Structural Concrete - Hassoun

432 Chapter 12 Slender Columns
Example 12.1
The column section shown in Fig. 12.5 carries an axial load P D = 136 K and a moment M D = 116 K⋅ft
due to dead load and an axial load P L = 110 K and a moment M L = 93 K⋅ft due to live load. The column
is part of a frame that is braced against sidesway and bent in single curvature about its major axis. The
unsupported length of the column is l c = 19 ft, and the moments at both ends of the column are equal.
Check the adequacy of the column using f ′ c = 4ksi and f y = 60 ksi.
4 no. 9
4 no. 9
Solution
1. Calculate factored loads:
Figure 12.5 Example 12.1.
P u = 1.2P D + 1.6P L = 1.2 × 136 + 1.6 × 110 = 339.2K
M u = 1.2M D + 1.6M L = 1.2 × 116 + 1.6 × 93 = 288K ⋅ ft
e = M u 288 × 12
= = 10.2in.
P u 339.2
2. Check if the column is long. Because the frame is braced against sidesway, assume K = 1.0,
r = 0.3h = 0.3 × 22 = 6.6 in., and l u = 19 ft.
Kl u 1 × 19 × 12
= = 34.5
r 6.6
For braced columns, if Kl u /r ≤ 34 − 12 M 1 /M 2 , slenderness effect may be neglected. Given end
moments M 1 = M 2 and M 1 /M 2 positive for single curvature,
Right − hand side = 34 − 12 M 1
= 34 − 12 × 1 = 22
M 2
Because Kl u /r = 34.5 > 22, slenderness effect must be considered
3. Calculate EI from Eq.12.10:
a. Calculate E c :
E c = 57, 000 √ √
f c ′ = 57, 000 4000 = 3605ksi
E s = 29, 000ksi