By symmetry, M

_{4} = M

_{1} and M

_{3} = M

_{2}

Apply three-moment equation between spans (0) and (1)

$M_0L_0 + 2M_1(L_0 + L_1) + M_2L_1 + \dfrac{6A_0\bar{a}_0}{L_0} + \dfrac{6A_1\bar{b}_1}{L_1} = 0$

$0 + 2M_1(0 + 10) + M_2(10) + 0 + \frac{1}{4}(100)(10^3) = 0$

$20M_1 + 10M_2 + 25\,000 = 0$

$20M_1 + 10M_2 = -25\,000$ ← Equation (1)

Apply three-moment equation between spans (1) and (2)

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

$M_1(10) + 2M_2(10 + 12) + M_2(12) + \frac{1}{4}(100)(10^3)$

$+ \left[ \frac{1}{4}(100)(12^3) + \frac{3}{8}(800)(12^2) \right] = 0$

$10M_1 + 56M_2 + 25\,000 + 86\,400 = 0$

$10M_1 + 56M_2 = -111\,400$ ← Equation (2)

From equations (1) and (2)

$M_1 = -280.39 ~ \text{lb}\cdot\text{ft} = M_4$ *answer*

$M_2 = -1939.22 ~ \text{lb}\cdot\text{ft} = M_3$ *answer*

Simple beam reactions

$V_{1L} = V_{1R} = \frac{1}{2}(10)(100) = 500 ~ \text{lb}$

$V_{2L} + V_{2R} = \frac{1}{2}[ \, 12(100) + 800 \, ] = 1000 ~ \text{lb}$

End moment reactions (couple)

${R_1}' = (1939.22 - 280.39) / 10 = 165.88 ~ \text{lb}$

${R_2}' = (1939.22 - 1939.22) / 12 = 0$

Support reactions

$R_1 = R_4 = 334.12 ~ \text{lb}$ *answer*

$R_2 = R_3 = 665.88 + 1000 = 1665.88 ~ \text{lb}$ *answer*

From the shear diagram

$\dfrac{x}{334.12} = \dfrac{10}{334.12 + 665.88}$

$x = 3.3412 ~ \text{ft}$

Moment at distance x from fixed supports

$M_x = M_1 + \text{Area in the shear diagram}$

$M_x = -280.39 + \frac{1}{2}(3.3412)(334.12)$

$M_x = 277.79 ~ \text{lb}\cdot\text{ft}$

Moment at the midspan of middle span

$M_{mid} = M_2 + \text{Area in the shear diagram}$

$M_{mid} = -1939.22 + \frac{1}{2}(1000 + 400)(6)$

$M_{mid} = 2260.78 ~ \text{lb}\cdot\text{ft}$

Thus,

$M_{max(+)} = 2260.78 ~ \text{lb}\cdot\text{ft}$ *answer*