Solution to Problem 583 | Design for Flexure and Shear

$12R_1 = W(6) + 2W(3)$

$R_1 = W$
 

$\Sigma M_{R1} = 0$

$12R_2 = W(6) + 2W(9)$

$R_2 = 2W$
 

Based on allowable bending stress:
$f_b = \dfrac{6M}{bd^2}$

$1500 = \dfrac{6(45W/8)(12)}{6(10^3)}$

$W = 2\,222.22 \, \text{ lb}$
 

Based on allowable shearing stress:
$f_v = \dfrac{3V}{2bd}$

$120 = \dfrac{3(2W)}{2(6)(10)}$

$W = 2400 \, \text{ lb}$
 

For safe value of W, use W = 2 222.22 lb.           answer