$\Sigma M_{R1} = 0$
$6R_2 = \frac{1}{2}(4)(600)(\frac{4}{3})$
$R_2 = \frac{800}{3} \, \text{ N}$
$EI \, t_{A/B} = (Area_{AB}) \, \bar{X}_A$
$EI \, t_{A/B} = \frac{1}{2}(6)(1600)(2) - \frac{1}{4}(4)(1600)(\frac{4}{5})$
$EI \, t_{A/B} = 8320 \, \text{N}\cdot\text{m}^3$
$EI \, t_{M/B} = (Area_{MB}) \, \bar{X}_M$
$EI \, t_{M/B} = \frac{1}{2}(3)(800)(1) - \frac{1}{4}(1)(25)(\frac{1}{5})$
$EI \, t_{M/B} = 1198.75 \, \text{N}\cdot\text{m}^3$
By ratio and proportion:
$\dfrac{\delta_m + t_{M/B}}{3} = \dfrac{t_{A/B}}{6}$
$\delta_m + t_{M/B} = \frac{1}{2}t_{A/B}$
$EI \, \delta_m + EI \, t_{M/B} = EI \, \frac{1}{2}t_{A/B}$
$EI \, \delta_m + 1198.75 = EI \, \frac{1}{2}(8320)$
$EI \, \delta_m = 2961.25 \, \text{N}\cdot\text{m}^3$ answer
