Simple shear reactions due to loadings
$2V_1 = 100(10)$
$V_1 = 500 ~ \text{lb}$
$2V_2 = (10)(160)$
$V_2 = 400 ~ \text{lb}$
$2V_3 = 2(400)$
$V_3 = 400 ~ \text{lb}$
Couple reactions due to end-moments
$10{R_1}' = M_2 = 980$
${R_1}' = 98 ~ \text{lb}$
$10{R_2}' = M_3 - M_2 = 1082 - 980$
${R_2}' = 10.2 ~ \text{lb}$
$12{R_3}' = M_3 = 1082$
${R_3}' = 90.2 ~ \text{lb}$
Reactions (see figure below)
$R_1 = 402 ~ \text{lb}$ answer
$R_2 = 598 + 389.8 = 987.8 ~ \text{lb}$ answer
$R_3 = 410.2 + 490.2 = 900.4 ~ \text{lb}$ answer
$R_4 = 309.8 ~ \text{lb}$ answer
$\dfrac{x_1}{402} = \dfrac{10}{402 + 598}$
$x_1 = 4.02 ~ \text{ft}$
$\dfrac{{x_2}^2}{389.8} = \dfrac{5^2}{389.8 + 10.2}$
$x_2 = 4.94 ~ \text{ft}$
$M_A = \frac{1}{2}x_1 (402) = \frac{1}{2}(4.02)(402)$
$M_A = 808.02 ~ \text{lb}\cdot\text{ft}$
$M_B = M_2 + \frac{2}{3}x_2 (389.8) = -980 + \frac{2}{3}(4.94)(389.8)$
$M_B = 302.66 ~ \text{lb}\cdot\text{ft}$
$M_C = M_3 + 3(490.2) = -1082 + 3(490.2)$
$M_C = 388.6 ~ \text{lb}\cdot\text{ft}$
$M_D = \Sigma M_{\text{to the right of }D} = 309.8(3)$
$M_D = 929.4 ~ \text{lb}\cdot\text{ft}$
Thus,
$M_{max(+)} = 930 ~ \text{lb}\cdot\text{ft}$ answer
