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

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Transcribed Image Text: For the cantilever beam and loading shown, using the area moment method, determine the deflection at the free end in mm measure. Use three decimal places. Express your final answer in absolute value. Assume E = 200 GPa and I = 6.3617 x 10-7 m4. 1 KN A B || 2 KN/m 0.40 m 0.60 m

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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) ...