Problem 885 | Continuous Beam by Moment Distribution Method

$K_{12} = \dfrac{60}{10} = 6$

$K_{23} = \dfrac{60}{10} = 6$

$K_{23} = \dfrac{60}{12} = 5$

$DF_{12} = 0$

$DF_{21} = \dfrac{6}{6 + 6} = 1/2$

$DF_{23} = \dfrac{6}{6 + 6} = 1/2$

$DF_{32} = \dfrac{6}{6 + 5} = 6/11$

$DF_{34} = \dfrac{5}{6 + 5} = 5/11$

$DF_{43} = 0$

$\begin{align}
FEM_{21} & = \dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\
& = 833.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$\begin{align}
FEM_{23} & = -\dfrac{5w_oL^2}{96} \\
& = -\dfrac{5(160)(10^2)}{96} \\
& = -\dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\
& = -833.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$\begin{align}
FEM_{32} & = \dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\
& = 833.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$\begin{align}
FEM_{34} & = -\sum \dfrac{Pab^2}{L^2} \\
& = -\dfrac{400(3)(9^2)}{12^2} - \dfrac{400(9)(3^2)}{12^2} \\
& = -900 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$FEM_{43} = 900 ~ \text{lb}\cdot\text{ft}$