ENGINEERING BULLETIN

Shear Wall-Frame Interaction BY

A DESIGN AID

lain A. MacLeod

PORTLAND CEMENT I ’ ASSOCIATION IEI

FOREWORO

Shear walls and frames in combination normally providethe required stiffness and strength to withstand lateral loadsin high-rise buildings. Apartment buildings up to 70 storieshigh have successfully utilized this concept. [n certain casesthe walls me much stiffer than the frames and thus takemost of the lateral load. For this reason, the participationof the frame in resisting lateral load is often ignored. TMsmay not always be a conservative procedure and it istherefore important that the effect of the frames beconsidered,

In this publication procedures are recommended foranalyzing shear wall-frame interaction problems. A newsimplified method for assessing the effect of the frames isalso presented.

COVER PHOTO: Lincoln Park Tower% Chicago, 111.Architects andEngineers: Dubin, Dubin, Black & Mo.touswm; Structural Em+neer: Eugene A, D.bin; General Contractor: James McHughComtr.cti.n Co.–.]! of Chimw,

@ PcmlandC.mnr A.wciadon 1970

Reprinted1974, 1981,1983,1985,1998

CONTENTS

Notation .

1. Behavior of Structures Under L?teral Lnad .

l.l. Behavior Difference Between Frames and Shear Walls

1.2. Behavior of Floor Slabs .

1.3. Effect of Torsion

1.4. Effect of Openings in Shear Walls

2, Simplified Methods of Estimating Lateral Lnad Distribution

2.1. Use of Charts from References 6 and 7

2.2. The Component Stiffness Method-Equation C

2.3. The Component Stiffness Method for Structures with Torsion

3. Calculation of Moments and Shears Within the Structure

4. More Accurate Analysis

4.1. Numerical Accuracy .

4.2. Plane Frame Computer program

4.3. Space Frame Computer Progmm

References . . . . . . . . . . . . . . . . . . . .

3

5

5

6

6

6

7

7

7

12

Is

15

Is

16

16

17

This publication is based on the facts, tests, and authorities stated herein. It is intendedfor the use of professional personnel competent to evaluate the significance and limitwtions of the reported findings and who will accept responsibility for the application of thematerial it contains. Obviously, the Portland Cement Association disclaims any and allresponsibility for application of the stated principles or for the accuracy of any of thesources other than work performed or information developed by the Association.

NOTATION

,4, = area of columns

A cl, Ac2 = areas of wall sections

a = distance of applied load from origin–see Fig. 9

B = width of frame

b = clear span of beams

C = column width

D = depth of beam

E= Young’s modulus (subscript indicatesstructural system)

Fx, Fm, Fn, F,= functions usedin Equations Aand B(Table l), dependent onthe type ofloading

lb at top of frameg= ratio , as used in

1, at bottom of frame

Table ~

H = total height of wall

h = height of column, i ,e, story height

Ib=mmwmt of inertia of a connectingbeam

1,, , IC2 = moments of inertia of wall sections

[W = moment of inertia of wall

KB=rotational stiffness of shear wall sup-port

Kf=lateral point load applied at top offrame to cause unit deflection in itsline of action

K,= factor representing both Kw and Kf

Kw=lateral point load applied at the topof a shear wall to cause unit defletition in its Iineofaction; also specifi-cally tbe stiffness of a shear wallwithout openin~–see Eq. (1)

K ~. = stiffness of a shear wall with open-ings–see Eq. (2)

!?= distance between centroidal axes ofwalls or columns, or span of beams

~ = ~ or ~ j“ equations of Table 1

A at top of framen = ratioA ~t bottom of frame, as used in

.Table I

P= interaction load at the top of theframe (subscript denotes frame repre-sented as spring)

R = support reaction coefficient for apropped cantilever

IC at top of frame , ~~ “~ed ins = ratio

Ic at bottom of frame

Table 1

W= total applied lateral load (subscriptdenotes load on specific wall)

~ = a variable used for shear wall withopenings–see Section I.4

Tw = dimensionless parameter which re.Iates the rotational stiffness of thewall to that of the foundation, i.e.

KBHratio —

4EW1W

A = total deflection at top of structure

AA = top deflection due to column axiafdeformation

AB = top deflection due to bending defor-mation

8 = a deformation at top of structurewith torsion—see Section 2.3

A = ratio of column to beam stiffness

(-e+p = variable used for shear wall with

openings–see Section 1.4

%wtlon number

~

Consider behavior of structures under lateral load

II

~

Uw a simplified method to estimate distribution of

NOtOrsiOn With torsion

-’* &

5’Calculate moments and shears in members

[

4.2

Notorsion With torsion

a==

Pig, 1, F1OWdia.framof analysisprocedure.

4

Shear

1. BEHAVIOR OF STRUCTURESUNOER LATERAL LOAO

A suggested procedure for lateral load analysis of high-risebuildings is illustrated by tbe flow diagram in Fig. 1 andthis publication, as a design aid, will follow that patternsection by section. The following is a general introductionto the behavior of structures under lateral load.

1.1. Behavior Difference Between Frames and Shear Walls

A rigid frame, an interconnection of vertical columns andhorizontal beams, bends predominantly in a shear modeasshown in Fig. 2(a). A shear wall deflects predominantly in abending mode, i.e. as a cantilever, as illustrated in Fig. 2(b).Elevator shafts, stairwells, and reinforced concrete wallsnormally exhibit this behavior.

It is not always easy to differentiate between modes ofdeformation. For example, a shear wall weakened by a row(or rows) of openings can tend to act like a rigid frame andconversely an infllled frame will tend to deflect in abending mode.

When all vertical units of a structure exhibit the same

ht@raCth)I’i A DESIGNAID

behavior under lateral load, i.e. if they are all rigid framesor all shear walls, the analysis is comparatively simple. Theload can be distributed to the units directly in proportionto their stiffnesses. The difference in behavior under lateralload, in combination with the in-plane rigidity of the floorslabs, causes nonuniform interacting forces to develop whenwalls and frames are present (Fig. 2 [c]). Thk makes theanalysis more difficult.

(.1 RIGID FRAME lb] SHEAR WALLSHEAR MODE BENDING MODE

DEFORMATION DEFoRMATION

{c) INTERCONNECTED FRAMEAND SHEAR WALL

(EQuAL DEFLECT IONS ATEACH STORY LEVEL)

Fig. 2 Defomnmion modes.

1.2. Behavior of Floor Slabs

For analysis, the floor slabs are normally considered to befully rigid within their own planes. This means that therewill be no relative movement between the vertical units ateach story level. In-plane deformation can be taken intoaccount but is seldom important.

Floor slabs bending out of plane contribute to thelateral stability of a structure by acting as beams betweenvertical members, and hence flat plate structures, forexample, act like rigid frames. However, the strength ofcolumn-slab joints in flat plate structures must be checked

Fig, 3, Wallwith a singlerow of openings.

1.4. Effect of Openings in Sberrr Wells

a useful parameter for assessing the effect of the openings isM, where

4(

12fb !2

)

/4,1 +.4,2~.

~ Icl +IC2 + ACIAC2

H = total height of wallIb = moment of inertia of a connecting beamh = story heightb = clear span of beams!?= distance between centroidal axes of the wall

sectionsI ,,, IC2 = moments of inertia of wall sectionsACI, AC2 = areas of wall sections

W’a,l , ;.,, 2(’cl} W

I.) SHEAR WALL WITHSINGLE Row OF OPENINGS

(c%%%?”

~M*b+... ,, w,,,,

(b] IDEALIZATION FORSHEAR CONNECTION METHOD

(c) FRAME IDEALIZATION

carefully if they are to be used in resisting lateral load.

1.3. Effect of Torsion

The effect of torsion should be considered if the layout isunsymmetrical or if the stiff vertical units are close to thecenter of the structure, Some earthquake codes require thata structure be capable of resisting a specified torsionalloading even if the applied lateral load theoretically doesnot cause torsion~ 1J* The importance of torsion may be

asseseed by comparing results from simplified methodswithout torsion (Sections 2.1 and 2,2) and with torsion(Section 2.3).

Openings can have unimportant effect onthe behavior ofshear walls, The openings are normally invertical rows andin the common case ofasingle rowofopenings (Fig, 3[a]),

.S. pers.ript numbers in parentheses desienate references o.

p.s. 17.

FormH that is greater than 8, the wall tends to behavelike a single cantilever, For Iowaff, e.g. less than 4, thebehavior ismorelike twoconnected walls and frame actionis more prominent.

With a single mwof openings the effect ofopeningsonthe stiffness can be assessed by comparing

3EIKw = -$

and

3E(IC, + I, ~)K w. = ff3&

(1)

(2)

where Kw and Kwo = stiffnesses of the wall without andwith openings, respectively

E = Young’s modulus of elasticity

IW = moment of inertia of wall withoutoneninm

(~., +A.,)U., +1.,)p=l+

A., A.. b2.,. .Coull and Choudhurytzl include a plot of K4 against afffor different values of p,

Tbe stiffness and tbe distribution of stress in a wall willnormally be appreciably affected by the presence of

6

openings. The shear connection methodtz’5 J is suitable forband calculation (although the use of a computer to do thecalculations is desirable) and can be used for more complex

~~problems than that illustrated in Fig. 3(b). Coull andChoudhury’s chartsfz ,3) are particularly useful for prob-lems with a single row or two symmetrical rows ofopenings.

Shear walls with openings can be idealized as illustratedin Fig. 3(c), using a plane frame analysis program (Section4.2).

‘For analysis of walls with openings, axial deformationof the wall sections should be included.

2. SIMPLIFIED METHOOS OF ESTIMATINGLATERAL LOAD DISTRIBUTION

If torsion is not considered, two simplified methods ofdetermining the interaction of frames and shear walls are:

. to use the charts given by Khan and Sbarounis(6)or PCA’S Advanced Engineering Bulletin No.14,f7) and

. to use Equation C (Table 2).

proportioned frame may be defined as one that has pointsof contra flexure at all nddbeam sections under lateral load.Any frame whose bay widths and member properties arereasombly constant across the width of tbe frame may beconsidered to be proportioned. The frame being propor-tioned and the column axial deformation being negligibleare basic assumptions made in reducing a multi-bay frameto a single- or three-bay frame,

The procedure for reducing tbe number gf bays is: ateach story level, sum_aU cofynn inertias (Ie) and beamrotational sJiffnesses (lb/!). (lC is the moment of inertia ofa cnhmm; Ib and !?are the moment of inertia and the spanof a beam, respectively.) The equivalent stiffnesses to beused in the substitute frames thus become

1== Zid/2 and Ib/!? = ~-b/k?

for the frame in Fig. 4(a) and

10= ZTC[6 and Ib/t =~~~l!i

for the frame in Fig. 4(b). This procedure is clearlydescribed by Khan and Sbarounis.

Having thus reduced the problem, the shears on theframe, moments on the shear wall, and deflection can befound using the charts.

2.2. The Component Stiffneee Method-Equation C

2.1. Use nf Charts from References 6 and 7

In order to use these charts the structure must be reduced~~to a single frame and a singfe wall by addition of the

properties of the separate vertical units. In both referencesthe stiffnesses (Iw) of all the shear walls are summed to givean equivalent single wall. For the frames, Khan andSbarnunis use a singfe-bay frame whereas Advanced Engi-neering Bulletin No. 14 uses a three-bay frame. Theseidealizations are illustrated in Fig. 4.

Both the frames in Fig. 4 are “pmportinned.” A

f

P-

F,wne

1.

b

L

h..,W.11

L,,cl

(b) REDUCED STRUCTLWEFOR REF. 7

F& 4. Reduced structure for Referemxw 6 md Z

The component stiffness method has more flexibility thanthe charts method referred to above, but it lacks accuracy ifthe wall is more flexible than the frame (KW/Kf < 1; seebelow),

The r?uzin assumption is that the frame takes constantshear, i.e. that the interactir.m force between the frame andwall can be represented by a concentrated force at the top.This is a reasonable assumption for preliminary analysis,especially when the frame tends to be flexible in compari-son with the wall.

Consider the single-bay frame and shear wall loadedin-plane by the uniformly distributed load shown in Fig.5(a). If tbe frame shear is assumed to be constant, thesystem can be treated as a wall supported at the top by aspring (Fig. 5 [c]), The spring stiffness K is defined as the

{lateral point load applied at the top oft e frame to cause““it deflection in its line of action. Kf can be calculatedusing Equations A and B of Table 1 and the top deflectionequation in the Table 2 notation.

Kw is defined as the Iateraf point load required to causeunit deflect ion at the top of the wall (similarly to Kf). ASdiscussed in Section 1.4, Eq. (1) can be used for wallswithout openings and Eq. (2) for walls with a single row ofopenings.

Table 2 (Equation C) gives relationships between P/W,

7w, and KJKf fOr different JOading cases. Si~l~ expres-sions for other load cases can easilv be established. P is theinteraction load at the top of the-frame, i.e. the constantshear; W is the total applied lateral load; and 7W is adimensionless parameter which relates the rotational stiff-

Table 1. Equations B and A for Top Deflection of Rigid Regular Frames

EQUATION B–for Bending deformation:

,,~:c, [~,(l-W+Fg(l -,c,32qAB=—

where AB = deflection at top of frame due to bending of members

W= total lateraf load

b = story height

H = total height

E = Young’s modulus (subscript denotes structural system)

ZIC = sum of moments of inertia of columns at first-story level

F,, Fr = functions ofs and g, dependent on the type of loading

~ = ratio Ic at top of frame

I

~ at bottom of frame finear variation of Ic and Ibc

Ib attop of framewith height. If Jl varies, use

g = ratioI, at bottom of frame

EI instead of I.

0.= $, where C is cohmm width and !2is distance between col-

umn centerlines

~= Z(E I/h)~, I.e. summation over width of structure at first.2Z(.5JJ!2)

story ~e;el

Ib =’moment of inertia of beam at bottom of structure

EQUATION A–for ~xial deformation:

where AA = deflection at top of frame due to axial deformation of ex

terior columns

Fn = function of n, dependent on the type of loading

Area of exterior column at top of framen = ratio

Area of exterior column at bottom of frame

(linear variation of ,4C with height)

,4C = area of exterior columns at first-story level

B = totaf width of frame

TOTAL DEFLECTION A= AB + AA

NOTESFor accuracy, the following rxmditions should be satisfied:

1. Qand lb should not vary ,acrow the frame,2. [ should not vary acro$s the frmne, except that 1, for an interior column

%$ ould be twice that of an exterior @wnn.3. Columns should h.,. points of io.trai%xure at nddhel@t.4, Story hei ht should he constant,

J5. /c, lb, a. A. should vary linearly with height.6.k ,< 5,7, AA should be smell compared with ALI,

Rewo.nable results can be expected in many cases which satisfy the above con.diti.ns only ,approximatety,

Eq..rio.s A and B both tend to o.erathn.te deflection,

3.0

2.0

Fm 1,5

I.0 I,0

2131/2

o. ~,0,5 1,0

m=sorm. g

Loadcondition

Point loadat top

Uniformlydistributed

Triangular(earthquake)

Loadcondition

Point loadat top

Uniformlydistributed

Triangular(earthquake)

F$(m=s)or Ft(m=g) Iloge m

n Ii

3 ~2-—+2m –—-logenl

loge m + 2 2

m-1 (m- 1)3

1-4n+3n2-2n210gen

(1 - fl)3

2-9n+18n2- lln3+6n310gen

6(1 - n)4

(~ 210gef2+ 5(l-n+10gen)3 n-1 (n- 1)2

9

+ ~-6“:n:,;33‘“g’n3n2 ~3

-;+3n - ~+; -logefl

+(n- 1)4

25 4n3 ~4

)

-fi+4n- 3n2+T-T-10ge n+

(n- l)s

ness of the wall to that of the foundation, i.e.

KBH‘rw=—

4EJW

where KB is the rotational stiffness Of the shear wailsupport. If the rotation at the base of the shear wall is to beneglected, the terms with Yw in Equation C shoufd be

~l”o Sflffrm,s -K,

/r- L,nk Bar,

ToldLo.i

Ijlil

“,

lW

/ ‘-From’shearW.(I

w

Ly;eldin, s“,,.,+

(01 STRUCTURE [b) INTERACTION (c1 FRAME MODELEDAT TOP ONLy BY SPRING

Fig. 5. Idealization for Equation C.

omitted. However, the effect of shear wall base rotation cansignificantly affect the distribution of load between shearwalls and frames, and Equation C can be used as a simplemethod of assessing this factor.

Table 2. Equation C

Loadcondition I Ecytion C

1+--$Point load E= w

at top w 3K

1’47. ‘y

T {)3*1+1-

Uniformly $=‘fW

distributed ,+ L+%@w Kf

~+~20 27W

Triangular ;=(earthquake) 3 KW

1+~+~w

Problems involving several frames and walls may bereduced to that of a singfe wall and frame as described inSection 2.1. Alternatively, Kfor KW foreach verticaf unitmaybe calculated separately and the results summed. XKfand ZKW are then used instead of Kf and KW in f@atiOnc.

Studies on shear wall-frame interaction normally usethree parameters to detine bebavior; namely, h,Iw, andXIc, where

EcIc/hA=—

E#bl!2

By using Kfi which is a function Of A and ~c, bebaviOr canbe discussed in terms of only two Variables, Kf and KW.Thk simulifiea the Dhvsical interuretationof the behavior.Also tb~ paramet~r -P/W is u~efuf for estimating tbeeffectiveness of tbe frame (or frames)in comparison withthe shear wafl(s) in resisting lateral load and for assessingthe effect of various assumptions in analysis.

Accuracy

When a frame and wall are interconnected as shown in Fig.5(a), maximum shear on the frame tends to occur tOwardsmidbeigbt (see Fig. 8). Equation C can underestimatemaximum frame shear by as much as 30 percent in this.“.area. Therefore, when calculating moments in theimme, ltis worthwhile to increase the calculated value of F’by 30percent.

If Kw/Kfis less than 1, the use of Equation Cis not

recommended and the use of chartsce,~j produces moreaccurate resufts.

NOTATION

P= interaction force at top

W= totafapplied lateral load

KBH‘w= 4EWIW

KB= rotational stiffness ofshear wall support

H= total height of wall

E= Young’smodulus (subscript denotes structural system)

IW= moment ofinertia of wall

KW = ~ (with constant ~w)

K,= point loadattop of frame tocause unit deflection inPits line of action ;i.e., ~or — since top de flee-

A AB+AAtion A. ~

Kf

NOTE:, For frame properties, see Table 3

‘Fig.6, Example structure for component stiffness method–notorsion.

Table 3. Moments of Inertia (in.4)of Frame Members, Example Structure*

story

1098-/65“432I

Ib

13,80013,80013,80013,80013,80013,80013,80013,80013,80013,800

——

2,5404,4307,3909,270

11,52014,14017,08021,68024,77026,270

*E = 3,000 ksi for wall and frame members.

Example Problem

This problem is definedin Fig. 6and Table 3, and we willfollow the flow diagram in Fig. 1. Results will be comparedwith those of a more accurate analysis in Example I byGoldbergX8J

A. D18TR3BUTELOADS TO THE VERTfCAL UNITS

use Equation C (Table 2), Foundation rotation of the shear

wall will not be considered; therefore, the terms with Twcan be ignored. Since there are two walls and seven frames,use ZKw and ZKf in Equation C, which becOmes

P 3/8

ii=‘+’%’

calculate K,

calculate Kf for a single frame, using Equation B frOm

Table 1. The finite widths of the beams and columns arenot considered here so that 19, = On = 0. Equaf iOn B thus

since~b= 13,800 in.4 (see Table 3),ZIC =4X 26,270 =105,080 in.,4

H= 126 ft,, average h=12.6 ft.,!2= Zo.oft.,

105,080 20‘=2X 12.6X3X 13,800 =2’02’

* = 0.0965 and thus ~ = 2.54 (see Table l),s = 26,270

g= land thus Fg=l,

it follows that

AB 12.62X126X123

7=12X3X 103X 105,080 X(2”54+1 ‘2x2”02)

= 0.0602 in./kip (computer value is 0,0607 in./kip).

Estimate A/P due to axial deformation of columns.Assuming column areas of the order of 400 sq.in. and n =

0.5, from Equation A (Table 1) we find that Fn = 0.77 and

AA H3 x 0.77= 1263 X I 23 x 0.77T = ECACB2 3X 103 x400X602X 12Z

= 0.00426 in./kip.AA

Hence ~ = 0.071; i.e., top deflection of a frame due to

column axial deformation would be approximately 7percent of that due to bending. Since column axialdeformation was neglected for the analysis given byGoldberg, it is also neglected here and

~=l=L = 16.6 kip/in.*f AB 0.0602

XKf = 7 X 16.6 = l16kip/in.

Calculate Kw3X3 XI03 6X603X123KW.~=1263x 123 x 12

= 4S6 kip/in,

XKW = 2 X 4S6 = 972 kip/in

P— would he used were column axial deformation*Kf = AB + AA

to be considered.

CaIcuk+te PI W

Calculate Loads on Units

W = 9 X S7.6 + 28.8 = 547 kips

P = 547 X 0.04= 21.9 kips

Ptoeachframe =2&= 3.1 klps

P to each wall= ~= 10.9 kips

That is, loadings are as shown in Fig. 7.

calculate Top Deflection

~=; =L!6. -0.187 in. (see Table 2)f

B. CAf.CULA~ MOMSNT8IN MEMBERS

Moment in Wall

Moment at base of structure for one.half of the appliedloading is 18,835 kip.ft. (see Fig. 6). Therefore, for eachwall

max. moment = 18,835- PH = 18,835- 10.9 X 126= 17,474 kip.ft.

Moments in Frame Columns

Assume an interior column takes twice the shear as an

exterior column. Therefore,

sheiu on interior colwnn = ‘->

=+=1.03 kips

Maximum moment will occur towards midheight of the

frame and have a value of

1.03 X 0.5 X 12X 1.3= 8.0 kip.ft.

Tbe 0.5 factor is for point of contra flexure at column

31

R

[o) FRAME (7 Th”31

14.4

g

10.9

28,828828 B28828,828,828.828,8288

(b) WALL [2 Thu31

NOTE: Loads are in kips

Fig. 7. Loading on units–no torsion.

midheight. The 1.3 factor allows for the fact that maximumframe shear will be underestimated by approximately 30percent.

C. COMPARISON W’TH MORE ACCURATEANALYSIS

Goldberg(s) takes account of in. lane deformation of thefloor slabs. Webster’s discussion(9 ? giv~$ ~~s”lts with ~i@d

SHEAR ON FRAME ( KIPS)

Fig. 8. Comparison of frame shearestimates.

Table 4. Accuracy of F.qustion C in Comparisonwith Frame Analysis*

I (1)

zDesign Framevalues anal ysis

Moment atbase ofshear wall,kip.ft 17,526

Topdeflection,in. 0.173

(2)

Equation C

17,474

0.187

(3)

Percent cliff. =(2)- (l)X loo

F

-0.3

+8.1

*Frame malysis in Example 1 by Goldbem (8)

floors. Therefore, comparison with the latter results showsthe effect of the constant shear assumption only. Fig. 8shows the difference between the calculated shears on atypical frame. Other results are given in Table 4 and showsatisfactory a~eement.

w

W.1 1

s

w

B‘w% _Spring supports

[walls only)

Wtqid beam

+ w

(b) DISTRIBUTION OF W

TO WALLS

PLAN WALLS FRAMEs

(al STRUCTURE (c) LOADING ON UNITS

PLAN

(d] IDEALIZATION

A%rin9s to represent the action of the fr.mes ~w

v

Q

Q?$

+

\

L‘w

RW (At TOP)

Q

~J-:;;:r,,‘– Rigid beom

oW ( ~st,ib,tedl

PLAN PLAN

(0) NO TOP MOVEMENT (f) TOP LOADING ONLY

Fig. 9. Component stiffness method with torsion.

2.3. The Component Stiffoess Method for Structureswith Torsion

By making the assumption that the frames take constantshear (as for Equation C), the structure shown in Fig. 9(a)may be idealized as in Fig. 9(d). Two degees of freedomwhich correspond to the deformations A and 8 at the top ofthe structure (Fig. 9 [f]) can be assigned and the structuresolved by the stiffness method, The analysis is carried outby adding the results from Systems 1 and 2 (Fig. 9 [e andf]). Lateral movement at the top of System 1 (Fig. 9 [e]) is

prevented by a force that equals R W, where R is the

support reaction coefficient for a propped cantilever (e.g.,for a uniformly distributed load, R = 3/8) and W is totallateral load. R W acts in the line of and in the oppositedirection to the resultant of W at the top of the structure.For analysis of System 2 (Fig. 9 [f]), the roof slab can beconsidered as a rigid beam on spring supports. The springstiffnesses are KW and K, as defined in Section 2.2.

The equations for determining the unknowns A and Oare set up as follows: Transverse equilibrium gives

ZKi(A + X8) ‘R W

and therefore

ZKiA + 2CKiX10= R W (3)

By taking moments about O,

XKi(A + X#+)A’i= R Wa

and therefore

2KiXiA + ZKiXfO = R Wa (4)

The above Ki represents both the wall stiffnesses, KW, andthe frame stiffnesses, Kfi and u is the distance Of appliedload from origin. For delineation of X and a, see Fig. 9(f).

Having solved Eqs. (3) and (4) for A and 6, the loads onthe springs are

F’i= Ki(A + Xi6) (5)

If there are only two walls, simple statics gives thedistribution of the total lateral load, W. With more thantwo walls, Eqs. (3) and (4) must be re-established,neglecting the frame springs and solving with Was the onlyloading (Fig. 9[b]). The resulting deformations are ficti-tious but can be used to calculate W,, i.e. the proportion ofWto each wall,accordingto Eq. (5) bysubstituting WiforPi.

In other wbrds, the process of analysis for more thantwo walls is:

A. Analyze System 1.I. Find the propped cantilever reaction coeffi-

cient, R, for the given loading.2. Calculate the portion of W tributary to each

wall by applying the system of Fig. 9(b).In both these calculations the frames should beignored.

B. Analyze System 2 as a rigid beam on springsupports (include the frames) to find the toploads, Pi, on each unit as shown in Fig. 9(f).

C. Add results for the two systems.

Care is needed with thesignsofthc forces. On a givenunit, RWiisalways opposite indirection to Wi, and positivePi from Eq. (5)wiObe inthesame directional W. The finalIoadingson the units are illustrated in Fig. 9(c). Situationscoul,d occur where some values of Pi would be in theOppOsite direction to W, e.g. when the effect of torsion ispronounced.

Theabove approach can be extended tocover problemswhere deformation in the longitudinal direction is alsopossible. Under these circumstances three simultaneousequations have to be solved.

Accuracy

The main assumptions which could produce inaccurateresults fmm this procedure are:

. Constant frame shear. Thisassumption wiO tend tobe more reliable when thestiffnesses of the fmmesare low compared with those of the walls. Inproblems with torsion, however, in addition torelative stiffness, the relative locations of the unitsare important andmayaffcct the accuracy.

. Same beha~ior of walls. The method of assessing

the proportion of distributed Imd to the shearwalls described previously is accurate only when allthe walls exhibit the same behavior under lateralload, e.g. if they all have fixed bases and uniformproperties with height. It may be rewonable,asafurther approximation, to de fine KWstrictly as theload at the top to cause unit deflection in its line ofaction, taking account of nonrigid foundation,variation of properties with height, and openings(see Eq. [2]). These factors alsoaffect thevalueofR.

The behavior ofashear wall. frame structure with torsionis bigMy complex and only rough accuracy should beexpected from the procedure outlined previously. As withall simplified methods, unusual situations can occur forwhich the accuracy will be unpredictable.

Example Calculation with Torsion

In Fig. 10(a) Wall 2 has been moved to the position ofFrame 5 of Fig. 6(a) and Frames 5,6, and 7 have each beenmoved one bay to the right. This is not a practical systembut serves here to illustrate the method and the effect oftorsion in general. The steps outlined previously arefollowed in this example calculation.

A. ANALYZESYSTEM 1

Establish R

For uniformly distributed load, R = 3/8 and W = 547 kips(see Section 2.2.A). Therefore,

RW= 3/8X 547 = 205 kipS

Distribute W to the Walls

For this problem with only two walls, W can be distributedby simple statics:

w .547 X241 —= lo9kips

120W2 = 547 L 109 = 438!@

With more than two walls, an,analysis as shown in Fig.9(b) is required. To illustrate the procedure, the calcula-tions required for such analysis are set out in Table 5 as“Analysis of System 1.“ Columns (l) through (5) ofTable 5 are used to calculate the coefficients for theequilibrium in Eqs. (3) and (4). For System I the framesare ignored and, with W substituted for R W, theseequations become

11 ,664A, + 699,SOOfl ~ = 547699,800A1 + 83,976,0000, = 547 X 96

These equations are solved to give Al and @, as in columns(6) and (7). Columns (8), (9), and (1 O) are for Eq. (5),which gives the values of Wi as calculated by statics above.

B. ANALYZE SYSTEM 2

Equations (3) and (4) must be re-established with the frame

Tab

le5.

Exa

mpl

eC

alcu

k4io

nkC

ompo

nent

Stif

fnes

sM

etho

dw

ithT

orsi

on”

Cal

cula

tion

ofco

effi

cien

tof

stif

fnes

sI

Ana

lysi

sof

Syst

em1

Ana

lysi

sof

Syst

em2

Tra

ns-

vers

est

iff-

ness

,K

i

cO

Or-

dim

tedi

s-ta

nce,

xi (2) o 120 24 48 72 96 144

168

192

—

Rot

a-tio

nal

Dis

-C

oeff

i-st

iff-

tanc

eci

ent,

ness

,D

efle

c.sq

uare

,K

Jix:

(X1U

3)(%

3)‘y

’

(3)

(4)

(5)

(6)

o0

00.

0187

14,4

0069

9.8

83,9

760.

0187

—69

9.8

83,9

76–

576

4.8

115

–2,

304

9.6

459

–

Tzr

Dis

trib

.lo

adcm

shea

rw

alls

,V

i=K

iX\,+

x,J3

1)

(lo) 10

943

8

stem

!3s

ist-

ing

>rc

e,

/W

i=

/8~

Rot

a-tio

n,

(X20

3)

>ef

lec-

tion,

A2

x10

3]

(12)

7.34

7.34 —

7.34

7.34

7.34

7.34

7.34

7.34

7.34 —

lefl

ec

tion

Rot

a-du

eto

Sum

oftio

n,an

gle

defl

ec-

0,ch

ange

,tio

n,:x

103)

x,$,

Al+

X#l

due

tcan

gle

:han

gfX

,82

:x10

3

(14)

Sum

ofde

flec

-tio

n,,2

+X

,82

(X10

3)

(15) 7.34

22.8

0

—

10.4

313

.52

16.6

119

.70

25.8

928

.98

32.0

7

—

Lat

eral

load

onsp

ring

,D

,=K

.X

<2+

iie2

Poin

tlo

ad,n

unit

Itto

p,D

-RW

.i< (1

7)

stru

c-tu

ral

unit

E(7)

(s)

(9)

0.47

20

0.01

870.

472

0.05

660.

0753

——

,—

—— —

——

——

—— —

——

——

—

——

(1)

[11)

40.8

64.4

(13)

(16)

Wal

l1

wal

l2

5,83

2.(

5,83

2.(

0.12

90.

129

0

15.4

642

.813

3.0

2.0

-31.

4

Subt

ota

11,6

64

Fram

e1

Fram

e2

Fram

e3

Fram

e4

Fram

e5

Fram

e6

Fram

e7

Tot

al

199.

:19

9.:

199.

219

9.:

199:

199.

:19

9.2

13,0

58

0.12

90.

129

0.12

90.

129

0.12

90.

129

0.12

9

3.09

6.18

9.27

12.3

618

.55

21.6

424

.73

2.08

2.69

3.31

3.92

5.17

5.77

6.39

—

2.1

2.7

3.3

3.9

— — — — —

— — — — —

5,18

414

.31,

0321

–9,

216

19.1

20,7

3628

.7&

lz

28,2

2433

.55.

622

~–

5.2

5.8

6.4

‘All

un

its

inH

p.,

feet

,o

rra

dia

ns

., “ . “, ,. .

(o) PROBLEM (b) SYSTEM ( [c) SYSTEM 2[No FcmmNo T,, Mw, mmt) (TOP L,odi”Q 0“1,1

NOTE: Dimensions and properties are given in Fig. 6 and Table 3

Fig, 10, ExmvJ/e structure for component stiffness method– with torsion.

stiffnesses included and with R W as the applied loading.Using columns (1) through (5) of Table 5 now gives

13,058A2 + 848,0000 * = 205

848,000A2 + 104,513,00002 = 205 X 96

These equations are solved to give AZ and O* as in COIUIIMS

(12) and (13) of Table 5. Columns (14) and (15) are used inEq. (5) to calculate Pi, which is given in column (1 6).

C. ADD RESULTS FOR SYSTEMS 1 AND 2

The final top loadings are given in column(17) of Table 5.This column, in combination with column (10), gives thedistribution of load to each unit. Note that, due to thetorsion, the top load on Wall 1 is positive, i.e. in thedirection of W. Also, the inaximum frame load (for Frame7) is more than doubled by moving the position of theshear wall.

3. CALCULATION OF MOMENTS ANO SHEARSWITHIN THE STRUCTURE

By use of the methods described in Section 2, theproportion of the lateral load tributary to each vertical unitcan be estimated. Since the shear walls are treated asvertical cantilevers, the shears and moments acting at anycross-section cm be calculated using statics. For shear wallswith openings, see Section 1.4.

For frames, a simple method of estimating the momentsand shedrs on each column and beam is described inContinuity in Concrete Building Frames. t 1‘j This distribu.tion is based on relative column.beam stiffnesses. Alterna.tively (and more ~pproximately), the total shear at each

story can be distributed so that the shear on an interiorcolumn is twice that on an exterior column. By assumingpoints of contraflexure at column midheight, the columnand beam moments can be calculated. Corrections can bemade for points of contraflexure being off-center in thelower and upper stories.( 11 J

It is important to judge whether the moments and shearsas calculated by the simplified analysis are significant in thedesign. The simplified analysis may be acceptable if theeffect of Pateral load to the frames or to the structure as awhole is small compared with the effect of vertical Ioadlng.In many cases a more rigorous analysis will be necessaryand the insight into behavior gained from the simplifiedanalysis will be useful in deciding what refinements arerequired.

4. MORE ACCURATE ANALYSIS

When further analysis is required, the use of a computer isadvisable. Several different approaches have been pro-grammed for analysis of high-rise buildings. In the following, attention is given only to methods for which computerprograms are readily available.( 12, 13J

4.1. Numerical Accuracy

When using computer programs, it is important to checkthat the form of the structure does not cause inaccuracy inthe solution of the simultaneous equations. Thk canhappen when the difference between stiffnesses of parts ofthe structure is large (e.g. when members are given high

U“’+, [F,ow) U“;, 2

I r (Shea W,(I md From,)

---b---[0) SIMPLIFIED PLAN OF STRUCTURE (bl ~o;vA;~~Ly~;wING CONNECTION OF UNITS

FiG 11. [dedizalion for plane frame analysis,

finite values in order to simulate “infinite rigidity”) orwhen the structure is very large. Unsatisfactory results of an

equilibrium check on the structure (normally give” with theoutput) are normally caused by such “ill conditioning.”

This behavior is not easy to predict since the error will be afunction of the number of significant figures usedin thecomputations as well as the form of the structure.

4.2. Plane Frame Computer Program

A plane frame computer program can be used for theanalysis of shear wall-frame structures provided in-planedeformation of floor slabs and torsion can be neglected,The basic approach is illustrated in Fig. 11. The verticalunits are connected at each floor level by “linkbars” whichsimulate the effect of the floor slabs intransmitting load intheir own plane.

Notes

I. The “linkbars’’s houldbep in-connecteda ndaxiallys tiff(although high finite stiffness can be troublesome, asnoted above).

2. Axial deformation of the beams may be neglected.3. Axial deformation of the columns and shear walls should

be neglected only iftheuser is confident that this doesnot appreciably affect the stiffness of the structure orparts of the structure.

Column axial deformation will have a more promi.nent effect in tafl slender frames or in frames with stiffbeams.

Although the error in deflection due to neglectingaxial deformation may be high, the resulting forces andmoments throughout the structure will not normally beaffected to the same extent. The major effect of thecolumn axial deformation will be to increase the mo-

ments and shears in the upper stories of a rigid frame.The ratio AA/AB as calculated using Equations A and

B (Table 1) isa useful measurcof the effect of columnaxial deformation on stiffness. If AA/AB is less than0.05, it is probably safe to neglect this effect.

4. When computer stcxage islimited, itmaybenccessarytoreduce the number of bays for analysis, using thesummotion procedure described in Section 2.1. Thisshould not be done ifaxid deformation is important,

5. The finite widths of the shear walls orcolumns can havean important effect on the stiffness of a frame and onbeams that are connected to shear walls, Equation B canbe used to assess the effect on stiffness.* The best way toaccount for finite width is to assume that the beams arefufly rigid over the widths of the columns (Fig. 3 [c]),Ability to do this is not a common feature in plane frameprograms. Alternatively, a framework type of analogymay be used to model the action of shear walls.(14.16J

6.It miy be necessary toinclude foundation movementsinthe analysis.

7. The effect of openings in shear walls, if important,should be accounted for in the analysis. See Section 1,4.

4.3. Space Frame Computar Program

If the effect of torsion is important (see Section 1,3), aspace frame computer program should be used. While thiswill significantly increase the data input, the solution time,and the amount of output in comparison with those of aplane fmmeanalysis, the value of theresults obtained mayjustify the additional effort.

Notes 2,3, 5,6, and70nplane frame analysis (Section4.2) are equally valid for space frames. Floor slabs can betreated as ftdly rigid within their ownplanes, but since thisis not rigorously possible with conventional programs, the

●The effect of th~ finite de~ths of the beam may also beassessed using Equation B.

MEMBER PROPERTIES FOR . = ,2

1,= .017 O,LS

1

For l,- Plan. ~endin~l.. .1,

~ > .52 (02-.2) ,LLO

AOZ,52($-,202)t L &~ + 0’—---+~-+

&d:

FLOO

Fig. 12. Idealization forin.plane stiffness of f700r slabs.

alternative is to model the floors as trusses or frameworks.Yettram and Husain’s framework anafog.yf 14, as illustratedin Fig. 12 is one way to do this. Use Of this analogy willtend to overestimate the stiffness of the floors.

REFERENCES

l. Blume, J, A.; Newmark, N. M.; and Corning, L. H.;Design of Multistory Reinforced Concrete Buildings forEarthquake Motions, Portland Cement Association,Skokie, Ill., 1961, page 71.

2. COUO, Alexander, and Choudhury, J. R., ’’Analysis ofCoupled Shear WalJs: Journal of the American Con-cretelnstitute, Sept. 1967; fioceedings, Vol. 64, pages587-593.

3. COUO, Alexander, and Choudhury, J. R., ’’Stresses andDeflections uncoupled Shear WaOs,’’ ACI.Jourmd,Feb.1967; Proceedings, Vol. 64, pages 6S.72.

4. Beck, Hubert, “Contribution to tbe Analysis ofCoupled Shear Walk,” ACI Jownal, Aug. 1962; R’o-ceedings, Vol. 59, pages 1055-1069.

5. Rosman, Riko, “Approximate Analysis of Shear WallsSubject to f.ateral fmads,” AC1 Journal, June 1964;fioceedings, Vol. 61, pages 717-732.

6.. Rhan, F. R., and Sbarounis, J. A., ’’Interaction of ShearWalls and Fmmes,’’ fioceedings, American Society ofCivil Engineers, Vol. 90, ST3, 1964, pages 285.335.

7. Design of Combined Frames and Shear Wells, AdvancedEngineering Bulletin No. 14, PCA, 1965.

8, Goldberg, J. E., “Anafysis of Multistory Buildings

Considering Shear Wall and Floor Deformations,’’ TallBuildings, Pergamon Press, Long Island City, N.Y.,1967, pages 349-373.

9. Webster, J. A., Dkcussion of Reference 8, Tal/Build-ings, Pergamnn Press, 1967, page 374.

]O. Continuit.y in Concrete Buildinz Fmme$, Fourth Edi.tion, PCA, 1959.

11. Frame Constants for Lateral Loads on MultistoryConcrete Buildings, Advanced Engineering Bulletin No.5. PCA. 1962.

12. ‘3 TRESS”A User’s Manual, M.I.T. Press, Cambridge,Mass., 1964.

13. Gouwens, A. J., “Lateral Lnad Analysis of MultistoryFrames with Shear Walls,” Computer Progam, PCA,1968.

14. Yettram, A, L,, and Husain, H. M., “Pfane FrameworkMethod for Plates in Extension,’’ Journal of Engineer-ing Mechanics Division, ASCE, Vol. 92, Feb. 1962,pages 157-168.

15. Hrennikoff, A., %lution of Problems of Elasticity bythe Framework Method,” Journal of Applied Me-chanics, American Society of Mechanical Engineers,Vol. 8, Dec. 1941, pages A169-A175.

16. Grinter, L. E., “Statistical State of Stress by GridAnal ysis,” Numerical Methods of Analysis in En@eer-ing, The MacMillan Company, New York, N.Y., 1949.

17

r ------------------------------ -------------------It

EII

1

KEY WORDS: high-rise building, shear walls, rigid frames, bending mode,shear mode, lateral load, point load, stiffness method, shear connectionmethod, torsion.

ABSTRACT: A shear wall deflects inabending mode; arigidframe bendsina shear mode. A simple method for analyzing shear wall-frame interactionspresent ed in this publication.

REFERENCE: MacLeod, Iain A., Sheav Wall-Frame Interaction-A DesignAid (EB066.01 D), Portland Cement Association, 1970.

L---------. ------------------------------------------------J

mPORTLAND CEMENT I I ASSOCIATION

Printed In U,S,A

5420 Old Orchard Road, Skokie, [11inois 60077.1083

EBL?66.01 D