From

Case No. 7 of

Summary of Beam Loadings, deflection at the center is

$\delta = \dfrac{Pb}{48EI}(3L^2 - 4b^2) \text{ when } a \gt b$

Thus, for Fig. P-685

EI δ_{midspan} = EI δ_{midspan} due to 100 lb force + EI δ_{midspan} due to 80 lb force

$EI \, \delta_{midspan} = \sum\dfrac{Pb}{48}(3L^2 - 4b^2)$

$EI \, \delta_{midspan} = \dfrac{100(2)}{48}[ \, 3(9^2) - 4(2^2) \, ] + \dfrac{80(3)}{48}[ \, 3(9^2) - 4(3^2) \, ]$

$EI \, \delta_{midspan} = 945.83 + 1035$

$EI \, \delta_{midspan} = 1980.83 \, \text{ lb}\cdot\text{ft}^3$ *answer*