Solution to Problem 580 | Design for Flexure and Shear

$M_{max} = \frac{1}{2}L \, (\frac{1}{2}P) = \frac{1}{4}PL$
 

From flexure formula:
$f_b = \dfrac{6M}{bh^2} = \dfrac{6(\frac{1}{4}PL)}{bh^2}$

$f_b = \dfrac{3PL}{2bh^2}$

$bh = \dfrac{3PL}{2f_b\,h}$
 

From shear stress formula:
$f_b = \dfrac{3V}{2bh} = \dfrac{3(\frac{1}{2}P)}{2\left( \dfrac{3PL}{2f_b\,h} \right)}$

$f_v = \dfrac{f_b \, h}{2L}$           answer