Page 181 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.45

Calculate the factored loads.
Load combinations:
1. U = 1.4D = 1.4(180) = 252 lb/ft
2. U = 1.2D + 1.6L = 1.2(180) + 1.6(1200) = 2136 lb/ft (governs)

w = 2.136 k/ft (governs)
u
.
(
M = wL 2 = 2 136 16) 2 = 68 35. k-ft
u
u
8 8
From Eq. (4.43)
φM = φ f bd 2 ω − .625
′
[
)]
(
0
1
m
n
Substitute the following into the above equation:
φM = M = 68.35 k-ft, b = 9.0 in., d = 20 in. (given)
n u
2
(68.35)(12) = 0.9[(2.5)(9)(20) w (1 − 0.625w)]
820.2 = 8100w (1− 0.625w)
2
0.625w – w + 0.10126 = 0
−
)
(.
(.
ω = 1 ± 1 4 0 625 0 10126) = 0 10864
.
2 0 625)
(.
f
ω = ρ y (4.40 repeated)
′ f
m
Therefore,
⎛ 60 ⎞
ρ
0 10864 = ⎜ ⎟
.
⎝ 25⎠
.
which yields r = 0.00453.
From Eq. (4.36)
A = rbd = 0.00453(9)(20) = 0.82 in. 2
s
Try one No. 9 bar, A = 1.0 in. 2
s
2
Calculate φM for the beam with A = 1.0 in. . From Eq 4.9, assuming that steel
n s
has yielded
Af
a = sy (4.9 repeated)
.
08 fb ′
m
(10 60. )( )
a = = 333 in.
.
080 (25. )(900. )
.