Problem 882 | Continuous Beam by Moment Distribution Method | Strength of Materials Review at MATHalino

Problem 882
Compute the moments over the supports of the beam shown in Fig. P-849.

Solution 882

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Beam Stiffness (Let I = 3)

Formula:

$K = \dfrac{I}{L}$

$K_{12} = 3/3 = 1$

$K_{23} = 3/3 = 1$

Distribution Factors

Formula:

$DF = \dfrac{K}{\Sigma K}$

$DF_{12} = 0$

$DF_{21} = \dfrac{1}{1 + 1} = 0.5$

$DF_{23} = \dfrac{1}{1 + 1} = 0.5$

$DF_{32} = 0$

Fixed End Moments

$\begin{align} FEM_{12} & = -\dfrac{w_{o1}{L_1}^2}{30} \\ & = -\dfrac{2000(3^2)}{30} \\ & = -600 ~ \text{N}\cdot\text{m} \end{align}$

$\begin{align} FEM_{21} & = \dfrac{w_{o1}{L_1}^2}{20} \\ & = \dfrac{2000(3^2)}{20} \\ & = 900 ~ \text{N}\cdot\text{m} \end{align}$

$\begin{align} FEM_{23} & = -\dfrac{w_{o1}{L_2}^2}{20} - \dfrac{w_{o2}{L_2}^2}{30} \\ & = -\dfrac{2000(3^2)}{20} - \dfrac{4000(3^2)}{30} \\ & = -2100 ~ \text{N}\cdot\text{m} \end{align}$

$\begin{align} FEM_{32} & = \dfrac{w_{o1}{L_2}^2}{30} + \dfrac{w_{o2}{L_2}^2}{20} \\ & = \dfrac{2000(3^2)}{30} + \dfrac{4000(3^2)}{20} \\ & = 2400 ~ \text{N}\cdot\text{m} \end{align}$

Joint	1		2	2		3
DF	0		0.5	0.5		0
FEM	-600		900	-2100		2400
	300	←	600	600	→	300
SUM	-300		1500	-1500		2700

Problem 882 | Continuous Beam by Moment Distribution Method

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