Answer
$\begin{aligned} & N_C=265 \mathrm{lb} \\ & V_C=-649 \mathrm{lb} \\ & M_C=-4.23 \mathrm{kip} \cdot \mathrm{ft} \\ & N_D=-265 \mathrm{lb} \\ & V_D=637 \mathrm{lb} \\ & M_D=-3.18 \mathrm{kip} \cdot \mathrm{ft}\end{aligned}$
Work Step by Step
Entire beam:
$
\begin{aligned}
↺+\Sigma M_A-0 ; \quad-150(5)-600(7.5)+B_y(15)-\frac{12}{13}(690)(25)-0 \\
B_y=1411.54 \mathrm{lb}
\end{aligned}
$
Segment $C B D$ :
$
\begin{array}{ll}
\pm \Sigma F_x=0 ; & -N_C-\frac{5}{13}(690)-0 \\
& N_C=-265 \mathrm{lb} \\
+\uparrow \Sigma F_y=0 ; & V_C-6-120+1411.54-690\left(\frac{12}{13}\right)-0 \\
& V_C--648.62--649 \mathrm{lb} \\
& -6(1)-120(1.5)+1411.54(3) \\
& -\frac{12}{13}(690)(13)-M_C-0 \\
& M_C=-4231.38 \mathrm{lb} \cdot \mathrm{ft}=-4.23 \mathrm{kip} \cdot \mathrm{ft}
\end{array}
$
Segment $D$ :
$
\begin{array}{ll}
\pm \Sigma F_x=0 ; & -N_D-\frac{5}{13}(690)-0 \\
& N_D=-265 \mathrm{lb} \\
+\uparrow \Sigma F_y=0 ; & V_D-\frac{12}{13}(690)-0 \\
& V_D=637 \mathrm{lb} \\
↺+\Sigma M_D=0 ; & -M_D-690\left(\frac{12}{13}\right)(5)-0 \\
& M_D=-3.18 \mathrm{kip} \cdot \mathrm{ft}
\end{array}
$
