Page 154 - Design of Reinforced Masonry Structures

4.18 CHAPTER FOUR

9.63'' (10'' nominal) Solution
For a 10-in. nominal lightweight CMU with a grout
3
weight of 140 lb/ft , the dead weight = 93 lb/ft depth (h) of
the beam (Table A.19).
Self-weight of the beam,

⎛ 40 ⎞
D = 93 = 310 lb/ft
⎝ 12 ⎠
40'' d = 34''
L = 1600 lb/ft
Load combinations:
1. U = 1.4 D = 1.4 (310) = 434 lb/ft
2#6 2. U = 1.2 D + 1.6 L = 1.2 (310) + 1.6 (1600)
= 2932 lb/ft > 434 lb/ft
FIGURE E4.2 Beam cross w = 2932 lb/ft (governs)
section for Example 4.2. u
Effective span = 15 ft 8 in. = 15.67 ft
(.
M = wL 2 = 2932 15 67) 2 = 89 994, lb-ft ≈ 90 0. k-ft
u
u
8 8
Calculate the nominal strength of the beam. Assume that tension reinforcement has
yielded (to be verified later).
2
A = 0.88 in. (two No. 6 bars, App. Table A.9), b = 9.63 in. (10 in. nominal)
s
Calculate the depth of compression block, a, from Eq. (4.9):
Af
a = sy (4.9)
080 fb ′ m
.
a = (088 60. )( ) = 457 in
.
.
.
080 (15. )(963. )
Calculate the nominal strength from Eq. (4.12):
⎛ a⎞
M = A f d − ⎟ (4.12)
s y ⎜
⎝
n
⎠ 2
The centroid of reinforcing bars is at 6 in. from the bottom, d = 40 – 6 = 34 in.
⎛
M = (. )088 60 34 − . 457 ⎞ ⎠
( )
⎝
n
2
= 1674 .5 kip-in = 139 .555 k-ft
Check the M / V d ratio, where d = d = 34 in. (see discussion in Section 4.10).

u
u v
v
2 932 15 67)
V = wL = (. ) ( . = 22 97 kips
u
.
u
2 2
d = d = 34 in.
v
M (90 )( )
12
u = = .138 > .10
Vd (22 .97 )(34 )
uv