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Solution: - step(1) \rightarrow Reactions at supportf B D \rightarrow\begin{array}{r}\sum f_{y}=0 \rightarrow V_{A}=\frac{\omega_{0} L}{2} \\\Sigma M=0 \rightarrow M_{A}=\frac{\omega_{0} L}{2} \times \frac{3 L}{4} \\M_{A}=\frac{3 \omega_{0} L^{2}}{8}\end{array}Step (II) \rightarrow Moment equation at distance x from pt.AM(x)=\frac{\omega_{0} L}{2} x-\frac{3 \omega_{0} L^{2}}{8}\langle x\rangle^{0}-\frac{\omega_{0}}{2}\left\langle x-\frac{L}{2}\right\rangle^{2}Using double entegration methodE I \frac{d^{2} y}{d x^{2}}=M(x)\begin{array}{l}E I \frac{d^{2} y}{d x^{2}}=\frac{\omega_{0} L}{2} x-\frac{3 \omega_{0} L^{2}}{8}\langle x\rangle^{0}+\frac{\omega_{0}}{2}\left\langle x-\frac{L}{2}\right\rangle^{2} \\\text { EI } \frac{d y}{d x}=\frac{\omega_{0} L x^{2}}{4}-\frac{3 \omega_{0} L^{2}}{d}(x)-\frac{\omega_{0}}{6}\left\langle x-\frac{L}{2}\right\rangle^{3}+c_{1} \\E I y=\frac{\omega_{0} L x^{3}}{12}-\frac{3 \omega_{0} L^{2} x^{2}}{16}-\frac{\omega_{0}}{24}\left\langle x-\frac{L}{2}\right\rangle^{4}+G x+c_{2}-\end{array}Boundry condition \rightarrow\frac{d y}{d x}=\theta=0 and y=A=0 at x=0in equation (1) &(11) \rightarrow\begin{array}{l}c_{1}=0 \\c_{2}=0\end{array}Now slope at c \rightarrow V_{c}^{\prime} \rightarrow x=L in eq(1)\begin{array}{c}E I \frac{d y}{d x}=E I \theta_{c}=\frac{\omega_{0} L L^{2}}{4}-\frac{3 \omega_{0} L^{2} \times L}{8}-\frac{\omega_{0}}{6}\left(L-\frac{L}{2}\right)^{3} \\\Rightarrow E I V_{c}^{\prime}=-\frac{7}{48} \omega_{0} L^{3} \\\Rightarrow V_{c}{ }^{\prime}=-\frac{7 \omega_{0} L^{3}}{48 E I}\end{array}Ansdiflection atc V_{c} \rightarrow x=L in eq (11) -\begin{array}{l}E I y=E I v_{c}=\frac{\omega_{0} L X L^{3}}{12}-\frac{3 \omega_{0} L^{2} \times L^{2}}{16}-\frac{\omega_{0}}{94}\left(L-\frac{L}{2}\right)^{4} \\V_{c}=\frac{-41 \omega_{0} L^{4}}{384 E I}\end{array} Feel free to comment and ask clarifications. Thank you. ...