124

Horizontal Distribution of Forces to Individual Shear Walls

Interaction of Shear Walls with Each Other

In the shown figure the slabs act as horizontal diaphragms extending between

cantilever walls and they are expected to ensure that the positions of the walls,

relative to each other, don't change during lateral displacement of the floors.

The flexural resistance of rectangular walls with respect to their weak axes

may be neglected in lateral load analysis.

The distribution of the total seismic load, xF or yF among all cantilever walls

may be approximated by the following expressions.

ixixix FFF

iyiyiy FFF

wwhheerree::

ixF == load induced in wall by inter-story translation only, in x-direction

iyF = load induced in wall by inter-story translation only, in y-direction

ixF" = load induced in wall by inter-story torsion only, in x-direction

iyF" = load induced in wall by inter-story torsion only, in y-direction

ixF = total external load to be resisted by a wall, in x-direction

iyF = total external load to be resisted by a wall, in y-direction

To obtain ixF and iyF ' , the forces xF and yF are distributed to the individual

shear walls in proportion to their rigidities.

125

The force resisted by wall i due to inter-story translation, in x-direction, is

given by

iy

iyx

ixI

IFF

The force resisted by wall i due to inter-story translation, in y-direction, is

given by

ix

ixy

iyI

IFF

where:

xF = total external load to be resisted by all walls, in x-direction

yF = total external load to be resisted by all walls, in y-direction

ixI = second moment of area of a wall section about x axis

iyI = second moment of areas of a wall section about y axis

ixI = total second moment of areas of all walls in x-direction

iyI = total second moment of area of all walls in y-direction

The force resisted by wall i due to inter-story torsion, in x-direction, is given

by

iyiixi

iyiyx

ixIyIx

IytorsionaccidentaleFF

22

The force resisted by wall i due to inter-story torsion, in y-direction, is given

by

iyiixi

ixixy

iyIyIx

IxtorsionaccidentaleFF

22

where:

ix = x-coordinate of a wall with respect to the center of rigidity C.R of the

lateral load resisting system

iy = y-coordinate of a wall with respect to the center of rigidity C.R of the

lateral load resisting system

xe = eccentricity resulting from non-coincidence of center of gravity C.G and

center of rigidity C.R, in x-direction

ye = eccentricity resulting from non-coincidence of center of gravity C.G and

center of rigidity C.R, in y-direction

126

7x

3 =

21

m

Example (4):

A seven-story building frame system with shear walls has the dimensions

shown in the Figure. Eight shear walls, each 3 m long and 0.2 m thick, are

used as a lateral force resisting system. Determine the forces acting on shear

wall G.

Building layout

Solution:

Neglecting moments of inertia about weak axes, second moments of area of

each of the shear walls about y-axis are given by

43

45.012

32.0mIIII HyGyByAy

A B

D

F

H G

E

C

6m

6m

6m

4.5

m

4.5

m

4.5

m

4.5

m

127

Total second moments of area about y-axis are given by

44

1

8.1445.0 mIi

iy

Second moments of area of each of the shear walls about x-axis are given by

43

45.012

32.0mIIII FxExDxCx

Total second moments of area about x-axis are given by

44

1

8.1445.0 mIi

ix

To locate the center of rigidity C.R, the distance from the origin to the C.R

y in the y-direction is given by

m

I

yI

y

i

iy

i

iiy

25.118.1

5.445.0218245.04

1

4

1

The distance from the origin to the C.R in the x-direction x is given by

128

m

I

xI

x

i

ix

i

iix

0.98.1

1845.024

1

4

1

Thus, the eccentricity in y-direction mey 25.20.925.11

The eccentricity in x-direction mex 0.00.90.9

Torsion caused by eccentricity ye , 1T xF25.2

Torsion caused by accidental eccentricity , 2T xx FF 9.01805.0

Total torsion, 21 TT xx FF 9.025.2

iy

iyx

ixI

IFF

xx

HxGxBxAx FF

FFFF 25.08.1

45.0

iyiixi

iyi

ixIyIx

IyTTF

22

21

xx

xxHxGxBxAx

FF

FFFFFF

9.025.20133.0

945.02945.0275.645.0275.645.02

45.075.69.025.22222

xF018.0 (For walls A and B as shown in the figure below)

xF042.0 (For walls G and H as shown in the figure below)

So, the forces acting on shear wall G are given by the following expression:

x

xx

F

FF

292.0

042.025.0

129

Example (5):

For the building shown in Example (3), evaluate the seismic force acting on shear

wall "A".

Solution:

43 45.012/32.0 mIDx

43 067.112/42.0 mIEx

m

I

xI

x

i

ix

i

iix

55.10067.145.0

15067.102

1

2

1

mex 05.35.755.10

43 45.012/32.0 mIII CyByAy

m

I

yI

y

i

iy

i

iiy

33.845.03

1545.01045.003

1

3

1

mey 83.05.733.8

130

Horizontal distribution of forces for wall A:

AxAxAx FFF

xxAx FFF 333.0)45.0(3

45.0

2222222 67.645.067.145.033.845.045.4067.155.1045.045.067.6

T

IyIx

IyTF

iyiixi

Ayi

Ax

TFAx 02426.0

Maximum eccentricity= 0.83 + 0.05 (15) = 1.58 m

Minimum eccentricity= 0.83 - 0.05 (15) = 0.08 m xxAx FFF 00194.008.002426.0.min,

xAx FF 00194.0333.0

or,

xAx FF 33.0

131

Classification of Structural Walls According To Seismic Risk

According to Chapter 2 and of ACI 318-14, structural walls are defined as being

walls proportioned to resist combinations of shears, moments, and axial forces in the

plane of the wall; a shear wall is a structural wall.

Reinforced concrete structural walls are categorized as follows:

1- Ordinary reinforced concrete structural walls: They are walls complying with the requirements of Chapter 11.

2- Special reinforced concrete structural walls: They are walls complying with the requirements of 18.2.3 through 18.2.8 and 18.10.

Special Provisions for Earthquake Resistance

Design requirements for earthquake-resistant structures in ACI 318 are determined by the SDC to which the structure is assigned.

According to Clause 5.2.2 of ACI 318-14, Seismic Design Categories (SDCs) shall be in accordance with the general building code.

Table R5.2.2 correlates SDC to seismic risk terminology used in ACI 318.

132

Design of Ordinary Shear Walls

The shear wall is designed as a cantilever beam fixed at the base, to transfer load to

the foundation. Shear forces, bending moments and axial loads are maximums at the

base of the wall.

Types of Reinforcement:

To control cracking, shear reinforcement is required in the transverse and

longitudinal directions, to resist in-plane shear forces.

The vertical reinforcement in the wall serves as flexural reinforcement. If large

moment capacity is required, additional reinforcement can be placed at the ends of

the wall within the section itself, or within enlargements at the ends. The heavily

reinforced or enlarged sections are called boundary elements.

ACI Table R18.2 summarizes the applicability of provisions of chapter 18 in terms of

various seismic design categories.

133

In-Plane Shear Strength:

According to ACI 11.5.1.1, design of cross sections subjected to shear are based on

un VV (1)

where uV is the factored force at the section considered and nV is the nominal shear

strength computed from ACI 11.5.4.4,or

scn VVV (2)

where cV is nominal shear strength provided by concrete and sV is nominal shear

strength provided by shear reinforcement.

Based on ACI 11.5.4.3, max,nV at any horizontal section for shear in plane of the wall

is not to be taken greater than

134

dhfV cn 65.2max, (3)

where h is thickness of wall, and d is the effective depth in the direction of

bending, may be taken as wl8.0 , where wl is length of wall considered in direction of

shear force, as stated in ACI 11.5.4.2. A larger value of d , equal to the distance from

extreme compression fiber to center of force of all reinforcement in tension, be

permitted if the center of tension is calculated by a strain compatibility analysis.

Based on ACI 11.5.4.6, the shear strength provided by concrete cV is given by any of

the following equations, as applicable.

For axial compression, Eqn. (4) is applicable

dhfV cc 53.0 (4)

For axial tension, Eqn. (5) is applicable

dhfA

NV c

g

uc

35153.0

(5)

where gA is the gross area of wall section and uN is the factored axial tension force

in Eqn. (5).

ACI 11.9.6 specifies that a more detailed analysis is permitted to evaluate cV as

follows, where cV is the lesser of the two values shown in Eqns. (6) and (7).

w

ucc

l

dNdhfV

4'88.0 (6)

hdl

V

M

hl

Nfl

fVw

u

u

w

ucw

cc

2

2.0'33.0

'16.0 (7)

Where uN is positive for compression and negative for tension. If 2// wuu lVM is negative, Eqn. (7) is not applicable.

135

Shear Reinforcement:

A- If in-plane shear force uV is less than 2/cV , minimum wall longitudinal and

transverse shear reinforcement shall be in accordance with ACI Table 11.6.1.

B- If in-plane shear force uV is more than or equal to 2/cV , minimum shear

reinforcement in the longitudinal direction, l shall be provided, based on ACI

11.6.2.

0025.00025.05.250.00025.0

t

w

wl

l

h (8)

The above value need not exceed t given in Table 11.6.1.

Minimum shear reinforcement in the transverse direction, t shall be at least 0.0025.

C- According to ACI 11.5.4.8 when the factored shear force uV exceeds cV ,

transverse shear reinforcement must be provided according to the following equation.

S

dfAV

ytv

s

(9)

Where v

A is area of transverse shear reinforcement within a distance S . Longitudinal

shear reinforcement, l is provided as in case (B), shown above.

Spacing of Transverse Reinforcement:

Based on ACI 11.7.3.1, spacing of transverse reinforcement is not to exceed the

smallest of cmhlw 45,3,5/ .

136

Spacing of Longitudinal Reinforcement:

Based on ACI 11.7.2.1, spacing of transverse reinforcement is not to exceed the

smallest of cmhlw 45,3,3/ .

Critical Section for Shear:

The critical section for shear is taken at a distance equal to half the wall length 2/wl ,

or half the wall height 2/wh , whichever is less. Sections between the base of the wall

and the critical section are to be designed for the shear at the critical section, as

specified in ACI 11.5.4.7.

Shear wall Reinforcement

137

Design for Flexure:

The wall must be designed to resist the bending moment at the base and the axial

force produced by the wall weight or the vertical loads it carries. Thus, it is

considered as a beam-column.

For rectangular shear walls containing uniformly distributed vertical reinforcement

and subjected to an axial load smaller than that producing balanced failure, the

following equation, developed by Cardenas and Magura in ACI SP-36 in 1973, can

be used to determine the approximate moment capacity of the wall.

wys

uwysu

l

C

fA

PlfAM 115.0

Where:

185.02

wl

C

cw

ys

fhl

fA

and

cw

u

fhl

P

C distance from the extreme compression fiber to the neutral axis

sA = total area of vertical reinforcement

wl = horizontal length of wall

uP = factored axial compressive load

yf = yield strength of reinforcement

= strength reduction factor for bending

Additional Reinforcement around Openings:

In addition to the required transverse and longitudinal reinforcement explained

earlier, ACI 11.7.5.1 states that not less than mm162 bars are provided around

all window and door openings in both directions in walls having two layers of

reinforcement. In walls having a single layer of reinforcement in both

directions,

mm161 is to be provided. Such bars are to be extended to develop

yf in tension at the corners of the openings.

Additional reinforcement around wall openings

138

Example (6):

For shear wall 'A' in example (5), design the reinforcement required for shear and

flexure using ACI 318-14 for reinforced concrete design (ordinary shear wall).

Use 22 /4200 and /300 cmkgfcmkgf yc .

Solution:

From example 5, xAx FF 33.0

Critical section for shear is located at a distance not more than the smaller

of 2/

2/

w

w

h

l, i.e., at 1.5 m from the base of the wall.

1- Design for shear:

Check for maximum nominal shear force

139

dh'f65.2V cmax,n

tons32.2201000/3008.02030065.2 KOtonstonsVu .16.1524.16532.22075.0max,

dh'f53.0V cc tons06.441000/3008.02030053.0Vc

tons045.3306.4475.0Vc tons523.162/045.332/Vc

In zones 1 through 8, 2/VV cu

1-1 Transverse shear reinforcement:

0020.0t

ofsmaller the2 S

cm

cmh

cmlw

45

603

605/

or cmS 45max,2

cmcmS

Ah

S

A tt /04.0 and 200020.00020.0 2

22

For two curtains of reinforcement and trying 10 mm bars

max,22

2

25.39 , 04.0785.02

ScmSS

O.K

Use 10 mm bars @ 30cm.

1-2 Longitudinal shear reinforcement:

0012.0l

ofsmaller the1 S

cm

cmh

cmlw

45

603

1003/

or cmS 45max,1

For two curtains of reinforcement, and trying 10 mm bars

11

0.7852 200012.00012.0

Sh

S

Al

And max,11 42.65 ScmS

Use 10mm bars @ 30cm.

Note that flexural reinforcement is expected to control.

140

2- Design for flexure and axial loads:

wys

uwysu

l

c1

fA

P1lfA5.0M

Where:

1w 85.02l

c

,

'fhl

fA

cw

ys and

'fhl

p

cw

u

For the vertical shear reinforcement of 10 mm @ 30cm, 2s cm28.17A ,

836.028030070

05.085.0 ,

04032.030020300

420028.17

'fhl

fA

cw

ys

, u

u

cw

u P00055.030020300

1000P

'fhl

P ,

79124.0P00055.004032.0

836.085.004032.02

P00055.004032.0

l

c uu

w

For zone 1:

Neglecting dead load supported by the shear wall and considering own weight of the

wall only, the load combination to be considered is

EDS QDS 2.09.0 , or

Vertical load= (0.90 – (0.2)(0.107) ) D = 0.88 D

tonsPu 68.315.22432.088.0

072346.079124.0

68.3100055.004032.0

wl

c

mtmtMu .47.265.55.130 , i.e. boundary elements are required at wall ends

mtM u .92.13455.13047.265'

2, 99.14785.02

26642009.0

10000092.134cmA additionals

Use 8 16 mm bars in each of the two boundary elements.

For zone 2:

tonsPu 72.275.22132.088.0

069685.079124.0

72.2700055.004032.0

wl

c

141

mtmtMu .99.219.96.125 , i.e. boundary elements are required at wall ends

mtM u .03.9496.12599.219'

2, 92.10785.02

26642009.0

10000003.94cmA additionals

Use 8 14 mm bars in each of the two boundary elements.

For zone 3: tonsPu 76.235.21832.088.0

066933.0

79124.0

76.2300055.004032.0

wl

c

mtmtMu .41.175.35.121 , i.e. boundary elements are required at wall ends

mtM u .06.5435.12141.175'

2, 95.6785.02

26642009.0

10000006.54cmA additionals

Use 6 14 mm bars in each of the two boundary elements.

For zone 4:

tonsPu 80.195.21532.088.0

064182.079124.0

80.1900055.004032.0

wl

c

mtmtMu .90.132.70.116 , i.e. boundary elements are required at wall ends

mtM u .20.1670.1169.132'

2, 18.3785.02

26642009.0

1000002.16cmA additionals

Use 3 14 mm bars in each of the two boundary elements.

142

For Zone 5

tonsPu 84.155.21232.088.0

061431.079124.0

84.1500055.004032.0

wl

c

mtmtMu .78.93.03.112 , i.e. no boundary elements are required at wall ends

For Zone 6 tonsPu 88.115.2932.088.0

058679.0

79124.0

88.1100055.004032.0

wl

c

mtmtMu .43.59.33.107 , i.e. no boundary elements are required at wall ends

For Zone 7 tonsPu 92.75.2632.088.0

055928.0

79124.0

92.700055.004032.0

wl

c

mtmtMu .32.31.59.102 , i.e. no boundary elements are required at wall ends

For Zone 8 tonsPu 96.35.2332.088.0

053177.0

79124.0

96.300055.004032.0

wl

c

mtmtMu .98.10.83.97 , i.e. no boundary elements are required at wall ends