Page 411 - Design of Reinforced Masonry Structures

6.62 CHAPTER SIX

Because M < M , the wall is not cracked at the service load. Therefore, use
ser
cr
Eq. (6.21) to calculate deflection.
(. )(
ser
δ = 5Mh 2 = 518 82 20 ×12) 2 = . 014 in.
ser
(
48EI 48 1800)(443 . )
3
(
mg
Maximum allowable d = 0.007h = 0.007(20 × 12) = 1.68 in. > 0.14 in.
s
The service load deflection (0.14 in.) is much smaller than the permissible deflection
(1.68 in.). Hence, the service load deflection is OK without P-∆ effects.
Calculate P-∆ effects.
Additional M caused by d = P d = 1.81(0.14) = 0.252 k-in.
ser ser u ser
Calculate moments due to factored loads.
2
.
wL 2 ⎛ e ⎞ 0 042 20 )( 12) ⎛ 4 ⎞
(
M = u + P ⎜ ⎟ = +181. ⎜ ⎟
u
8 u ⎝ ⎠ 2 8 ⎝ ⎝ 2 ⎠

= 28 82 k-in. > M = 18 96 k-in.
.
.
cr
Because M > M , the wall is cracked; therefore, use Eq. (6.23) to calculate d .
u
cr
u
Calculate the moment of inertia of the cracked section, where I is given by Eq. (6.2).
cr
−
(
I = bc 3 + nA d c) 2 (6.23 repeated)
se
cr
3
Try No. 4 bar @ 24 in. on center, the reinforcement per unit width of wall,
2
A = 0.10 in. (Table A.23)
s
P + A f 1 81 + 0 10 60)
.
.
(
A = u s y = = 013. in. 2
se
f 60
y
E 29 000
,
n = s = = 16 11 (or from Table A.14)
.
E 1800
m
A 013
.
.
ρ = se = = 0 00284
bd ()( .
12 3 81)
rn = (0.00284)(16.11) = 0.046
From Table A.15, for nr = 0.046, k = 0.2608

c = kd = (0.2608)(3.81) ≈ 1.0 in. < 1.25 in. (shell thickness)
Hence, the centroidal axis lies in the shell and the section can be treated as a rectangular
section.
bc 3
I = + nA d c) 2
−
(
cr
3 se
()( 3
12 1 0 . )
)( .
= + 16 11 0 1(.)( . 333 81 1 0 . )− 2
3
= 20 54 in. 4
.