Problem 881 | Continuous Beam by Moment Distribution Method

$K_{AB} = 0$

$K_{BC} = \dfrac{30}{6} \times \dfrac{3}{4} = 3.75$

$K_{CD} = \dfrac{30}{10} = 3.0$

$DF_{BA} = 0$

$DF_{BC} = 1.0$

$DF_{CB} = \dfrac{3.75}{3.75 + 3.0} = 5/9$

$DF_{CD} = \dfrac{3.0}{3.75 + 3.0} = 4/9$

$DF_{DC} = 0$

$\begin{align}
FEM_{BC} & = -\dfrac{w_oL^2}{12} \\
& = -\dfrac{50(6^2)}{12} \\
& = -150 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$FEM_{CB} = 150 ~ \text{lb}\cdot\text{ft}$
 

$\begin{align}
FEM_{CD} & = -\dfrac{w_oL^2}{30} \\
& = -\dfrac{60(10^2)}{30} \\
& = -200 ~ \text{lb}\cdot\text{ft}
\end{align}$
 

$\begin{align}
FEM_{DC} & = \dfrac{w_oL^2}{20} \\
& = \dfrac{60(10^2)}{20} \\
& = 300 ~ \text{lb}\cdot\text{ft}
\end{align}$