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solutiou!As per molnent Area theorem 1ir The diffreve of clopes between the any two paints is Egvalto the area of \left(\frac{m}{E R}\right. )diggram between the two painta."So first we will draw Bendingmoment (m) Diagram. For simplicity of calculation, I will draw berdirg moment diagram separately for forces 1 \mathrm{KM} and 2 \mathrm{kM} / \mathrm{m} and then apply method of superposition to add them.Calculatior.Bendizg morent at A Due to 1 \mathrm{KH} load =1 \times 0=0Bending momert at C Due to 1 \mathrm{kH} loud =1 \times(\cdot 4+6)=1 \mathrm{kN}-\mathrm{m}Berding momertat B due to 2ktrm lund =0Bending moment ot C due to 21 \mathrm{~km} / \mathrm{m} land =2 \times 0.6 \times \frac{0.6}{2}=0.36 \mathrm{krm}Nou Add the Area of \frac{m}{E I} Diagram Dre to loud 1 \mathrm{KM} and 2 \mathrm{~km} / \mathrm{m}.Herle combined Area of \frac{m}{E I} Diagram, as per principle of Superposition,\begin{array}{l}A=\frac{1}{2} \times \frac{1}{E I} \times 1+0.6 \times \frac{0.36}{\varepsilon I} \times \frac{1}{3} \\\Rightarrow A=\frac{0.572}{E I} \\\end{array}But As per moment area theorem 1, we know that this area A is Equal to Diftrerce of slobe between paints A and C.\Rightarrow \quad \theta_{A}-\theta_{C}=A=\frac{0.572}{E I}But Q_{C}=0 because at fined हnd C Ho Rotation will take Place.\begin{array}{l}\Rightarrow \quad \theta_{A}=\frac{0.572}{E I} \\E=200 \mathrm{GPp}=200 \times 10^{3} \mathrm{mPa} \\I=6.3617 \times 10^{-7} \mathrm{m4} \\=6.3617 \times 10^{-7} \times 10^{12} \mathrm{~mm} 4 \\\Rightarrow Q_{A}=\frac{0.572 \times 106 \times 10^{3}\left(\mathrm{M}-\mathrm{mm}^{2}\right)}{200 \times 10^{3} \times 6.3617 \times 10^{5}\left(\mathrm{M}-\mathrm{mm}^{2}\right)} \\=6.3617 \times 10^{5} \mathrm{~mm} 4 \\\Rightarrow Q A=4.4956 \times 10^{-3} \mathrm{rad} \\\end{array}\begin{aligned}Q_{A} & =4.4956 \times 10^{-3} \times \frac{180}{\pi} \text { degree } \\\Rightarrow Q_{A} & =0.2576 \text { degree }\end{aligned}Ans ...