Solution to Problem 120 Shear Stress

For member AB:
 

 

Length, $L_{AB} = \sqrt{4^2 + 4^2} = 5.66 \, \text{ft}$

Weight, $W_{AB} = 5.66(200) = 1\,132 \, \text{lb}$
 

$\Sigma M_A = 0$

$4 R_{BH} + 4 R_{BV} = 2 W_{AB}$

$4 R_{BH} + 4 R_{BV} = 2(1132)$

$R_{BH} + R_{BV} = 566$       → Equation (1)
 

For member BC:
 

 

Length, $L_{BC} = \sqrt{3^2 + 6^2} = 6.71 \, \text{ft}$

Weight, $W_{BC} = 6.71(200) = W_{BC} = 1342 \, \text{lb}$
 

$\Sigma M_C = 0$

$6R_{BH} = 1.5W_{BC} + 3R_{BV}$

$6R_{BH} - 3R_{BV} = 1.5(1342)$

$2R_{BH} - R_{BV} = 671$       → Equation (2)
 

Add equations (1) and (2)
$R_{BH} + R_{BV} = 566$       → Equation (1)

$2R_{BH} - R_{BV} = 671$       → Equation (2)

$3R_{BH} = 1237$

$R_{BH} = 412.33 \, \text{lb}$
 

From equation (1):
$412.33 + R_{BV} = 566$

$R_{BV} = 153.67 \, \text{lb}$
 

From the FBD of member AB
$\Sigma F_H = 0$

$R_{AH} = R_{BH} = 412.33 \, \text{lb}$
 

$\Sigma F_V = 0$

$R_{AV} + R_{BV} = W_{AB}$

$R_{AV} + 153.67 = 1132$

$R_{AV} = 978.33 \, \text{lb}$
 

$R_A = \sqrt{R_{AH}^2 + R_{AV}^2}$

$R_A = \sqrt{412.33^2 + 978.33^2}$

$R_A = 1061.67 \, \text{lb}$       → shear force of pin at A
 

$V = \tau A$

$1061.67 = 5000 (\frac{1}{4} \pi d^2)$

$d = 0.520 \, \text{in}$           answer