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
