Couple Shear Wall

Coupled shear wall In coupled shear wall, walls are connected by bending resistantelements .

Presence of moment resisting connections greatly increase thestiffness of the wall system.

Behavior of couple shear wall When shear wall is connected by pin ended links

If shear walls is connected by pin-ended links that transmit onlyaxial forces between them.

Total external moment will be resisted individually by each wallseparately

maximum tensile and compressive stresses on opposite edges

When shear wall is connected by Rigid beams

If the walls are connected by rigid beams then applied momentwill be resisted by the two walls acting as a single composite unit.

Max. tensile and compressive stresses occurs at the oppositeextreme edges.

The practical solution of a pair of wall connected by flexible beam willbe between these two extreme cases.

When the wall deflect under the action of the lateral loads , theconnecting beam ends are forced to rotate and displace vertically.

So beams are bend in double curvature and resist free bending of thewalls.

Behavior of laterally loaded coupled system M=M1+M2+N*l

Methods of analysis Approximate method

This method is simpler and more co-operative for hand calculation.This method is restricted to regular structure and regular load system.Continuous medium method.

Exact methodsthis method deal with irregular structures and complex loadings.To execute this method we need service of digital computer.Finite element method.

The continuous medium method

Basic assumption made in analysis are as follows

The properties of the walls and connecting beams do not change overthe height, and the story heights are constant.

Plane sections before bending remain plane after bending for allstructural members.

The discrete set of connecting beams , each of flexural rigidity EIb maybe replaced by an equivalent continuous connecting medium offlexural rigidity EIb/h per unit height, where h is story height.

The walls deflects equally horizontally, as a result of the high inplanerigidity of surrounding floor slabs and the axial stiffness of theconnecting beams. Slopes of the walls are everywhere equal alongthe height.

The discrete set of axial forces , shear forces and bending moments inthe connecting beams may than be replaced by equivalentcontinuous distribution of intensity n,q and m respectively per unitheight.

Internal forces in couple shear wall In particular if the connecting medium is assumed cut along thevertical line of contraflexure, the only forces acting there are shearforce of intensity q per unit height and axial force of intensity n perunit height.

So the axial force N in each wall at any level z will then be equal to theintegral of the shear in connecting medium above that level.

Internal forces in couple shear wall

Consider the condition of vertical compatibility along the cut line of the contraflexureas shown in figure.

Relative vertical displacements will occur at the cut ends of the cantilever due tofollowing four basic actions

Rotation of wall cross section due to bending

Bending & Shearing deformations of the connectingbeams under the action of the shear force.

The effect of shear deformations in the connecting beams mayreadily be included by replacing the true flexural rigidity EIb by anequivalent flexural rigidity EIc.

GA is the shearing rigidity. is the shape factor for shear. equal to 1.2 in the case of rectangularsections.

Axial deformation of wall under action of axial force N

Any Vertical or rotational relative displacement at the base

In the original deflected structure there can be no relative verticaldisplacement on the line of contraflexure of the connecting beams.

The last term will be zero in common case of rigid base

On considering free bending due to externally applied moment M andreverse bending due to shear and axial forces in connecting medium ,the moment curvature relationship for two walls are at any level givenby below equation

Where Ma is moment caused by axial forces in connecting beams.

I=I1+I2 and A=A1+A2

Axial forces in walls

Shear forces in connecting members

Wall moments

Deflections

Significance of parameter kH

Example of coupled shear wall

Step 1:Determine the areas and second moment of areaof connecting beamWall properties:-

I1 = 1/12 * 53 * 0.3 =3.125 m4I2 = 1/12 * 73 * 0.3 =8.575 m4

I = I1 + I2 = 11.7 m4

A1 = 5 * 0.3 =1.5 m2A2 = 7 * 0.3 = 2.1 m2

A = A1 + A2 = 3.6 m2

For connecting beams, Assuming that entire cross section is effective

Ib = 1/12 * 0.43 * 0.3 = 1.6 * 10-3 m4

The second moment of area is reduced to include sheardeformations.

Assuming Poissons ratio v=0.15 for concreteSo shear modulus

G= E/(2*(1+v))=E/2.3

Because of reduction effective second moment of area

r = 12* 2.3 *1.6 * 10-3 * 1.2 / (22 * 0.4 *0.3)r = 0.1104

Ic = 1.6 * 10-3 /(1+0.1104) = 1.441 * 10-3 m4

Step 2 : Determine the structural parameter k, and kH

K2 = 1+ (3.6 *11.7)/(1.5*2.1*82) =1.2089k= 1.0995

2 = (12 * 1.4*10-3 * 82 )/(2.23 * 2.8 * 11.7)=3.1725 *10-3 = 0.05633 m-1 kH=1.0995 *0.05633 *56 =3.468

Step 3:Calculate wind moment at base level and percentage ofmoment at this level carried by individual cantilever action(k1)and composite cantilever action (k2) Total base moment =1/2* 16.5 * 562 =25.872 kN.m Now for base level z=0 so z/H=0

k1=42% k2=58%

Portion of base moment due to individual cantilever action=0.42*25,872 =10,866 kN.m

So moment on wall 1, M1= (3.125/11.7)*10,866 =2902 kN.mmoment on wall 2, M2= (8.575/11.7)*10,866=7964 kN.m

Portion of base moment due to composite cantilever action=0.58 * 25,872 = 15,006 kN.m

Step 4: calculate second moment of area Ig of the compositecross section. Calculate the stresses at the extreme fibers of thewalls , using ordinary beam theory.

Ig= 3.125 + 8.575 + (1.5*2.1*82)/3.6 =67.7 m4

Using ordinary beam theory, the stresses at the salient points A,B,Cand D are , on adding the stressed due to individual and compositecantilever stresses.

Taking tensile stress as positive.

sa=(2902*2.5)/3.125 +(15,006*7.167)/67.7 =3910 kN/m2 sb=-(2902*2.5)/3.125 +(15,006*2.167)/67.7 =-1841 kN/m2 sc=(7964*3.5)/8.575 +(15,006*0.167)/67.7 =3288 kN/m2 sd=-(7964*3.5)/8.575 - (15,006*6.83)/67.7 =-4765 kN/m2

Step 5: Determine the maximum shear force factor andhence the shear force For value of maximum shear force factor kH =3.468, F2(max)=0.381at level z/H=0.39.

Maximum shear flow qmax

qmax =16.5*56*0.381/(1.2092*8)=36.39 kN/m So maximum possible shear in any connecting beamQmax=qmax*h=36.39*2.8=101.9 kN Maximum possible moment in connecting beamMmax=Qmax*(b/2)=101.9*1=101.9 kN.m

Step 6: Determine the deflection factor F3 and maximumlateral deflection at top of the structure For kH =3.468 and k=1.0995 the value of maximum deflection factoris 0.333.

Max deflection at top of the structure

Assuming that the dynamic modulus of elasticity =36 kN/mm2 Ymax=(16.5*564*0.333)/(8*36*106*11.7)=0.016 or 16mm

Step 7: Axial force and bending moment

Bending moment