$EI = 200,000(60.8 \times 10^6)$
$EI = 1.216 \times 10^{13} ~ \text{N}\cdot \text{mm}^2$
Part (1)
$\delta = \dfrac{PL^3}{3EI}$
$\delta = \dfrac{200(1.75^3)(1000^4)}{3(1.216 \times 10^{13})}$
$\delta = 29.38 ~ \text{mm}$ ← answer
Part (2)
$EI \, t_{C/A} = (\text{Area}_{AC}) \cdot \bar{X}_C = 0$
$\frac{1}{2}(3.5)(3.5R)\left[ \frac{2}{3}(3.5) \right] - \frac{1}{2}(1.75)(350)\left[ 1.75 + \frac{2}{3}(1.75) \right] = 0$
$14.291R = 893.229$
$R = 62.5 ~ \text{kN}$ ← answer
Another Solution for Part (2)
Using Superposition Method
$\dfrac{R(3.5^3)}{3EI} = \dfrac{200(1.75^3)}{3EI} + 1.75 \left[ \dfrac{200(1.75^2)}{2EI} \right]$
$\dfrac{343}{24}R = \dfrac{42,875}{48}$
$R = 62.5 ~ \text{kN}$ ← (okay!)
Part (3)
$EI \, t_{B/A} = (\text{Area}_{AB}) \cdot \bar{X}_B$
$EI \, t_{B/A} = \frac{1}{2}(1.75)(3.5R)\left[ \frac{2}{3}(1.75) \right] + \frac{1}{2}(1.75)(1.75R)\left[ \frac{1}{3}(1.75) \right] - \frac{1}{2}(1.75)(350)\left[ \frac{2}{3}(1.75) \right]$
$EI \, t_{B/A} = 3.573R + 0.893R - 357.292$
$EI \, t_{B/A} = 4.466R - 357.292$
$(1.216 \times 10^{13})(-9.5) = (4.466R - 357.292)(1000^4)$
$R = 54.14 ~ \text{kN}$ ← answer
