Solution to Problem 574 | Horizontal Shearing Stress

$f_v = \dfrac{VQ}{Ib}$
 

 

Where
$b = x$

$Q = \frac{1}{2}xy\,(\frac{1}{2}h - \frac{2}{3}y) = \frac{1}{2}xy \left( \dfrac{3h - 4y}{6} \right)$

$Q = \frac{1}{12}xy(3h - 4y)$
 

$f_v = \dfrac{V \, [ \, \frac{1}{12}xy(3h - 4y) \,]}{Ix} = \dfrac{V}{12I}(3hy - 4y^2)$

$\dfrac{df_v}{dy} = \dfrac{V}{12I}(3h - 8y) = 0$

$3h = 8y$

$y = \frac{3}{8}h$
 

Location of maximum horizontal shearing stress:
$d = \frac{1}{2}h - y = \frac{1}{2}h - \frac{3}{8}h$

$d = \frac{1}{8}h$           answer