Page 303 - Design of Reinforced Masonry Structures

COLUMNS 5.23

Calculate factored loads.
Load Combinations:
1. U = 1.4D = 1.4(200) = 280 kips
2. U = 1.2D + 1.6L = 1.2(200) + 1.6(300) = 720 > 280 kips
P = 720 kips (governs)
u
fP = 720 kips
n
For a nominal 24 × 24 in. CMU,
A = 23.625 × 23.625 = 558 in. 2
n
h = 28 ft b = t = 23.635 in.
r = 0.289t = 0.289(23.625) = 6.83 in.
)( )
h (28 12
= = 40 .19 < 99
r . 683
For h ≤ 99,
r
⎤
⎡ ( ) ⎥ ( ) 2
2
40 19
.
h
1
C = 1 − =− = 08 .776 (5.13 repeated)
⎢
P ⎣ 140 r ⎦ 140
Alternatively, C can be found directly from Table A.16 as follows:
P
)( )
h (28 12
= = 14 .22
t 23 .625
By interpolation from Table A.16,
C = 0.880 – (0.880 – 0.862)(0.22) = 0.876
P
From Eq. (5.12)
φP = φ(.080 )[.080 f A − A + f A C P
′
]
(
)
y
n
st
m
n
st
720 = (0.9)(0.08)[0.80 (1.5)(558 – A ) + (60)(A )](0.876)
st
st
A = 8.03 in. 2
st
Try eight No. 9 bars,
2
A = 8.0 in. ≈ 8.03 in. 2
st
Check longitudinal reinforcement compliance with the code.
ρ = A st = 80
.
= 0 0143.
.
A
n 558 0
r max = 0.04 r min = 0.0025 r provided = 0.0143
0.0025 < 0.0143 < 0.04 OK
The longitudinal reinforcement complies with the code requirements.
Lateral ties: Provide /8-in.-diameter lateral ties. Tie spacing should be the smallest of
3
1. Sixteen times the of longitudinal bar diameter = 16 (1.0) = 16 in. (governs).
3
2. Forty eight times the lateral tie diameter = 48( /8) = 18 in.
3. Least column cross-sectional dimension = 24 in.