Page 244 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.107

Solution
2
Given: ′ f = 2000 lb/in. , length of support, c = 3 ft, length of span = 30 ft, height
m
of beam = 15 ft, width of beam = 11.625 in. (12 in. nominal), uniform load = 1500 +
3500 = 5000 lb/ft = 416.67 lb/in.
a. Check compressive stress at support.
⎡ ⎛ h ⎞ ⎤ ⎡ ⎛ ( 15))( )12 ⎞ 3 ⎤
3
2 2
.
(
.
F = 020 f ′ 1 − ⎥ = 0 20 2000 1 ⎢ − ⎜ ⎟ ⎥ = 387 lb/in.
)
m ⎢
a
11
⎣ ⎝ 48 t ⎠ ⎦ ⎣ ⎢ ⎝ (48 )( .625 )⎠ ⎦ ⎥
Actually compressive is calculated on area equal to the wall thickness times the
length of bearing plus twice the panel thickness.
33
Actual compressive stress = (5000 )( ) = 239 lb/in. 2 < F
5 [
(.625 3 12)( ) + 2 11 625)] a
.
(
11
= 387 lb/inn. 2 OK
Commentary: If we calculate the allowable compressive stress based on MSJC-08, its
value would be as follows:
The radius of gyration for a 12 in. nominal wall, r = 0.289t = 0.289 (11.625) = 3.36 in.
12
h ()()
15
= = 53 .57 < 99
r . 336
⎡ ⎛ h ⎞ ⎤ ⎡ ⎛ 53..57 ⎞ ⎤
2
2
F = 025 f ′ 1 − ⎥ = 0 25 2000 1 − ⎥ = 427 lb/in. 2 > 239 lb/in. 2 OK
⎢
.
)
m ⎢
. (
a r ⎠ ⎝ 140 ⎠
⎣ ⎝ 140 ⎦ ⎣ ⎦
b. Calculate flexural stress and the reinforcement requirement:
For continuous span beams, ε = c = 3 = 1
l 30 10
β = h = 15 = 1
l 30 2
⎛ w⎞
Flexural stress = (coefficient from Table 4.9) ⎜ ⎟
⎝ b ⎠
where
w = load per inch of wall = 416.67 lb/in.
From Table 4.9, for b = ½, Coefficient = +1.05 (top of beam)
Coefficient = −1.31 (bottom of beam)
⎛ w⎞ ⎛ 416 67⎞
.
2
Flexural stress (top) = 1.05 ⎜ ⎟ = 1.05 ⎜ ⎝ 11 625 ⎠ ⎟ = 38 lb/in. (compression)
b ⎠
⎝
.
Allowable compressive stress = 0.33 ′ f = 0.33 (2000)
m
2
2
= 660 lb/in. > 38 lb/ft OK
⎛ w⎞ ⎛ 416 67⎞
.
2
Flexural stress (bottom) = −1.31 ⎜ ⎟ = −1.31 ⎜ ⎟ = −47 lb/in. (tension)
⎝ b ⎠ ⎝ 11 625 ⎠
.