$\Sigma M_{R2} = 0$
$9R_1 + 20(4)(2) = \frac{1}{2}(6)(60)(5)$
$R_1 = \frac{740}{9} ~ \text{lb}$
$\Sigma M_{R1} = 0$
$9R_2 = \frac{1}{2}(6)(60)(4) + 20(4)(11)$
$R_2 = \frac{1600}{9} ~ \text{lb}$
$M_A = M_D = 0$
$M_B = 6(\frac{740}{9}) - \frac{1}{2}(6)(60)(2) = \frac{400}{3} ~ \text{lb}\cdot\text{ft}$
$M_C = -20(4)(2) = -160 ~ \text{lb}\cdot\text{ft}$
At x = 6 ft (labeled as point B), use span A-B-C
$M_A L_{AB} + 2M_B(L_{AB} + L_{BC}) + M_C L_{BC} + \left( \dfrac{6A_1\bar{a}}{L} \right)_{AB} + \left( \dfrac{6A_2\bar{b}}{L} \right)_{CB}$
$= 6EI \left( \dfrac{h_{BA}}{L_{AB}} + \dfrac{h_{BC}}{L_{BC}} \right)$
$0 + 2(\frac{400}{3})(6 + 3) - 160(3) + \dfrac{8(60)(6^3)}{60} + 0 = 6EI \left( \dfrac{h_B}{6} + \dfrac{h_B}{3} \right)$
$3648 = 3EI \, h_B$
$EI \, h_B = 1216$ ← the positive sign indicates that A and C are above B
$\delta_B = \dfrac{1216}{EI} ~ \text{lb}\cdot\text{ft}^3 ~ \text{downward}$ answer
At the end of overhang (labeled as point D), use span B-C-D
$M_B L_{BC} + 2M_C (L_{BC} + L_{CD}) + M_D L_{CD} + \left( \dfrac{6A \bar{a}}{L} \right)_{BC} + \left( \dfrac{6A_2\bar{b}_2}{L_2} \right)_{DC}$
$= 6EI \left( \dfrac{h_{CB}}{L_{BC}} + \dfrac{h_{CD}}{L_{CD}} \right)$
Note: point B is below point C
$\frac{400}{3}(3) + 2(-160)(3 + 4) + 0 + 0 + \frac{1}{4}(20)(4^3) = 6EI \left( \dfrac{-1216}{3EI} + \dfrac{h_D}{4} \right)$
$-1520 = -2432 + \frac{3}{2}EI \, h_D$
$EI \, h_D = 608$ ← point D is above point C
$\delta_D = \dfrac{608}{EI} ~ \text{lb}\cdot\text{ft}^3 ~ \text{upward}$ answer
