Beam Stiffness (Let I = 30)
Formula: $K = \dfrac{I}{L}$
$K_{AB} = 0$
$K_{BC} = \dfrac{30}{6} \times \dfrac{3}{4} = 3.75$
$K_{CD} = \dfrac{30}{10} = 3.0$
Distribution Factors
Formula: $DF = \dfrac{K}{\Sigma K}$
$DF_{BA} = 0$
$DF_{BC} = 1.0$
$DF_{CB} = \dfrac{3.75}{3.75 + 3.0} = 5/9$
$DF_{CD} = \dfrac{3.0}{3.75 + 3.0} = 4/9$
$DF_{DC} = 0$
Fixed End Moments
$\begin{align}
FEM_{BA} & = 50(4)(2) \\
& = 400 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{BC} & = \dfrac{w_oL^2}{12} \\
& = \dfrac{50(6^2)}{12} \\
& = 150 ~ \text{lb}\cdot\text{ft}
\end{align}$
$FEM_{CB} = 150 ~ \text{lb}\cdot\text{ft}$
$\begin{align}
FEM_{CD} & = \dfrac{w_oL^2}{30} \\
& = \dfrac{60(10^2)}{30} \\
& = 200 ~ \text{lb}\cdot\text{ft}
\end{align}$
$\begin{align}
FEM_{DC} & = \dfrac{w_oL^2}{20} \\
& = \dfrac{60(10^2)}{20} \\
& = 300 ~ \text{lb}\cdot\text{ft}
\end{align}$
Joint 
A 
B 
B 

C 
C 

D 
DF 

0 
1.0 

5/9 
4/9 

0 
FEM 

400 
150 

150 
200 

300 


0 
250 
→ 
125 








97.22 
77.78 
→ 
38.89 
SUM 

400 
400 

122.22 
122.22 

338.89 
Answer:
M_{B} = 400 lb·ft
M_{C} = 122.22 lb·ft
M_{D} = 338.89 lb·ft
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