Problem 884 | Continuous Beam by Moment Distribution Method

$K_{01} = 0$

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

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

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

$DF_{10} = 0$

$DF_{12} = 1.0$

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

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

$DF_{32} = \dfrac{3}{3 + 4} = \dfrac{3}{7}$

$DF_{34} = \dfrac{4}{3 + 4} = \dfrac{4}{7}$

$DF_{43} = 0$

$\begin{align}
FEM_{12} & = -\dfrac{w_oL^2}{12} \\
& = -\dfrac{4(4^2)}{12} \\
& = -\dfrac{16}{3} ~ \text{kN}\cdot\text{m} \\
& = -5333.33 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{21} & = \dfrac{16}{3} ~ \text{kN}\cdot\text{m} \\
& = 5333.33 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{23} & = -\sum \dfrac{Pab^2}{L^2} \\
& = -\dfrac{2(1)(3^2)}{4^2} - \dfrac{2(3)(1^2)}{4^2} \\
& = -1.5 ~ \text{kN}\cdot\text{m} \\
& = -1500 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{32} & = -1.5 ~ \text{kN}\cdot\text{m} \\
& = -1500 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{34} & = -\dfrac{w_oL^2}{30} \\
& = -\dfrac{6(3^2)}{30} \\
& = -1.8 ~ \text{kN}\cdot\text{m} \\
& = -1800 ~ \text{N}\cdot\text{m}
\end{align}$
 

$\begin{align}
FEM_{43} & = \dfrac{w_oL^2}{20} \\
& = \dfrac{6(3^2)}{20} \\
& = 2.7 ~ \text{kN}\cdot\text{m} \\
& = 2700 ~ \text{N}\cdot\text{m}
\end{align}$