Problem 879 | Continuous Beam by Moment Distribution Method

$K_{12} = \dfrac{12}{12} \times \dfrac{3}{4} = 0.75$

$K_{23} = \dfrac{12}{12} = 1$

$K_{34} = \dfrac{12}{12} \times \dfrac{3}{4} = 0.75$

$K_{45} = 0$

$DF_{12} = 1.0$

$DF_{21} = \dfrac{0.75}{0.75 + 1} = 3/7$

$DF_{23} = \dfrac{1}{0.75 + 1} = 4/7$

$DF_{32} = \dfrac{1}{1 + 0.75} = 4/7$

$DF_{34} = \dfrac{0.75}{1 + 0.75} = 3/7$

$DF_{43} = 1.0$

$DF_{45} = 0.0$

$\begin{align}
FEM_{21} & = \dfrac{w_oL^2}{20} \\
& = \dfrac{150(12^2)}{20} \\
& = 1080 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$\begin{align}
FEM_{23} & = -\sum \dfrac{Pab^2}{L^2} \\
& = -\dfrac{400(3)(9^2)}{12^2} - \dfrac{400(8)(4^2)}{12^2} \\
& = -\dfrac{2825}{3} ~ \text{lb}\cdot\text{ft} \\
& = -941.67 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$\begin{align}
FEM_{32} & = \sum \dfrac{Pa^2b}{L^2} \\
& = \dfrac{400(3^2)(9)}{12^2} + \dfrac{400(8^2)(4)}{12^2} \\
& = \dfrac{2275}{3} ~ \text{lb}\cdot\text{ft} \\
& = 758.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$\begin{align}
FEM_{34} & = -\dfrac{w_oL^2}{12} \\
& = -\dfrac{50(12^2)}{12} \\
& = -600 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$FEM_{43} = 600 ~ \text{lb}\cdot\text{ft}$
 

$\begin{align}
FEM_{45} & = -50(4)(2) \\
& = -400 ~ \text{lb}\cdot\text{ft}
\end{align}$