By symmetry, R

_{1} = R

_{4}, R

_{2} = R

_{3}, and M

_{2} = M

_{3}
Apply three-moment equation to spans (1) and (2)

$M_1 L_1 + 2M_2(L_1 + L_2) + M_3 L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2}$

Where,

$M_1 = 0$
$M_3 = M_2 = M$

$L_1 = L_2 = L$

$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{6A_2\bar{b}_2}{L_2} = \frac{1}{4}w_o L^3$

Thus,

$0 + 2M(L + L) + ML + \frac{1}{4}w_o L^3 + \frac{1}{4}w_o L^3 = 0$

$5LM = -\frac{1}{2}w_o L^3$

$M = -\frac{1}{10}w_o L^2$

Hence,

$M_2 = -\frac{1}{10}w_o L^2 = -0.1 w_o L^2$ *answer*

$M_3 = -\frac{1}{10}w_o L^2 = -0.1 w_o L^2$ *answer*

From the first span

Simple reactions due to loadings

$V_1 = \frac{1}{2}w_o L$

Couple reaction due to end moment

${R_1}' = \dfrac{M_2}{L} = \dfrac{\frac{1}{10}w_o L^2}{L} = \frac{1}{10}w_o L$

Thus,

$R_{1L} = V_1 - {R_1}' = \frac{1}{2}w_o L - \frac{1}{10}w_o L$

$R_{1L} = \frac{2}{5}w_o L$

$R_{1R} = V_1 + {R_1}' = \frac{1}{2}w_o L + \frac{1}{10}w_o L$

$R_{1R} = \frac{3}{5}w_o L$

From the second span

Simple reactions due to loadings

$V_2 = \frac{1}{2}w_o L$

Couple reaction due to end moment

${R_2}' = \dfrac{M_3 - M_2}{L} = \dfrac{M_2 - M_2}{L} = 0$

Thus,

$R_{2L} = V_{2R} - {R_2}' = \frac{1}{2}w_o L - 0$

$R_{2L} = \frac{1}{2}w_o L$

From the load diagram below

$R_1 = R_4 = \frac{2}{5}w_o L$

$R_1 = R_4 = 0.4w_o L$ *answer*

$R_2 = R_3 = \frac{3}{5}w_o L + \frac{1}{2}w_o L = \frac{11}{10}w_o L$

$R_2 = R_3 = 1.1w_o L$ *answer*

From the shear diagram

$\dfrac{x}{0.4w_o L} = \dfrac{L - x}{0.6w_o L}$

$0.6x = 0.4L - 0.4x$

$x = 0.4L$

At distance x on the 1st span

$M_x = \frac{1}{2}x(0.4w_o L) = \frac{1}{2}(0.4L)(0.4w_o L)$

$M_x = 0.08w_o L^2$

At the midspan of the middle span

$M_m = M_2 + \frac{1}{2}(0.5L)(0.5w_o L) = -0.1w_o L^2 + 0.125w_o L^2$

$M_m = 0.025w_o L^2$

Hence,

$M_{max\,(+)} = M_x = 0.08w_o L^2$ *answer*