$\Sigma M_{R2} = 0$
$3R_1 = 600 + \frac{1}{2}(3)(900)(1)$
$R_1 = 650 \, \text{ N}$
$\Sigma M_{R2} = 0$
$3R_2 + 600 = \frac{1}{2}(3)(900)(2)$
$R_2 = 700 \, \text{ N}$
$EI \, t_{C/B} = (Area_{BC}) \, \bar{X}_C$
$EI \, t_{C/B} = \frac{1}{2}(3)(1950)(1) - 3(600)(\frac{3}{2}) - \frac{1}{4}(3)(1350)(\frac{3}{5})$
$EI \, t_{C/B} = -382.5 \, \text{ N}\cdot\text{m}^3$
$EI \, t_{A/B} = (Area_{AB}) \, \bar{X}_A$
$EI \, t_{A/B} = - 1(600)(\frac{1}{2})$
$EI \, t_{A/B} = -300 \, \text{ N}\cdot\text{m}^3$
The negative signs above indicates only the location of elastic curve relative to the reference tangent. It does not indicate magnitude. It shows that the elastic curve is below the reference tangent at points A and C.
By ratio and proportion
$\dfrac{\delta_A - t_{A/B}}{1} = \dfrac{t_{C/B}}{3}$
$\delta_A = \frac{1}{3}t_{C/B} + t_{A/B}$
$EI \, \delta_A = \frac{1}{3}EI \, t_{C/B} + EI \, t_{A/B}$
$EI \, \delta_A = \frac{1}{3}(382.5) + 300$
$EI \, \delta_A = 427.5 \, \text{ N}\cdot\text{m}^3$ answer
