Mechanics of Materials by Timothy A. Philpot, P1.2

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:

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

\sigma =\frac{P}{A}

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

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

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

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 

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

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

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

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 

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

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

t_{min}=0.20835\:in

t_{min}=0.208\:in

 

 

 

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