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 
Answer:
M_{1} = M_{4} = 280.39 lb·ft
M_{2} = M_{3} = 1939.22 lb·ft
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 Download the attached file below.