Due to the 400 N concentrated force alone
$P = 400 ~ \text{N}$
$L = 4 ~ \text{m}$
$\alpha L = 3 ~ \text{m, thus,} ~ \alpha = \frac{3}{4}$
$M_1 = -\dfrac{3PL}{16} \cdot \dfrac{2 + 3\alpha}{3 + 3\alpha} = -\dfrac{3(400)(4)}{16} \cdot \dfrac{2 + 3(\frac{3}{4})}{3 + 3(\frac{3}{4})}$
$M_1 = -242.86 ~ \text{N}\cdot\text{m}$
$M_2 = -\dfrac{3PL}{16} \cdot \dfrac{2}{3 + 3\alpha} = -\dfrac{3(400)(4)}{16} \cdot \dfrac{2}{3 + 3(\frac{3}{4})}$
$M_2 = -114.28 ~ \text{N}\cdot\text{m}$
$M_3 = \dfrac{3PL}{16} \cdot \dfrac{1}{3 + 3\alpha} = \dfrac{3(400)(4)}{16} \cdot \dfrac{1}{3 + 3(\frac{3}{4})}$
$M_3 = 57.14 ~ \text{N}\cdot\text{m}$
Due to the 600 N/m uniform load alone
$w_o = 600 ~ \text{N/m}$
$L = 3 ~ \text{m}$
$\alpha L = 4 ~ \text{m, thus,} ~ \alpha = \frac{4}{3}$
$M_1 = \dfrac{w_o L^2}{8} \cdot \dfrac{1}{3 + 3\alpha} = \dfrac{600(3^2)}{8} \cdot \dfrac{1}{3 + 3(\frac{4}{3})}$
$M_1 = 96.43 ~ \text{N}\cdot\text{m}$
$M_2 = -\dfrac{w_o L^2}{8} \cdot \dfrac{2}{3 + 3\alpha} = -\dfrac{600(3^2)}{8} \cdot \dfrac{2}{3 + 3(\frac{4}{3})}$
$M_2 = -192.86 ~ \text{N}\cdot\text{m}$
$M_3 = -\dfrac{w_o L^2}{8} \cdot \dfrac{2 + 3\alpha}{3 + 3\alpha}= -\dfrac{600(3^2)}{8} \cdot \dfrac{2 + 3(\frac{4}{3})}{3 + 3(\frac{4}{3})}$
$M_3 = -578.57 ~ \text{N}\cdot\text{m}$
Thus,
$M_1 = -242.86 + 96.43 = -146.43 ~ \text{N}\cdot\text{m}$ answer
$M_2 = -114.28 - 192.86 = -307.14 ~ \text{N}\cdot\text{m}$ answer
$M_3 = 57.14 - 578.57 = -521.43 ~ \text{N}\cdot\text{m}$ answer
