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The detailed solution is given below: The FBD of the beam A B is shoun belaw:\begin{aligned}\Sigma F_{x}=0 \Rightarrow & A_{A}=F \cos \theta=F \times \frac{3}{5} \\& \therefore 5 H_{A}=3 F \\\Sigma F_{y}=0 \Rightarrow & V_{A}+F \sin \alpha=30000 \Rightarrow V_{A}+\frac{4}{5} F=30000\end{aligned}\begin{aligned}\sum M_{A}=0 \Rightarrow & 30000 \times 2-(F \sin \theta \times 6)=0 \\& \Rightarrow 60000=F \times \frac{4}{5} \times 6 \\& \Rightarrow F=60000 \times 5 \\\quad V_{A}=30000 & =12500 \mathrm{~N}\end{aligned}From (2), V_{A}=30000-\frac{4}{5} F=30000-(4 / 5 \times 12500) \Rightarrow V_{A}=20000 \mathrm{~N}From (1) H_{A}=3 / 5 \mathrm{~F} \Rightarrow H_{A}=7500 \mathrm{~N}(13) Avenge sheah stues at A=\frac{R A}{2 A \text { rea of pinat } A} (due to double sheac)\begin{aligned}& 2 \text { Area of pinatA } \\= & \frac{\sqrt{7500^{2}+20000^{2}}}{2 \pi\left(10^{2}\right)} \quad\left(\gamma_{A}=10 \mathrm{~mm}\right) \\= & 37.99 \mathrm{MPa} \quad\left(\gamma_{B}=15 \mathrm{~mm}\right) \\= & F / \pi r_{b}^{2} \\= & 12500 / \pi\left(15^{2}\right) \quad A A 2 \mathrm{MAa} \\= & 17.68 \mathrm{MRa}\end{aligned}(14)\text { Anenage shean streys at } B(15) Shean load at pin B=F=12500 \mathrm{~N} ...