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 |
Answer:
M1 = -300 N·m
M2 = -1500 N·m
M3 = -2700 N·m
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