Answer
$\begin{aligned} & N_C=0 \\ & V_C=-1.50 \mathrm{kN} \\ & M_C=13.5 \mathrm{kN} \cdot \mathrm{m}\end{aligned}$
Work Step by Step
Support Reactions.
$
\begin{aligned}
& \zeta+\Sigma M_A=0 ; \quad B_y(6)-\frac{1}{2}(6)(6)(2)=0 \quad B_y=6.00 \mathrm{kN} \\
& \pm \Sigma F_x=0 ; \quad B_x=0 \\
&
\end{aligned}
$
Internal Loadings. Referring to the $F B D$ of right segment of the beam sectioned through $C$, Fig.
$
\begin{array}{lll}
\rightarrow \Sigma F_x=0 ; & N_C=0 \\
+\uparrow \Sigma F_y=0 ; & V_C+6.00-\frac{1}{2}(3)(3)=0 &\\ V_C=-1.50 \mathrm{kN} \\\\
\zeta+\Sigma M_C=0 ; & 6.00(3)-\frac{1}{2}(3)(3)(1)-M_C=0 & \\M_C=13.5 \mathrm{kN} \cdot \mathrm{m}
\end{array}
$
