Answer
$\begin{aligned} & V_C=-4.00 \mathrm{kip} \\ & M_C=24.0 \mathrm{kip} \cdot \mathrm{ft}\end{aligned}$
Work Step by Step
Support Reactions. Referring to the $F B D$ of the entire beam shown in Fig. $a$,
$
↺+\Sigma M_A=0 ; \quad B_y(12)-\frac{1}{2}(4)(6)(4)=0 \quad B_y=4.00 \mathrm{kip}\\
$
Internal Loadings.
$
\begin{array}{lll}
+\uparrow \Sigma F_y=0 ; & V_C+4.00=0 & V_C=-4.00 \mathrm{kip} \\
↺+\Sigma M_C=0 ; & 4.00(6)-M_C=0 & M_C=24.0 \mathrm{kip} \cdot \mathrm{ft}
\end{array}
$
