Answer
$\begin{aligned} & N_C=0 \\ & V_C=-386 \mathrm{lb} \\ & M_C=-857 \mathrm{lb} \cdot \mathrm{ft} \\ & N_D=0 \\ & V_D=300 \mathrm{lb} \\ & M_D=-600 \mathrm{lb} \cdot \mathrm{ft}\end{aligned}$
Work Step by Step
Support Reactions:
$
\begin{aligned}
& ↺+\Sigma M_B=0 ; \quad 500(8)-300(8)-A_y(14)-0 \\
& A_y=114.29 \mathrm{lb}
\end{aligned}
$
Intermal Forces:
$
\begin{aligned}
& \text { 士 } \Sigma F_x=0 \quad N_C=0 \\
& +\uparrow \Sigma F_y=0 ; \quad 114.29-500-V_C=0 \quad V_C=-386 \mathrm{lb} \\
& ↺+\Sigma M_C=0 \\
& M_C+500(4)-114.29(10)-0 \\
& M_C=-857 \mathrm{lb} \cdot \mathrm{ft} \\
&
\end{aligned}
$
Applying the equations of equilibrium to segment ED we have
$
\begin{gathered}
\rightarrow \Sigma F_s=0 ; \quad N_D-0 \\
+1 \Sigma F_y=0 ; \quad V_D-300=0 \quad V_D=300 \mathrm{lb} \\
↺+\Sigma M_D=0 ; \quad-M_D-300(2)=0 \quad M_D=-600 \mathrm{lb} \cdot \mathrm{ft}
\end{gathered}
$
