$\Sigma M_{R2} = 0$
$8R_1 = 200(4)(10) + 200(4)(2)$
$R_1 = 1200 \text{lb}$
$\Sigma_{R1} = 0$
$8R_2 + 200(4)(2) = 200(4)(6)$
$R_2 = 400 \, \text{lb}$
$M_A = M_D = 0$
$M_B = -200(4)(2) = -1600 \, \text{lb}\cdot\text{ft}$
$M_C = 400(4) - 200(4)(2) = 0$
For the deflection at A, use span A-B-D
$L_1 = 4 \text{ft}$ ← span A to B
$L_2 = 8 \text{ft}$ ← span B to D
$M_A L_1 + 2M_B(L_1 + L_2) + M_DL_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_A}{L_1} + \dfrac{h_D}{L_2} \right)$
$0 + 2(-1600)(4 + 8) + 0 + \dfrac{200(4^3)}{4} + \dfrac{200(4^2)}{4(8)}[ \, 2(8^2) - 4^2 \, ] = 6EI \left( \dfrac{h_A}{4} + 0 \right)$
$-38,400 + 3,200 + 11,200 = 1.5EI \, h_A$
$EI \, h_A = -16,000$
The negative sign indicates that point A is below point B, hence,
$\delta_A = \dfrac{16,000}{EI} ~ \text{ downward}$ answer
For the deflection at C, use span B-C-D
$L_1 = 4 \text{ft}$ ← span B to C
$L_2 = 4 \text{ft}$ ← span C to D
$M_B L_1 + 2M_C(L_1 + L_2) + M_DL_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_B}{L_1} + \dfrac{h_D}{L_2} \right)$
$-1600(4) + 0 + 0 + 0 + \dfrac{200(4^3)}{4} = 6EI \left( \dfrac{h_C}{L_1} + \dfrac{h_C}{L_2} \right)$
$-3200 = 3EI \, h_C$
$EI \, h_C = -1066.67$
The negative sign indicates that points B and D are below point C, hence,
$\delta_C = \dfrac{1066.67}{EI} ~ \text{ upward}$ answer
