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Q.1)Dording moments.(i) A+A:-D \cdot M_{A}=0(ii)\begin{aligned}A+B:-B \cdot M D & =-1 \times 0.40 \\& =-0.40 \mathrm{kNm} .\end{aligned}(iii)\begin{aligned}\text { At C:- } \nabla \cdot M_{C} & =-1 \times 1.0-2 \times \frac{0.60^{2}}{2} \\B \cdot M_{C} & =-1.36 \mathrm{kNm}(A) .\end{aligned}Finding Areas & controids.(i) For are bitwer D and C.Toking a section x-x at o distara x from point C in sigmont BC and cakulating momont about tho pection in toims of x\begin{array}{l}M_{x}=2 \cdot 2 x-1.76-\frac{2 x^{2}}{2}=2.2 x-1.76-x^{2} \quad(0-0.60) \\\text { Ared }\left(A_{1}\right):-\int_{0}^{L} \frac{M x}{E I} d x=\frac{1}{E I}\left[\int_{0}^{0.6}\left(2.2 x-1.76-x^{2}\right) d x\right] \\A_{1}=-\frac{0.492}{E I}\end{array}\begin{aligned}\text { Certroid }\left(C_{1}\right) & =\int_{0}^{L} \frac{M \cdot x}{E I} d x \\& =\frac{1}{E I}\left[\int_{0}^{0}\left(2.2 x^{2}-1 . J 6 x-x^{J}\right) d x\right] \\& =0.241 \mathrm{~m} \text { from } C \\& =0.759 \mathrm{~m} \text { jrom } \mathrm{A}\end{aligned}For area betwier A and D\text { Are }\left(A_{2}\right)=-\frac{1}{2} \times \frac{0.40}{E I} \times 0.40=\frac{-0.08}{E_{I}}Controid \left(C_{2}\right)=\frac{2}{\nabla} \times 0.40=0.267 \mathrm{~m} from \mathrm{A}Uring principle of moment areo method,\begin{array}{l}t_{A C}=f_{A}=\text { momont of } \frac{M}{E I} \text { diagrom about } A \\f_{A}=\frac{-4.492}{E I} \times 0.759-\frac{0.08}{E I} \times 0.267 \\f_{A}=-\frac{0.795}{E I}(t)=\frac{0.795}{200 \times 10^{6} \times 6.7617 \times 10^{-7}} \\=3.104 \mathrm{~mm}(\$) \\\end{array}Deflection at frec end:- \delta_{A}=0.104 \mathrm{~mm}(b) ...