Civil Engineering Reference
In-Depth Information
3"
3 #10 (3.79 in.
2
)
14"
20"
3 #10 (3.79 in.
2
)
12"
3"
Figure 10.20
In Example 10.7 the nominal load that the short column of Figure 10.20 can support at
an eccentricity of 10 in. with respect to the
x
axis is determined. If we plot on the interac-
tion diagram the intersection point of
K
n
and
R
n
for a particular column and draw a straight
line from that point to the lower left corner or origin of the figure, we will have a line with
a constant
e
/
h
. For the column of Example 10.6
e
/
h
0.5. Therefore a line is plot-
ted from the origin through a set of assumed values for
K
n
and
R
n
in the proportion of 10/20
to each other. In this case,
K
n
was set equal to 0.8 and
R
n
10/20
0.4. Next a line
was drawn from that intersection point to the origin of the diagram as shown in Figure
10.16. Finally, the intersection of this line with
0.5
0.8
g
(0.0316 in this example) was deter-
mined, and the value of
K
n
or
R
n
was read. This latter value enables us to compute
P
n
.
EXAMPLE 10.7
Using the appropriate interaction curves, determine the value of
P
n
for the short tied column shown
in Figure 10.20 if
e
x
10
. Assume
c
f
4000 psi and
f
y
60,000 psi.
e
10
20
h
0.500
SOLUTION
g
(2)(3.79)
(12)(20)
0.0316
14
20
0.700
Plotting a straight line through the origin and the intersection of assumed values of
K
n
and
R
n
(say 0.8 and 0.4, respectively).
For
g
of 0.0316 we read the value of
R
n
P
n
e
R
n
c
A
g
h
0.24
f
P
n
(0.24)(4)(12
20)(20)
460.8k
10
When the usual column is subjected to axial load and moment, it seems reasonable
to assume initially that
0.65 for tied columns and 0.70 for spiral columns. It is to be
