Page 240 - Design of Reinforced Masonry Structures

DESIGN OF REINFORCED MASONRY BEAMS 4.103

Wall thickness = 8"
h = 24'

2' 2'

L = 24'

FIGURE E4.28 Wall panel supported over bearing pads.

Solution
2
Given: ′ f = 1500 lb/ft , wall thickness, b = t = 7.625 in. (8 in. nominal), wall
m
height, h = 24 ft, effective span L = 24 ft, length of support = 2.0 ft., dead weight of
wall = 100 lb/ft 2
Gravity load on wall = 2500 lb/ft
Maximum permitted bearing length of supports
= 0.1 (distance between vertical supporting edge members)
= 0.1(24)
= 2.4 ft > 2.0 ft OK

Allowable stress design approach (Ref. 4.18):
a. Calculate allowable compressive stress:
⎡ ⎛ h ⎞ ⎤ ⎡ ⎛ ( 24))( )12 ⎞ 3 ⎤
3
m ⎢
2
F = 020 f ′ 1 − ⎥ = 0 20 1500 1 ⎢ − ⎜ ⎟ ⎥ = 154 lb/ft
)
.
(
.
a t ⎠ ⎝ )⎠
⎣ ⎝ 48 ⎦ ⎣ ⎢ (48 )( .625 ⎦ ⎥
7
Actual compressive stress is calculated on area equal to the wall thickness times the
length of bearing plus twice the panel thickness.
(
wL 2500 24)
Reaction at supports = = = 30 000 lb
2 2
,
,
Actual compressive stress = 30 000 = 144 lb/ft 2 < F = 154 lb/ft 2 OK
4
(. [12 + 2 7 625) ] a
7 625)
(.
Commentary: If we calculate the allowable compressive stress based on MSJC-08, its
value would be as follows:
The radius of gyration for an 8 in. nominal wall,
r = 0.289t = 0.289 (7.625) = 2.2 in.
12
h (24 )( )
= = 130 .91 > 99
r . 22