Question For the cantilever beam and loading shown, using the area moment method, determine the slope at the free end in degree measure. Use three decimal places. Express your final answer in absolute value. Assume E = 200 GPa and I = 6.3617x 10-7m4. 1 KN TA III, 2 KN/m B с 0.40 m 0.60 m

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 ...