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:
M_{2} = 1077.27 lb·ft
M_{3} = 640 lb·ft
Download the Excel File
The table above was done in MS Excel. You may download the file for your reference. Here is the link:
Why is it I=60? is it given?
You can assume any number for I, you can even use just I with no number. No it is not given. The use of 60 is I think the LCM of spans.