$R_1 = R_2 = \frac{1}{2} \left[ \frac{1}{2}(12)(240) \right] = 720 ~ \text{lb}$
$M_B = M_D = -\frac{1}{2}(3)(120)(1) = -180 ~ \text{lb}\cdot\text{ft}$
$M_C = -\frac{1}{2}(6)(240)(2) + 3(720) = 720 ~ \text{lb}\cdot\text{ft}$
Apply three-moment equation to span A-B-D
$M_A L_{AB} + 2M_B(L_{AB} + L_{BD}) + M_D L_{BD} + \left( \dfrac{6A\bar{a}}{L} \right)_{AB} + \left( \dfrac{6A\bar{b}}{L} \right)_{DB}$
$= 6EI \left( \dfrac{h_{BA}}{L_{AB}} + \dfrac{h_{BD}}{L_{BD}} \right)$
$0 + 2(-180)(3 + 6) + (-180)(6) + \dfrac{8w_o L^3}{60} + \left[ \dfrac{5w_o L^3}{32} + \dfrac{w_o L^3}{4} \right]$
$= 6EI \left( \dfrac{h_{BA}}{3} + 0 \right)$
$-3240 - 1080 + \dfrac{8(120)(3^3)}{60} + \left[ \dfrac{5(120)(6^3)}{32} + \dfrac{120(6^3)}{4} \right] = 2EI \, h_{BA}$
$-3240 - 1080 + 432 + [4050 + 6480] = 2EI \, h_{BA}$
$2EI \, h_{BA} = 6642$
$EI \, h_{BA} = 3321 ~ \text{lb}\cdot\text{ft}^3$
The positive sign indicates that point A is above point B. Thus,
$EI \, \delta = 3321 ~ \text{lb}\cdot\text{ft}^3 ~ \text{up at ends}$ answer
Apply three-moment equation to span B-C-D
$M_B L_{BC} + 2M_C(L_{BC} + L_{CD}) + M_D L_{CD} + \left( \dfrac{6A\bar{a}}{L} \right)_{BC} + \left( \dfrac{6A\bar{b}}{L} \right)_{DC}$
$= 6EI \left( \dfrac{h_{CB}}{L_{BC}} + \dfrac{h_{CD}}{L_{CD}} \right)$
$-180(3) + 2(720)(3 + 3) + (-180)(3) + \left[ \dfrac{8w_o L^3}{60} + \dfrac{w_o L^3}{4} \right] + \left[ \dfrac{8w_o L^3}{60} + \dfrac{w_o L^3}{4} \right]$
$= 6EI \left( \dfrac{h_C}{3} + \dfrac{h_C}{3} \right)$
$-540 + 8640 - 540 + \left[ \dfrac{8(120)(3^3)}{60} + \dfrac{120(3^3)}{4} \right] + \left[ \dfrac{8(120)(3^3)}{60} + \dfrac{120(3^3)}{4} \right]$
$= 4EI \, h_C$
$-540 + 8640 - 540 + [432 + 810] + [432 + 810] = 4EI \, h_C$
$4EI \, h_C = 10,044$
$EI \, h_C = 2511 ~ \text{lb}\cdot\text{ft}^3$
The positive sign indicates that points B and D are above point C. Thus,
$EI \, \delta = 2,511 ~ \text{lb}\cdot\text{ft}^3 ~ \text{down at midspan}$ answer
