Question 11.5: A steel column is constructed from a W 10 × 60 wide-flange s...

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

which 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

(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

which 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

which 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