Structural Concrete - Hassoun

11.11 Load Capacity of Circular Columns 381
Since ε t < 0.002, then φ = 0.65.
φP n = 0.65(1459) =948.3K
φM n = 0.65(729.5) =474 K ⋅ ft
Note: If side bars are neglected, then
P b = 733.8 + 359.4 − 381 = 712.2 K
P n (at e = 6in.) =1027.75 + 359.4 − 114.2 = 1273 K
If side bars are considered, the increase in P b is about 2% and that in P n is about 14.6%.
11.11 LOAD CAPACITY OF CIRCULAR COLUMNS
11.11.1 Balanced Condition
The values of the balanced load P b and the balanced moment M b for circular sections can be determined
using the equations of equilibrium, as was done in the case of rectangular sections. The bars
in a circular section are arranged in such a way that their distance from the axis of plastic centroid
varies, depending on the number of bars in the section. The main problem is to find the depth of
the compressive block a and the stresses in the reinforcing bars. The following example explains
the analysis of circular sections under balanced conditions. A similar procedure can be adopted to
analyze sections when tension or compression controls.
Example 11.9
Determine the balanced load P b and the balanced moment M b for the 16-in. diameter circular spiral
column reinforced with eight no. 9 bars shown in Fig. 11.14. Given: f c ′ = 4ksiandf y = 60 ksi.
S = 8 − 2.5 = 5.5in.
S 1 = S cos 22.5deg = 5.1 in.
S 2 = S cos 67.5deg = 2.1 in.
d = 8 + 5.1 = 13.1in.
S 3 = 1.85 in.
Solution
S 4 = 4.85 in.
1. Because the reinforcement bars are symmetrical about the axis A–A passing through the center
of the circle, the plastic centroid lies on that axis.
2. Determine the location of the neutral axis:
d t = 13.1 in.
ε y = f y
E S
(E S = 29,000 ksi)
c b
d t
=
0.003
0.003 + ε y
=
87
c b = (13.1) =7.75 in.
87 + 60
a b = 0.85 × 7.75 = 6.59 in.
0.003
= 87
0.003 + f y ∕E S 87 + f y