Civil Engineering Reference
In-Depth Information
f
s
f
y
By interpolation between Graphs A.11 and A.12
g
is found to equal 0.0235 and
1.0.
A
g
(0.0235)(314)
7.38 in.
2
Use 8 #9 bars
8.00 in.
2
Same notes as for Examples 10.3 and 10.4.
In Example 10.6 it is desired to select a 14-in. wide column with approximately 2%
steel. This is done by trying different column depths and then determining the steel per-
centage required in each case.
EXAMPLE 10.6
Design a 14-in. wide rectangular short tied column with bars only in the two end faces for
P
u
500
c
k,
M
u
250 ft-k,
f
4000 psi, and
f
y
60,000 psi. Select a column with approximately 2% steel.
e
M
u
P
u
(12)(250)
SOLUTION
6.00
500
P
n
P
u
500
0.65
769.2 k
Trying several column sizes (14
20, 14
22, 14
24) and determining reinforcing.
Trial sizes
14
20
14
22
14
24
P
n
K
n
0.687
0.624
0.572
c
A
g
f
P
n
e
R
n
0.206
0.170
0.143
c
A
g
h
f
h
2
2.50
h
0.750
0.773
0.792
g
by interpolation
0.0315
0.0202
0.0109
Use 14
22 column
A
g
(0.0202)(14
22)
6.22 in.
2
Use 8 #8 bars
6.28 in.
2
Same notes as for Examples 10.3 and 10.4.
One more illustration of the use of the ACI interaction diagrams is presented with
Example 10.7. In this example, the nominal column load
P
n
at a given eccentricity which
a column can support is determined.
With reference to the ACI interaction curves, the reader should carefully note that the
value of
R
n
(which is ) for a particular column, equals
e
/
h
times the value of
K
n
for that column. This fact needs to be understood when the user desires to de-
termine the nominal load that a column can support at a given eccentricity.
c
A
g
h
P
n
e
/
f
c
A
g
)
(
P
n
/
f
