Part (1):
$M = \dfrac{wL^2}{8} = \dfrac{10(9^2)}{8}$
$M = 101.25 ~ \text{kN}\cdot\text{m}$
$f_b = \dfrac{Mc}{I} = \dfrac{101.25(1000^2)(306/2)}{145 \times 10^6}$
$f_b = 106.84 \text{MPa}$ ← answer
Part (2):
$\left( \dfrac{5wL^4}{384EI} - \dfrac{R_2 L^3}{48EI} \right)_\text{beam} = \left( \dfrac{R_2 H}{AE} \right)_\text{column}$
Since Ebeam = Ecolumn = 200 GPa, we can cancel out E from both sides of the equation.
$\dfrac{5wL^4}{384I} - \dfrac{R_2 L^3}{48I} = \dfrac{R_2 H}{A}$
$\dfrac{5(10)(9^4)(1000^4)}{384(145 \times 10^6)} - \dfrac{R_2(9^3)(1000^4)}{48(145 \times 10^6)} = \dfrac{R_2(4)(1000^2)}{\frac{1}{4}\pi (200^2 - 180^2)}$
$5,891,702.59 - 104,741.38R_2 = 670.13R_2$
$105,411.51R_2 = 5,891,702.59$
$R_2 = 55.89 \text{ kN}$
$f_c = \dfrac{R_2}{A} = \dfrac{55.89(1000)}{\frac{1}{4}\pi (200^2 - 180^2)}$
$f_c = 9.36 ~ \text{MPa}$ ← answer
Part (3):
$\Sigma F_v = 0$
$2R_1 + R_2 = 10(9)$
$2R_1 + 55.89 = 90$
$R_1 = 17.055 ~ \text{kN}$
$M_2 = 4.5R_1 - 10(4.5)(4.5/2)$
$M_2 = 4.5(17.055) - 101.25$
$M_2 = -24.5025 ~ \text{kN} \cdot \text{m}$
$f_b = \dfrac{Mc}{I} = \dfrac{24.5025(1000^2)(306/2)}{145 \times 10^6}$
$f_b = 25.85 ~ \text{MPa}$ ← answer
