Page 186 - Design of Reinforced Masonry Structures

4.50 CHAPTER FOUR

Assume r = 0.50r . For ′ f = 4000 psi and Grade 60 reinforcement, 0.50r = 0.0117
b
b
m
(Table A.11). From Eq. (4.40)
′ f ⎛ 60 ⎞
ω = ρ y = 0 0117 = 0 1755.
.
′ f ⎝ 4 ⎠
m
[
)(
.
φk = 0 9. (4000 00 1755 1. ) − ( 0 625 0 1755)]
)
.
(
n
= 562 5 psi
.
Alternatively, we could have determined f k from Table A.13. For r = 0.0117 and
n
′ f = 4000 psi, from interpolation,
m
f k = 533 + (575 − 533)(0.7) = 562.4 psi
n
From Eq. (4.47),
⎛ bd 2 ⎞ ⎛ (. ) ( 2
763 20) ⎞
φM = φk n ⎜ ⎟ = 562 4( . ) ⎜ ⎟ = 143 .04 k-ft
n ⎠ ⎝ 12,,000 ⎠
⎝12 000,
From Eq. (4.48)
.
,
,
(
bd = 12 000 M u = 12 000 63 05) = 1345 in.
3 3
2
φ k n 562 4
.
with b = 7.63 in.,
d = 1345 = 13 3 in.
.
.
763
With 4-in. cover, h reqd = 13.3 + 4 = 17.3 in. Try a nominal 8 × 24 in. beam. With the
centroid of reinforcement at 4 in. from the tension face of the beam, d = 24 – 4 = 20 in.
Calculate the required amount of reinforcement assuming d = 20 in. From Eq. (4.13)
⎛ a⎞
M =φ M = φ A f d − ⎟ ⎠ 2
s y ⎜
⎝
n
u
⎛ a⎞
Assume d − ⎟ ≈095. d [Eq. (4.78)] and estimate A .
⎜
⎝
s
⎠ 2
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. , or two No. 5 bars, A = 0.88 in. . Of these two
s
s
choices, try one No. 8 bar (a better choice). Calculate φM with d = 20 in. and A =
n s
2
0.79 in. . From Eq. (4.9)
Af (. ) (
079 60)
a = sy = = 1.994 in.
.
080 fb ′ m 080 40 763)
.
(
.
)
(
.