strength of material books | Engineering Mathematics and Sciences

A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution:

We are given the following values:

\begin{align*} \text{Outside Diameter}, D & = 2.50\ \text{in} \\ \text{Axial Load}, P & = 27\ \text{kips} \\ \text{Maximum Axial Stress}, \sigma & = 18\ \text{ksi}\\ \text{Inside Diameter}, d & = D-2t \end{align*}

We are solving for the unknown wall thickness of the tube, $t$ .

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

\begin{align*} \sigma & =\frac{P}{A} \\ A_{\text{min}} & =\frac{P}{\sigma } \\ A_{\text{min}} & = \frac{27\:\text{kips}}{18\:\text{ksi}} \\ A_{\text{min}} & = 1.500\:\text{in}^2 \end{align*}

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

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, $d$

\begin{align*} A & =\frac{\pi }{4}\left(D^2-d^2\right) \\ A & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ 1.500\ \text{in}^2 & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ 4 \left( 1.500\ \text{in}^2 \right) & = \pi \left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} & = \left(2.50\ \text{in}\right)^2-d^2 \\ d^{2} & = \left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} \\ d & = \sqrt{\left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi}} \\ d & = 2.08330\ \text{in} \end{align*}

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

\begin{align*} t & =\frac{D-d}{2} \\ t & = \frac{2.50\:\text{in}-2.08330\:\text{in}}{2} \\ t & = 0.208\ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}