :

Triangular shape

10'

10'

- - - - --;,.

.'"

;

0', _1.75

I

!

"' :

0;\... 1.45

F

, -..

!

.'

/

!

02" , .0B

' 0'

F

'-..

' 2'

!",

~r-r---r7Y-.,-"'77',*:"

[!J

"

Actual shape

Determine if a Type 1a vertical irregularity from Table 12.3-2 (StiffnessSoft Story Irregularity) exists in the first story

Calculations and Discussion

[L]

i

/.' 0It_ 0.71

Code Reference

To determine if this is a Type 1a vertical irregularity (Stiffness-Soft Story

Irregularity) there are two tests I. The lateral story stiffness is less than 70 percent of that ofthe story above.

2. The lateral story stiffness is less than 80 percent of the average stiffness of the three stories above.

42

2006 IBC Structural/Seismic Design Manual, Vol. I

Example 4 • Venicallrregularity Type 1

1

I

I

§12. 3.2.2

If the stiffness of the story meets at least one of the two criteria above, the structure is deemed to have a soft story, and a modal analysis (§12.9) is generally required by Table 12.6- I. The definition of soft story in the code compares values of the lateral stiffness of individual stories. Generally , it is not practical to use stiffness properties unless these can be easily determ ined. There are many structural configurations where the evaluat ion of story stiffness is complex and is often not an available output from computer programs. Recogni zing that the basic intent of this irregularity check is to determ ine if the lateral-force distribution will differ significantly from the pattern prescribed by §12.8.3, which assumes a prescribed shape for the first dynamic mode of response, this type of irregularity can also be determined by comparing values of drift ratios due to the prescr ibed lateral forces. This deformat ion comparison may even be more effective than the stiffness comparison because the shape of the first mode shape is often closely approximated by the structure displacements due to the specified §12.8.3 force pattern . Floor level displacements and corresponding story-drift ratios are directly available from computer programs. To compare displacements rather than stiffness, it is necessary to use the reciprocal of the limitin g percentage ratios of 70 and 80 percent as they apply to story stiffness, or reverse their applicability to the story or stories above. The following example shows this equivalent use of the displacement propert ies.

I

From the given displacements , story drifts and the story-drift rat io's values are determi ned. The story-drift ratio is the story drift divided by the story height. These story-drift ra tios will be used for the required comparisons because they better represent the changes in the slope of the mode shape when there are significant differences in interstory heights. (Note: story displacements can be used if the story heights are nearly equal.)

I

In terms of the calculated story-drift ratios, the soft story occurs when one of the following cond itions exists.

I I I I I I

I

s

When 70 percent of ---"- exceeds h, or

s,. - Ii,. h,

When 80 percent of -Ii,' exceeds -I [(0,.• -Ii,.) , + (0,• - Ii,_. ) + (0, • -Ii,)] .• h, 3 h, h, h, the story-drift ratios arc determi ned as

~= o ..

=

h,

h,

t.,

0,. - 0"

h,

h,

- '=

(0.7 1- 0) 144

= 0.00493

= (1.08 - 0.71) = 0.00308 120

~ = Ii,. - 0,. = (1.45 - 1.08) = 0.00308 h,

h,

120

2006 IBC Struc tural/Seismic Design Manual. Vol. I

43

§12. 3.2.2

Ex ample 4 •

_t:J._, =

Iz,

Vertical Irreg ularity Type 1

Ii" - Ii"

s,

=

1

(1.75-1.45) = 0.00250 120

~(0.00308 + 0.00308 + 0.00250) = 3

0.00289

Checking the 70-percent requirement: O.70(1i,,) hi

= 0.70(0 .00493) = 0.00345 > 0.00308 ... NG

: . Soft story exists. . . Note that 70 percent of first story drift is larger than second story drift. Alternately: 0.00493 > (0.00308 x 1.30 = 0.0040) . . . thus soft story. Also note that structural irregularities of Types Ia, Ib, or 2 in Table 12.3-2 do not apply where no story-drift ratio under design lateral force is greater than 130 percent of the story-drift ratio of the next story above, §12.3.2.2, Exception 1. Checking the 80-percent requirement:

0.80(~) = 0.80(0.00493) = 0.00394 > 0.00289 ... NG h, .', Soft story exists. . . condition Ia Alternately: 0.00493 > (0.00289 x 1.20 = 0.00347) . .. thus soft story. Check for extreme soft story, (Vertical Structural Irregularity, Type l b) Checking the 60-percent requirement: 0.60(0.00493) = 0.002958 < 0.00308 . . . o. k. Alternately: 0.00493 > (0.00308 x 1.4 = 0.004312) .. . o.k. Checking the 70-percent requirement: 0.70 (0.00493) = 0.003451 > 0.00289 . .. NG Al ternately: 0.00493 > (0.00289 x 1.3 = 0.00375) . .. NG Thus: Stiffness-Extreme Soft Story exists - condition lb.

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2006 IBC S tructural/Seism ic Design Manual, Vol. I

I I I I I I I I I

Examp le 4

tI

Vertical Irr egularity Type 1

§12.3.2.2

Recall from Table 12.3-2 for Ib , extreme soft story, reference §12.3.3. 1. This building is SOC 0 , and is permitted, Structures having SDC E or F and also having vertical irregularity Type Ib shall not be permitted.

Commentary

Section 12.8.6 requires that story drifts be computed using the maximum inelastic response displac ements b.r , which include the deflection amplification factor Cd

s = Cdb.

rr

x

(Eq 12.8-15)

I

However, for the purpose of the story drift, or story-drift ratio, comparisons needed for softstory determination, the displacement bxe due to the design seismic forces can be used as in this example. In the exampl e above, only the first story was checked for possible soft-story vertical irregularity . In practice, all stories must be checked, unless a modal analysis is performed. It is often convenient to create tables to facilitate this exerc ise, see Tables 4.1 and 4.2. Table 4. I Soft-Story Status ln

Leve l

Sto ry Displacement

Story Drift

Story-drift Rat io

0 .8x (S torydrift Ratio )

0 .7x (S tory drift Ratio)

Avg . of S tory- drift Ratio of Next 3 Stories

Soft Story Status la

5

2.02 in

0.27 in

0.00225

0.00180

0.00 158

No

4

1.75

0 .30

0.00250

0.00200

0.00175

No

3 2

1.45

0.37

0.00 246

0.00216

1.08

0. 37 0.71

0 .00308 0 .0030 8

0.0024 6

0.00 21 6

0.0026 1

No No

0.00493

0.00 394

0.00345

0 .00289

Yes

Avg, of Story-drift Ratio of Next 3 Stories

Soft Story Status lb

0.71

T able 4.2 Soft-Story Status Ib

Level

S tory Disp lacement

S tory Drift

Story-drift ratio

0 .7x (S torydrift Ratio)

0 .6x (Storydr ift Ratio)

5

2.02 in

0.27 in

0 .00 225

0.001 58

0 .00 135

No

4

1.75

0 .30

0 .00250

0.00175

0.00150

No

3

1.45

0 .37

0 .00 30 8

0.002 16

0.001 85

2

1.08

0.37

0.00308

0.71

0 .71

0.00493

0.002 16 0.00345

0.00185 0.00296

No 0 .0026 1

No

0.00289

Ye s

2006 IBC Structural/Seismic Design Manual, Vol. I

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§12.3.2.2

Example 5 •

Vertical Irregularity Type 2

...' a!nple 5 ;V~rtical lrregularity Type 2

§12.3~2.2

The five-story special moment frame office building has a heavy utility equipment installation at Level 2. This results in the floor weight distribution shown below. Ws = 90 k

W,= 110k

W, = 110 k

W,= 100k

[!J .

Determine if there is a Type 2 vertical weight (mass) irregularity .

,

:Calci!li!tions and Discussion

Code Reference

A weight, or mass, vertical irregularity is considered to exist when the effective mass of any story is more than 150 percent of the effective mass of an adjacent story. However, this requirement does not apply to the roof if the roof is lighter than the floor below. Note that it does apply if the roof is heavier than the floor below. Checking the effective mass of Level 2 against the effective mass of Levels 1 and 3 At Levell 1.5 X WI

= 1.5(100 kips) = 150 kips

At Level 3 1.5

X

W3 = 1.5(110 kips) = 165 kips

Wz = 170 kips > 150 kips .. Weight irregularity exists. 46

2006 IBC Structural/Seismic Design Manual, Vol. I

1 1

1

Example 5 • Vertical Irreg ularity Type 2

I

§ 12.3 .2 .2

, • ..,

.Commentary As in the case of vertical irregularity Type la or Ib, this Type 2 irregularity also results in a primary mode shape that can be substantially different from the triangular shape and lateral load distribution given by § 12.8.3. Consequently, the appropriate load distribution must be determined by the modal analysis procedure of § 12.9, unless the irregular structure is not more than two stories and is Occupancy Category l or II (see Table 12.6-1).

1

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§12.3.2 .2

Ex ample 6 •

Vertical Irregularity Type 3

~

ample 6 Vertic al Irregularity Typ e 3

§ 12.3.2.2

The lateral-foree-resisting system of the five-story specia l moment frame building shown below has a 25-foot setback at the third, fourth, and fifth stories.

4@2S' -100'

I.....

.....

Level

5

4

3

2

'/

[!J

.... ...-

DDD DDD DDD D.DDD "/

//"/

//

/

/ / , /

//, /

Determine if a Type 3 vertical irregularity (vertical geometric irregularity) exists

CalcuJ~tiC?ns and Discussion

Code Reference

A vertical geometric irregularity is considered to exist where the horizontal dimension of the lateral-foree-resisting system in any story is more than 130 percent of that in the adjacent story. One-story penthou ses are not subject to this requirement. In this example, the setback of Level 3 must be checked. The ratios of the two levels are Width of Level 2 = (lOa ft) Width of Le vel 3 (75 ft)

= 1.33

133 percent > 130 percent .'. Vertical geom etric irregulari ty exists.

48

2006 IBC Structural/Seismic Des ign Manual, Vol . I

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Example 6 • Vertical Irregularity Typ e 3

§12.3.2 .2

,. Commentary The more than l3 0-percent change in width of the lateral-force-resisting system between adjacent stories could result in a primary mode shape that is substantially different from the shape assumed for proper applications of Equation 12.8- 11. If the change is a decrease in width of the upper adjacent story (the usual situation), the mode shape difference can be mitigated by designing for an increased stiffness in the story with a reduced width.

)

Similarl y, if the width decrease is in the lower adjacent story (the unusu al situation), the Type la soft-story irregularity can be avoided by a proportional increase in the stiffness of the lower story. However, when the width decrease is in the lower story, there could be an overturning moment-load-transfer discontinuity that would require a dynamic analysis per Table 12.6-1. Note that if the frame elements in the bay between lines 4 and 5 were not included as part of the designated lateral-force-resisting system, the vertical geometric irregularity would not exist.

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§12. 3.2.2

Ex amp le 7

11

Vertic al Irre g u lar i ty Type 4

,Ex, mple 7 § 12.3.2.2

Vertical Irregularity Typ e 4

A concrete building has the building frame system shown below. The shear wall between lines A and B has an in-plane offset from the shear wall between lines C and D.

rrrT 3@25'=75'

Level

5 12'

4 12'

IE< ----~

'DO' e:

3 25'

12'

2

--- 0 00

L--

Shear wa ll

50'

12'

Shear wall

1 12'

// '/

[!J

/ ,

'

/

//

'/

/ /

"/

Determine if there is a Type 4 vertical irregularity (in-plane discontinuity) in the verticallateral-force-resisting element

Calculations and Discussion

Code Reference

A Type 4 vertical irregularity exists when there is an in-plane offset of the lateral-forceresisting elements greater than the length of those elements . In this examp le, the left side of the upper shear wall (between lines A and B) is offset 50 feet from the left side of the lower shea r wall (between lines C and D). This 50-foot offset is greater than the 25-foot length of the offset wall clements . : . In-plane discontinuity exists .

50

2006 IBC Structural/Seismic Design Manual, Vol . I

Examp le 7 • Vertical Irregularity Type 4

s:

§12.3.2.2

'"

C;ommentary The intent of this irregularity check is to provide correction offorce transfer or loadpath deficiencies. It should be noted that any in-plane offset, even those less than or equal to the length or bay width of the resisting element, can result in an overturning momentload-transfer discontinuity that requires the application of §12.3.3.3. When the offset exceeds the length of the resisting element, there is also a shear transfer discontinuity that requires application of § 12.3.3.4 for the strength of collector elements along the offset. In this example, the columns under wall A-B are subject to the prov isions of § 12.3.3.3, and the collector element between lines Band C at Level 2 is subject to the provisions of § 12.3.3.4.

I I 1

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§12.3.2.2

Exa mple 8 •

Ve rtIc al Irregu l ar ity Type 5a

IE. ampleB Verfic a l l r r eg ularity Type 5a

§12.3.2,,2

A concrete bearin g-wall building has the typical transve rse shear-wall configuration sho wn be low. All walls in this direction are identical, and the individu al piers have the shear contribution given below. Then , V, is the nominal shear strength calcu lated in accordance with Chapter 19, and Vm is defined herein as the shear corresponding to the development of the "nominal flexure strength also calculated in accordance with Chapter 19." Note that VII/ is not defined in ACI or Chapter 19. Level

J

PIER

\In

V",

1

20 kips 30 15 80 15

30 kips 40 10 120 10

2 3 4 5

[!J

Determine if a Type 5 vertical irregularity (discontinuity in capacity- weakstory) condition exists

Calculations and Discussion

Code Reference

A Type Sa weak-story discontinuity in capacity exists when the story strength is less than 80 percent of that in the story above. The story strength is the total strength of all seismic-force-resisting elements shari ng the story shear for the direction under consideration. Using the smaller values of VII and VII/ given for each pier, the story strengths are First story strength = 20 + 30 + 10 = 60 kips Second story strength

= 80 + 10 = 90 kips

Check if first-story strength is less than 80 percent of that of the second story. 60 kips < 0.8(90) = 72 kips :. Weak story condition exists.

52

2006 IBC Structural/Seismic Design Manual, Vol. I

Example 8 • VerlicallrregularJty Type Sa

§12.3 .2.2

Check if first-story strength is less than 65 percent of that of the second story (Irregularity Type 5b). 60 kips < 0.65(90 kips) = 58.5 kips :. 60 kips > 58.5 kips . . Therefore the lower story is not an extreme soft story, Irregularity Type 5b.

Commentary

This irregularity check is to detect any concentration of inelastic behavior in one supporting story that can lead to the loss of vertical load capacity. Elements subject to this check are the shear-wall piers (where the shear contribution is the lower of either the shear at development of the flexural strength , or the shear strength), bracing members and their connections, and frame columns. Frame columns with weak column-strong beam condit ions have a shear contribution equal to that developed when the top and bottom of the column are at flexural capacity. Where there is a strong column-weak beam condition, the column shear resistance contribution should be the shear corresponding to the development of the adjoining beam yield hinges and the column base connection capacity. In any case, the column shear contribution shall not exceed the column shear capacity. An extreme weak story is prohibited (under §12.3.3.1) for structures more than two stories or 30 feet in height if the "weak story" has a calculated strength ofless than 80 percent of the story above . A weak-story condition is absolutely prohibited in SDC E and F.

I

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§12.3.3.1

Example 9

II

Vertical Irregularity Type 5a

Example 9 Verticallrregulaljty Type Sa

§ 12.3.3.1

A five-story building has a steel special moment-resisting frame (SMRF). The frame consists ofW24 beams and W14 columns with the following member strength properties. o

A

Beams at Levels I and 2: Mllb = ZF." = 250 kip-ft Columns on lines Band C at both levels: M" c = 250 kip-ft at axial loading of 1.2PD + 0.5PL Column base connections at grade (based on grade-beam strength): Jvf"GB = 100 kip-ft In addition, assume for the purposes of illustration only, that the columns have been designed such that a strong beam-weak column condi tion is permitted.

Level 5

12'

.

4

-.

12' 3

~

.. -'

..

.

12" 2 12'

.....

..

1 14'

//

"/

/ /

/

//

./

///

Determine if a Type 5 vertical irregularity (discontinuity in capac ity-weak story) cond ition exists in the first story.

[!J

Determine first-story strength

[!J

Determine second -story strength

~ Determine if weak-story exists at first story Calculations and Discussion

Code Reference

A Type 5 weak-story discontinuity in capacity exists when the story strength is less than 80 percent of that of the story above (where it is less than 65 percent, an extreme weak story exists) . The story strength is consi dered to be the total strength of all seismic-foree-resisting elements that share the story shea r for the directi on under consideration. To determine if a weak story exists in the first story, the sums of the column shears in the first and second stories-when the member moment capacities are developed by lateral loading-must be determined and compared. In this example, it is assumed that the beam moments at a beam-column joint are 54

2006 IBC Stru ctur al/Se ismic Des ig n Manual, Vol. I

Example 9 • Vertical Irregularity Type Sa

§12.3.3.1

distributed equally to the sections of the columns directly above and below the joint. Given below are the calculations for first and second stories.

[!J

Determine first story strength Columns A and D must be checked for strong column-weak beam considerations

....

200

2Mc = 400 > M;

= 250

~)

: . Strong column-weak beam condition exists.

--...

250

FOR MOMENT

200

Next, the shear in each column must be determined. Note moment capacity of beam (25012) governs over moment capacity of column (200) to determine shear

a

v

Clear height = 14 ft - 2 ft

125~

M. /2

125

=125 kip-ft

L) I

250

'-'

= 12 ft

-125+100 - -- = 187-k' .) "iPS 12

• o

v

FOR SHEAR Mf=100kip·ft

200 .........

Checking columns Band C for strong column-weak beam considerations

250(+)250

2Mc = 400 < 2Jvfb = 500

200"-

FOR MOMENT

:. Strong beam-weak column condition exists.

a

Next, the shear in each column must be determined. Note moment capacity of column governs over v moment capacity of beam to determine shear.

200.. . . . . Me = 200 kip-ft

200(+) 200 200'-' 200 r '

Clear height = 14 ft - 2 ft

VB= Vc=

= 12 ft

200 + 100 = 25.0 kips 12

100 .....J

M

V

4

G

First story strength = VA + VB + VD = 2(18.75) + 2(25.0)

=100 klp-ft

FOR SHEAR

= 87.5 kips

2006 IBC Structural/Seismic Design Manual, Vol. I

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§12.3.3.1

~

Example 9 •

Ver tical Irregu larity Type 5a

Determine second story strength Columns A and D mu st be checked for st rong column-w eak beam at Level 2

FOR MOMEN T

:. strong column-weak beam condition exis ts.

"""" 200

v Mb I 2= 125 kip-ft

'-'

125

Clear height

=

12 ft - 2 ft = 10ft

125 r-FOR SHEAR

V." --

tr

-

" D-

125 + 125 -?5 0 k'IpS - _. 10 -J

125

Mb I 2= 125 kip-ft

v

• 125"

J

Check ing columns B and C for strong column-weak beam con siderations

2Mc = 400 < 2Mb = 500 : . Strong beam-weak column con dition exists . FOR MOMENT

v Me = 200 kip-ft

Clear height

=

12 ft - 2 ft

=

200( +

IO ft 10'

VB = Vc = 200 + 200 = 40.0 kips 10 Me =200 klp-ft

•

Second story strength = V-I + VB + Vc+ VD + 2(25.0) + 2(40.0) = 130.0 kips

2006 IBC Structural/Seismic Design Manual, Vol . I

)200

200 ........

200 J1"""'

v

56

200 ........

200 '-"

FOR SHEAR

Ex ampl e 9

~ Determine

~

Vertical Ir regulari ty Type Sa

§12 .3 . 3. 1

if weak story exist s at first story

First story strength = 87.5 kips Second story strength = 130.0 kips

I

87.5 < 0.80(130)

=

104

(T 12.3-2, Item 5a)

:. Weak story condition in first story exists.

I

I

,

I I \

I

I I I

I I

2006 IBC Structural/Seismic Design Manual, Vol. I

57

§ 12. 3.2.1

In tr oduction 10 Ho r/zonla /lrregularitles

Introduction to Horizontal Irregularities

1 §12.3.2.1

Horizontal structura l irregularities are identified in Table 12.3-1. There are five types of horizontal irregularities: la.

Tor sional Irregularity - to be considered when diaphragms are not flexible as determined in §12.3.1.2

lb .

Extreme Torsional Irregularity - to be considered when diaphragms are not flexib le as determined in §12.3.1.2

2.

Re-entrant Comer Irregularity.

3.

Diaphragm Discontinuity Irregularity.

4.

Out-of-plane Offsets Irregularity.

5.

Nonparallel Systems - Irregularity.

These irregularities can be categorized as being either special response condition s or cases of irregular load path. Types Ia, Ib, 2, 3, and 5 are special response conditions:

Type 1a and 1b. When the ratio of maximum story drift to average story drift exceeds the given limit, there is the potential for an unbalance in the inelastic deformation demands at the two extreme sides ofa story. As a consequence, the equivalent stiffness of the side having maximum deformation will be reduced, and the eccentr icity between the centers of mass and rigidity will be increased along with the corresponding torsions. An amplification factor Ax is to be applied to the accidental torsion M'a to represent the effects of this unbalanced stiffness, § I2.8.4. I to 12.8.4.3.

Type 2. The opening and closing deformation response or flapping action of the projecting legs of the building plan adjace nt to re-entrant comers can result in concentrated forces at the comer point. Elements must be provided to transfer these forces into the diaphragms.

Type 3. Excessive openings in a diaphragm can result in a flexible diaphragm response along with force concentrations and load path deficienci es at the boundari es of the openings. Elements must be provided to transfer the forces into the diaphragm and the structural system. Type 4. The out-of-plane offset irregularity represents the irregular load path category. In this case, shears and overturning moments must be transferred from the level above the offset to the level below the offset, and there is a horizontal offset in the load path for the shears. Type 5. The response deform ations and load patterns on a system with nonparallel lateral-force-resisting elements can have significant differences from those of a regular system. Further analysis of deformation and load behavior may be necessary.

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Example 10 a Horizontal Irregularity Type 1a and Type 1b

ri3fnple 1 oui on a/Irregulari ty Type 1a a nd ype'lb

§12.3.2.1

§ 12.3,,2 .

A three-story special moment-resisting frame building has rigid floor diaphragms . Unde r code-prescribed seismic forces, including the effects of acc idental torsion, it has the follow ing elastic displacements OXl! at Levels I and 2. = 1.20 in

OR,:!.

= 1.90 in

OL,2 = 1.00 in

OR, l

= 1.20 in

OL.:!

OR,2

Level

----------------->

3

- ------7

OR ,1

2

[}J

I I I

Determine if a Type 1a or Type 1b torsional irregularity exists at the second story If it does:

~ Compute the torsional amplification factor Ax for Level 2

C€l'cu'ations and Discussion

Code Reference

A Type 1a torsional irregu larity is con sidered to exist whe n the maxi mum story drift, includ ing accidental tors ion effects, at one end of the structure transverse to an axis is more than 1.2 times the average of the story drifts of the two ends of the structure, see § 12.8.6 for story drift determination

2006 IBC Structural/S eismic Design Man ual, Vol. I

59

§12.J.2.1

ITI

Example 10 . Horizon'al Irreg ula rit y Type 1a an d Typ e 1b

Determine if a Type 1a torsional irregularity exists at the second story Referri ng to the above figure showing the displacements bJe due to the prescribed lateral forces, this irregularity check is defined in terms of story drift D.x = (bx - bx-d at ends R (right) and L (left) of the structu re. Torsional irregularity exists at Leve l x when

where

T 12.3-1

I I I

I Determining story drifts at Level 2

D.L,2 = 1.20 - 1.00 = 0.20 in D.R.2 = 1.90 - 1.20 = 0.70 in A

U al'g

=

0.20 + 0.70 = 0 45 . . In 2

Checking 1.2 criteria

D. """ = 0.7 = 1.55 > 1.2 D. a,-, 0.45 :. Tors ional irregularity exis ts - Type Ia. Check for extreme torsional irregulari ty

D. ---"!!!!.

. -lrregu Ian-ty exists . - T ype Ib. = -0_70 - = I .55 . .. thus, extreme torsion D. .", 0.45

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Example 10 . Horizontal Irr egularity Type 1a and Type 1b

~ Compute amplification factor Ax for Leve l 2

1

§12.3.2.1

§12 .8.4.3

When torsional irregularity exists at a Level x, the accidental torsional moment M'n must be increased by an amplification factor Ax . This must be done for each level, and each level may have a different Ax value. In this example, A., is computed for Level 2. Note that Ax is a function of the displacements as opposed to/versus the drift. 2

4 ' .'

=

(jmtI.' (

(!BC Eq 16-44)

1.260 ' 8 )

bma., = 1.90 in... (b R.2 ) b

= bL,] avg

+ b R., 2

=

1.30 + 1.90 = 1.60 in 2

A , = ( 1.90 )2= 0.98 < 1.0 ... Note Ax shall not be less than 1.0 1.2(1.60) :. use Ax = 1.0.

Commentary

I I I I I I

In §12.8.4.3, there is the pro vision that the more severe loading shall be considered. The interpretation of this for the case of the story drift and displacements to be used for the average values I'l.b",.g and bm·g is as follows . The most severe condition is when both bR,X and bL,Xare computed for the same accidental center-o f-mass displacement that causes the maximum displacement bmax. For the condition shown in this example where b RX = bma.n the centers-of-mass at all levels should be displaced by the accidental eccentricity to the right side R, and both bR,X and bL..r should be evaluated for this load condition. Table 12.3-1 triggers a number of special design requirem ents for torsionally irregular struc tures. In fact, if irregularity Type Ib (Extreme Tors ional Irregularity) is present, § 12.3.3.1 is triggered, which prohibits such structures for SOC E or F. It is important to recognize that torsionai irregularity is defined in terms of story drift I'l.." while the evaluatio n of A.r by Equation 12.8-14 is, in terms of displacements bxc • There can be instances where the story-drift values indicate torsional irregul arity and where the related displacement values produce an Ax value less than 1.0. This result is not the intent of the provision, and the value of Ax used to determine acci dental torsion should not be less than 1.0. The displacement and story-drift values should be obtained by the equivalent lateral-force method with the code-prescribed lateral forces. Theoret ically, if the dynam ic analysis pro cedure were to be used, the values of I'l.ma.' and I'l.m.g would have to be found for each dynamic mode, then combined by the appropriate SRSS or CQC procedures, and then scaled to the code-prescribed base shear. However, in view of the complexity of this determination and the judgmental nature of the 1.2 factor, it is reasoned that the equivalent static force method is sufficiently accurate to detect torsional irregu larity and evaluate the Ax factor. 2006

lac Stru ctural/Se ismic D esign Manu al,

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§12.3 .2.1

Example 10 . Horizontal Irregularity Typ e 1a and Typ e 1b

J

If the dynamic analys is procedure is either elected or required, then § 12.7.3 requires the use of a three-dimensional model if there are any irregularities. For cases oflarge eccentricity and low torsional rigidity, the static force procedure can resul t in a negative displacement on one side and a positive on the other. For example, this occurs if Du = - DAD in. and DR.3 = 1.80 in. The value of Dm'g in Equation 12.8- 14 should be calculated as the algebraic average .

= (- 40) + 1.80 = l AO = 0.70 in 2

2

When dy namic analysis is used, the algebraic average value Dm'g should be found for each mode, and the individual modal results must be properly combined to determ ine the total response value for Dm .g •

1

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Exam ple 11 • Horizontal Irregularity Type 2

Example 11 orizontallrregularity T}'J e 2

§12.3.2. 1

§12.3.2.. 1

The plan configuration of a ten-story special moment frame building is as shown below.

G

I

1
0.100 V,h"C d (363.0)(20 ft)(l2)(5 .5)

: . P-delta effects must be considered.

Check for 8 :'0 8max using the given

= 0.5 =

8 lII ax

~Cd

0.5

~=

0.80

= 0. 1136

(0.80)(5 .5)

(Eq 12.8-17)

0.103 < 0.1136 ... o.k. Final design story drift and story shear in first story

§12.8.7

When 8 > 0.10, the initial design story drift and design story shear must be augmented by the incremental factor ad related to P-deita effects ad=

~= 1-8

\.0 = 1.\15 1-0.103

The final design story drift in the first story is I'J. 1 =adI'J. = (1.\15)(3.96) = 4.415 in

The final design story shear is VI

76

= adVI = (1.\15)(363.0) = 404.7 kips

2006 IBC Structural/Seism ic Design Manual, Vol. I

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Example 16 • P-delta Effects

§12.B.7

Check for story- drift compliance in the first story

1

Allowable story drift /:; ,,110'" = 0.020 hi /:;0110'"

§12. 8.7 T 12.12-1

= 0.020(20 ft)( 12) = 4.80 in

/:;; = 4.415 < 4.80 in .. . o.k.

, Commentary In § 12.8.7 the P-delta effects on the design story drift and the design story shear are evaluated by the follow ing procedure:

I.

Given the initial design story drift /:;" = 0.,- 0".1 at story x: compute for each story x the stability coefficient 8x given by Equation 12.8- 16. For each story where 8., is equal to, or greater than 0.10, compute the corresponding incremental factor relating to P -delta effects ad = 1/( I - 8.,). This factor accounts for the multiplier effect due to the initial story drift /:;., leading to another increment of story drift, leading to another story drift, which would lead to yet another increment, etc. Thus both the drift and the shear in the story would be increased by a factor equal to the series of I + 8 + 8 2 + 8 3 + ---, which converges to 1(1 - 8) = ad. As a resu lt the initial story drift /:;" and story shear V, need to be multiplied by the factor ad to represent the total final P-delta effect.

2.

The fina l resulting story drift IJ. ~ = ad IJ.., need s to comply with the drift limitations of §12 .12.

3.

In each story requiring consideration of P-delta effects the initial story shears are increased to ad V" . The structural elements must be designed to resist the resulting final story shears, overturning moments and element actions.

I I

I I I

,':=

Some computer programs for frame analysis state that P-delta effects are incl uded directly in the analysis. The engineer should verify that the total gravity load employed and the method used in these programs will provide results that are essentially equivalent to the augmented story shear method described above. The provisions in §§ 12.8.6 and 12.8.7 for the evaluation of the final story drifts state that the final story drift shall be ad times the initial drift IJ.. However, in a multi-story building having 8 > 0.1 in more than one story, the initial story shears in these stories are increased by the ad factor. This is equivalent to an added latera l load equal to (a d-I) V, applied to each story level having 8 > 0.1. Therefore the new story drifts in the stories below would be inc reased not only by their own ad but by the added lateral load effect from the stories above; thus , the fina l drifts should be found by a new analysis with the added lateral loads equal to (ad - I) V" along with the initi al lateral loads on the frame. 2006 la c Structural/S eism ic Design Manual, Vol. I

77

1

Example 17 • Seis mic Bas e Shear

§12.8.1

]

ii~ample 1 7 §12.8.1

S,e ismic Base Shear

]

Find the design base shear for a 5-story steel special moment-resisting frame building shown below . The following information is given. Seismic Design Category D S DS =

0.45g

I I I

60'

SOl = 0.28g J = 1.0

R =8 W = 1626 kips 11" = 60 feet

- - - '-- - '-- - '-- - '--

To solve this example, follow these steps.

[L]

Determine the structure period

~ Determine the seismic response coefficient

c,

~ Determine seismic base shear Calculations and Discussion

[I]

Determine the structure period

Code Reference

§12.8.2.1

The appropriate fundamental period To is to be used. Cr for steel moment-resisting frames is 0.035.

T; =

CrUd :; = 0.035(60)~ = 0.D75 sec

~ Determine the seismic response coefficient Cs

(Eq 12.8-7)

§12.8.1

The design value of C, is the smaller value of

C = S os = (0.45) = 0 0561 s

(R) J

78

(8) . 1.0

200 6 IBC Structural/ Se ismic Design Manual, Vol. I

(Eq 12.8-2)

I I I I I I I I

Example 18

~

D

Approximate Fundamental Period

§12.8.2.1

Concrete special moment frame (SMF) structure Height of the tallest part of the building is 33 feet, and this is used to determine per iod. Roof pentho uses are generally not considered in determining hi!> but heights of setbacks are included. However, if the setback represents more than a 130-percent change in the lateral force system dimension, there is a vertical geometric irregularity (Table 12.3-2). Taller structu res, more than five stories or 65 fee t in height, require dynamic analysis for this type of irregularity.

Setback

k--J 33'

-'-

- '- - '-

-

L-

_

L.-

_

L.-

h« = 33 feet

CT = 0.016 ; x = 0.9 To = CT(hnY = 0.0 16(33)°·9 = 0.37 sec

[!J

Steel eccentric braced frame (EBF)

44'

EBF structures use the C, for the "all other buildings" category C T = 0.030 ; x = 0.75 T = CT(hn}T: = 0.030(44)°·75 = 0.5 1 sec

~

Masonry shear wall building

29'

29'

I~~f"

~ TYP' 60'


z-:

(Eq 12.8-6)

2006 IBC Structural/Seism ic D esign Manual, Vol. I

79

§12.8.2.1

Example 18 • Approximate Fundame ntal Period

'. ample 18 Appro~imate Fundamental Period

§ 12.8.2.1

Determine the period for each of the structures shown below usi ng the appropriate fundament al period formula (Eq 12.8-7) The coefficient CT and the exponent x are dependent on the type of structural system used.

[!J

Steel special moment frame (SMF) structure

[!J

Concrete special moment frame (SMF) structure

@J

Steel eccentric braced frame (EBF)

~ Masonry shear wall building

~

Tilt-up building

Code Reference

Calculatipns and Discussion

[!J

Steel special moment frame (SMF) structure

§12.8.2.1

Height of the structure above its base is 96 feet. The additional 22-foot depth of the basement is not considered in determining 11" for period calculation.

C r = 0.028; x

= 0. 8

To = Cr (I1"r = 0.028(96)°·8 = 1.08 sec

-,-

96' Supe rslructu re

Grad e

, 22' _L

Note : In the SEAOC Blue Book, base is defmed as the level at which earthq uake motions are considered to be imparted, or the level at which the structure, as a dynamic vibrator, is supported. For this structure the solution is the same.

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2006 IBC Structuraf/Seismic Design Manual, Vol . I

Basemenl

Example 23 • Vertica l Distributio n of Seismic Forc e

1

Exampl e 23 · Vertical Distribution of Seismic Force

§12.B.3

§ 12 .8.3

A nine-story building has a moment-resisting steel frame for a lateral-force-resisting system. Find the vertical distribution oflatera l forces F x • The following information IS given.

I

IV = 3762 kips C, = 0.062 R = 8.0

level

= 3.0 = 1.0 T = 1.06 sec

Q"

1

9

?I fI T I (

27'

0) EO

27'

):

Story weight

2 14k

12' 8

4 OSk

7

4 QSk

6

4 O~

5

5 54k

' 2' ' 2' ' 2'

,

4 22k

3

4 22k

2

440k

12'

'2'

I

' 2' ' 2' 4 65k

20' //

" 1' /

/

1' /

~ / / /I' /

Total

I

I

3762 kips

To solve this example, follow these steps.

I I I I

'/

[!J

Determine V

~

Find Fx at each level

~ Find the distribution exponent k ~ Determine vertical force distribution Code R...-ference

:Ca/culations and Discussion

[!J

Determine V

§12.8-1

This is the total design lateral force or shear at the base of the structure. It is determined as follows

v = C, IV = 0.062 (376zk) = 233.8 kips

(Eq 12.8-1)

2006 IBC Structura l/Se;smlc Design Man ual, Vol. I

93

§12. 8.3

~

Ex ample 23 •

Find

I I

Ver tic al Dis tr ibu tion of Seis mic Force

r, at each level

The vertica l distribution of seismic forces is determined as (Eq 12.8- 11 )

]

where

c

= I'X

lVxhl

(Eq 12.8-12)

n

2: IV; hi

I ]

i- I

Since there are nine levels above the ground, 11 = 9 Thus: F = 233.8w,l1; x

• ~

,

LJ lV /Ii

I

i- I

§12.8.3 Find the distribution exponent k The distribution exponent k is equal to 1.0 for buildings having a period of T s 0.5 seconds, and is equal to 2.0 for buildings having a period oi T>: 2.5. For intermediate value of the building period, k is determined by linear interpo lation. Thus:

2.5 2.0

...

1.5

." e

1.28

'C ~

0-

1.0 0.5

o

o

0.5

1.0

1.06

1.5

2.0

2.5

I I I I I I I I

Exponent, k

J 94

2006 IBC Struc tural/S eism ic Design Manu al, Vol. I

I

Examp le 23 • Verti cal Distribution of Seismic Force

I

§12 .B.3

Now: for

T

k

= 1.06 sec = 1.0 + (1.06 _ 0.5) (

I ) 2.5 -0.5

= 1.28 k

Use:

@J

= 1.28

Equation 12.8-12 is solved in the table below given V= 233.8 kips and k = 1.28

".f•

I

I I I I I I I I

U'.•

Level X

ii,

( ft)

(kips)

9 8 7 6 5 4 3 2

116 ft 104 92 80 68 56

439 382 326 273

214 405 405 405 584

44

32 20

222 173 127 84 46

•

wxhx

W./I.~

C~ , = - I

LW/l i

F. = C••V (kips)

kip-It

422 440 465

93.946 154.710 132.030 110.565 129.648 73.006 53.594 36.960 21,390

0. 116 0. 192 0.169 0. 137 0. 161 0.09 1 0.067 0.04 6 0.027

27.3 44.8 38.3 32. 1 37.6 21.2 15.5 10.8 6.2

~ =3762

~ = 806.289

1.004

233.2

422

F/ w. = Su 0.127 0.1 II 0.094 0.079 0.064 0.050 0.037 0.024 0.013

Commelltary Note that certain types of vertical irregularity can result in a dynamic response hav ing a load distribution significantly different from that given in this section . Table 12.6- I lists the minimum allowable analysis procedures for seismic design . Redundancy requ irements must also be evaluated once the type oflateral-force-resisting system to be used is specified, because this may require modification of the building framing system and vertical distribution of horizontal forces as a result of changes in building period T. Often, the horizontal forces at each floor level are increased when p is greater than 1.0. Th is is done to simplify the analysis of the framing members. The horizontal forces need not be increase d at each floor level whe n p is greater than 1.0, provided that, when stre ss check ing the individual mem bers of the lateral-foree-resisting system, the seismic forces are factored by p. When checking building drift, p = 1.0 (§12.3.4.1) shall be used.

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95

§ 12. 8. 3

Exampl e 23 •

Vertica l Distr ibution of Se ism ic Forc e

Structu res that have a vertical irregularity of Type Ia, Ib, 2, or 3 in Table 12.6-1, or plan irregular ities of Type l a or Ib in Tab le 12.6-1, and having a height exceeding five stories or 65 feet may have significantly different force distributions. Structures exceeding 240 feet in height shall require dynamic analysis. The configuration and final design of this structure must be checked for these irregularities. Most structural analysis programs used today perform this calculation, and it is rarely necessary to manually perform the calculations shown above . However, it is recommended that these calculations be performe d to confirm the computer analysis and to gain insight to structural behavio r. Note that ( So )max is approximately twice C , and S" = r 1.2 0.5 1

. . Torsional irregularity Type la exists - Note: if li ma.,1liao'g is larger than 1.4, then torsional irregu larity Type 1b exists.

~ Determine the amplification factor A x Because a torsiona l irregularity exists, §12.8.4.3 requires that the second story torsional moment be ampli fied by the following factor. In this example, beca use the only source of torsion is the accidental eccentricity, the amplification factor will be used to calculate a new and increased accidental eccentricity , as shown below.

Ax

(Eq 12.8-14)

Where:

the average story displacement is computed as =

1.44+0.75 = 1.10 in 2

=

1.44 ((1 .2)(1.10) )= 1.19 in

'

~ New accidental torsion eccentricity Since Az (i.e., Ax for the second story) is greater than unity, a second analysis for torsion must be performed using the new accidental eccentricity.

e"cc

10 4

I I

= (1.19)(4.0 ft) = 4.76 ft

2006 IBC Structural/S eismic De sign Manual, Vol. I

I I I

1

Example 25 • Amplification

ct Accidental Torsion

§12.8.4.3

,9 gmmentary Example calculati ons were given for the second story. In practice, each story requi res an evaluation of the most severe element actions and a check for the torsional irregularity condition. If torsional irregularity exists and Ax is greater than 1.0 at any level (or levels) , a second torsional analysis must be performed using the new accidental eccentricities. However, it is 110/ required to find the resulting new Ax values and repeat the process a second or third time (until the Ax converges to a constant or reaches the limit of 3.0). The results of the first analysis with the use ofA., are sufficient for design purposes. While this example involves wall shear evaluation, the same procedure applies to the determination of the most severe element actions for any other lateral-foree-resisting system having rigid diaphragms.

I I I I I

When the dynamic analysis method of §12.9 is used, all the requirements of horizontal shear distribution, given in §12.8.4, including torsion calculations that may be accounted for by displacing the calculated centers of mass of each level (§12.8.4.1 and §12.8.4.2) also apply . However, §12.9.5 states that amplification of accidental torsion, need not be amplified by Ax where accidental torsional effects are included in the dynamic analysis model. Only the accidental torsion is required to be amplified if torsional irregularity exists . Also note that Ax is not required to exceed 3.0.

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105

§12.3.3.3

Example 26 •

Elements Supporting Discontinuous Systems

ain~/e26

lements Supporting Discontinuous Systems A reinforced concrete building has the lateral-foree-resisting system shown below. Shear walls at the first-floor level are discontinuous between lines A and Band lines C and D. The following information is given. Seismic Design Category C S DS= 1.10

T 12.2-1

Ordinary reinforced> concrete shear wall (ORCSW) building frame system : R = 5 and Q o = 2.5

Note: ORCSW not permitted in SDC D, E, or F.

§ 12.4.2.3

Office building live load: use factor of 0.5 on L

L~.f

Axial loads on column C D = 40 kips L = 20 kips QE = 100 kips

4 ,---------.,....----,

12'

....,.. ~

Shear wall

12'

12' Column C 24" x 24"

f c =4000 12'

Determine the following for column C.

[!J

Required strength

~ Detailing requirements

106

2006 IBC Structural/Seismic Design Manual, Vol. I

psi

Exam plo 26 • Elements Supporting Discontinuous Systems

• Calculations and Discussion

§12.J.J.J

Code Reference

This examp le demonstrates the loading criteria and detail ing required for elements supporting discontinued or offset elements of a seismic-force-resisting system .

Required strength Because of the discontinuous configuration of the shear wall at the first story, the first story columns on lines A and D must support the wall elements above this level. Column C on line D is treat ed in this example. Because of symmetry, the column on line A would have identica l requirements.

I

Section 12.3.3.3 requires that the column shall have a design strength to resi st special seismic load combination of § 12.4.3.2

Pu = 1.2D + 0.5L + 1.0Em

§ 12.4.2.3 (Comb. 5)

P" = 0.9D + 1.0Em

§ 12.4.2 .3 (Comb. 7)

E.. = Q" QE+ 0.2 SDS D = 2.5( 100) + 0.2( 1.10)(40) = 259 kips

§ 12.4.3.2 (Comb. 5)

E.. = Q" QE - 0.2 S DS D = 2.5( 100) - 0.2( 1.10)(40) = 24 1 kips

§ 12.4.3 .2 (Comb. 7)

where

or

1

Substituting the values of dead, live, and seismic loads

I

I I I I I I

+ 0.5 (20) + 259 = 317 kips co mpression

P"

= 1.2 (40)

P"

= 0.9 (40) - 0.5 (241) = -205 kips tension ·

and

2006 IBC Stru ctural/Seis mic Design Manual. Vol. I

107

§12.3.3.3

Examp l e 2 6

ff

Elements Supporting Discon tinuous Systems

Cotpmentary To transfer the shears from walls A-B and C-D to the first-story wall B-C, collector beams A-B and C-D are required at Levell . These would have to be designed according to the requirements of § 12.10.2. The load requirements of § 12.3.3.3 and relat ed sections of the relevant materials chapters apply to the following vertical irregularities and vertical elements.

1. Discontinuous shear wall. The wall at left has a Type 4 vertical structural irregularity. Note that only the column needs to resist the special load combi nations since it supports the shear wa ll. 1,14"_ _- r t --Column

2. Discontinuous column. Th is frame has a Type 4 vertica l structural irregularity.

DDD

DDD Transfer girder

108

2006 IBC Structural/Seismic Design Manual, Vol. I

Example 26 • Elements Supporting Discontinuous Systems

3. Out-of-plane offset. The wall on Line A at the first story is discontinuous. This structure has a Type 4 plan structural irregularity, and §12.3.3.3 applies to the supporting columns. The portion of the diaphragm transferring shear (i.e., transfer diaphragm) to the offset wall must be designed per the requirements of §12.3.3.4. Note that the transfer diaphragm and the offset shear wall are subject to the p factor, but not to the special load combinations.

§12.3.3.3

Oisconllnued wall

It should be noted that for any of the supporting columns shown above, the load demand Em of § 12.4.3.2 Equations 5 and 7 need not exceed the maximum force that can be transferred to the element by the lateral-foree-resisting system.

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109

fI" '-:-rt:

-

1

Example 27 • Elements Supporting Discontinuous Walls or Frames

§12.3.3.3

1

r

·& r'!'!amp le 2 7 .". ~~/emen ts Supporting Discontinuous Walls

orFrames

§12.3.3.3

This example illustrates the application of the requirements of § 12.3.3.3 for the allowable stress design of elements that support a discont inuous lateral-foree-resisting system. In this example, a light-framed bearing-wall building with plywood shear panels has a Type 4 vertical structural irregularity in one of its shear walls, as shown below. The following information is given. Seismic Design Category C S DS = l.IO R =6.5 no = 3.0 Cd =4 /I = 0.5 Axial loads on the timber column under the discontinuous portion of the shear wall are

Ughl framed wall with plywood sheathing

Timber column

Applicable load combinations

~ Required column design strength

[1J

Code Reference

Applicable load combinations For vertical irregularity Type 4, § 12.3.3.3 requires that the timber column have the design strength to resist the special seismic load combinations of § 12.4.3.2 . This is required for both allowable stress design and strength design. For strength design the applicable load combinations for allowable strength design are

7. (0.9 - 0.2SDS)D + QoE Appl icable load combinations for allowable strength design are:

110

2006 IBC Structural/Seismic Design Manual, Vol. I

J

I

Determine the following.

CalclJlations and Discussion

I I

I

Dead D = 6.0 kips Live L = 3.0 kips Seismic Q£ = ±7.0 kips

[!J

1

I I I I I I I I I I

Example 27 • Ele ments Supp or ting Discontinu ous Waifs or Fram es

§ 12.3 .3 .3

5. (1.0 + 0.105 80S) D + 0.525 + QoQE + 0.75L 6. (0.8 - 0.14 80s) D + 0.7 QuQE

1 Required column desig n streng th (strength design) In this shear wall, the timber column carries only axial loads. The appropriate dead, live, and seismic loads are determined as D = 6.0 kips L

= 3.0 kips

Em = n, QE + 0.2 80S D = 3.0(7.0)+ 0.2 (1.10) (6.0) = 22.3 kips J

or

Em = Q o QE- 0.2 80S D = 3.0(7.0) - 0.2 (!.IO) (6.0) = 19.7 kips

For the required strength design -strength check, both load combinations must be checked.

P

I I I I I I I I I

=

1.2D + L + Em

P = 1.2 (6.0) + 0.5 (3.0) + 22.3 = 31.0 kips . . . (compression) P

=

0.9D - 1.0Em

P = 0.9 (6.0) - 1.0 (19.7) =-14.3 kips .. . (tension) The load factor on L in combination 5 is permi tted to equal 0.5 for all occupancies in whic h L; is less than or equal to 100 psf, with the exception of garages or areas occupied as places of public assembly. Commentary

For strength design, the tim ber column must be checked for a compression load of31.0 kips and a tension load of 14.3 kips. In making an allowable stress design check, § 12.4.3.3 permits use of an allowable stress increase of 1.2. The 1.2 stress increase may be combined with the duration ofload increase described in the NO S. The resulting design strength = (1.2)(1.0)( 1.33) (allowable stress desig n). This also applies to the mechanical hold-down element required to resist the tension load.

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111

§12.3.3.3

Example 27 • Elements Supporting Discontinuous Walls

or Frames

The purpose of the design-strength check is to confirm the ability of the column to carry higher and more realistic loads required by the discontinuity in the shear wall at the first floor. This is done by increasing the normal seismic load in the column QE by the factor Q o = 3.0 to calculate the maximum seismic load effect Em (§ 12.4.3).

I 1

I I I I I

I I I I I

112

2006 IBC Structural/Seismic Design Manual, Vol. I

E::ample 28 • Soil Pr es s ur e At Foundation

le2 oil Pressure At Foundations

§§2A; 12.13 .4

III

§§2.4; 12.13.

Geotechnical investiga tion reports usua lly prov ide soil-bearing pressures on an allowable stress design basis while seismic forces in ASCE /SEI 7-05 and most concrete design (ACI/318-05, § 15.2.2 and R 15.2), are on a strength design basis. The purpose of this exam ple is to illustrate footing design in this situation. A spread footing supports a reinforced concrete column. The soil classification at the site is sand (SW). The following information is given .

r

Seismic Design Category C SDS = 1.0, / = 1.0 P = 1.0 for structural system PD = 50 kips PD includes the footing and imposed soil weight) PL = 30 kips P E = ± 40 kips, VE = 25 kips, (these are the QE loads due to base shear V) Snow load S = 0 Wind load W < QE /1.4

Grade

4'

)J

The seismic loads are from an equivalent lateral analysis. The loads given above follow the sign convention shown in the figure. Perform the followi ng tasks.

I}J

Determine the design criteria and allowable bearing pressure

~ Determine footing size ~ Determine soil pressure reactions for strength design of the footing section

2006 IBC Structural/S eism ic Design Manual, Vol. I

11 3

§§2.4; 12.13.4

Example 28 • Soil Pressure At Foundation

:Calculations and Discussion

[L]

Code Reference

Determine the design criteria and allowable bearing pressure

§2.4

The seismic-force reactions on the footing are based on strength design. However, allowable stress design may be used for sizing the foundation using the load combinations given in §2.4.1.

D + 0.7£

(Comb. 5)

D + 0.75 (0.7£ + L )

(Comb. 6)

0.6D + 0.7£

(Comb. 7)

Section 12.13 .4 permits reduction of overturning effects at the foundation-soil interface by 25 percent (if an equivalent lateral for ce analysis is used) or 10 percent (if modal analysis is used ). Therefore , for the soil pres sure the seism ic effe ct is reduced

D + 0.75(0.7E)

(Comb. 5)

D + 0.75[0.7(0.75)£ + L]

(Comb. 6)

0.6D + 0.7(0.75)£

(Comb . 7)

Because foundation investigation reports for bu ild ings typically specify bearing pre ssures on an allowable stress design basis, crit eria for determining footing size are also on this basis.

The earthquake loads to be resisted are specified in §12.4.2 by

£=£,, +£,.

(Eq 12.4- I)

Per § 12.4 .2.2, £ 1' = 0 for determ ining soil p ressure. Equation 12.4 - I reduces to (Eq 12.4-3)

For the san d class of material and footing depth of 4 feet, the allowable gross foun dati on pressure pa from a site-specific geotechnical investigation recommendation is

p a = 2.40 ksf for sustained loa ds and pa = 3.20 ksf for trans ient loads, such as seismic.

114

2006 IB C S tr u c tu r al /Se ism i c Des ig n Manual, Vol. I

§§ 2.4; 12.13.4

Example 28 • Soli Pres sure At Foundation

[3J

1 1

Determine foot ing size

P

=

D + 0.75(0.7E) = 50 + 0.75(0.7)(40) = 56 kips

(Comb. 5)

P = D + 0.75[0.7(0.75)£ + L]

(Comb. 6)

= 50 + 0.75[0.7(0.75)40 + 30) = 88 kips

P = 0.6D + 0.7(0.75)£

(Comb. 7)

= 0.6(50) + 0.7(0.75)(-40) = 9 kips

Equation 6 governs. The requ ired footing size is 88 kips/3.20 ksf = 27.5 sf Use 5 ft, 3-in-square footing. A = 27.6 sf

~ Determine soil pressure reactions for strength design of footing For the design of the concrete elements, strength design is used. The reduction in overturning does not apply, and the vertical seismic load effec t is included P

= I.3D + 0.5L + E =

I I I I I

§2.3.2 (Comb. 5)

1.2(50) + 0.5(30) + 40 + 0.2( 1.0)(50) = 11 5k

A uniform pressure of 115k/27.6 sf = 4.17 ksf should be used to determine the internal forces of the footing. (Note that if the footing also resisted moments, the pressure would not be uniform.) The other seismic load combination is

P =0.9D +£

§2.3.2 (Comb. 7)

= 0.9(50) - 40 + 0.2(1.0)50 = -5k Note that this indicates uplift will occur. ASCE/SEI 7-05 does not require that foundation stability be maintained using strength-level seismic forces. This combination is only used here to determine internal forces of concrete elements of the foundat ion. As it results in no internal forces, it may be neglected.

I 2006 IBC StructuraVSeismic D esign Manu al, Vol. I

11 5

§12.8.6

Example 29 • Drift

Example 29 Drift

§12"B.6

A four-story special moment-resisting frame (SMRF) building has the typical floor plan as shown below. The typical elevation of Lines A through D is also shown, and the structure does not have horizontal irregularity Types 1a or lb. The following information is given .

Occupancy Importance Category I Seismic Design Category 0 1 = 1.0 Cd = 5.5 T = 0.60 sec

Seism ic force

Typical fluor plan

Level

4

3

2

DDD DDD DDD

12'

12'

12'

12'

Ty pical Elevati on

The following are the deflections (computed from static analysis - effects of P-delta have been checked) bxe at the center-of-mass of each floor level. These values include both translational and torsional (with accidental eccentricity) effects. As required by § 12.8.6.2, b.~c has been determined in accordance with design forces based on the computed fundamental perio d without the upper limit (CI/Ta ) of § 12.8.2.

11 6

20 06 IBC Structura l/Se IsmIc Design Man ua l, Vol . I

1

Example 29 • Drift

Level

0" 1.51 in 1.03

4

1

§ 12.B.6

3 2

.63 .30

For each floor-level center-of-mass, determine the following.

[L]

Maximum inelastic response deflection 05.

~ Design story drift ~ in story 3 ~ Check story 3 for story-drift limit Calculations and Discussion

[!J

Code Reference

Maximum inelastic response deflection Ox

§12.B

These are determined using the Ose values and the Cd factor

o = CA, I

= 5.56" = 5.50 1.0

.r

CEq 12.8-15)

se

The refore: Level

I I I I I I

4 3 2

1

0.(1' 1.51 in 1.03 0.63 0.30

6, 8.31 in 5.67 3.47 1.65

~ Design story drift ~ in story 3 due to Ox Story 3 is located between Levels 2 and 3. Thu s:

~J =

5.67 - 3.47 = 2.20 in

1

I

2006 IBC Structural/Seismic D esign Manual, Vol. I

117

§12.8.6

@J

I

Example 29 • Drift

§12.12.1

1

For this fou r-story building with Occupancy Importance Category I, § 12.12.1, Table 12.12-1 requires that the design story drift /1 shall not exceed 0.025 times the story height.

1

For story 3

J

Check story 3 for story-drift limit

/1J

=

2.20 in

]

Story-drift limit = 0.025 (144) = 3.60 in > 2.20 in :. Story drift is within the limit.

I I I I I I I I I )

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2006 IBC Structural/Seismic Design Manual, Vol. 1

I I I I

Example 3 0 • Stor.! Drift Limitations

§12 .1 2

lEJralnple 30

Story Drift Limitations

*12.12

For the design of new buildings, the code places limits on the design story drifts, /).. The limits are based on the design earthquake displacement or deflection Ox and not the elastic response deflections ext! corresponding to the design lateral forces of § 12.8. In the example give n below, a four-story steel special mo ment-res isting frame (SMF) structure has the design force deflections oxt! as shown. These have been determined according to § 12.8, using a static, elastic analysis.

Occupancy Category I

Level 4

Seismic Design Category D

12'

= 1.0

12'

1

Cd

~

f

~

D Deflected shape '

6. e 2.44 In

3

1.91

2

1.36

12'

= 5.5

0.79 16'

P = 1.3

0

Determine the foll ow ing.

[}J

Design earthquake deflections Ox

~ Compare design story drifts with the limit value Code Reference

Calculations and Discussion

[}J

Design earthquake deflections

ax

§12 .8.6

The design earthquake deflections Ox are determined from the following =

Cd° ,rr

(E q 12.8- 15)

1 =

5.5c5.

tr

1.0

=

5.50 oft!

2006 IB C St ructural/SeIsmi c Des ign Manual, Vol . I

1 19

Example 30 a Story Drift Limitations

§12.12

@J

Compare story drifts with the limit value

§12.12

For this four-story building in Occupancy Category I, § 12.12, Table 12.2-1 requires that the calculated design story drift shall not exceed 0.025 time s the story height. For SMF in SDC D, E, and F, this limit is reduced by

p per §12.12.1.1:

!1alp = 0.0251111.3 = 0.019211 Determine drift limit at each level Levels 4 , 3, and 2

!1 S; 0.019211

= 0.0192 (12 ft x 12 in/ft) = 2.76 in

Levell

!1 S; 0.019211 = 0.0192 (16 ft x 12 in/ft) = 3.68 in

For b. = Cl.. - Cl.._I, check actual design story drifts against limits

Level x

C."

4 3 2 1

2.08 in 1.62 1.13 0.65

Ox 11.43 in 8.92 6.24 3.59

D.

Limit

2.51 in

2.76

2.68 2.65 3.59

2.76 2.76 3.68

Status o.k. o.k. o.k. o.k.

Therefore: The story drift limits of § 12.12 are satisfied. Note that use of the drift limit of 0.02511 requires interior and exterior wall systems to be detail to accommodate this drift per Table 12.12-1

. , Whenever the dynamic analysis procedure of § 12.9 is used, story drift should be determined as the modal combination of the story-drift value for each mode. Determination of story drift from the difference of the combined mode deflections may produce erroneous results because differences in the combined modal displacements can be less than the corresponding combined modal story drift.

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Example 31 • Vertical Sei smi c Load Effect

Exal te 31 Vertical Seismic Load Effect

§12.4.2.2

§12.4.2.2

Find the vertical seismic load effect, E.-, on the non-prestressed canti lever beam shown below. The follow ing information is given. Seismic Design Category D Beam unit we ight = 200 plf SDS=

1.0

Find the following for strength design.

[!J

Upward seismic forces on beam

~ Beam end reactions Calculations and Discussion

[!J

Code Reference

§12.4.2.2

Upward seismic forces on beam For SOC 0 , the design of hori zonta l cantilever beams must consider 1. The governing load combination including E as defined in §12.4.2

E

= E" + E\O

(Eq 12.4-1) (Eq 12.4-2) (Eq 12.4-3) (Eq 12.4-4)

E" = 0.2SDsD QE = 0 for verti cal load, giving E

=0 -

0.2(1.0) D =- 0.2D

where the negative sign is for an upward action.

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§12..1.2.2

I

Example 31 • Vertical Seismic Load Effect

The governing load combination including the upward seismic effect from §2.3.2, (7) is qe

=

0.9D + 1.0E

=

1

0.9D + (- 0.2D)

=0.7D

1

= 0.7(200 plf) =

I I I I I I I

140 plf downward

:. no net upward load. The governing load combination including the downward seismic effect from §2.3.2, (5) is qe = 1.2D + 1.0E + L + 0.28

= 1.2D + 1.0(0.2)(1.0)D + 0 + 0 =l.4D = 1.4 (200 pit)

= 280 plf downward :. this is the maximum downward load on the beam.

2. A minimum net upward seismic force. The terminology of "net upward seismic force" is intended to specify that gravity load effects cannot be considered to reduce the effects of the vertical seismi c forces and that the beam must have the strength to resist the actions caused by this net upward force without consideration of any dead loads. This force is computed as 0.2 times the dead load qs = - 0.2WD = - 0.2(200) = - 40 plf

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2006 IBC Structural/Seismic Design Manual, Vol. I

J

J §12.4.2.2

I I

1

EKample J 1 " Vertical Seismic Load Effect

§12.4.2.2

Beam end reactions for upward force of 40 plf

J J

v,J = qEl! = 40 plf(lO ft) = 400 lb

J

M.4= qe ; = 40~0)2

=

2000 lb/ft

The beam must have strengths .pll;, and .pM. to resist these actions, and the actions due to the applicable gravity load combinations.

I

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§11.4.5

I

Examp le 32 • Desig n Response Spe ctrum

Exam . ", p l e 32 '!.esign Response Spectrum

§11.4.5

Determine the general design response spectrum for a site where the followi ng spectra l response acceleration parameters have been evaluated accordin g to the general procedure of §11 .4.

I

I

0.45g SOl = 0.28g TL = 8 sec

S DS=

[!J

I

Determine design response spectrum

J 'Calculations and Discussion

Code Reference

Section I J .4.5 provides the equations for the 5-percent damped accele ration response spectrum Sa for the period T intervals of

os T::: To, and T > T, To and T, arc calculated as

---""- =

T, =

S DI S DS

124

0.2(0.28) 0.45

=

I I I

O.I2 sec

= 0.28 = 0.62 sec 0.45

2006 IBC Structural/Seism ic Design Man ual, Vol. I

I I I I I I I I

Examp le 32 " Des ign Response Spectrum

§11.4.5

The spectral accelerations are calculated as 1.

For the interval 0:::: T:::: To

Sa

= 0.6

SDS T+ OASDS

= 0.6

g (0.45 ) T + 0.4(OA5g)

=

2.

3.

0.12

[2.25T + 0.18]g

For To < T ::::

Sa

=

(Eq 11 .4-5)

T"

r.

SDS = 0.45g

For r.
0.3 SosIpWp = 0.3 Wp .. . o.k.

pU

I I I

= 1.0

At lower rod connect ion level

z

=ZL

= 37 ft

= 0.4(1.0) [1 + 2 (37)] If' 2.5

60

= 0.357 Wp > 0.3 IVp

• • •

p

o.k.

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155

§13.5.3

Ex ampl e 42 •

1

Ex terior Nonstructural Wall Elements : Precast Pane l

Fp u + Fp L = (0.411 + 0.357) 2 2

The required average, Fp

w

J

I'

= 0.384Wp = 0.384(14.) =

5.53 kips

This force is applied at the panel centroid C and acts horizontally in either the out-ofplane or the in-plane direction.

Combined dead and seismic forces on panel and co nnections

§13.5.2

There are two seismic load conditions to be considered: out-of-plane and in-plane . These are shown below as concentrated forces , In this examp le, Combination 5 of §2.3.2, 1.2D + QE, is the controlling load combina tion,

~ Dea d load, seismic out-of-plane, and vertical seismic for ces Panel connec tion reactions due to factored dead load, out-of-plane seismic forces, and vertical seismic forces are calculated as follows:

,

g'

I~

_ _-

.._.. ..

5' I

_ _ _..-

5'

.. .. ..

,

f- .•_ .. _ ••_ •• -J

0~

,It

,

..

g'

I.-

-

f- •. _ .. _ .. _ ..... t -

-

Fp = 5.5 3 kips

1.2Wp + O.2Wp = 1.4Wp = 1.4 (14.4) =20 .16 kips

Each bracket and rod connection takes the following axial load due to the out-of-plane force Fp at center-of-mass

Ps + PR = Fp = 5.53 = 1.38 kips

4

4

where Pe is the bracket force and P R is the rod force.

156

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I I I I I I I I I I I I I

Examplo 42

1

IR

Exterior Nonstructural Wall Elements : Precast Panel

§13.5.3

Each bracket takes the following downward in-plane shear force due to vertical loads I AWp

20.16 . = - - = 10.08 kips

VB = - -

2

2

Note that each rod, because it carries only axial forces, has no in-plane, dead, or seismic loading.

~

Dead load, seismic in-plane, and vertical seismic forces Panel connection reactions due to factored dead load, in-plane seismic forces , and vertical seismic forces are calculated as 9'

5'

9'

,/

c

F, = 5.53 kips

5'

1.4W,= 20.16 kips

Each bracket takes the following in-plane horizontal shear force due to lateral seismic load

r,

H B =-

2

I I I I I

5.53 ki = =2.77 IpS

2

Each bracket takes the following upward or downward shear force due to the reversible lateral seism ic load

FB -- -5(F- p ) 18

-

-

5(5.53) -- ± I .54 kiIpS 18

Each bracket takes the following downward force due to vertical loads: RB =

1.4Wp

2

20.16 . = - - = 10.08 kips

2

Under the in-plane seismic loading , each rod carries no force. 20D6 IBC Structural/Seismic Design Manual, Vol. I

157

§13.5.3

Example 42 •

Exterior Nonstructural Wall Elem ents : Precast Pan el

1 @J

Design forces for the brackets

~ Body of connection

I

Under §13.5.3 and Tab le 13.5.1 the body of the connection must be designed for = 1.0 and Rp = 2.5. These are the up and Rp values used for the determination of Fp .

Up

Therefore, there is no need to change the load actions due to this force. The bracket must be designed to resist the following sets of load actions.

I

PB = ± 1.38 axial load together with VB = 10.08 kips downward shear

I

and H B = ± 2.77 kips horizontal shear together with

FB+ RB= 1.54 + 10.08 = 11.62 kips downward shear

~ Fasteners Under § 13.5.3, Item d., and Table 13.5.1, fasteners must be designed for u p = 1.25 and Rp = 1.0. Thus, it is necessary to multiply the Fp load actions by ( 1.25)(2.5) = 3.125 because these values were based on ap = 1.0 and Rp = 2.5. Fasteners must be designed to resist

I J

(3.125) P B= 3.125(1.38) = 4.31 kips axial load together with VB

=

10.08 kips downward shear

and

3.125HB= 3.125(2.77) = 8.66 kips horizontal shear together with 3.125FB+ RB = 3.125(1.54) + 10.08 = 14.89 kips downward shear

~ Design forces f or t he rods ~ Body of connection The body of the connection must be designed to resist a force based on ap = 1.0 and R p = 2.5 P R = 1.39 kips axia l load

1 58

20 06 IBC S tru ctural/Seismic Design Manua', Vol . I

I I I I I I I

1 1

1

Exampl e 42 s Exter ior Nonst ructura l Wall Elements: Precast P anel

[!J

§13.5.3

F asteners Fasteners in the connecting system must be des igned to resist a force based on ap = 1.25 and Rp = 1.0 (3. I 25)PR = 3.125(1.38) = 4.31 kips axial load

r I J

I I

I

I I I

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159

§12.1.3

Example 43 •

Beam Horizontal Tie Force

EXa m p le 43 'Bea m H orizonta l Tie Force

§12. 1.3

Th is example illustrates use of the beam inter-connection requirement of §12.1.3. The requirement is to ensure that important parts of a structure are "tied together." Find the minimum required tie capacity for the connection between the two simple beams shown in the example below, The following information is given. Seismic Design Category D SDS

p~~~~~~~ k"

= 1.0

Pin support "p"

Dead Load D = 6 kip/ft Live Load L = 4 kip/ft

[!J

Determine tie force

~ Determine horizontal support force at "P" Calculations and Discussion

[!J

Cod~ Reference

Determine tie force Requirements for ties and continuity are specified in §12,1.3 . For this particular example, it is required to determine the "tie force" for design of the horizontal tie interconnecting the two simply supported beams. This force is designated as Fp , given by the greater value of Fp

= 0.133

SDSW p

or F p = 0.05H'p

where wp is the weight of the smaller (shorter) beam Wp =

160

40 ft (D) = 40(6) = 240 kips

2006 IBC Structural/Seismic Design Manual, Vol. I

Example 43 • Beam Hori zontal Tie Force

1

§12 .1.3

For S DS = 1.0, the controll ing tie force is Fp = 0.133( 1.0)(240) = 31.9 kips

~

Determine horizontal support force at "P" Section 12.1.4 requires a horizontal support force for each beam equal to 5 percent of the dead plus live load reaction. Given a sliding bearing at the left support of the 40-foot beam, the required design force at the pin support "P" is

I

H=0.05(6 klf +4 kIf)

(~O) = 10 kips

I

I I I I I I I

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161

Example 44 • Collector Elemen ts

§12 .10.2

§12.10.2 Collectors "collect" forces and carry them to vertical shear-resisting elements. Collectors are sometimes called drag struts. The purpose of this example is to show the determination of the maximum seismic force for design of collector elements. In the example below, a tilt-up building, with special reinforced concrete shear walls and a panelized wood roof, has a partial interior shear wall on Line 2. A collector is necessary to "collect" the diaphragm loads tributary to Line 2 and bring them to the shear wall. The following information is given .

2

3 100'

Occupancy Category I 50'

Seismic Design Category D

50'

RP=;===ir:====r===u....... Tributary roof area for cctec tor

--l-_-u--Colleclor

R =5.5 no = 2.5 1 = 1.0 8 DS = 1.20 Roof dead load = 15 psf Wall height = 30 ft, no parapet Wall weight = 113 psf

lI=j::============!l Roofplan

k""Shear walls

Nole: Roof framing, except collector, not shown .

By inspection, for the one-story shear wall build ing, Equation 12.8-2 will govern. Base shear = V=

W=

S

(Eq 12.8-2)

DS W = 0.2 l 8W R structure weight above one half hi

Interior shear wall

50'

Collector

Determine the following. Elevation Section A-A

[!J

Collector unfactored force at tie to wall

~. Special seismic load of §12.4.3.2 at tie to wall

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1

Example 44 • Collector Elements

Calculations and Discussion

C!.J

§12.10.2

Code Reference

Collector unfactored force at tie to wall

§12.10.2

The seismic forc e in the collector is made up of two parts : I) the tribut ary out-of-plane wall forces, and 2) the tributary roof diaphragm force. The paneli zed wood roof has been determined to be flexible ; thus the tribut ary roof area is taken as the IOO-foot by 50-foot area shown on the roof plan above. Seism ic forces for collector design are determined from Equat ion 12.10-1 used for diaphragm design. This equat ion reduces to the following for a sing le story structure.

F, =-w,

WI

Fp l max

=

P

0.3 Sp,lWpx =

O.30Wpx

Fp l min = 0.15 Sp,lWp. = 0.15Wp T T

=

=

1.2 S p,

R

= V

W x

design force at roof

= structure wei ght above one half hi = W =

weight tributary to the collector element

giving:

I I I I I

I

V

Fp 1

= -Wpl =

Wpl

=

tributary roof and out-of-plane wall weight

Wp l

=

15 psf(lOO)(50) + 113 psf (3 0) (100) = 75,000 + 169,500 = 244.5 kips 2

W

0.218Wp l

: . Fp1 = 0.218(244.5) = 53.3 kips. Note: This force corresponds to the diaphragm design forces calcu lated using §12.10.1. These forces are compared to the diaphragm shear strength ; including the shea r strength of connection between the diaphragm and collector. The design of the collector and its connections requires that the axial forces be amplified as shown below.

Spec ial seismic load of §12.4.3 .2 at tie to wall

§12.10.2

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163

1

Ex amp le 44 • Coll ec tor El em ents

§12.1 0.2

Given the force Fpl specified by Equation 12.10-1, the collector elements, splices , and their connections to resisting clements shall have the design strength to resist the earthquake loads as defi ned in the Specia l Load Combinations of §12.4.3.2. The governing load combination is 1.2 D + 0.5L + Em

§2.3.2 (Comb . 5)

J

I 1

where (Eq 12.4-5) Here, Q£ is the horizontal collector desig n force

Fpl =

53.3 kips, and

n oQ£ = 2.5(53.3) = 133.25 kips axial tension and compressio n load

I J

0.2 SDsD = 0.2( 1.0)D = 0.2D vertical load

The strength design of the collector and its connections must resist the following load components. n "Q£ = 2.5(53.3) = 133.25 kips axial tension and compression load and vertica l downward load equal to 1.2D + 0.5L + 0.2 D = 1.4 D + 0.5 L

with

I

D = (50 ft + 50 ft)(50 ft)(l5 psf) = 2250 Ib L

= (50 ft + 50 ft)(50 ft) (0.5)(10 psf) = 750 Ib

The resultin g total factored vertical load is 1.4(2250) + 0.5(750)

=

3525 lb

which is applied as a uniform distributed load w = 3525/50 ft = 70.5 plf on the 50-foot length of the collector element.

:'l

Commentary

,.

Note that §12.4.3.1 specifies that the term noQ£ in Equation 12.4-7 need not exceed the maximum force that can be delivered by the lateral-force-resisting system as determined by rational analysis . For example, the overturning moment capacity of the shear wall can limit the required strength of the collector and its connection to the shear wall.

1 64

I

200 6 IBC St ru ctural/Seismic Desig n Manu al, Vol. /

I I I I I

Out-at-Plene Wall Anch orage of Concrete or Masonry Walls to FlexIble Dia phragms

Example 45

§12. 11.2 § 12. 11.2. 1

am le45 Out-of..Plane Wall Anchorage of Concrete or Masonry Walls to Flexible Diaphragms §12.. 11..2 and 12.11.2.1 For the tilt-up wall panel shown below, the seismic force required for the design of the wall anchorage to the flexible roo f diaphragm is to be determined. This will be done for a representative I-foot width of wa ll. The following information is given.

Top of parapet

4'

Occu pancy Importance Category I

F. n: l• +---+

Roof

Seismic Design Category D

J = 1.0 SDS = 1.0 Panel thickness = 8 in Normal weight concrete CI50 pet)

20'

. Assumed pin support

~ ~ Ground

Determine the following.

[!J

Design criteria

~ Wall anchorage force Calculations and Discussion

[!J

Code Reference

Design criteria

§12.11.2.1

Because of the frequent failure of wall/roof ties in past earthquakes, the code requires that the force used to desig n wall anchorage to flexible diaphragms be greater than that given in §12. I 1.2.1 for the desig n of the wall panel sections . The following equation is to be used to determine anchor design forces, with minimum limit given in § 12.11.2.

Fp

=

CEq 12.11-1)

0.8 SDS!ElVl\'

2:: 400 8Ds/ lblft 2:: 280 lb/ ft of wall where

WI\'

is the weight ofa I-foot width of wa ll that is tributary to the anch or.

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1 65

§12.11.2 §12. 11 .2. 1

~

Ex am ple 4 5 • Out-ot-Pten e Wall Anchorage of Concrete or Ma so nry Walls to Fl exible Diaphragm s

I

Wall anchorage force The tributary wall weight is one-half of the weight between the roof and base p lus all the weight above the roof.

IV w

= 150C~)(4 ft + 10 ft)(I

ft) = 1400 lb/ft

)

I

For the given values of Sos = 1.0 and 1= 1.0 , Equation 12. 11-) gives Fp = 0.8( 1.0)( 1.0)wp = 1.2wp

= 0.8(1400 )

)

= 1120 Ib/ft > 400 (1.0)(1.0) = 400 Ib/ft . . . o.k.

> 280 Ib/ft .. . o.k.

:. F"",." = Fp = I 120 Ib/ft This is the QE load in the seismic load combinations.

)

I I I

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Example 46 • Wall Anchorage to Flexible Diaphragms

ample 4 6 all nchorage to Flexible

iap ragms

§12 .11.2.1

§12.11.2.1

Th is example illustrates use of the allowable stress design proced ure for the design of stee l and wood elements of the wall anchorage system in a building with a flexible roof diaphragm. Th e drawing below shows a tilt-up wall panel that is connected near its top to a flexible roof diaphragm. The anchorage force has been calculated per § 12.11.2.1 as Fonch = 1680 lb/ft. The wall anchorage connections to the roof are to be provided at 4 feet on center.

Wall panel

Wall-roof tic detail

Determine the strength' design requirement s for the followi ng.

[IJ

Design force for premanufactured steel anchorage element

~ Design force for wood subpurlin tie element Calculations and Discussion

[IJ

Code Reference

Design force for premanufactured steel anchorage element. The task is to desig n the steel anchorage elements (i.e., hold-downs) that connect the tilt-up wall pan el to the wood subpurlins of the roof diaphragm. The ancho rage consists of two hold-down elements , one on each side of the subpurlin . The manufacturer's catalog provides allowable capacity values for earthquake loading for a given type and size of hold-down element.

The steel hold-down elements of the anchorage system resist only the axial anchorage load and there are no dead or live load effects.

2006 IBC Structural/Se ismic Design Manual, Vol . I

167

§12.1 1.2.1

Example 46 •

Wall An ch orage to Flexib le D i aph ragms

For the 4-foot spacing, the strength design axial load is

J J

E = QE = PE = Faae" (4) = (1680)(4) = ± 6720 Ib This example, uses the ASD load combinations of §2.4, where the applicable seismic load combinations permi t 0.7£ to be resisted with an increase in allowable stress based on duration (i.e., the Cd duration factor for wood) .

]

The allowabl e stress design axial load requirement for each pair of hold-down elements is 0.7£ = 0.7PE 0.7(6720) = ± 4800 lb From the manufacturer's catalog , select a hold-down element having a capacity of at least

J 4800 lb = 2400 Ib 2 The hold-down detail must provide both tensile and compress ive resistance for this load. Whenever hold-downs are used in pairs, as shown in the wall-roof tie detail above, the throug h-bol ts in the subpurlin must be checked for double shear bearing. Also, the paired anchorage embedment in the wall is likely to involve an overlapping pull-out cone condition in the concrete : refer to ACI 3 I8 Append ix D for design requirements. When single-sided hold-downs are used, these must consider the effects of eccentricity. Generally, double hold-downs are preferred, but single-sided hold-downs are often used with all eccentricities fully considered.

~ Design force for wood subpurlin tie element The strength design axial load on the wood element of the wall anchorage system is

PE = (1680)(4) = ± 6720 lb Using the seismic load combinations of §2.4, select the wood element such that the allowable capacity of the element, for the combined bend ing and axial stress including dead and live load effects, can support a ± axial load of

0.7PE = 0.7(6720) = 4800 lb applied at the anchored end.

16 8

2006 IB C S tr uctural/Seismic Design Man ual, Vol . I

I I

I I

I I I I I I

Example 46

1

l:f

Wall Anchorage to Flexible Diap hragms

§12.11.2.1

1-

Commentary For comparison , the forces acting on wood, co ncrete, and steel elements are shown below. For wood, the load is divided by the dura tion fac tor Cd of 1.0 to permit comparison. For stee l, the load is increased by 1.4 per §12.11.2.2.

ASD

Material

Wood

I I

Concrete

Steel

0.8SDsIW 1.6

0.5 SoslW

(0.35 SoslW)

0.8 SoslW

N/A

1.4(0.8 SoslW) = 1.12 SoslW

(0 .78 SoslW)

I I I I I I I

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§12. 10.1.1

Ex ample 4 7

Q

De term i nation of Diaphragm Force Fp ::: Lowrise

~ mple 4 7

!#

ermination of Diaphragm Force f px : Lowrise

§12.'10. 1.1

This exampl e illustrates determination of the diaphragm design force Fpx of Equation 12.10-1, for the design of the roof diaphragm of a single-story building. A single-story tilt-up bui lding with special reinforced concrete shear walls and a panelized wood roof is shown below. This type of roof construction can generally be shown to behave per flexible diaphragm assumptions.

cp

r

200 '

Normal wall

b

o

The following information is given . Occupancy Importance Category I Seismi c Design Category 0 Roof plan

1.0 S DS = 1.0 R =5.0 P = 1.0 Diaphragm weight = 15 psf Wall weig ht = 80 psf

J

=

Roof diaphragm

Elevation through building

Find the following.

[!J 170

Diaphragm design force at the roof

2006 IBC Structural/Seismic Design Manual, Vol. I

Examp le 47 • Det ermin ation of Diaphragm Forc e F px: Lowrise

Calculations and Discussion

[!J

§12.10.1. 1

Code Reference

Diaphragm design force at the roof

§12.10.1.1

§12.10.1.1 requires that the design seismic force for diaphragms be determined by n

Fp:r =

1

2: F, 2:

~= l t,'px n

(Eq 12.10- 1)

lV j

i-.t

with limits of 0.2

SDs/Wp.t :::: Fp.t:::: 0.4 SDs/Wp.,

which for S DS =

I I I I I I I I I

1.0 and 1 = 1.0

are

For a short period single story building, Equation 12.10-1 becomes (see commentary below for derivation)

with the given val ues of SDS = 1.0, R = 5.0 and, for a l-foot-w ide strip Wpl

F pl

= weight of diaphragm

+ weight of one-half height of normal walls

=

IOO( 15) + 2( I0)(80) = 3100 Ib/ft

=

(1.0)(1.0) 5.0 11'p.t = 0.2

Wpl

= 0.2(3 100) = 620 lb/ft

Check limits: 0.2wp.t < 0.2wp ) < O.4wp.t •. . o.k.

Note that the redundancy factor of p is to be applied to the Q£ load actions due to Fp l (such as chord forces and diaphragm shear loads in the diaphrag m).

2006 IBC Structura l/Seismic Design Manua l, Vol. I

17 1

§12.10.1. 1

1

Example 47 • Determ ination of Diaphragm Force Fpx: Lo wrise

Commentary I. The we ight W p., includ es the weight of the diaphragm plu s the tributary weight of elem ents nonnal to the diaphragm that are one-half story height belo w and above the diaphragm level. Walls parallel to the direction of the seismic forces are usually not considered in the.detennination of the tributary roof weight because these walls do not obta in support, in the direction of the force, from the roof diaphragm. 2. The sing le-story building version of Equation 16-65 is derived as follows . m

2: F, Fps

= ~w •• px

2:

(Eq 12.10-1)

Wi

F X = CIX V =

]

I I I

v • n

w,h,

2: W/l:

(Eq 12.8-11 )

i.. 1

where

C,_., =

- J1.(•

}\'.r

for short period of < 0.5 sec (k = J .0) .

For a single-story building, i

=

I , x = 1, and

11

= I

11';

=

I I I

I

2:

(Eq 12.8-12)

W

i- I

and Equation 12.8- 11 gives

F) =

W/I,

V= V

w,lz,

I I 1

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2006 IBC Structural/Seismic Des ign Manual, Vol. I

I I I

1

~

Examp le 47

Determ ination of Diaphragm Forc e Fpx: Lowrise

§12. 10. 1. 1

where

1 I

1

V= C W = 50 S ! IV

(Eq 12.8-1 and 12.8-2)

R

5

Finally, for the single story building, Equation 12.10-1 is F

F,

P1

= -

IV

11'

1'1

= -

V

IV

lV

s!

1'1

5 = -0 - 11'

R

1'1

I I I

\

I

2006 IBC Structural/Seismic D esign Manual, Vol. I

17 3

§12.10.1

Example 48

D

Determ ination of Diaphragm Force F px : H ighrise

ample 48 ,De termina tion of Diaphragm Force Fp x : Highrise t-

§12. 0.1

This example illustrates determination of the diaphragm design force Fpx of Equation 12. 10- 1 for a representative floor of a multi-story building. The nine-story moment frame bui lding shown below has the tabulated design seismic forces P.r:. These were determined from Equations 12.8-11 and 12.8-12 , the design base shear. The following information is given. Seismic Design Category 0

W = 3,762 kips C, = 0.06215 8Ds = 1.0 P = 1.3 I = 1.0 T = 1.06 sec V = CW= 233.8 kips k = 2 for Eq 12.8-12

Level

1
;

2 1

30 20

30 20

100 200

3000 4000

0.429 0.571

300

7000

1.00

c,

Story Shear

F,

v,

k

Level

Note: 11., in feet Wx in kips = WxlzJk

Story Force

51.25 68.45

51.35 119.7

Sa 0.516 0.342

119.7

fx= c, (1 19.7 kips) ~

k

1-rr,lzx

Sa = Fx / fV, .. nPSa = effective story acceleration

2006 IBC StructuraUSe;sm;c De sign Manual. Vol. I

187

§15.7.6

Example 53 •

Tank with S up p or t ed Bottom

, ~ample 53

.!In

With Supported Bottom

§15.7.6

A small liquid storage tank is supported on a concrete slab. The tank does not contain toxic or explosive substances. The following informati on is given. SDS=

I W

I3J 'C~/c;yla tions

[!J

1.20

= 1.0 Weight of tank and max imum normal operating contents = 120 kips = 0.50 inch =

20'

Slab Grade

Find the design base shear

and Discussion

§15.7.6

Code Reference

The tank is a nonbu ilding structure, and seismic requirements for tanks with supported bottoms are given in §15.7.6. This secti on requires that seismic forces be determined using the procedures of §15.4.2. The period may be computed by other rational methods, similar to Example 51

where = 20 ft L D = 10ft LID = 20/10 = 2.0 w = W/L = 120,000 Ib/20 = 6000 plf = 0.50 in I 6000(10) wd 1,440,000 (0.50/12) t

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2006 IBC Struc tu r al/S ei smi c D es ig n Manua l, Vol . I

1

Example 53 • Tank With Supported Bottom

Now:

1

T

§15. 7. 6

7.65 X 10-6 (2.0)2 (1,440 ,000) = 0.0367 sec < 0.06 . .. rigid =

Thu s, rigid nonbuilding structure, § 15.4.2 The lateral force shall be obtai ned as follows

V =0.3SDsIW=0.36W =

(Eq 15.4-5)

0.36 ( 120) = 43.2 kips

where

SDS= 1.20 I = 1.00 W = 120 kips The design lateral seismic force is to be applied at the center-of-mass of the tank and its contents. Note that the center-of-mass of the contents and of the tank do not normally coincide. The distribution of forces vertically shall be in accordance with § 12.8.3.

Commentary

I

The procedures above are intend ed for tanks that have relati vely small diameters (less than 20 feet) and where the forces generated by fluid-sloshing modes are small. For large diameter tanks , the effects of sloshing must be considered. Refer to American Wate r Works Association Standard ANSI! AWWA D100 "Welded Steel Tanks for Water Storage," or American Petroleum Institute Standard 650, "Welded Steel Tanks for Oil Storage" for more detailed guidance.

I I I

I I

2006 lac Structural/Seismic Design Manual. Vol. I

189

IBC §180B.2.23.1

Example 54 •

Pile Interconn ections

ample 54 Pile Interconnections

IBC'§1808.2.23.1

A two-story masonry bearing wall structure has a pile foundation, Piles are located around the perimeter of the building. The foundation plan of the building is shown below. The following information is given. Original grade

Seismic Design Category 0 J = 1.0 SDS = 1.0 Pile cap size: 3 feet square by 2 feet deep Grade beam: I foot 6 inches by 2 feet Allowable lateral bearing = 200 psf per foot of depth below natural grade, for the very dense granular soil at the site.

2'·0"

2'· 0"

Section A-A: Typi cal pile cap Pile Cap

Dead Load

Reduced Live Load

3 10

46 kips 58

16 kips 16

t

~ A

a 100

Negative Posi tive

None Required None Required 10 sf -36.&

I

20 sf -30.5

12 sf -35.5

None Required

+4 .7

x 1.00

x 1.00

x 1.00

+4.7*

x 1.00

x 1.00

-15.&

sf

Negati ve

None Required

-15.&

x 1.00

None Required

< 10 sf

Posi tive Negat ive

14.6 -15.8

x 1.00 x 1.00

x 1.00 x 1.00

x 1.00 x 1.00

14.6 - 15.&

+ 14. 1

x 1.00

x 1.00

x 1.00

+ 14. 1

- 15.3

x 1.00

x 1.00

x 1.00

- 15.3

+ 14.6 -1 9.5

x 1.00 x 1.00

x 1.00 x 1.00

x 1.00 x 1.00

+ 14.6 -1 9.5

+14.1

x 1.00

x 1.00

x 1.00

+14. 1

- 1&. 6

x 1.00

x 1.00

x 1.00

- 18.6

Positive

Stud

17.3 sf

Siding

< 10 sf

Negative Positive Neg ative

None Required 20 sf 10 sf 17.3 sf + 14.6 + 13.9 + 14. 1 10 s f 20 sf 17.3 sf - 15.& - 15.1 -15.3

None Required

None Required 10 s f 20 sf 17.3 sf Posi tive Int + 14.6 + 13.9 + 14.1 Stud 17.3 sf (4) 17.3 sf 10 s f 20 sf Negative - 19.5 - 1&.2 -1 &. 6 • Note. A minimum pressure of 10 psf 15 required per§6.4.2.2.1

198

x 1.00

-1 3.3

-;;

:::

+5.&

None Required None Required

Int (4)

+5 .9· -14.6

None Required

> 100

Siding

x 1.00 x 1.00

Positive Negative Positive

Joist

(3)

x 1.00 x 1.00

Neg ative

Positive Comer

x 1.00 x 1.00

sf

12 sf

Screw

20 sf 12 sf 10 sf +5.6 +5.9 +5. & 10sf 20 sf 12 sf - 14.6 - 14.2 - 14.5 No ne Req uired

+5.9 -14.6

x 1.00

Roof Deck

Deck

Positive

None Required Non e Required

2006 IBC StructurallSelsmlc Design Manual, Vol. I

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1

I )

)

1

Examp / a 55 •

Si mplified Win d Load s o n 2· S tory B uil din gs

§ 6.4

The component and cladding pressures should be applied as described in Figure 6-3 and as shown in the diagram below.

O

Interior Zones

Q

End Zones

Roofs- Zone 1f\Nalis- Zone 4

'.,; Roofs- Zone 21Walls - Zone 5 Comer Zones Roofs- Zone 3

I 1 1

I

2006 IBC StructuraUSeismic Design Manual, Vol. I

199

Example 56 •

§6.4

Simplifi ed Wind Loads on Low R ise Buildings

Per §6.4. 1.1 , for conform ing low-rise bui ldings, wind loads can be determined using simplified provisions.

The following information is given .

A

B

c

/

1 -

I

2 -

Main windforce-res isting system

3-story office build ing located in urban/subu rban area ofNW Texas - situated on fiat ground b

(0

Typ

3 -

1
C = - 0.5 50 p B L

-

q" = q " '

ll 6 fi

F6-6

= 18.2 psf

q"GCp = 18.2 (0.85)(- 0.5) = -7.74 psf

I I I I I I

Side walls

c,

=-0.7

F 6-6

q"GCp = 18.2 (0.85)(- 0.7) = -1 0.8 psf Roof 11 6 = =2.3 > 1.0 L 50 h

-

c,

= - 1.3 x 0.8 (Area Reduction Factor) = 1.04

F 6-6

q"GCp = 18.2 psf(0.85)(xI.04) = x 16. 1 psf

lid·1Design wind loads Main wind-force-resisting system Rigid building

208

2006 IBC Structura l/Seismic Design Man ual, Vol . I

§6.5.12 §6.5.l 2.2 §6.5. 12.2.1

I I I I I I I I

Example 58. Floor Vibra:ions

CEq 6-17) Windward wall qh(GCp i ) = (18.2)(0. 18) =3.28 psf(±) Ii

0-15 ft

20 25 30

40 50

60 70

80

90 100 116

p = CJ=GCp - Q1J(GCp ;)

Case 1 shown

10. 1 10.7 11.2 11.6 12.4 13.0 13.5

14.0 14.5 14.8 / Sample Calculation 15 I P = 12.4 - 1B.2(-0.18) = 15.7 Case 1 .

15.7

12.4 - 18.2(+0.18)

=9.1 Case 2

Leeward wall p = q"GCp - qh (GCp i ) p = - 7.74 -1 8.2(-0.1 8) = - 4.5 psf

Case 1

p = - 7.74 -1 8.2(0.18) = -I 1.0 psf

Case 2

Side walls

= -10.8 - 18.2(0.18) = - 14. I psf Roof

= -1 6.1 -18.2(0.18) = - 19.4 psf

2006 IBC Structural/Seismic Design Manual, Vol. I

209

§6.5

Example 57 a Wind Loads - Analytical Procedure

11

e.1

Design wind loads - graphically r--r--r---r---,. 19.4 ps f Cas e 1 15.7 psI Case 2 9.1 psI

Wind

.....----:J-f---'--L---L.---l-~

11.0 psf

~I--~

4.5 psf Case 1 11.0 psI Case 2

14.1 psf

Plan Wind -----,)

Elevation Case 1: Internal Pressure Inwa rd Case 2: Internal Pressu re Outward

Verify projected load is greater than 10 psf 10.1 + 11.0 = 21.1> 10 psf. . .o.k.

§6.1.4.1

To obtain frame loads, multiply pressures by tributary width = 50/2 = 25 ft or perform Rigid Diaphragm Analysis

210

2006 IBC Structural/Seismic Design Manual, Vol. I