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*