Concrete Shear Wall Frame Interactions

ENGINEERING

BULLETIN

Shear Wall-Frame Interaction A DESIGN AID BYlain A. MacLeod

PORTLAND CEMENT IEII

’ ASSOCIATION

FOREWORO

Shear walls and frames in combination normally provide the required stiffness and strength to withstand lateral loads in high-rise buildings. Apartment buildings up to 70 stories high have successfully utilized this concept. [n certain cases the walls me much stiffer than the frames and thus take most of the lateral load. For this reason, the participation of the frame in resisting lateral load is often ignored. TMs may not always be a conservative procedure and it is therefore important that the effect of the frames be considered, In this publication procedures are recommended for analyzing shear wall-frame interaction problems. A new simplified method for assessing the effect of the frames is also presented.

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

@ PcmlandC.mnr A.wciadon 1970 Reprinted1974, 1981,1983,1985,1998

CONTENTS

Notation

.

1. Behavior of Structures

3

Under L?teral Lnad

5

.

l.l. Behavior Difference Between Frames and Shear Walls

5

1.2. Behavior of Floor Slabs

6

.

1.3. Effect of Torsion

6

1.4. Effect of Openings in Shear Walls

6 7

2, Simplified Methods of Estimating Lateral Lnad Distribution

7

2.1. Use of Charts from References 6 and 7

7

2.2. The Component

Stiffness Method-Equation

C

2.3. The Component

Stiffness Method for Structures with Torsion

12

3. Calculation of Moments and Shears Within the Structure

Is

4. More Accurate Analysis

15

4.1. Numerical Accuracy

.

Is

4.2. Plane Frame Computer program

16

4.3. Space Frame Computer Progmm

16

References

.

.

.

.

.

.

. .

.

. .

.

.

. .

.

. .

.

.

17

This publication is based on the facts, tests, and authorities stated herein. It is intended for the use of professional personnel competent to evaluate the significance and limitw tions of the reported findings and who will accept responsibility for the application of the material it contains. Obviously, the Portland Cement Association disclaims any and all responsibility for application of the stated principles or for the accuracy of any of the sources 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

D = depth of beam

s = ratio

E= Young’s modulus (subscript indicates structural system) functions usedin Equations Aand B (Table l), dependent onthe type of loading

g= ratio

P= interaction load at the top of the frame (subscript denotes frame represented as spring) R = support reaction coefficient propped cantilever

C = column width

Fx, Fm, Fn, F,=

A at top of frame ~t bottom of frame, as used in . Table I

n = ratioA

lb at top of frame

IC at top of frame

for a

, ~~ “~ed in

Ic at bottom of frame Table 1

W = total applied lateral load (subscript denotes load on specific wall)

, as used in

1, at bottom of frame Table ~

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

H = total height of wall h = height of column, i ,e, story height Ib=mmwmt

of inertia

of a connecting

beam 1,, , IC2 = moments of inertia of wall sections [W = moment of inertia of wall KB=rotational port

stiffness of shear wall sup-

Kf=lateral point load applied at top of frame to cause unit deflection in its line of action K,= factor representing

both Kw and Kf

Kw=lateral point load applied at the top of a shear wall to cause unit defleti tion in its Iineofaction; also specifically tbe stiffness of a shear wall without openin~–see Eq. (1) K ~. = stiffness of a shear wall with openings–see Eq. (2) !?= distance between centroidal axes of walls or columns, or span of beams ~ = ~ or ~ j“ equations of Table 1

Tw = dimensionless parameter which re. Iates the rotational stiffness of the wall to that of the foundation, i.e. KBH ratio — 4EW1W A = total deflection at top of structure AA = top deflection deformation

due to column

axiaf

AB = top deflection mation

due to bending defor-

8 = a deformation at top of structure with torsion—see Section 2.3 A = ratio of column to beam stiffness

(-e+

p = variable used for shear openings–see Section 1.4

wall with

%wtlon number

Consider behavior of structures under lateral load ~

I I

Uw a simplified method to estimate distribution of

~ With torsion

NOtOrsiOn

-’*

&

5’

Calculate moments and shears in members

Notorsion

4.2 [

Pig, 1, F1OWdia.framof analysisprocedure.

4

a==

With torsion

Shear

ht@raCth)I’i

1. BEHAVIOR OF STRUCTURES UNOER LATERAL LOAO

A suggested procedure for lateral load analysis of high-rise buildings is illustrated by tbe flow diagram in Fig. 1 and this publication, as a design aid, will follow that pattern section by section. The following is a general introduction to the behavior of structures under lateral load.

1.1. Behavior

Difference

Between

A DESIGNAID

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

Frames and Shear Walls

A rigid frame, an interconnection of vertical columns and horizontal beams, bends predominantly in a shear modeas shown in Fig. 2(a). A shear wall deflects predominantly in a bending mode, i.e. as a cantilever, as illustrated in Fig. 2(b). Elevator shafts, stairwells, and reinforced concrete walls normally exhibit this behavior. It is not always easy to differentiate between modes of deformation. For example, a shear wall weakened by a row (or rows) of openings can tend to act like a rigid frame and conversely an infllled frame will tend to deflect in a bending mode. When all vertical units of a structure exhibit the same

(.1 RIGID FRAME SHEAR MODE DEFORMATION

lb] SHEAR WALL BENDING MODE DEFoRMATION

Fig. 2 Defomnmion modes.

{c)

INTERCONNECTED FRAME AND SHEAR WALL (EQuAL DEFLECT IONS AT EACH STORY LEVEL)

1.2. Behavior of Floor Slabs

For analysis, the floor slabs are normally considered to be fully rigid within their own planes. This means that there will be no relative movement between the vertical units at each story level. In-plane deformation can be taken into account but is seldom important. Floor slabs bending out of plane contribute to the lateral stability of a structure by acting as beams between vertical members, and hence flat plate structures, for example, act like rigid frames. However, the strength of column-slab joints in flat plate structures must be checked

a useful parameter for assessing the effect of the openings is M, where /4,1 +.4,2 12fb !2 ~. ~ Icl +IC2 + ACIAC2 ) 4( H = total height of wall Ib = moment of inertia of a connecting beam h = story height b = clear span of beams !?= distance between centroidal axes of the wall sections I ,,, IC2 = moments of inertia of wall sections ACI, AC2 = areas of wall sections

(c%%%?”

W’a, l, (’cl}

M

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

;.,, 2 W

I.) SHEAR WALL WITH SINGLE Row OF OPENINGS

(b] IDEALIZATION SHEAR CONNECTION

FOR METHOD

(c) FRAME

IDEALIZATION

Fig, 3, Wallwith a single row of openings. FormH that is greater than 8, the wall tends to behave like a single cantilever, For Iowaff, e.g. less than 4, the behavior ismorelike twoconnected walls and frame action is more prominent. With a single mwof openings the effect ofopeningson the stiffness can be assessed by comparing

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 is unsymmetrical or if the stiff vertical units are close to the center of the structure, Some earthquake codes require that a structure be capable of resisting a specified torsional loading even if the applied lateral load theoretically does not cause torsion~ 1J * The importance of torsion may be asseseed by comparing results from simplified methods without torsion (Sections 2.1 and 2,2) and with torsion (Section 2.3).

1.4. Effect of Openings in Sberrr Wells

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

.S. p.s.

6

pers.ript 17.

numbers

in

parentheses

desienate

references

o.

3EI Kw = -$

(1)

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

(2)

ff3&

where Kw and Kwo = stiffnesses of the wall without and with openings, respectively E = Young’s modulus of elasticity IW = moment of inertia of wall without oneninm

p=l+

(~.,

+A.,)U.,

+1.,)

A.,.,. A.. . b2 include a plot of K4 against aff

Coull and Choudhurytzl for different values of p, Tbe stiffness and tbe distribution normally be appreciably affected

of stress in a wall will by the presence of

openings. The shear connection methodtz’5 J is suitable for band calculation (although the use of a computer to do the calculations is desirable) and can be used for more complex ~~problems than that illustrated in Fig. 3(b). Coull and Choudhury’s chartsfz ,3) are particularly useful for problems with a single row or two symmetrical rows of openings. Shear walls with openings can be idealized as illustrated in Fig. 3(c), using a plane frame analysis program (Section 4.2). ‘For analysis of walls with openings, axial deformation of the wall sections should be included.

proportioned frame may be defined as one that has points of contra flexure at all nddbeam sections under lateral load. Any frame whose bay widths and member properties are reasombly constant across the width of tbe frame may be considered to be proportioned. The frame being proportioned and the column axial deformation being negligible are basic assumptions made in reducing a multi-bay frame to a single- or three-bay frame, The procedure for reducing tbe number gf bays is: at each story level, sum_aU cofynn inertias (Ie) and beam rotational sJiffnesses (lb/!). (lC is the moment of inertia of a cnhmm; Ib and !?are the moment of inertia and the span of a beam, respectively.) The equivalent stiffnesses to be used in the substitute frames thus become 1== Zid/2 and Ib/!? = ~-b/k?

2. SIMPLIFIED LATERAL

METHOOS OF ESTIMATING LOAD DISTRIBUTION

for the frame in Fig. 4(a) and 10= ZTC[6 and Ib/t =~~~l!i

If torsion is not considered, two simplified methods of determining 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).

for the frame in Fig. 4(b). This procedure is clearly described by Khan and Sbarounis. Having thus reduced the problem, the shears on the frame, moments on the shear wall, and deflection can be found 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 references the stiffnesses (Iw) of all the shear walls are summed to give an equivalent single wall. For the frames, Khan and Sbarnunis use a singfe-bay frame whereas Advanced Engineering Bulletin No. 14 uses a three-bay frame. These idealizations are illustrated in Fig. 4. Both the frames in Fig. 4 are “pmportinned.” A

P-

h.., W.11

F,wne

1.

b

f

L

L,,cl (b) REDUCED STRUCTLWE FOR REF. 7

F& 4. Reduced structure for Referemxw 6 md Z

The component stiffness method has more flexibility than the charts method referred to above, but it lacks accuracy if the wall is more flexible than the frame (KW/Kf < 1; see below), The r?uzin assumption is that the frame takes constant shear, i.e. that the interactir.m force between the frame and wall 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 comparison with the wall. Consider the single-bay frame and shear wall loaded in-plane by the uniformly distributed load shown in Fig. 5(a). If tbe frame shear is assumed to be constant, the system can be treated as a wall supported at the top by a spring (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 calculated using Equations A and B of Table 1 and the top deflection equation in the Table 2 notation. Kw is defined as the Iateraf point load required to cause unit deflect ion at the top of the wall (similarly to Kf). AS discussed in Section 1.4, Eq. (1) can be used for walls without openings and Eq. (2) for walls with a single row of openings. Table 2 (Equation C) gives relationships between P/W, 7w, and KJKf fOr different JOading cases. Si~l~ expressions for other load cases can easilv be established. P is the interaction load at the top of the-frame, i.e. the constant shear; W is the total applied lateral load; and 7W is a dimensionless parameter which relates the rotational stiff-

Table 1. Equations

B and A for Top Deflection of Rigid Regular Frames

EQUATION B–for Bending deformation: 3.0

AB=—

,,~:c,

[~,(l-W+Fg(l

-,c,32q

where AB = deflection at top of frame due to bending of members W= total lateraf load

2.0

b = story height

Fm 1,5

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

I.0

I,0 213 1/2

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 ~ at bottom of frame finear variation of Ic and Ib c with height. If Jl varies, use Ib attop of frame g = ratio EI instead of I. I, at bottom of frame I

0.= $, where C is cohmm

width and !2is distance between col-

o. ~,

I.e. summation

over width of structure at first.

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

g

Load condition Point load at top

umn centerlines ~= Z(E I/h) ~, 2Z(.5JJ!2) story ~e;el

1,0

0,5 m=sorm.

F$(m=s)or

I

Ft(m=g)

I

m

loge

n

i

Uniformly distributed

Triangular (earthquake)

~2 –—-logenl

3

-—+2m loge m +

2

2 (m- 1)3

m-1 EQUATION A–for ~xial deformation: Load condition where AA = deflection at top of frame due to axial deformation

of ex

terior columns Fn = function of n, dependent n = ratio

on the type of loading

Area of exterior column at top of frame

Area of exterior column at bottom of frame (linear variation of ,4C with height)

Point load at top

1-4n+3n2-2n210gen (1 - fl)3

Uniformly distributed

2-9n+18n2-

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

B = totaf width of frame

TOTAL DEFLECTION

9 + ~-

A= AB + AA

NOTES For accuracy, the following rxmditions should be satisfied: 1. Q and 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, 5. /c, lb, a. J 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,

6(1 - n)4

(

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

lln3+6n310gen

Triangular (earthquake)

6“ :n:,;33 ‘“g’ n

-;+3n +

3n2 - ~+; (n-

25 -fi+4n-

~3

-logefl 1)4 4n3

3n2+T-T-10ge

~4

n

+ (n-

l)s

)

ness of the wall to that of the foundation,

Problems involving several frames and walls may be reduced to that of a singfe wall and frame as described in Section 2.1. Alternatively, Kfor KW foreach verticaf unit maybe calculated separately and the results summed. XKf and ZKW are then used instead of Kf and KW in f@atiOn c. Studies on shear wall-frame interaction normally use three parameters to detine bebavior; namely, h,Iw, and XIc, where EcIc/h A=— E#bl!2

i.e.

KBH ‘rw=— 4EJW where KB is the rotational stiffness Of the shear wail support. If the rotation at the base of the shear wall is to be neglected, the terms with Yw in Equation C shoufd be

~l”o /r-

Sflffrm,s

-K,

L,nk Bar,

Told Lo.i

By using Kfi which is a function Of A and ~c, bebaviOr can be 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 tbe effectiveness of tbe frame (or frames)in comparison with the shear wafl(s) in resisting lateral load and for assessing the effect of various assumptions in analysis.

w

“,

lW

Ijlil /

Ly;eldin,

‘-From’

shearW.(I s“,,.,+

[b)

(01 STRUCTURE

INTERACTION AT TOP ONLy

Accuracy (c1 FRAME MODELED BY SPRING

When a frame and wall are interconnected as shown in Fig. 5(a), maximum shear on the frame tends to occur tOwards midbeigbt (see Fig. 8). Equation C can underestimate maximum frame shear by as much as 30 percent in this .“. area. Therefore, when calculating moments in theimme, lt is worthwhile to increase the calculated value of F’by 30 percent. If Kw/Kfis less than 1, the use of Equation Cis not recommended and the use of chartsce,~j produces more accurate resufts.

Fig. 5. Idealization for Equation C.

omitted. However, the effect of shear wall base rotation can significantly affect the distribution of load between shear walls and frames, and Equation C can be used as a simple method of assessing this factor. Table 2. Equation C Load condition

I

C

Ecytion

NOTATION

1+--$ Point load at top

w

E= w

3K

‘w=

1’47.

T Uniformly distributed

Triangular (earthquake)

‘fW

L+% @w

Kf

KBH 4EWIW

E= Young’smodulus IW= moment ofinertia KW = ~ K,=

~+~ 20 27W ;= 3 1+~+~

lateral load

KB= rotational stiffness ofshear H= total height of wall

{) ,+

force at top

‘y

3 *1+1-

$=

P= interaction W= totafapplied

KW w

wall support

(subscript denotes structural system) of wall

(with constant ~w)

point loadattop

of frame tocause unit deflection in since top de fleeits line of action ;i.e., ~or — P A AB+AA tion A. ~ Kf

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

3/8 ‘+’%’

calculate K, calculate Kf for a single frame, using Equation B frOm Table 1. The finite widths of the beams and columns are not 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 ‘=2X 12.6X3X

20 13,800 =2’02’

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

For

frame

properties,

see

Table

land

thus Fg=l,

3

it follows

that

AB ‘Fig.6, Example structure for component stiffness method–no

7=12X3X

torsion.

12.62X126X123 103X 105,080 X(2”54+1

= 0.0602 in./kip (computer Table 3. Moments of Inertia (in.4) of Frame Members, Example Structure*

10 9 8 -/ 6 5 “4 3 2 I *E

= 3,000

AA

13,800 13,800 13,800 13,800 13,800 13,800 13,800 13,800 13,800 13,800

2,540 4,430 7,390 9,270 11,520 14,140 17,080 21,680 24,770 26,270

—— ksi for

wall

T

H3 x 0.77= = ECACB2

1263 X I 23 x 0.77 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 7 percent of that due to bending. Since column axial deformation was neglected for the analysis given by Goldberg, it is also neglected here and

~=l=L f

and frame members.

AB

0.0602

= 16.6 kip/in.*

XKf = 7 X 16.6 = l16kip/in. Calculate Kw

Example Problem

3X3 XI03 6X603X123 1263x 123 x 12

KW.~= This problem is definedin Fig. 6and Table 3, and we will follow the flow diagram in Fig. 1. Results will be compared with those of a more accurate analysis in Example I by GoldbergX8J

= 4S6 kip/in, XKW = 2 X 4S6 = 972 kip/in

—

A. D18TR3BUTELOADS TO THE VERTfCAL UNITS *Kf

use Equation C (Table 2), Foundation

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

Ib

story

‘2x2”02)

rotation of the shear

= AB

P +

to be considered.

AA

would

he

used

were

column

axial

deformation

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

midheight.

CaIcuk+te PI W

C. COMPARISON W’TH MORE ACCURATE ANALYSIS Goldberg(s) takes account of in. lane deformation of the floor slabs. Webster’s discussion(9 ? giv~$ ~~s”lts with ~i@d

Calculate Loads on Units W = 9 X S7.6 + 28.8 = 547 kips P = 547 X 0.04= Ptoeachframe

21.9 kips

=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 applied loading is 18,835 kip.ft. (see Fig. 6). Therefore, for each wall max. moment = 18,835- PH = 18,835- 10.9 X 126 = 17,474 kip.ft.

SHEAR ON FRAME

( KIPS)

Fig. 8. Comparison of frame shear estimates. Moments in Frame Columns Assume

an interior

exterior

column.

column

takes

twice

the

sheiu on interior

as an

moment

will

colwnn

= ‘-

>

occur

towards

kips midheight

of

the

frame and have a value of 1.03 X 0.5 X 12X

Tbe 0.5 factor

is for point

31

1.3=

8.0 kip.ft.

of contra flexure at column 10.9

14.4 28,8 288 28 B 288 28,8 28,8 28.8 28,8

[o)

g

NOTE:

FRAME

(7 Th”31

Loads are in kips

Fig. 7. Loading on units–no torsion.

(b) WALL

z I

(2)

(3)

Frame anal ysis

Equation C

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

Moment at base of shear wall, kip.ft

17,526

17,474

-0.3

Top deflection, in.

0.173

0.187

+8.1

Design values

(1)

*Frame malysis in Example 1 by Goldbem (8)

288 R

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

=+=1.03 Maximum

shear

Therefore,

[2 Thu31

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

w

B

_Spring

‘w%

+

(b)

supports

[walls

Wtqid

only)

beam

w

DISTRIBUTION TO

OF

W

WALLS

W.1 1 s

w WALLS

PLAN (al

(c)

STRUCTURE

A%rin9s

to

represent the

FRAMEs

LOADING

ON

of the

fr.mes

action

UNITS

~w

Q

Q?$

+

\

L

v

‘w

RW (At

~J-:;;:r,,

TOP)

‘– o

Q PLAN (d]

IDEALIZATION

NO

beom

W ( ~st,ib,tedl PLAN

(0)

Rigid

TOP

PLAN MOVEMENT

(f)

TOP

LOADING

ONLY

Fig. 9. Component stiffness method with torsion.

2.3. The Component with Torsion

Stiffoess Method for Structures

By making the assumption that the frames take constant shear (as for Equation C), the structure shown in Fig. 9(a) may be idealized as in Fig. 9(d). Two degees of freedom which correspond to the deformations A and 8 at the top of the structure (Fig. 9 [f]) can be assigned and the structure solved by the stiffness method, The analysis is carried out by adding the results from Systems 1 and 2 (Fig. 9 [e and f]). 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 total lateral load. R W acts in the line of and in the opposite direction to the resultant of W at the top of the structure. For analysis of System 2 (Fig. 9 [f]), the roof slab can be considered as a rigid beam on spring supports. The spring stiffnesses are KW and K, as defined in Section 2.2. The equations for determining the unknowns A and O are 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, and the frame stiffnesses, Kfi and u is the distance Of applied load from origin. For delineation of X and a, see Fig. 9(f). Having solved Eqs. (3) and (4) for A and 6, the loads on the springs are F’i= Ki(A + Xi6) (5) If there are only two walls, simple statics gives the distribution of the total lateral load, W. With more than two walls, Eqs. (3) and (4) must be re-established, neglecting the frame springs and solving with Was the only loading (Fig. 9[b]). The resulting deformations are fictitious but can be used to calculate W,, i.e. the proportion of Wto each wall,accordingto Eq. (5) bysubstituting Wifor Pi. In other wbrds, the process of analysis for more than two

walls

is:

A. Analyze System 1. I. Find the propped cantilever reaction coefficient, 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 be ignored. B. Analyze System 2 as a rigid beam on spring supports (include the frames) to find the top loads, 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 given unit, RWiisalways opposite indirection to Wi, and positive Pi from Eq. (5)wiObe inthesame directional W. The final Ioadingson the units are illustrated in Fig. 9(c). Situations coul,d occur where some values of Pi would be in the OppOsite direction to W, e.g. when the effect of torsion is pronounced. Theabove approach can be extended tocover problems where deformation in the longitudinal direction is also possible. Under these circumstances three simultaneous equations have to be solved.

Accuracy The main assumptions which could produce inaccurate results fmm this procedure are: . Constant frame shear. Thisassumption wiO tend to be more reliable when thestiffnesses of the fmmes are low compared with those of the walls. In problems with torsion, however, in addition to relative stiffness, the relative locations of the units are important andmayaffcct the accuracy. . Same beha~ior of walls. The method of assessing

the proportion of distributed Imd to the shear walls described previously is accurate only when all the walls exhibit the same behavior under lateral load, e.g. if they all have fixed bases and uniform properties with height. It may be rewonable,asa further approximation, to de fine KWstrictly as the load at the top to cause unit deflection in its line of action, taking account of nonrigid foundation, variation of properties with height, and openings (see Eq. [2]). These factors alsoaffect thevalueof R. The behavior ofashear wall. frame structure with torsion is bigMy complex and only rough accuracy should be expected from the procedure outlined previously. As with all simplified methods, unusual situations can occur for which the accuracy will be unpredictable.

Example Calculation with Torsion In Fig. 10(a) Wall 2 has been moved to the position of Frame 5 of Fig. 6(a) and Frames 5,6, and 7 have each been moved one bay to the right. This is not a practical system but serves here to illustrate the method and the effect of torsion in general. The steps outlined previously are followed in this example calculation. A. ANALYZE SYSTEM 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 distributed by simple statics: w .547 X24 lo9kips 1 —=120 W2 = 547 L 109 = 438!@ With more than two walls, an,analysis as shown in Fig. 9(b) is required. To illustrate the procedure, the calculations required for such analysis are set out in Table 5 as “Analysis of System 1.“ Columns (l) through (5) of Table 5 are used to calculate the coefficients for the equilibrium in Eqs. (3) and (4). For System I the frames are ignored and, with W substituted for R W, these equations become 11 ,664A, + 699,SOOfl ~ = 547 699,800A1 + 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

13,058

‘All units in Hp.,

Total

Frame Frame Frame Frame Frame Frame Frame

feet,

Rotational stiffness,

of stiffness

4.8 9.6 14.3 19.1 28.7 33.5

699.8

—

576 2,304 5,184 9,216 20,736 28,224

0 699.8

(4)

o 14,400

(3)

– – –

z –

&l 5.622 ~

–

0.0187 0.0187

(6)

‘y’

Deflec.

115 459 1,0321

83,976

0 83,976

(5)

I

(s)

(9)

(lo)

Distrib. load cm shear walls, Vi= KiX \,+x,J31)

—

— — — — — — —

—

—

—

— —

— —

0 0.0566

0.472 0.472

,—

—

—

— —

0.0187 0.0753

—

— — — —

109 438

— —

—

— —

40.8 64.4

[11)

/8 ~

/ Wi=

>rce,

ing

3sist-

stem!

E

(7)

tion Rotadue to Sum of tion, angle deflec0, change, tion, :x 103) x,$, Al+ X#l

Tzr

Analysis of System 1

Stiffness Method with Torsion”

CoeffiDistance cient, KJi square, (X1U3) (%3) x:

of coefficient

or radians

—

24 48 72 96 144 168 192

11,664

Subtota

199.: 199.: 199.2 199.: 199: 199.: 199.2

o 120

5,832.( 5,832.(

Wall 1 wall 2

1 2 3 4 5 6 7

(2)

(1)

structural unit

cOOr-

dimte distance, xi

Transverse stiffness, Ki

Calculation

Table 5. Example Calcuk4ionkComponent

—

7.34 7.34 7.34 7.34 7.34 7.34 7.34

—

7.34 7.34

(12)

>eflection, A2 x 103]

0.129 0.129 0.129 0.129 0.129 0.129 0.129

0.129 0.129

(13)

(X203)

Rotation,

3.09 6.18 9.27 12.36 18.55 21.64 24.73

15.46

0

(14)

X,82 :x103

due tc angle :hangf

leflec

—

10.43 13.52 16.61 19.70 25.89 28.98 32.07

—

7.34 22.80

(15)

Sum of deflection, ,2+X,82 (X103)

Analysis of System 2

—

2.08 2.69 3.31 3.92 5.17 5.77 6.39

42.8 133.0

(16)

<2+iie2

Lateral load on spring, D,=K.X

2.1 2.7 3.3 3.9 5.2 5.8 6.4

2.0 -31.4

(17)

Point load ,n unit It top, Di< -RW.

.,

(o)

“

.

“,

,.

.

PROBLEM

(b)

SYSTEM

[No FcmmNo

NOTE:

Dimensions

and

properties

Fig, 10, ExmvJ/e structure for

are

component

(

T,,

given

[c)

Mw, mmt)

in

stiffness

Fig.

6

SYSTEM

(TOP L,odi”Q

and

method–

Table

with

2 0“1,1

3

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 in Eq. (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 the distribution of load to each unit. Note that, due to the torsion, the top load on Wall 1 is positive, i.e. in the direction of W. Also, the inaximum frame load (for Frame 7) is more than doubled by moving the position of the shear wall.

story can be distributed so that the shear on an interior column is twice that on an exterior column. By assuming points of contraflexure at column midheight, the column and beam moments can be calculated. Corrections can be made for points of contraflexure being off-center in the lower and upper stories.( 11 J It is important to judge whether the moments and shears as calculated by the simplified analysis are significant in the design. The simplified analysis may be acceptable if the effect of Pateral load to the frames or to the structure as a whole is small compared with the effect of vertical Ioadlng. In many cases a more rigorous analysis will be necessary and the insight into behavior gained from the simplified analysis will be useful in deciding what refinements are required.

4. MORE

3. CALCULATION WITHIN

OF MOMENTS ANO SHEARS THE STRUCTURE

By use of the methods described in Section 2, the proportion of the lateral load tributary to each vertical unit can be estimated. Since the shear walls are treated as vertical cantilevers, the shears and moments acting at any cross-section cm be calculated using statics. For shear walls with openings, see Section 1.4. For frames, a simple method of estimating the moments and shedrs on each column and beam is described in Continuity 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

ACCURATE

ANALYSIS

When further analysis is required, the use of a computer is advisable. Several different approaches have been programmed for analysis of high-rise buildings. In the follow ing, attention is given only to methods for which computer programs are readily available.( 12, 13 J

4.1. Numerical

Accuracy

When using computer programs, it is important to check that the form of the structure does not cause inaccuracy in the solution of the simultaneous equations. Thk can happen when the difference between stiffnesses of parts of the structure is large (e.g. when members are given high

U“’+,

[F,ow)

I

r

U“;, 2 (Shea

W,(I

md From,)

---b--[0)

SIMPLIFIED

PLAN

FiG 11. [dedizalion

OF

for

STRUCTURE

plane

frame

(bl

~o;vA;~~Ly~;wING

CONNECTION

UNITS

analysis,

finite values in order to simulate “infinite rigidity”) or when the structure is very large. Unsatisfactory results of an equilibrium check on the structure (normally give” with the output) are normally caused by such “ill conditioning.” This behavior is not easy to predict since the error will be a function of the number of significant figures usedin the computations as well as the form of the structure.

4.2. Plane Frame Computer

OF

Program

A plane frame computer program can be used for the analysis of shear wall-frame structures provided in-plane deformation of floor slabs and torsion can be neglected, The basic approach is illustrated in Fig. 11. The vertical units are connected at each floor level by “linkbars” which simulate the effect of the floor slabs intransmitting load in their own plane.

Notes I. The “linkbars’’s houldbep in-connecteda ndaxiallys tiff (although high finite stiffness can be troublesome, as noted 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 does not appreciably affect the stiffness of the structure or parts of the structure. Column axial deformation will have a more promi. nent effect in tafl slender frames or in frames with stiff beams. Although the error in deflection due to neglecting axial deformation may be high, the resulting forces and moments throughout the structure will not normally be affected to the same extent. The major effect of the column 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 column axial deformation on stiffness. If AA/AB is less than 0.05, it is probably safe to neglect this effect. 4. When computer stcxage islimited, itmaybenccessaryto reduce the number of bays for analysis, using the summotion procedure described in Section 2.1. This should not be done ifaxid deformation is important, 5. The finite widths of the shear walls orcolumns can have an important effect on the stiffness of a frame and on beams that are connected to shear walls, Equation B can be used to assess the effect on stiffness.* The best way to account for finite width is to assume that the beams are fufly rigid over the widths of the columns (Fig. 3 [c]), Ability to do this is not a common feature in plane frame programs. Alternatively, a framework type of analogy may be used to model the action of shear walls.(14.16J 6.It miy be necessary toinclude foundation movementsin the 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), a space frame computer program should be used. While this will significantly increase the data input, the solution time, and the amount of output in comparison with those of a plane fmmeanalysis, the value of theresults obtained may justify the additional effort. Notes 2,3, 5,6, and70nplane frame analysis (Section 4.2) are equally valid for space frames. Floor slabs can be treated as ftdly rigid within their ownplanes, but since this is not rigorously possible with conventional programs, the ●The effect of th~ finite de~ths of the beam may also be assessed using Equation

B.

MEMBER

PROPERTIES

1,=

.017

l..

.1,

FOR . = ,2

O,LS For

~

l,-

Plan.

~endin~

1

> .52 (02-.2) LO

AOZ,52($-,202)t

,L 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 illustrated in Fig. 12 is one way to do this. Use Of this analogy will tend 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 for Earthquake Motions, Portland Cement Association, Skokie, Ill., 1961, page 71. 2. COUO, Alexander, and Choudhury, J. R., ’’Analysis of Coupled Shear WalJs: Journal of the American Concretelnstitute, Sept. 1967; fioceedings, Vol. 64, pages 587-593. 3. COUO, Alexander, and Choudhury, J. R., ’’Stresses and Deflections uncoupled Shear WaOs,’’ ACI.Jourmd,Feb. 1967; Proceedings, Vol. 64, pages 6S.72. 4. Beck, Hubert, “Contribution to tbe Analysis of Coupled Shear Walk,” ACI Jownal, Aug. 1962; R’oceedings, Vol. 59, pages 1055-1069. 5. Rosman, Riko, “Approximate Analysis of Shear Walls Subject to f.ateral fmads,” AC1 Journal, June 1964; fioceedings, Vol. 61, pages 717-732. 6.. Rhan, F. R., and Sbarounis, J. A., ’’Interaction of Shear Walls and Fmmes,’’ fioceedings, American Society of Civil Engineers, Vol. 90, ST3, 1964, pages 285.335. 7. Design of Combined Frames and Shear Wells, Advanced Engineering Bulletin No. 14, PCA, 1965. 8, Goldberg, J. E., “Anafysis of Multistory Buildings

Considering Shear Wall and Floor Deformations,’’ Tall Buildings, Pergamon Press, Long Island City, N. Y., 1967, pages 349-373. 9. Webster, J. A., Dkcussion of Reference 8, Tal/Buildings, Pergamnn Press, 1967, page 374. ]O. Continuit.y in Concrete Buildinz Fmme$, Fourth Edi. tion,

PCA,

1959.

11. Frame Constants for Lateral Loads on Multistory Concrete 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 Multistory Frames with Shear Walls,” Computer Progam, PCA, 1968. 14. Yettram, A, L,, and Husain, H. M., “Pfane Framework Method for Plates in Extension,’’ Journal of Engineering Mechanics Division, ASCE, Vol. 92, Feb. 1962, pages 157-168. 15. Hrennikoff, A., %lution of Problems of Elasticity by the Framework Method,” Journal of Applied Mechanics, American Society of Mechanical Engineers, Vol. 8, Dec. 1941, pages A169-A175. 16. Grinter, L. E., “Statistical State of Stress by Grid Anal ysis,” Numerical Methods of Analysis in En@eering, The MacMillan Company, New York, N. Y., 1949.

17

r -----------------------------I t I

------------------E

I

1

KEY WORDS: high-rise building, shear walls, rigid frames, bending mode, shear mode, lateral load, point load, stiffness method, shear connection method, torsion. ABSTRACT: A shear wall deflects inabending mode; arigidframe bendsin a shear mode. A simple method for analyzing shear wall-frame interactions present ed in this publication. REFERENCE: MacLeod, Iain A., Sheav Wall-Frame Interaction-A Aid (EB066.01 D), Portland Cement Association, 1970.

L---------.

------------------------------------------------J

Design

PORTLAND 5420

Printed

In U,S,A

CEMENT

Old Orchard

m I

Road, Skokie,

I

ASSOCIATION

[11inois 60077.1083

EBL?66.01 D