-
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