Civil Engineering Reference
In-Depth Information
Value of
falls between
values for Graphs A.3 and A.4. Therefore interpolating between the
two as follows:
0.70
0.75
0.80
g
0.0220
0.0202
0.0185
A
s
g
bh
(0.0202)(14)(20)
5.66 in.
2
Use 6 #9 bars
6.00 in.
2
Notes
(a)
Note that
0.65 as initially assumed since the graphs used show
f
s
f
y
is
1.0 and thus
t
0.002.
(b)
Code requirements must be checked as in Example 9.1. (See Figure 10.25 to understand.)
EXAMPLE 10.4
c
Design a short square column for the following conditions:
P
u
600 k,
M
u
125 ft-k,
4000
f
psi, and
f
y
60,000 psi. Place the bars uniformly around all four faces of the column.
c
SOLUTION
Assume the column will have an average compression stress
about
0.6
f
2400 psi.
600
2.400
250 in.
2
A
g
required
Try a 16
16-in. column (
A
g
256 in.
2
) with the bar arrangement shown in Figure 10.18.
e
M
u
P
u
(12)(125)
2.50
600
P
n
P
u
600
0.65
923.1 k
P
n
923.1
(4)(16
K
n
c
A
g
16)
0.901
f
P
n
e
(600)(2.50)
(4. 0)(16
16)(16)
R
n
c
A
g
h
0.0916
f
11
16
0.6875
Interpolating between values given in Graphs A.6 and A.7.
0.600
0.6875
0.700
g
0.0240
0.0227
0.0225
A
s
(0.0227)(16)(16)
5.81 in.
2
Use eight #8 bars
6.28 in.
2
