$\Sigma M_{D} = 0$
$8R_1 = 200(5) + 400(1)$
$R_1 = 175 \, \text{ lb}$
$\Sigma M_{A} = 0$
$8R_2 = 200(3) + 400(7)$
$R_2 = 425 \, \text{ lb}$
By ratio and proportion
$\dfrac{y_C}{1} = \dfrac{2125}{5}$
$y_C = 425 \, \text{ lb}\cdot\text{ft}$
From the conjugate beam
$\Sigma M_D = 0$
$8F_1 + \frac{1}{2}(4)(1600)[1 + \frac{2}{3}(4)] = \frac{1}{2}(3)(525)[5 + \frac{1}{3}(3)] + \frac{1}{2}(5)(2125)[\frac{2}{3}(5)]$
$F_1 = 1337.5 \, \text{ lb}\cdot\text{ft}^2$
$\Sigma M_A = 0$
$8F_2 + \frac{1}{2}(4)(1600)[\frac{1}{3}(4)] = \frac{1}{2}(3)(525)[\frac{2}{3}(3)] + \frac{1}{2}(5)(2125)[3 + \frac{1}{3}(5)]$
$F_2 = 1562.5 \, \text{ lb}\cdot\text{ft}^2$
Consider the section to the left of B in conjugate beam
$M_B = \frac{1}{2}(3)(525)[\frac{1}{3}(3)] - 3F_1$
$MB = 787.5 - 3(1337.5)$
$M_B = -3225 \, \text{ lb}\cdot\text{ft}^3$
Thus, the deflection at B is
$EI ~ \delta_B = M_B$
$EI ~ \delta_B = 3225 \, \text{ lb}\cdot\text{ft}^3$ answer
Consider the section to the right of C in conjugate beam
$M_C = \frac{1}{2}(1)(y_C)[\frac{1}{3}(1)] - 1F_2$
$M_C = \frac{1}{2}(1)(425)[\frac{1}{3}(1)] - 1(1562.5)$
$M_C = -1491.67 \, \text{ lb}\cdot\text{ft}^3$
Thus, the deflection at C is
$EI ~ \delta_C = M_C$
$EI ~ \delta_C = -1491.67 \, \text{ lb}\cdot\text{ft}^3$
$EI ~ \delta_C = 1491.67 \, \text{ lb}\cdot\text{ft}^3$ downward answer
