Page 407 - Design of Reinforced Masonry Structures

6.58 CHAPTER SIX

Because M < M , the wall is not cracked at the service load condition. Therefore,
ser
cr
use Eq. (6.22) to calculate deflection.
(. )(
ser
δ = 5Mh 2 = 515 18 20 ×12) 2 = . 0 114 in.
ser
(
(
48EI 48 1800)(443 . )
3
mg
Maximum allowable d = 0.007h = 0.007(20 × 12) = 1.68 in. > 0.114 in.
s
The service load deflection (0.136 in.) is much smaller than the permissible deflec-
tion (1.68 in.). Hence, the service load deflection is OK without P-∆ effects. Calculate
P-∆ effects.
Additional M caused by d = P d = 1.328(0.114) = 0.15 k-in.
ser
ser
ser
u
Calculate moments due to factored loads.
⎛
2
e ⎞
(
(
.
M = wL 2 + P ⎜ ⎟ = 0 042 20 12) + 032. ⎛ 4 ⎞
⎜ ⎜ ⎟
u
u u ⎝ 2 ⎠
8 ⎝ ⎠ 2 8

+
.
= 25 2 0 64 = 25 84 k-in. > M cr = 18 96 k-in.
.
.
.
Because M > M , the wall is cracked; therefore, use Eq. (6.23) to calculate d .
u
u
cr
Calculate the moment of inertia of the cracked section, I , given by Eq. (6.24):
cr
−
2
(
I = bc 3 + nA d c)
cr
se
3
For No. 6 bar @ 24 in. on center, the reinforcement per unit width of wall,
2
A = 0.20 in. (Table A.23)
s
P + A f 1 328 + 0 20 60)
.
(
.
A = u s y = = 0 222. in. 2
se
f y 60
E 29 000
,
n = s = = 16 11. (or from Table A.14)
E 1800
m
.
.
ρ = A se = 0 222 = 0 0048
bd ()( .
12 3 81)

rn = (0.0048)(16.11) = 0.078
k = (ρ n + ρ n − ρ n
2
)
2
2
= ( . 0 078 ) + ( . 0 078 ) − . 0 078
2
= . 0 3255
(Alternatively, from Table A.15, nr = 0.078 and k = 0.3246 ≈ 0.325 in.)
c = kd = (0.325)(3.81) = 1.24 in. < 1.25 in. (shell thickness)