Page 410 - Design of Reinforced Masonry Structures

WALLS UNDER GRAVITY AND TRANSVERSE LOADS 6.61

Hence, seismic load governs.
Calculate service loads according to ASCE 7-05 provisions for ASD load combinations.
Load combinations to be considered are :
1. Dead load + roof live load (ASD Load Combination 3)
2. Dead load + wind load (ASD Load Combination 5a)
3. Dead load + 0.7 earthquake (ASD Load Combination 5b)
4. Dead load + 0.75(0.7E) + 0.75L (ASD Load Combination 6)
r
Service loads: Superimposed loads: P = 0.35 + 0.24 = 0.59 k/ft of wall
f
Factored loads:
P = 1.2D + 1.6L = 1.2 (0.35) + 1.6 (0.24) = 0.804 ≈ 0.80 k/ft of wall
f r
Wall weight at midheight P = 1.2 (0.084) (10) = 1.008 k/ft
uw
P = P + P = 0.80 + 1.008 = 1.81 k/ft = 1810 lb/ft
uf
uw
u
2
Lateral load on wall due to wind w = 1.6 (20) = 32 lb/ft of wall area
u
2
2
Lateral load on wall due to earthquake w = 42 lb/ft of wall area > 32 lb/ft (wind)
u
Therefore, seismic lateral load governs.
Area of cross section of 1-ft length of wall:
Wall thickness t = 7.625 in. (8 in. nominal)
A = 12(7.625) = 91.5 in. 2
g
Moment of inertia of gross section,
12 7 625)
4
I = bt 3 = ()( . 3 = 443 3.in.
g
12 12
Check PA and compare it with 0 05. ′ f .
u g m
P 1810
2
.
u = = 20 lb/in
.
A g 91 5
.
005 ′ f = 0.05(2000) = 100 lb/in. 2
m
PA < 0 05. ′ f , hence, special conditions do not apply.
u g m
Calculate deflection, d . First, check if M is less than or greater than M .
ser ser cr

⎛
4 ⎞
e ⎞
u
.
M = wL 2 + P ⎜ ⎟ = 0 294 ⎛ 20 2 ⎞ ⎟ + 059. ⎛ ⎜ ⎟ = 18 82 k-in.
⎜
.
ser u ⎠ ⎝ 2 2 ⎠
8 ⎝ ⎠ 2 ⎝ 8
M = f S
cr r
2
f = 163 lb/in. (MSJC-08 Table 3.1.8.2)
r
12 7 625)
S = bd 2 = ()( . 2 = 116 3.in.
3
6 6
M = (0.163)(116.3) = 18.96 k-in. > M = 18.82 k-in.
cr
ser