Page 305 - Design of Reinforced Masonry Structures

COLUMNS 5.25

h/t = 20(12)/16 = 15 < 30 OK
A = 15.625 (15.625) = 244 in. 2
n
b = t = 15.625 in.
r = 0.289t = 0.289(15.625) = 4.52 in.
12
h (20 )( )
= = 53 .1 < 99
r . 452
h
For ≤ 99,
r
2
⎤
⎡ ( ) ⎥ ( ) 2
h
53 1 .
C = 1 − =− = 08566 (5.13 repeated)
1
.
⎢
P ⎣ 140 r ⎦ 140
Alternatively, C can be found directly from Table A.16 as follows:
P
)( )
h
= (20 12 = 15 .36
t 15 .625
By interpolation from Table A.16,
C = 0.862 – (0.862 – 0.843)(0.36) = 0.855
P
From Eq. (5.12)
′
(
φP = φ(.080 )[.080 f A − A + f A C
)
]
n m n st y st P
620 = (0.9)(0.80)[0.80 (1.8)(244 – A ) + (60)(A )](0.856)
st
st
A = 11.18 in. 2
st
Try eight No. 11 bars,
2
2
A = 12.5 in. > A st, reqd = 11.18 in. OK
st
(Note: Bars larger than No. 9 are not permitted by MSJC Code for strength design of
masonry. However, No. 11 bars have been used here solely for illustrative purposes.
Check longitudinal reinforcement compliance with the code.
.
ρ = A st = 12 5 = 0 0512. > ρ max = 0 04. NG
A 244
n

The longitudinal reinforcement provided is greater than permitted by the code. This indi-
cates that a larger column size is required (or higher strength masonry needs to be specified
for the column see Example 5.6). The next larger square CMU column would be a nominal
24 × 24 in. Calculate the area of longitudinal reinforcement required for this column.
For a nominal 24 × 24 in. CMU,
A = 23.625(23.625) = 558 in. 2
n
h = 20 ft b = t = 23.635 in.
h/t = 20(12)/24 = 10 < 30 OK
r = 0.289t = 0.289(23.625) = 6.83 in.