Page 531 - Design of Reinforced Masonry Structures

SHEAR WALLS 7.93

2
′ f = 2000 lb/ft , and f = 60 ksi. Determine the horizontal and vertical reinforcement
m y
requirements for this wall.
Solution
Given:
Length of the shear wall, L = 8 ft = 96 in.
w
Thickness of shear wall t = 7.625 in. (8-in. nominal)
Story height = 18 ft
According to IBC 2006 Section 2106.5.2, the segment of the wall from its base to
L = 8 ft above the base is to be provided entirely by the reinforcement. The nominal
w
shear strength in this region is given by
V = V = A r f
n
n n y
s
V = total design factored shear calculated at a height above the base equal to the
s
smaller of
h = h s = 18 = 9ft
v
2 2
or h = L w = 8 = 4ft
v
2 2
h = 4 ft (governs)
v
Weight of the wall above h = 4 ft:
v
W wall = (18 − 4)(8)(84) = 9408 ≈ 9.4 kips

The seismic force due to weight of wall
V = C W = 0.18 (9.4) = 1.69 kips
w
s
w
V = V + V = 15 + 1.69 = 16.69 kips
w
w
d
Provide No. 3 Grade 60 bars at 16-in. on centers vertically. From Eq. (7.79)
V = A r f y
n n
n
= L f A /s
v
w y
= (96)(60)(0.11)/16
= 39.6 kips
The design shear strength of the wall is not less than 1.25 times the shear correspond-
ing to its nominal flexural strength. Therefore, the strength reduction factor for shear,
f = 0.8
fV = (0.8)(39.6) = 31.68 kips > V = 16.69 kips (satisfactory)
u
n
The minimum horizontal steel area is specified to be 0.0007A (MSJC-08
g
Section 1.17.3.2.6).
A v, min = 0.0007A g
2
= 0.0007(7.625)(12) = 0.064 in. /ft