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(1)a.\theta=\tan ^{-1}(3 / 4)for B C.\theta=63.5^{\circ} \text {. }\begin{array}{l}F_{B C}=P \cos \theta \quad \text { fores on } B C \text { are compressive } \\=50 \times \cos 63.5 \\\text { fores. } \\f_{B C}=30 \mathrm{kN} \text {. } \\\operatorname{Strues}_{B C}=f_{B C} / A \\A=\pi v^{2} \\A=\pi \times \frac{25^{2}}{4} \\A=490.625 \times 10^{-6} \mathrm{~m}^{2} \text {. } \\\operatorname{Stm}_{B C}=0.061 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \\\end{array}(a).for A C.\begin{array}{l}\text { Strus } A C=0.076 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \\\end{array}(b)(Diameter) D=20 \mathrm{~mm}. =20 \times 10^{-3} \mathrm{~m}.Shear stress (C)=P / A\begin{aligned}A & =\frac{\pi d^{2}}{4} \\\therefore \quad C & =\frac{30 \times 10^{3} \times 4 \times 10^{6}}{\pi \times 20 \times 20} \\C & =0.095 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\end{aligned} ...