Solution to Problem 119 Shear Stress

From the FBD:
$\Sigma M_C = 0$

$0.25R_{BV} = 0.25(40 \sin 35^{\circ}) + 0.2(40 \cos 35^{\circ})$

$R_{BV} = 49.156 \, \text{kN}$
 

 

$\Sigma F_H = 0$

$R_{BH} = 40 \cos 35^{\circ}$

$R_{BH} = 32.766 \, \text{kN}$
 

$R_B = \sqrt{{R_{BH}}^2 + {R_{BV}}^2}$

$R_B = \sqrt{32.766^2 + 49.156^2}$

$R_B = 59.076 \, \text{kN}$       → shear force of pin at B
 

$V_B = \tau_B \, A$       → double shear

$59.076 (1000) = \tau_B \bigl\{ \, 2 \left[ \, \frac{1}{4} \pi (20^2) \, \right] \, \bigr\}$

$\tau_B = 94.02 \, \text{ MPa}$           answer