Problem 1-2|Stress | Mechanics of Materials| Ninth Edition| R.C. Hibbeler|

Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 500-lb load is applied along the centroidal axis of the member.

Problem 1-2

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Mechanics of Materials 3rd Edition 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.


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 


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