$E = 1.5 \times 10^6 ~ \text{psi} = \left( 1.5 \times 10^6 \dfrac{\text{lb}}{\text{in.}^2} \right)\left( \dfrac{12 ~ \text{in.}}{1 ~ \text{ft.}} \right)^2 = 216 \times 10^6 ~ \text{psf}$
$I = 115.2 ~ \text{in.}^4 = \left( 115.2 ~ \text{in.}^4 \right)\left( \dfrac{1 ~ \text{ft.}}{12 ~ \text{in.}} \right)^4 = 1/180 ~ \text{ft.}^4$
$k = 300 ~ \text{lb/in.} = \left( 300 \dfrac{\text{lb}}{\text{in.}} \right)\left( \dfrac{12 ~ \text{in.}}{1 ~ \text{ft.}} \right) = 3600 ~ \text{lb/ft.}$
$6EI = 6(216 \times 10^6)(1/180) = 7.2 \times 10^6 ~ \text{lb}\cdot\text{ft.}^2$
$\delta_1 = R_1 / k = R_1 / 3600$
$M_2 = 10R_1 - 200(10)(5) = 10R_1 - 10,000$
Apply three-moment equation to span 1-2-3
$M_1 L_{12} + 2M_2(L_{12} + L_{23}) + M_3L_{23} + \left( \dfrac{6A\bar{a}}{L} \right)_{12} + \left( \dfrac{6A\bar{b}}{L} \right)_{32}$
$= 6EI \left( \dfrac{h_{21}}{L_{12}} + \dfrac{h_{23}}{L_{23}} \right)$
Note: for point 1 below point 2, h21 = -δ1
$0 + 2M_2(10 + 10) + 10M_3 + \dfrac{200(10^3)}{4} + \dfrac{3(1600)(10^2)}{8}$
$= (7.2 \times 10^6)\left( \dfrac{-\delta_1}{10} + 0 \right)$
$40M_2 + 10M_3 + 50,000 + 60,000 = -720,000\delta_1$
$40(10R_1 - 10,000) + 10M_3 + 110,000 = -720,000(R_1/3600)$
$400R_1 - 400,000 + 10M_3 + 110,000 = -200R_1$
$600R_1 + 10M_3 = 290,000$ ← Equation (1)
Apply three-moment equation to span 2-3-4
$M_2 L_{23} + 2M_3(L_{23} + L_{34}) + M_4L_{34} + \left( \dfrac{6A\bar{a}}{L} \right)_{23} + \left( \dfrac{6A\bar{b}}{L} \right)_{43} = 0$
$(10R_1 - 10,000)10 + 2M_3(10 + 0) + 0 + \dfrac{3(1600)(10^2)}{8} + 0 = 0$
$100R_1 - 100,000 + 20M_3 + 60,000 = 0$
$100R_1 + 20M_3 = 40,000$ ← Equation (2)
From Equation (1) and Equation (2)
$R_1 = 5,400/11 ~ \text{lb} = 490.91 ~ \text{lb}$ answer
$M_3 = -5,000/11 ~ \text{lb}\cdot\text{ft.} = -454.54 ~ \text{lb}\cdot\text{ft.}$
$\delta_1 = R_1 / 3600 = 0.136 ~ \text{ft.} = 1.64 ~ \text{in.}$ answer
