Beam Stiffness (Let I = 60)
Formula: $K = \dfrac{I}{L}$
$K_{12} = \dfrac{60}{10} = 6$
$K_{23} = \dfrac{60}{10} = 6$
$K_{23} = \dfrac{60}{12} = 5$
Distribution Factors
Formula: $DF = \dfrac{K}{\Sigma K}$
$DF_{12} = 0$
$DF_{21} = \dfrac{6}{6 + 6} = 1/2$
$DF_{23} = \dfrac{6}{6 + 6} = 1/2$
$DF_{32} = \dfrac{6}{6 + 5} = 6/11$
$DF_{34} = \dfrac{5}{6 + 5} = 5/11$
$DF_{43} = 0$
Fixed End Moments
$\begin{align}
FEM_{12} & = \dfrac{w_oL^2}{12} \\
& = \dfrac{100(10^2)}{12} \\
& = \dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\
& = 833.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{21} & = \dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\
& = 833.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{23} & = \dfrac{5w_oL^2}{96} \\
& = \dfrac{5(160)(10^2)}{96} \\
& = \dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\
& = 833.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{32} & = \dfrac{2500}{3} ~ \text{lb}\cdot\text{ft} \\
& = 833.33 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{34} & = \sum \dfrac{Pab^2}{L^2} \\
& = \dfrac{400(3)(9^2)}{12^2}  \dfrac{400(9)(3^2)}{12^2} \\
& = 900 ~ \text{lb}\cdot\text{ft}
\end{align}$
$FEM_{43} = 900 ~ \text{lb}\cdot\text{ft}$
Joint 
1 

2 
2 

3 
3 

4 
DF 
0 

1/2 
1/2 

6/11 
5/11 

0 
FEM 
833.33 

833.33 
833.33 

833.33 
900 

900 




18.18 
← 
36.36 
30.30 
→ 
15.15 

4.55 
← 
9.09 
9.09 
→ 
4.55 







1.24 
← 
2.48 
2.07 
→ 
1.03 

0.31 
← 
0.62 
0.62 
→ 
0.31 







0.08 
← 
0.17 
0.14 
→ 
0.07 

0.02 
← 
0.04 
0.04 
→ 
0.02 







0.01 
← 
0.01 
0.01 
→ 
0.00 

0.00 
← 
0.00 
0.00 





SUM 
838.31 

823.58 
823.58 

867.48 
867.48 

916.26 
Answer:
M_{1} = 838.21 lb·ft
M_{2} = 823.58 lb·ft
M_{3} = 867.48 lb·ft
M_{4} = 916.26 lb·ft
Download the Microsoft Excel File
 Download the attached file below.