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\begin{array}{l} \Sigma F_{y}=0 \quad R A+R_{B}=\frac{1}{2} \times 4 \times 6 \\=12 \mathrm{FH} \\\Sigma M_{A}=0 \quad R B \times 6-\left[\frac{1}{2} \times 4 \times 6\right] \times(6 \times Y / 3)=0\end{array}\Sigma M_{A}=0 \quad R B \times 6-\left[\frac{1}{2} \times 4 \times 6\right] \times(6 \times Y / 3)=0\begin{array}{l}R_{B}=4 \mathrm{KN} \\R A=12-4=8 \mathrm{kN} .\end{array}Internal loading at point D\begin{array}{l}M=4 \times 1.5-\left(\frac{1}{2} \times 1.5 \times 1\right) \frac{1.5}{3} \\M=5.625 \mathrm{kN}-m \\V=\text { shear force } \\V=\frac{1}{2} \times 1.5-4 \\V=-3.25\end{array} ...