Combined Axial and Bending
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