From Span 1-2
$R_1 = 500 - 98 = 402 \, \text{lb}$
Moment at the center of the first span
$M_A = 5R_1 - 100(5)(2.5)$
$M_A = 5(402) - 1250$
$M_A = 760 ~ \text{lb}\cdot\text{ft}$
Apply three-moment equation to span A-2-3
$M_A L_{A2} + 2M_2(L_{A2} + L_{23}) + M_3L_{23} + \left( \dfrac{6A\bar{a}}{L} \right)_{A2} + \left( \dfrac{6A\bar{b}}{L} \right)_{32}$
$= 6EI \left( \dfrac{h_{2A}}{L_{A2}} + \dfrac{h_{23}}{L_{23}} \right)$
$760(5) + 2(-980)(5 + 10) + (-1082)(10) + \dfrac{w_o L^3}{4} + \dfrac{5w_o L^3}{32} = 6EI \left( \dfrac{h_A}{5} + 0 \right)$
$3800 - 29,400 - 10,820 + \dfrac{100(5^3)}{4} + \dfrac{5(160)(10^3)}{32} = \frac{6}{5}EI \, h_A$
$3,800 - 29,400 - 10,820 + 3,125 + 25,000 = \frac{6}{5}EI \, h_A$
$\frac{6}{5}EI \, h_A = -8,295$
$EI \, h_A = -6,912.5 ~ \text{lb}\cdot\text{ft}^3$
The negative sign means that point A is below point 2. Thus,
$EI \, \delta = 6,912.5 ~ \text{lb}\cdot\text{ft}^3 ~ \text{down}$ answer
