Page 538 - Design of Reinforced Masonry Structures

7.100 CHAPTER SEVEN

Solution
Given:
Length of the shear wall L = 18 ft. 8 in. = 224 in.
w
Thickness of shear wall t = 7.625 in. (8-in. nominal)
The depth of compression block a would be determined by an iterative procedure.
There are 10 vertical bars in the wall. Assume that one bar (nearest the compression face
of wall) is in compression and other nine bars are in tension, and assume also that all
nine bars have yielded so that the stress in each is f (= 60 ksi).
y
Trial 1: Ignore compression force in the reinforcing bar in the compression zone of
the wall as well as the area displaced by the compression reinforcing bars.
′
C = . 080 f at
max m
= . 0 80 ( . )( .625 a ) = 12 .2 a kips
7
0
2
For 9 No. 6 bars,
A = 3.98 in. 2
s
T max = A f = 3.98 (60)
s y
= 3.98 (60) = 238.8 kips
Equating C to T , we have
max max
12.2a = 238.8 kips
a = 238 8 . = 19 57 kips
.
12 2 .
The calculated value of a = 19.5 in. ignores the compression force in the bar nearest
the compression face. Obviously, the correct value of a would be smaller than 19.57 in.
Trial 2: Assume a = 18 in.
c = a = 18 = 22 5.in.
080 080
.
.
Calculate forces in masonry and reinforcing bars based on compatibility of strains
as shown in Fig. E7.17B.
⎛ 18 5 ⎞
.
ε = ⎜ ⎟ (. ) 0 00206 < ε = . f
0 0025 = .
0 002007 ( < f )
s1 ⎝ 22 5 ⎠ y s y
.
15 ⎞
⎛ .
ε = ⎜ ⎟ (. ) 0 00017 < ε = .
0 002077 ( f < f )
0 0025 = .
22 5 ⎠
s2 ⎝ y s y
.
22.5'' 1.5'' 24''
4''
e s3
e s2
e m = 0.0025 e s1
FIGURE E7.17B Strain distribution for Trial 2 iteration.