Structural Concrete - Hassoun

11.13 Design of Columns under Eccentric Loading 391
Figure 11.18 Column sections of (a) Example 11.11 and (b) Example 11.12.
Example 11.12
Use the charts to determine the column strength, φP n , of the short column shown in Fig. 11.18b acting
at an eccentricity e = 12 in. Use f c ′ = 5ksiandf y = 60 ksi.
Solution
1. Properties of the section: H = 24 in., γh = 24 – 2 × 2.5 = 19 in. (distance between tension and
compression steel). γ = 19/24 = 0.79, and ρ = 8(1.27)/(14 × 24) = 0.030.
2. Since e < d, assume compression-controlled section. Let ε t = 0.002, f s /f y = 1.0, and φ = 0.65.
From the charts of Fig. 11.17, get K n = 0.36 = P n /(5 × 14 × 24). Then P n = 605 K.
3. Check assumption for compression-controlled section: For K n = 0.36, R n = K n (e/h) = 0.36 (12/24)
= 0.18. From charts, get ρ = 0.018 < 0.03. Therefore, P n > 605 K (to use ρ = 0.03).
4. Second trial: Let ε t = 0.0015, f s = 0.0015 (29,000) = 43.5 ksi.
f s
= 43.5
f y 60 = 0.725 ρ = 0.03 K n = 0.44
P n
0.44 =
P
5 × 14 × 24 n = 740 K
( )
5. Check assumption: For K n = 0.44, R n = 0.44 12
= 0.22. From charts, ρ = 0.03 as given.
24
Therefore, P n = 740 K.
φP n = 0.65(740) =480 K and φM n = 0.65(740) =480 K ⋅ ft
By analysis, φP n = 485 K (which is close to 480 K ⋅ ft).
11.13 DESIGN OF COLUMNS UNDER ECCENTRIC LOADING
In the previous sections, the analysis, behavior, and the load–interaction diagram of columns
subjected to an axial load and bending moment were discussed. The design of columns is more
complicated because the external load and moment, P u and M u , are given and it is required to