Question 11.8: An aluminum tube (alloy 2014-T6) with an effective length L ...
An aluminum tube (alloy 2014-T6) with an effective length L = 405 mm is compressed by an axial force P = 22 kN (Fig. 11-40).
Determine the minimum required outer diameter d if the thickness t equals one-tenth the outer diameter.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1, 2. Conceptualize, Categorize: Use the Aluminum Association formulas for alloy 2014-T6 given as Eqs. (11-88c and d) for analyzing this column. However, you must make an initial guess as to which formula is applicable, because each formula applies to a different range of slenderness ratios. Assume that the slenderness ratio of the tube is less than 55, in which case Eq. (11-88c) with K = 1 applies:
\sigma_{\text {allow }}=\left[213-1.577\left(\frac{K L}{r}\right)\right] MPa \quad 0 \leq \frac{K L}{r} \leq 55 (11-88c)
\sigma_{\text {allow }}=\frac{3.81 \times 10^{5} MPa }{\left(\frac{K L}{r}\right)^{2}} \quad \frac{K L}{r} \geq 55 (11-88d)
\sigma_{\text {allow }}=213-1.577\left(\frac{L}{r}\right) MPa (a)
3. Analyze: In this equation, replace the allowable stress by the actual stress P/A, that is, by the axial load divided by the cross-sectional area. The cross- sectional area is
A=\frac{\pi}{4}\left[d^{2}-(d-2 t)^{2}\right]=\frac{\pi}{4}\left[d^{2}-(0.8 d)^{2}\right]=0.2827 d^{2} (b)
Therefore, the stress P/A is
\frac{P}{A}=\frac{22,000 N }{0.2827 d^{2}}=\frac{77,821}{d^{2}}in which P/A has units of Newtons per square millimeter (MPa) and d has units of mm. Substitute into Eq. (a) to get
\frac{77,821}{d^{2}}=213-1.577\left(\frac{L}{r}\right) MPa (c)
The slenderness ratio L/r also can be expressed in terms of the diameter d. First, find the moment of inertia and radius of gyration of the cross section:
I=\frac{\pi}{64}\left[d^{4}-(d-2 t)^{4}\right]=\frac{\pi}{64}\left[d^{4}-(0.8 d)^{4}\right]=0.02898 d^{4}r=\sqrt{\frac{I}{A}}=\sqrt{\frac{0.02898 d^{4}}{0.2827 d^{2}}}=0.3202 d
Therefore, the slenderness ratio is
\frac{L}{r}=\frac{405 mm }{0.3202 d}=\frac{1265.8}{d} (d)
where (as before) the diameter d has units of millimeters.
Substitute into Eq. (c) to obtain the following equation, in which d is the only unknown quantity:
With a little rearranging, this equation becomes
213.0 d^{2}-1996.1666 d-77,821.0=0and solving for the outer diameter gives
d = 24.4 mm
4. Finalize: This result is satisfactory provided the slenderness ratio is less than 55, as required for Eq. (a) to be valid. To verify that this is the case, calculate the slenderness ratio from Eq. (d):
\frac{L}{r}=\frac{1265.8 mm }{d}=\frac{1265.8 mm }{24.4 m }=51.9Therefore, the solution is valid, and the minimum required diameter is
d_{\min }=24.4 mm