Page 542 - Design of Reinforced Masonry Structures

7.104 CHAPTER SEVEN

C
108'' L 108''
4''
C 6.38K
C s 26.4 kips
8'' 28'' 24'' 24'' 24'' 12'' 12'' 24'' 24'' 24'' 24''
4''
FIGURE E7.17E Force diagram.

Check that M ≥ 1.3 M
n cr
M = S f
cr n r
Thickness of wall = 7.625 in.
Length of wall L = 18 ft 8 in. = 224 in.
w
7 625 224)
S = tL 2 w = (. )( 2 = 63 675, in. 3
n
6 6
2
f = 163 lb/in. (MSJC-08 Table 3.1.8.2, Type M or S Portland cement/lime or mortar
r
cement for flexural tensile stress normal to bed joints in fully grouted masonry in run-
ning or stack bond)
63
M = (,765 )(163 ) = 860 .8 k-ft
cr
12 ,000
1.3M = 1.3(860.8) = 1119 k-ft
cr
M = 2297 k-ft > 1.3M = 1119 k-ft
n cr
fM = 0.9(2297) = 2067 k-ft
n
Check strain in the extreme tension reinforcing bars.
e m = 0.0025
c = 20 in. d − c = 220 − 20 in.

20'' From strain distribution diagram (Fig. E7.17F),
−
ε s = dc = 200 = 10
ε m c 20
e = 10 e = 10 (0.0025) = 0.025
m
s
200''
4e = 4(0.00207) = 0.00828
y
e > 4e y
s
The percentage of steel in the wall is
e s
⎛ 442 ⎞
.
.
FIGURE E7.17F Strain dis- ρ = ⎜ ⎟ ⎠ ( 100) = 0 259 percent
)( .
(
tribution diagram. ⎝ 224 7 65)