Beam Stiffness (Let
I = 60)
Formula: $K = \dfrac{I}{L}$
$K_{AB} = \dfrac{60}{12} \times \dfrac{3}{4} = 3.75$
$K_{BC} = \dfrac{60}{10} \times \dfrac{3}{4} = 4.5$
Distribution Factors
Formula: $DF = \dfrac{K}{\Sigma K}$
$DF_{AB} = DF_{CB} = 1.0$
$DF_{BA} = \dfrac{3.75}{3.75 + 4.5} = \dfrac{5}{11}$
$DF_{BC} = \dfrac{4.5}{3.75 + 4.5} = \dfrac{6}{11}$
Fixed End Moments
$\begin{align}
FEM_{AB} & = -\sum \dfrac{Pab^2}{L^2} \\
& = -\dfrac{300(4)(8^2)}{12^2} - \dfrac{300(8)(4^2)}{12^2}\\
& = -800 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{BA} & = \sum \dfrac{Pa^2b}{L^2} \\
& = \dfrac{300(4^2)(8)}{12^2} + \dfrac{300(8^2)(4)}{12^2}\\
& = 800 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{BC} & = -\dfrac{w_o L^2}{20} - \dfrac{w_o L^2}{12} \\
& = -\dfrac{75(10^2)}{20} - \dfrac{60(10^2)}{12}\\
& = -875 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{CB} & = \dfrac{w_o L^2}{30} + \dfrac{w_o L^2}{12} \\
& = \dfrac{75(10^2)}{30} + \dfrac{60(10^2)}{12}\\
& = 750 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{CD} & = -\frac{1}{2}(8)(60)(\frac{8}{3} \\
& = -640 ~ \text{lb}\cdot\text{ft}
\end{align}$
| Joint |
A |
|
B |
B |
|
C |
C |
D |
| DF |
1.0 |
|
5/11 |
6/11 |
|
1 |
0 |
|
| FEM |
-800 |
|
800 |
-875 |
|
750 |
-640 |
|
| |
800 |
→ |
400 |
-55 |
← |
-110 |
0 |
|
| |
|
|
-122.73 |
-147.27 |
|
|
|
|
| SUM |
0 |
|
1077.27 |
-1077.27 |
|
640 |
-640 |
|
Answer:
M2 = -1077.27 lb·ft
M3 = -640 lb·ft
Download the Excel File
The table above was done in MS Excel. You may download the file for your reference.
- See the attached file below.
