Civil Engineering Reference
In-Depth Information
Figure 10.6
M
0 about tensile steel
(623.7)(9.50)
M
n
(582.62)(15.38)
(169.8)(19.00)
0
M
n
6261.3 in.- k
521.8 ft- k
In this manner, a series of
P
n
and
M
n
values are determined to correspond with a
strain of
0.003 on the compression edge and varying strains on the far column edge. The
resulting values are plotted on a curve as shown in Figure 10.8.
A few remarks are made here concerning the extreme points on this curve. One
end of the curve will correspond to the case where
P
n
is at its maximum value and
M
n
is zero. For this case,
P
n
is determined as in Chapter 9 for the axially loaded column of
Example 10.2.
c
(
A
g
A
s
)
A
s
f
y
P
n
0.85
f
(0.85)(4.0)(14
24
6.00)
(6.00)(60)
1482 k
On the other end of the curve,
M
n
is determined for the case where
P
n
is zero. This is
the procedure used for a doubly reinforced member as previously described in Chapter 5.
For the column of Example 10.2,
M
n
is equal to 297 ft-k.
A column normally fails by either tension or compression. In between the two ex-
tremes lies the so-called balanced load condition, where we may have a simultaneous ten-
sion and compression failure. In Chapter 3 the term
balanced section
was used in
referring to a section whose compression concrete strain reached 0.003 at the same time
as the tensile steel reached its yield strain at
f
y
/
E
s
. In a beam, this situation theoretically
occurred when the steel percentage equaled
b
.
For columns the definition of balanced loading is the same as it was for beams—that
is, a column that has a strain of 0.003 on its compression side at the same time that its ten-
