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, x F or y F among all cantilever walls may be approximated by the following expressions. ix ix ix F F F iy iy iy F F F where: ix F = load induced in wall by inter-story translation only, in x-direction iy F = load induced in wall by inter-story translation only, in y-direction ix F " = load induced in wall by inter-story torsion only, in x-direction iy F " = load induced in wall by inter-story torsion only, in y-direction ix F = total external load to be resisted by a wall, in x-direction iy F = total external load to be resisted by a wall, in y-direction To obtain ix F and iy F ' , the forces x F and y F are distributed to the individual shear walls in proportion to their rigidities.
Horizontal Distribution of Forces to Individual Shear Walls
Apr 05, 2023
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Example 1Interaction 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
ix I
IF F
The force resisted by wall i due to inter-story translation, in y-direction, is
given by
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
22
The force resisted by wall i due to inter-story torsion, in y-direction, is given
by
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
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
4
3
Total second moments of area about y-axis are given by
4 4
iy
Second moments of area of each of the shear walls about x-axis are given by
4
3
Total second moments of area about x-axis are given by
4 4
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
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
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
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
Example (5):
For the building shown in Example (3), evaluate the seismic force acting on shear
wall "A".
AxAxAx FFF
Minimum eccentricity= 0.83 - 0.05 (15) = 0.08 m xxAx FFF 00194.008.002426.0.min,
xAx FF 00194.0333.0
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.
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
(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
dhf A
(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
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
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
ytv
s
(9)
Where vA 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/ .
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.
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
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:
139
1-1 Transverse shear reinforcement:
22
For two curtains of reinforcement and trying 10 mm bars
max,22
2
1-2 Longitudinal shear reinforcement:
140
'fhl
p
cw
u
For the vertical shear reinforcement of 10 mm @ 30cm, 2 s cm28.17A ,
836.0280300 70
05.0 85.0 ,
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
tonsPu 68.315.22432.088.0
072346.0 79124.0
mtmtMu .47.265.55.130 , i.e. boundary elements are required at wall ends
mtM u .92.13455.13047.265'
10000092.134 cmA additionals
Use 8 16 mm bars in each of the two boundary elements.
For zone 2:
mtmtMu .99.219.96.125 , i.e. boundary elements are required at wall ends
mtM u .03.9496.12599.219'
10000003.94 cmA 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
mtmtMu .41.175.35.121 , i.e. boundary elements are required at wall ends
mtM u .06.5435.12141.175'
10000006.54 cmA additionals
Use 6 14 mm bars in each of the two boundary elements.
For zone 4:
mtmtMu .90.132.70.116 , i.e. boundary elements are required at wall ends
mtM u .20.1670.1169.132'
1000002.16 cmA additionals
Use 3 14 mm bars in each of the two boundary elements.
142
wl
c
mtmtMu .78.93.03.112 , i.e. no boundary elements are required at wall ends
For Zone 6
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
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
ix I
IF F
The force resisted by wall i due to inter-story translation, in y-direction, is
given by
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
22
The force resisted by wall i due to inter-story torsion, in y-direction, is given
by
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
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
4
3
Total second moments of area about y-axis are given by
4 4
iy
Second moments of area of each of the shear walls about x-axis are given by
4
3
Total second moments of area about x-axis are given by
4 4
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
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
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
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
Example (5):
For the building shown in Example (3), evaluate the seismic force acting on shear
wall "A".
AxAxAx FFF
Minimum eccentricity= 0.83 - 0.05 (15) = 0.08 m xxAx FFF 00194.008.002426.0.min,
xAx FF 00194.0333.0
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.
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
(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
dhf A
(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
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
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
ytv
s
(9)
Where vA 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/ .
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.
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
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:
139
1-1 Transverse shear reinforcement:
22
For two curtains of reinforcement and trying 10 mm bars
max,22
2
1-2 Longitudinal shear reinforcement:
140
'fhl
p
cw
u
For the vertical shear reinforcement of 10 mm @ 30cm, 2 s cm28.17A ,
836.0280300 70
05.0 85.0 ,
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
tonsPu 68.315.22432.088.0
072346.0 79124.0
mtmtMu .47.265.55.130 , i.e. boundary elements are required at wall ends
mtM u .92.13455.13047.265'
10000092.134 cmA additionals
Use 8 16 mm bars in each of the two boundary elements.
For zone 2:
mtmtMu .99.219.96.125 , i.e. boundary elements are required at wall ends
mtM u .03.9496.12599.219'
10000003.94 cmA 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
mtmtMu .41.175.35.121 , i.e. boundary elements are required at wall ends
mtM u .06.5435.12141.175'
10000006.54 cmA additionals
Use 6 14 mm bars in each of the two boundary elements.
For zone 4:
mtmtMu .90.132.70.116 , i.e. boundary elements are required at wall ends
mtM u .20.1670.1169.132'
1000002.16 cmA additionals
Use 3 14 mm bars in each of the two boundary elements.
142
wl
c
mtmtMu .78.93.03.112 , i.e. no boundary elements are required at wall ends
For Zone 6
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
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