ibc 2006 structural seismic design manua whit examplesl 1 220p

COD APPLICATUO EXAMPLES

']

']

11']

'1:I,)

:)r I,I,1

I.1

IIIII

J

III

III

I

II

Table of Contents

CopyrightlPublisher/Editor/Disclaimer ii

Preface VlJ

Acknowledgments ix

Suggestions for Improvement / Errata Notifi cation xi

Introduction I

How to Use This Document 2

Notation 3Definitions 18

EXAMPLE DESCRIPTION ASCEISEI 7-05 PAGE

Example i Classification/Importance Factors §11.5-1 25

Seismic Design Category §11.6 26Example I Earthquake Load Combinations:

Strength Design §12.4.2.3 27

Example 2 Comb inations of Loads §2.4 32

Example 3 Design Spectral Respon se Accelerations §11.4 36Introduction to Vertical Irregularities §12.3.2.2 41

Example 4 Vertical Irregularity Type la and Type Ib §12.3.2.2 42

Example 5 Vertical Irregularity Type 2 §12.3.2.2 46

Example 6 Vertical Irregularity Type 3 §12.3.2.2 48

Example 7 Vertical Irregularity Type 4 §12.3.2.2 50Example 8 Vertical Irregular ity Type 5a §12.3.2.2 52

Example 9 Vertical Irregularity Type 5a §12.3.3.1 54

Introduction to Horizontal Irregularities §12.3.2.1 58

Example 10 Horizontal Irregularity Type Ia and Type Ib .. §12.3.2.1.. 59

Example I I Horizontal Irregularity Type 2 §12.3.2.1 63

Example 12 Horizontal Irregularity Type 3 §12.3.2.1 65

Example 13 Horizontal Irregularity Type 4 §12.3.2.1 67Example 14 Horizontal Irregularity Type 5 §12.3.2.1 68Example 15 Redundancy Factor p §12.3.4 69

Example 16 P-delta Effects §12.8.7 74

Example 17 Seismic Base Shear §12.8.1... 78

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

EXAMPLE DESCRIPTION ASCE/SEI 7-05 PAGE

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

Example 18Example 19Example 20Example 21

Example 22

Example 23Example 24Example 25Example 26Example 27

Example 28

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

Example 36Example 37Example 38

Example 39

Example 40Example 4 1

Example 42Example 43Example 44Example 45

Example 46Example 47

Example 48

Example 49

Appro ximate Fundamental Period §12.8.2.I 80Simplified Alternative Structural Design Procedure §I2.14 83

Combination of Structural Systems: Vertical §I2.2.3.I 86Combination of Framing Systems:in Different Directions §I2.2.2 90Combin ation of Structural Systems:Along the Same Axis §I 2.2.3.2 92Vertical Distribution of Seismic Force §12.8.3 93Horizontal Distribution of Shear §12.8.4 97Amplification of Accident al Torsion §I2.8.4.3 102Elements Supporting Discontinuous Systems §I2.3.3.3 106Elements Supporting Discontinuous Walls or Frames §12.3.3.3 I 10

Soil Pressure at Foundations §2.4§I2.I3.4 I I3

Drift §12.8.6 I 16Story Drift Lim itations §12.12 I 19

Vertical Seismic Load Effect. §12.4.2.2 121Design Response Spectrum §11.4.5 124Dual Systems §12.2.5. I 126

Lateral Forces for One-Story Wall Panels §12.11 129Out-of-Plane Seismic Forces for Two-Story Wall Panel §12.11. I

§I2.11.2 133Rigid Equipment.. §I3.3.1 137

Flexible Equipment §13.3.1 140Relative Motion of Equipment Attachments §I3.3.2 143Deformation Compatibility for Seismic DesignCategories D, E, and F .· § I2.12.4 145Adjoining Rigid Elements §12.7.4 148Exterior Elements: Wall Panel §I3.5 .3 150Exterior Nonstructural Wall Elements: Precast Panel. §13.5.3 153Beam Horizontal Tie Force §12.1.3 160Collector Elements §12.10.2 162

Out-of-Plane Wall Anchorage of Concrete or Masonry §12.11.2Walls to Flexible Diaphragms §12.11.2.1 165Wall Anchorage to Flexible Diaphragms §12.11.2.1 167Determination of Diaphragm Force Fpx :

Lowrise §12.10 .1.1 170

Determination of Diaphragm Force Fpx :

Highrise §12.10.1 174Building Separations §12.12.3 176

IIIIIIIIIIIIIIII

1

Table of Conten ts

EXAMPLE DESCRIPTION ASCE/SEI 7-05 PAGE

I

IIIIII

III

Example 50

Example 51

Example 52


Example 55


Flexible Nonbuilding Structure §15.5 179

Lateral Force on Nonbuilding Structure §15.0 182

Rigid No nbuilding Structure §15.4.2 186

Tank With Supported Bottom §15.7.6 188

Pile Interconnections IBC §1808.2.23.1 190

Simplified Wind Loads on 2-Story Buildings §6.4 193

Simplified Wind Loads on Low-Rise Buildings §6.4 200Wind Loads - Analytica l Procedure §6.5 205

2006 IBC Structural/Seismic Design Man ual, Vol. I V

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

I

I

I

II

IIIIIII

I

II1

IIIII

Preface

This document is the initial volume in the three-volume 20061BC Structural/Seismic DesignManual, It has been developed by the Structural Engineers Association of California(SEAOC) with funding provided by SEAOC. Its purpose is to provide guidance on theinterpretation 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 RecommendedLateral Force Requirements and Commentary (also called the Blue Book).

The 2006 lBC Structural/Seismic Design Manual was developed to fill a void that existsbetween the commentary of the Blue Book, which explains the basis for the code provisions,and everyday structural engineering design practice . The 2006 lBC Structural/SeismicDesign Manual illustrates how the provisions of the code are used. Volume 1: CodeApplication Examples, provides step-by-step examples for using individual code provisions,such as computing base shear or building period. Volumes 1I and lIl: Building DesignExamples, 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 realisticdesign.

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

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

SEAOC, NCSEA and ICC intend to update the 2006 lBC Structural/Seismic Design Manualwith each edition of the building code.

Jon P. Kiland and Rafael SabelliProject Managers

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

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

1

J

J

IIIII

Acknowledgments

The 2006 IBC Structural/Seismic Design Manual Volume J was written by a group of highlyqualified structural engineers. They were selected by a steering committee set up by theSEAOC Board of Directors and were chosen for their knowledge and experience withstructural engineering practice and seismic design. The consultants for Volumes I, II, and IIIare:

Jon P. Kiland, Co-Project ManagerRafael Sabell i, Co-Project ManagerDouglas S. ThompsonDan WerdowatzMatt Eatherton

John W. LawsonJoe MaffeiKevin MooreStephen Kerr

A number of SEAOC members and other structural engineers helped check the examples inthis 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 thenumerical computations.

I ' _.

'l,. ,- I

•• L...

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

ICC

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

Sugges tion s for Impro vem ent

Inkeep ing with SEAOC's and NCSEA's Mission Statements: "to adva nce the structuralengineering profession" and "to provide structural engineers with the most currentinformation and tools to improve their practice," SEAOC and NCSEA plan to update thisdocument as structural/seismic requirements change and new research and betterunderstand ing of building performance in earthquakes becomes ava ilable.

Comments and suggestions for improvements are welcome and shou ld be sent tothe following:

Structural Engineers Association of Cal ifornia (SEAOC)Attention : Executive Director14 14 K Street, Suite 260Sacramento, California 95814Telephone: (9 16) 447-1198 ; Fax: (916) 932-2209E-mail: leeiWseaoc.org; Web address: www.seaoc .org

IIIII

x

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

2006 IBC Structural/Seismic DesIgn Manual, Vol. I

I

IIIIJ

II

I

1 Introduction

I1

IIIIIIII

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

The 2006 !BC has had structural provisions removed from its text and has referenced severalnational standards documents for structural design provisions. The primary referenceddocument is ASCE/SEI 7-05, which contains the "Minimum Design Loads for Buildings andOther Structures." ASCE/SEI 7-05 is referenced for load and deformation design demands onstructural elements, National Material design standards (such as ACI, AISC, MSJC andNOS) are then referenced to take the structural load demands from ASCE/SEI 7-05 andperform specific materia l designs.

Volume I presents 58 examples that illustrate the application of specific structural/seismicprovisions of the !Be. Each example is a separate problem, or group of problems, and dealsprimarily with a single code provision. Each example begins with a description of theproblem to be solved and a statement of given information. The problem is solved throughthe normal sequence of steps, each of which is illustrated in full. Appropriate code referencesfor 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 buildingsillustrat 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 concreteshear wall building.

While the 2006 lBC Structural/Seismic Design Manual is based on the 2006 !BC, there aresome provision ofSEAOC's 2005 Recommended Lateral Force Provisions and Commentary(Blue Book) that are app licable. When differences between the !BC and Blue Book aresignificant they are brought to the attention of the reader.

The 2006 lBC Structural/Seism ic Design Manual is intended for use by practicing structuralengineers and structural designers, building departments, other plan review agencies , andstructural engineering students.

2006 IBC Structural/Seismic Design Manual, Vol. I 1

Ho w to Use This Do cum ent

• :JC . . '>;

The various code applicat ion examples of Volume I are organized by topic consistent withprevious editions. To find an example for a particular provision of the code, look at theupper, outer come r of each page, or in the table of contents.

Generally , the ASCE/SEI 7-05 notat ion is used throughout. Some other notation is defined inthe followi ng pages, or in the examples.

Reference to ASCE/SEI 7-05 sections and formulas is abbreviated. For example, "ASCE/SEI7-05 §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" isunderstood 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 inthe right-hand margin under the category Code Reference.

Generally, the examples are presented in the following format. First, there is a statementof the example to be solved, including given information, diagrams, and sketches. This isfollowed by the "Calculations and Discussion" section, which provides the solution to theexample and appropriate discussion to assist the reader. Finally, many of the examples havea third section designated "Commentary." In this section, comments and discussionon the example and related material are made. Commentary is intended to provide a betterunderstanding of the example and/or to offer guidance to the reader on use of the informationgenerated 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 thereferenced Standard ASCE/SEI 7-05 unless otherwise indicated. Occasionally, reference ismade to other 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.

2 2006 lac Structural/Seismic Design Manual, Vol. I

]

1

IIII

IIIIIII)

II

Notation

The following notations are used in this document. These are generally consistent with thoseused in ASCE/SEI 7-05 and other Standards such as ACI and AISC. Some new notationshave also been added. The reader is cautio ned that the same notation may be used more thanonce and may carry entire ly different meanings in different situations, For example, E canmean the tabulated elastic modulus under the AISC definition (steel) or it can mean theearthquake load under §12.4.2 of ASCE/SEI 7-05 . When the same notation is used in two ormore 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 anchor, in square inches

the combined effective area, in square feet, of the shear wallsin the first story of the structure

AI

Ag =

Ag =

Ai =

IIIIIIIII

A ch

Ae

=

=

=

=

cross-sectional area of a struct ural member measured out-to-out oftransverse reinforcement

net area of concrete section bounded by web thickness and length ofsection in the directio n of shear force considered

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

flange area

gross area of section

the gross area of that wall in which ADis identified

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

area of the load-carrying foundation

the effective area of the projection of an assumed concrete failuresurface upon the surface from which the anchor protrudes , in squareinches

area of non-prestressed tension rein forcement

2006 lac Stru ctural/Seismic Design Manual, Vol. I 3

Nota tion

Ash = total cross-sectional area of transverse reinforcement (including ]supplementary crossties) having a spacing s" and crossing a sectionwith 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 IAs, = area of link stiffener IAT = tributary area

Av = area of shear reinforcement within a distance s, or area of shearreinforcement perpendicular to flexural tension reinforcement within adistance s for deep flexural members

A,'J = required area of leg reinforcement in each group of diagonal bars in adiagonally reinforced coupling beam

Avr = area of shear-friction reinforcement

IAll' = (web) link web area

Aw = (weld) effective cross-sectional area of the weld IA., = the torsional amplification factor at Level x - §12.8.4.3

Ia = (concrete) depth of equivalent rectangu lar stress block

a = (concrete spandrel) shear span, distance between concentrated load Iand face of supports

ae = coefficient defin ing the relative contribution of concrete strength to Iwall strength

ad = incrementa l factor relating to the P-delta' ef fects as determined in I§12.8.7

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

4 2006 IBC Structural/S eismic Design Manual, Vol. I

I Notation

bit member width-thickness ratio

Cd = deflection amplification factor as given in Tables 12.2-1 or 15.4-1 or15.4-2

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, snow thermal factor

C1'X = vertical distribution factor - §12.8.3

c distance from extreme compression fiber to neutral axis of a flexuralmember

D dead load, the effect of dead load

I De = the length, in feet, of a shear wall in the first story in thedirection parallel to the applied forces

I Dh gross weight of helicopter

Dp = relative seismic displacement that a component must be designed to

I accommodate - §13.3.2

d effective depth of section (distance from extreme compression fiber to

I centroid of tension reinforcement)

db = (anchor bolt) anchor shank diameter

I db = (concrete) bar diameter

I d, = column panel zone depth

E = (steel) modulus of elasticity

IE combined effect of horizontal and vertical earthquake-induced forces

I(§12.4)

seismic load effect including overstrength factors (§§12.4.3.2 andEm =

I12.14.2.2.2)


N otation I£1 = flexural sti ffness of compression member 1s, = modules of elasticity of concrete , in psi

IE., = (concrete) modulus of elastic ity of reinforcement

e = EBF link length

F = load due to fluids

Fa = site coefficient defined in §11 .4.3

Fa = axial compressive stress that would be permitted if axial force aloneexisted

Fa = flood load IFb = bending stress that would be permitted if bending moment alone Iexisted

FaM = nominal strength of the base material to be welded IFexx = classification number of weld metal (minimum specified streng th)

IFi,Fu,P, = portion of seismic base shear, V, induced at Level i.n, or x asdetermined in §12.8.3.

IFp = seismic force, induced by the parts being connected, centered at thecomponent's center of gravity and distributed relative to the

Icompon ent's mass distribution, as determined in §12.8.3

Fp, = the diaphragm desig n force

IF" specified minimum tensile strength, ksi=

F" = through-thickness weld stresses at the beam-column interface IFill = minimum specified tensile strength of the anchor

IF,. long period site coefficient (at 1.0 second period) see §11.4.3=

F, = the design lateral force applied at Level x Ir, = the lateral force induced at any Level i - §12.8.3

IFw = (stee l LRFD) nominal strength of the weld electrode material

FII' = (steel ASD) allowable weld stress I6 2006 1BC Structura l/Seis mic Design Manual, Vo/./

notation I

he = assumed web depth for stability

h;, hn,h, = height in feet above the base to Level i , 11or x, respectively Ih, = height in feet of the roof above the base Ihsx = 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 accordance with §11.5.1 II = moment of inertia of section resisting externally applied factored loads

I a = moment of inertia of cracked section transformed to concrete

I g = (concrete, neglecting reinforcement) moment of inertia of grossconcrete section about centroidal axis

t; = moment of inertia of reinforcement about centroidal axis of member Icross section .

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

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

t, = component importance factor that is either 1.00 or 1.5, as determined Iin §13.3.1

K = (steel) effecti ve length factor for prismatic member Ik = a distribution exponent - §12.8.3

IL live load, except rooflive load, including any permitted live load=reduction (i.e, reduced design live load). Live load related internal

Imoments or forces. Concentrated impact loads

Lo = unreduced design live load

ILb = (steel) unbraced beam length for determining allowable bending stress

Lp = limiting laterally unbraced length for full plastic flexural strength, Iuniform moment case

L, = roof live load including any permitted live load reduction I8 20061BC Structural/Seismic Design Manual, Vol. I

Notation

Ie (steel RBS) length of radius cut in beam flange for reduced beam

1section (RBS) design

t: length of a compression member in a frame, measured from center tocenter of the joints in the frame

/;, = distance from column centerl ine to centerline of hinge for reducedbending strength (RBS) connection design

I" = clear span measured face-to-face of supports

I" unsupported length of compression member

Ill' = length of entire wall, or of segment of wall considered, in direction ofshear force

Level i

Leveln =

Level,r =

level of the structure referred to by the subscript i." i = I" designates the first level above the base

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

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

M = (steel) maximum factored moment

factored moment to be used for design of compression member

IIIIII

Mer =

"1;- =

moment at centerline of column

moment at which flexural cracking occurs in response to externallyapplied loads

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

moment at face of column

(concrete ) modified moment

(steel) maximum moment that can be resisted by the member in theabsence of axial load

(steel) nominal moment strength at section

(concrete) required plastic moment strength of shearhead cross section

2006 IBC Stru ctural/Seismic Design Manual, Vol. J 9

Notation

u, = (steel) nominal plastic flexural strength, FyZ

Mpa = nominal plastic flexural strength modified by axial load

1u; = nominal plastic flexural strength using expected yield strength of steel

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

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

M, = (concrete) moment due to loads causing appreciable sway

M, = torsional moment IM,a = accidental torsional moment

M" = (concrete) factored moment at section

IM" = (steel) required flexural strength on a member or joint

M,•. = moment corresponding to onset of yielding at the extreme fiber from Ian elastic stress distribution

MI = smaller factored end moment on a compression member, positive if Imember is bent in single curvature, negative if bent in doublecurvature

IM) larger factored end moment on compression member, always positive=

N = number of stories IP = ponding load

IP = (steel) factored axial load

P = (wind) design wind pressure IPDL, i» ,r.; = unfactored axial load in frame member I

Pb = nominal axial load strength at balanced strain conditions

Pbl = connection force for design of column continuity plates Ir , = (concrete) critical load


Notation

Pc = (concrete anchorage) design tensile strength

P" = nominal axial load strength at given eccentricity, or nominalaxial strength of a column

Po = nominal axial load strength at zero eccentricity

P si = FyA

p" (concrete) factored axial load, or factored axial load at giveneccentricity

p" = (steel) nominal axial strength ofa column, or required axial strengthon 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),Ag

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

I R = rain load

R The response modification factor from Table 12.2-1

IR" nominal strength

1Rp = component response modification factor that varies from J.00 to 3.50

as set forth in Table J3.5- J or Table J3.6-1

I R" required strength

IR)' = ratio of expected yield strength F)'cto the minimum specified yield

strength Fy

IR, Rl R2 = live load reduction in percent - !Be §§1607.9.2/1607. J1.2

r rate of reduction equal to 0.08 percent for floors

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

Ir = radius of gyration about y axisy


Notation

S = snow load ISo = design spectral response acceleration

I= 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 shortperiod (i.e., 0.2 seconds) = (2/3) S,«.. - §11.4.4

S, = Mapped, MCE, 5% damped, spectral acceleration parameter at shortperiods (i.e., 0.2 seconds) as determined by §11.4.1

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

SI = Mapped, MCE, 5% damped, spectral acceleration parameter for al-s econd period as determined in §11.4.1

s'IIS = MCE, 5% damped, spectral response acce leration parameter for short Iperiods (i.e., 0.2 seconds) = FoS,. adjusted for site class effects

IS,I/I = MCE, 5% damped , spectral response acceleration parameter forl-second period = F•.SJ, adjusted for site class effects

ISRBS = section modulus at the reduced beam section (RBS)

S = spacing of shear or torsion reinforcement in direction parallel to Ilongitudinal reinforcement, or spacing of transverse reinforcementmeasured along the longitud inal axis

IT = self-straining force arising from contracti on or expansion resultingfrom temperature change, shrinkage, moisture change, creep in

Icomponent materials, movement due to differential settlement orcombinations thereof

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

To = approximate fundamental period as determined in accordance with I§12.8.2.1

To = 0.2 (SOl 1Sos) IT, = SOl 1Sos IIf = thickness of flange

tw = thickness of web

12 2006 IBC Structural/Seismic D esign Manual, Vol. I

1 Notation

I; ratio of expected yield strength F,.<to the minimum specified yieldstrength F."

U = required strength to resist factored loads or related intemal momentsand forces

V the tota l design seismic lateral force or shear at the base of the buildingor structure

Vc = (concrete) nominal shear strength provided by concrete

I Vc (concrete anchorage) design shear strength

VDL, Vu , V,ei, = unfactored shear in frame member

v,,, = shear correspond ing to the development of the "nominal flexural

I strength - calculated in accordance with Chapter 19"

v, (concrete) nominal shear strength at sectionI

V" (steel) nomina l shear strength of a member

Vp = (steel) shear strength of an active link

Vpa = nominal shear streng th of an active link modified by the axial loadmagnitude

IVpx the portion of the seismic shear force at the level of the diaphragm,

required to be transferred to the components of the vertical seismic-lateral-force-resisting system because of the offsets or changes in

Istiffuess of the components above or below the diaphragm

V, = (concrete) nominal shear streng th provided by shear reinforcement

I V, (steel) shear strength of member, 0.55 Fidt

V" = (concrete anchorage) required shear strength from factored loads

IV" (concrete) factored shear force at section=

I v, = (loads) factored horizontal shear in a story

I V" (steel) required shear strength on a member

v. = the seismic design story shear (force) in story x, (i.e.• between Level xandx-I )

2006 IBC Structural/Seismic Design Manual, Vol. I 1 3

Notation Iw = the total effective seismic dead load (weight) defined in §12.7.2

and §12. l4.8. l

W = (wind) load due to wind pressure 1WI' = component operating weig ht

l1'c = weights of concrete, in pcf

11';, 11-'-r = that portion of W located at or assigned to Level i or x, respectively

wp = the weigh t of the smaller portion of the structure Iwp = the weight of the diaphragm and other elemen ts of the structure

tributary to the diaphragm IlVp:c = the weigh t of the diaphragm and elements tributary thereto at

Level x, includ ing applicable portions of other loads defined in 1

§12.7.2 I

) VII' = weight of the wall tributary to the anchor I11,= = column panel zone width

IX = height of upper support attachment at Level x as measured from thebase

y = height of lower support attachm ent at Level Yas measured from thebase

IZ (steel) plastic section modulus=

z = height in structure at point of attachment of component, §13.3.1 IZeDs = plastic section modu lus at the reduced beam section (RBS)

I(concrete) capacity-reduction or strength-reduction factor<I> =

<l>b = (steel) resistance factor for flexure I<l>c = (steel) resistance factor for compression I<1>,.. = resistance factor for shear strength of panel-zone of beam-to-

column connections Ia = (concrete) angle between the diagonal reinforcement and the

longitudinal axis of a diagonally reinforced coupling beam

14 2006 IBC Stru ctural/Seismic Design Manu al, Vol. I

Notation

1a,~ (steel) centroid locations of gusset connection for braced frame

diagonal

Uc coefficient defining the relative contribution of concrete strengthto wall strength

~c ratio of long side to short side of concentrated load or reaction

I area

= the ratio of shear demand to shear capacity for the story betweenLevel x and x- I

P = a redundancy factor determined in accordance with §12.3.4

P = (conc rete) ratio of nonprestressed tension reinforcement (As/bd)

Ph reinforcement ratio producing balanced strain conditions

I pn = ratio of area of distributed reinforcement parallel to the plane of

J Ac" to gross concrete area perpendicular to the reinforcement

ps = ratio of volume of spiral reinforcement to total volume of core(out-to-out of spirals) ofa spirally reinforced compressionmember

I p" ratio of area of distr ibuted reinforcement perpendicular to theplane ofA«,to gross concrete area Ac"

I = lightweight aggregate concrete factor; 1.0 for normal-weightconcrete, 0,75 for "all lightweight" concre te, and 0,85 for "sand-

I lightweight" concrete

Ap = limiting slenderness parameter for compact element

Ifo = length of radius cut in beam flange for reduced beam section

I(RBS) connection design

distance from column centerline to centerline of hinge for RBSflo

Iconnection design

f n = clear span measu red face-to-face of supports

I f u = unsupported length of comp ression member

f ll' length of entire wall or of segment of wall considered in directionof shear force

2006 IBC Structural/Seismic Design Manua l, Vol. I 1 5

Notation II

1-1 coefficient offriction

8 = design story drift, shall be computed as the differences of the 1deflections at the center of mass at the top and bottom or the storyunder consideration. Note: Where ASD is used, 8 shall be 1computed using earthquake forces without dividing by 1.4, see§12.12

I8 = design story drift

8 a allowable story drift, as obtained from Table 12.12-1 for any story I8aA = allowable story drift for structure A

I8 aB allowable story drift for structure B=

no = system overstrength factor as given in Table 12.2-1 IOx = inelastic deflections of Level x - §12.8.6

IO,WE the average of the displacements at the extreme points of the structure

at Level x IONl4X = the maximum displacement at Level x

OX4 = deflection at structure Level x of structure A Is, = the deflections determined by an elastic analysis of the seismic-force- Iresisting system

OM = maximum of Ox IOAlI,OM2 = displacements of the adjacent building where 0M2 is at same level as

OAII IOl:4 deflection at structure level y of structure A

IOrB = deflection at structure level y of structure B

8 stability coefficient - §12.8.7 II

16 20061BC Structural/Seismic Design Manual, Vol. I

I

IIIIII

Definitions

Active Fault/Active Fault Trace. A fault for which there is an average historic slip rate ofI mm per year or more and geologic evidence of seismic activity within Holocene (past I 1,000years) times. Active fault traces are designated by the appropriate regulatory agency and/orregistered design professional subject to identification by a geologic report.

Allowable Stress Design. A method of proportioning structural members, such that elasticallycomputed stresses produced in the members by nominal loads do not exceed specified allowablestresses (also called working stress design).

Attachments, Seismic. Means by which components and their supports are secured orconnected to the seismic-foree-resisting system of the structure. Such attachments includeanchor bolts, welded connections and mechan ical fasteners.

Balcony, Exterior. An exterior floor projecting from and supported by a structure withoutadditional independent supports .

Base. The level at which the horizontal seismic ground motions are considered to be impartedto the structure.

Base Shear. Total design lateral force or shear at the base.

Boundary Elements. Chords and collectors at diaphragm and shear wall edges, interioropenings, discontinuities, and re-entrant corners.

Boundary Members. Portions along wall and diaphragm edges strengthened by longitudinaland transverse reinforcement and/or structural steel members.

Brittle. Systems, members, materials and connections that do not exhibit significan t energydissipation capacity in the inelastic range.

Cantilevered Column System. A structural system relying on column elements that cantileverfrom a fixed base and have minimal rotational resistance capacity at the top with lateral forcesapplied essentially at the top and are used for lateral resistance.

Collector. A diaphragm or shear wall element parallel to the applied load that collects andtransfers shear forces to the vertical-foree-resisting elements or distributes forces within adiaphragm or shear wall.

Component. A part or element of an architectural, electrical, mechanical, or structural system.Component, equipment. A mechanical or electrical component or element that is partof a mechanical and/or electrical system within or without a building system.Component, flexible. Component, including its attachments, having a fundamentalperiod greater than 0.06 second .


Definitions

Component, rigid. Component, including its attachments, having a fundamentalperiod less than or equal to 0.06 second .

Confined Region. The portion of a reinforced concrete component in which the concrete isconfined by closely spaced special transverse reinforcement restraining the concrete indirections perpendicular to the applied stress.

Coupling Beam. A beam that is used to connect adjacent concrete wall piers to make them acttogether as a unit to resist lateral forces. .

Dead Loads. The weight of materials of construction incorporated into the building, includingbut not limited to walls, floors, roofs, ceilings, stairways, built-in partitions, finishes, cladding,and other similarly incorporated architectural and structural items, and fixed service equipment,including the weight of cranes.

Deck. An exterior floor supported on at least two opposing sides by an adjacent structure, and/orposts, piers, or other independent supports .

Deformability. The ratio of the ultimate deformation to the limit deformation,High deformability element. An element whose deformability is not less than 3.5 whensubjected to four fully reversed cycles at the limit deformation,Limited deformability element. An element that is neither a low deformability nor ahigh deformability element.Low deformability element. An element whose deformability is 1.5 or less.

Deformation.Limit deformation. Two times the initial deformation that occurs at a load equal to 40percent of the maximum strength.Ultimate deformation. The deformation at which failure occurs and which shall bedeemed to occur if the sustainable load reduces to 80 percent or less of the maximumstrength.

Design Earthquake. The earthquake effects that are 2/3 of MCE earthquake effects.

Design Strength. The product of the nominal strength and a resistance factor (or strengthreduction factor).

Designated Seismic System. Those architectural, electrical, and mechanical systems and theircomponents that require design in accordance with Chapter 13 that have a componentimportance factor, lp , greater than 1.0.

Diaphragm, Flexible. A diaphragm is flexible for the purpose of distribution of story shear andtorsional moment when the lateral deformation of the diaphragm is more than two times theaverage story drift of the associated story, determined by comparing the computed maximum inplane deflection of the diaphragm itself under lateral force with the story drift of adjoiningvertical lateral-force-resisting elements under equivalent tributary lateral force.

Diaphragm, Rigid. A diaphragm that does not conform to the definition of flexible diaphragm.

18 2006 IBC Structural/Seismic Design Manual, Vol. 1

1

I1

IIIJ

IIIIIIIIIII

1

IIIIII

Definitions

Displacement.Design Displacement. The design earthquake lateral displacement, exclud ingadditional displacement due to actual and accidental torsion , required for design of theisolation system.Total Design Displacement. The design earthquake lateral displacement, includingadditional displacement due to actual and accidental torsion. required for design of theisolation system.Total Maximum Displacement. The maximum considered earthquake lateraldisplacement, including additional displacement due to actual and accidental torsion,required for verification of the stability of the isolation system or elements thereof,design of building separations, and vertical load testing of isolator unit prototype.

Displacement Restraint System. A collect ion of structural elements that limits lateraldisplacement of seismically isolated structu res due to the maximum considered earthquake.

Duration of Load. The period of continuous application of a given load, or the aggregate ofperiods of intermittent applications of the same load.

Effective Damping. The value of equivalent viscous damping corresponding to energydissipated during cyclic response of the isolation system.

Effective Stiffness. The value of the lateral force in the isolation system, or an elementthereof, divided by the corresponding lateral displacement.

ElementDuctile element. An element capable of sustaining large cyclic deformations beyond theattainment of its strength.Limited ductile element. An element that is capable of sustaining moderate cyclicdeformations beyond the attainment of nominal strength without significant loss ofstrength.Nonductile element. An element having a mode of failure that results in an abrupt lossof resistance when the element is deformed beyond the deformation corresponding to thedevelopment of its nominal strength . Nonductile elements cannot reliably sustainsignificant deformation beyond that attained at their nominal strength.

Equipment Support. Those structural members or assemblies of members or manufacturedelements, including braces, frames, lugs, snubbers, hangers, or saddles that transmit gravity loadand operating load between the equipment and the structure.

Essential Facilities. Buildings and other structures that are intended to remain operational in theevent of extreme environmental loading from flood, wind, snow, or earthquakes.

Factored Load. The product of a nominal load and a load factor.

Flexible Equipment Connections. Those connections between equipment components thatpermit rotational and/or translational movement without degradation of perform ance.

2006 lac Structural/Seismic Design Manual, Vol. I 19

Definitions

Frame.Braced frame. An essentially vertical truss, or its equivalent , of the concentric oreccentric type that is provided in a building frame system or dual frame system to resistshear.Concentrically braced frame (CB F). A braced frame in which the members aresubjected primarily to axial forces.Eccentrically braced frame (EBF). A diagonally braced frame in which at least oneend of each brace frames into a beam a short distance from a beam-column or fromanother diagonal brace.Ordinary concentrically braced frame (OCBF). A steel concentrically braced framein which members and connections are designed for moderate ductility.Special concentrically braced frame (SCBF). A steel or composite steel and concreteconcentrically braced frame in which members and connections are designed for ductilebehavior.

Frame, Moment.Intermediate moment frame (IMF). A moment frame in which members and joints arecapable of resisting forces by flexure as well as along the axis of the members.Ordinary moment frame (OMF). A moment frame in which members and joints arecapable of resisting forces by flexure as well as along the axis ofthe members.Special moment frame (SMF). A moment frame in which members and joints arecapable of resisting forces by flexure as well as along the axis of the members.

Frame System.Building frame system. A structural system with an essentially complete space framesystem providing support for vertical loads. Seismic force resistance is provided by shearwalls or braced frames.Dual frame system. A structural system with an essentially complete space framesystem providing support for vertical loads. Seismic force resistance is provided by amoment-resisting frame and shear walls or braced frames.Space frame system. A structural system composed of interconnected members, otherthan bearing walls , that is capable of supporting vertical loads and that also may provideresistance to seismic forces .

Gravity Load (W). The total dead load and applicable portions of other loads as defined in§§ 12.7.2 and 12.14.8.1.

Hazardous Contents. A material that is highly toxic or potentially explosive and in sufficientquantity to pose a significant life-safety threat to the general public if an uncontrolled releasewere to occur.

Impact Load. The load resulting from moving machinery, elevators, craneways, vehicles, andother similar forces and kinetic loads, pressure, and possible surcharge from fixed or movingloads.

Importance Factor. A factor assigned to each structure according to its occupancy categoryas prescribed in §11.5.1.

20 2006 IBC Structural/Seismic Design Manual, Vol. I

)

I1)

1

IIIIIIIIIIIIII

I

IIIIIII

Definitions

Inverted Pendulum-type Structures. Structures that have a large portion of their massconcentrated near the top and, thus, have essentially one degree of freedom in horizontaltranslation. The structures are usually T-shaped with a single column supporting the beams orframing at the top.

Isolation Interface. The boundary between the upper portion of the structure, which isisolated , and the lower portion of the structure, which moves rigidly with the ground.

Isolation System. The collection of structural elements that includes individual isolator units,structural elements that transfer force between elements of the isolation system andconnections to other structural elements.

Isolator Unit. A horizontally flexible and vertically stiff structural element of the isolationsystem that permits large lateral deformations under design seismic load. An isolator unit maybe used either as part of or in addition to the weight-supporting system of the building.

Joint. A portion ofa column bounded by the highest and lowest surfaces of the other membersframing into it.

Limit State. A condition beyond which'a structure or member becomes unfit for service and isjudged to be no longer useful for its intended function (serviceability limit state) or to be unsafe(strength limit state).

Live Loads. Those loads produced by the use and occupancy of the building or other structureand do not include construction or environmental loads such as wind load, snow load, rain load,earthquake load, flood load,ordead load.

Live Loads (Roof), Those loads produced I) during maintenance by workers, equipment, andmaterials; and 2) during the life of the structure by movable objects such as planters and bypeople .

Load and Resistance Factor Design (LRFD). A method of proportioning structural membersand their connections using load and resistance factors such that no applicable limit state isreached when the structure is subjected to appropriate load combinations. The term "LRFD" isused in the design of steel and wood structures.

Load Factor. A factor that accounts for deviations of the actual load from the nomin al load, foruncertainties in the analysis that transforms the load into a load effect, and for the probabilitythat more than one extreme load will occur simultaneously,

Loads. Forces or other actions that result from the weight of building materials, occupants andtheir possessions, environmental effect , differential movement, and restrained dimensionalchanges. Permanent loads are those loads in which variations over time are rare or of smallmagnitude. Other loads are variable loads (see also "Nominal loads").

Loads Effects. Forces and deformations produced in structural members by the applied loads.


Definitions

Maximum Considered Earthquake. The most severe earthquake effects considered by thiscode.

Nominal Loads. The magnitudes of the loads specified in this chapter (dead, live, soil, wind,snow, rain, flood, and earthquake.)

Nonbuilding Structure. A structure, other than a building, constructed of a type included inChapter 15 and within the limits of §15.1.1.

Other Structures. Structures, other than buildings, for which loads are specified in thischapter.

P-delta Effect. The second order effect on shears, axial forces and moments of framemembers induced by axial loads on a laterally displaced building frame.

Panel (Part of a Structure). The section of a floor, wall, or roof located between thesupporting frame of two adjacent rows of columns and girders or column bands of floor orroof construction.

Resistance Factor. A factor that accounts for deviations of the actual strength from thenominal strength and the manner and consequences of failure (also called strength reductionfactor).

Seismic Design Category. A classification assigned to a structure based on its occupancycategory and the severity of the design earthquake ground motion at the site, see §11.4.

Seismic-fo rce-resisting system. The part of the structural system that has been considered inthe design to provide the required resistance to the seismic forces prescribed herein.

Seism ic Forces. The assumed forces prescribed herein, related to the response of the structureto earthquake motions, to be used in the design of the structure and its components.

Seismic Response Coefficient . Coefficient C" as determined from §12.8.

Shallow Anc hors. Shallow anchors are those with embedme ntlength-to-diameter ratios ofless than 8.

Shear Pa nel. A floor, roof, or wall component sheathed to act as a shear wall or diaphragm.

Shear Wa ll. A wall designed to resist lateral forces parallel to the plane of the wall.

Shear Wa ll-frame Interactive System. A structural system that uses combination s of shearwalls and frames designed to resist lateral forces in proportion to their rigidities, consideringinteraction between shear walls and frames on all levels.

Site Class . A classification assigned to a site based on the types of soils present and theirengineering properties as defined in §11.4.2.

2 2 200 6 IBC Structural/Seis mic Design Manua l, Vol. 1

IIIIII

IIIIIIIII

11

Dennltions

Site Coefficients. The values of Faand F"

indicated in Tables 11.4-1 and 11.4-2, respect ively.

Special Transverse Reinforcement. Reinforcement composed of spirals, closed stirrups, orhoops and supplementary cross-ties provided to restrain the concrete and qualify the portion ofthe component, where used, as a confined region.

Story Drift Ratio. The story drift divided by the story height.

Strength, Nominal. The capacity of a structure or member to resist the effects of loads, asdetermined by computations using specified material strengths and dimensions and formulasderived from accepted principles of structural mechanics or by field tests or laboratory tests ofscaled models, allowing for modeling effects and differences between laboratory and fieldconditions.

Strength Design. A method of proportion ing structural members such that the computedforces produced in the members by factored loads do not exceed the member design strength(also called load and resistance factor design.) The term "strength design" is used in the designof concrete and masonry structural elements.

Strength Required. Strength of a member, cross section, or connection required to resistfactored loads or related internal moments and forces in such combinations as stipulated bythese provisions.

Torsiona l Force Distributio n. The distribu tion of horizontal seismic forces through a rigiddiaphragm when the center of mass of the structure at the level under consideration does notcoincide with the center of rigidity (sometimes referred to as a diaphragm rotation).

To ughness. The ability of a material to absorb energy without losing significant strength .

Wall, Load-bearing. Any wall meeting either of the follow ing classifications:I. Any metal or wood stud wall that supports more than 100 pounds per linear foot

(1459 N/m) of vertica l load in addition to its own weight.2. Any masonry or concrete wall that supports more than 200 pounds per linear foot

(2919 N/m) of vertical load in addition to its own weight.

Wall, Non load-bearing. Any wall that is not a load-bearing wall.

Wind-res traint Seismic System. The collection of structural elements that provides restraintof the seismic-isolated-structure for wind loads. The wind-restraint system may be either anintegral part of isolator units or a separate device.

2006 IBC Structural/Sei smic Design Manual, Vol. I 23

Definitio ns

24 2006 IBC StructurallSe lsmlc Design Manual, Vol. I

]

I

IIIII

Example i • Classlficationllmportance Factors/Seismic Des ign Categ ory §11.5-1§11.6

Determine the importan ce factors and the seismic design category for a faci lity given thefollowing information.

Type of occupancy - Elementary School with capacity greater than 250

S DS = 1.17SOl = 0.70SI = 0.75

Determine the following.

[!J Building category and importance factors for general occupancy and forone build ing to be used for emergency shelter

[!J Seismic Design Category (SOC)

IC~/c~d/~Jions and Discussion

[!J Building category and importance factors.

Cd~e Referenc.~

From Table 1\.5- 1, " Importance Factors ," for the given occupancy category, the generalcategory is II. The occupancy category is used to determine the "Se ismic Design Category,"§11 .6- 1. The one building to be used for an emergency shelter is Category IV.

The importance fac tors for seismic loads are from Table 11 .5-1. Importance factors for snowloads are from Table 7-4. Importance factors for wind loads are from Table 6-1.

CategoryIIIV

SeismicFactor f

\.01.5

SnowFactor 1

\.01.20

WindFactor f

1.0\.15

2006 IBC Structural/Se ismic Des ig n Manu al, Vol. I 25

§11.S· 1 Example i • Cla ssification/Importance Fa ctors Seismic Des ign Category§ 11. 6

~ Seismic Design Category

All structures are assigned to a Seismic Design Category (SDC) based on their OccupancyCategory and the spectral response acceleration coefficients So< and SOl, irrespec tive of thefundamental period of vibration of the structure T. Each building and structure shall beassigned to the most severe SDC in accordance with Table 11.6- I or I 1.6-2 as follows.

Table 2.1 Occupancy Category vs Seismic Design Category

Nature of Occupancy Table I 1.6-1 Table 11.6-2 SDCOccupancy Category SDS SDC SOl SDC USE*

School II 1.I7 D* 0.70 D* E

Emergency IV 1.17 0 * 0.70 0 * FShelterRecall: SI = 0.75% for this table

*Note that for Occupancy Categories I, II, and III having S, equal to or greater than 0.75 (recall Sj =

0.75), the building shall be assigned to SDC E. Also for Occupancy Category IV having S,~ 0.75,the building shall be assigned to SDC F.


J

]

]

J

IIIIIIIIIIII1

I

Ex amp le 1 • Earth quake Load Co m bi nations: St rength Design §12. 4.2 .3

This example demonstrates the application of the strength design load combinations thatinvolve the seismi c load E given in §12.4;2.3 . This will be done for the moment-resistingframe structure shown below.

8Ds = 1.10I = 1.0P = 1.3I I = 0.5

Snow load S = 0

A B

, / // / r r • / r r r >

c

D

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

Dead Load Live Load Left-to-Right Right-to-LeftD L Seism ic Load Seismic Load

(--+QI;;) (-- QI;;)

Beam Mome nt at A - 100 kip -ft -50 kip-ft +120 kip- ft -120 kip-ft

Column C-D Axial Load +90 kips +40 kips +110 kips - 110 kips

Column Moment at C +40 kip-ft +20 kip-ft +160 kip-ft -) 60 kip-It

Sign Convention: Positive moment induces flexural tension on the bottom side of a beam andat the right side of a column. Positi ve axial load induces compression . Note that for theparticular location of Column C-D, the seismic Axial Load and Moment at C are bothpositive for 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 Strength design seismic load combinations (Comb.)

[!J Strength design moments at beam end A for seismic load combinations

[!J Strength design interaction pairs of axial load and moment for thedesign 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)

10.9D + 1.0E (Comb. 7)

where for a given type of load action such as moment M or axial load P IE=E,,+E,. (Eq 12.4-1) IE; =PQE (Eq 12.4-3)

E,.=0.2SDSD (Eq 12.4-4) ICombined, these yield I

E =PQE+ 0.2SDSD (Eq 12.4-3)

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

E =PQE - 0.2SDSD Iwhen the algebraic sign, ±, of QE is taken as opposite to that for D. IFor the given values of: p =1.3, SDS= 1.10, the load combinations are I

1.2D + 1.3QE+ (0.2)( I.I)D + 0.5L =1.42D + 1.3QE+ 0.5L (Comb. 5) Iwhen the signs of QEand D are the same, and

I1.2D +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 .I

0.9D + 1.3QE + (0.2)(1.1)D =1.I2D + 1.3QE (Comb. 7) Iwhen the signs of QE and D are the same, and I

I28 2006 IBC Structural/Seismic Design Manual, Vol. I

I

I1

]

I

IIIIII

I

Example 1 • Earthquake Lo ad Combinations: Strength Design §1 2.4.2.3

0.9 D + 1.3QE- (0 .2)( 1.l)D =0.68D + 1.3Q£ (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

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

0.68D + 1.3QE

with D = MD = - 100 and QE = 120

MA = 0.68(-100) + 1.3(120) = 88 kip-ft

: . Beam section at A must be designed for

MA = - 323 kip-ft and + 88 kip-ft

2006 IBC Structural/Se ismic Design Manual, Vol. I 29

§12.4.2.3 Example 1 • Earthquake Load Combinations: Strength Design

[!J Strength design interaction pairs of axial load and moment for thedesign of column section at C for seismic load combinations

II

30

The seismic load combinations using the definitions of E given by Equations 12.4-1through 12.4-4 can be used for the design requirement of a single action such as themoment at beam end A, but they cannot be used for interactive pairs of actions suchas the axial load and moment at the column section C. These pairs must occursimultaneously because of a common load combination. For example , both the axialload and the moment must be due to a common direction of the lateral seismicloading and a common sense of the vertical seismic acceleration effect represented by0.2 SDsD. There can be cases where the axial load algebraic signs are the same for QEand D, while the moment algebraic signs are different. This condition would prohibitthe 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 thelateral seismic load effect QE, and the independent reversible property of the verticalseismic 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 publicassembIy areas])

For the specific values of p = 1.3 and SDS = l.l0, the load combinations provide thefollowing values for MA, and the interaction pair Pc and Me. Note that the interactionpair Pc and Me must occur simultaneously at a specific load combination of gravityload, and lateral and vertical seismic load effects. The interaction design of thecolumn section must satisfy all of the eight pairs ofPe and Me from the seismic load

2006 IBC Structural/Seismic Design Manual, Vol. I

IIIIIIIIIIIIII

II

Example 1 a Earthquake Load Co mbin ations : Strength Des ig n §1 2.4 .2 .3

combinations along with the pairs from the gravity load combinations and wind loadcombinations.

Combination MA kip-ft Pc kips and Me kip-ft

1.420 + 1.3 (Q i;) + O.5L -35 +268.8 and +242. 8O.98D + 1.3 (QE) + O.5L -9 +26.8 and +225.21.42D - 1.3 (Qd + 0.5L -299 +229.2 and -109.20.98D - 1.3 (Qd + 0.5L -255 - 12,8 and · -126. 81.12D + 1.3(Qd +20 +22 1.8 and +220 .8

0.68D + 1.3 (QE) +64 + 182.2 and +203 .2

1.12D - 1.3 (Qd -244 -20.2 and -131.20.68D - 1.3 (QE) -200 -59 .8 and - 148.8

The governing values are und erlined for MA [same as determined in Part (2) ] and for theinteraction pairs ofPc 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 vide anautomatic method of considering the individual algebraic signs of the load actions, thedirection of the lateral seismic load, and the independent ± action of 0.2 SDCD. There is noneed to use the "same sign" and "opposite sign" limitations of Equations 12.4-2 and 12.4-3since all possible com binations are represented . Thi s is important for interactive pa irs ofactions that must be evaluated for a common load combination.

When the Modal Response Spectrum Analysis' procedure of §12.9 is used, the algebraic signsof seismic load actions are lost because of the process of combining the individual modalresponses. The signs to be used for an interaction pair of actions due to a given direction oflateral loading can be obtained from the primary mode response where the prim ary mode isthe mode having the largest participation fac tor for the given direction of lateral seismicloading. Or, alternatively, the signs can be obtained from the equivalent lateral forceprocedure of §12.8.


§2.4 Example 2 • Combinations of Loads

-:

The code permits the use of allowable stress design for the design of wood members andtheir fastenings (ASCE/SEI 7-05 §2.4 and §12.4.2.3). Section 2.4 defines the basic loadcombinations for allowable stress design.

This example illustrates the application ofthis method for the plywood shear wall shownbelow. The wall is a bearing and shear wall in a light wood framed build ing.

The following information is given.

Seismic Design Category B

J = 1.0P = 1.0

5DS =0.3

E = Ell = bQ£ = 4 kips (seismicforce due to the base sheardetermined from §12.4.2)

Grav ity loadsDead lVD = 0.3 kif (tr ibutary dead

load, including weight ofwall

Live lIIL = (roof load supported byother elements)

Gravity loads

ITnTTTTmPlywood

shear wall

Shear Wall Elevati on

Moment arm from center of post to center of hold-down boltL = 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 Tfor the following load combinations.

[!J Basic allowable stress design


1I r- ,.

Example 2 • Combinations of Loads

. .,.,.~..

§2.4

[IJ Basic allowable stress design §12.4.2.3

The governing load combinations for basic allowable stress design are Basic ASDCombinations 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

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

I

1

IIII\

II

II

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

and E = PQE - 0.2SosD

=QE- 0.06D when D and QEhave oppos ite sense

For ASD Basic Combination 5 the load combination is:

D +0.7E

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

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

and D(1.0)+0.7(-0.6D-Qd

= 0.958D - O.7QE for D and QEwith opposite sense


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

= 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


0.6D + 0.7E

(Eq 12.4-1)

(Eq 12.4-3)

(Eq 12.4-4)

(Comb . 5)

(Comb. 6)

(Comb. 8)


§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 notinvolved, 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 combinationwould be

1.042D + 0.7QE

~ Required unit shear capacity q

Base shear and the resulting element seismic forces QE determined under §12.8.1are on a strength design basis. For allowable stress design, QE must be factored by 0.7as 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

= 0.7QE = 0.7(4000) = 280 Ifq L 10ft P

This unit shear is used to determine the plywood thickness and nailing requirementsfrom lBe Table 2306.4.1, which gives allowable shear values for short-time durationloads 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

1

34 2006 IBC Structural/Seismic Des;gn·Manual, Vol. I

II

I1

IIIIII

I

Example 2 • Comb/nations of Loads §2.4

into 2x members with 6-inch spacing of fasteners at panel edges; allowable shear of340 plf.

Required hold-down tensile capacity T

Taking moments about point 0 at center of post at right side of wall withE" =oQ£=4000 Ib, the value of the hold-down tension force T due to horizontalseismic forces is computed

0.558(300 pit) I0 ft(5 ft -~ ) - o.7(4000 Ib)(9 ft) + T(9.4 ft) = 02(12)

Thus :

8125.88 Ib ft - 25,200 Ib ft + 9.4 ft(T) = 0

T = 1816.39 Ib tension

Similarly the boundary element compression capacity is computed

1.042(300 pit) lOft (5 ft - ~ ) + 0.7 (4000 Ib)(9 ft) - C(9.4 ft) = 02(12)

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

c =4295 Ib compression

The tension value is used for the selection of the pre-manufactured hold-downelements. 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 Durationfactor for resisting seismic loads. This is not considered a stress increase (although ithas the same effect). Therefore, the catalog "1.33 x allowable" capacity values maybe used to select the appropriate hold-down element.

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

E = pQ£+ 0.2 SDsD

and use ± E in the load combinations.


§11.4 Des ign Spectral Response AcceleraOons

For a given building site, the maximum conside red earthquake spectral responseaccelerations S, at short periods, and S) at I-second period are given by the accelerat ioncontour maps in §22. This example illustrates the general procedure for determin ing thedesign spectral response parameters Sosand SDl from the mapped values of Ss and 8). Theparameters Sos and So, are used to calculate the design base shear in §12.8 and the DesignResponse Spectrum in §11.4.5. -

Note that by far the most accurate, easiest, and most effic ient way to obtain the spectraldesign values is to use the USGS website iwww.eqhazmaps.usgs.govr. Given the longitudeand latitude of the site, the website provides va lues of Ss and S). The site longitude andlatitude 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 determined that a building site near Sacramento, California

is located at Latitude 38.123° North and Longitude - 121.123 D (or 121.123D west). The soilprofile is Site Class D.

Determin e the following.

[!.J Maximum considered earthquake spectral response accelerations s,and Sl

!TI Site coefficients and adjusted maximum considered earthquake spectralresponse acceleration parameters SMS and SMl

[!J Design spectral response acceleration parameters Sos and SOl

~ 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


I

IIIIIIIII

III

Design Spec tral Response Accelerations §11.4

. "",.. Code.8eference· "

[!J Maximum considered earthquake spectral response accelerations §11.4.1

For the given position (Near Sonora - NW of Sacramento, California) of 38° North

(Latitude = 38.123°) and 121.123° West (Longitu de =- 121.123'), USGS provides thevalues of

5s = 46.2%g = 0.462g

5, = 20.3%g = O.203g

~ Site coefficients and adjusted maximum considered earthquake spectralresponse accelerations §1 1.4.3

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

The adjusted maximum conside red earthqu ake spectral response accelerations (based 011

§11.4.3) are also given on the CD ROM as follows

IIIIIII

Fa= 1.58

F,.=1.99

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

S'/I =F,S, =1.99(O.203g) =0.404g

TII.4-1

T 11.4-2

(EqI1.4-1)

(Eq 11.4-2)

2006 lac Structural/Seismic Design Manu al, Vol. I 37

§ 11.4 D esign Sp ectral Response Acc el erations

~ Design spectral response acceleration parameters

2 2S DS =- SMS =- (0.73g) =0.49g

3 3

'J ?SOl =~ S.I{I =~ (0.404g) =0.27g

I§1l.4.4 I

(Eq 11.4-3) I1

(Eq 11.4-4)

For periods greater than or equal to Toand less than or equal to T" the design spectralresponse acceleration So shall be taken equal to Sos

For periods greater than T:" and less than TL , the design spectral response acceleration Sashall be given by

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

38

General procedure response spectrum

SDSSo = 0.6- T + 0.4 Sosr:

Sa = (SOI) / T

Where: To = 0.20 (SOl / Sos)

= 0.2 (0.27 /0.49)

= O. I I sec

T, = SOl / 50s

= 0.27 / 0.49

= 0.55 sec

Tt. = 8 sec


§1l .4.5

(Eq 11 .4-5)

(Eq 11.4-6)

(F 22-15)

I1

IIII

IIIIIII

Thus:

T =Period

0.000.110.550.801.001.201.401.602.00

ScJg

0.180.490.490.340.270.230.190.170.135

Des ign Spectral Response Accelera ti ons §11 A

Computation for Sa0.4 (0.49)0.490.27 /0.550.27 / 0.80.27 / 1.000.27 / 1.20.27 / 1.40.27 / 1.60.27 /2.0

... .....

S. in g 's

0.5

S DS = 0.49g

0.4

0.3

0.20.18

0.1

... ...... -... - ... -

oo 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

To=0.11 sec T, =0.55 sec

General Procedure Response Spectrum


Calculation of seismic response coefficient Cs (Recall Soil Site Class D,f = 1.0 and T= 0.60) §12.8.1

§11.4

40

Design Spectral Response Accelerations

The seismic response coefficient shall be determined by

C, = SDS I (RIl)

= 0.49 I (6.0/1.0)

= 0.08 2 ... Governs

The value of C, need not exceed

C,. = SDI I (RIlE) T

= 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, = 0.5S1I (RIl)


(Eq 12.8-2)

(Eq 12.8-3)

(Eq 12.8-5)

1

I1

IIIIIIIIIIIIIIII

-,':

Introduction to Vertical Irregul arities

'.

§12.3.2.2

§12.3.2.2

1

Table 12.3-2 defines vertical structural irregularities and assigns analysis and designprocedures to each type and seismic design category. These irregularit ies can be divided intotwo categories. The first, dynamic force-distribution irreg ularities, which are Types Ia, Ib, 2,and 3. The second, irregularities in load path or force transfer, which are Types 4 and 5. Thevertical irregularities are

Ia. Stiffness Soft Story Irregularity

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 ionirregularities shall be analyzed using the dynamic analysis procedure (or moda l analysisprocedure) prescribed in §12.7. (Refer to Table 12.6.1) Structure Description 3. The vert icalforce distribution provided by §12.8.3 may be assumed to be adequate for structures lackingvertical irregulari ty Types Ia, Ib, 2, and 3. However, stiffness and mass discontinuities maysignificantly affect the vertical distribution of forces and, for this reason the modal analysisprocedure, which can account for these discontinuities, is necessary.

Although designers may opt to use the dynamic analysis procedure and bypass checks forirregularity Types Ia, Ib, 2, and 3, the reference sections listed in Table 12.3-2 should still bechecked for limitations and design requirements. Note that §12.3.3.1 prohibits structures withvertical irregularity Types Ib, Sa, or 5b for Seismic Design Categories E and F.

Regular structures are assumed to have a reasonably uniform distribution of inelasticbehavior in elements throughout the lateral- force-resisting system. When vertical irregu larityTypes 4 and 5 exist. there is the possibility of having localized concentrat ions of excessiveinelastic deformations due to the irregular load path or weak story. In this case, the codeprescribes addi tiona l strengthening to correct the deficienc ies for structures in certain seismicdesign categories (SDCs) . In the case of vertical irregularity Type 5b, limits are placed on thebuilding height for all SDCs exceptS DC A.

2006 lac Structural/S eismic Design Manu al, Vol. I 41

§12.3.2.2 Ex ample 4 • Vertic al1rregularlty Typ e 1 1~ ,

&ample4.Vertical Irregularity Type 1a and Type 1b § 12.3.2 .2

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

Ft + Fs

10'

10'

10'

' 0'

' 2'

~ [';'==::;-1;::' =::::::;11- - - - - --;,. 0" . 2.02i:

F, -.. / :

DD.• 0', _1.75

Triangular .'"shape ; I

F, -.. "' : !

DD / · 0;\... 1.45F .' /, -.. ! !

DD ./ , 02",.0B

F i "'-.. /.' 0.I,' It_ 0.71!",~r-r---r7Y-.,-"'77',*:" Actual shape

J

II

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

Calculations and Discussion CodeReference

[L] 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 threestories above.


Example 4 • Venicallrregularity Type 1 §12.3.2.2

1 If the stiffness of the story meets at least one of the two criteria above, the structure isdeemed to have a soft story, and a modal analysis (§12.9) is generally required by Table12.6- I.

I

I

The definit ion of soft story in the code compares values of the lateral stiffness of individualstories. Generally , it is not practical to use stiffness properties unless these can be easilydetermined. There are many structural configurations where the evaluat ion of story stiffnessis complex and is often not an available output from computer programs. Recogni zing thatthe basic intent of this irregularity check is to determine if the lateral-force distribution willdiffer significantly from the pattern prescribed by §12.8.3, which assumes a prescribed shapefor the first dynamic mode of response, this type of irregularity can also be determined bycomparing values of drift ratios due to the prescr ibed lateral forces. This deformat ioncomparison may even be more effective than the stiffness comparison because the shape ofthe first mode shape is often closely approximated by the structure displacements due to thespecified §12.8.3 force pattern . Floor level displacements and corresponding story-drift ratiosare directly available from computer programs. To compare displacements rather thanstiffness, it is necessary to use the reciprocal of the limiting percentage ratios of 70 and 80percent as they apply to story stiffness, or reverse their applicability to the story or storiesabove. 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 determined.The story-drift ratio is the story drift divided by the story height. These story-drift ra tios willbe used for the required comparisons because they better represent the changes in the slopeof the mode shape when there are significant differences in interstory heights. (Note: storydisplacements 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 followingcond itions exists.

or

t. , 0,. - 0" = (1.08 - 0.71) = 0.00308- '=h, h, 120

~= Ii,. - 0,. = (1.45 - 1.08) = 0.00308h, h, 120

s s - IiWhen 70 percent of ---"- exceeds ,. ,.

h, h,

Ii, I [(0, -Ii,.) (0, - Ii, ) (0, -Ii,)]When 80 percent of - ' exceeds - .• , + • _. + • .•h, 3 h, h, h,

= 0.00493(0.7 1- 0)

=144

~= o..h, h,

the story-drift ratios arc determi ned as

IIIIIII

2006 IBC Structural/Seismic Design Manual. Vol. I 4 3

§12. 3.2.2 Example 4 • Vertical Irreg ularity Type 1 1

_t:J._, = Ii" - Ii" =

Iz, s,(1.75-1.45) = 0.00250

120

44

~(0.00308 + 0.00308 + 0.00250) = 0.002893

Checking the 70-percent requirement:

O.70(1i,,) =0.70(0 .00493) =0.00345 > 0.00308 ... NGhi

: . 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 notapply where no story-drift ratio under design lateral force is greater than 130 percentof the story-drift ratio of the next story above, §12.3.2.2, Exception 1.


0.80(~) = 0.80(0.00493) = 0.00394 > 0.00289 ... NGh,

.', 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 lb)


0.60(0.00493) = 0.002958 < 0.00308 . . . o. k.

Alternately: 0.00493 > (0.00308 x 1.4 = 0.004312) .. . o.k.


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.


I

IIIIIIII

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 isSOC 0 , and is permitted, Structures having SDC E or F and also having vertical irregularityType Ib shall not be permitted.

Commentary

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

s = Cdb.rrx I (Eq 12.8-15)

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 inthis example. In the example above, only the first story was checked for possible soft-storyvertical irregularity . In practice, all stories must be checked, unless a modal analysis isperformed. It is often convenient to create tables to facilitate this exerc ise, see Tables 4.1and 4.2.

Table 4. I Soft-Story Status ln

Sto ry Story Story-drift 0 .8x (S tory- 0 .7x (Story - Avg . of Story-drift RatioLeve l Displacement Drift Rat io drift Ratio ) drift Ratio) of Next 3 Stories

5 2.02 in 0.27 in 0.00225 0.00180 0.00 158

4 1.75 0.30 0.00250 0.00200 0.00175

3 1.45 0.37 0.00308 0.00 246 0.00216

2 1.08 0.37 0 .00308 0.0024 6 0.00 216 0.0026 1

0.71 0.71 0.00493 0.00 394 0.00345 0.00289

T able 4.2 Soft-Story Status Ib

Story Story Story-drift 0 .7x (S tory- 0.6x (Story- Avg, of Story-drift RatioLevel Disp lacement Drift ratio drift Ratio) dr ift Ratio) of Next 3 Stories

5 2.02 in 0.27 in 0.00 225 0.001 58 0.00 135

4 1.75 0.30 0.00250 0.00175 0.00150

3 1.45 0.37 0.00 30 8 0.002 16 0.001 85

2 1.08 0.37 0.00308 0.002 16 0.00185 0.0026 1

0.71 0 .71 0.00493 0.00345 0.00296 0.00289

Soft StoryStatus la

No

No

No

No

Yes

Soft StoryStatus lb

No

No

No

No

Yes


§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 equipmentinstallation 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 anystory is more than 150 percent of the effective mass of an adjacent story. However, thisrequirement does not apply to the roof if the roof is lighter than the floor below. Note that itdoes 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.


11

1

1

I

IIIIIII

I

Example 5 • Vertical Irregularity Type 2 §12.3 .2 .2

I , • ..,

.Commentary

As in the case of vertical irregularity Type la or Ib, this Type 2 irregularity also results in aprimary mode shape that can be substantially different from the triangular shape and lateralload distribution given by §12.8.3. Consequently, the appropriate load distribution must bedetermined by the modal analysis procedure of §12.9, unless the irregular structure is notmore than two stories and is Occupancy Category l or II (see Table 12.6-1).

2006 lac Structural/Seismic Design Manual, Vol. I 47

§12.3.2 .2 Ex ample 6 • Vertical Irregularity Type 3

~

ample 6Vertical Irregularity Type 3 § 12.3.2.2

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

4@2S' -100'I ..... - ......... ...-

5 DDDDDD

3 DDD2 D.DDD'/ "/ //"/ // / / / , / //, /

Level

4

[!J Determine if a Type 3 vertical irregularity (vertical geometric irregularity)

exists

CalcuJ~tiC?ns and Discussion Code Reference

A vertical geometric irregulari ty is considered to exist where the horizontal dimension of thelateral-foree-resisting system in any story is more than 130 percent of that in the adjacentstory. One-story penthouses 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) = 1.33Width of Level 3 (75 ft)

133 percent > 130 percent

.'. Vertical geom etric irregulari ty exists.

48 2006 IBC Structural/Seismic Des ign Manual, Vol . I

II

)

IIII

Example 6 • Vertical Irregularity Type 3 §12.3.2 .2

, .Commentary

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

Similarly, if the width decrease is in the lower adjacent story (the unusual situation), theType la soft-story irregularity can be avoided by a proportional increase in the stiffness ofthe lower story. However, when the width decrease is in the lower story, there could be anoverturning moment-load-transfer discontinuity that would require a dynamic analysis perTable 12.6-1.

Note that if the frame elements in the bay between lines 4 and 5 were not included as part ofthe designated lateral-force-resisting system, the vertical geometric irregularity would notexist.

2006 IBC Structural/Seismic Design Manual, Vol.J 49

§12. 3.2.2 Ex ample 7 11 Vertical Irre gu lar i ty Type 4

,Ex, mple 7Vertical Irregularity Type 4 § 12.3.2.2

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

3@25'=75'r r r T

Level IE-< ----~

12'

12'

12'

12'

12'

5

'DO' - L--4

0---e:3

0025'

250'

1

// '/ / , ' / / / '/ / / " /

Shear wall

Shear wall

[!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 leftside of the upper shear wall (between lines A and B) is offset 50 feet from the leftside of the lower shear wall (between lines C and D). This 50-foot offset is greaterthan the 25-foot length of the offset wall clements .

: . In-plane discontinuity exists .

50 2006 IBC Structural/Seismic Design Manual, Vol . I

II1

III

I

I

Examp le 7 • Vertical Irregularity Type 4 §12.3.2.2

s: '"C;ommentary

The intent of this irregularity check is to provide correction offorce transfer or load-path deficiencies. It should be noted that any in-plane offset, even those less than or equalto 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 offsetexceeds the length of the resisting element, there is also a shear transfer discontinuitythat requires application of §12.3.3.4 for the strength of collector elements along theoffset. 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 theprovisions of §12.3.3.4.

2006 IBC StructlirallSuismic Design Manual. Vol. I 51

§12.3.2.2 Exa mple 8 • VertIc al Irregu lar ity Type 5a

IE. ampleBVerfic a l l r reg ularity Type 5a §12.3.2,,2

A concrete bearing-wall building has the typical transverse shear-wall configurationshown below. All walls in this direction are identical, and the individual piers havethe shear contribution given below. Then, V, is the nominal shear strength calcu lated inaccordance with Chapter 19, and Vm is defined herein as the shear corresponding to thedevelopment of the "nominal flexure strength also calculated in accordance with Chapter19." Note that VII/ is not defined in ACI or Chapter 19.

Level J

PIER \I V",n

1 20 kips 30 kips

2 30 40

3 15 10

4 80 120

5 15 10

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


A Type Sa weak-story discontinuity in capacity exists when the story strength is lessthan 80 percent of that in the story above. The story strength is the total strength of allseismic-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.


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 (IrregularityType 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 onesupporting story that can lead to the loss of vertical load capacity. Elementssubject to this check are the shear-wall piers (where the shear contribution isthe lower of either the shear at development of the flexural strength , or theshear strength), bracing members and their connections, and frame columns.Frame columns with weak column-strong beam condit ions have a shearcontribution equal to that developed when the top and bottom of the columnare at flexural capacity. Where there is a strong column-weak beam condition,the column shear resistance contribution should be the shear corresponding tothe development of the adjoining beam yield hinges and the column baseconnection capacity. In any case, the column shear contribution shall notexceed the column shear capacity.

An extreme weak story is prohibited (under §12.3.3.1) for structures more than twostories or 30 feet in height if the "weak story" has a calculated strength oflessthan 80 percent of the story above . A weak-story condition is absolutely prohibited inSDC E and F.

I

IIII

I2006 IBC Slructural/Seismic Design Manual. Vol. I 53

§12.3.3.1 Example 9 II Vertical Irregularity Type 5a

Example 9Verticallrregulaljty Type Sa § 12.3.3.1

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

5

4 . -.

3 ~ .. -' .. .

2

1 .. ......

// "/ / / / // . / ///

Beams at Levels I and 2:Mllb = ZF."= 250 kip-ft

Columns on lines Band C atboth levels:

M" c=250 kip-ft ataxial 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 purposesof illustration only, that the columnshave been designed such that astrong beam-weak column condi tionis permitted.

12'

12'

12"

12'

14'

Level

A o

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


A Type 5 weak-story discontinuity in capacity exists when the story strength is less than 80percent of that of the story above (where it is less than 65 percent, an extreme weak storyexists) . The story strength is consi dered to be the total strength of all seismic-foree-resistingelements that share the story shear for the direction under consideration.

To determine if a weak story exists in the first story, the sums of the column shearsin the first and second stories-when the member moment capacities are developedby 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 ctural/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 considerations200

2Mc =400 > M; = 250

: . Strong column-weak beam condition exists.

....~) 250

--... FOR MOMENT200

v

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

a M. /2 =125 kip-ft

125~

L) 250

125 I'-'

Clear height = 14 ft - 2 ft = 12 ft

125+100 187-k'- --- = .) "iPS12 v

• o Mf=100kip·ftFOR SHEAR

Checking columns Band C for strong column-weak beamconsiderations

2Mc = 400 < 2Jvfb = 500

200 .........

250(+)250

200"-

:. Strong beam-weak column condition exists.

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

moment capacity of beam to determine shear.

Clear height = 14 ft - 2 ft = 12 ft

FOR MOMENT

200 ..........

a 200(+) 200Me = 200 kip-ft

200'-'

200 r'

VB= Vc= 200 + 100 = 25.0 kips12 V

4M =100 klp-ft

G100 .....J

FOR SHEAR

First story strength = VA + VB + VD = 2(18.75) + 2(25.0) =87.5 kips


§12.3.3.1 Example 9 • Ver tical Irregularity Type 5a

~ Determine second story strength

Columns A and D must be checked for st rong column-weak beam at Level 2

:. strong column-weak beam condition exis ts.FOR MOMENT

'-'125

Mb I 2= 125 kip-ftv

""""200

Clear height = 12 ft - 2 ft = 10ft 125r--

V - tr - 125 + 125 -?5 0 k'" - " D - - _. IpS. 10

-J125

FOR SHEAR

v•

Mb I 2= 125 kip-ft

125" J

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

2Mc = 400 < 2Mb= 500

: . Strong beam-weak column condition exists .

vFOR MOMENT

Me = 200 kip-ft

Clear height = 12 ft - 2 ft = IO ft10'

200 ........

200(+ )200200 ........

VB = Vc = 200 + 200 = 40.0 kips10

200 J1"""' FOR SHEAR

v•

Me =200 klp-ft200 '-"

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


Example 9 ~ Vertical Ir regulari ty Type Sa §12 .3 .3. 1

~ Determine if weak story exists at first story

First story strength = 87.5 kips

Second story strength = 130.0 kips

I

I

I,

87.5 < 0.80(130) = 104

:. Weak story condition in first story exists.

(T 12.3-2, Item 5a)

II\

IIII

II


§ 12. 3.2.1 In troduction 10 Ho r/zonla /lrregularitles

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

la. Torsional Irregularity - to be considered when diaphragms are not flexible asdetermined in §12.3.1.2

lb . Extreme Torsional Irregularity - to be considered when diaphragms are notflexib 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.

Introduction toHorizontal Irregularities §12.3.2.1

1

1

]

I

58

These irregularities can be categorized as being either special response conditionsor 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 thegiven limit, there is the potential for an unbalance in the inelastic deformation demands at thetwo extreme sides ofa story. As a consequence, the equivalent stiffness of the sidehaving maximum deformation will be reduced, and the eccentr icity between thecenters 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 theeffects 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 theprojecting legs of the building plan adjacent to re-entrant comers can result inconcentrated forces at the comer point. Elements must be provided to transferthese forces into the diaphragms.

Type 3. Excessive openings in a diaphragm can result in a flexible diaphragmresponse along with force concentrations and load path deficienci es at theboundari es of the openings. Elements must be provided to transfer the forcesinto the diaphragm and the structural system.

Type 4. The out-of-plane offset irregularity represents the irregular load path category. In thiscase, shears and overturning moments must be transferred from the level above the offset tothe 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 nonparallellateral-force-resisting elements can have significant differences from those of a regularsystem. Further analysis of deformation and load behavior may be necessary.

2006 IBC Stru ctural/Seismic Design Manu al, Vol. I

II

IIIIIIII

Example 10 a Horizontal Irregularity Type 1a and Type 1b §12.3.2.1

ri3fnple 1oui on a/Irregulari ty Type 1a and ype'lb § 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 hasthe follow ing elastic displacements OXl! at Levels I and 2.

OL.:! = 1.20 in OR,:!. = 1.90 in

A Type 1a torsional irregu larity is considered to exist when the maximum storydrift, includ ing accidental tors ion effects, at one end of the structure transverse toan axis is more than 1.2 times the average of the story drifts of the two ends of thestructure, see §12.8.6 for story drift determination

[}J Determine if a Type 1a or Type 1b torsional irregularity exists at thesecond story

If it does:

~ Compute the torsional amplification factor Ax for Level 2II

I

OL,2 = 1.00 in OR,l = 1.20 in

Level

3

2

C€l'cu'ations and Discussion

OR,2

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

OR,1- ------7

Code Reference

2006 IBC Structural/S eismic Design Manual, Vol. I 59

§12.J.2.1 Example 10 . Horizon'al Irreg ula rit y Type 1a and Typ e 1b

ITI Determine if a Type 1a torsional irregularity exists at the second story

Referring to the above figure showing the displacements bJe due to theprescribed lateral forces, this irregularity check is defined in terms of storydrift D.x =(bx - bx-d at ends R (right) and L (left) of the structu re. Torsionalirregularity exists at Leve l x when

where

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 = 0.20 + 0.70 = 0 45 .U al'g . In

2

Checking 1.2 criteria

D. """ = 0.7 = 1.55 > 1.2D. a,-, 0.45

:. Tors ional irregularity exis ts - Type Ia.

Check for extreme torsional irregulari ty

D. 0_70 I 55 h . - I - . T Ib---"!!!!. = - - = . . . .t us, extreme torsion lrregu an ty exists - ype .D. .", 0.45


T 12.3-1 II

I

IIIIIIIIII

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

~ Compute amplification factor Ax for Leve l 2

§12.3.2.1

§12.8.4.3

1

When torsional irregularity exists at a Level x, the accidental torsional moment M'nmust beincreased by an amplification factor Ax. This must be done for each level, and each level mayhave 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.'' .' 1.26

0' 8

bma., = 1.90 in... (bR.2)

b = bL,] + bR., = 1.30 + 1.90 = 1.60 inavg 2 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

(!BC Eq 16-44)

IIIIII

In §12.8.4.3, there is the provision that the more severe loading shall be considered. Theinterpretation of this for the case of the story drift and displacements to be used for theaverage values I'l.b",.g and bm·g is as follows . The most severe condition is when both bR,X andbL,Xare computed for the same accidental center-o f-mass displacement that causes themaximum displacement bmax. For the condition shown in this example where bRX = bma.n thecenters-of-mass at all levels should be displaced by the accidental eccentricity to the rightside R, and both bR,Xand bL..rshould be evaluated for this load condition.

Table 12.3-1 triggers a number of special design requirements for torsionally irregularstructures. In fact, if irregularity Type Ib (Extreme Torsional Irregularity) is present,§12.3.3.1 is triggered, which prohibits such structures for SOC E or F. It is important torecognize that torsionai irregularity is defined in terms of story drift I'l.." while the evaluatio nof A.r by Equation 12.8-14 is, in terms of displacements bxc • There can be instances where thestory-drift values indicate torsional irregularity and where the related displacement valuesproduce an Ax value less than 1.0. This result is not the intent of the provision, and the valueof Ax used to determine accidental torsion should not be less than 1.0.

The displacement and story-drift values should be obtained by the equivalent lateral-forcemethod with the code-prescribed lateral forces. Theoret ically, if the dynamic analysisprocedure were to be used, the values of I'l.ma.' and I'l.m.g would have to be found for eachdynamic mode, then combined by the appropriate SRSS or CQC procedures, and then scaledto the code-prescribed base shear. However, in view of the complexity of this determinationand the judgmental nature of the 1.2 factor, it is reasoned that the equivalent static forcemethod is sufficiently accurate to detect torsional irregularity and evaluate the Ax factor.

2006 lac Structural/Se ismic Design Manual, Vol. I 61

§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 useof a three-dimensional model if there are any irregularities.

For cases oflarge eccentricity and low torsional rigidity, the static force procedure can resul tin a negative displacement on one side and a positive on the other. For example, this occurs ifDu = - DAD in. and DR.3 = 1.80 in. The value of Dm'g in Equation 12.8- 14 should be calculatedas the algebraic average .

= (- 40) + 1.80 = l AO =0.70 in2 2

When dynamic analysis is used, the algebraic average value Dm'g should be found for eachmode, and the individual modal results must be properly combined to determ ine the totalresponse value for Dm .g•


1

IIIIIIIIIII

Exam ple 11 • Horizontal Irregularity Type 2 §12.3.2.1

Example 11orizontallrregularity T}'J e 2 §12.3.2.. 1

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

0-

0-~oN

@)c<)

0-

8-

GI1<

0) ®I I

4 @l2S= 100'

[!J Determine if there is a Type 2 re-entrant corner irregularity


A Type 2 re-entrant comer irregularity exists when the plan configuration of a structure andits lateral -foree-resisti ng system contain re-entrant corners, where both projections of thestructure beyond a re-entran t comer are greater than 15 percent of the plan dimension of thestructure in the direction considered.

The plan configuration of this building, and its lateral-force-resisting system, has re-entrantcomer dimensions as shown. For the sides on line I , the projection beyond the re-entrantcomer is

100 ft - 75ft = 25 ft

This is 25 or 25 percent of the 100-ft plan dimension . . . More than 15 percent.100

For the sides on line E, the projection is

60 ft - 40 ft = 20 ft

2006 IBC Structura l/Se is mic p esign Manual, Vol . I 63

§1 2.3.2.1 Exam ple 11 • Ho riz ontal Irregularity Typ e 2

This is 20 or 33.3 percent of the 60-ft plan dimension . . . More than 15 percent.~ .

Since both projections exceed 15 percent , there is a re-entrant comer irregularity.

Re-entrant comer irregularity exists.

Commentary

Whenever the Type 2 re-entrant comer irregularity exists, see the diaphragm designrequirements of §12.3.3.4 for SDC D, E, and F.

64 2006 IBC Structural/Seismic Design Manual, Vol. J

IIII

Example 12 IZ Horizontal Irregular i ty Type 3 §12.3.2.1

• a;8V»ple 12Horizonfal lrregularity ype 3 §12.3.2 .1

A five-story concrete building has a bearing wall system located around the perimeterof the buil ding. Lateral forces are resisted by the bearing walls acting as shear wa lls.The floor plan of the second floor of the building is shown below. The symmetricallyplaced open area in the diaphragm is for an atrium, and has dimensions of 40 feet by 75 feet.All diaphragms above the second floor are without significant openings.

? ~ ~ ?r

125'

1IE75'

~®-®--

bCD

®- "

®-Second floor plan

[!J Determine if a Type 3 diaphragm discontinuity irregularity exists at the secondfloor level


A Type 3 diaphragm discontinu ity irregularity exists when diaphragms have abruptdiscontinuities or variations in stiffness, including cutout or open areas comprising more than50 percent of the gross enclosed area of the diaphragm, or changes in effect ive diaphragmstiffness of more than 50 percen t from one story to the next.

The first check is for gross area

Gro ss enclosed area of the diaphragm is 80 ft x 125 ft = 10,000 sq ft

Area of opening is 40 ft x 75ft = 3000 sq ft

50 percent of gross area = 0.5(10,000) = 5000 sq ft

3000 < 5000 sq ft

" No diaphragm discontinuity irregularity exists.

2006 IBC Structural/Se ismIc Design Manual, Vol. I 65

§ 12. 3. 2. 1 Exampl e 12 • Horizonrallrregu/arit y Type 3

The second check is for stiffness.

The sti ffuess of the second floor diaphragm with its opening must be compared withthe stiffness of the solid diaphragm at the third floor. If the change in stiffness exceeds50 percent, a diaphragm discontinu ity irregularity exists for the structure.

This comparison can be performed as follows.

Find the simple beam mid -span deflec tions L12 and L1J for the diaphragms at Levels2 and 3, respectively, due to a common distri buted load IV such as I kif.

w = 1 kif

-.

t:.~"'" ...., -------- - ~ - - . >

- ----- - - - - - - -- .. ..... ~ Deflected shape

w =1 kif

II

II}

I

t:.~ ------ -------~

If L12> 1.5L1J, there is diaphragm discontinuity.

66 200 6 IBC Structural/Se;smic Design Manual, Vol. I

Denected shape

Example 13 • Horizontal Irregularity Type 4 §12.3.2.1

.... an Ie -13rizontallrregularity TjIj e 4 §12.3.2.1

A four-story building has a concrete shear wall lateral-force-resisting system in abui lding frame system configuration. The plan configuration of the shear walls is.shown below.

10'

10'

10'

10'

Eleva tion Line E

oIII

0)- "2 in

'"@'"

oI.

®I

Typical floor plan

0) @ 0I I<E 25' ~

4 @25' o l 00' < 1

Ground (first) floor plan

[!J Determine it there is a Type 4 out-at-plane offset irregularity between the firstand second stories


An out-of-plane offset plan irregularity exists when there are discontinuities in a lateralforce path. For example: out-of-plane offsets of vertical lateral- force-resisting elements suchas shear walls . The first story shear wall on line 0 has a 25-foot out-of-plane offset to theshear wall on line E at the second story and above. This constitutes an out-of-plane offsetirregularity, and the referenced sections in Tab le 12.3.2.1 apply to the design.

: . Offset irregu larity exists.

2006 IBC Structura l/S eis mic Desig n Manual, Vo l. I 67

§12.3.2.1 Exa mple 14 • Horizontal Irregularity Type 5

Example 14H orizon tal Irregularity Type 5 §12.3.2.1

A ten-story building has the floor plan shown below at all levels. Special momentresisting frames are located on the perimeter of the building on lines 1,4, A, and F.

0-0-~

inN

~@;~M

Typical floor plan

[!J Determine if a Type 5 nonparallel system irregularity exists


A Type 5 nonparallel system irregularity is considered to exist when the verticallateral-force-resisting elements are not parallel to or symmetric about the majororthogonal axes ofthe build ing's lateral-foree-resisting system.

The vertical lateral-foree-resisting frame elements located on line F are notparallel to the major orthogonal axes of the building (i.e., lines 4 and A).Therefore a nonparallel system irregularity exists, and the referenced sectionin Table 12.3-1 applies to the design, see §12.5.3, §12.7-3, and Table 12.6-1.

:. A nonparallel system irregularity exists.

A 3-dimens ional dynamic analysis is recommended.


1

I

III\

II

Example 15 • Reliability/Redundancy Coefficient p § 12.3.4

·Ei.rample 15

Redundancy Factor p §12.3.4

The calculation of the redundancy factor p has changed considerably between earliercodes (1997 UBC; 2000 and 2003 1BC; ASCE/SEI 7-02) and the ASC E/SEI 7-05 . Thecalculation is in some ways simpler, although it nevertheless requires some effort forconditions that do not compl y with prescriptive requirements (unless the full penalty istaken, as described below) .

ASCE/SEI 7-05 permits the redundancy factor to be taken as 1.0 in the followingcircumstances (§12.3.4.1):

I. Struc tures assigned to Seismic Design Category B or C. (Note that the loadcombinations that include the redundancy factor are not used for Seismic DesignCategory A.)

2. Drift calculation and P-delta effects.

3. Design of nonstructural components.

4. Design of nonbuilding structures that are not similar to buildings.

5. Design of collector elements, splices and their connections for which the loadcombinations with overstrength factor of §12.4.3.2 are used.

6. Desig n of members or connections where the load combinat ions with overstrengthof §12.4.3.2 are required for des ign.

7. Diaphragm loads determined using Eq. 12.10-1 (note that this does not apply toforces transferred through a diaphragm, such as due to an out-of-plane offset inthe seismic load resisting system, and the higher p factor may apply as otherwiserequired).

8. Structures with damping systems designed in accordance wi th 18.

Additionally, §12.3.4.2 ident ifies two other conditions in which p may be taken as 1.0.Note that the criteria for these condi tions need only be met at floor levels in which morethan 35-percent of the base shea r is being resisted; for the top level or levels of tallerstructures, the cond itions need not be met. The factor may be taken as 1.0 when either ofthe conditions listed below is met. In all other conditions, p is taken as 1.0. There is nolonger a calcu lated p factor between the minimum and maximum value s.


§12.3.4 Example 15 II Reliability/Redundancy Coefficient p

Condition I12.3.4.2(a) Configurations in which the removal of one element (as described belowin the summary of Table 12.3-3) will not result in an increase of more than 33-percentreduction in story shear strength or in an extreme torsional irregularity (as defined inTable 12.3-1).

Summary of Table 12.3-3

Removal of one element is defined as:

1. The removal of a brace (braced frames).

2. Loss of moment resistance at the beam-to-column connections at both ends of asingle beam (moment frames).

3. Removal ofa shear wall or wall pier with a height-to-length ratio greater than 1.0(shear wall systems).

4. Loss of moment resistance at the base connections of any single cantilevercolumn (cantilever column systems).

5. For other systems, such as seismically isolated structures, no prescriptiverequirements are given, allowing p to be taken as 1.0.

Condition II12.3.4.2(b) Configurations with no plan irregularities at any level and with sufficientperimeter braced frames, moment frames, or shearwalls. Sufficient perimeter bracing isdefined as at least two bays of seismic force-resisting perimeter framing on each side ofthe structure in each orthogonal direction. For shear wall systems the number of bays iscalculated as the length of shear wall divided by the story height (two times the length ofshear wall divided by the story height for light-framed construction).

EXAMPLETo illustrate the application of the method for establishing the redundancy factor, thestructure shown in Figure 15.1 will be analyzed.

Wall E Wall FStiffness K. Stiffness Kf

WaliA WalleStiffness Ko Stiffness x,

WailSStiffness Kn

Wall G Wall HStiffness Kg Stiffness x,

WaliDStiffness K!

Figure 15-1


Example 15 " Reliability/Redundancy Coefficient p

Given information:

SDCDOne story, concrete shearwall building

All walls have the same nominal shear strength , R"

The story height is 18 feet.

The length of each shear wall is 15 feet.

§12.3.4

I)

I

IIII

For purposes of the required strength of the walls, the redundancy factor must bedetermined and used in Equation 12.4-3 to determine the horizontal seismic load effect.None of the conditions listed in §12.3.4.1 apply, and thus §12.3.4.2 must be used todetermine whether is 1.0 or 1.3.

Because there are two bays of shear wall on each of the perimeter lines of resistance andthe building is completely regular, §12.3.4.2(b) might allow a factor of 1.0. However, thelength of each shear-wall bay is less than the story height, the number of bays as definedby §12.3.4.2(b) is less than two, and thus the configuration does not automatically qualifyfor a redundancy factor of 1.0. The configuration will therefore be analyzed using themethod outlined in §12.3.4.2(a), namely, by removing a wall and assessing the effect onstory shear strength and on building torsion. In this example Wall C will be removed.Because of the symmetry of the system, the removal of one wall covers the cases of theremoval of each of the other walls. In a more typical system, a separate check would needto be performed for several (or even all) of the walls.

The effect on story shear strength can be considered in at least two ways. The mostconventional way to calculate the modified story shear strength is based on the modifiedelastic distribution of forces and the capacity of the most heavily stressed wall. Such ananalysis of the structure with all four bays present shows that the seismic forces in eachline of resistance (including the effects of accidental torsion) are 52.5-percent of the baseshear, with each bay on each line resisting 26.25-percent; this distribution is shown inFigure 15.2(a). If the stiffness of one line of resistance is reduced by half, the designseismic forces change to 42-percent resisted on the weaker line and on the stronger line;this distribution is shown in Figure 15.2(b). Thus the increase in the force on the mostheavily loaded bay is 42%/26.25% = 1.6, and the reduced force level causing yielding ofthat wall is 1/1.6 = 62.5%. Using this method, then, the effect on story drift is assessed tobe a decrease in capacity of 100% - 62.5% = 37.5%, and thus the configuration wouldnot qualify for a p factor of 1.0.

2006 IBC Strucrural/Seismic Design Manual, Vol. I 7 1

§12.3.4 Example 15 • Reliability/Redundancy Coefficient p

2.5% 2.5% 6.5% 6.5%

23.75% 1 ~26.25% 31% 1

-$- -$-

23.75% 1 t ~ 26.25% 31% t t ~ 42 %

2.5%

(a)

2.5%

Figure 15-2

6.5%

(b)

6.5%

While this is an acceptable method of demonstrating compliance with the conditionsjustifying a factor p of 1.0, this method is not required. A more direct method ofestablishing story shear capacity is to utilize a plastic mechanism analysis. This is themethod envisioned by the committee that authored the redundancy provision, and it ismore consistent with the principles of seismic design (i.e., considering strength and limitstates, rather than elastic design). In this method of analysis, the story shear capacitybefore removal of a wall is the sum of the capacities of the 4 walls resisting the seismicforce in the direction under consideration (provided that the orthogonal walls havesufficient strength to resist the torsion, which in this case is only the accidental torsion).This is shown in Figure 15.3(a), where Rn denotes the capacity of the wall. If one wall isremoved, the story shear capacity is the sum of the capacities of the 3 remaining wallsresisting the seismic force in the direction under consideration; again, the orthogonalwalls must be checked for the forces resulting from building torsion, which in this case issubstantial. This is shown in Figure l5.3(b). Thus the reduction in capacity is only 25percent. The resulting building torsional forces must be resisted by the frames in theorthogonal direction. This interpretation of the story shear capacity has been endorsed bythe SEAOC Seismology Committee.

e. = 50/a Rtf n;> 5% R. Ru = 32.5% R. Ru = 32.5% R.

R. ~ ~ n, R. ~

-$- -$-

R. ~ j t ~ R. R. ~ t ~ R.

n,= 5% R. Ru = 5% R.

(a)

Ru = 32.5% R. e, = 32.5% R.

(b)

Figure 15-3


Examp le 15 • Reliab ility/Re dundancy Coefficient p §12.3.4

I1

)

To qualify for a factor of 1.0, the system with one wall removed must also be checkedfor an extreme torsional irregularity as defined in Table 12.3-1. For the example, usingthe plastic mechanism analysis, the deflection in the direction ofloading is R,,1Kn• Theadditional deflection at each perimeter line due to rotation is 0.325RnlKn• This is less thanthe 40-percent maximum that is allowed by Table 12.3-1 before an extreme torsionalirregularity is deemed to exist. Thus, the configuration qualifies for a p factor of 1.0.

2006 IBC Strucr ural/Seismic Design Ma nual, Vol. I 73

§12.8.7 Example 16 • P-delta Effec ts

. xample 16P.-d Ita ERects §12.8.7

In high-rise building design, 'important secondary moments and addi tional story driftscan be developed in the lateral-force-resisting system by P-delta effec ts. P-deltaeffects are the result of the axial load P in a column being moved laterally byhorizontal displacements, thereby causing additional secondary column and girdermoments. The purpose of this example is to illustrate the procedure that must be usedto check the overall stabili ty of the frame system for such effects.

A IS-story building has a steel specia l moment frame (SMF).

T/ / " " " " "

R = 8

Cc/ = 5.5

1 = 1.0

At the firs t story , hi = 20'

l:.D = W = 8643 kipsl:.L = 3850 kipsVI = V = 0.042W = 363.0 kips, ~ = 0.80h, = 20 ftDeflection at level x = I due to seismic base shear V (without P-delta effects)Ol e = 0.00311 1 = 0.72 in


Seismic Design Category D

Seismic Use Group I

Determ ine the following.

[!J Initial design story drift l:!. in first story

~ P-delta criteria for the building

~ Check the first story for P-delta requirements

~ Final design story drift and story shear in first story

~ Check for story drift compliance in first story

74 2006 IBC Structural/Seismic Des ign Manual, Vol. I

Example 16 • P-del ta Effects §12 .B.7

,,Calculations and Discussion

OJ Initial design story drift !:J. in first story

At story x = I, the preliminary desig n story drift is

where

b) = Cdb'r = 5.5(0.72) = 3.96 inI 1.0

Now: !:J. = 3.96

Code Reference

§12 .8.6

(Eq 12.8-15)

§12 .8.7

J

This value is termed initial because it may need to be increased by the incremental factorGd = 1.0/(1-6) as determined in Part []] of this example.

~ P-delta criteria for the building

P-delta effects must be considered whenever the ratio of secondary momentsto primary moments exceeds 10 percent. This ratio is defined as stabilitycoefficient 6

6= (Eq 12.8- I6)

IIIIII

where

6 = stability coefficient for story x

P-, = total design vertica l load on all columns in story x(Note: no factor above 1.0 is required)

!:J. = initial design story drift in story x occurring simultaneously with CdT!,

v, = seismic shear force in story x

lis.< = height of story x

C: = deflection ampl ification factor in Table 12.2-1 (given = 5.5)

P-delta effects must be considered when 6> o.I0

2006 IBC Structural/S eismic D esign Manual, Vol. I 75

Section 12.8.7 requires that the total vertical load P l at the first story be considered thetotal dead J:.D plus floor live J:.L and snow load S above the first story. These loads areunfactored for determination of P-delta effects.

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

§ 12 .8. 7

76

Example 16 • P·delta Effects

Check P-delta requirements for the first story

Using S = 0 for the building site,

PI = 8643 + 3850 = 12,493 kips

For story x = I,

8 , = ?,I'J. = (8643 + 3850)(3.96) = 0. 103 > 0.100V,h"C d (363.0)(20 ft)(l2)(5 .5)

: . P-delta effects must be considered.

Check for 8 :'0 8max using the given ~ = 0.80

8 = 0.5 = 0.5 = 0. 1136lIIax ~Cd (0.80)(5 .5)

0.103 < 0.1136 ... o.k.

Final design story drift and story shear in first story

ad=~ = \.0 = 1.\151-8 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 = adVI = (1.\15)(363.0) = 404.7 kips


§12.8.7

(Eq 12.8-17)

§12.8.7

I1

II

J

III

IIIIIIII

I

1

Example 16 • P-delta Effects

Check for story-drift compliance in the first story

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

/:;0110'" =0.020(20 ft)( 12) =4.80 in

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

§12.B.7

§12.8.7

T 12.12-1

,Commentary

In § 12.8.7 the P-delta effects on the design story drift and the design story shear areevaluated by the follow ing procedure:

I

I

I.

2.

Given the initial design story drift /:;" = 0.,- 0".1 at story x: compute for each story xthe stability coefficient 8x given by Equation 12.8-16. For each story where 8., isequal to, or greater than 0.10, compute the corresponding incremental factor relatingto P-delta effects ad = 1/( I - 8.,). This factor accounts for the multiplier effect due tothe initial story drift /:;., leading to another increment of story drift, leading to anotherstory drift, which would lead to yet another increment, etc. Thus both the drift and theshear in the story would be increased by a factor equal to the series of I + 8 + 82 + 83

+ ---, which converges to 1(1 - 8) = ad. As a resu lt the initial story drift /:;" and storyshear V, need to be multiplied by the factor ad to represent the total final P-deltaeffect.

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

III

3. In each story requiring consideration ofP-delta effects the initial story shears areincreased to ,':=ad V" . The structural elements must be designed to resist theresulting final story shears, overturning moments and element actions.

Some computer programs for frame analysis state that P-delta effects are incl uded directly inthe analysis. The engineer should verify that the total gravity load employed and the methodused in these programs will provide results that are essentially equivalent to the augmentedstory shear method described above.

The provisions in §§12.8.6 and 12.8.7 for the evaluation of the final story drifts state that thefinal 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 storyshears in these stories are increased by the ad factor. This is equivalent to an added latera lload equal to (ad-I) V, applied to each story level having 8 > 0.1. Therefore the new storydrifts in the stories below would be inc reased not only by their own ad but by the addedlateral load effect from the stories above; thus , the fina l drifts should be found by a newanalysis with the added lateral loads equal to (ad - I) V" along with the initi al lateral loads onthe frame.

2006 la c Structural/S eismic Design Manual, Vol. I 77

ii~ample 17S,eismic Base Shear

§12.8.1 Example 17 • Seis mic Bas e Shear

§12.8.1

1]

]Find the design base shear for a 5-story steel special moment-resisting frame buildingshown below.



S DS = 0.45gSOl = 0.28gJ = 1.0R = 8W = 1626 kips11" = 60 feet

To solve this example, follow these steps.

60'

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

III

[I] Determine the structure period §12.8.2.1

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

[L] Determine the structure period

~ Determine the seismic response coefficient c,~ Determine seismic base shear

Calculations and Discussion

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

~ Determine the seismic response coefficient Cs

The design value of C, is the smaller value of

C = S os = (0.45) = 0 0561s (R) (8) . -

J 1.0

78 200 6 IBC Structural/Seismic Design Manual, Vol. I

Code Reference

(Eq 12.8-7)

§12.8.1

(Eq 12.8-2)

IIIIIIII

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, andthis is used to determine per iod. Roof penthouses aregenerally not considered in determining hi!> but heightsof setbacks are included. However, if the setbackrepresents more than a 130-percent change in thelateral force system dimension, there is a vertical 33'

geometric irregular ity (Table 12.3-2). Tallerstructu res, more than five stories or 65 fee t in height,require dynamic analysis for this type of irregularity.

h« = 33 feet

Setback

k--J

-'- - '- - '- - L- _ L.- _ L.-

CT = 0.016 ; x = 0.9

To = CT(hnY= 0.016(33)°·9= 0.37 sec

[!J Steel eccentric braced frame (EBF)

EBF structures use the C, for the "all otherbuildings" category

CT = 0.030 ; x =0.75

T = CT(hn}T: = 0.030(44)°·75 = 0.51 sec

~ Masonry shear wall building

44'

29'

IE

~TYP'

60'

Front wall elevation

29'

I~~f"45'

<

Back wall elevation

For this structure, CTmay be taken as 0.020 and x may be taken as 0.75, the values for "allother buildings"

To= CT(h,Y = 0.020(29)°·75= 0.25 sec

2006 IBC Structural/Seismic Design Manual, Vol. I 8 1

§12. 8.2.1 Exa mple 18 • Approximate Fu nd amental Period

~ Tilt-up building

Consider a tilt-up building 150 feet by 200 feet in plan that has a panelized wood roof andthe typical wall elevation shown below.

20'I~I• 4 - 15'typ~ '- . . .

4t- HI "" ,~'" zo' '" IE )

Typical wall elevation

CT = 0.020; X = 0.75

T = CT(hn)' = 0.020(20)°·75= 0.19 sec

This type of structural system has relatively rigid walls and a flexible roof diaphragm.The code formula for period does not take into consideration the fact that the realperiod of the building is highly dependent on the roof diaphragm construction. Thus ,the period computed above is not a good estimate of the rea l fundamental period ofthis type of building. It is acceptable, however, for use in determining design baseshear.

comm'fmtaf)'.

The fundamental period T of the building may also be established by analyticalprocedures with the limitation given in §12.8.2.

8 2 2006 la c Structural/Seismic Design Man ual, Vol. I

1

,)

I

I

IIIII)

II

Example 19 II Simplified Alternative Structural Design Procedure §12.14

Exam Ie 19Simplified I ernettve Structura l Design Procedure § 12.14

Determine the seismic base shear and the seismic lateral forces for a three-story woodstructural panel wall building using the simplified alternative structural design procedure.


Occupancy Category I

S DS = 1.0

R = 6X

W = 750 kips

T f ~ EffectiveSeismic

I( 20' + 20' 1 Weight, w,Level

150 kips

12'

300 kips

12'

300 kips

12'


[}J Check applicability of simplified alternative method

~ Determine seismic base shear

~ Determine seismic lateral forces at each level


[}J Check applicability of simplified alternative method §12.14.1.1

Light-framed cons truction not more than three stories, or other buildings with bearing wallsor building frame systems not more than three stories, can use the simplified alternat ivemethod when general conditions are satisfied.

2006 lBC Structural/Se ismic Des ign Manual, Vol. I 83

The following is a comparison of simplified base shear with standard design baseshear. The standard method of determining the seismic base shear is

§12.14 Example 19 • Simplified Alternative Structura l Des ign Procedure


1.2(1.0)(750 kips)

6X

= 138.5 kips

@J Determine seismic lateral forces at each level

WF =-" V

x W

FI = 300 (138.5) = 55.4 kips750

F2 = ~~~ (138.5) = 55.4 kips

F 3 = ISO (138.5) = 27.7 kips750

Commentary

v=csw

where


§12.14.7.1

(Eq 12.14-9)

§12.14.7.2

(Eq 12.14-10)

§12.8

(Eq 12.8-1)

(Eq 12.8-2)

]

IIIIIIIIIIIII

1

Example 19 • Simplified Alrernative Structural Design Procedure

The distribution of seismic forces over the height of the structure is

§12.14

1where

Fx = C.~V

n

2: w,h;i.. 1

(Eq 12.8-11)

(IBe Eq 16-42)

1Note: distribution exponent k = 1.0 for structures having a period of 0.5 second orless.

Level .r h, U', w,h."'xlix

F, Fju '.}: wi"i

3 36 fl 150 kips 5,400 kip-It 0.333 38.4 kips 0.278

2 24 300 7 .200 0.444 5 1.2 0.185

I I 12 300 3.600 0.222 25.6 0.093

L II',", 16.200 L ~ 115.4

The seismic base shear Vand lateral forces F" at each level except the roof are all lessthan the simplified method, see table below. The principal advantage of the simplifiedmethod is that period T need not be calculated and design story drift (). may be takenas I percent of the story height, §12.14.7.5.

Comparison of Simplified VS Standard

Lateral Force F.% Difference

Level x Simplified Standard

3 27.7 kips 38.4 kips 72

2 55.4 51.2 10855.4 25.6 216

Total 138.5 115.2 120


§12.2.3.1 Example 20 • Combination of Structural Systems: Vertical

!~ample 20

'Combination of Structural Systems: Vertical '12.2.3.1

In structural engineering practice, it is sometimes necessary to design buildings thathave a vertical combination of different lateral-force-resisting systems. For example,the bottom part of the structure may be a rigid frame and the top part may be a braced frameor shear wall. This example illustrates use of the requirements of §12.2.3.1 to determine theapplicable response modification coefficient R system overstrength factor no and deflectionamplification factor Cd values for combined vertical systems.

For the three systems shown below, determine the required R coefficient, no factor, Cdfactor, and related design base shear requirements.

.Calculations and Discussion Code Reference

[!J Steel Special concentrically braced frame (SCBF) over steel special

moment frame (SMF)

Seismic Design Category C

Ordinary steel concentricallybraced frame

R =6.0nu = 2.0Cd = 5.0P = 1.0

Special steel moment frameR = 8.0no= 3.0C« = 5.5P = 1.0

This combined system falls under vertical combinations of §12.2.3.1. Because the rigidframing system is above the flexible framing system, the exception for a two-stage analysisin §12.2.3.1 cannot be used. Therefore, the structure in this direction must use the lowestR = 6.0 and the largest no = 3.0. Recall that lithe floor and roof diaphragms could beconsidered to be flexible, nu would be 2.5, per footnote g, Table 12.2-1.

86 20D6 IBC Structural/Seismic Design Manual, Vol. I

Example 20 If CombInation of Structural Systems: Vertical §12.2.3.1

~ Ordinary reinforced concrete shear wall (ORCSW) over special reinforcedconcrete moment frame (SRCMF)

Seismic Design Category C

Ord inary reinforced concreteshear wall (non-bearing)

R =5.00 =2.5Cd = 4.5P = 1.0

Special reinforced concretemoment frame

R = 8.0.00 = 3.0Cd = 5.5P = 1.0

Th is combined system falls under vertical combinations of §12.2.3. 1. Because therigid portio n is above the flexible portion, a two-stage analysis cannot be used.Therefore, the structure in this direction must use the lowest, R = 5.0, and the largest,.00 = 3.0. Also note that ordinary reinforced concre te shear wall systems are not permittedabove 35 feet in SOC 0 , E, or F (Table 12.2-1 ).

[!J Concrete SRCMF over a concrete building frame system

~ Applicable criteria.This is a vertical combination of a flexible system over a more rigid system.Under §12.2.3.1, a two-stage static analysi s may be used, provided the structuresconform to the following four requirements.

Seismic Design Category B

Concrete special reinforced concretemome nt-frameR =8.0.00 =3.0C« =5.5P = 1.3Stiffness upper portion = 175 kip-inT upper = 0.55 secT eoll/billed =0.56 sec

Shear walls

2006 IBC Structu ral/Se is mic Design Manual. Vol. I 87

10,000 kip-in > 10(175) = 1750 kip-in ... o.k.

Check requirements of §12.2.3.1 for a two-stage analysis

Also note R is different for bearing wall systems versus building frame systems for specialreinforced concrete shear walls, see Table 12.2-1.

b. Period of entire structure is not greater than l.l times the period of upper structureconsidered a separate structure fixed at the base.

J

1

J

1

I

IIIII

Example 20 • Combination of Structural Systems: Vertical

a. The stiffness of the lower portion is at least 10 times the stiffness of the upperportion. For multi-story upper or lower portions , the stiffness should be thestiffness of the first mode.

Special reinforced concrete shear wallR =6.0no=2.5Cd =5P = 1.0Stiffness = 10,000 kip-inrlo"er = 0.03 sec

§12.2.3.1

0.56 sec < 1.1 (.55) = 0.61 sec . . . o.k.

88

c. Flexible upper portion supported on the rigid lower portion shall be designed as aseparate structure using appropriate values of Rand p.

d. Rigid lower portion shall be designed as a separate structure using appropriatevalues of Rand p. Reactions from the upper structure shall be determined fromanalysis of the upper structure amplified by the ratio ofRip of the upper structureover Rip of the lower structure. This ratio shall not be less than 1.0.

2006 lac Structural/Seismic Design Manual, Vol. J

IIIIIII

Example 20 • Combination of Structural Systems: Vertical §12.2.3.1

~ Design procedures for upper and lower structures

Design upper SRCMF using

R =8.0n =3.0p = 1.3

Design lower portion of the building framesystem for the combined effects of amplifiedreactions from the upper portion and lateralforces due to the base shear for the lowerportion of the structure (using R = 6.0,Q =2.5, and p =1.0 for the lowerportion).

--.......r VIrum.

8.0 /1.3Amplified Vllilm • =- - Vllilme =1.03 Vllilme

6.0/1.0

The reactions from the upper portion shall be determined from the analysis of the upper portionamplified by the ratio of (Rip) for the upper portion over (Rip) of the lower portion.

Note that for the basic seismic load combinations the factor p must still be applied to forcescorresponding to Vlower.



!TI Determine the R value for each direction

Examp le 21") : ' .

gombination of F!"aming Systems in Differ~nt Directions §12.2.2

IJ

J

I1

II

IIIIIIIIIII

_-- Shear wall

---G

Example 21 • Combination of Fram ing Systems in Different Dir ections

Typ ical floor Plan

Lines I , 2, and 3 are special reinforced concrete moment framesR =8.0, QQ= 3.0, Cd = 5.5 Table 12.2-1 (C5)

Lines A and D are special reinforced concrete shear walls (bearing wall system)R = 5, Q" = 2.5, Cd= 5, Table 12.2-1 (AI)

Determine the R, Cd, and Q " values for each direction.

This example illustrates the determination of response modification coefficient R, systemover strength factor QQ, and deflection amplification factor Cdvalues for a building that hasdifferent seismic framing systems along different axes (i.e., directions) ofthc building.

In this example, a three-story building has concrete shear walls in one direction andconcrete moment frames in the other. Floors are concrete slab, and the building is SDC Dand Occupancy Category I.

§12.2.2

Example 21 • Combination of Framing Systems in Different Directions §12.2.2

.Calculations and Discussion., Y ,.

Code Reference

I

The provisions of §12.2.2 require that where different seismic-foree-resisting systems areused along the two orthogonal axes of the structure, the appropriate response modificationcoefficient R, system overstrength factor Q o' and deflection ampl ification factor Cd for eachsystem shall be used.

Use R =5.0, Qo =2.5, and Cs=5 for the north-south direction.

and R = 8.0, Qo = 3.0, and Cd= 5.5 for the east-west direction.

~'.

Commentary

Note that since this is SDC D, ordinary reinforced concrete shear walls are not permitted.


§12.2.3.2 Example 22 " Combination of Structural Systems: Along the Same Axis

.. ~ .. -EX~mple22

COmbination of Structural Systems:, , long the Same Axis §12.2.3.2

Occasionally, it is necessary or convenient to have different structural systems in the samedirection. This example shows how the response modification coefficient R value isdetermined in such a situation.

A one-story steel frame structure has the roof plan shown below. The structure is assigned toSeismic Use Group 1.

-,'- -- -'- -'

Roof Plan

Lines 1 and 4 are ordinary steel moment frames: R = 3.5

Lines 2 and 3 are special steel concentrically braced frames: R = 6.0

[!J Determine the R value for the N/S direction


When a combination of structural systems is used in the same direction, §12.2.3.2 requiresthat (except for dual systems and shear wall-frame interactive systems) the value ofR usedshall not be greater than the least value of any system utilized in that direction.

: . Use R = 3.5 for entire structure.

Commentgl'Y

An exception is given for light frame , flexible diaphragm buildings of Occupancy Category Ior Il two stories or less in height. However, to qualify as a flex ible diaphragm, the lateraldeformation of the diaphragm must be more than two times the average story drift of theassociated story; see definition in §12.3.1.3.


1

•

Example 17 • Seismic Base Shear

and

C- SOl _ (0.28) _,--i)-() -0.0467 for T S. TL

R ~ (0.75)/ 1.0

but shall not be less than

C, = 0.01

§12.B.l

(Eq 12.8-3)

(Eq 12.8-4)

(Eq 12.8-5)

In addition, for structu res located where S, is equal to or greater than 0.6g, C, shall not beless than

ComtnimtcJry

The So, value of 0.28g given in this example is based on an S, value of 0.2I g. If the S,value were to have been equal or greater than 0.6g, then the lower bound on C, is

I

IIII

C = 0.5S,

s ( ~)

: . Design value of C, =0.0467


The seismic base shear is given by

v =C,W

= 0.0467(1626 kips)

= 75.9 kips

C0.5/S,

> - -s z-: R

(Eq 12.8-6)

§12.8.1

(Eq 12.8-1)

(Eq 12.8-6)


§12.8.2.1 Example 18 • Approximate Fundame ntal Period

'. ample 18Appro~imate Fundamental Period § 12.8.2.1

Determine the period for each of the structures shown below using the appropriatefundamental 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

Calculatipns and Discussion

[!J Steel special moment frame (SMF) structure

Code Reference

§12.8.2.1

Basemenl

Superslructu re

-,-

96'

e

,22'

_L

Grad

To= Cr (I1"r = 0.028(96)°·8= 1.08 sec

Cr = 0.028; x = 0.8

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

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

80 2006 IBC Structuraf/Seismic Design Manual, Vol . I

Example 23 • Vertica l Distributio n of Seismic Forc e §12.B.3

1 Example 23 ·Vertical Distribution of Seismic Force § 12 .8.3

A nine-story building has a moment-resisting steel frame for a lateral-force-resistingsystem. Find the vertical distribution oflatera l forces Fx •

' 2'

' 2'

'2'

' 2'

' 2'

' 2'

12'

20'

12'22k

22k

54k

40k

65k

2

4

4

4

5

4

4

4

4

// " 1' / / 1' / ~ / / /I' / '/

8 OSk

? f T

3

2

5

7 QSk

6 O~

,

I 27' I 27' I Storylevel ( 0) EO ): weight

9 14k

IV = 3762 kipsC, = 0.062R = 8.0Q" = 3.01 = 1.0T = 1.06 sec

The following informationIS given.

I

I

Total 3762 kips

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

IIIII

I


[!J Determine V

~ Find Fx at each level

~ Find the distribution exponent k

~ Determine vertical force distribution

:Ca/culations and Discussion

[!J Determine V

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

Code R...-ference

§12.8-1

(Eq 12.8-1)

2006 IBC Structural/Se;smlc Design Man ual, Vol. I 93

§12. 8.3 Example 23 • Ver tic al Dis tr ibu tion of Seismic Force

~ Find r, at each level

The vertica l distribution of seismic forces is determined as

(Eq 12.8- 11 )

II]

where

c = lVxhlI'X n

2: IV; hii- I

(Eq 12.8-12) I]

Since there are nine levels above the ground, 11 = 9

Thus:

Find the distribution exponent k §12.8.3The 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 thebuilding period, k is determined by linear interpo lation.

2006 IBC Struc tural/S eismic Design Manual, Vol. I

Exponent, k

IIIIIIIIIJ

I

2.52.01.0 1.51.06

0.5oo

2.5

2.0

... 1.5

." 1.28e'C~ 1.00-

0.5

F = 233.8w,l1;x •

~ ,LJ lV /Iii- I

Thus:

94

Examp le 23 • Vertical Distribution of Seismic Force

I Now:

for T =1.06 sec

k =1.0 + (1.06 _ 0.5) ( I )2.5 -0.5

=1.28

Use: k =1.28

§12 .B.3

I

IIIIIIII

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

• • W./I .~".f U'.• wxhx C~ , = --I F. = C••VLevel X ii, ( ft) (kips) kip-It LW/l i (kips) F/ w. = Su

9 116 ft 439 214 93.946 0.116 27.3 0.127

8 104 382 405 154.710 0.192 44.8 0.1 II

7 92 326 405 132.030 0.169 38.3 0.094

6 80 273 405 110.565 0.137 32.1 0.079

5 68 222 584 129.648 0.161 37.6 0.064

4 56 173 422 73.006 0.09 1 21.2 0.050

3 44 127 422 53.594 0.067 15.5 0.037

2 32 84 440 36.960 0.04 6 10.8 0.024

20 46 465 21,390 0.027 6.2 0.013

~ =3762 ~ = 806.289 1.004 233.2

Commelltary

Note that certain types of vertical irregularity can result in a dynamic response hav ing a loaddistribution significantly different from that given in this section . Table 12.6- I lists theminimum allowable analysis procedures for seismic design . Redundancy requ irements mustalso 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 verticaldistribution 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 isis done to simplify the analysis of the framing members. The horizontal forces need not beincreased at each floor level when p is greater than 1.0, provided that, when stress check ingthe individual mem bers of the lateral-foree-resisting system, the seismic forces are factoredby p. When checking building drift, p = 1.0 (§12.3.4.1) shall be used.

2006 lac Structural/Seismic Design Manual. Vol. I 95

§ 12. 8.3 Exampl e 23 • Vertica l Distr ibution of Se ism ic Forc e

Structures that have a vertical irregularity ofType Ia, Ib, 2, or 3 in Table 12.6-1, or planirregular ities of Type l a or Ib in Table 12.6-1, and having a height exceeding five stories or65 feet may have significantly different force distributions. Structures exceeding 240 feet inheight shall require dynamic analysis. The configuration and final design of this structuremust be checked for these irregularities. Most structural analysis programs used todayperform this calculation, and it is rarely necessary to manually perform the calculationsshown above. However, it is recommended that these calculations be performed to confirmthe computer analysis and to gain insight to structural behavio r. Note that ( So )max is

approximately twice C , and S" = r <pSa from a modal analysis.

96 2006 IBC Structural/Seism ic Des ign Manual, Vol. I

IIII

IIIIrIIIIIIII

Example 24 • HOlizontal Distribution of Shear §12.BA

1 EKamp!e 24:Horizontal Distribution of Shear §12.8.4

A single-story building has a rigid roof diaphragm. See appendix to this example for aprocedure for the distribution of lateral forces in structures with rigid diaphragms andcross walls and/or frames of any orientation . Lateral forces in both directions are resistedby shear walls. The mass of the roof can be considered to be uniformly distributed, and inthis example, the weight of the walls is neglected. In actual pract ice, particularly withconcrete shear walls, the weight of the walls should be included in the determination ofthe center-of-mass (CM).


Design base shear: V = 100 kips in north-south direction

Wall rigidi ties: RA = 300 kip/inRB = 100 kip/inRc =RD =200 kip/in

Center-of-mass: Xm= 40 ft,Ym = 20 ft

Analyze for seismic forces in north -south direction.

r y0

-, Shear wallbelow•• .' XR

A

~40' ",,

Roofdiaphragm

YR

X)

c

IEBO'

)1

Roof plan

2006 IBC Structural/Seismic Design Manual. Vol. I 9 7

§12.B.4 Example 24 • Hor/zontal Dist ribution of Shear


I

OJ Eccentricity and rigidity properties

[3J Direct shear in walls A and B

[I] Plan irregularity requirements

~ Torsional shear in walls A and B

~ Total shear in walls A and B


OJ Eccentricity and rigidity properties

The rigidity of the structure in the direction of applied force is the sum ofthe rigidities of walls parallel to this force .

~ode Reference

§12 .8.4.1

III

98

R = RA + RB = 300 + 100 = 400 kip/in

The centers ofrigidity (CR) along the x and y axes are

eccentricity e = Xm - XR = 40 - 20 = 20 ft

Torsional rigidity about the center of rigidity is de termine d as

J =RA (20)2+ RB (60/ + Rc (20)2 +RD(20)2

=300 (20i + 100 (60)2 +200 (20)2 + 200 (20i = 64 x 104 (kip/in) fe

The seismic force V app lied at the CM is equivalent to having Vapplied at the CRtogether with a counter-clockwise torsion T. With the requirements for accidentaleccentricity es«, the total shear on walls A and B can be found by the addition of thedirect and torsional load cases .

2006 IBC Structural/Seismic Design Manual, Vol. 1

IIIIIIIIIII

Examp le 24 • Horizontal Distribution of Shear §1 2.8A

/ r rr: / /

20'

CR

• ...; ;-

20'20' 60'

/ / /

VC

A

VO.A D

Direct shear contribution

VO,S VT,A D VT,O VT, a4

A l ~/

faCR~ T= V(e:: eecc)

~VT,C C

Torsional shear contribution

~ Direct shear in walls A and B

R 300 .V = A X (V) = x 100 = 75.0 kips

0.11 RA

+ RB 300 + 100

V = RD X (V) = 100 x 100 = 25.0 kipsD.n R + R 300 + 100

A n

@J Plan irregularity requirements

The determination of torsional irregularity, Items la and lb in Table 12.3-1, requiresthe evaluation of the story drifts in walls A and B. This evaluation must include accidentaltorsion caused by an eccentricity of 5 percent of the building dimension.

eacc + 0.05 (80 ft) = 4.0 ft

For the determination of torsional irregularity, the initial most severe tors ional shears,V' and corresponding story drifts (so as to produce the lowest value of the averagestory drift) will result from the largest eccentrici ty e + eacc • These are

v' = V (e+e"cJ(xR)(RA ) = 100(20 + 4)(20)(300) = 22.5 ki sT.A j 64 x 10~ P

~' = V(e+eacJ(80 -xR)(RB ) = 100(20+4)(60)(100) = 22.5 ki sT.B j 64 x 1O~ P

2006 IBC Structural/Seismic Design Manual, Vol . I 99

§12.8.4 Example 24 • Hor izo ntal Distribution of Sh ear I

The initial total shears are

V~ = VD.A - V;.A = 75.0 - 22.5 = 52.5 kips

V; = VD.B + \~.B = 25.0 + 22.5 = 47.5 kips

(NOTE: This is not the design force for Wall A, as accidental eccentricity here is used toreduce the force).

The resul ting displacem ents b', which for this single-story building are also the storydrift values, are

b' = V~ = 52.5 = 0.18 inRA 300

)

)

= b~ = 0.48 in

Section 12.8.4.3 requires the evaluation and application of the torsional amplificationfactor

= 0.18 + 0.48 = 0.33 in2

= V~ = 47.5 = 0.48 inRB 100 I

III

IIII

(IBC Eq 16-44)

b"4' = 0.49 = 1.45 > 1.4b.,., 0.33

:. Extreme Torsional Irregularity Type Ib exists. (See Example 26) AssumingSDC D, structural modeling must include 3 dimensions per §12.7.3, anddiaphragm shear transfer forces to collectors must be increased 25 percent per§12.3.3.4.

A, = ( b"4' )' = ( 0.48 ) ' = 1.47 < 3.0. 1.2." 1.2(0.33)

Note: the factor Ax is not calculated iteratively (i.e., it is not recalculated with amplifiedtorsion).

~ Torsional shears in walls A and B

To account for the effects of torsional irregularity, §12.8.4.2 requires that the accidentaltorsional moment, Ve.cc, be multiplied by the torsional amplification factor Ax.


Example 24 • Horizontal Distribution of Shear

The most severe total shears result from the use of V [e - .{,ed CC ] for VT,A and

V [e + A,eacc] for VT•B

§12.B.4

V T.A =

V T,B =

100 kips[(20 - 1.47 x 4]20(~00 kip/in) = 13.24 kips. 64 x 104 (kip/in jft '

100 kips[(20 + 1.47 x 4]60(100 kip/in) = 24 3 k'" . IpS

64 x 10 (kip/injft '

II

IIIII

I

~ Total shear in walls A and B

Total shear in each wall is the algebraic sum ofthc direct and torsiona l shearcomponents

v:, = VD_, - VT.A = 75.0 - 13.2 = 61.8 kips

VB = VD.B + V T,B = 25.0 + 24.3 = 49.3 kips

r

COmmentary

Section 12.8.4.2 requires that the most severe load combination for each element shall beconsidered for design . This load combination involves the direct and torsional shears,and the "most severe" condition is as follows.

I . Where the torsional shear has the same sense, and is therefore added to the direct shear,the torsional shear shall be calculated using actual eccentricity plus the accidentaleccentricity to give the largest additive torsional shear.

2. Where the torsional shear has the opposite sense to that of the direct shear and is to besubtracted, the torsional shear must be based on the actua l eccentricity minus theaccidental eccentricity to give the smallest subtractive shear.

The §12.8.4.3 requirement to multiply only the accidental torsional moment by Ax differssignificantly from the 2000 !Be. It restores the requirements of the 1997 UBC and 1999 BlueBook.


§12.8.4.3 Example 25 • Amplification of Accidental Tors ion

§ 12 .8.4.3

This example illustrates how to include the effects of accidental eccentricity in the lateralforce analysis of a multi-story building. The structure is a five-story reinforced concretebuilding frame system. A three-dimensional rigid diaphragm model has been formu latedfor the evaluation of element actions and deformations due to prescribed loading conditions.Shear walls resist lateral forces in both directions.

f f f4 @J20'=80'

A

®-gII

bN

@@-M

IN0

Floor plan at Level x

The lateral seismic forces Fein the north-south direction, structure dimensions, andaccidental eccentricity eacc for each Level x are given below.

Level.\" F. L. X es «: =0.05L.r

5 110.0 kips 80.0 ft 24.2 ft ± 4.0 ft

4 82.8 80.0 25.1 ±4.03 65.1 80.0 27.8 ±4.0

2 42. 1 80.0 30.3 ±4.0

I 23.0 80.0 31.5 ±4.0

In addition, for the given lateral seismic forces F, a computer analysis provides thefollowing results for the second story. Separate values are given for the applicationof the forces F, at the centers of mass and the ±O.05Lx displacements as required by§12.8.4.2. In this example, it is assumed for simplicity that the location of the centerof-mass CAtf.e is congruent with the center of rigidity at the level in question , resultingin zero inherent at torsion.


I .

Example 25 II Amplification of Accidental Torsion §12.8.4.3

I Wall shear 1'"Wall shear V.

Story drift II,

Story drift II,

Level 2 displacement 0"

Level 2 displacement 58

x ,~

185.0 kips

115.0 kips

0.35 in

0.62

0 .80

1.31

Force F~ Position

X rl - e... r

196.0 kips

104.0 kips

0.37 in0.56

0.85

1.1 8

X r~ + 00 e...-.:

174.0 kips

126.0 kips

0.33 in0.68

0.75

1.44

For the second story, find the following.

[L] Maximum force in shear walls A and 8

~ Check if torsional irregularity exists

@J Determine the amplification factor Ax

~ New accidental torsion eccentricity

VB = 126.0 kips

VA = 196.0 kips

Check if torsional irregularity exists

The building may have a torsional irregularity Type I (Table 12.3-1). The following is acheck of the story drifts.

CodeReference

= 0.68 + 0.33 = 0.51 in2

= 0.68 in~lJla:c


[!J Maximum force in shear walls A and 8

The maximum force in each shear wall is a result of direct shear, inherent torsion (centerof mass not being congruent with center of rigidity) and the contribution due to accidentaltorsion. As mentioned above, in this example it is assumed that accidenta l eccentricity isthe only source of torsional mome nt at this floor level. From the above table, it isdetermined that

IIIIIII


§12.8.4.3 Example 2S • Amplification of Accidenta l Torsion

li"",. = 0.68 = 1.33 > 1.2«; 0.51

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

~ Determine the amplification factor A x

Because a torsiona l irregularity exists, §12.8.4.3 requires that the second storytorsional moment be amplified by the following factor. In this example, because the onlysource of torsion is the accidental eccentricity, the amplification factor will be used tocalculate a new and increased accidental eccentricity , as shown below.

J

1

II

Ax (Eq 12.8-14)

Where:

the average story displacement is computed as

= 1.44+0.75 = 1.10 in2

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

~ New accidental torsion eccentricity

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

e"cc = (1.19)(4.0 ft) =4.76 ft


III

1

,9 gmmentary

Example 25 • Amplification ct Accidental Torsion §12.8.4.3

IIIII

Example calculations were given for the second story. In practice, each story requiresan evaluation of the most severe element actions and a check for the torsional irregularitycondition.

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 Axvalues and repeat the process asecond or third time (until the Ax converges to a constant or reaches the limit of 3.0). Theresults 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 thedetermination of the most severe element actions for any other lateral-foree-resistingsystem having rigid diaphragms.

When the dynamic analysis method of §12.9 is used, all the requirements of horizontal sheardistribution, given in §12.8.4, including torsion calculations that may be accounted for bydisplacing 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 Axwhere accidental torsional effects are included in the dynamic analysis model. Only theaccidental torsion is required to be amplified if torsional irregularity exists . Also note that Axis not required to exceed 3.0.


§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 Bandlines C and D.


Seismic Design Category CS DS= 1.10

Ordinary reinforced> concrete shear wall (ORCSW) buildingframe system: R = 5 and Qo = 2.5

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

Office building live load: use factor of 0.5 on L

T 12.2-1

§12.4.2.3

Axial loads on column CD = 40 kipsL = 20 kipsQE = 100 kips

12'

12'

12'

12'

L~.f4 ,---------.,....----,

....,..~ Shear wall

Column C24" x 24"

f c =4000 psi

Determine the following for column C.

[!J Required strength

~ Detailing requirements


Examplo 26 • Elements Supporting Discontinuous Systems §12.J.J.J

•Calculations and Discussion Code Reference

I

This examp le demonstrates the loading criteria and detail ing required for elementssupporting 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 firststory columns on lines A and D must support the wall elements above this level. ColumnC on line D is treat ed in this example. Because of symmetry, the column on line A wouldhave identica l requirements.

Section 12.3.3.3 requires that the column shall have a design strength to resist specialseismic load combination of §12.4.3.2

where

Pu = 1.2D+ 0.5L + 1.0Em

P" = 0.9D + 1.0Em

§12.4.2.3 (Comb. 5)

§12.4.2 .3 (Comb. 7)

1I

IIIIII

E.. = Q" QE+ 0.2 SDS D = 2.5( 100) + 0.2( 1.10)(40) = 259 kips §12.4.3.2 (Comb. 5)

or 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)

Substituting the values of dead, live, and seismic loads

P" = 1.2 (40) + 0.5 (20) + 259 = 317 kips compression

and

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

2006 IBC Structural/Seis mic Design Manual. Vol. I 107

§12.3.3.3 Examp l e 26 ff Elements Supporting Discontinuous Systems

Cotpmentary

To transfer the shears from walls A-B and C-D to the first-story wall B-C, collectorbeams A-B and C-D are required at Levell . These would have to be designedaccording to the requirements of §12.10.2.

The load requirements of §12.3.3.3 and related sections of the relevant materials chaptersapply to the following vertical irregularities and vertical elements.

1. Discontinuous shear wall. Thewall at left has a Type 4 verticalstructural irregularity. Note thatonly the column needs to resistthe special load combinationssince it supports the shear wall.

2. Discontinuous column. Th is framehas a Type 4 vertica l structuralirregularity.


1,14"__-rt--Column

DDDDDD

Transfergirder

Example 26 • Elements Supporting Discontinuous Systems

3. Out-of-plane offset. The wall onLine A at the first story isdiscontinuous. This structure hasa Type 4 plan structuralirregularity, and §12.3.3.3applies to the supportingcolumns. The portion of thediaphragm transferring shear(i.e., transfer diaphragm) to theoffset wall must be designed perthe requirements of §12.3.3.4.Note that the transfer diaphragmand the offset shear wall aresubject to the p factor, but not tothe special load combinations.

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

§12.3.3.3

Oisconllnuedwall


7. (0.9 - 0.2SDS)D + QoE

Appl icable load combinations for allowable strength design are:

1

11

IIIJ

IIIIIIIIIII

Ughl framed wallwith plywoodsheathing

Timber column

§12.3.3.3

Code Reference

Example 27 • Elements Supporting Discontinuous Walls or Frames

Dead D = 6.0 kipsLive L = 3.0 kipsSeismic Q£ = ±7.0 kips



Axial loads on the timber column under thediscontinuous portion of the shear wall are

Seismic Design Category CS DS = l.IOR =6.5no = 3.0Cd =4/I = 0.5

[!J Applicable load combinations

~ Required column design strength

This example illustrates the application of the requirements of §12.3.3.3 for the allowablestress design of elements that support a discont inuous lateral-foree-resisting system.

§12.3.3.3

In this example, a light-framed bearing-wall building with plywood shear panels has aType 4 vertical structural irregularity in one of its shear walls, as shown below.

CalclJlations and Discussion

[1J Applicable load combinations

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

fI" '-:-rt: - r

·& ample 2 7r'!'! .".

~~/ements Supporting Discontinuous Walls orFrames


1

Example 27 • Ele ments Supporting Discontinuous Waifs or Frames § 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

Required column design strength (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

J or

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

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

IIIIIIIII

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

P = 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 inwhich L; is less than or equal to 100 psf, with the exception of garages or areas occupiedas places of public assembly.

Commentary

For strength design, the timber column must be checked for a compression load of31.0 kipsand a tension load of 14.3 kips.

In making an allowable stress design check, §12.4.3.3 permits use of an allowable stressincrease of 1.2. The 1.2 stress increase may be combined with the duration ofload increasedescribed in the NOS. The resulting design strength = (1.2)(1.0)( 1.33) (allowable stressdesig n). This also applies to the mechanical hold-down element required to resist the tensionload.


§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 carryhigher and more realistic loads required by the discontinuity in the shear wall at the firstfloor. This is done by increasing the normal seismic load in the column QE by the factorQo = 3.0 to calculate the maximum seismic load effect Em (§12.4.3).


I1

IIIIIIIIII

E::ample 28 • Soil Pressur e At Foundation §§2A; 12.13.4

III le2oil Pressure At Foundations §§2.4; 12.13.

Geotechnical investiga tion reports usua lly provide soil-bearing pressures on an allowablestress 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 thisexample is to illustrate footing design in this situation.

A spread footing supports a reinforced concrete column. The soil classificationat the site is sand (SW).

The following information is given .

Seismic Design Category CSDS = 1.0, / = 1.0P = 1.0 for structural systemPD = 50 kips

PD includes the footing and imposedsoil weight)

PL = 30 kipsPE = ± 40 kips, VE = 25 kips,

(these are the QE loads due to base shear V)Snow load S = 0Wind load W < QE /1.4

The seismic loads are from an equivalent lateral analysis.

r)J

4'

Grade

The loads given above follow the sign convention shown in the figure.

Perform the following tasks.

I}J Determine the design criteria and allowable bearing pressure

~ Determine footing size

~ Determine soil pressure reactions for strength design of the footingsection

2006 IBC Structural/S eismic Design Manual, Vol. I 11 3

§§2.4; 12.13.4 Example 28 • Soil Pressure At Foundation

:Calculations and Discussion Code Reference

[L] Determine the design criteria and allowable bearing pressure §2.4

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

D + 0.7£

D + 0.75 (0.7£ + L)

0.6D + 0.7£

(Comb. 5)

(Comb. 6)

(Comb. 7)

Section 12.13 .4 permits reduction of overturning effects at the foundation-soil interface by25 percent (if an equivalent lateral for ce analysis is used) or 10 percent (if modal analysis isused ). Therefore , for the soil pres sure the seismic effe ct is reduced

D + 0.75(0.7E)

D + 0.75[0.7(0.75)£ + L]

0.6D + 0.7(0.75)£

Because foundation investigation reports for buildings typically specify bearingpre ssures on an allowable stress design basis, criter ia for determining footing sizeare also on this basis.

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

(Comb. 5)

(Comb. 6)

(Comb . 7)

£=£,, +£,.

Per §12.4 .2.2, £ 1' = 0 for determ ining soil pressure. Equation 12.4- I reduces to

(Eq 12.4- I)

(Eq 12.4-3)

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

pa=2.40 ksf for sustained loads and

pa = 3.20 ksf for trans ient loads, such as seismic.

114 2006 IBC S tr uc tu r al /Se ism i c Design Manual, Vol. I

11

Example 28 • Soli Pres sure At Foundation

[3J Determine footing size

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

P = D + 0.75[0.7(0.75)£ + L]

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

P = 0.6D + 0.7(0.75)£

= 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 sfUse 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 inoverturning does not apply, and the vertical seismic load effect is included

§§ 2.4; 12.13.4

(Comb. 5)

(Comb. 6)

(Comb. 7)

A uniform pressure of 115k/27.6 sf = 4.17 ksf should be used to determine the internalforces of the footing. (Note that if the footing also resisted moments, the pressure wouldnot be uniform.)

Note that this indicates uplift will occur. ASCE/SEI 7-05 does not require that foundationstability be maintained using strength-level seismic forces. This combination is onlyused here to determine internal forces of concrete elements of the foundat ion. As itresults in no internal forces, it may be neglected.

IIIIII

P = I.3D + 0.5L + E

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

The other seismic load combination is

P =0.9D +£

= 0.9(50) - 40 + 0.2(1.0)50 = -5k

§2.3.2 (Comb. 5)

§2.3.2 (Comb. 7)

2006 IBC StructuraVSeismic Design Manu al, Vol. I 11 5

§12.8.6 Example 29 • Drift

Example 29Drift §12"B.6

A four-story special moment-resisting frame (SMRF) building has the typical floorplan as shown below. The typical elevation ofLines A through D is also shown, andthe structure does not have horizontal irregularity Types 1a or lb.


Occupancy Importance Category I

Seismic Design Category 0

1 = 1.0Cd = 5.5T = 0.60 sec

Seismic force

Typical fluor plan

Level

4

DDD 12'

3

DDD 12'

2

DDD 12'

12'

Typical Elevation

The following are the deflections (computed from static analysis - effects of P-delta havebeen checked) bxe at the center-of-mass of each floor level. These values include bothtranslational 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 computedfundamental 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 §12.B.6

Level 0"

4 1.51 in

1 3 1.03

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


[!J Maximum inelastic response deflection Ox

These are determined using the Osevalues and the Cd factor

o = CA, = 5.56" = 5.50.r I 1.0 se

Therefore:

Code Reference

§12.B

CEq 12.8-15)

IIIIII1

I

Level 0.(1' 6,

4 1.51 in 8.31 in3 1.03 5.672 0.63 3.471 0.30 1.65

~ Design story drift ~ in story 3 due to Ox

Story 3 is located between Levels 2 and 3.

Thus: ~J = 5.67 - 3.47 = 2.20 in


§12.8.6 Example 29 • Drift

@J Check story 3 for story-drift limit §12.12.1

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

For story 3

/1J = 2.20 in

Story-drift limit = 0.025 (144) = 3.60 in > 2.20 in

:. Story drift is within the limit.


I1

1

J

]

IIIIIIIII)

IIII

Example 30 • Stor.! Drift Limitations §12.1 2

lEJralnple 30Story 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 Oxand notthe 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 determinedaccording to §12.8, using a static, elastic analysis.

Level ~ f ~D

Occupancy Category I Deflected 6.eshape '

Seismic Design Category D4 2.44 In

12'

3 1.91

1 = 1.0 12'

2 1.36

Cd = 5.5 12'

0.79

= 1.3 16'P 0

Determine the foll owing.

[}J Design earthquake deflections Ox

~ Compare design story drifts with the limit value


[}J Design earthquake deflections axThe design earthquake deflections Ox are determined from the following

Code Reference

§12.8.6

= Cd° ,rr1

(Eq 12.8- 15)

= 5.5c5.tr = 5.501.0 oft!

2006 IBC St ructural/SeIsmic Des ign Manual, Vol . I 119

§12.12 Example 30 a Story Drift Limitations

@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 requiresthat 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." Ox D. Limit Status

4 2.08 in 11.43 in 2.51 in 2.76 o.k.

3 1.62 8.92 2.68 2.76 o.k.

2 1.13 6.24 2.65 2.76 o.k.

1 0.65 3.59 3.59 3.68 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 systemsto 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 determinedas the modal combination of the story-drift value for each mode.Determination of story drift from the difference of the combined mode deflections mayproduce erroneous results because differences in the combined modal displacements can beless than the corresponding combined modal story drift.


Example 31 • Vertical Seismic Load Effect §12.4.2.2

Exal te 31Vertical Seismic Load Effect §12.4.2.2

Find the vertical seismic load effect, E.-, on the non-prestressed canti lever beam shownbelow.



Beam unit weight = 200 plfSDS= 1.0

Find the following for strength design.

[!J Upward seismic forces on beam

~ Beam end reactions


[!J 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

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.

Code Reference

§12.4.2.2

(Eq 12.4-1)

(Eq 12.4-2)

(Eq 12.4-3)

(Eq 12.4-4)

2006 IBC Structural/Se ismic Desig n Manual, Vol . I 121

§12..1.2.2 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 = 0.9D + (- 0.2D)

=0.7D

= 0.7(200 plf)

= 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 upwardseismic force" is intended to specify that gravity load effects cannot beconsidered to reduce the effects of the vertical seismic forces and that thebeam must have the strength to resist the actions caused by this net upwardforce without consideration of any dead loads. This force is computed as0.2 times the dead load

qs = - 0.2WD = - 0.2(200) = - 40 plf §12.4.2.2

I

1

1

IIIIIIIJ

J

II


1

J

J

J

I

IIIIII

EKample J 1 " Vertical Seismic Load Effect

Beam end reactions for upward force of 40 plf

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

M.4= qe ; = 40~0)2 = 2000 lb/ft

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

§12.4.2.2

20061BC Structural/Seismic Design Manual, Vol.J 123

§11.4.5 Example 32 • Design Response Spe ctrum I

Examp le 32. ",

'!.esign Response Spectrum §11.4.5 IDetermine the general design response spectrum for a site where the followingspectra l response acceleration parameters have been evaluated according to the generalprocedure of §11 .4.

S DS= 0.45gSOl = 0.28gTL = 8 sec

[!J Determine design response spectrum

I

I

J

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

'Calculations and Discussion

os T::: To, and T > T,

Toand T, arc calculated as

---""- = 0.2(0.28) = O.I2 sec0.45

T, = S DI = 0.28 = 0.62 secS DS 0.45


Code Reference

III

IIIIIIII

Examp le 32 " Des ign Response Spectrum

The spectral accelerations are calculated as

1. For the interval 0:::: T:::: To

Sa = 0.6 SDS T+ OASDST"

= 0.6 (0.45g

) T + 0.4(OA5g)0.12

= [2.25T + 0.18]g

2. For To < T :::: r.

Sa = SDS = 0.45g

3. For r. < T:::: TL

§11.4.5

(Eq 11 .4-5)

S- SOl _ 0.28

a - - ---gT T

4. For T~ TL

s - S TL _ 2.24ga - DI T2 - ~

nsc Eq 16-21)

From this information the elastic design response spectrum for the site can be drawn asshown in Figure 33.1 below, per Figure 11.4-1, in ASCE/SEI 7-05

So

0.45g

0.28g

0.18g

III! ~0.28g

: I T_ _ __ ~ L _

To=0.12 sec Ts=0.62 sec 1.0'--_--'- .1...-_-4- .1..----. T(sec)

TL = 8 sec

Figure 33.1 Elastic response spectrum

2006 IBC Structural/S eis mic Des ign Manual, Vol . , 125

This example illustrates the determination of design lateral forces for the two basicelements of a dual system. §12.2.5.1 prescribes the following features for a dual system .

§12.2.5.1 . example 33 II Dual Systems

§ 12.2.5.1

I1I1

In present practice, the frame element design loads for a dual system are usually aresult of a computer analysis of the combined frame-shear wall system.

J

IIIIIIIIIIII

Moment frame

Point A

~ V=400kips

I. Resistance to lateral load is provided by the combination of the moment frames and byshear walls or braced frames. Recall that the moment-resisting frames provided must beable to resist at least 25 percent of the design forces.

2. The two systems are designed to resist the total design base shear in proportion totheir relative rigidities.

Design Base ShearV = 400 kipsQE = MQE = 53.0 kip-ft

In this example, the Equivalent Lateral-Foree-Procedure of §12.8 has been used to determinethe seismic demand QE at point A in the dual system of the building shown below.This is the beam moment MQE . Shear wall


Determine the following for the moment frame system.

Seismic Design Category Dp = 1.0J = 1.0

From the results of the computer analysisLV shear walls = 355 kipsL V columns = 45Total design base shear = 400 kips

[!J Design criteria

[!J Seismic design moment at A = M'QE


OJ Design criteria

According to the two listed requirements, the moment frame must be designed for thegreater value of either the QE value due to the design base shear V loading on the

combined frame - shear wall system, or the Q~ value resulting from at least 25 percentof the design forces. Th is 25-percent requirement may be interpreted in two ways.

CodeReference

11J

, -:Calculations and Discussion

Example 33 • Dual Systems § 12.2.5.1

J

~ Q~ may be found by an equivalent lateral-force analysis of the independentmoment frame using 25 percent of the design base shear V.

~ Q~ may be found by factoring the combined frame-shear wall system Q~value such that Q~ corresponds to the action that would occur if the portionof the base shear resisted by the moment frame VFwere to be at least equal to 25percent of the design base shear V.

~ Seismic Design Moment at A = M'QE

It is elected to use the factored QE (option b) listed above, because this procedure includesthe interaction effects between the frame and the shear wall

From the combined frame-shear wall analysis with forces due to the design base shearV = 400 kips, the portion VFofthe base shear resisted by the moment-frame is equalto the sum of the first story frame column shears in the direction ofloading. For thisexample, assume that

The required values QEcorresponding to a frame base shear resistance equal to 25percent of V is given by

and the seismic design moment at A is

M~E = O.2~~400) (53.0) = 117.8 kip-ft

2006 IBC Structural/Seismic Des ign Manual, Vol. I 127

§ 12 .2 .5. 1

;.Comrriel1Jary

Example 33 • Du al System s I

Use of a dual system has the advantage of providing the structure with an independentvertical load-carrying system capable of resisting 25 percent of the design base shear, whileat the same time the primary system, either shear wall or braced frame, carries itsproportio nal share of the design base shear. For this configuration, the code permits use of alarger R value for the primary system than would be permitted without the 25-percent framesystem.

Design Criterion la involving the design of the moment frame independent from the shearwall or bracing system for 25 percent of the design base shear should be considered for highrise buildings . The slender configuration of the shear walls or bracing systems can actua llyload the moment frame at the upper levels of the combined model, and excessively largemoment frame design actions would result from the use of Design Criterion Ib, where these

. ld b I . I' db O.25Vlarge aclions wou e mu lip ie y - -VF

128 2006 1BC Strucrural/Selsmic Design Manual, Vol. I

I1

IIIIIIIIIIIII)

Example 34 • Lateral Forces for One- Story Wall Pan els §12.1 1

xa Ie 34eteret orees for'Olle-Star all Panels 12.11

This example illustrates the determination of the total design lateral seismic forceon a tilt-up wall panel supported at its base and at the roof diaphragm level. Note that thepanel is a bearing wall and shear wall.

For the tilt-up wall panel shown bel ow, determine the out-of-plane sei smic forcesrequired for the design of the wall section. This is usually done for a representative1-foot width of the wall length, assuming a uniformly distributed out-of-planeloading.

The fo llow ing information is given.


I = 1.0SDS= l.Og

Panel thickness = 8 inchesNormal weight con crete ( 150 pe t)

Determ ine the followi ng.

Till-Up panel

Top of parapet

4'

Roof

20'

Ground

[!J Out-of-plane force for wall panel design

~ Shear and moment diagrams for wall panel design

~ Loading, shear and moment diagrams for parapet design

:Calculafions and Discussion

[!J Out-of-plane force for wall panel design

Under §12. I 1.1, the design lateral loading is determined using

Fp = 0.40 SDsIwp ~ 0.1 wp

where wp is the we ight of the wal l.

Code Reference

§12.11

Per §12.1 1.2, the force must be taken as no less than 400 lb/ft SDsI, nor less than 280 Iblft

2006 IB C Structu ral/S eis mic Design Manual, Vol. I 12 9

§12.11 Ex ample 34 :I Lat eral Forc es for One-Story Wall Panels

Note that if the diaphragm is flexible , §12.l1.2. 1 requ ires the anchorage force (but not thewall force) to be increased.

The force Fp is considered to be applied at the mid-height (centroid) of the panel, butthis must be uniformly distributed between the base and the.top of parapet.

For the given SDS = 1.0 and J = 1.0, the wall panel seismi c force is

Fp = OAO(1.0)(1.0)w = OAOw

The weight of the panel between the base and the top of the parapet is

w\\' = L~) (150) (24) = 2400 lb per foot of width

Fp = 0040 (2400) = 950 Ib/ft

Fp > 400 lb/ft SDsI = 400(1)(1) = 400 lb/ft

Fp > 280 lb/ft

The force Fp is the total force on the panel. It acts at the centroid. For design of thepanel for out-of-plane forces, Fp must be expressed as a distributed load,(p

r = 960 Ib/ft =40.0 lflftJP 24 ft P

~ Shear and moment diagrams for wall panel design

Using the uniformly distributed load,(p, the loading, shear, and moment diagrams aredetermined for a unit width of panel. The 40.0 plf/ft uniform loading is also applied tothe parapet. See step 3, below, for the parapet design load.

40.0 plllft

4'

RR

20'

Rs

Loading

384

Shear (Ib/ft)

·3 20

Moment Ib-ftIft

1843


I

Example 34 • Lateral Forces for One· Stoi"y Wall Pane ls §12.11

When the uniform load is also applied to the parapet, the total force on the panel is

40.0 plf/ft (24 ft) = 960 lbft

The reaction at the roof and base are calculated as

RR= 960(12) = 576 Ib/ft20

RR = 960 - 576 =385 lb/ft

The shears and moments are the QEload actions for strength design. Note that the reactionat the roof RR is not necessarily the face used for wall-to-roof anchorage design, seersc §1620.2.1.

~ Loading, shear and moment diagrams for parapet design §13.3.1

This section requ ires that the design force for parapets (note that parapets are classified asarchitectural components) be determined by Equation 13.3-1 with the Table 13.5-1 values of

ap = 2.5 and Rp = 2.5

for the unbraced cantilever parapet portion of the wall panel.

The parapet is considered an element with an attachment elevation at the rooflevel

z=h

T 13.5-1

IIIII

The weight of the parapet is

Wp = (1~)(150)(4) =400 lb per foot of width

The concentrated force applied at the mid-height (centroid) of the parapet is

F = 0.4 (2.5)(1.0)(1.0) (1 + 2 20)wp 2.5 20 p

Fp = 1.2Wp = 1.2 (400) = 480 lb/ft < 1.6 SDsIpWp = 640 lb/f] . . . o.k.

and > 0.3 SDslpWp . . . o.k.

(Eq 13.3-1)

(Eq 13.3-2)

(Eq 13.3-3)

2006 la c Structural/Seismic Design Manual, Vol. I 131

§12.11 Example 34 • Lateral Forces for One-Story Wall Panels

The equivalent uniform seismic force is

h = 480 = 120 plf/ft for parapet design4

120 plrJft

4'

Loading

480

Shear (Ib/ft)

-960

Moment (lb-fUft)

••f I ..... ..,. 'I

Note that for a large portion ofthe lower south-east region of the USA (Texas, Arkansas, Louisiana,Mississippi, Alabama, Georgia, and Florida) the minimum wind forces may govern over the seismicforces.


Exampl e 35 • Out-of-Plan e Seismic Forces for Two-Story Wall Panel §12. 11.1§12.11.2

Example 35 Out-of-Plane Seismic!Forces for Two-Story Wall Panel §12_11.1 and 12.11.2

This example illustrates determination of out-of-plane seismic forces for the designof the two-story tilt-up wall pane l shown below. A typical solid panel (no door or windowopenings) is assume d. Walls span from floor to floor to roof. The typica l wall panel in thisbuilding has no pilasters and the tilt-up walls are bearing walls. Th e roof consists of 1-1/2inch, 20-gage metal deck ing on open web steel joists and has been determined to be aflexible diaphragm. The seco nd floor consists of I-inch, IS-gage compo site decking with a 2II2-inch ligh tweight conc rete topp ing. This is considered a rig id diaphragm.

Th e following information is given.


S DS = 1.0J = 1.0

Wall weight = WI/' = 113 psf38'

Wallpanel

2'

20'

Wan section

16'

II

III

I

Determ ine the fo llowing .

[L] Out-of-plane forces for wall panel design

[3J Out-of-plane forces for wall anchorage design


[L] Out-of-plane forces for wall panel design §12.11.1

Requirements for out-of-p lane seismic forces are speci fied in §1620.1.7

r, = 0040 Soslw; 2:0.111'",

= OAO( 1.0)( 1.0)11'",= OAOw", = OAO(113)

= 45 .2 psf


§12. 11.1§12. 11.2

Examp le 35 " Out-of-Plan e Seismic Forces for Two-Story Wall Panel

For a representative 1-foot-wide strip of wall length, Fp is appl ied as a uniform load

/p =Fp( I ft) =45.2 plf

~ 2' ...~

r

~

~

~~ 20'~~

~~~

~ 16'....~....

.... ~

For the purpose ofwall design, the required shears and moments may be evaluatedby using reaction va lues based on the tributary area for the l-ft strip

(16 ft ) 1"

R1 - 2-YP= 8(45.2) = 362 1b

R, [C:) + (2;)}';' ~ 18(45.2) ~ 814 Ib

Note that the 2-foot-high parapet must be des igned for the seismic force Fp specifiedin §13.3.1, with height z at parapet centroid 37 ft, Gp = 2.5 and Rp = 2.5

~ Out-at-plane torces tor wall anchorage design

[!;] Anchorage force for the flexible roof diaphragm

r, = 0.8 8Ds/wl\'

where W w is the we ight of the wall tributary to the anchor

w,,~ [(2~ft) +2ft](1 13 psf) ~ 1356 pl f

Fp = 0.8( 1.0)(1.0)(1356) = 1085 plf

§12.11.2.1

(Eq 12.11-1)

The design force per anchor is Fp times the anchor spacing. For exampleif the spacing is at 4 feet, the anchor mus t be designed for (10 85) (4 ft) = 4340 lb.


1

E1Camp/~ 35 " Out·of·Plane Solsmic Forces for Two-Story Wall Panel

~ Anchorage force for the rigid second floor diaphragm .

For the case of rigid diaphragms the anchorage force is given by the greaterof the following:

a. The force set forth in §12.11.1.

b. A force of 400 SDsI (pit).

c. 280 (pit) of wall.

§12.11.1§12.11. 2

§12.11.2

)

II1

II

z = 16 ft = the height of the anchorage of the rigid diaphragm attachment,and Wp is the weight of the wall tributary to the anchor

Wp = [(2~ft) +C62

fl )](113 pst) = 2034 plf

F = 0.4(1.0)(1.0)(1.0) [I + 2 (.!i)] IVp 2.5 36 p

= 0.302Wp = 0.302(2034) = 615 plf


§12.11.1§12.11 .2

Commentary

Example 35 • Out-at-Plane Seismic Forces for Two·Story Wall Panel 1

For flexible or rigid diaphragms for all seismic design categories (SDCs), the seismic out-ofplane forces for the design of the wall are not dependent on the height of the wall inrelati onship to the total height of the building, §12.11.

For flexible diaphragms of SDCs A and B, the seismic anchorage forces are given in§12.11.2 and for SDCs C, D, E, and F, the seismic anchorage forces are given in §12.11.2.1.

For rigid diaphragms of SDCs A and B, the seismic anchorage forces are given in §12.11.2.

For rigid diaphragms of SDCs C, D, E, and F, the seismic anchorage forces are given in§12.11.2.


II

IIIIIIIIIIII

Example 36 • RIgid Equipment § 13. 3.1

Example 36Rigid Equipmen §13.3.1

This example illustrates determination of the design seismic force for the attachmentsof rigid equipment (see commentary). Attachment, as used in the code, means thosecomponents, including anchorage, bracing, and support mountings, that "attach" theequipment to the structure.

The three-story building structure shown below has rigid electrical equipment supportedon nonductile porcelain insulators that provide anchorage to the structure. Identicalequipment is located at the base and at the roof of the building.

2

12'

Wp ~NOndUClj le allachmenls~ shallow expansion anchors

Roof r-------~--.....Level

5 DS = 1.1Ip = 1.0Wp = 10 kips


12'

12'

Find the following.

[!J Design criteria

~ Design lateral seismic force at base

~ Design lateral seismic force at roof

Ca/cu/~t(ons and Discussion Code Reference

[!J Design criteria §13.3-1

The total des ign lateral seismic component force to be transferred to the structureis determined from

(Eq 13.3-1)

2006 IBC Structural/Seismic Design Manua l, Vol. I 137

§13.3 .1 Examp le 36 • Rig id Equipment

Values of Qp and Rp are given in Table 13.6-1. Also note that for shallowexpansion anchors Rp = 1.5, see §13.4.2.


Zx =0

F = 0.4(1.0)(1.1)(1 0 kips) [1+ 2 (~)] = 2.93 ki sp (1.5/1.0) 36 P

Also §13.3. 1 has a requi rementthat Fp be not less than 0.3 SDS Jp W p

Check r, s 0.3 SDs l p Wp = 0.3 (1. 1) (1.0) 10 = 3.3 kips

: . Fp = 3.3 kips . . . Equation 13.3-3 governs


Zx = h; = 36 ft

F = 0.4(1 .0)(1.1)(1 0 kips) [1+ 2 (36)] = 8.8 ki sp (1.5/1.0) 36 P

Section 13.3.1 states that Fp need not exceed 1.6 SDS Jp Wp

Check Fp S 1.6 SDS Ip Wp = 1.6 (1.1) (1.0) 10 = 17.6 kips

:. Fp = 8.8 kips ... Equation 16-67 governs.

pommentary

T 13.6-1

§13.3-1

(Eq 13.3-3)

(Eq 13.3-2)

The definition of a rigid component (e.g. , item of equipment) is given in §11.2. Rigidequipment (including its attachments; anchorages, bracing, and support mountings)that has a period less than or equal to 0.06 seconds.


Example 36 • Rigid Equipm ent §13.3 .1

The fundamental period Tp for mechanical and electrical equipment shall be determined bythe formula given in §13.6.2

(Eq 13.6-1)

I

Where:g = acceleration of gravity in inches/sec/Kp = stiffness of resilient support systemTp = component fundamental periodIVp = component operating weight

The component anchorage design force Fp (i.e., the force in the connected part)is a function of l/Rp , where Rp = 1.5 for shallow anchors, (see §13.4.2).

Generally , only equipment such as anchorage or attachments or components need bedesigned for seismic forces. This is discussed in §13.1.4. Where equipment, which can beeither flexible or rigid, comes mounted on a supporting frame that is part of the manufacturedunit, the supporting frame must also meet the seismic design requirements of §13.

Note that §13.2.5 allows testing as an alternative to the analytical methods of §13. Testingshould comply with ICC-ES AC I56.

Section 13.1.3 requires a component importance factor greate r than 1.0 (lp = 1.5) for the following .

IIIIII

•••

Life safety component required to function after an earthquakeComponents of hazardous materialsOccupancy Category IV components needed for continued operation of the facility


§ 13.3.1 Ex amp le 37 " Flexible Equipment

§ 13.3.1

This example illustrates determination of the design seismic force for the attachmentsof flexible equipment, see commentary. Attachment as used in the code means thosecomponents, including anchorage, bracing, and support mountings, that "attach" theequipment to the structure.

The three-story building structure shown below has flexible air-handling equipmentsupported by a ductile anchorage system. Anchor bolts in the floor slab meet theembedment length requirements. Identical equipment is located at the base and atthe roof of the building.

12'

12'

achmenlsLevel

~'/DUClileall

Roof

2

1

'~wp

/ / / , / / / / /

SDS= 1. IIp = 1.0W; = 10 kips


Find the following.

[!J Design criteria



'Calculations and Discussion Code Reference

[!J Design criteria §13.3.1

The tota l design lateral seismic component force to be transferred to the structure isdetermined from

CEq 13.3-1)

140 2006 IBC Str uc tura l/Seis mic Desig n Manua l, Vo l . I

Exampl e 37 • Flexible Equipment § 13.3.1

Values of ap and Rp are given in Table 13.6-1. Since the equipment is flexible andhas limited defonnability elements and attachm ents

1

IIIIIII

ap = 2.5, Rp = 2.5


z = 0

F = 0.4(2.5)(1.1)(10 kips) [I +2~] =4.4 ki sp (2.5/1.0) 36 p

Section 13.3.1 has a requirement that Fp be not less than 0.3 SvslpWp

Chec k r, ~ 0.3 SvslpWp= 0.3( 1.1)( 1.0)(10) = 3.3 kips

: . Fp = 4.4 kips . .. Eq 13.3- 1 governs .


z = h = 36 ft

F = 0.4(2.5)(1.1)(1 0 kips) [I + 2 36] = 13.2 ki 5p (2.5/1.0) 36 p

Sect ion 13.3.1 states that Fp need not exceed 1.6 Sos JpWp

Check Fp :s 1.6 SvsWp = 1.6(1.1)(10) = 17.6 kips

: . Fp = 13.2 kips ... Eq 13.3-1 governs.

T 13.6-1

(Eq 13.3-3)

CEq 13.3-2)

20061BC Structural/Se ismic Design Manual, Vol. / 14 1

§13.3.1 Example 37 • Flexible Equipment 1

'commentary

A component importance factor greater than 1.0 (Ip = 1.5) is required for the following.

Also note that §13.2.1 requires that, "Architectural, mechanical, and electrical componentssupports and attachments shall comply with the sections referenced in Table 13.2-1."

It should be noted that the component anchorage design force, Fp (i.e., the force in theconnected part), is a function of l/Rp , where anchorage of any kind is shallow (see §13.4.2).

IIIIIIIIIIIIIIII

Life safety component required to function after an earthquakeComponents of hazardous materialsOccupancy Category IV components needed for continued operation of the facility

•••

The definition of flexible equipment is given in §11 .2. Flexible equipment (including itsattachments anchorages, bracing, and support mountings), has a period greater than 0.06second .

Generally, only equipment anchorage or components need be designed for seismic forces.Where the equipment, which can be either flexible or rigid, comes mounted on a supportingframe that is part of the manufactured unit, then the supporting frame must also meet theseismic design requirements of §13.3.

Those architectural, mechanical, and electrical systems and their components that are part ofa designated seismic system, as defined in §13.2.1, shall be qualified by either test orcalculation. A certificate of compliance shall be submitted to both the registered designprofessional in responsible charge of the design of the designated seismic system and thebuilding official for review and approval. ICC ES has published Acceptance Criteria (AC156) that addresses the qualification test to satisfy the referenced code requirements.


Example 38 II Relative Mo tion of Eq uipment Attachments

Exa o11J Ie 38

e a ive otion 0 S uipment ttachments

§13.3.2

13.3.2

Section 13.3.2 requires that the design of equipment attachments in buildings have theeffects of the relative displacement of attachment points considered in the lateral forcedesign. This example illustrates appl ication ofthis requirement.

A unique control panel fram e is attached to the floor framing at Levels 2 and 3of the special steel moment frame building shown below.


Seismic Design Category DOccupancy Category II,

(}xAe = I.OR inbyAe 0.72 inR = 8.0Cd = 5.56 aA = O.015hx

Panel frame: EI = lOx 104 kip-in'

Determi ne the following :

Level

4

,~r"",12'

3

12'

s.,2

Level v12'

1 . .12' Deflected

-Y..shape

Code Reference

~ Story drift to be considered

~ Induced moment and shear in frame


~ Story drift to be considered

Sec tion 13.3.2 requires that equipment attachments be designed for effects induced byDp (re lative seismic disp lacement). This is determined as follows.

whe re

Dp = bxA - byA = 1.98 in

b xA = b x,.w C d = 5.94 inbxAe=1.08 inC d = 5.50

Governs (Eq 13.3-5)


§13.3.2 Example 38 • Relative Motion of Equipment Attachments

6.1'..1('= 0.72 inCd = 5.50

Note that Dp is not required to be taken as greater than

!1 6.48= (x - y) ---E.t.L = (432 - 288) - = 2.26

hn 432

where

x = 36 ft x 12 = 432 iny =24ftxI2=288in!1a:l = 0.0 15 h« = 0.015 (432) = 6.48 inhsx = 36 ft x 12=432 in

Thus : Dp = 1.98 in

~ Induced moment and shear in frame

(Eq 13.3-6)

§13.3.2

A liberal estimate of the moment and shear can be made using the following equations.

2Mv=-

H

M

v

_ 6EIDp _ 6(10 x 104)(1.98) - 5792 ki .

- H2 - (144)2 - . Ip-m

- 2M _ 57,29 - 0795 ki- - - -- - . ipSH 72

Dp

l<--tf

M~(~_: HI

I

I Mv ...~

6EIDM=--P

H2

The attachment details, including the body and anchorage of connectors, should follow theapplicable requirements of §13.4. For example, if the anchorage is provided by shallowanchor bolts, then Rp = 1.5.

When anchorage is constructed of nonductile materials, Rp = 1.0. One example of anonductile anchorage is the use of adhesive . Adhesive is a "glued" attachment (e.g.,attachment of pedestal legs for a raised computer floor). It should be noted that attachment byadhesive is not the same as anchor bolts set in a drilled hole with an epoxy type adhesive.


1 Example 39 • Deformation Compatibility for Seismic Design Cat eg or ies D, E, and F § 12. 12.4

J

.1:iiample 39 Deformation Compatibility for:Seis..mic Design Categories 0 , E, and F - ". §12. 12.4

A two-level concrete parking struct ure has the space frame shown be low. Thedesignated lateral-force-resist ing system consists of a two-bay speci al rein forcedconcrete moment- fram e (SRCMF) located on each side of the structure. The secondlevel gravity load-bearing system is a post-ten sioned flat plate slab supported onordinary rein forced concrete columns,

fffff0-~~=

0-

Plan at second level

[!J Moment in ordinary column

~ Detailing requirements for ordinary column

5

III\

II



lixc = 0.57 inR = 8.0Cd = 5.5Column section = 12 in x 12 inColumn clear height = 12 ftConcrete E; = 3 x 10J ksi

1 = 1.25

Find the following.

f f ?'- SRCMF ~

r:l l J"jVElevation L10e E


§ 12. 12.4 Exam ple 39 • Deformation Compatib ility for Seismic Design Categories D, E. an d F 1

[!J Moment in ordinary column §12.12.4

Section 2 1. JJ of ACI 318-05 specifies requirements for frame members that are notpart of the designated lateral forc e-resisting system. The ordinary columns locatedin the perimeter frames, and the interior flat plate/column system, fall under theserequirements and must be checked for the moments induced by the maximuminelastic response displacement. For this example, the columns on line E will beevaluated.

~

;..t;alcula.t!ons and Discussion CodeReference

II

= ei l " = 5.5(0.57) = 2.51 inI 1.25

(Eq 12.8-15) J

The moment induced in the ordinary column due to the maximum inelastic responsedisplacement Ox on line E must be determined.

For purposes of this example, a fixed-fixed condition is used for simplicity. In actualapplications, column moment is usually determined from a frame analysis.

The cracked sec tion moment of inertia Ie can be approximated as 50 percent of thegross section 19 . Section 2 1.11 of ACI 31 8-05 impl ies that the stiffness of elements thatare part of the lateral-foree-resisting system shall be reduced - a common approach is touse one half of the gross section properties. This requirement also applies to elementsthat are not part of the lateral-foree-resisting system,

1

IIIIIIIIIIII

= bd3

= 12 (12) 3 = 1728 in4

12 12

= 12 ft x 12 inches = 144 in

I . 4= .s. = 864 In2

= 6(3 x 103)(864 )(2.51) = 1883 kip-in

(144)2

h

M eol

Detailing requirements for ordinary column.

Section 21.11.1 ofACI 318-05 requires that frame members, such as the column, that areassumed not to be part of the late ral-foree-resisting sys tem must be detailed according toACI §2 1. 11.2 or §2 1.11.3, depending on the magnitude of the moments induced by ox.

200 6 IBC Stru ctura l/ Se is mic Design Manual, Vol. I146

E~ample 39 " Deformation Com patibility ior Seismic Design Categories D, E. an d F

Gommentary

§12.12.4

1

1

In actual applications, the flat plate slab must be checked for flexure and punching shear dueto gravity loads and the frame analysis actions induced by ox.

Note that this example problem shows only one way to configure this structure - that is tocombine a ductile SRCMRF with an ordinary, or non-ductile, interior column. ACIrequirements for this configuration stress that the non-duc tile interior column must resist thestructure lateral deformation by strength alone.

However, the code also permits an alternative way to configure this structure - by combiningthe ductile SRCMRF with ductile interio r columns. In this configuration, if interior concretecolumns are detailed according to the requirements of ACI 3 I8 §21.11.3, then designmoments resulting from lateral structure seismic displacements need not be calculated forthat column at all.

2006 (BC Structural/Seismic Design Manual, Vo/./ 147

§ 12. 7.4 EXllmple 40 • Adj oining Rig id El em ent s 1

Example 40Adj oining !}ligid Elements §12.7.4

The concrete special reinforced concrete moment-resisting frame (SRCMF) shown below isrestrained by the partial height infill wall that is not considered to be a part ofthe seismicforce-res isting system. The infill is solid masonry and has no prov ision for an expansion jo intat the column faces. The design story drift t; was computed according to the procedure givenin §12.8.6.

If

[::LJ Deformation compatibility criteria

~ Approximate column shear

[::LJ Deformation compatibility criteria §12.7.4

The infi ll wall, which is not required by the desig n to be part of the latera l-force-resistingsystem, is an adjo ining rigid element. Under §12.7.4, it must be shown that the adjo iningrigid element, in this case the masonry infill wall, must not impair the vertical- or lateralload-resis ting abil ity of the SRCMF columns. Thus, the columns must be checked forability to withsta nd the t; displacement of 2.5 inches whi le being simultaneously restrainedby the 6-foot-high infill walls.


t; =2.5 in

Column properties

f: =3000 psiE, = 3 x 103 ksiAc = 144 in4

t, =854 in4

Determine the following .


Inrlll wall

Typical elevation

Code Reference

II

IIIIII

148 2006 IBC Str uctural/ Se ismic D esign Manual, Vol . I

1

I,

IIIIII

Example 40 • Adjoining Rigid Elements §1'.7.4

[f] Approximate co lumn shear

Column shear will be determined from the frame design story drift, 1:>. . For purposesof the example, the expression for shear due to a fixed-fixed condition will be used forsimplicity. Also note the restrained column height is 6 ft or 72 inches.

v = l2E,Ill. = 12(3 x 103)(854)(2

.5) = 205.9 ki scol 11 3 (72)3 P

Column clear height = 72 in

Because the SRCMF is the primary lateral-foree-resisting system, I:>. has been determinedby neglecting the stiffness of the rigid masonry.

The induced column shear stress is '1;,., = 1447 psi. This is approximately 26ff:Ac

and would result in column shear failure. Therefore, a gap must be provided betwee n thecolumn faces and the infill walls. Alternately, it would be necessary to either design thecolumn for the induced shears and moments caused by the infill wall, or demonstrate thatthe wall will fail before the column is damaged. Generally, it is far easier (and more reliable)to provide a gap sufficie ntly wide to accommodate 1:>..

For this example, with the restraining wall height equal to one half the column height, thegap should be greater than or equal to 1:>./2 = 1.25 in. If this were provided, the columnclear height would be 144 inches, with resulting column shear

, 12(3 x 103)(854

)(2.5) 25 7 ki Thi . h h f h . d IV I = 3 = . IpS. IS IS one-erg tot e restrame co umnc" (144)

shear of205 kips, and corresponds to a column shear stress of approximately 3.3-JJ:.

,Commentary

It is also possib le that the restraint of the infill walls could cause an irregularity, such as abuilding torsional irregularity . This should be evaluated if such restraints are present.

2006 IBC Struc tural/Seismic Design Manual. Vol. t 149

§13.5.3 Exa mp le 4 1 • Exter io r Elements : Wall Panel

This example illustrates the determinat ion of the design lateral se ismic forc e Fp

on an exterior element of a building , in this case an exterior wall panel.

A five-story moment frame building is shown below. The cladding on the exteriorof the bu ilding consists of precast reinforced concrete wall panels.

§13.5.3

The following information is given.Level

5

Seismic Design Category D 12'

4 Typical

I = 1.0 exlerior

S DS= 1.012' panel

Panel size: I I ft 11 in by 19 ft 11 in 3

Panel thickness: 6in 12'

Panel weight: Wp = 14.4 kips2

12'

12'

Find the following.

[!J Design criteria

[I] Design lateral seismic force on a panel at the fourth story

@J Design latera l seismic force on a panel at the first story

Cijlculations and Discussion

[!J Design criteria

Code Reference

§13.5.3

For design of exterior elements, such as the wall panels on a building, that areattached to the building at two levels, design lateral seismic forc es are determinedfrom Equation 13.3-4. The panels are attached at the two elevations ZL and z., 'Th e intent of the code is to provide a value of Fp that represents the average of theacc eleration inputs from the two attachment locations. This can be taken as the averageof the two ~} values at z equal to ZL and z" .


1 Example 41 • Exterior Elements: Wall Panel

O.4n S I [ , ]= P OS J' 1+ 2 '::- ~, > 0.3 SDslpWRp 11

= 1.0, Rp = 2.5

§13.S.3

(Eq 13.3-1)

T 13.5-1

1

I

~ Design lateral seismic force on a panel at the fourth story

Assuming connections are I foot above and below the nominal 12-foot panel height

Z" =47 ft

ZL = 37 ft

11 = ft

= 0.4(1.0)(1.0)(1.0) [I +2(47)] 11' =0.4 11W2.5 60 p P

F = 0.4 (1.0)(1.0)(1.0) [I+2(37)] W = 0.357 W:pL 2.5 60 p P

Fp~ = Fpu + FpL = (0.411 + 0.357) w2 2 p

IIIII

Fp~ = 0.384Wp = (0.384)(14.4) = 5.53 kips

Check: Fp4 > 0.3 SDslpH'p =0.3( 1.0)(1.0)Wp =0.3Wp . . . o.k.

Check: Fp4 ~ 1.6 SDslpWp = 1.6(1.0)(1.0)Wp = 1.6Wp. . . o.k.

~ Design lateral seismic force on a panel at the first story

The following are known.

Z" =Ilft

ZL = 0

h =60 ft

(Eq 13.3-3)

(Eq 13.3-2)

2006 IBC Structu ral/Seism ic Design Manual, Vol. I 151

§13.5 .3 Example 41 • Exter ior Elemen ts: Wall Panel

Fpu = 0.4(1.0)(1.0)(1.0) [I + 2 (~)]W = 0.219if':2.5 2.5 60 p p

Check that Fpu is greater than 0.3 SvsIpWp

Fpu = 0.3(1.0)(1.0)Wp = 0.30Wp . . . not o.k.

Also FpL< Fpu < 0.30 TVp

: . use FpL= Fpu= O. 30 TVp

Fpl = Fpu + FpL = 0.30Wp= (0.30)(14.4) = 4.32 kips2

11

1

1

J

I

~ " , 'f 'h "-(:ommentary

Note that the design of the panel may be controlled by non-seismic load conditions of thefabrication process, transportation, and installation . Also note that the forces induced bydisp lacement Dp from Equation 13.3-5 need to be checked per §13.3.2.1.

152 2006 IBC Structural/Seism ic Design Manual, Vol. I

I

IIII

Ex ample 42 Q Ex terior Nonstructural Wall Elements: Precast Panel §13.5.3

xample42Exterior ons ruc turei all Elements: reces Panel §13.5.3

This example illustrates the det ermination of the total design seismic lateral force for thedesign of the conne ctions of an exterior wall panel to a building.

An exterior nonbearing panel is located at the fourth story of a five-story moment framebuilding. The panel support system is shown below, where the pair of upper brackets mustprov ide resistance to out-of-plane wind and seismic forces and in-plane vertical andhorizontal forces. The panel is supported vertically from these brackets. The lower pair ofrod connections provides res istance to only the out-of-plane forces.


SDS = 1.0fp = 1.0fi = 0.5Height to roof, h = 60 ftPanel weight = 14.4 kipsP = 1.0 per §12.3.4.1(3).Panel live load, L = 0

Find the following.

[!.J Strength design load combinations

~ Lateral seismic force at center-of-mass C of panel

~ Combined dead and seismic forces on panel and connections

@J Design forces for the brackets

~ Design forces for the rods

12'

-'C~!~ulations and Dis_cussifJn Code Re(~relJce

[!.J Strength design load combinations §2.3.2

For design of the panel connections to the building, the applicable strength designload combinations are

2006 IBC Structural/Se ismic Des ign Manual, Vol. I 1 5 3

§1J.5.J Example 42 • Ex terior Nonstructural Wall Elements : Precast Panel J

a) 1.2D + 1.0E + )1" L = I.2D + 1.0E (Comb. 5)

where, with E = pQ£+ 0.2 SDSD (Eq 12A-I) I= 1.0Q£ + 0.2(1.0)D

= Q£ +0.2D

As SDS = 1.0g, the equation reduces to

lAD + Q£ , for Q£ and D with same signs and type of load action.

J

Ib) 0.9D + 1.0£ (Comb. 7)

with E = pQ£ + 0.2 SsoD (Eq 12A-I )

= Q£ +0.2D1J


In the seismic load combinations, Q£ is the load action on the connection due tothe lateral load Fp applied either in-plane or out-of-plane at the panel center-of-massper §13.3.

This combination need not be considered because the rod connections resist only the Q£axial load, and the bracket connections have shear resistance capacity independent of thedirection of the Q£ shear load: for example, upward resistance is equal to downwardresistance. Therefore, this load combination is satisfied by lAD + Q£ for Q£ and D withthe same signs.

IIIIIIIIIII

(Comb. 7)

(Eq 12A-2)

l.lD + Q£ , for Q£ and D with same signs.

0.70D + Q£, for Q£ and D with opposite signs.

This combination need not be considered since it is less than lAD + Q£.

with E = pQ£ - 0.2 SsoD

c) 0.9D + 1.0Q£

As Sos = 1.0g, the equation reduces to

As SDS = LOg, the equati on reduces to

I Example 42 • Exteri or Nonstructural Wall Elements: Precasl Panel §13 . 5.3

1~ Lateral seismic force at cen ter-o f-mass C of panel

Section 13.5.3, Item d., requires that the connection seismic load actions be determined bythe force Fp given by §13.3.1 applied to the center-of-mass of the wall panel. The values ofRp and op are given in Table 13.5-1 for the body and fastene rs of the connection elements.

To represen t the average seismic acceleration on the panel, Fp will be determined as theaverage of the Fp values for the upper bracket elevation level, z,,, and for the lower rodelevation elevati ons, ZL. For the higher story levels of the building, this average Fp

would be essentially equal to the Fp value using Z = z, at the panel center-of-mass elevation.However, this use of elevations z = Zc may not be valid for the lower story levels becauseof the limitation of

(Eq 13.3-3)

With the given values of Sos= 1.0, and Ip = 1.0

O.4opSDsl p ( z)F = 1+ 2- IVp n, 11 p

(Eq 13.3-1)

r, 2: 0.3 SosIpIVp = 0.3(1.0)( 1.0)IVp = 0.3Wp (Eq 13.3-3)

Op = 1.0 and Rp = 2.5, for body of connection T 13.5-1

= Zu = 47 ft

At lower rod connect ion level

IVp = weigh t panel = 14.4 kips

At upper bracket connection level

=ZL = 37 ft

= 0.4(1.0) [1 + 2 (37)] If'2.5 60 p

= 0.4 11 Wp> 0.3 SosIpWp= 0.3Wp .. . o.k.

z

F = 0.4(1.0) [I + 2 (47)] IVpU 2.5 60 p

1

IIIIII

= 0.357 Wp > 0.3 IVp • • • o.k.

2006 IB C Structural/Seismic Design Manu al, Vol . I 155

Exampl e 42 • Ex terior Nonstructural Wall Elements : Precast Pane l§13.5.3

The required average, FpFpu + FpL = (0.411 + 0.357) w

2 2 I'

1J

=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.

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 combination,

~ 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:

2006 IBC Str uctural/S eismic D esign Manual, Vol. I

Combined dead and seismic forces on panel and co nnections IIIIIIIIIIIIII

§13.5.2

Fp = 5.5 3 kips

.4 (14.4) =20 .16 kips

g' , I.- g' ,

.._.._.._..- f- .•_ .. _ ••_ ••-J-I~

5'

0~,ItI

5' -, .._.._.._..- f- •. _ .. _ .. _ ..... t-

1.2Wp + O.2Wp = 1.4Wp = 1

where Pe is the bracket force and PR is the rod force.

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

Ps + PR = Fp = 5.53 = 1.38 kips4 4

156

1

Examplo 42 IR Exterior Nonstructural Wall Elements : Precast Panel §13.5.3

Each bracket takes the following downward in-plane shear force due tovertical loads

I AWp 20.16 .VB = - - = -- = 10.08 kips

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 seismicforces , and vertical seismic forces are calculated as

9' 9'

5'c ,/

5'

1.4W,= 20.16 kips

F, = 5.53 kips

IIIII

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

r, 5.53 kiHB =- = - =2.77 IpS2 2

Each bracket takes the following upward or downward shear force due tothe reversible lateral seismic load

F - 5(Fp ) - 5(5.53) - I 54 kiB - - - - - ± . IpS

18 18

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

1.4Wp 20.16 .RB = = -- = 10.08 kips

2 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 Elements: Precast Pan el

1@J Design forces for the brackets

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

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.

PB = ±1.38 axial load together with

VB = 10.08 kips downward shear

and

HB = ± 2.77 kips horizontal shear together with

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

~ FastenersUnder §13.5.3, Item d., and Table 13.5.1, fasteners must be designed forup = 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

(3.125) PB= 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 for the rods

~ Body of connectionThe body of the connection must be designed to resist a force based on ap = 1.0and Rp = 2.5

PR = 1.39 kips axia l load

1 58 20 06 IBC Stru ctural/Seismic Design Manua', Vol . I

I

I

I

IJ

IIIIIII

11

1

rIJ

I

I

IIII

I

Example 42 s Exter ior Nonstructural Wall Elements: Precast Panel §13.5.3

[!J Fasteners

Fasteners in the connecting system must be des igned to resist a force based onap = 1.25 and Rp = 1.0

(3. I25)PR= 3.125(1.38) = 4.31 kips axial load


§12.1.3 Example 43 • Beam Horizontal Tie Force

EXamp le 43'Beam Horizontal 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 simplebeams shown in the example below,



SDS = 1.0

Dead Load D = 6 kip/ft

Live Load L = 4 kip/ft

Pin support "p"

p~~~~~~~ k"

Cod~Reference

[!J Determine tie force

~ Determine horizontal support force at "P"


[!J Determine tie force

Requirements for ties and continuity are specified in §12,1.3 . For this particularexample, it is required to determine the "tie force" for design of the horizontal tieinterconnecting the two simply supported beams. This force is designated as Fp ,

given by the greater value of

Fp = 0.133 SDSWp

or

Fp = 0.05H'p

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

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


1 For SDS = 1.0, the controll ing tie force is

Fp = 0.133( 1.0)(240) = 31.9 kips

Example 43 • Beam Horizontal Tie Force §12.1.3

~ Determine horizontal support force at "P"

Section 12.1.4 requires a horizontal support force for each beam equal to 5 percent of thedead 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

IIIIIII

I2006 IBC Structural/Seismic Design Manual, Vol. I 161

§12 .10.2 Example 44 • Collector Elemen ts

§12.10.2

Collectors "collect" forces and carry them to vertical shear-resisting elements. Collectors aresometimes called drag struts. The purpose of this example is to show the determination of themaximum seismic force for design of collector elements. In the example below, a tilt-upbuilding, with special reinforced concrete shear walls and a panelized wood roof, has apartial interior shear wall on Line 2. A collector is necessary to "collect" the diaphragm loadstributary to Line 2 and bring them to the shear wall.

3

Nole: Roof framing, exceptcollector, not shown .

100'

50'

2

Roofplan

50'

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

--l-_-u--Colleclor

k""Shear wallslI=j::============!l

Occupancy Category I



R =5.5no= 2.51 = 1.08DS = 1.20

Roof dead load = 15 psfWall height = 30 ft, no parapetWall weight = 113 psf

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

SBase shear = V= DS W = 0.2 l 8W

RW= structure weight above one half hi

(Eq 12.8-2)

Interiorshear wall

Collector

50'

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

162 2006 IBC Structural/SeIsmic Design Manual, Vol. I

1 Example 44 • Collector Elements §12.10.2


C!.J Collector unfactored force at tie to wall

CodeReference

§12.10.2

The seismic force in the collector is made up of two parts : I) the tributary out-of-planewall forces, and 2) the tributary roof diaphragm force. The paneli zed wood roof has beendetermined to be flexible ; thus the tributary roof area is taken as the IOO-foot by 50-footarea shown on the roof plan above. Seism ic forces for collector design are determined fromEquat ion 12.10-1 used for diaphragm design. This equat ion reduces to the following for asing le story structure.

F,=-w,W PI

Fp l max = 0.3 Sp,lWpx = O.30Wpx

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

= 1.2 S p, W = VR x

= design force at roof= structure weight above one halfhi = W= weight tributary to the collector element

giving:

Wpl = tributary roof and out-of-plane wall weight

: . Fp1 =0.218(244.5) = 53.3 kips.

VFp1 = -Wpl = 0.218Wp l

W

Wp l = 15 psf(lOO)(50) + 113 psf (320) (100) = 75,000 + 169,500 = 244.5 kips

§12.10.2

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 shearstrength of connection between the diaphragm and collector. The design of thecollector and its connections requires that the axial forces be amplified as shownbelow.

Spec ial seismic load of §12.4.3.2 at tie to wall

IIIII

II

2006 IBC Structural/Seismic Design Man ual. Vol . I 163

Given the force Fpl specified by Equation 12.10-1, the collector elements, splices , and theirconnections to resisting clements shall have the design strength to resist the earthquake loadsas defined in the Specia l Load Combinations of §12.4.3.2.

§12.10.2 Examp le 44 • Coll ec tor Elements

The governing load combination is

1.2 D + 0.5L + Em §2.3.2 (Comb. 5)

1

J

I1

where

Here, Q£ is the horizontal collector design force Fpl = 53.3 kips, and

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

(Eq 12.4-5)

IJ

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 loadcomponents.

n "Q£ = 2.5(53.3) = 133.25 kips axial tension and compression loadand vertica l downward load equal to

1.2D + 0.5L + 0.2 D = 1.4 D + 0.5 L

II

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

The resulting total factored vertical load is

1.4(2250) + 0.5(750) = 3525 lb

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

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

IIIII

, .

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

:'lCommentary

164 2006 IBC St ru ctural/Seismic Design Manu al, Vol. /

Example 45 Out-at-Plene Wall Anchorage of Concrete or Masonry Walls to FlexIble Dia phragms §12. 11.2§12.11.2. 1

am le45Out-of..Plane Wall Anchorage of Concrete orMasonry 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 ofthe wall anchorage to the flexible roof diaphragm is to be determined. This will bedone for a representative I-foot width of wall.


Occu pancy Importance Category I


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


[!J Design criteria

~ Wall anchorage force


F. n: l• +---+

. Assumed pin support~ ~

Top of parapet

4'

Roof

20'

Ground

Code Reference

[!J Design criteria §12.11.2.1

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

Fp = 0.8 SDS!ElVl\'

2:: 400 8Ds/ lblft

2:: 280 lb/ ft of wall

where WI\' is the weight ofa I-foot width of wall that is tributary to the anchor.

CEq 12.11-1)

2006 IBC Structural/S eis mic DesIgn Manual, Vol. I 165

§12.11.2§12. 11.2. 1

Ex ample 45 • Out-ot-Pten e Wall Anchorage of Concrete or Ma sonry 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 allthe weight above the roof.

IVw = 150C~) (4 ft + 10 ft)(I ft) = 1400 lb/ft

For the given values ofSos = 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.

166 2006 IBC Structural/Seismic Design Manua l, Vol. I

)

I

)

)

IIIIIIIII

Example 46 • Wall Anchorage to Flexible Diaphragms §12.11.2.1

ample 4 6all nchorage to Flexible iap ragms §12.11.2.1

Th is example illustrates use of the allowable stress design procedure for the design ofstee l and wood elements of the wall anchorage system in a building with a flexible roofdiaphragm.

The drawing below shows a tilt-up wall panel that is connected near its top to a flexibleroof 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


[IJ Design force for premanufactured steel anchorage element.

The task is to design the steel anchorage elements (i.e., hold-downs) that connect the tilt-upwall panel to the wood subpurlins of the roof diaphragm. The anchorage consists of twohold-down elements , one on each side of the subpurlin .The manufacturer's catalog provides allowable capacity values for earthquake loading for agiven type and size of hold-down element.

The steel hold-down elements of the anchorage system resist only the axial anchorageload 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 chorage to Flexib le Diaph ragms J

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

E = QE= PE= Faae" (4) = (1680)(4) = ± 6720 Ib

This example, uses the ASD load combinations of §2.4, where the applicable seismic loadcombinations permi t 0.7£ to be resisted with an increase in allowable stress based onduration (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

4800 lb = 2400 Ib2

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, thethrough-bol ts in the subpurlin must be checked for double shear bearing. Also, the pairedanchorage embedment in the wall is likely to involve an overlapping pull-out conecondition 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 oftenused 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 theallowable capacity of the element, for the combined bending and axial stress includingdead 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 IBC Str uctural/Seismic Design Manual, Vol . I

J

]

J

I

III

IIIIII

1 1

Commentary

Example 46 l:f Wall Anchorage to Flexible Diap hragms §12.11.2.1

For comparison , the forces acting on wood, concrete, and steel elements are shown below. For wood,the load is divided by the dura tion fac tor Cdof 1.0 to permit comparison. For stee l, the load isincreased by 1.4 per §12.11.2.2.

Material

Wood0.8SDsIW

1.60.5 SoslW

ASD

(0.35 SoslW)

II

IIIIIII

Concrete

Steel

0.8 SoslW

1.4(0.8 SoslW) = 1.12 SoslW

N/A

(0 .78 SoslW)

2006 IB C Structu ra l/Seismic Design Manual, Vol. I 169

§12. 10.1.1 Example 4 7 Q Determ ination of Diaphragm Force Fp::: Lowrise

~ mple 4 7

!# ermination of Diaphragm Force f px : Lowrise §12.'10.1.1

This example illustrates determination of the diaphragm design force Fpx ofEquation 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 panelizedwood roof is shown below. This type of roof construction can generally be shown tobehave per flexible diaphragm assumptions.

bo

cp 200'rNormal wall

Occupancy Importance Category I

The following information isgiven .

Seismic Design Category 0Roof plan

Roof diaphragm

J = 1.0SDS= 1.0R =5.0P = 1.0Diaphragm weight = 15 psfWall weight = 80 psf

Elevation through building

Find the following.

[!J Diaphragm design force at the roof


Examp le 47 • Determination of Diaphragm Force Fpx: Lowrise §12.10.1. 1


[!J 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

1

n

2: F,F = ~= lt,'p:r n px

2: lV j

i-.t

with limits of

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

which for

S DS = 1.0 and 1 = 1.0

(Eq 12.10-1)

IIIIIIIII

are

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

with the given values of SDS = 1.0, R = 5.0

and, for a l-foot-wide strip

Wpl = weight of diaphragm + weight of one-half height of normal walls

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

(1.0)(1.0)F p l = 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 dueto Fp l (such as chord forces and diaphragm shear loads in the diaphragm).

2006 IBC Structura l/Seismic Design Manua l, Vol. I 1 7 1

§12.10.1.1 Example 47 • Determination of Diaphragm Force Fpx: Lowrise 1

Commentary

I. The weight Wp., includes the weight of the diaphragm plus the tributary weightof elements nonnal to the diaphragm that are one-half story height below andabove the diaphragm level. Walls parallel to the direction of the seismic forces areusually not considered in the.detennination of the tributary roof weight becausethese walls do not obta in support, in the direction of the force, from the roofdiaphragm.

2. The sing le-story building version of Equation 16-65 is derived as follows .

m

2: F,F = ~wps •• px

2: Wi

(Eq 12.10-1)

]

III

v •F = C V = w,h,

X IX n

2: W/l:i..1

(Eq 12.8-11 )

i = I , x = 1, and 11 = I

For a single-story building,

and Equation 12.8- 11 gives

III

II1

III

(Eq 12.8-12)for short period of < 0.5 sec (k = J.0) .- J •}\'.r 1.(

I

2: 11'; = Wi- I

F) = W/I, V = Vw,lz,

where C,_., =

172 2006 IBC Structural/Seismic Des ign Manual, Vol. I

1where

Example 47 ~ Determ ination of Diaphragm Forc e Fpx: Lowrise §12.10.1. 1

1I1I

II

\

I

V= C W = 50 S! IV5 R

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

F, V 50 s!F 1 = - 11' = - lV = -- 11'P IV 1'1 IV 1'1 R 1'1

(Eq 12.8-1 and 12.8-2)


§12.10.1 Example 48 D Determination of Diaphragm Force Fpx : Highrise

ample 48t-

,De termination of Diaphragm Force Fpx : Highrise §12. 0.1

This example illustrates determination of the diaphragm design force Fpx of Equation12. 10-1 for a representative floor of a multi-story building.

The nine-story moment frame bui lding shown below has the tabulated design seismicforces P.r:. These were determined from Equations 12.8-11 and 12.8-12, the designbase shear.


Level 1< 27' * 27':J Story'1 1 Weight, kips


W = 3,762 kipsC, = 0.062158Ds = 1.0P = 1.3I = 1.0T = 1.06 secV = CW= 233.8 kipsk = 2 for Eq 12.8-12

12

12'

12'

12'

12'

20 '

9

8

7

6

5

4

3

2

1

/ / ' // / //// // / / / ' /

214

405

405

405

584

422

422

440

465

kC = WJ l" F

I

Level x h (ft) h2 I\' kips II'h vr tFx =Cl·.rVL I\'.h. IV, I

9 11 6 13456 214 2879584 0.153 35.8 0.167

8 104 10816 405 4380480 0.233 54.4 0.134

7 92 8464 405 3427920 0.183 42.8 0.106

6 80 6400 405 2592000 0.138 32.3 0.079

5 68 4624 584 2700416 0.144 33.7 0.058

4 56 3136 422 1323392 0.071 16.6 0.039

3 44 1936 422 816992 0.044 10.3 0.024

2 32 1024 440 450560 0.024 5.6 0.013

I 20 400 465 186000 0.010 2.3 0.005

Totals: 3,762 18,757,344 233.8

174 2006 IB C St ructural/Seismic Des ig n Man ua l, Vol. I

Example 48 • Delerminallon of Diaphragm Forc e FpK: Highrise §1Z.10.1

1 ~Calqulations and iiiscus~ion Code Reference

1

1

J

I,

IIIII

[I] Diaphragm force at Level 7 §12.10-1

Seismic forces on the floor and roof diaphragm are specified in §12.10-1. The followingequation is used to determine the diaphragm force Fpx at Level x

(Eq 12.10-1)

Section 12.10.1.1 also has the following limits on Fpx

For Level 7, x = 7

F = (42.8 + 54.4 + 35.8)(405) = (0.130)(405) =52.6 ki sp7 (405 + 405 + 214) P

Check limits:

0.2 SDsIwpx = 0.2wpx

= 0.2(405) = 81.1 kips > 52.6 kips ... not o.k.

0.4 SDsIwpx = O.4wpx

= 0.4(405) = 121.5 kips > 52.6 kips . . . o.k.

:. Fp7= 81.1 kips...minimum value (0.2 SDsIwpx) governs.

Note that the redundancy factor, in this example p = 1.3, is to be applied to the load Q£ dueto F P.f (such as chord forces and floor-to-frame shear connections). Also note that Equation12.10-1 will always govern for the design of the diaphragm versus Equation 12.8-12.

2006 IBC Structural/Seismic Design Manual, Vol. I 1 7 5

§12.12.3 Example 49 • Building Separations

i#.a.~ple 49SUilding Separations.. §12.12.3

Building separations are necessary to prevent or reduce the possibility of two adjacentstructures impacting during an earthquake. Requirements for building separations aregiven in §12.12.3. In this example, the static displacements oxe due to the prescribedlateral forces of §12.8 and information about each structure are given below. Note thatthe displacements given are at the plan view edges of the building.

Structure 2

2 0.75 inI 0.35o 0

R= 6

C,/=5

1.38 in.1.000.47

o

Structure J

Levelr 0..

2

o

3

.--.-----r-1~ ~aralion""'~ II

Structure 1 Structure 2

Find the required separations for the following situations.

[!J Separations within the same building

[!J Separation from an adjacent building on the same property

~ Separation from an adjacent building on another property

Calculatiqns and Discussion Code Reference

[!J Separation within the same building §12.12.3

Expansion joints are often used to break a large building or an irregular building intotwo or more parts above the foundation level. This effectively creates separate structureswithin the same building. The code requires that the structures be separated by theamount OMf

where

OMI = maximum inelastic displacement of Structure 1

0,"12 = ~aximum inelastic displacement of Structure 2


J Example 49 ., Building Separ<Jtlons

where

s _ « ) _ CJ 6,n.U/d- Ux max - - /-

The required separation is determined in the following two steps.

§ 12.12.J

(Eq 12.8-1)

I

~ Determine ine lastic displacements of each structure §I2.8.6To determine the min imum separation between parts of the same buildingthat are separated by an expansion joint, the maximum inelastic floordisp lacements 6" must be determined for each structure. These are atlevel, x= 2

For Structure I

~ Separation from an adjacent building on the same property

If Structures I and 2 above are adjacent, individual buildings on the same property, thesolution is the same as that shown above in Step I . The code makes no distinction betweenan "internal" separation in the same building and the separation required between twoadjacent buildings on the same property.

III

I

II

s, - C~2J _ 5.5(1.0) - 5 5 .U '\(I - - /- - 1.0 -. in

For Structure 2

6M2

= CJ/6 22 = 5.0(0.75) = 3.75 in1.0

~ Dete rmine the required separation

The requi red separation is determined from the individual maximuminelastic disp lacements of each structure as

6MT = 6,\11 + 6M2 = 5.5 + 3.75 = 9.25 in

6U T = 9.25 in


(Eq 12.8-15)

(Eq 12.8-15)

§I 2.I2.3

177

912.12.3 Example 49 • Building Separations

~ Separation from an adjacent building on another property §12.12.3

If Struc ture I is a building under design and Structure 2 is an existing bu ilding on anadjoining property, we would genera lly not have information about the seismicdisplacements of Structu re 2. Often even basic information about the struc tura l system ofStruc ture 2 may not be known. In this case, separa tion must be based only on info rmationabout Struc ture I. The larges t elastic displacement of Struc ture I is 0)<= 1.38 inches andoccurs at the roof (Level 3). Th e maximum inelastic displacement is calculated as

J

J

Structure I must be set back 7.59 inches from the property line, unless a smallerseparation is just ified by a rati onal analys is based on maximum ground motion s.Such an analys is is difficul t to perform, and is generally not required except in veryspecial cases.

0 ,/ = CdO J, = 5.5(1.38) = 7.59 inr I 1.0

(Eq 12.8-15)

I

J

I


IIIIIIII

Example 50Flexible onbuildiJ1Jg' TUC UTe

Ex ample 50 • Fl exible Nonbuilding Structure § 1 5. 5

§15,,5

A tall steel bin tower is supporte d by a concrete found ation. The tower sits onsymmetrically braced legs


Seismic Design Category DWeight of towe r and maximumnormal operating contents = 150 kips

Occupancy Category IIISite Class D

I = 1.25 per Table 11.5-1S, = 1.70, S, = 0.65SDS = 1.20, SD/ = 0.65

The stiffness of the supporti ngtower is 8.30 kip/in

Code Reference

Determ ine the following.

[TI Period of vibration

~ Design base shear

~ Vertical distribution of seismic forces

~ Overturning moment at base

,calculations and Discussion

[TI Period of vibration.

For calculation purposes, the mass is assum ed to be located at the top of the tower.The period must be determined by §15.4.4.

/fn 150 kips/(386 kip/in/sec") 1 36T = 2n - = 2n =. sec

k 8.30 kip/in

Because the period is greater than .06 second, the vessel does not qual ify as a rigidnonbuilding structu re and thus is considered flexible. See §15.4.2.

2006 IBC Structura l/Se ism ic Desig n Man ual, Vol. I 1 7 9

§15.5 Example 50 • Flexible Nonbuildlng Structure

It should be noted that the value of the period, T, should not be calculated using anyof the approximate methods in §12.8.2.1, nor is it intended to be subject to the limitationspresented in §12.8.2. This is because the approximate method presented is intended forbuildings and is not applicable to structural systems that differ significantly from typicalbuilding configurations and characteristics . Refe r to Section CI 09.1.4 of the 1999 SEAOCBlue Book for further discussion.

~ Design base shear

The des ign base shear for nonbuilding structures is calculated from the same expressions asfor buildings. These are given in §12.8.1. In addit ion, nonbuilding structures, such as thevessel, must also sat isfy the requirements of §15.5.

v = C, W= 0.217 (150 kips) = 32.55 kips

where

II]

I

C = SDS =0.50, R /I

SDS= 1.2R = 3.0Q o = 2.0CD = 2.5I = 1.25

The value of C, computed in accordance with Eq. 12.8-2 need not exceed

SC = DI =0.199s CRt!)T

whereSOl =0.65R = 3.0I = 1.25T = 1.36 sec

But C, shall not be taken less than

C, = 0.01

whereSDS= 1.20I = 1.25


(Eq 12.8-2)

T 15.4-2T 15.4-2T 15.4-2T 11.5-1

(Eq 12.8-2)

(Eq 12.8-5)

IIIIIIIIIII

1 Example 50 • Fl exible Nonbuilding Structure

Note that for this tower, because the I-second spectral response SIis equal to 0.65, (S, ~ 0.60g), the value of the seismic response coefficient C,shall not be taken as less than

§ 15. 5

IIIIIII

c, = 0.5S, =0.135 (Eq 12.8-6)(RI l)

whereSI =0.65R =3.0[ = 1.25

Thus: C, =0.199 governs

Also note that if this tower (Occupancy Category 11) were located on a site with mappedmaximum considered earthquake spectral response accele ration at I- second period S"equal to or greater than 0.75g, it would be assigned to SDC E (§11.6). Thu s, the heightwould be limited to 100 ft per Table 15.4-2.

2006 IBC Structural/Seism ic Design Manual, Vol. I 181

§15.0 Example 51 • Lateral Force on Nonbui/ding Structure

I;xaniple 51Lateral Force on Nonbuilding Structure,,-- §15.0

A nonbuilding structure with a special reinforced concrete moment frame (SRCMF)supports some rigid aggregate storage bins. Weights U~ and W2 include the maximum

normal operating weights of the storage bins and contents as well as the tributaryframe weight. See §15.4.1.1 and Table 11.5. 1


Occupancy Importance Category 1"1 = 1.0

[!J Design base shear

~ Vertical distribution of seismic forces

Site Class DSMS= 2.0, Sf = 2.0S Ail = 1.5, SI = 1.0S DS = 1.33SOl = 1.00T =2.0 secTV = 300 kips


.C~/culations and Discussion

F, _--.

Level

2

15'

30'

Code Reference

[!J Design base shear. §15.4

Because this is a flexible structure, (i.e., the period T > 0.06 sec, see §15.4.2, and thestructu re is similar to a building, see §15.4.1) the general expressions for design base sheargiven in §12.4 and §15.4 must be used. Note that an intermediate reinforced concretemoment frame (lRCMF) building structure is not permitted for SDC D, E, or F per Tab le15.4. 1. Also note that the value for R is 8 for normal design of an SRCMF.

The total base shear in a given direction is determined from

V= CsW

where

C, = S DS = (1.33) =0.166(R I I) (8.0)1(1 .0)


(Eq 12.8-1)

(Eq 12.8-2)

J

Exa mple 51 ., Lat eral Force on NOll buJlding Structure §15.0

where

5DS =1.33R =8I = 1.0

The value of C., computed in accordance with Equation 12.8-2 need not exceed

J

I

C, = SOl = (1.0) for T :::J L=0.063(R II)T (8/1.0)2.0

where

5D/ = 1.0R = 8.0I = 1.0T =2.0

Check T'S TL= > TL= 12.0 sec

The value of C, shall not be taken less than

C = 0.5S, = (0.5)(1 .0) = 0.063

x (R I I) (..!-)1.0

where

(Eq 12.8-3)

(Region 1, F 22-16)

(Eq 12.8-6)

V= C,rV= (0.063)(300) = 18.9 kips

r, = c, V= C.~ (18.9 kips)

Vertical distribution of seismic forces

The design base shear must be distributed over the height of the structu re in thesame manner as that for a building structure.

I\

III

5, = 1.0R =8I = 1.0T =2.0

Thus: C, = 0.063

Note 5, 2: 0.6g

Equat ions 12.8-3 and 12.8-6 govern.

(Eq 12.8-1)

§12 .8-2

(Eq 16-41)

2006 lac Structural/Se ismic Design Manual, Vol. I 183

k = 1.0 for T ~ 0.50 secand k = 2.0 for T 2: 2.50 secand k = interpolate between 1 and 2.5 sec

Example 51 • Lateral Force on Nonbulldlng Structure

Thus:

IIIIII

I]

JCEq 12.8-11)

J

]

J

J

II

T2.52.01.0.5o

o

1.01--_ _ -

K

2.0

n

'" W il lL "j- I

C"., =

k = 1.0 + 1.0 ( 2.0 -0.5) = 1.752.5-0.5

where

where

Now for T =2.0 sec

§15.0


11

Example 51 • Lateral Force on Nonbuilding Structure §15.0

Story Shears (k =1.75)

Story Story

Height Weight Force Shear

J Iz:rI W, I

C,·x F, V, SaLevel !t.t W.Jl.f

2 45 781.85 200 156369.45 0.803 15.17 15.17 0.0761 30 384 .56 100 38455.83 0.197 3.73 18.9 0.037

300 194825.28 1.00 18.9

J

I

IIII

Note: k = 1.75It, in feetW, in kips

c, = W,h; / 194825.28F, = C,., (18.9)Sa = F,T / W, " rep Sa

= effective story acceleration


§ 15.4 .2 Example 52. Rigid NonbuiJding Structure 1

'~ample 52'''igid Nonbuilding_ Structure §15.4.2 )

The code has special requi rements for the determination of seismic forces for designof rigid nonbuilding structures. In this example, rigid ore crushing equipment is supportedby a massive concrete pedestal and seismic design forces are to be determined.


30'

20'

Grade

SOS = 1.33I = 1.0T = 0.02 secW EQUlPAfENT = 100 kipsW SUPPORT = 200 kips


[!J Design base shear

[3J Vert ical distribution of seismic forcesI

J

186

Design base shear

For rigid nonbuilding structures, Equation 15.4-5 is used to determine designbase shear.

V = O.3Sos I W= 0.3 (1.33) (1.0) W= 0.399W

= 0.399 (100 + 200) = 119.7 kips

Vertical distribution of seismic forces

The force shall be distributed with height in accordance with §12.8.3

Fx = C,,,V = C,'x (119.7 kips)


§15.4.2

(Eq 15.4-5)

(Eq 12.8-1 1)

IIIIIIIII

1 Example 52 • Rigid Nonbuilding Structure §15. 4.2

III

(Eq 12.8-12)

Story StoryHeight Weight Force Shear

Level Izxk W, k

Sah.t W.Jlx e ll.>; F, v,2 30 30 100 3000 0.429 51.25 51.35 0.5161 20 20 200 4000 0.571 68.45 119.7 0.342

300 7000 1.00 119.7

Note: 11., in feet fx= c, (1 19.7 kips)Wx in kips Sa = Fx / fV, .. nPSa

k ~ k = effective story accelerationc, = WxlzJ 1-rr,lzx

2006 IBC StructuraUSe;sm;c Design Manual. Vol. I 187

§15.7.6 Example 53 • Tank with Sup p or ted 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 containtoxic or explosive substances.

The following informati on is given.

SDS= 1.20I = 1.0W = Weight of tank and

maximum normaloperating contents

= 120 kips= 0.50 inch

I3J Find the design base shear

Slab

Grade

20'

§15.7.6

'C~/c;ylations and Discussion Code Reference

[!J The tank is a nonbu ilding structure, and seismic requirements for tanks with

supported bottoms are given in §15.7.6. This section requires that seismicforces be determined using the procedures of §15.4.2.

The period may be computed by other rational methods, similar to Example 51

where

LDLIDwI

wd

t

= 20 ft= 10ft= 20/10 = 2.0= W/L = 120,000 Ib/20 = 6000 plf= 0.50 in

6000(10)1,440,000

(0.50/12)

188 2006 IBC Struc tu r al/Seismi c Des ig n Manua l, Vol . I

Thu s, rigid nonbuilding structure, §15.4.2

1

1

Now: T

Example 53 • Tank With Supported Bottom

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

§15. 7.6

The lateral force shall be obtained as follows

V =0.3SDsIW=0.36W= 0.36 (120) = 43.2 kips

whereSDS= 1.20I = 1.00W = 120 kips

(Eq 15.4-5)

II

IIII

The design lateral seismic force is to be applied at the center-of-mass of the tank and itscontents. Note that the center-of-mass of the contents and of the tank do not normallycoincide. The distribution of forces vertically shall be in accordance with §12.8.3.

Commentary

The procedures above are intended for tanks that have relati vely small diameters (lessthan 20 feet) and where the forces generated by fluid-sloshing modes are small. For largediameter tanks , the effects of sloshing must be considered. Refer to American Wate r WorksAssociation Standard ANSI!AWWA D100 "Welded Steel Tanks for Water Storage," orAmerican Petroleum Institute Standard 650, "Welded Steel Tanks for Oil Storage" for moredetailed guidance.

2006 lac Structural/Seismic Design Manual. Vol. I 189

IBC §180B.2.23.1 Example 54 • Pile Interconn ections

ample 54Pile Interconnections IBC'§1808.2.23.1

A two-story masonry bearing wall structure has a pile foundation, Piles are locatedaround the perimeter of the building. The foundation plan of the building is shownbelow.

The following information is given.Original grade

Seismic Design Category 0J = 1.0SDS = 1.0

Pile cap size: 3 feet square by 2 feet deepGrade beam: I foot 6 inches by 2 feetAllowable lateral bearing = 200 psfper 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 capPile Dead Reduced Seismic QE

Cap Load Live Load N/S E/W

3 46 kips 16 kips 14 kips 0

10 58 16 14 0

o11

o109

o

f f f r4 11iI 2S' = 100 '

:=lA0 0 0

2 3 4 5A

~ tA

a<0II

®- a'"@)N

C6

Foundation plan


[!J Interconnection requirements

~ Interconnection force between pile caps 3 and 10

[!J Required "tie" restraint between pile caps 3 and 10

19 0 2006 IBC Stru c tu r al/Seism ic Design Ma n ua l, Vol. I

..Calculations and Discussion

[!J Interconnection requirements

Example 54 § Pilc Interconnections IBC §1BOB.2.23.1

Code Reference

IBC §1808.2

I

IIIIIII

The code requires that individual pile caps of every structure subject to seismic forces beinterconnected with ties. This is specified in §1808.2.23.1. The ties must be capable ofresisting in tension and compression a minimum horizontal tie force equal to 10 percent ofthe larger column vertical load. The column vertical load is to be considered the dead,reduced live, and seismic loads on the pile cap. An exception to §1808.2.23.1 allows use of"equivalent restraint" which , in this example, is provided by the confinement of very densegranular soil at the site .

Interconnect ion force between pile caps 3 and 10

Maximum loads on each pile cap under E/W seismic forces are

Pile cap 3 =46 + 16 + 0 =62 kips

Pile cap 10 = 58 + 16 + 0 = 74 kips

Minimum horizontal tie force Sos II 0 = 0.10 times the largest column vertical load

P = 0.10 (74) = 7.40 kips

[!J Required "tie" restraint between pile caps 3 and 10

The choices are to add a grade beam (i.e., tie beam) connecting pile caps 3 and 10, or to try touse passive pressure restraint on the pile cap in lieu of a grade beam. The latter is consideredan "equivalent restraint" (by soil confinement or bearing pressure) under the exception torae §1808.2.23.1.

For the allowable lateral bearing = 200 psf per foot of depth below natural grade, the passivepressure resistance is

. [2(200) + 4(200)]Passive pressure = (2 ft) = 1200 plf

2. 7400lbs

Required length = = 6.2 ft1200 plf

This is greater than 3'-0" pile cap wid th, but pile cap and a tributary length ofN/S gradebeam on either side of the pile cap may be designed to resist tie forces using the passivepressure. This system is shown below and, if this is properly designed , no grade beambetween pile caps 3 and 10 (or similar caps) is required.

2006 IB C Structural/Seismic Design Manual, Vol. I 191

IBC §1808. 2. 23 .1 Example 54 • Pile Interconnections

1,200 plf

~--4 800 psflf!

-.-•.....

::::~::::......

~:::::::::::~

:: : : .~: ::: : ::::~--'<-

2'·0"

6.2'E--------,Equivalent restraint system in plan Section 8-8: Grade beam

Normally, buildings on pile foundations are required to have interconnecting ties betweenpile caps. This is particularly true in the case of high-rise buildings and buildings with heavyvertical loads on individual pile caps. Ties are essential in tall buildings. Ties are alsonecessary when the site soil conditions are so poor that lateral movements, or geotechnicalhazards, such as liquefaction, are possible.

In the design of relatively lightweight one- and two-story buildings, the exception to theinterconnecting tie requirement of §1808.2.23.1 may permit a more economical foundationdesign. However, when interconnecting ties are omitted, a geotechnical engineer shouldconfirm the appropriateness of this decision, and the project specifications should call forthe back-fill and compaction methods necessary to provide required passive pressureresistance.


1~l

I

Example 55 D Simplified Wind Loads on 2aStory BUildings

The following is an example of the simplified wind load procedure of ASCE/SEI 7-05.

Calculate the wind loads on the following building.

Dimensions: 100 ft wide by 120 ft long by 25 ft high (2 stories - 13 ft and 12 ft) .

Wind Speed: Located in Minneapolis, Minnesota - 90 mph zone .

Importance: The facility is an office building with no special functions - Therefore thebuilding category in Table I-I is Category II.

§6.4

F 6-1

Exposure:

Enclosure:

Suburban office park surrounded by trees and typical suburban constructionon all sides - Therefore the exposure category is B. §6.5.6

The building has no unusual openings in the envelope, nor is it in a hurricaneregion, so no concerns for wind-borne debris - Classify as Enclosed. §6-2

Topography: Height of adjacent hills is less than 60 feet - Wind speed-up effects not a concern.(§6.5.7.1.5) x; = 1.0

Structure: The structure is an X-braced steel frame with evenly distributed braces on all fourexterior walls. The second floor is concrete slab on metal form deck on steel floorbeams. The roof is metal roof deck on steel joists on steel joist girders.

I

IIIII

Design Method:To utilize ASCE/SEI 7-05 Simplified Procedure (Method 1) all ofthe followingcriteria must be met.

1) With no breaks in the roof or floor (structural separations) the diaphragms aresimple, as defined in §6-2

2) The building height is less than 60 feet and least horizontal dimensions3) The building is enclosed and not prone to wind-borne debris4) The building is regular shaped5) The building is rigid with a period less than I second6) The site is not subject to wind speed-up effects7) The building is symmetrical8) For a building with well distributed MWFRS torsional load case in note 5 of

Figure 6-10 will not govern the design. Therefore design by Method 1 §6.4


§6.4 Example 55 • Simplified Wind Loads on 2·Story Buf/dings

25'


[!J Main wind force

~ MWFRS end zone width

~ MWFRS design wind pressures

~ Components and cladding

~ Edge Strip

~ Design wind pressure on components

[1J Main Wind-Force Resisting System-MWFRS (Lateral Load Structural Frame)

Using Method 1 §6.4, the simplified design wind pressure ps is the product of the basesimpl ified design pressure ps30 taken from Figure 6-2 and multiplied by the Height andExposure Adjustment Factor Afrom Figure 6-2, the Topographic Factor K=I from §6.5.7, andby the Importance Factor J from Table 6-1. The equation for ps is shown in §6.4.2. I Eq 6-1.

~ Calculate the MWFRS End Zone Width

End Zone =2a, so first calculate a, the Edge Strip Width.


Example 55 • Simp lified Wind Loa ds on 2·Story Building s §6.4

1Edge Strip =a =Lesser of:

But not less than:

Therefore :

• 10% of the least horizontal dimension = 0.10 x 100 ft= 10ft

· 40% of the eave height = 0040 x 25 ft = lOft·4% of the least horizontal dimension = 0.04 x 100 ft =4

ft• 3ft

a = 10ft, so the End Zone =2a = 2 x 10 ft = 20 ft

J

~ Calculate the MWFRS design wind pressure

Using Equation 6-1:p, = AKjp,3oLook up the base pressures P,30 from Figure 6-2 then modify for height, exposure,topography, and importance factor. No interpolation is required because the flat roof anglefalls in the row of "O to 5." With the mean roof height of 25 feet and the exposure being "B",the Height and Exposure Adjustment Factor A from Figure 6-2 = 1.0. Since the building siteis level from §6.5.7, K=r= 1.0. For a building Category II as defined in Table I- I, theImportance Factor I = 1.0. -

Tran svers e MWFRS - 90 mph, Exposure B, Height 25.0 rt

P IJ O A K:1 1 p,Type Zone Surface Label

Roof Angle HI. & Exp. Topographic Import. Design0'" to 5'" Factor Factor Factor Pressure

EndW all A 12.8 A 1.00 A 1.00 A 1.00 - 12.8 psfRoof B No Roof Projection for Flat RoofsHorizWall C 8.5 A 1.00 I A 1.00 IA 1.00 I ~ 8.5 psf

IntRoof D Nn Roof Proiection for Flat Roofs

EndWind E -15.4 A 1.00 A 1.00 A 1.00 - -15.4 psf

Ven Lee F -8.8 x 1.00 A 1.00 A 1.00 - -8.8 psfWind G -10.7 A 1.00 A 1.00 A 1.00 = -10.7 psf

IntLee H -6.8 A 1.00 A 1.00 A 1.00 ·6.8 psf-

Longitudinal MW FRS - 90 mph , Exposure B, Height 25.0 rt

P 130 A x; 1 p,Type Zone Surface Label

Base HI. & Exp, Topographic Import. DesignPress. Factor Factor Factor Pressure

EndWall A 12.8 A 1.00 x 1.00 A 1.00 - 12.8 psf

Horiz Roof B No Roof Projection in Lonaitudi nal DirectionWall C 8.5 A 1.00 I A 1.00 I )" 1.00 = 8.5 psf

IntRoof D No Roof Pro 'ecticn in Lonzitudinal Direction

EndWind E ·15.4 A 1.00 ). 1.00 A 1.00 - -15.4 psf

Vert Lee F -8.8 A 1.00 A 1.00 A 1.00 - -8.8 psfWind G -10 .7 A 1.00 A 1.00 A 1.00 = -10.7 psf

IntLee H -6.8 A 1.00 A 1.00 A 1.00 - -6.8 psf


§6.4 Example 55 • Simplified Wind Loads on 2-Story Buildings

Apply the pressure s to the building as described in Figure 6-2. The designations of"Transverse" and "Longitudinal" are keyed to the direction of the MWFRS being evaluated.When the resisting system being designed is perpendicular to the ridge line of the gable orhip roof, its direction is classified as "Transverse." When it is parallel to the ridge , it isclassified as "Longitudinal." When the roof is flat (slope ~5 · ) , and thus has no ridge line, theloading diagram becomes the same in each direction, as shown in the following diagram. Theloading diagrams shown should be mirrored about each axis of the building until each of thefour comers has been the "reference comer" as shown for each load case.

Design wind pressures p,usin g Eq 6-]

In addition, the minimum load case from §6.4.2.1 .1 must also be checked. App ly a load of 10psf on the building projection on a vertica l plane normal to the wind. In other words , create aload case with all horizontal zones equal to 10 psf, and all vertical zones equal to O. Checkthis load case as an independent case, do not combine with the case from §6.4.2.1. It shouldbe applied in each direction as well.


1

1

EYsmple 55 Q Simp lified Wind Loads on 2-Sr ory Buil din gs §6.4

I1

i< :

~f"'enc.Corner

LBeingEvacuated

I

IIIIIII

Minimum design wind loading

~ Components and Cladding (Everything except the Lateral Load StructuralFrame)

Accordi ng to §6.1.1, all "buildings .. ..and all components and cladding" must be designed forwind loads. Therefore, all parts of the exterior building envelope and any load paths, that arenot part of the main wind-force-resisting system (lateral frame), should be designed asComponents and Cladding (C&C) . For buildings such as this that qualify under §6.4.2.1, theC&C can be designed using §6.4.2.2, Eq 6-2.

~ Calculate the Edge Strip, a

Previously ca lculated in the MWFRS calcu lations, a = lOft

~ Calculate the design wind pressure on several components usingEquation 6·2

pnel =A. KztlP nel30

Look up the base pressures directly from Figure 6-3, then modify for Height, Exposure, Topographyand Importance Category. With the mean roof height of25 feet and the exposure being "B," theHeight and Exposure Adjustment Factor from Figure 6-3 = 1.00. Since the building is a leve l sitefrom §6.5.7, K, = 1.0, and for a Building Category II , the Importance Factor f lO' = 1.00.

2006 IBC Struc tural/Seismic Design Manual, Vol. I 197

§6.4 Example 55 • Simplifi ed Wind Loads oh 2· Story Bu ild;ngs ]

C & C - 90 mph, Exposure B, Heigh t = 25.0 ft

Type Zone Item EfT Direction Interpolation Pnl:tJO x Ku I P JJO

Wind Base HI. & Topo. Import. DesignArea Press Exp. Factor Factor Pressure

FactorDeck Positive None Required +5.9 x 1.00 x 1.00 x 1.00 +5.9·Screw < 10 sf Ne gative None Required -14.6 x 1.00 x 1.00 x 1.00 -14.6

Positive 10 sf 20 sf 12 sf+5.& x 1.00 x 1.00 x 1.00 +5.&·

Roof +5.9 +5.6 +5.&

Int Deck 12 sf 10sf 20 sf 12 sf-14.5 x 1.00 x 1.00 1.00 - 14.5Negative x

( I) - 14.6 - 14.2 - 14.5

> 100Positive No ne Required +4 .7 x 1.00 x 1.00 x 1.00 +4.7·

Joistsf Negative None Required -13.3 x 1.00 x 1.00 x 1.00 -13.3

Deck Positive None Required +5.9 x 1.00 x 1.00 x 1.00 +5.9·'", Screw < 10 sf Negative None Required -24 .4 x 1.00 x 1.00 x 1.00 -24.4's Positive None Required +5.& x 1.00 x 1.00 x 1.00 +5 .& ·•0 Roof 10 sf I 20 sf 12 sf"-

Edge Deck 12 sf Negati ve -23.9 x 1.00 x 1.00 x 1.00 -23.90 -24.4 I -2 1.& -23.90~ (2) Positive None Required +4.7 x 1.00 x 1.00 x 1.00 +4.7·

Joist > 100sf Nega tive None Required -15.& x 1.00 x 1.00 x 1.00 - 15.&

Deck Positive None Required +5.9 x 1.00 x 1.00 x 1.00 +5.9·Screw < 10 sf Negative None Required -36 .& x 1.00 x 1.00 x 1.00 -36.8

Positive None Required +5.& x 1.00 x 1.00 x 1.00 +5.&·Roof 10 sf 20 sf 12 sf

Comer Deck 12 sf Negative-36.& I -30.5

-35.5 x 1.00 x 1.00 x 1.00 -35 .5-35.5

(3)> 100

Positive None Required +4.7 x 1.00 x 1.00 x 1.00 +4.7*JOiSl

sf Negati ve None Required -15.& x 1.00 x 1.00 x 1.00 -15.&

Positive None Required 14.6 x 1.00 x 1.00 x 1.00 14.6Siding < 10 sf Negat ive None Required -15.8 x 1.00 x 1.00 x 1.00 - 15.&

10 sf 20 sf 17.3 sf+ 14. 1 x 1.00 x 1.00 1.00 + 14.1

Int Positive x+ 14.6 +13.9 +14. 1

(4) Stud 17.3 sf 10 sf 20 sf 17.3 sf- 15.3 x 1.00 x 1.00 1.00 - 15.3Negative x

-;; - 15.& - 15.1 -15.3::: Positive None Required + 14.6 x 1.00 x 1.00 x 1.00 + 14.6

Siding < 10 sf Neg ative None Required -19.5 x 1.00 x 1.00 x 1.00 -19.5

Positive10 sf 20 sf 17.3 sf +14.1 x 1.00 x 1.00 x 1.00 +14.1

IntStud 17.3 sf

+ 14.6 +13.9 + 14.1(4)

Negative10 sf 20 sf 17.3 sf

- 1&.6 x 1.00 x 1.00 x 1.00 - 18.6- 19.5 - 1&.2 -1 &.6

• Note. A minimum pressure of 10 psf 15 required per§6.4.2.2.1

198 2006 IBC StructurallSelsmlc Design Manual, Vol. I

1

1

I)

)

1 Examp/a 55 • Simplified Wind Load s on 2· Story Buil dings

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

§ 6.4

I1

1I

O Interior ZonesRoofs- Zone 1f\Nalis- Zone 4

Q End Zones'.,; Roofs- Zone 21Walls - Zone 5

Comer ZonesRoofs- Zone 3

2006 IBC StructuraUSeismic Design Manual, Vol. I 199

§6.4 Example 56 • Simplifi ed Wind Loads on Low Rise Buildings

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


A B c

1 -

2 -

/I

Typ

Main windforce-res istingsystem

b(0

3-story office build inglocated in urban/suburbanarea ofNW Texas - situatedon fiat ground

3 -

1<100'

PLANFlexibleDiaphragmTyp

W II II" nseel typ/' a mu 10

spaced 5 ~

)1/

Longitudinal Elevation Tra nsve rse Eleva tion


[}J Wind loads on MWFRS at Grid A

~ Wind loads on second-story wall mullion

200 20D6 IBC Structural/Seismic Design Manual, Vol. I

I1

Examp le 56 g Simplified Win d Loa ds on Low Rise Buildings §6A

[!J Wind loads on MWFRS at Grid A

11a.1Check applicability of simplified provisions §6.4.1 .1

I . Simple diaphragm building(See definition under "building, simple diaphragm")

Yes §6.2

2. Low rise building(Mean roof height < 60 ft and building widt h)

Yes §6.2

3. Building enclosed Yes §6.2

4. Regular shape Yes §6.2

5. Not flexible (II) > l hz) (T < I sec) Yes §6.2

T=O.I N=0.I(3)=0.3 secN = Number of Stories

6. No special wind characteristics Yes

7. Flat, gabled or hipped roof Yes

NW Texas basic wind speed = 90 mphThe desig n professional should contact the local building department toconfirm design wind speed .

Topographic factor K, = 1.0

Heig ht and exposure adjustment 'A.See §6.5 .6 for exposure category definitions

Example bu ilding in urban/suburban area is considered exposure BMean roof height (h) = 35 ft (see defin ition §6.2) (8 < 10")Adjustment factor from Figure 6-2, 8 = 1.05

F 6-2

§6.5.7

F 6-la

Note 5, F 6-10Yes

Therefore, simplified provisions are app licable

8. Torsional irregularities not a concern

Determine basic parameters

IIIIIIII

2006 IBC Structural/Se ismic Design Ma nual, Vol. J 201

Check minimum requirement:Horizontal load Eq 6-1 = (14.4 psf*12 ft + 9.6 psf*(25-12))*35 ft = 10.42 kipsMin load §6.1.4. 1 = {I0 psf* 25 ft)*35 ft = 8.75 kips < 10.42

: . 6.1.4.1 does not govern

11d·1 Determine end zone dimensions Note 10, F 6-2

Edge Strip a = 0.10 (60) = 6 ft . . . Governsor

= 0.40 (35) = 14 ftbut not less than

~ 0.04 (60) = 2.4 ftor

~3ft

End Zone 2a = 12 ft F 6-2

Horizontal Loads Vertical LoadsEnd Zone Int. Zone End Zone lnt. Zone

Load Roof A 8 C D E F G HV Dir. Angle Wall Roof Wall Roof WW LW WW LW

Roof Roof Roof Roof90 oto 12.8 -6.7 8.5 -4.0 -15.4 -8.8 -10.7 - 6.8

mph Transverse 5"17.8 -4.7 11.9 -2.6 -15.4 -10.7 - 10.7 - 8.1

20"

Interpolating: Forexample, roof angle 7.6" 13.7 -6.4 9.1 - 3.8 - 15.4 -9.1 -10.7 -7.0= arctan 10 = 7.6" (use 0) (use 0)

§6.4

202

Exam ple 56 • Simplified Wind Loads on Low Rise Buildings

Importance Factor J = 1.0(Category II Build ing from Table 1-1)

11c.1Obtain tabulated loads

Simplified Design Wind Pressure P.dO (psf)

11e.1Determine load on MWFRS at Grid A

Forces determined using Eq 6-1 ps = AK; JP.,3D

Horizontal load at wall :In end zone [A] = (1.05)(1.0)(1.0)(13.7 pst) = 14.4 psfIn interior zone [C] = (1.05)(1.0)(1.0)(9.1 pst) = 9.6 psf

Per §6.1.4.1, check 10 psf minimum over projected area of vertical plane

2006 IBC Str uc tural/ Seismic D esign Ma nual, Vol. I

T 6-1

F 6-2

§6.4.2.1

]

I1

IIIIIIIIIIIIII

II

Example 56 8 Simplified Wind Loads on Low Rise Bu ildings §6.4

Horizonta l point loads to frame:Roof Load

(5 ft tributary ht) VR=(14.4 psf" 12 ft + 9.6 psfl'(25 ft - 2 ft» 5 ft = 1488 Ib3cd Floor Load

(10 ft tributary ht) V3 = (14.4 psf" 12 ft + 9.6 psfl'( 25 ft - 2 £1» 10 ft = 2976 Ib2nd Floor Load

(12 .5 ft tributary ht) V2= (14.4 psf" 12 ft + 9.6 psfl'(25 ft - 2 ft) 12.5 ft = 3720 Ib

Note: Forces to Grid A are shown based on a tributary basis that is conservative forthe analysis of Grid A. Alternatively, the forces could be distribu ted to gr ids A and Cby applying the loads as a simple span beam.

Vertical load at roof:Windward Roof - In end zone [E] = (1.05)( 1.0)( 1.0)(-15.4 pst) = -16.2 psf

In interior zone [G] = (1.05)( 1.0)( 1.0)(-10.7 psf) = -11.2 psf

Leew ard Roof-In end zone [F] = (1.05)( 1.0)( 1.0)(-9. 1 psf) = -9 .56 psfIn interior zon e [H] = ( 1.05)( 1.0)( 1.0)(-7 .0 psf) = -7.35 psf

Vertical uniform loads to frame:Windward: (16 .2 psf)( 12 ft) + (9.56 psf)(25 - 12) = 340 plf= .34 kif upliftLeeward: (11.2 psf)(12 ft) + (7.35 psf)(25 - 12) = 210 plf = .21 kif uplift

Note : Forces applied to Grid A are shown as a distributed load along the framelength . A more detai led analysis of forces based on roof framing would include asmaller distributed load and upli ft point loads at locations where beams frame into thegrid A moment frame at grids I, 2, and 3.

0.34 kif

VR = 1.49 k > Trib. HI. 10 ft/2 = 5 It

V3 = z.se' > 10 ft/2 + 10 ft/2 = 10 n

V2 = 3.72k ) 10 ft/2 + 15 ft/2 = 12.5 ft

Elevation


§6.4 Example 56 • Simplified Wind Loads on Low Rise Buildings

cr C( cr0-r r:J l

Load Cases: D 9.6 psI x TribHIS§6.4.2.1

14.4 psI x Trib HI

10 psI x Trib HI 5 §6.4.2 .1 .1

Plan

[!J Wind loads on second story wall mullion

~ Determine zone of mullion F 6-3

Interior of wall area - Zone 4

Effective wind area = 5 ft (10 ft) = 50 sq ft

Wind Loads ps = AKztl Pnel30 §6.4.2 .2 (Eq 6-2)

pne130 = 13.0 psfpositive= -14.3 psfnegative (suction) F 6-3

p, = (1.05)(1.0)(1.0)(13.0 pSfposilive)(5 ft tributary) = 68.5 plfp s = (1.05)(1.0)(1.0)(-14.3 pSfnegalive)(5 ft tributary) = 75 plf

2nd floor

3'· floor

r~

(

=,/

(

=,/

..... lJ75 plf or 68.5 plf

204 2006 IBC Structural/Seismic Design Manua l, Vol. I

Ex ample 58 . Floor Vibrations

A 9-sto ry building has a moment-resisting frame for a lateral force-resisting system.

Find the lateral forces on the frame due to wind.

Office build ing 50 ft by 50 ft inplan with MWFRS at exterior.Located in an urban/suburbanarea ofN.W. Texas

Determ ine:

[TI Wind loads on MWFRS

50'

~ I1

4

12'

12'

12'

12'

12'

12'

3"12'

12'

Elevation

Icai~ulationsJ!n(f Discussion

I1

I

[TI Wind loads on MWFRS

11a.1 Determine basic wind speed

Ut ilize ASCE/SEI 7-05 §6

Use meth od 2 analytical procedure

Chapter 6

§6.5

2006 IB C Structural/S eismic Design Manual, Vol. I 205

§6.5 Example 57 • Wind Loads - Analytical Procedure

Confirm building is regular shaped and not subject to across wind loading, vortex 1shedding, instability due to galloping or flutter ; or does not have a site location forwhich channeling effects or buffet ing in wake of upwind obstructions warrant special Iconditions §6.5.1

Design procedure §6.5.3 IBasic wind speed V = 90 mph §6.5.4,

F 6-1 Ilib·1 Determine velocity pressure

Wind directionality factor Kd= 0.85 §6.5.4.4,(applies when using load combina tions T6-4in ASCE/SEI 7-05 §2.3 and §2.4)

Importance factor I = 1.00 §6.5.5,(Structural Category II, Table 1-1) T 6-1

Exposure Category B §6.5.6 IVelocity pressure coeff K= (Case 2) §6.5.6.6,

T 6-3 IExposure 8

" Case 2

I0· 15 fl 0.5720 0.6225 0.6630 0.70 I40 0.7650 0.8160 0.8570 0.89 I80 0.9390 0.96

100 0.99

I11 6 1.03 • By Interpolation120 1.04

Topographic factor KZ1 = I §6.5.7 I(example building on flat land, no nearby hills)

Gust effect factor G §6.5.8 I9-story buildingNatural period = 0.1(9) = 0.9 sec §9.5.5.3.2

1(Eq 9.5.5.3.2-la) INatural frequency = - = 1.1 Hz > 1.0

0.9Therefore: Rigid structure §6.2

IG= 0.85 §6.5.8.1

206 2006 IBC Structural/S eismic Design Ma nual, Vol. I

Enclosure ClassificationExample building enclosed

Velo city Pressureq==O.00256K2K2kKdV2

/

= 0.00256K=KrK2V 2/

=O.00256K=( 1.0)(0.85)(90)2( 1.0)

Example 58 a Floor Vibra t ions

§6.5.9

§6.5. 10Eq 6- 15

0-15 ft202530405060708090

100116

10.0 psf10.911.612.313.414.315.015.716.416.917.418.2

11 c.1 Determine pressure and force coefficients

Internal pressure coefficients - GCpi

GCpi = ±0.18 Case 1: Internal Pressure InwardCase 2: Internal Pressure Outward

External pressure coefficients - Cp

For example building, monoslope roof'B = 0

§6.5.11

§6.5.11.1,F 6-5

§6.5.11.2,F 6-6

L

Elevation

L

Plan

(Note: Internal pressuresmust be added to orsubtracted from externalpressures typical


§6.5 Example 57 • Wind Lo ads - Analytical Procedure I

Windward wall C; = 0.8 F 6-6

III ,GC = , (0 .85)(0 .8)

0· 15 n 6.80 I20 7.4125 7.8930 8.3640 9.11 I50 9.7260 10.270 10.7

I80 11.290 11.5

100 11.8116 12.4 I

Leeward wall IL 50

F6-6- = - = 1 ---> C = - 0.5B 50 p Iq" = q " ' ll 6 fi = 18.2 psf

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

c, =-0.7 F 6-6

Iq"GCp = 18.2 (0.85)(- 0.7) = -1 0.8 psf

Roof Ih 11 6- = - =2.3 > 1.0L 50

Ic, = - 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 Ilid·1Design wind loads §6.5.12 I

Main wind-force-resisting system §6.5.l 2.2 IRigid building §6.5.12.2.1

I208 2006 IBC Structura l/Seismic Design Manual, Vol . I I

Example 58. Floor Vibra:ions

Windward wall

qh(GCp i) = (18.2)(0. 18) =3.28 psf(±)

CEq 6-17)

Ii

0-15 ft202530405060708090

100116

Leeward wallp =q"GCp - qh (GCp i)

p = CJ=GCp - Q1J(GCp;) Case 1 shown

10. 110.711.211.612.413.013.514.014.514.8/ Sample Calculation15 I P = 12.4 - 1B.2(-0.18) = 15.7 Case 1

. 12.4 - 18.2(+0.18) =9.1 Case 215.7

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 psfRoof

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


§6.5 Example 57 a Wind Loads - Analytical Procedure

11 e.1 Design wind loads - graphically

r--r--r---r---,. 19.4 psf

4.5 psf Case 111.0 psI Case 2

Cas e 1 15.7 psI .....----:J-f---'--L---L.---l-~

Case 2 9.1 psI

Wind-----,)

11.0 psf

Plan

14.1 psf

Wind~I--~

Elevation

Case 1: Internal Pressure InwardCase 2: Internal Pressu re Outward

Verify projected load is greater than 10 psf10.1 + 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 RigidDiaphragm Analysis


ibc 2006 structural seismic design manua whit examplesl 1 220p

Documents

type ib example

strength design example

exam ple example

comb inations of loads

i table of conten ts

pdelta effects example

redu ndancy factor p

vertical irregularity