Page 532 - Design of Reinforced Masonry Structures

7.94 CHAPTER SEVEN

With No. 3 bars at 16 in. on centers,
1
01
1
2
A = (. )( ) = .0825 in. > 0.064 in. /ft (required)
2
2
0
v
16
Therefore, horizontal reinforcement provided is satisfactory.
Check maximum allowable bar spacing: Maximum spacing of horizontal reinforce-
ment is specified to be (MSJC-08 Section 1.17.3.2.6) as the smaller of
1. One-third of the wall length,
812)
( ) (
s = L /3 = = 32 in.
w
3
2. One-third of the wall height,
18 12
s = h w = ()() = 72 in.
3 3
s = 48 in. (governs)
Spacing provided = 16 in., satisfactory.
Shear above the plastic hinge zone: The strength-level shear on the shear wall at a
height h v
V = 16.69 kips (calculated earlier)
u
The strength-level moment associated with V is
u
M = V (h − h )
u
u
v
= 16.69 (18 − 4) = 234 kip-ft
The strength-level axial load associated with V is
u
P = (1.2 + 0.2S )D
u
D
where D = diaphragm dead load plus weight of wall
P = [1.2 + 0.2 (0.925)](0.5 × 8 + 9.4)
u
= 18.6 kips
M 234
u = = 175 10. > . use 1.0 (maximum)
Vd v (. ) (
16 69 8)
u
In the wall segment above the plastic region, the shear strength equals the sum of the

contributions from masonry and reinforcement,
V = V + V s
m
n
⎡ ⎛ M ⎞ ⎤
⎢
V = 4 1 75− . ⎜ u ⎟ ⎥ A n f m ′ + . u
0 25P
n
⎣ ⎝ Vd v ⎠ ⎦
u
×
+
×
= (4 1− .75 1 . )( .63 96 )( 2000 ) 0 . ( ,600
25
)
18
0
7
= 78,354 lb
fV = 0.8(78,354) = 62,683 lb > V = 16,690 lb satisfactory
m
u