Page 299 - Design of Reinforced Masonry Structures

COLUMNS 5.19

h = 20 ft b = t = 18.0 in.
r = 0.289t = 0.289(18.0) = 5.2 in.
h (20 )( )
12
= = 46 .15 < 99
r . 52
h
For ≤ 99,
r
2
⎡ ( ) ⎥ ( ) 2
⎤
.
h
46 15
C = 1 − =− = 08 .991 (5.13 repeated)
1
⎢
P ⎣ 140 r ⎦ 140
Alternatively, C can be found directly from Table A.16 as follows:
P
h (20 )( )
12
= = 13 .33
t 18
By interpolation from Table A.16,
C = 0.897 – (0.897 – 0.880)(0.33) = 0.891
P
From Eq. (5.11)
′
]
(
)
φP = φ(.080 )[.080 f A − A + f A C
n m n st y st P
= 0.9(0.80)[0.80 (2.5)(324 – 6.28) + (60)(6.28)](0.891)
= 649 kips
Check loads on column. Calculate factored loads.
Load Combinations:
1. U = 1.4D = 1.4(250) = 350 kips
2. U = 1.2D + 1.6L = 1.2(250) + 1.6(200) = 620 > 350 kips
P = 620 kips (governs)
u
fP = 649 kips > P = 620 kips OK
n
u
Check the longitudinal reinforcement compliance with the code.
.
ρ = A st = 628 = 0 0194.
A n 324
r max = 0.04 r min = 0.0025 r provided = 0.0194
0.0025 < 0.0194 < 0.04 OK

The longitudinal reinforcement complies with the code requirements.
fP = 649 kips > P = 620 kips
n
u
The column can support the imposed service loads.
Example 5.4 Nominal strength of a rectangular CMU column.
A nominal 16 × 24 in. CMU column having an effective height of 20 ft is reinforced
with six No. 9 Grade 60 bars (Fig. E5.4). It carries a service dead load of 150 kips and
a service live load of 225 kips. ′ f = 1800 psi. Determine fP for the column and check
m n
if the column can support the imposed service loads.