From the load diagram
$\Sigma M_{R1} = 0$
$6R_2 = \frac{1}{2}(4)(600)[\frac{1}{3}(4)]$
$R_2 = 266.67 \, \text{ N}$
$\dfrac{y}{1} = \dfrac{600}{4}$
$y = 150 \, \text{ N/m}$
From the moment diagram
$a = 3R_2 = 3(266.67)$
$a = 800 \, \text{ N}\cdot\text{m}$
$b = -\frac{1}{2}(1)(y)[\frac{1}{3}(1)]$
$b = -\frac{1}{6}y = -\frac{1}{6}(150)$
$b = -25 \, \text{ N}\cdot\text{m}$
From the conjugate beam
$\Sigma M_{F1} = 0$
$6F_2 + \frac{1}{4}(4)(1600)[\frac{1}{5}(4)] = \frac{1}{2}(6)(1600)[\frac{1}{3}(6)]$
$F_2 = 1386.67 \, \text{ N}\cdot\text{m}^2$
$M_{midspan} = \frac{1}{2}(3a)[\frac{1}{3}(3)] - \frac{1}{4}(1b)[\frac{1}{5}(1)] - 3F_2$
$M_{midspan} = \frac{1}{2}(3)(800)[\frac{1}{3}(3)] - \frac{1}{4}(1)(25)[\frac{1}{5}(1)] - 3(1386.67)$
$M_{midspan} = -2961.25 \, \text{ N}\cdot\text{m}^3$
Thus, the deflection at the midspan is
$EI \, \delta_m = M_{midspan}$
$EI \, \delta_m = -2961.25 \, \text{ N}\cdot\text{m}^3$
$EI \, \delta_m = 2961.25 \, \text{ N}\cdot\text{m}^3$ below the neutral axis answer
