Page 408 - Design of Reinforced Masonry Structures

WALLS UNDER GRAVITY AND TRANSVERSE LOADS 6.59

Hence, the neutral axis lies in the shell, and the section can be treated as a rectangular
section.
−
I = bc 3 + nA ( d c) 2
cr se
3
12 1 24)
= ()( . 3 + 16 11 0.2222 3 81 1 24)− . 2
)( .
(.)(
3
= 31 25 in 4
.
First iteration:
)
δ = 5 Mh 2 + 5( M ser − M h 2
cr
cr
u
48 EI 48 EI
mg m cr
−
×
(
(.
)
(
.
.
= 518 96620 12)( × ) 2 + 5 2584 1896 20012) 2
48 1800 443 3) 48 1800 31 25)
(
.
(
(
)( .
)
0 143 0 734
= . + .
.
= . < h = 168 in. OK
0 877 in. < 000.
Calculate moment due to new d .
u
M = 25.84 + 1.328(0.877) = 27.0 k-in.
u2
The new value of moment, 27.0 k-in., is very close to the first value of 25.84 k-in.;
therefore, further iteration is not necessary. Calculate the nominal moment, M .
n
a ⎞
M = ( A f + P ⎜ ) ⎛ ⎝ d − ⎟ ⎠ 2 (6.18, MSJC-08/3-28)
s y
n
u
where
P + A f 1 328 + 0 20 60)
(
.
.
a = u s y = = 0 694 in. < 1.25 in. (shell thickness)
.
.
008 fb ′ m 080 20 12)
2
.
(
.
)
(
Hence, the compression block lies in the shell, and the section can be treated as a
rectangular section.
⎛ a ⎞
M = ( A f + P ⎜ ) ⎝ d − ⎟ ⎠ 2
n
u
s y
⎛ 0 694 ⎞
.
] ⎜
= [( .20 )(60 ) + .328 3 .881− ⎟
1
0
⎝ 2 ⎠
= 46 15 k-in.
.
fM = (0.9)(46.15) = 41.54 k-in. > M = 27.0 k-in. OK
u2
n
Therefore, the wall is adequately reinforced.