Solution to Problem 514 | Flexure Formula

$R_B = 500 \, \text{N}$
 

At section a-a:
$\cos \theta = \dfrac{x}{3} = \dfrac{4}{5}$

$x = 2.4 \, \text{m}$
 

$M = xR_A - 200x\,(x/2)$

$M = 2.4(500) - 200(2.4)(2.4/2)$

$M = 624 \, \text{N}\cdot\text{m}$
 

$f_b = \dfrac{Mc}{I} = \dfrac{624(1000)(50/2)}{\dfrac{50(50^3)}{12}}$

$f_b = 29.952 \, \text{MPa}$           answer