Solution to Problem 568 | Horizontal Shearing Stress

$f_v = \dfrac{VQ}{Ib}$
 

Where
$f_v = \tau$

$Q = Ay = \frac{1}{2}\pi r^2 \left( \dfrac{4r}{3\pi} \right) = \frac{2}{3}r^3$

$I = \frac{1}{4}\pi \, r^4$

$b = 2r$
 

Thus,
$\tau = \dfrac{V(\frac{2}{3}r^3)}{\frac{1}{4}\pi r^4(2r)} = \dfrac{8Vr^3}{6\pi r^5}$

$\tau = \frac{4}{3}(V \, / \, \pi r^2)$ (okay!)