Combined axial compression and bending

$$\sigma = -\dfrac{P}{A} \pm \dfrac{Mc}{I}$$

Combined axial tension and bending

$$\sigma = \dfrac{P}{A} \pm \dfrac{Mc}{I}$$

For the flexure quantity $Mc/I$, use (+) for fibers in tension and (-) for fibers in compression.

**Problem 902**

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 902**

For *P* applied off-center:

$\sigma = -\dfrac{P}{A} - \dfrac{6M}{bd^2}$

$\sigma = -\dfrac{P}{0.5^2} - \dfrac{6(0.5P)}{0.5(0.5^2)}$

$\sigma = -28P$

For *P* applied axially:

$\sigma = -\dfrac{P}{0.5^2}$

$\sigma = -4P$

$\text{Ratio} = \dfrac{-28P}{-4P}$

$\text{Ratio} = \dfrac{7}{1}$ *answer*

## Comments

## the answer in the first cass

the answer in the first cass should be 28P not 24P ..... rest is fine

## Thanks

Thanks. Necessary corrections were applied to the page. Thank you very much.