Beam Stiffness (Let I = 10)
Formula: $K = \dfrac{I}{L}$
$K_{BA} = 0$
$K_{BC} = \dfrac{10}{10} \times \dfrac{3}{4} = 0.75$
$K_{CD} = \dfrac{10}{10} = 1.0$
Distribution Factors
Formula: $DF = \dfrac{K}{\Sigma K}$
$DF_{BA} = 0$
$DF_{BC} = 1.0$
$DF_{CB} = \dfrac{0.75}{1.0 + 0.75} = 3/7$
$DF_{CD} = \dfrac{1.0}{1.0 + 0.75} = 4/7$
$DF_{DC} = 0$
Fixed End Moments
$\begin{align}
FEM_{BA} & = 500(6)(3) \\
& = 9000 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{BC} & = -\dfrac{w_oL^2}{12} \\
& = -\dfrac{500(10^2)}{12} \\
& = -\dfrac{12,500}{3} ~ \text{lb}\cdot\text{ft} \\
& = -4166.67 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{CB} & = \dfrac{12,500}{3} ~ \text{lb}\cdot\text{ft} \\
& = 4166.67 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{CD} & = -\dfrac{Pab^2}{L^2} \\
& = -\dfrac{4000(7)(3^2)}{10^2} \\
& = -2520 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{DC} & = \dfrac{Pa^2b}{L^2} \\
& = \dfrac{4000(7^2)(3)}{10^2} \\
& = 5880 ~ \text{lb}\cdot\text{ft}
\end{align}$
| Joint |
A |
B |
B |
|
C |
C |
|
D |
| DF |
|
0 |
1.0 |
|
3/7 |
4/7 |
|
0 |
| FEM |
|
9000 |
-4166.67 |
|
4166.67 |
-2520 |
|
5880 |
| |
|
0 |
-4833.33 |
→ |
-2416.67 |
|
|
|
| |
|
|
|
|
330.00 |
440.00 |
→ |
220.00 |
| SUM |
|
9000 |
-9000 |
|
2080 |
-2080 |
|
6100 |
Answer:
MB = -9000 lb·ft
MC = -2080 lb·ft
MD = -6100 lb·ft
Download the MS Excel File of the Table
- See the attached file below.
