## Solution:

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

$\displaystyle \sigma =\frac{P}{A}$

$\displaystyle A_{min}=\frac{P}{\sigma }$

$\displaystyle A_{min}=\frac{27\:kips}{18\:ksi}$

$\displaystyle A_{min}=1.500\:in.^2$

The cross-sectional area of the aluminum tube is given by

$\displaystyle A=\frac{\pi }{4}\left(D^2-d^2\right)$

Set this expression equal to the minimum area and solve for the maximum inside diameter

$\displaystyle \frac{\pi }{4}\left[\left(2.50\:in\right)^2-d^2\right]=1.500\:in^2$

$\displaystyle \left(2.50\:in\right)^2-d^2=\frac{4}{\pi }\left(1.500\:in^2\right)$

$\displaystyle \left(2.50\:in\right)^2-\frac{4}{\pi }\left(1.500\:in^2\right)=d^2$

$\displaystyle d_{max}=2.08330\:in$

The outside diameter D, the inside diameter d, and the wall thickness t are related by

$D=d+2t$

Therefore, the minimum wall thickness required for the aluminum tube is

$\displaystyle t_{min}=\frac{D-d}{2}$

$\displaystyle t_{min}=\frac{2.50\:in-2.08330\:in}{2}$

$\displaystyle t_{min}=0.20835\:in$

$\displaystyle t_{min}=0.208\:in$