Question 11.7: Find the minimum required thickness tmin for a steel pipe co...
Find the minimum required thickness t_{min} for a steel pipe column with a length of L = 12 ft and outer diameter of d = 6.5 in. supporting an axial load of P = 54 kips (Fig. 11-39). The column is fixed at the base and free at the top. (Use E = 29,000 ksi and σ_{Y} = 36 ksi.)
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize: Use the AISC formulas of Eqs. (11-82) through (11-86) when analyzing this column. Since the column has fixed and free end conditions, the effective length is
\sigma_{ e }=\frac{\pi^{2} E}{\left(\frac{K L}{r}\right)^{2}} (11-82)
\sigma_{\text {allow }}=\frac{\sigma_{ cr }}{1.67} (11-86)
L_{e}=K L=2(12 ft )=24 ftAlso, the critical slenderness ratio of Eq. (11-83) is
\left(\frac{K L}{r}\right)_{c}=4.71 \sqrt{\frac{E}{\sigma_{ Y }}}=4.71 \sqrt{\frac{29,000 ksi }{36 ksi }}=133.7 (a)
3, 4. Analyze, Finalize:
First trial: To determine the required thickness of the column, use a trial- and-error method. Start by assuming a trial value t = 0.5 in. Then the moment of inertia of the cross-sectional area is
I=\frac{\pi}{64}\left[d^{4}-(d-2 t)^{4}\right]=\frac{\pi}{64}\left[(6.5 in.)^{4}-(5.5 in.)^{4}\right]=42.706 in ^{4}Also, the cross-sectional area and radius of gyration are
A=\frac{\pi}{4}\left[d^{2}-(d-2 t)^{2}\right]=\frac{\pi}{4}\left[(6.5 \text { in. })^{2}-(5.5 \text { in. })^{2}\right]=9.425 in ^{2}and
r=\sqrt{\frac{I}{A}}=\sqrt{\frac{42.706 in ^{4}}{9.425 in ^{2}}}=2.129 in.
Therefore, the slenderness ratio of the column is
\frac{K L}{r}=\frac{2(144 in.)}{2.129 in.}=135.3This ratio is larger than the critical slenderness ratio of Eq. (a), so obtain the allowable stress from Eqs. (11-82), (11-85), and (11-86):
\frac{K L}{r}>4.71 \sqrt{\frac{E}{\sigma_{ Y }}} (11-85)
\sigma_{\text {allow }}=\frac{0.877 \sigma_{e}}{1.67}=\frac{0.877\left[\frac{\pi^{2} E}{\left(\frac{K L}{r}\right)^{2}}\right]}{1.67}=\frac{0.877\left[\frac{\pi^{2}(29,000 ksi )}{135.3^{2}}\right]}{1.67}=8.211 ksiThus, the allowable axial load is
P_{\text {allow }}=\sigma_{\text {allow }} A=8.211 ksi \left(9.425 in ^{2}\right)=77.4 \text { kips }Since this load is greater than the required load of 54 kips, try a smaller value of the thickness t.
Additional trials: Performing similar calculations for t = 0.25 in. and t = 0.375 in. gives the results:
t=0.25 \text { in. } \quad P_{\text {allow }}=43.5 kipst=0.375 in. \quad P_{\text {allow }}=61.5 kips
t=0.5 in. \quad P_{\text {allow }}=77.4 kips
Interpolate to find that t = 0.32 in., which corresponds to a load of 54 kips. Therefore, the required thickness of the pipe column is
t_{\min }=0.32 in