Page 184 - Design of Reinforced Masonry Structures

4.48 CHAPTER FOUR

,
bd = 12 000 M u
2
φ k
n
.
(
,
= 12 000 63 05) = 2710 5 iin. 3
.
279 14
.
.
.
d = 2710 5 = 2710 5 = 18 85 in.
.
b 763
.
With the assumed value of total beam depth, h, equal 24 in., the design depth
d would be 24 − 4 in. (cover) = 20 > 18.85 in. So, d = 20 in. as assumed is
adequate. Calculate the required amount of reinforcement based on d = 20 in.
From Eq. (4.13)
⎛ a⎞
φM = φA f d − ⎟
s y ⎜
n ⎝ ⎠ 2
⎛ a⎞
Assume d − ⎟ ≈095. d [Eq. (4.78)] and estimate the area of tension reinforce-
⎜
⎝
ment, A . s ⎠ 2
φM = M = 63 05 k-ft
.
n u
(63.05)(12) = 0.9A (60)[(0.95)(20)]
s
A = 0.74 in. 2
s
2
2
Try one No. 8 bar, A = 0.79 in. . Calculate M with d = 20 in. and A = 0.79 in. .
s u s
From Eq. (4.9)
Af (. ) (
079 60)
a = sy = = 3.888 in.
)
.
(
080 fb ′ 080 20 763)
(
.
.
.
m
From Eq. (4.13),
φM = φA f ⎛ d − a ⎞
n s y ⎝ ⎠ 2
⎛
.
)( )
= 09 . (079 60 20 − 388 ⎞ ⎠
.
⎝
= = 770 4 k-in. = 64.2 k-ft 2
.
φM = 64 2 k-ft > M = 63 05 k-ft OK
.
.
n u
Check the M / V d ratio.
u v
u
3 142 12 67)
V = wL = (. )( . = 19 9 . kips
u
u
2 2
63
M (.005 12)( )
>
u = = 19 10
.
.
19 9 20)
Vd (. ) (
u v