Question 11.6: Find the minimum required thickness tmin for a steel pipe co...
Find the minimum required thickness t_{\min} for a steel pipe column of length L = 3.6 m and outer diameter d = 160 mm supporting an axial load P = 240 kN (Fig. 11-38). The column is fixed at the base and free at the top. (Use E = 200 GPa and \sigma_{Y} = 250 MPa.)
We will use the AISC formulas (Eqs. 11-79 through 11-82) when analyzing this column. Since the column has fixed-free end conditions, the effective length is
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)
\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)
L_{e} = KL = 2(3.6 m) = 7.2 m
Also, 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}(200 GPa)}{250 MPa}}=125.7 (b)
First trial. To determine the required thickness of the column, we will use a trial-and-error method. Let us start by assuming a trial value t = 7.0 mm. Then the moment of inertia of the cross-sectional area is
I=\frac{\pi}{64}\left[d^{4}-(d-2t)^{4}\right]=\frac{\pi}{64}\left[(160 mm)^{4}-(146 mm)^{4}\right]=9.866\times10^{6} mm^{4}Also, the cross-sectional area and radius of gyration are
A=\frac{\pi}{4}\left[d^{2}-(d-2t)^{2}\right]=\frac{\pi}{4}\left[(160 mm)^{2}-(146 mm)^{2}\right]=3365 mm^{2}r=\sqrt{\frac{I}{A}}=\sqrt{\frac{9.866\times10^{6} mm^{4}}{3365 mm^{2}}}=54.15 mm
Therefore, the slenderness ratio of the column is
\frac{KL}{r}=\frac{2(3.6 m)}{54.15 mm}=133.0Since this ratio is larger than the critical slenderness ratio (Eq. b), we obtain the factor of safety and the allowable stress from Eqs. (11-80) and (11-82):
n_{2}=\frac{23}{12}\approx1.92 \frac{KL}{r}\geq\left(\frac{KL}{r}\right)_{c} (11-80)
n_{2}=1.92
\frac{\sigma_{allow}}{\sigma_{Y}}=\frac{(KL/r)^{2}_{c}}{2n_{2}(KL/r)^{2}}=\frac{(125.7)^{2}}{2(1.92)(133.0)^{2}}=0.2326
\sigma_{allow} = 0.2326\sigma_{Y} = 0.2326(250 MPa) = 58.15 MPa
Thus, the allowable axial load is
P_{allow} = \sigma_{allow}A = (58.15 MPa)(3365 mm²) = 196 kN
Since this load is less than the required load of 240 kN, we must try a larger value of the thickness t.
Additional trials. Performing similar calculations for t = 8 mm and t = 9 mm, we get the following results:
t = 7.0 mm P_{allow} = 196 kN
t = 8.0 mm P_{allow} = 220 kN
t = 9.0 mm P_{allow} = 243 kN
By interpolation, we see that t = 8.9 mm corresponds to a load of 240 kN. Therefore, the required thickness of the pipe column is
t_{\min} = 8.9 mm