If the cross section of the femur at section a–a can be approximated as a circular tube as shown, determine the maximum normal stress developed on the cross section at section a–a due to the load of 75 lb.

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Accepted Answer

Internal Loadings: Considering the equilibrium for the free-body diagram of the femur's upper segment, Fig. a,

[latex]+\uparrow \Sigma F_{y}=0 \quad N-75=0 \quad N=75 \mathrm{lb}[/latex]

[latex]\curvearrowleft +\Sigma M_{O}=0 ; \quad M-75(2)=0 \quad M=150 \mathrm{lb} \cdot \mathrm{in}[/latex]

Section Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the femur's cross section are

[latex]A=\pi\left(1^{2}-0.5^{2}\right)=0.75 \pi \mathrm{in}^{2}[/latex]

[latex]I=\frac{\pi}{4}\left(1^{4}-0.5^{4}\right)=0.234375 \pi \mathrm{in}^{4}[/latex]

Normal Stress: The normal stress is a combination of axial and bending stress. Thus,

[latex]\sigma=\frac{N}{A}+\frac{M y}{I}[/latex]

By inspection, the maximum normal stress is in compression.

[latex]\sigma_{\max }=\frac{-75}{0.75 \pi}-\frac{150(1)}{0.234375 \pi}=-236 \mathrm{psi}=236 \mathrm{psi}(C)[/latex]

Question 8.82: If the cross section of the femur at section a–a can be appr...

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