Page 162 - Civil Engineering Formulas

COLUMN FORMULAS 99

The axial-load capacity P u kip (N), of short, rectangular members subject to
axial load and bending may be determined from
P u (0.85 f c ba A s f y A s f s ) (3.34)
a
P u e 0.85 f c ba d 2 A s f y (d d ) (3.35)

where e eccentricity, in (mm), of axial load at end of member with respect to
centroid of tensile reinforcement, calculated by conventional methods
of frame analysis
b width of compression face, in (mm)
a depth of equivalent rectangular compressive-stress distribution, in
(mm)
2
2
A s area of compressive reinforcement, in (mm )
2
2
A s area of tension reinforcement, in (mm )
d distance from extreme compression surface to centroid of tensile
reinforcement, in (mm)
d distance from extreme compression surface to centroid of compres-
sion reinforcement, in (mm)
f s tensile stress in steel, ksi (MPa)
The two preceding equations assume that a does not exceed the column
depth, that reinforcement is in one or two faces parallel to axis of bending, and
that reinforcement in any face is located at about the same distance from the
axis of bending. Whether the compression steel actually yields at ultimate
strength, as assumed in these and the following equations, can be verified by
strain compatibility calculations. That is, when the concrete crushes, the strain
in the compression steel, 0.003 (c – d )/c, must be larger than the strain when
the steel starts to yield, f y /E s . In this case, c is the distance, in (mm), from the
extreme compression surface to the neutral axis and E s is the modulus of elastic-
ity of the steel, ksi (MPa).
The load, P b for balanced conditions can be computed from the preceding P u
equation with f s f y and
a a b (3.36)
1 c b
87,000 1 d

87,000 f y
The balanced moment, in. kip (k . Nm), can be obtained from
(3.37)
M b P b e b
0.85 f c ba b d d a b

2

A s f y (d d d ) A s f y d