Question 11.7: An aluminum tube (alloy 2014-T6) with an effective length L ...
An aluminum tube (alloy 2014-T6) with an effective length L = 16.0 in. is compressed by an axial force P = 5.0 k (Fig. 11-39).
Determine the minimum required outer diameter d if the thickness t equals one-tenth the outer diameter.
We will use the Aluminum Association formulas for alloy 2014-T6 (Eqs. 11-84a and b) for analyzing this column. However, we must make an initial guess as to which formula is applicable, because each formula applies to a different range of slenderness ratios. Let us assume that the slenderness ratio of the tube is less than 55, in which case we use Eq. (11-84a) with K = 1:
\sigma_{allow}=30.7-0.23\left(\frac{KL}{r}\right)ksi 0\leq \frac{KL}{r}\leq55 (11-84a)
\sigma_{allow}=\frac{54,000 ksi}{(KL/r)^{2}} \frac{KL}{r}\geq55 (11-84b)
\sigma_{allow}=30.7-0.23\left(\frac{L}{r}\right)ksi (c)
In this equation, we can 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}-\left(d-2t\right)^{2}\right]=\frac{\pi}{4}\left[d^{2}-\left(0.8d\right)^{2}\right]=0.2827d^{2} (d)
Therefore, the stress P/A is
\frac{P}{A}=\frac{5.0 k}{0.2827d^{2}}=\frac{17.69}{d^{2}}in which P/A has units of kips per square inch (ksi) and d has units of inches (in.). Substituting into Eq. (c), we get
\frac{17.69}{d^{2}}=30.7-0.23\left(\frac{L}{r}\right)ksi (e)
The slenderness ratio L/r can also be expressed in terms of the diameter d. First, we find the moment of inertia and radius of gyration of the cross section:
I=\frac{\pi}{64}\left[d^{4}-\left(d-2t\right)^{4}\right]=\frac{\pi}{64}\left[d^{4}-(0.8d)^{4}\right]=0.02898d^{4}r=\sqrt{\frac{I}{A}}=\sqrt{\frac{0.02898d^{4}}{0.2827d^{2}}}=0.3202d
Therefore, the slenderness ratio is
\frac{L}{r}=\frac{16.0 in.}{0.3202d}=\frac{49.97 in.}{d} (f)
where (as before) the diameter d has units of inches.
Substituting into Eq. (e), we obtain the following equation, in which d is the only unknown quantity:
With a little rearranging, this equation becomes
30.7d² – 11.49d – 17.69 = 0
from which we find
d = 0.97 in.
This result is satisfactory provided the slenderness ratio is less than 55, as required for Eq. (c) to be valid. To verify that this is the case, we calculate the slenderness ratio from Eq. (f):
\frac{L}{r}=\frac{49.97 in.}{d}=\frac{49.97 in.}{0.97 in.}=51.5Therefore, the solution is valid, and the minimum required diameter is
d_{\min}=0.97 in.