Problem 883 | Continuous Beam by Moment Distribution Method | Strength of Materials Review at MATHalino

Problem 883
Compute the moments over the supports of the beam shown in Fig. P-853.

Solution 883

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Beam Stiffness (Let I = 60)

Formula:

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

$K_{12} = 60/10 = 6$

$K_{23} = 60/12 = 5$

$K_{34} = 60/10 = 6$

Distribution Factors

Formula:

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

$DF_{12} = 0$

$DF_{21} = \dfrac{6}{6 + 5} = 6/11$

$DF_{23} = \dfrac{5}{6 + 5} = 5/11$

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

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

$DF_{43} = 0$

Fixed End Moments

$\begin{align} FEM_{12} & = -\dfrac{w_oL^2}{12} \\ & = -\dfrac{100(10^2)}{12} \\ & = -\dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\ & = -833.33 ~ \text{lb}\cdot\text{ft} \end{align}$

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

$\begin{align} FEM_{23} & = -\dfrac{w_oL^2}{12} - \dfrac{PL}{8} \\ & = -\dfrac{100(12^2)}{12} - \dfrac{800(12)}{8} \\ & = -2400 ~ \text{lb}\cdot\text{ft} \end{align}$

$FEM_{32} = 2400 ~ \text{lb}\cdot\text{ft}$

$\begin{align} FEM_{34} & = -\dfrac{w_oL^2}{12} \\ & = -\dfrac{100(10^2)}{12} \\ & = -\dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\ & = -833.33 ~ \text{lb}\cdot\text{ft} \end{align}$

$\begin{align} FEM_{43} & = \dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\ & = 833.33 ~ \text{lb}\cdot\text{ft} \end{align}$

Joint	1		2	2		3	3		4
DF	0		6/11	5/11		5/11	5/11		0
FEM	-833.33		833.33	-2400		2400	-833.33		833.33
	427.27	←	854.55	712.12	→	356.06
				-436.98	←	-873.97	-1048.76	→	-524.38
	119.18	←	238.35	198.63	→	99.31
				-22.57	←	-45.14	-54.17	→	-27.09
	6.16	←	12.31	10.26	→	5.13
				-1.17	←	-2.33	-2.80	→	-1.40
	0.32	←	0.64	0.53	→	0.26
				-0.06	←	-0.12	-0.14	→	-0.07
	0.02	←	0.03	0.03	→	0.01
				0.00	←	-0.01	-0.01	→	0.00
	0.00	←	0.00	0.00
SUM	-280.39		1939.22	-1939.22		1939.22	-1939.22		280.39

Problem 883 | Continuous Beam by Moment Distribution Method

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