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SC 342 DESIGN OF SHEAR WALL

STRUCTURES

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1.0: WALLS IN BUILDINGS 1.1. Function, types and loads on walls

A wall is defined in BS 8110: Part 1, clause 1.2.4, as a vertical load bearing member whose length exceeds four times its thickness. This definition distinguishes a wall from a column.All buildings contain walls the function of which is to carry loads, enclose and divide space, exclude weather and retain heat.Walls may be classified into the following classes1.Internal non-load bearing walls of blockwork or light movable partitions that divide space only2.External curtain walls that carry self weight and lateral wind loads

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3. External and internal infill walls in framed structures that may be designed to provide stability to the building but do not carry vertical building loads; external walls would also carry lateral wind loads

4. Load bearing walls designed to carry vertical building loads and horizontal lateral load and in-plane wind loads and provide stability.

Type 4 may be considered as structural concrete walls, i.e. shear walls, which can further be classified as:

(a) A braced wall: Is a wall where reactions to lateral forces are provided by lateral supports such as floors or cross-walls

(b) An un-braced wall: Is a wall providing its own lateral stability such as cantilever wall

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(c) A stocky wall: Is a wall where the effective height divided by the thickness, le/h, does not exceed 15 for a braced wall or 10 for an un-braced wall

(d) A slender wall: Is a wall other than a stocky wall

1.2: Shear wallsWhat are shear walls?are vertical structural elements with a dual role of resisting both the gravity and lateral loads.

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1.2.1 Structural features

wall thickness varies from 150 mm to 400 mm, depending on the number of stories, building age, and thermal insulation requirements.

walls generally start at foundation level and are continuous throughout the building height

wall must be symmetrical located in plan to reduce ill effects of twists in buildings (Figure 1)

Shear walls are more effective when located along exterior perimeter of the building – such a layout increases resistance of the building to twisting

6Figure 1: Reinforced Concrete shear walls in buildings

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RC buildings without shear walls structures

Damage of beam-column joint

Soft and weak storey

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1.2.2 Advantages of shear walls

are easy to construct, because reinforcement detailing of walls is relatively straight-forward and therefore easily implemented at site.

are efficient, both in terms of construction cost and effectiveness in minimizing lateral damage in structural and non-structural elements (like glass windows and building contents)

provide large strength and stiffness to buildings in the direction of their orientation, which significantly reduces lateral sway of the building and thereby reduces damage to structure and its contents

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1.2.3 Disadvantages of shear walls

shear walls present barriers, which may interfere with architectural and services requirement.

lateral load resistance in shear wall buildings is usually concentrated on a few walls rather than on large number of columns.

since shear walls carry large lateral forces, the overturning effects on them are large. Thus, design of their foundations requires special attention.

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1.3: Basic considerations

Checking the stability of a building against horizontal forces is imperative, particularly wind loads, which acts in the form of pressure or suction on the exterior walls, window or façade elements. From there they are led into the floor slabs, which are acting as nearly rigid diaphragms. These diaphragm slabs are supported horizontally by the stiffening shear walls, which are acting like vertical cantilevers clamped into the foundation slab.

Foundation uplift should be avoided by increasing the size of the wall footing. If footing size becomes excessive, possible solutions are:

(a) mat foundation(b) pile foundation(c) coupled shear

walls

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NOTE: A conservative assumption is to neglect columns and girders for horizontal load analysis, although they do contribute to the horizontal stability of the building.

x

1

2

3

w (k N /m ²)w (k N /m ²)

z

y

l

C o lu m n s - n eg lec ted

Figure 1.0: Columns and beams neglected

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The wind load acts as a surface load (kN/m2), which usually increases with the altitude above ground level. For design purposes it can be assumed to be constant. The substitute constant wind load can be determined in such a way that the bending moment at foundation level is equal to that one which belongs to the actual wind load (Figure 1.2a).

For consideration of the whole building as a cantilever beam, it is helpful to determine the line loads, in order to compute the lumped moments and shear forces of the whole building (Figure 1.2b), thus:

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p

p

z ,o

y ,o

1

2

3

p

p

p

1

2

3

w

w

y

z

yz

a

b

Figure 1.2a: Plan

awp

bwp

yOy

zOz

,

,

Line loads

Area loads

wx, wy in kN/m2

in kN/m

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x

z o r y

p

i

l

p

V [k N ]i M [k N m ]p i

2 /2

+ -

i

i

i

l li i

i

Figure 1.2b: Moment and shear force diagram

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1.4: Overall geometry of the shear walls

Shear walls are rectangular in cross-section, i.e., one dimension of the cross-section is much larger than the other.

L- and U-shaped sections are also used.

A thin-walled hollow RC shafts around the elevator core of buildings also act as shear walls.

Figure 1.3: Typical shear walls

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1.3: Structural layout of the shear walls

The major considerations in selecting a structural system for a multistory building with shear walls are: The building must have enough rigidity to withstand

the service loads without excessive deflections or vibrations,

It is desirable that the wall be loaded with enough vertical load to resist any uplift of parts of the wall foundations due to lateral walls,

The locations of frames and walls should minimize tensional deformations of the building about the vertical axis of the building,

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The walls must have adequate strength in shear and in combined flexure and axial loads,

The wall must have enough cover on the reinforcement, and

Multiple walls should be used to minimize demands on footings.

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1.5: Stable and unstable system

1.5.1: Statically determinate shear wall system

A shear wall system is called ‘statically determinate’ if only three (wall-type) shear walls are acting together in one floor plan. For a statically determinate system the wind load distribution key can be determined very easily by applying conditions of equilibrium only. For the load case wz in Fig. 1.2a we obtain the following:

0p

2/ppp

3

0,z21

For the load case wy we see clearly that the force p3 has to balance py,o, viz. p3 = py,o

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The remaining moment, i.e. 2/apM 0,ye

has to be taken by a couple of forces formed by p’1 and p’2

bp2/ap

0pp'20,y

'2

'1

Thus;b2/appp

b2/app

0,y'2

'1

0,y'2

The shares p1, p2, p3 are line loads (kN/m) which produce shear forces and bending moments in the walls (i.e. Figure 1.2b)

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1.5.2: Statically indeterminate shear wall system

When there are more than three walls then the problem is statically indeterminate and, as a rule, the stiffnesses have to be determined for a proper solution. Statically determinate systems are seldom to be found in practice, most structures are statically indeterminate.

For better comprehension of the general solution it might be recommendable to differentiate the indeterminate system into those with a symmetric arrangement of the shear walls (symmetric systems) and those with an arbitrary arrangement (non-symmetric systems).

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z

y

U n stab le sy s tem

N eg lig ib le lo adre s is tan ce iny - d irec tio n

(a )

(b )

(c )

(d )

(e )

(f)S tab le sy s tem

Figure 1.4: Stable and unstable shear wall systems

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Generally, the relative lateral deflection in any one storey should not exceed the storey height divided by 500.

The figure below shows the deflected profiles for a shear wall and a rigid frame.

• One way to limit the sway of buildings and provide stability is to increase the section sizes of the members to create a rigid, moment-resisting frame. However, this method increases storey heights, thus increasing the building cost. It is rarely used for more than 7 or 8 storeys.

• Another way is to provide stiff, shear resisting walls liked to a flexible frame. These can be external walls or internal walls around lift shafts and stair wells (a core) or sometimes both are provided.

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• STRUCTURAL FORMS OF SHEAR WALLS:• Monolithic shear walls are classified as short, squat or

cantilever according to their height to depth ratio

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• POSITIONING OF SHEAR WALLS:The shape and plan position of the shear wall influences the behavior

of the structure considerably. Structurally, the best position for the shear walls is in the centre of each half of the building. This is rarely practical, however, since it dictates the utilization of the space, so they are positioned at the ends.

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This shape and position of the walls give good flexural stiffness in the short direction, but relies on the stiffness of the frame in the other direction.

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This arrangement provides good flexural stiffness in both directions, but may cause problems from restraint or shrinkage. As does this arrangement with a single core, but which does not have the problem from restraint of shrinkage.

However, this arrangement lacks the good torsional stiffness of the previous arrangements due to the eccentricity of the core.If the core remains in this position then it must be designed explicitly for the torsion. It is far preferable to adopt a symmetrical arrangement to avoid this.

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1.5.2.1: Symmetric system

If a horizontal load acts in the axis of symmetry of a symmetric system, the displacements of every shear wall in every floor are the same, because the diaphragms force all elements to a uniform and affined bending behavior (Figure 1.5). If we denote the displacements of the origin of the coordinate system by wo and the displacement of a single wall by wi we can describe this fact by:

w(x) = wo(x) = w1(x) = w2(x) = … wi(x) = wn(x) ....(1.1)

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W

Figure 1.5: Symmetric system - shear walls

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If all displacements in a certain level x are equal, the individual load-effects can be obtained from the total ones according to the ratio of individual bending stiffness to total bending stiffness (basic equation of the deformation method):

i

n)x(,0)x(,n EI

EIMM

....(1.2)

where Mn,(x) = bending moment in shear wall n at the level x EIn = bending stiffness in shear wall n at the level x

iEI = sum of all single bending stiffness

M0,(x) = total bending moment at level x due to exterior horizontal loads

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1.5.3 Non-Symmetric Systems

We distinguish the following coordinate systems, as shown in Figure 1.6.

A general one, with ,axeszy

wherethe origin of which lies at an arbitrary point

O

A special one, with ,axeszy where

the origin of which lies at the shear center So of the lumped beam

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The longitudinal axis of the building is the x-axis. Displacements in y- direction are called v, displacements in z-direction are called w. Forces get an index y or z according to the direction of the adjacent axis. The coordinates ўo and žo of the shear center are not yet known; they have to be calculated. The index 0 (zero) will be used, if any property of the lumped system is to be described. As counterpart to the single or individual element we use the expression ”total” or ”lumped”, if the lumped system is meant.

In a first step we consider the case ”pure bending” i.e. all forces are acting on the shear center, and in a second step the cases ”pure torsion” i.e. the torsional moments are obtained by multiplying the horizontal resultant force by the eccentricity e.

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y

O

z

S

z

y

y

z(w)

(v)

Pz,o

e o

b/2 b/2

b

y

x

z

P

P

z,o

y,o

Q

Q

z

y

Figure 1.6: Statical system of a “Lumped” beam – Unsymmetrical shear walls

The sum of all deviational moments is zero: 0EIi,yz

The torsional stiffness of a single shear wall is zero: 0EI,0GIi,wi,T

All cross-sections and load functions are constant over the height.

assumptions

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1.5.3.1 Pure Bending

We assume that a shear center So of the lumped system exists with the property that a horizontal force acting on S will yield to pure translational displacements without rotation of the floor plan. Afterwards this assumption has to be proved. The shear center So of the system is comparable with the shear center Si of a single beam. We shall see that the shear center is at the same time the pivot for torsional moments, which also corresponds to the well-known relations of the individual beam. In a first step we shift the horizontal force into the shear center and study the case of pure bending. We obtain the following relations for the load case Pyo:

Geometry: All displacements wi (x) are equal: w1(x) = w2(x) = wi(x) = w0(x) ....(1.1) Equilibrium: The sum of all partial forces Pz,i must be equal to the external force Pz,o: Pz,1 + Pz,2 + Pz,3 + ..........+ Pz,i + .....+ Pz,n = Pz,0 ....(1.2)

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Differential equation of a beam, using Equ. (1.1):

)x(wEI

P)x(w 4

01,y

1,z41

)x(wEI

P)x(w 4

02,y

2,z42 ....(1.3)

)x(wEI

P)x(w 4

0i,y

i,z4i

Combining Equ. (1.2) and Equ. (1.3) leads to

0,zn,yi,y2,y1,y40 PEI........EI.......EIEI)x(w

or

i,y

0,z40 EI

P)x(w ....(1.4)

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yy

s s

V Vz ,i z ,o

o

o

i

i

o

Oy

z

Figure 1.6: Determination of shear centre

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If we know according to Equ. (1.4), the individual (and total) displacement functions w(x) or their derivatives wIV(x) we can by the help of Equ. (1.3), compute the partial forces Pz,i; which are to be assigned to the individual shear walls i:

i,y

1,y0,z

401,y1,z EI

EIPwEIP

i,y

i,y0,z

40i,yi,z EI

EIPwEIP ....(1.5)

i,y

n,y0,z

40n,yn,z EI

EIPwEIP

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The individual load effects in a single beam can now easily be computed. An especially plausible form is obtained, if they are expressed in terms of the total load effects. The total load effects are: Total shear force )xL(P)x(Q 0,z0,z ....(1.6)

Total bending moment 2

xLP)x(M

2

0,z0,y

....(1.7)

Where L = height of the structure x = height of point in consideration Equs (1.6) and (1.7) can either be obtained by integration of the differential equation of the beam or, more easily, by a single application of the equilibrium conditions. The individual load effects are:

)xL(P)x(Q i,zi,z ....(1.8)

2

xLP)x(M

2

i,zi,y

....(1.9)

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A comparison of the Equ. (1.6)/(1.3) and (1.8)/(1.4) leads to

i,y

n,y

0,z

n,z

0,z

n,z

EI

EI

P

P

)x(Q

)x(Q ....(2.0a)

i,y

n,y

0,z

n,z

0,z

n,z

EI

EI

P

P

)x(M

)x(M ....(2.0b)

A result proves that in a statically indeterminate structure the load effects are distributed according to the ratio of the individual and total stiffness. The shear center can now be easily computed. If no rotation of the lumped cross section is allowed, the resultant of all shear forces Qi must aim at the shear center. Thus an equation is available for the determination of the coordinates of the shear center (see Fig. 1.6):

_

ii,z0,z' yQyQ ....(2.1)

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Differentiation of Eqn (2.1) leads to

i'

i,z0'

0,z yQyQ

....(2.2) Using Eqn. (2.3)

i,z'

i,z

0,z'

0,z

PQ

PQ

....(2.3)

we obtain from Eqn.(2.2)

ii,z00,z yPyP

0,z

ii,z0 P

yPy

....(2.4)

In equation (2.4) we substitute Pz.i by Eqn (1.5); this leads to

i,y

ii,y0,z

0,z0 EI

yEIP

P

1y

i,y

ii,y0 EI

yEIy ....(2.4a)

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According to the above shown derivation we get the second needed coordinate of S

i,z

ii,z0 EI

zEIz

....(2.5)

The interpretation of Eqn 2.5 might be the shear center of the lumped beam corresponds to the gravity center, if instead of the areas of bending stiffness is applied in the corresponding formulae.

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1.5.3.2 Pure TorsionAs the shear center usually is not identical with the point of application of the horizontal load resultant the eccentricity e leads to an additional twisting moment

e.Qm 0,z0,x ....(2.6) The sign of mx,o is shown in Fig. 1.8, it leads to additional shear forces in the individual shear walls. As we have assumed the torsional stiffness to be zero, we do not obtain individual torsional moments. Their influence is only to be taken into account, if a considerable torsional stiffness is given, e.g. in form of tubes or cores. The total torsional moment Mx,o (x) is obtained from the twisting moment Mx,o by an equilibrium consideration

)xl(m)x(M 0,x0,x ....(2.7)

d x

x

m

M x ,o

x ,o

x ,ox ,oM + M

Figure 1.7: Sign of the torsional moment

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The total torsional moment must be in equilibrium with the individual torsional moments caused by the individual shear forces Qy,i and Qz,i and their lever arms to the point of reference. It is useful to take the shear center as the point of reference, because it simplifies the formulae. The equilibrium conditions yield to

ii,yii,z0,x z).x(Qy).x(Q)x(M ....(2.8) The individual shear forces shall be expressed in terms of derivatives of the rotation χo(x) (see Fig. 1.8) of the lumped system:

i0i

i0i

z)x(

y)x()x(w

….(2.9)

Introduction of Equ. (2.9) into the differential equation of the beam leads to

)x(ZEI)x(EI)x(Q iiioii,z

iiiii,Zi,y

….(3.0) )x(YEI)x(EI)x(Q iii

oii,yiiiii,yi,y

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Equ. (3.0) in combination with Equ. (2.8) yields to

2;i,z

2ii,y

iiioo,x ZEIyEI)x()x(M

)()(

2;,

2,

)(0,

iziiy

xxiiio ZEIyEI

Mx ….(3.1)

y

s s

Vz ,i

o

o

i

i

o

Oy

z

y

z i

z o

z i

iy

Vy ,i

y

z

M x ,o

ow (x )

Figure 1.8: Rotation of the floor around the shear center

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Equ. (3.1) can be used to determine the displacements of the individual shear walls. Normally the displacements are of no interest; what is needed are the internal forces of the individual walls, e.g. Qy,n; Qz,n; My,n; Mz,n; which can be computed if the partial loads py,n and Pz,n are known. This can easily be obtained by introducing Equ. (3.1) into Equ. (3.2) which leads to the individual shear forces Qy,n and Qz,n:

Differentiation of Equ. (3.2) leads to the individual forces Py,n and Pz,n, integration to the individual bending moments My,n and Mz,n.

)x(M.)ZEIyEI(

ZEI)x(Q

)x(M.)ZEIyEI(

ZEI)x(Q

o,x2ii,z

2ii,y

nn,Z

n,z

o,x2ii,z

2ii,y

nn,Zn,y

….(3.2)

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o,x2ii,z

zii,y

nn,zn,z

o,x2ii,z

zii,y

nn,zn,y

m)zEIyEI(

yEIq

m)zEIyEI(

ZEIq

….(3.3)

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The signs depend on the definition of shear forces and moments. This varies from text-book to text-book; in this chapter the signs of the above mentioned internal forces are defined in Fig. 4.5.

2

o,xzii,zii,y

nn,zn,y

2

o,xzii,zii,y

nn,zn,z

2

xlm

)zEIyEI(

yEIM

2

xlm

)zEIyEI(

ZEIM

….(3.4)