By symmetry, M
_{4} = M
_{1} and M
_{3} = M
_{2}
Apply three-moment equation between spans (0) and (1)
$M_0L_0 + 2M_1(L_0 + L_1) + M_2L_1 + \dfrac{6A_0\bar{a}_0}{L_0} + \dfrac{6A_1\bar{b}_1}{L_1} = 0$
$0 + 2M_1(0 + \alpha L) + M_2(\alpha L) + 0 + 0 = 0$
$2\alpha L \, M_1 + \alpha L \, M_2 = 0$
$M_1 = -\frac{1}{2}M_2$
Apply three-moment equation between spans (1) and (2)
$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$
$-\frac{1}{2}M_2(\alpha L) + 2M_2(\alpha L + L) + M_2(L) + 0 + \frac{1}{4}w_o L^3 = 0$
$-\frac{1}{2}\alpha L \, M_2 + 2\alpha L \, M_2 + 2L \, M_2 + L \, M_2 = -\frac{1}{4}w_o L^3$
$\frac{3}{2}\alpha L \, M_2 + 3L \, M_2 = -\frac{1}{4}w_o L^3$
$\frac{3}{2}\alpha \, M_2 + 3 \, M_2 = -\frac{1}{4}w_o L^2$
$\frac{3}{2}(\alpha + 2)M_2 = -\frac{1}{4}w_o L^2$
$M_2 = -\dfrac{w_o L^2}{4} \cdot \dfrac{2}{3(\alpha + 2)}$
$M_2 = -\dfrac{w_o L^2}{12} \cdot \dfrac{2}{\alpha + 2} = M_3$ answer
$M_1 = -\dfrac{1}{2}\left( -\dfrac{w_o L^2}{12} \cdot \dfrac{2}{\alpha + 2} \right)$
$M_1 = \dfrac{w_o L^2}{12} \cdot \dfrac{1}{\alpha + 2} = M_4$ answer