Compare the maximum stress in bent rod 1/2 in. square, where the load P is 1/2 in. off center as shown in Figure P-902, with the maximum stress if the rod were straight and the load applied axially.

 

Solution
For P applied off-center:

The most stressed are the bottom fibers, it is subjected to compression due to axial load and compression due to bending.
σ=−PA−6 Mbd 2$\sigma =-{\displaystyle \frac{P}{A}}-{\displaystyle \frac{\mathrm{6 }M}{b{\mathrm{d }}^2}}$

σ=−P 0.52−6(0.5 P)0.5(0.52)$\sigma =-{\displaystyle \frac{\mathrm{P }}{{0.5}^2}}-{\displaystyle \frac{6(\mathrm{0.5 }P)}{0.5({0.5}^2)}}$

σ=−28 P$\sigma =-\mathrm{28 }P$

 

For P applied axially:

σ=−PA$\sigma =-{\displaystyle \frac{P}{A}}$

σ=−P 0. 52$\sigma = - {\displaystyle \frac{\mathrm{P }}{{0.5}^2}}$

σ=−4 P$\sigma =-\mathrm{4 }P$

 

Ratio=−28 P−4 P $\text{Ratio}={\displaystyle \frac{-\mathrm{28 }P}{-\mathrm{4 }P}}$

Ratio=71 $\text{Ratio}={\displaystyle \frac{7}{1}}$       answer