Page 183 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.47

7.63'' (8'' nominal) Examples 4.15 Determination of beam depth and
reinforcement when only the beam width b is known.
An 8-in. wide CMU lintel with a clear span of 12 ft has to carry
a superimposed dead load of 850 lb/ft and a live load of 1200 lb/
ft. Use ′ f = 2000 psi and Grade 60 reinforcement (Fig. E4.15).
m
Determine the depth of the lintel and the area of reinforcement.
d
h The masonry would be built from normal weight CMU with a
3
grout weight of 140 lb/ft . Assume that length of bearing on each
A s support is 8 in.
Solution
FIGURE E4.15 Beam
cross section for Example Clear span = 12 ft
4.15.
Assuming the bearing width of 8 in. on each side of the beam,
Effective span = 12 ft + 8 in. = 12.67 ft
Assume h = 24 in. and the centroid of reinforcement as 4 in. from the bottom of the
CMU.
d = 24 – 4 = 20 in.
3
For normal weight 8-in. wide CMU with a grout weight of 140 lb/ft
Dead weight of lintel masonry = 84 lb/ft depth of the lintel (Table A.19)
⎛ 24 ⎞
Self-weight of the lintel = 84 ⎜ ⎟ = 168lb/ft
⎝ 12 ⎠
Total dead weight, D = 168 + 850 = 1018 lb/ft
Load combinations:
1. U = 1.4D = 1.4(1018) = 1425 lb/ft
2. U = 1.2D + 1.6L = 1.2(1018) + 1.6(1200) = 3142 lb/ft >1425 lb/ft
w = 3142 lb/ft = 3.142 k/ft (governs)
u
.
.
u
M = wL 2 = (3 142 )(12 67 ) 2 = 63 05. k-ft
u
8 8
For an 8-in. wide CMU, b = 7.63 in. From Eq. (4.48)
12 000 M
,
2
bd = u (M , k-ft, b and d in inches)
φ k u
n

From Eq. (4.45)
φ f ′
(
1
0
φk = [ m ω − .625 ω)]
n
Assume r = 0.5r b
′
For f = 2000 psi and Grade 60 reinforcement, 0.50r = 0.0058 (Table A.11).
m
b
From Eq. (4.40)
f ⎛ 60 000, ⎞
ω = ρ y = 0 0058 = 0 174.
.
′ f ⎝ 2000 ⎠
m
−
x
0
φk = 09.[((2000 )( .174 )(1 0 .625 0 .174 )]
n
= 279 .14 psi