coupled shear wall design

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Scientific Research and Essay Vol. 4 (4), pp. 328-345, April, 2009Available online at http://www.academicjournals.org/SREISSN 1992-2248 © 2009 Academic Journals

Full Length Research Paper

Non-planar coupled shear walls with stiffening beamsE. Emsen1*, C. D. Turkozer2, O. Aksogan2, R. Resatoglu3 and M. Bikçe4

1Department of Civil Engineering, Akdeniz University, 07058, Antalya, Turkey.

2Department of Civil Engineering, University of Cukurova, 01330, Adana, Turkey.

3Department of Civil Engineering, Near East University, Nicosia, North Cyprus, Turkey.

4Department of Civil Engineering, Mustafa Kemal University, 31024 Hatay, Turkey.

Accepted 17 February, 2009

This paper concerns itself with the static analysis of non-planar coupled shear walls with any number ofstiffening beams. Furthermore, the change of the heights of the stories and connecting beams from

region to region along the height are taken into consideration. The stiffening of coupled shear walls isrealized by placing high connecting beams at the levels of whole or partial stories used as storage orservice areas. The analysis is based on Vlasov’s theory of thin-walled beams and ContinuousConnection Method (CCM). In the analysis, the compatibility equation has been written at the midpointsof the connecting and stiffening beams. The method of analysis presented was compared with theSAP2000 structural analysis program. The results obtained showed good agreement, verifying theaccuracy of the proposed method, which can efficiently be used for the pre-design computations of tallbuildings.

Key words: Static analysis, continuous connection method, coupled shear wall, stiffening beam, non-planar,warping deformation.

INTRODUCTION

In multi-storey buildings made of reinforced concrete,lateral loads are often resisted by specially arrangedshear walls. Shear wall components may be planar, areusually located at the sides of the building or in the formof a core which houses staircases or elevator shafts.Weakening of shear walls in tall buildings by doors, win-dows and corridor openings is one of the most frequentlyencountered problems of structural engineering. Whenthe coupling action between the piers separated by open-ings becomes important, some of the external moment isresisted by the couple formed by the axial forces in the

walls due to the increase in the stiffness of the coupledsystem by the connecting beams. Actually, the defor-mation of a coupled shear wall subjected to lateral load-ing is not confined to its plane. Studies considering in-plane, out-of-plane and torsional deformations in theinvestigation of coupled shear walls are called non-planarcoupled shear wall analyses. In non-planar coupled shearwalls, both the flexural and torsional behaviours under

*Corresponding author. E-mail: [email protected]. Tel:+90-0242-3106385. Fax: +90-0242-3106306.

external loading have to be taken into account in theanalysis. When thin-walled structures are twisted, there isa so-called warping of the cross-section and theBernoulli-Navier hypothesis is violated. The warping oshear walls is greatly restrained by the floor slabs and thefoundations. A classical analysis of warping torsion re-quires the prior evaluation of the shear centre locationthe principal sectorial area diagram, the warping momentof inertia and the torsion constant (Zbirohowski, 1967).

When the height restrictions prevent connecting beamsfrom fulfilling their tasks of reducing the maximum tota

shear wall bending moments at the bottom and the maxi-mum lateral displacements at the top, beams with highmoments of inertia, called “stiffening beams”, are placedat certain heights to make up for this deficiency. Stif-fening of coupled shear walls decreases the lateral dis-placements, thus, rendering an increase in the height othe building possible. Hence, assigning some stories ofthe building as storages, service areas and the like andplacing high beams on those floors seems to be a logicasolution. Such coupled shear walls are called “stiffenedcoupled shear walls”. Such beams can be steel trussesor reinforced concrete beams of very high bending stiff-


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Emsen et al. 329

Z

X

Y

O

H

(a)

X

Z

2nd

pier1

s pier

Y

H

laminae

(b)

O

q

Figure 1. Non-planar coupled shear wall and equivalent structure.

ness.All of the analyses in the literature on stiffened coupled

shear walls concern themselves with planar coupledshear walls (Aksogan et al., 1993; Arslan et al., 2004;Aksogan et al., 2007). No study has been made, to theknowledge of the authors, concerning the static analysisof stiffened non-planar coupled shear walls, so far.

In the present work, the static analysis of non-planarcoupled shear walls with any number of stiffening beams,is carried out which is applicable for asymmetric struc-tural systems as well as symmetric ones on rigid founda-tion. The analysis is based on the Continuous ConnectionMethod (CCM), in conjunction with Vlasov’s theory (1961)of thin-walled beams, following an approach similar to theone used by Tso and Biswas (1973). In the CCM, theconnecting beams are assumed to have the same pro-perties and spacing along the entire height of the wall.The discrete system of connecting beams is replaced bycontinuous laminae of equivalent stiffness (Rosman,1964). CCM has been employed in the analysis and the

compatibility equation has been written at the mid-pointsof the connecting beams. For this purpose, the con-necting beams have been replaced by an equivalent la-yered medium. The axial force in the piers is determinedfrom the differential equation which is obtained by usingthe compatibility and equilibrium equations. Then, all rele-vant quantities of the problem are determined employingtheir expressions in terms of the axial force.

The present formulation is implemented with a FortranComputer program. Using this computer program anasymmetrical example has been solved and comparedwith the solutions found by the SAP2000 (MacLeod et al.,

1977; Wilson, 1997) structural analysis program and aperfect match has been observed.

ANALYSIS

In this study, based on Vlasov’s theory of thin walled

beams and the CCM, an approximate method is pre-sented for the analysis of non-planar coupled shear walstructures. The deformation of a coupled shear wall sub-

jected to a lateral loading is not always confined to itsplane. For this reason, the present analysis is a threedimensional analysis of coupled shear walls (Figure 1a)The CCM was developed by assuming that the discretesystem of connections, in the form of individual couplingbeams or floor slabs, could be replaced by continuouslaminae as shown in Figure 1b.

The basic assumptions of the CCM for non-planar cou-pled shear walls can be summarized as follows:

• The geometric and material properties are constan

throughout each region i along the height.• Vlasov’s theory for thin-walled beams of open section isvalid for each pier.

• The walls and beams are assumed to be linearly ela-stic.

• The outline of a transverse section of the coupled sheawall at a floor level remains unchanged in plan (due tothe rigid diaphragm assumption for floors). Moreover, theparts of the shear wall between floor levels are also as-sumed to satisfy this condition. Depending on the foregoing assumption, the axis of each connecting beam re-


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330 Sci. Res. Essays

Y

1sy

c

2gx

1gx

2sy

2gy

1gy

2sx

1sx

1X

1φ

connecting beam

S1

G1

S2

G2

2φ

2nd

pier1st pier

X

1Y 2X

2Y

Figure 2. Plan of non-planar coupled shear wall in region i.

mains straight in plan and does not change its length.Furthermore, the slope and curvature at the ends of aconnecting beam in the vertical plane are also assumedto be equal. Consequently, it can be proved in a straightforward manner that, depending on the fact that there areno vertical external forces on the connecting beams, theirmid-points are points of contraflexure.

• The discrete set of connecting beams with bendingstiffness EIci in region i are replaced by an equivalentcontinuous connecting medium of flexural rigidity EIci /hi

per unit length in the vertical direction.• The discrete shear forces in the connecting beams inregion i are replaced by an equivalent continuous shearflow function qi, per unit length in the vertical directionalong the mid-points of the connecting laminae.

• The torsional stiffness of the connecting beams isneglected.

• Bernoulli-Navier hypothesis is assumed to be valid forthe connecting and stiffening beams.

A non-planar coupled shear wall and its plan for oneregion are given in Figures 2-3 with global axes OX, OYand OZ, the origin being at the mid-point of the clearspan in the base plane. The X axis is parallel to thelongitudinal direction of the connecting beams. The Z axisis the vertical axis and the Y axis is such that it completesan orthogonal right-handed system of axes. During thedeformation, the outline of a transverse section of theshear wall remains unchanged (Figure 2).

Referring to the axes OX and OY, the coordinates of

the centroids of the piers are taken to be11 gg

y,x

and ( )22 gg

y,x , respectively. Throughout this study, the sub-

scripts 1 and 2 express the left and the right piers, res-

pectively. The subscript i ( )n1,2,...,i = , refers to the

number of a region. Similarly, the shear centers of the

piers are located at11 ss

y,x and ( )22 ss

y,x , respectively

The coordinates referred to the principal axes of pier

( )1,2 j = which are represented by ( ) j j Y,X , making anangle

jφ with the respective global axes are shown in

Figure 2. The jZ axis is parallel to the global Z axis.

The axial force in each pier is found by writing down thevertical force equilibrium equation for the part of one pieabove any horizontal cross-section as

=

−

=

++=

+

i

1t

t

1i

1t

z

z

t

z

z

iiV)dzq(dzqT

t

1t

i

( )n1,2,...,i = (1)

wherei

V is the shear force in ith stiffening beam. A cu

through the points of contra-flexure of laminae exposesthe shear flow qi. The vertical force equilibrium of a dzelement of one pier yields the relation

iiTq ′−= ( )n1,2,...,i = (2)

Where a prime denotes differentiation with respect to z.

Compatibility equations

While obtaining the compatibility equations in a non-pla-nar coupled shear wall analysis, it is assumed that alrows of connecting laminae will be cut through the midpoints, which are the points of zero moment.

The vertical displacement due to bending can be obtainedas the product of the slope at the section consideredand the perpendicular distance of the point on Z axis fromthe respective neutral axis. In addition, vertical displacement arises, also, due to the twisting of the piers, and is


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Emsen et al. 331

(i-1)

(i)

(i+1)

(n-1)

(n)

(1)

(2)

(3)

Z

X

Y

z1

z2

z3

zi-1

zi

zi+1

zn-1

zn

z

H

O

h1

h2

hi-1

hi

hn-1

hn

stiffened stories

hi+1

Figure 3. Non-planar stiffened coupled shear wall.

and is equal to the value of the twist at the section consi-dered, times the sectorial area, ω , at the point on Z axis.

Therefore, the total relative vertical displacement due tothe deflections and rotations of the piers, can be writtenas

( ) ( ) ( )1i12i2gi1gi2gi1gi2i1 1212

yvyvxuxu ωθ′−ωθ′−′−′+′−′=δ (3)

The first two terms in equation (3) represent the contribu-tions of the bending of the piers about the principal axesand the last represents the contribution of the twisting of

the piers.1

ω and2

ω are the sectorial areas at points just

to the left and the right of the cut for piers 1 and 2, res-pectively.

The local displacements ji

u , ji

v and jiθ ( 2,1 j = ,

n1,2,...,i = ) and 2 may be expressed in terms of the glo-

bal displacementsi

u ,i

v andiθ by substituting expres-

sions (4) into (3) and rearranging, yields

)d(bvauiiii1

+ωθ′+′+′=δ (5)

in which

21 ω−ω=ω (6)

12 ggxxa −=

12 ggyyb −=

11112222 gsgsgsgsyxxyxyyxd −+−= (7)

(7)For the compatibility of displacements, the relativevertical displacements of the cut ends must be equal tozero. Hence,

( ) +=

+−+ωθ′+′+′

+

n

1i j

z

z

j

21

iiidzT

A

1

A

1

E

1dbvau

j

1 j

0GA

ch2.1

EI12

ch

E

TdzT

A

1

A

1

E

1

ii1ic

i

c

3

ii

z

z

i

21

=

+

′+

+−

+

(8)


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(a)

1gi1 xu ′−

2Y

1X

2X

1Y

2gi2 xu ′

G1

G2

2Y

1X

2X1Y

2gi2yv ′

1gi1 yv ′−

G1

G2

2i2 ωθ′

1i1 ωθ′− S1

S2

(b) (c)

*i2δ

**i2δ

c/2 c/2

Ti Ti

Pi

*i3δ

**i3δ

Pi

c/2 c/2

Pi

*i4δ

**i4δ Pi

c/2 c/2

(d) (e) (f)

Figure 4. Relative vertical displacements at the mid-point of a lamina.

i1xvIE

1′′

a

1φ

2φ

b

i1yuIE

1′′

i2xvIE

2′′

X

Y

i2y uIE 2 ′′

G1

G2

Figure 5. Internal bending moments.

Differentiating this equation with respect to z and letting

i ji θ=θ

isi ji jyuu θ−= isi ji jxvv θ+= (4)

ii c

i

c

3

i

iGA

ch2.1

EI12

ch+=γ (9)

the following equation is obtained:

( ) 0E

TT

A

1

A

1

E

1dbvau

i

i

i

21

iii =γ ′′

+

+−+ωθ ′′+′′+′′

( )n1,2,...,i = (10)

The successive terms in (8) represent the relative vertical

displacements due to: (a) the bending about1

X and2

X

axes, (b) the bending about1

Y and2

Y axes, (c) the rota-

tions of the piers, (d) the axial deformation of the piers,(e) the bending deformation in the laminae, and (f) theshearing deformation in the piers, respectively, as seenin Figure 4.

In Figure 4, δ expresses the displacements pertainingto the left side with superscript (*) and to the right sidewith superscript (**) and the relative vertical displacementis found by adding the contributions from both sides.

Equilibrium equations

The coordinate system and positive directions of internalbending moments acting on the different components ofthe coupled shear wall are adapted as shown vectoriallyin Figure 5.

These internal moments, along with the couple pro-duced by the axial force, Ti, balance the external bending

momentsiEX

M andiEY

M . For the equilibrium of the mo-

ments about X and Y axes, the following relationshipscan be derived using Vlasov’s theory of thin walledbeams:

MbTcosvIsinuIcosvIsinuIEi2211 EXi2i2x2i2y1i1x1i1y

=−+φ′′+φ′′+φ′′+φ′′

(11)

( ) MaTsinvIcosuIsinvIcosuIEi2211 EYi2i2x2i2y1i1x1i1y

=−+φ′′−φ′′+φ′′−φ′′

(12)

In these equations, jx

I and jy

I are the second moments

of area of the cross-sections, and jxy

I is the product of

inertia transformation equations for moments of inertia,


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Emsen et al. 333

Y

2sx 1sx

2sy

1sy

1φ

S1

S2 2φ

X

i1yuIE

1′′

i1xvIE

1′′

i2xvIE

2

′′

i2yuIE

2′′

i11IE θ′′− ω

i22IE θ′′−ω

Figure 6. Internal bimoments and bending moments.

the moment components are found, in terms of theglobal of pier j (j=1, 2) about axes parallel to the global

axes and passing through the centroids. jx

I and jy

I are

the second moments of area of pier j ( 2,1 j = ) about its

respective principal axes. On substituting the local dis-placement expressions given before in (4) into expres-sions (11-12) and applying the well-known displacementsat the point on Z axis, as

bTIEvIEuIEMiiXiXiXYEXi

+θ′′+′′+′′=θ

( )n1,2,...,i = (13)

aTIEvIEuIEMiiYiXYiYEYi

+θ′′−′′+′′=θ

( )n1,2,...,i = (14)

where

21 xxXIII +=

21 yyYIII +=

21 xyxyXYIII += (15)

22112211 xysxysxsxsXIyIyIxIxI −−+=

θ (16)

22112211 xysxysysysYIxIxIyIyI −−+=

θ (17)

When non-planar coupled shear walls are rigidly con-strained at the base, the cross-sections of the piers donot warp uniformly along the height and internal stresses

evoke in the wall due to the bimoments. The internalforce resultants shown in Figure 6 are caused by biaxialbending. These bending moments form bimoments withthe resisting moments at the bottom of the wall and showup in the bimoment equilibrium equation.

In order to obtain the bimoment equilibrium equation,the coupled shear wall will be cut through by a horizontalplane such that an upper part is isolated from the lowerpart of the structure. The internal bimoments in the struc-ture consist of two parts; one contributed by the individualpiers (Figure 6) and the other due to the resistance of theconnecting laminae and stiffening baems (Figure 7).

In Figure 6, the internal bimoment expressions ji ji

IE θ′′− ω

(j=1,2, i=1,2,…,n), are caused by the non-uniform warp-ing of the cross-sections of the piers. It must be men-tioned that the computation of the internal bimomentscreated at the point on Z axis by the bending moments ofthe wall, are carried out after transferring the bendingmoments to the shear centers of the piers.

Leti

B be the resultant bimoment about Z axis, which

is due to the resistance offered by the piers. It can bewritten as (Figure 6):

( )1121 si1yi2i1i

yuIEIEIEB ′′+θ ′′−θ ′′−=ωω

) ) )22

1122

si2xsi1xsi2yxvIExvIEyuIE ′′−−′′+′′+ (18)

Using the geometric relationship, the resultant resistingbimoment of the piers for all regions in the structure arefound as:

iiXiYiIEvIEuIEB θ ′′−′′−′′=ωθθ

( )n1,2,...,i = (19)

in which

2221112211221121 xyssxyssy

2

sy

2

sx

2

sx

2

sIyx2Iyx2IyIyIxIxIII −−+++++=

ωωω

(20)

In addition, there are bending momentsi1xq

M ,i1yq

M ,

i2xqM , and

i2yqM and bimoments

i1qB and

i2qB , due to the

summation of the shear forces in the laminae, as shownin Figures 7-8. All quantities are related to the shearcenters of the piers, S1 and S2. In order to determine thebimoment due to the shear forces in the laminae, therelationship between the force components in the piers,may be determined from the free body diagram in Figure 7.

Leti

B be the resultant bimoment due to the additiona


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(H-z)

qi

Vi

q1

V1

Ti

Ti

i1yqM

i1xqM

i1qB

i2yqM

i2xqM

i2qB

(H-z)

Vi

V1

Ti

Ti

i1yqM

i1xqM

i1qB

i2yqM

i2xqM

i2qB

Figure 7. 3-D view of the additional internal bending moments and bimomentsdue to the shear forces in the connecting medium.

Y

2sx 1sx

2sy

1sy

1φ S1

S2 2φ

X

i1yqM

i1xqM

i2xqM

i2yqM

i1qB

2qB

Figure 8. Cross-sectional view of the additional internal bending moments and bimomentsdue to the shear flow in the connecting medium.

bending moments and bimoments about Z axis, asshown in Figure 8.

The shear forces in the laminae produce bimoments

on the piers. For each pier, the bimoment jiq

B is caused

by T i (instead of the summation of the shear flow in theconnecting medium, see equation (1)), and is equal to the

product of this force and the principal sectorial area j

ω of

the point of its application. Therefore, for the two piers,

1iqTB

i1ω−= 2iq TB i2 ω= (21)

The resultant bimoment,i

B , due to these additiona


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Emsen et al. 335

Y

2sx 1sx

2sy

1sy

1φ

S1

S22φ

X

i1xvIE

1′′′

i1y uIE 1 ′′′

i2y uIE 2 ′′′

i2xvIE

2′′′

i22i2 GJEI 2 θ′+θ ′′′− ω

i11i1 GJEI 1 θ′+θ ′′′− ω

Figure 9. Internal twisting moments and shear forces.

qi

Ti+dTi

dz

G1

G2

Ti+dTi

Ti

i1yQ

i1xQ

i1tqM

i1yQ

i1xQ

i1tqM Ti

i2yQ

i2xQ

i2yQ

i2xQ

i2tqM

i2tqM

Figure 10. 3-D view of the additional internal shear forces andtwisting moments due to the shear flow in the connecting medium.

bending moments and bimoments about Z axis, then,found as:

( ) ( ) ( ) ( )2i2

2i21i1

1i1i2i1 syq

sxqsyqsxqqqi yMxMyMxMBBB +−+−++= (22)

These additional bending moments and bimomentsacting on an element are shown in Figure 8. From the

consideration of equilibrium of the moments about1

X

and1

Y axes for pier 1, the following relations can be

obtained, respectively:

( ) 0yTM1

i1gixq

=+ ( ) 0xTM 1i1

giyq=−− (23)

1i1

gixqyTM −= 1

i1

giyqxTM −= (24)

Similar consideration for the other pier gives

2i2

gixqyTM = 2

i2

giyqxTM = (25)

Substituting (21,24-25) in (22), using (7) and simplifyingthe resultant bimoment, about the vertical axis throughpoint O, due to the component bending moments andbimoments is found as:

( )ii

TdB +ω−= ( )n1,2,...,i = (26)

Equating the external bimoment,iE

B , to the internal resis

ting bimoments, the bimoment equilibrium equation of theisolated part of the coupled shear wall above an arbitrary

horizontal plane is established. Finally, the foregoingequilibrium equations for all regions of the structure canbe written as follows:

iiEBBB

i+= (27)

( )iiiXiYE

TdIEvIEuIEBi

+ω−θ′′−′′−′′=ωθθ

( )n1,2,...,i = (28)

In order to obtain the twisting moment equilibriumequation, the coupled shear wall will be cut through by ahorizontal plane such that an upper free body diagram isisolated from the rest of the structure.

The internal twisting moments (torque) in the structureconsist of two parts; one contributed by the individuapiers as shown in Figure 9 and the other due to theresistance of the connecting laminae, in other words, the

differential effect of the shear flowi

q in the connecting

medium as shown in Figure 10.

In Figure 9, expressions ji j

JG θ′ (j=1,2, i=1,2,…,n) are

the St. Venant twisting moments. Expressions ji j

IE θ ′′′− ω

(j=1,2, i=1,2,…,n) are additional twisting moments due tothe non-uniform warping (Zbirohowski, 1967) of the piersalong the height. Furthermore, the total twisting moment


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Y

2sy

1sy

1φ

S1

S22

φ

X

i1yQ

i1xQ

2yQ

i1tqM

2xQ

i2tqM

1sx 2sx

Figure 11. Cross-sectional view with the additional internal twisting moments and

shear forces due to the shear flowi

q .

due to the shear forces in the cross-sections of the piersabout Z axis must also be considered.

Letit

M be the resultant torque about Z axis, which is

due to the resistance offered by the piers. It can bewritten as (Figure 9):

1121i si1yi2i1i22i11tyuIEIEIEGJGJM ′′′+θ ′′′−θ ′′′−θ′+θ′=

ωω

) ) )22

1122

si2xsi1xsi2yxvIExvIEyuIE ′′′−−′′′+′′′+ (29)

structure are found as:

iiiXiYt IEJGvIEuIEM i θ ′′′−θ′+′′′−′′′= ωθθ ( )n1,2,...,i = (30)

in which

21JJJ += (31)

In addition, there are shear forcesi1x

Q ,i1y

Q ,i2x

Q , and

i2yQ , and twisting moments

i1tqM and

i2tqM , developed in

the section due to the shear flow,i

q , in the laminae, as

shown in Figures 10-11.

The bimoments, due toi

q in the two piers, are:

1iz dzqdB i1 ω= 2iz dzqdB i2 ω−= (32)

These bimoments produce the flexural twisting mo-

mentsi1tq

M andi2tq

M , related to the shear centers S1 and

S2, respectively. Based on Vlasov’s theory of thin-walledbeams:

1i

z

tqq

dz

dBM i1

i1ω==

2i

z

tqq

dz

dBM i2

i2ω−== (33)

The resultant twisting moment,it

M , due to these addi-

tional torques and shear forces about Z axis is, thenfound as:

( ) ( ) ( ) ( )2i22i2

1i11i1i2i1i

sysxsysxtqtqtxQyQxQyQMMM +−−−−+=

(34)

From the consideration of equilibrium of the moments

about1

X and1

Y axes for pier 1, the following relations

can be obtained, respectively:

0xdzqdzQ 1i1

gix =+ 0ydzqdzQ

1i1 giy =+ (35)

1i1

gix xqQ −= 1i1 giy yqQ −= (36)

Similar consideration for the other pier gives

2i2

gixxqQ =

2i2 giyyqQ = (37)

Substituting (33,36-37) in (34), using (7) and simplifyingthe resultant twisting moment, about the vertical axisthrough point O, due to the component shears andtorques, yield

( )it

qdMi

+ω= ( )n1,2,...,i = (38)

Equating the external twisting moment,iEt

M , to the inter

nal resisting moments, with opposite sense, the equili-brium equation of the isolated part of the coupled shearwall above an arbitrary horizontal plane is establishedFinally, the foregoing equilibrium equation for all regionsof the structure can be written as follows:

iii ttEtMMM += (39)

( )iiiiiXiYEt

TdIEJGvIEuIEMi

′+ω−θ ′′′−θ′+′′′−′′′=ωθθ

( )n1,2,...,i = (40


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General solution for axial force

Using the compatibility equation (8) and the four equili-brium equations (13), (14), (28), and (40), the 4n un-

knowns of the problem, namelyi

u ,i

v ,iθ , and

iT , can be

found under the applied loadingsiEX

M ,iEY

M ,iE

B , and

iEtM . The elimination of iu , iv , and iθ from equations

(10,13-14,28,40) yields the following differential equation

fori

T :

( ) ( ) ( ) rKKIMTTT13EYii3ii2ii1 i

+′′−=β+′′β−′′′′βω

( ) ( ) rMKMKME

GJrKKIM

iiii Et4EX3EY24EX ′+++−′′−

ω (41)

where

ωγ =β I

ii1 2i

i2r

E

JG

A

I+

γ +=β ω

AE

JGi3 =β (42)

The geometrical quantities used in equation (41) aredefined as follows:

( )∆

+= θθ XXYYX

1

IIIIK

( )∆

+= θθ XYYXY

2

IIIIK

( )

∆

−= XYX

3

IbIaK

( )∆

−−= YXY

4

IbIaK

(43)

21KbKadr −++ω=

1Y2XKIKIII

θθωω −−=

43

21

bKaK

A

1

A

1

A

1++

+= ( )2XYYX III −=∆

In equation (41),iEX

M andiEY

M are the external bending

moments andiEt

M is the external twisting moment about

the respective global axes and can be written as:

( ) ( )

2

zHWzHPM Y

YEXi

−+−=

( ) ( )

2

zHWzHPM X

XEYi

−+−=

(44)

( ) ( ) ( )( ) ( )( )WXYWYXPXYPYXEt dzHWdzHWdPdPM i −+−−++−=

where, XP and YP are the concentrated forces, XW andY

W are the uniform loads in X and Y directions,PX

d and

PYd are the moment arms of the components of the con-

centrated force,WX

d andWY

d are the moment arms of

the distributed force components from the point on Z axis,respectively. Substituting expressions (44) in (41) andsolving the resulting differential equation, Ti is found asfollows:

[ ] [ ]zCoshDzSinhDTi1i2i1i1i

α+α=

Emsen et al. 337

[ ] [ ]zCoshDzSinhDi2i4i2i3

α+α+

( ){Y4X3i22

i3

WKWKGJ2E2

1+β

β+

( )([

X3Y4X3i3

WKHPK2PK2zHJG ++−β+

( ) )zWKWKWKHY4X3Y4

+−+

(X1XWYX3

WrKWrdWKIE2 +−−ω

)] }Y2YWXY4

WrKWrdWKI −++ω

( )n1,2,...,i = (45)

in which

β

ββ−β−β=α

i1

i3i1

2

i2i2

i12

4

β

ββ−β+β=α

i1

i3i1

2

i2i2

i22

4 (46)

Employing the boundary conditions to determine the inte-gration constants, Ti can be obtained in a straightforwardmanner.

Determination of the Shear Forces in the StiffeningBeams

Before writing down the boundary conditions, the sheaforces in the stiffening beams must be determined. Forthis purpose, compatibility equation (8) must be writtenboth for section i at level

iz and the stiffening beam i and

solved simultaneously. Thus, employing the definition

+

+

=

ii

ii

s

3

s

c

3

i

c

i

i

EI12

c

GA

c2.1

EI12

ch

GA

ch2.1

S (47)

the shear force in the ith

stiffening beam is found asfollows:

izziii

TSV=

′−= ( )n1,2,...,i = (48)

Boundary Conditions

To determine the integration constants in the single fourthorder differential equation (41) the following 4n boundaryconditions are employed:

1- The structure being rigidly fixed at the base ( 0z = )

0vu0zn0zn0zn

=θ=====

0vu0zn0zn0zn

=θ′=′=′===

(49)


11/18


Applying equation (8) at ( 0z = ) and using (49), the first

boundary condition is:

0T0zn =′=

(50)

2- From equation (40), substitutingnθ′′ ,

nθ ′′′ ,

nu ′′′ and

nv ′′′ in

terms ofn

T and putting 00zn =θ′=

, the second boundary

condition at the bottom ( 0z = ) is found as:

( )

′−

′′′γ −

′+′−′−+

∆

′−′−

θ 3nnn

n

4EX3EY

1XYEXXEY

YEtKTT

A

TKMKM

r

KIMIMIM

nn

nn

n

( )

′−

′′′γ −

′+′−′−−

∆

′−′+

θ 4nnn

n

4EX3EY

2XYEYYEX

XKTT

A

TKMKM

r

KIMIMI

nn

nn

( ) 0TdTA

TKMKM

r

Innn

n

4EX3EY nn=′+ω+

′′′γ −

′+′−′−+ ω (51)

3- From the equilibrium of the vertical forces in each pierin the uppermost region of the shear wall

1

H

z11

VdzqT += (52)

Applying this equation for the uppermost position

( Hz = ), the first term on the right dropping out,considering expression (48), the third boundary conditionis found as follows:

Hz

11Hz1TST

=

= ′−= (53)

4- Substituting1

u ′′ ,1

v ′′ and1θ′′ in the bimoment expression

(28) and applying it at the top ( Hz = ) the fourth boundarycondition is obtained as:

( )

′′γ −+−−+

∆

−−

θ 11

1

4EX3EY

1XYEXXEY

YET

A

TKMKM

r

KIMIMIB

11

11

1

] ( )

+−−−

∆

−+−

θA

TKMKM

r

KIMIMIKT 1

4EX3EY

2XYEYYEX

X31 11

11

) ]

′′γ −+−−+−′′γ − ω

11

1

4EX3EY4111T

A

TKMKM

r

IKTT

11

( ) 0Td1 =+ω+ (54)

5- From the vertical force equilibrium of one piece of the

stiffening beam in either of the piers at heighti

zz =

ii zziizz)1i(TVT

==− =+ (55)

Substituting (48) in (55), the fifth type boundary condition

is obtained as follows:

iiizzizziizz)1i(

TTST===−

=′− ( )n1,2,...,i = (56)

6- Applying the compatibility equation (8) for two

neighbouring regions (i) and (i−1) at height izz = , thefollowing equation is obtained as the sixth type boundarycondition:

( )

( )

( )

( )

( )

+′=

+′

=

−−

=−

−− ii

i

1i1i

ic

3

i

c

i

zzi

c

3

1i

c

1i

zz1i EI12

ch

GA

ch2.1T

EI12

ch

GA

ch2.1T

( )n2,3,...,i = (57)

7- Since the total twisting moments of the two

neighbouring regions (i) and (i−1), at heighti

zz = balance

each other

( )[ ] ( )[ ]iY1iYEtEt uIEuIEMM i1i ′′′+′′′−+− θ−θ−

( )[ ] ( )[ ]i1iiX1iX JGJGvIEvIE θ′+θ′−+′′′−′′′+ −θ−θ

( )[ ] ( ) ( ) ( )[ ] 0TdTdIEIE i1ii1i =′+ω−′+ω+θ ′′′−θ ′′′+ −ω−ω (58)

Expressing all unknown functions in terms of Ti and itsderivatives, after some rearrangements, the seventh typeboundary condition is found as:

( ) ( ) 0CCTCTCTCTC i3)1i(3ii21i)1i(2ii11i)1i(1 =−+′−′+′′′−′′′ −−−−−

( )n2,3,...,i = (59)

where

i

ii

iiii

Et

EY3EX42X1Y

EYYXEYXXYEXXYEXYXY

i3

2X1Y

4X3Yi2

iX2iY1i

i1

Mr

MKMK

r

KIKII

MIIMIIMIIMIIC

drA

I

rA

KI

rA

KIKIKIC

r

I

r

IK

r

IKC

+

′+′

−−−

∆

′+′−

∆

′+′=

+ω++−−−=

γ −

γ +

γ =

θθω

θθθθ

ωθθ

θθ

ωθθ

(60)

8- Since the total bimoments of the two neighbouring

regions (i) and (i−1), at heighti

zz = balance each other


12/18

( )[ ] ( )[ ] ( )[ ]iX1iXiY1iYEE vIEvIEuIEuIEBB i1i ′′−′′+′′+′′−+− θ−θθ−θ−

( )[ ] ( ) ( ) ( )[ ] 0TdTdIEIE i1ii1i =+ω−+ω+θ′′−θ′′+ −ω−ω (61)

Expressing the other unknown functions in terms of T i and its derivatives, after some necessary rearrange-

ments, the eighth type boundary condition is obtained as:

( ) ( ) 0CCTCTCTCTC i4)1i(4ii21i)1i(2ii11i)1i(1 =−+−+′′−′′ −−−−− (62)

Where

∆

+−

∆

+=

θθθθ iiii EYYXEYXXYEXXYEXYXY

i4

MIIMIIMIIMIIC

i

ii

E

EY3EX42X1Y Br

MKMK

r

KIKII+

+

−−− θθω (63)

To determine the integration constants D1i to D4i, the

boundary conditions at the top, bottom and between eachpair of consecutive regions are used. Substituting them in

expression (45), the general solution for T i ( )n,1,2,i …= can be found.

Determination of lateral displacements and rotationfunctions

The lateral displacement and rotation functions (i

u ,i

v ,

andiθ ) can be found using (13-14,40) as follows:

i2ii1ii

i

4EX3EXiGzGdzdzT

A

TKMKM

Er

1ii

++

γ ′′−+−−=θ

(64)

i2ii1

XEYXYEX

3i1iiRzRdzdz

IMIMKTKE

E

1u ii ++

∆+

∆−−θ′′=

(65)

i2ii1

YEXXYEY

4i2iiNzNdzdz

IMIMKTKE

E

1v ii ++

∆+

∆−−θ′′−=

(66)where boundary conditions (49) and the equivalence ofthe horizontal displacements and the respective slopesfor every pair of neighbouring regions at their commonboundary (z = zi) are used to determine the integration

constants.

Numerical results

In the example, the behaviour of the coupled shear wallwith four stiffening beams in Figures 12-13 is examined.Stiffening beams of 3.0 m height are placed at the levelsof ninth, thirteenth and seventeenth stories and a last onewith 2.0 m height at the top, as seen in Figure 12.

The sectorial area diagram of the cross-section of thestructure is given in Figure 14. The structure is solved

Emsen et al. 339

both by the present method using the CCM and by theSAP2000 structural analysis program (Wilson, 1997)using the frame method (MacLeod, 1977), for which themodel and its 3-D view are given in Figure 15.

The geometrical properties and the cross-sectionalview of the structure are given in Figures 12-13. The

total height of the shear wall is 72 m, the storey height is3 m and the thicknesses of the piers and the connectingbeams are shown in Figure 13. The height of the con-necting beams is 0.4m and the elasticity and shear-

moduli are E=2.85×106 kN/m

2 and G=1055556 kN/m

2,

respectively. The external loads act at the top of thestructure as shown in Figures 12-13. The lateral dis-placements at points on Z axis and the rotations, foundby the present program and the SAP2000 structural ana-lysis program, are compared, for the unstiffened andstiffened cases, in Figures 16-17.

Conclusions

In this study, non-planar shear walls coupled with connecting and stiffening beams are analyzed under lateral loading. The analysis is performed using Vlasov’s thin-walledbeam theory in conjunction with the CCM.

In the example, the behaviour of a coupled shear walwith four stiffening beams is examined. The resultsobtained have been compared with those of SAP2000and a good agreement has been observed.

As seen in Figures 16-17, the stiffening of coupledshear walls cause a decrease in the maximum displacement at the top and the maximum bending moment at thebottom of a building. Thus, by using such stiffeningbeams the heights of buildings can be increased moreThe stiffening of coupled shear walls is realized by placing high connecting beams at the levels of whole orpartial stories used as storage or service areas. Thenumber and levels of these high beams, to improve thestructural behaviour of the buildings, is up to the designengineer.

The method proposed in this study, has two main advantages, which are that the data preparation is much easiercompared to the equivalent frame method for non-planacoupled shear walls and the computation time needed ismuch shorter compared to other methods. Hence, themethod presented in this study is very useful for pre-design purposes while determining the dimensions of non

planar coupled shear wall structures.

ACKNOWLEDGEMENTS

The financial support of the Scientific Research ProjectsUnit of Akdeniz University is gratefully acknowledged.

Nomenclature: a, distance between the centroids of the piers

in X direction; j

A cross sectional area of the jth pier;ic

A , cross

sectional area of connecting beams in region i; b, distancebetween the centroids of the piers in Y direction; B, Bimoment;


13/18


Figure 12. Non-planar asymmetrical example structure.

Y

X

2 m 2 m 2 m 1.5 m2 m 1.5 m 2 m 2 m

2 m 4 m 1.5 m2 m 1.5 m 4 m

4 m

3 m

2 m

5 m

1

(-1, 0)

1

0.3 m

1

(1, 0)(-3, 0)

2

3

5

8

9

11 76

(-7, 0)

(-7, 4)

(-7, -3) (-5, -3)

(-3, -3)

10

(-5, 4) (-3, 4) (-1, 4)

4

(1, -3)

2

3

45

6 8

910

7

(4, 0)

(4, 2)

(2.5, 4)(1, 4)

(4, -3)

(2.5, -3) (8, -3)

(8, 2)

(6, 2)

0.3 m

17

28

410 5

9

6 5 6

3 4

3

2

7

8

9

0.3 m

0.2 m

0.3 m

0.2 m 0.3 m

0.3 m

0.3 m

0.3 m

(0, 0)

0.3 m

0.2 m

0.2 m

0.2 m

0.2 m

0.2 m

4 0 0 k N

1500 kN-m 5 0 0 k N

Figure 13. Cross-sectional view of the example structure.


14/18

Emsen et al. 341

ω1 = -2.393-1.541

0.163

3.571

_+

ω2 = 14.328

0.163

-4.327

-5.920

-5.920

-15.624

10.690

6.672

6.672

-3.032

-4.719

-9.014

3.873+

+

+

+

_

_

_

_

Iω1 = 152.169 m4

9.664

-8.691

-4.327 3.571

-4.719

-3.060

9.936

2.220

4.715

-4.435

4.654

4.654

17.982

+

+

+

+

_

_

_

_

_

_

_

+

Iω2 = 164.834 m4

_

+

+

+

_

Figure 14. The principal sectorial area diagram of the cross-section.

Figure 15. Frame model of the example structure and its 3-D view.


15/18


0

3

6

9

1215

18

21

24

27

30

33

36

39

42

45

48

51

5457

60

63

66

69

72

75

0 . 0 0

0 . 0 4

0 . 0 8

0 . 1 2

0 . 1 6

0 . 2 0

0 . 2 4

0 . 2 8

Lateral displacement in X direction (m)

S t o r e y h e i g h t ( m )

Present study (unstiffened)

SAP2000 (unstiffened)Present study (stiffened)

SAP2000 (stiffened)

0

3

6

9

12

15

18

21

24

27

30

33

36

39

42

45

48

51

54

57

60

63

66

69

72

75

0 . 0 0

0 . 0 4

0 . 0 8

0 . 1 2

0 . 1 6

0 . 2 0

0 . 2 4

0 . 2 8

Lateral displacement in Y direction (m)



SAP2000 (unstiffened)

Present study (stiffened)

SAP2000 (stiffened)

0

3

6

9

12

15

1821

24

27

30

33

36

39

42

45

48

51

54

57

60

63

66

69

72

75

- 0 . 0 3 6

- 0 . 0 3 0

- 0 . 0 2 4

- 0 . 0 1 8

- 0 . 0 1 2

- 0 . 0 0 6

0 . 0 0

0

Rotation about Z axis (rad)





SAP2000 (stiffened)

Figure 16. Comparison of the lateral displacements and rotations.


16/18

Emsen et al. 343

0

3

6

9

1215

18

21

24

27

30

33

36

39

42

45

48

51

54

57

60

63

66

69

72

75

0 5 0

0

1 0 0 0

1 5 0 0

2 0 0 0

2 5 0 0

3 0 0 0

3 5 0 0

4 0 0 0

Axial force (kN)





SAP2000 (stiffened)

0

3

6

9

12

15

18

21

24

27

30

33

36

39

42

45

48

51

5457

60

63

66

69

72

75

0 5 0

0 0

1 0 0 0

0

1 5 0 0

0

2 0 0 0

0

2 5 0 0

0

3 0 0 0

0

Bimoment ( kN-m )




2

0

3

6

9

12

15

18

21

24

27

30

33

36

39

42

45

48

51

54

57

60

63

66

69

72

75

0 5 0

0 0

1 0 0 0

0

1 5 0 0

0

2 0 0 0

0

2 5 0 0

0

3 0 0 0

0

3 5 0 0

0

Bending moment about X axis (kN-m)





SAP2000 (stiffened)

0

3

6

9

12

1518

21

24

27

30

33

36

39

42

45

48

51

54

57

6063

66

69

72

75

- 1 5 0 0 0

- 1 2 0 0 0

- 9 0 0 0

- 6 0 0 0

- 3 0 0 0 0

3 0 0 0

6 0 0 0

9 0 0 0

Bending moment about Y axis (kN-m)





SAP2000 (stiffened)

Figure 17. The axial forces, the total bimoments and the total shear wall bending moments along the height.


17/18


iEB , external bimoment value;

iB , resultant bimoment, about Z

axis, due to the resistance offered by the piers in region i;i

B

resultant bimoment, about Z axis, due to the componentbending moments and bimoments in region i; d, a geometricproperty; dPX , dPY, moment arms of the components of the

concentrated force; dWX , dWY, moment arms of the distributedforce components; E, elasticity modulus; G, shear modulus;

j j ggy,x , coordinates of the centroid of jth pier, referring to

global axes, X and Y, respectively; j j gg

y,x , coordinates of cen-

troid of the jth pier, referring to principal axes, X and Y , res-

pectively; H , total height of shear wall; hi, storey height inregion i; i, region number;

icI , moment of inertia of connecting

beams in region i; j j yx

I,I , moments of inertia of pier j w.r.t.

global X and Y axes, respectively; jxy

I , product of inertia of pier

j w.r.t. global X and Y axes in region i; j j yx

I,I , moments of

inertia of pier j w.r.t. principal X and Y axes, respectively; jxy

I ,

product of inertia of pier j w.r.t. principal X and Y axes; j

Iω

,

sectorial moment of inertia of pier j;ω

I , sum of the sectorial

moments of inertia of the two piers at the point on Z axis;ω

I , a

geometrical constant;X

Iθ

,Y

Iθ

, sectional properties; j, pier num-

ber; J j, St. Venant torsional constant (moment of inertia) of pier j; j, sum of the St. Venant torsional constants of the two piers;K1, K2, K3, K4, geometrical quantities related to moments of

inertia;ii EYEX

M,M , external bending moments in region i about

the respective axes due to the loading above the cross-section

considered; iEtM , external twisting moment in region i due to

the loading above the cross-section considered;it

M , resultant

torque, about Z axis, due to resistance offered by piersin region

i;it

M , resultant torque, about Z axis, due to component

shears and torques in region i; N, number of regions in verticaldirection; O(X,Y,Z), orthogonal system of global axes; PX , PY, concentrated forces in X and Y directions, respectively;

ji ji yxQ,Q , shear forces developed in a region due to shear

flow, qi, in the laminae; qi, shear flow in laminae per unit length

in region i; R, a geometrical constant; j

S , shear center of jth

pier; j j ss

y,x , coordinates of the shear center of the jth pier,

referring to global axes, X and Y, respectively; Ti, axial force inregion i; T, thickness of a thin-walled beam; ux , uy , uz , displacement components in the directions of orthogonal globalsystem of axes O(X,Y,Z); ui, horizontal global displacement of

the point on Z axis in X direction in region i; ji

u , horizontal

principal displacement of the center of jth pier in

jiX direction in

region i; vi, horizontal global displacement of the point on Z axis

global X and Y axes, respectively; jxy

I , product of inertia of pier j

w.r.t. global X and Y axes in region i; j j yx

I,I , moments of inertia

of pier j w.r.t. principal X and Y axes, respectively; jxy

I , produc

of inertia of pier j w.r.t. principal X and Y axes; j

Iω , sectorial mo

ment of inertia of pier j;ωI , sum of the sectorial moments

of inertia of the two piers at the point on Z axis;ω

I , a geo

metrical constant; XIθ , YIθ , sectional properties; j, pier number; J

St. Venant torsional constant (moment of inertia) of pier j; j, sumof the St. Venant torsional constants of the two piers; K1, K2, K3K4, geometrical quantities related to moments of inertia

ii EYEXM,M , external bending moments in region i about the

respective axes due to the loading above the cross-section

considered;iEt

M , external twisting moment in region i due to

the loading above the cross-section considered;it

M , resultan

torque, about Z axis, due to resistance offered by piersin

region i;it

M , resultant torque, about Z axis, due to

component shears and torques in region i; N, number of regions

in vertical direction; O(X,Y,Z), orthogonal system of global axesPX , PY, concentrated forces in X and Y directions, respectively

ji ji yxQ,Q , shear forces developed in a region due to shea

flow, qi, in the laminae; qi, shear flow in laminae per unit length

in region i; R, a geometrical constant; j

S , shear center of jt

pier; j j ss

y,x , coordinates of the shear center of the jth pier

refering to global axes, X and Y, respectively; T i, axial force inregion i; T, thickness of a thin-walled beam; ux , uy , uz displacement components in the directions of orthogonal globasystem of axes O(X,Y,Z); ui, horizontal global displacement o

the point on Z axis in X direction in region i; ji

u , horizonta

principal displacement of the center of jth pier in

jiX direction in

region i; vi, horizontal global displacement of the point on Zaxisin Y direction in region i;

jiv , horizontal principal displace

ment of the center of jth pier in

jiY direction in region i; Vi, shea

force in ith stiffening beam; Z, spatial coordinate measured

along the height of the structure; WX, uniform load in X directionWY, uniform load in Y direction; β1i , β2i , β3i, geometricaconstants; δ1i, the relative vertical displacement due to the bend

ing of the piers in X and Y directions and due to the warping of the piers; δ2i, the relative vertical displacement due to theaxial deformation of the piers caused by the induced axiaforces arising from the shear flow qi in the connecting mediumδ3i, the relative vertical displacement due to the bending of theconnecting beams; δ

4i, the relative vertical displacement due to

the shear deformations in the laminae; θI, rotational global displacement of the rigid diaphragm in region i; θ ji, rotationa

principal displacement of jth

pier about ji

Z direction in region i

zθ′ , angle of twist per unit length; ω j, sectorial area of pier j a

the point on Z axis; jφ , angle between the global axes and the

principal axes of jth pier;

iγ , a constant property of connecting

beams; , a geometrical constant.


18/18

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coupled shear wall design

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