Question 11.7: An aluminum tube (alloy 2014-T6) with an effective length L ......

We will use the Aluminum Association formulas for alloy 2014-T6 (Eqs. 11-86a, b, and c) 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 between 12 and 55, in which case we use Eq. (11-86b):

σ_{allow}=28\ ksi \quad \quad 0≤\frac{L}{r}≤12                 (11-86a)

σ_{allow}=30.7\ -\ 0.23\left(\frac{L}{r}\right)\ ksi \quad \quad 12≤\frac{L}{r}≤55                 (11-86b)

σ_{allow}=\frac{54,000\ ksi}{(L/r)^2}\quad \quad 55≤\frac{L}{r}             (11-86c)

σ_{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 axial load P = 5.0 k, and the cross-sectional area is

A=\frac{π}{4}[d^2\ -\ (d\ -\ 2t)^2]=\frac{π}{4}[d^2\ -\ (0.8d)^2] = 0.3827d²                     (d)

Therefore, the stress P/A is

\frac{P}{A}=\frac{5.0\ k}{0.2827d^2}=\frac{17.69}{d^2}

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{π}{64}[d^4\ -\ (d\ -\ 2t)^4]=\frac{π}{64}[d^4\ -\ (0.8d)^4]=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}

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:

\frac{17.69}{d^2}=30.7\ -\ 0.23\left(\frac{49.97}{d}\right)

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 between 12 and 55, as we assumed when we selected Eq. (c). To verify that this is the case, we calculate the slenderness ratio:

\frac{L}{r}=\frac{49.97\ in}{d}=\frac{49.97\ in.}{0.97\ in.} = 51.5

Therefore, the solution is valid, and the minimum required diameter is

d_{min} = 0.97 in.