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:
Outside Diameter , D = 2.50 in Axial Load , P = 27 kips Maximum Axial Stress , σ = 18 ksi Inside Diameter , d = D − 2 t \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*} Outside Diameter , D Axial Load , P Maximum Axial Stress , σ Inside Diameter , d = 2.50 in = 27 kips = 18 ksi = D − 2 t We are solving for the unknown wall thickness of the tube, t t 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
σ = P A A min = P σ A min = 27 kips 18 ksi A min = 1.500 in 2 \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*} σ A min A min A min = A P = σ P = 18 ksi 27 kips = 1.500 in 2 The cross-sectional area of the aluminum tube is given by
A = π 4 ( D 2 − d 2 ) A=\frac{\pi }{4}\left(D^2-d^2\right) A = 4 π ( D 2 − d 2 ) Set this expression equal to the minimum area and solve for the maximum inside diameter , d d d
A = π 4 ( D 2 − d 2 ) A = π 4 [ ( 2.50 in ) 2 − d 2 ] 1.500 in 2 = π 4 [ ( 2.50 in ) 2 − d 2 ] 4 ( 1.500 in 2 ) = π [ ( 2.50 in ) 2 − d 2 ] 4 ( 1.500 in 2 ) π = ( 2.50 in ) 2 − d 2 d 2 = ( 2.50 in 2 ) − 4 ( 1.500 in 2 ) π d = ( 2.50 in 2 ) − 4 ( 1.500 in 2 ) π d = 2.08330 in \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*} A A 1.500 in 2 4 ( 1.500 in 2 ) π 4 ( 1.500 in 2 ) d 2 d d = 4 π ( D 2 − d 2 ) = 4 π [ ( 2.50 in ) 2 − d 2 ] = 4 π [ ( 2.50 in ) 2 − d 2 ] = π [ ( 2.50 in ) 2 − d 2 ] = ( 2.50 in ) 2 − d 2 = ( 2.50 in 2 ) − π 4 ( 1.500 in 2 ) = ( 2.50 in 2 ) − π 4 ( 1.500 in 2 ) = 2.08330 in The outside diameter D D D d d d t t t
Therefore, the minimum wall thickness required for the aluminum tube is
t = D − d 2 t = 2.50 in − 2.08330 in 2 t = 0.208 in ( Answer ) \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*} t t t = 2 D − d = 2 2.50 in − 2.08330 in = 0.208 in ( Answer )
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