Support Added at the Midspan of Simple Beam to Prevent Excessive Deflection

$M = 101.25 ~ \text{kN}\cdot\text{m}$
 

$f_b = \dfrac{Mc}{I} = \dfrac{101.25(1000^2)(306/2)}{145 \times 10^6}$

$f_b = 106.84 \text{MPa}$   ←   answer
 

Part (2):
 

$\dfrac{R_2 L^3}{48EI} = \dfrac{5wL^4}{384EI}$

$R_2 = \frac{5}{8}wL = \frac{5}{8}(1)(9)$

$R_2 = 56.25 ~ \text{kN}$
 

$f_c = \dfrac{R_2}{A} = \dfrac{56.25(1000)}{\frac{1}{4}\pi (200^2 - 180^2)}$

$f_c = 9.42 ~ \text{MPa}$   ←   answer
 

Part (3):
$\Sigma F_v = 0$

$2R_1 + R_2 = 10(9)$

$2R_1 + 56.25 = 90$

$R_1 = 16.875 ~ \text{kN}$
 

$M_2 = 4.5R_1 - 10(4.5)(4.5/2) = 4.5(16.875) - 101.25$

$M_2 = -25.3125 ~ \text{kN}$
 

$f_b = \dfrac{Mc}{I} = \dfrac{25.3125(1000^2)(306/2}{145 \times 10^6}$

$f_b = 26.71 ~ \text{MPa}$   ←   answer