Page 300 - Design of Reinforced Masonry Structures

5.20 CHAPTER FIVE

5"
15
8

5"
23 h = 20'
8

6#9
FIGURE E5.4

Solution
2
Given: CMU column nominal 16 × 24 in., h = 20 ft, ′ f = 1800 psi, A = 6.0 in. (six No. 9
m st
Grade 60 bars), D =150 kips, L = 225 kips.
Check dimensions and h/t ratio for code compliance.
Nominal column width = 16 in. > 8 in. OK
Nominal column depth = 24 in. < 3(16) = 48 in. OK
h/t = 20(12)/16 = 15 < 30 OK

For a nominal 16 × 24 in. CMU,
A = (15.625)(23.625) = 369 in. 2
n
2
A = 6.0 in. (six No. 9 bars)
st
h = 20 ft t = 15.625 in.
r = 0.289t = 0.289(15.625) = 4.515 ≈ 4.52 in.
12
h (20 )( )
= = 53 .1 < 99
r . 452
h
For ≤ 99,
r 2 2
⎡
⎤
53 1 .
h
=−
1
C = 1 − ( ) ⎥ ( ) = 08 .556 (5.13 repeated)
⎢
P ⎣ 140 r ⎦ 140
Alternatively, C can be found directly from Table A.16 as follows:
P
h
12
= (20 )( ) = 15 .36
t 15 .625
By interpolation from Table A.16,
C = 0.862 – (0.862 – 0.843)(0.36) = 0.855
P