Solution to Problem 517 | Flexure Formula

$w = (3.5325 \, \text{kg/m})(9.81 \, \text{m/s}^2)$

$w = 34.65 \, \text{N/m}$
 

$R_L = R_R = 6w/2$

$R_L = R_R = 6(34.65)/2$

$R_L = R_R = 103.96 \, \text{N}$
 

For simply supported beam subjected to uniformly distributed load, the maximum moment will occur at the midspan. At midspan:
$M = 3(103.96) - 34.65(3)(3/2)$

$M = 155.955 \, \text{N}\cdot\text{m}$
 

$(\,f_b\,)_{max} = \dfrac{Mc}{I} = \dfrac{M(h/2)}{\dfrac{bh^3}{12}}$

$(\,f_b\,)_{max} = \dfrac{155.955(1000)(30/2)}{\dfrac{15(30^3)}{12}}$

$(\,f_b\,)_{max} = 69.31 \, \text{ MPa}$           answer