Page 415 - Design of Reinforced Masonry Structures

6.66 CHAPTER SIX

Solution
Estimate the area of masonry in compression. Assume that the depth of compression
block equals the face-shell thickness of the masonry unit. For nominal 8 × 8 × 16 in.
1
nominal concrete masonry unit, the face-shell is 1 / 4 in. thick.
1
1
t = 1 / 4 in. a = 1 / 4 in.
s
Effective width b is the smallest of
e
(a) Center-to-center of reinforcement = 32 in. (governs).
(b) Six times the nominal thickness of masonry unit = 6 × 8 = 48 in.
(c) 72 in.
Use b = 32 in.
e
Area of masonry in compression = b t = 32(1.25) = 40 in. 2
e s
Compression stress resultant:
′
.
C = 0 80. f ab = 0 80 2 0 40. ( . )( ) = 64 0 kips
m
Tensile force resultant,
T = A f = 0.60(60) = 36.0 kips < C = 64 kips
s y
1
Hence, the depth of the compression block is less than 1 / 4 in. (= t ) as assumed, and it
s
lies within the face-shell.
Accordingly, the section can be analyzed as a rectangular section. Check the capac-
ity of the section:
P + A f
M = u s y
n fb ′
080
.
m

.
.
(
P = 1 2 200.( ) +1 6 80) = 368 lb/ft = 0 368 k/ft
u
On a width of 32 in. (effective width),
⎛ 32 ⎞
P = 0 368. ⎜ ⎟= 098. kip
u
⎝12 ⎠
.
.
(
.
A = 0 98 + 0 60 60) = 072 in.
(20 32
080 . )( )
.
From Eq. (6.18),
⎛ a⎞
M = ( A f + P ⎜ ) d − ⎟
n s y u ⎝ 2 ⎠

×
8
09
M = (.06 60 + . ) ⎛ ⎜ .381 − . 072 ⎞ ⎟
n ⎝ 2 ⎠
= 127 .58 k-in. = 10 63 k-ft
.
.
The 10.63 k-ft moment is resisted by wall length of 32 in. (effective width). Therefore,
the nominal moment strength of wall per linear foot is
⎛ 12 ⎞
M = (. ) ⎜ ⎟ = . 3 99 k-ft
10
63
n ⎠
⎝ 32
fM = 0.9(3.99) = 3.59 k-ft per linear ft of the wall.
n