Question 11.5: A steel column is constructed from a W 10 × 60 wide-flange s...
A steel column is constructed from a W 10 × 60 wide-flange section (Fig. 11-37). Assume that the column has pin supports and may buckle in any direction. Also, assume that the steel has modulus of elasticity E = 29,000 ksi and yield stress \sigma_{Y} = 36 ksi.
(a) If the length of the column is L = 20 ft, what is the allowable axial load?
(b) If the column is subjected to an axial load P = 200 k, what is the maximum permissible length?
We will use the AISC formulas (Eqs. 11-79 through 11-82) when analyzing this column. Since the column has pin supports, the effective-length factor K = 1. Also, since the column will buckle about the weak axis of bending, we will use the smaller radius of gyration:
n_{1}=\frac{5}{3}+\frac{3(KL/r)}{8(KL/r)_{c}}-\frac{(KL/r)^{3}}{8(KL/r)^{3}_{c}} \frac{KL}{r}\leq\left(\frac{KL}{r}\right)_{c} (11-79)
n_2 = \frac{23}{12}≈ 1.92 \ \ \ \ \frac{KL}{r} ≥ \left( \frac{KL}{r} \right)_c (11-80)
\frac{\sigma_{allow}}{\sigma_{Y}}=\frac{1}{n_{1}}\left[1-\frac{(KL/r)^{2}}{2(KL/r)^{2}_{c}}\right] \frac{KL}{r}\leq\left(\frac{KL}{r}\right)_{c} (11-81)
\frac{\sigma_{allow}}{\sigma_{Y}}=\frac{(KL/r)^{2}_{c}}{2n_{2}(KL/r)^{2}} \frac{KL}{r}\geq\left(\frac{KL}{r}\right)_{c} (11-82)
r = 2.57 in.
as obtained from Table E-1, Appendix E. The critical slenderness ratio (Eq. 11-76) is
\left(\frac{KL}{r}\right)_{c}=\sqrt{\frac{2\pi^{2}E}{\sigma_{Y}}}=\sqrt{\frac{2\pi^{2}(29,000 ksi)}{36 ksi}}=126.1 (a)
(a) Allowable axial load. If the length L = 20 ft, the slenderness ratio of the column is
\frac{L}{r}=\frac{(20 ft)(12 in./ft)}{2.57 in.}=93.4which is less than the critical ratio (Eq. a). Therefore, we will use Eqs. (11-79) and (11-81) to obtain the factor of safety and allowable stress, respectively:
n_{1}=\frac{5}{3}+\frac{3(KL/r)}{8(KL/r)_{c}}-\frac{(KL/r)^{3}}{8(KL/r)^{3}_{c}}=\frac{5}{3}+\frac{3(93.4)}{8(126.1)}-\frac{(93.4)^{3}}{8(126.1)^{3}}=1.89
\frac{\sigma_{allow}}{\sigma_{Y}}=\frac{1}{n_{1}}\left[1-\frac{(KL/r)^{2}}{2(KL/r)^{2}_{c}}\right]=\frac{1}{1.89}\left[1-\frac{(93.4)^{2}}{2(126.1)^{2}}\right]=0.384
\sigma_{allow}=0.384\sigma_{Y}=0.384(36 ksi)=13.8 ksi
Since the cross-sectional area of the column is A = 17.6 in.² (from Table E-1), the allowable axial load is
P_{allow}=\sigma_{allow}A=(13.8 ksi)(17.6 in.^{2})=243 k(b) Maximum permissible length. To determine the maximum length when the axial load P = 200 k, we begin with an estimated value of the length and then use a trial-and-error procedure. Note that when the load P = 200 k, the maximum length is greater than 20 ft (because a length of 20 ft corresponds to an axial load of 243 k). Therefore, as a trial value, we will assume L = 25 ft. The corresponding slenderness ratio is
\frac{L}{r}=\frac{(25 ft)(12 in./ft)}{2.57 in.}=116.7which is less than the critical ratio. Therefore, we again use Eqs. (11-79) and (11-81) to obtain the factor of safety and allowable stress:
n_{1}=\frac{5}{3}+\frac{3(KL/r)}{8(KL/r)_{c}}-\frac{(KL/r)^{3}}{8(KL/r)^{3}_{c}}=\frac{5}{3}+\frac{3(116.7)}{8(126.1)}-\frac{(116.7)^{3}}{8(126.1)^{3}}=1.915
\frac{\sigma_{allow}}{\sigma_{Y}}=\frac{1}{n_{1}}\left[1-\frac{(KL/r)^{2}}{2(KL/r)^{2}_{c}}\right]=\frac{1}{1.915}\left[1-\frac{(116.7)^{2}}{2(126.1)^{2}}\right]= 0.299
\sigma_{allow}=0.299\sigma_{Y}=0.299(36 ksi)=10.8 ksi
Thus, the allowable axial load corresponding to a length L = 25 ft is
P_{allow}=\sigma_{allow}A=(10.8 ksi)(17.6 in.²) = 190 kwhich is less than the given load of 200 k. Therefore, the permissible length is less than 25 ft.
Performing similar calculations for L = 24.0 ft and L = 24.5 ft, we obtain the following results:
L = 24.0 ft P_{allow} = 201 k
L = 24.5 ft P_{allow} = 194 k
L = 25.0 ft P_{allow} = 190 k
Interpolating between these results, we see that a load of 200 k corresponds to a length of 24.1 ft. Thus, the maximum permissible length of the column is
L_{\max}=24.1 ft