Problem 877 | Continuous Beam by Moment Distribution Method

$K_{AB} = \dfrac{60}{12} \times \dfrac{3}{4} = 3.75$

$K_{BC} = \dfrac{60}{10} \times \dfrac{3}{4} = 4.5$

$DF_{AB} = DF_{CB} = 1.0$

$DF_{BA} = \dfrac{3.75}{3.75 + 4.5} = \dfrac{5}{11}$

$DF_{BC} = \dfrac{4.5}{3.75 + 4.5} = \dfrac{6}{11}$

$\begin{align} FEM_{BA} & = \sum \dfrac{Pa^2b}{L^2} \\ & = \dfrac{300(4^2)(8)}{12^2} + \dfrac{300(8^2)(4)}{12^2}\\ & = 800 ~ \text{lb}\cdot\text{ft} \end{align}$
 

$\begin{align} FEM_{BC} & = -\dfrac{w_o L^2}{20} - \dfrac{w_o L^2}{12} \\ & = -\dfrac{75(10^2)}{20} - \dfrac{60(10^2)}{12}\\ & = -875 ~ \text{lb}\cdot\text{ft} \end{align}$
 

$\begin{align} FEM_{CB} & = \dfrac{w_o L^2}{30} + \dfrac{w_o L^2}{12} \\ & = \dfrac{75(10^2)}{30} + \dfrac{60(10^2)}{12}\\ & = 750 ~ \text{lb}\cdot\text{ft} \end{align}$
 

$\begin{align} FEM_{CD} & = -\frac{1}{2}(8)(60)(\frac{8}{3} \\ & = -640 ~ \text{lb}\cdot\text{ft} \end{align}$

• See the attached file below.