Problem 878 | Continuous Beam by Moment Distribution Method

$K_{12} = \dfrac{12}{4} \times \dfrac{3}{4} = 2.25$

$K_{23} = \dfrac{12}{4} = 3$

$K_{34} = \dfrac{12}{3} \times \dfrac{3}{4} = 3$

$DF_{12} = 1.0$

$DF_{21} = \dfrac{2.25}{2.25 + 3} = 3/7$

$DF_{23} = \dfrac{3}{2.25 + 3} = 4/7$

$DF_{32} = \dfrac{3}{3 + 3} = 1/2$

$DF_{34} = \dfrac{3}{3 + 3} = 1/2$

$DF_{43} = 1.0$

$\begin{align}
FEM_{21} & = \dfrac{Ma}{L}\left( \dfrac{3b}{L} - 1 \right) \\
& = \dfrac{8(2)}{4}\left( \dfrac{3 \times 2}{4} - 1 \right) \\
& = 2 ~ \text{kN}\cdot\text{m} \\
& = 2000 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{23} & = -\dfrac{w_o L^2}{12} \\
& = -\dfrac{2(4^2)}{12} \\
& = -8/3 ~ \text{kN}\cdot\text{m} \\
& = -2666.67 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{32} & = 8/3 ~ \text{kN}\cdot\text{m} \\
& = 2666.67 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{34} &= -\dfrac{Pab^2}{L^2} \\
& = -\dfrac{2(1)(2^2)}{3^2} \\
& = -8/3 ~ \text{kN}\cdot\text{m} \\
& = -2666.67 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{43} &= \dfrac{Pa^2b}{L^2} \\
& = \dfrac{2(1^2)(4)}{3^2} \\
& = 4/3 ~ \text{kN}\cdot\text{m} \\
& = 1333.33 ~ \text{N}\cdot\text{m}
\end{align}$