Answer
$\begin{aligned} & N_D=0 \\ & V_D=3.00 \mathrm{kip} \\ & M_D=12.0 \mathrm{kip} \cdot \mathrm{ft} \\ & N_E=0 \\ & V_E=-8.00 \mathrm{kip} \\ & M_E=-20.0 \mathrm{kip} \cdot \mathrm{ft}\end{aligned}$
Work Step by Step
Support Reaction.
$
\begin{array}{ccc}
↺+\Sigma M_B=0 ; & 5(6)+6-A_y(12)=0 & A_y=3.00 \mathrm{kip} \\
↺+\Sigma M_A=0 ; & B_y(12)-5(6)+6=0 & B_y=2.00 \mathrm{kip} \\
\pm \Sigma F_X=0 & B_x=0 &
\end{array}
$
Internal Loading. Referring to the left segment of member $A B$ sectioned through $D$
$
\begin{array}{lll}
\pm \Sigma F_x=0 ; & N_D=0 & \\
+\uparrow \Sigma F_y=0 ; & 3.00-V_D=0 & V_D=3.00 \mathrm{kip} \\
↺+\Sigma M_D=0 ; & M_D+6-3.00(6)=0 & M_D=12.0 \mathrm{kip} \cdot \mathrm{ft}
\end{array}
$
Referring to the left segment of member $B C$ sectioned through $E$,
$
\begin{array}{llll}
\pm \Sigma F_x=0 ; & N_E=0 & & \\
+\uparrow \Sigma F_y=0 ; & -V_E-1.5(4)-2.00=0 & V_E=-8.00 \mathrm{kip} & \text { } \\
↺+\Sigma M_E=0 ; & M_E+1.5(4)(2)+2.00(4)=0 & M_E=-20.0 \mathrm{kip} \cdot \mathrm{ft} &
\end{array}
$
