Solution to Problem 520 | Flexure Formula

From the shear diagram:
$M_{max} = \frac{1}{2}(3)(3w)$

$M_{max} = 4.5w \, \text{N}\cdot\text{m}$
 

$(\,f_b\,)_{max} = \dfrac{M}{S}$

$120 = \dfrac{4.5w (1000)}{833 \times 10^3}$

$w = 22,213.33 \, \text{N/m}$

$w = 2,264.36 \, \text{kg/m}$
 

$w = DL + LL$

$2\,264.36 = 74 + LL$

$LL = 2,190.36 \, \text{kg/m}$

$LL = 21.5 \, \text{ kN/m}$           answer