Solution to Problem 582 | Design for Flexure and Shear

$\Sigma M_{R2} = 0$

$3R_1 = 5(2) + 2$

$R_1 = 4 \, \text{ kN}$
 

$\Sigma M_{R2} = 0$

$3R_2 + 2 = 5(1)$

$R_2 = 1 \, \text{ kN}$
 

Based on bending stress (square b = d):
$f_b = \dfrac{6M}{bd^2}$

$8 = \dfrac{6(4)(1000^2)}{d^3}$

$d = 144.22 \, \text{ mm}$
 

Based on shear stress (square b = d):
$f_v = \dfrac{3V}{2bd}$

$1 = \dfrac{3(4)(1000)}{2d^2}$

$d = 77.46 \, \text{ mm}$
 

Use 145 mm × 145 mm square beam           answer