Page 314 - Design of Reinforced Masonry Structures

5.34 CHAPTER FIVE

and live load are 100 and 175 kips, respectively. Assuming ′ f = 2500 psi, calculate the
m
moment-carrying capacity of the column about its minor axis (axis parallel to the long
side of the column).

Solution
Given: CMU column nominal 16 × 24 in., D = 100 kips, L = 175 kips, ′ f = 1800 psi.
m
2
f = 60 ksi, A = 2.41 in. (four No. 7 bars), h = 24 ft.
y
st
P = 400 kips (see calculations in Example 5.8).
u
Check dimensions and h/t ratio for code compliance.
Nominal column width = 16 > 8 in. OK
Nominal column depth = 24 in. < 3(16) = 48 in. OK
h/t = 24(12)/16 = 18 < 30 OK
For a nominal 16 × 24 in. CMU,
A = (15.625)(23.625) = 369.14 in. 2
n
h = 24 ft t = 15.635 in.
r = 0.289t = 0.289(15.625) = 4.516 in.
h (24 12
)( )
= = 63 .77 < 99
r . 4 516
h
For ≤ 99,
r
⎤
⎡ ( ) ⎥ ( ) 2
2
63 77
.
h
1
C = 1 − =− = 0.7793 (5.13 repeated)
⎢
P ⎣ 140 r ⎦ 140
Alternatively, C can be found directly from Table A.16 as follows:
P
12
h (24 )( )
= = 18 .43
t 15 .625
By interpolation from Table A.16,
C = 0.802 – (0.802 − 0.779)(0.43) = 0.792
P
From Eq. (5.12)
′
]
φP = φ(.080 )[.080 f A − A + f A C
(
)
y
P
st
st
n
m
n
= (0.9)(0.80)[0.80 (2.5)(369.41 – 2.41) + (60)(2.41)](0.793)
= 501.3 kips > P = 400 kips
u
Because fP (= 501.3 kips) is greater than P (= 400 kips), the column can carry
n
u
some bending moment.