Question YA 30 kN D B 3m 60 kNm Hc 3m* 3m A 3m Question 2: For the indeterminate beam shown above, determine the support reactions, slope at C and deflections at D and E using the Superposition Theorem. Make use of the information provided in the Table below. Assume flexural rigidity as EI. Beam and Loading Elastic Curve Maximum Deflection Slope at End Equation of Elastic Curve y = x (- 3L) y=- ** - 4L2° + 6132) Ko y-

Please do rate the answer and comment for any clarification.   due to point load 30 \mathrm{kN}\begin{array}{l}\Delta_{B_{30}}=-\left(\frac{P L^{3}}{3 E I}+\frac{P L^{2}}{2 E I} \times L_{B D}\right) \mid \begin{array}{l}\text { Similarly } \\\Delta_{50}=-\left(\frac{30 \times(3)^{5}}{3 E I}\right. \\+\frac{30 \times 3^{2} \times 9}{2 E I}\end{array} \\=-\left(\frac{30 \times(3)^{3}}{3 E I}+\frac{30 \times(3)^{2}}{2 E I} \times 3\right)^{2}=\frac{-1485}{E I} \\\text { due }+60 k v m \\\end{array}Acc, do Marmell reciprocal laue\Delta_{B} due to moment at E is equal to \Delta_{E} due to moment at B\begin{aligned} \Delta_{E_{6_{0}}} & =-\left(\frac{M L^{2}}{2 E I}+\frac{M L}{E I} \times L_{B E}\right) \\ & =-\left(\frac{60 \times(6)^{2}}{2 E I}+\frac{60 \times 6}{E I} \times 3\right) \\ \Delta_{B_{60}}=\Delta_{E_{\infty}} & =\frac{-2160}{E I}\end{aligned}\begin{aligned} \Delta_{c_{1}} & =-\left(\frac{M L^{2}}{2 E I}+\frac{M L}{E I} \times L_{E C}\right) \\ & =-\left(\frac{60 \times 9^{2}}{2 E I}+\frac{60 \times 9}{E I} \times 3\right) \\ & =-\frac{4050}{E I}\end{aligned} due to reaction at B\begin{aligned}\Delta_{B_{B}} & =\frac{R_{B}(6)^{3}}{3 E I} \\& =\frac{72 R_{B}}{E I} \\\Delta_{C_{B}} & =\frac{R_{B}(6)^{3}}{3 E I}+\frac{R_{B}(6)^{2}}{2 E I} \times 6 \\& =\frac{180 R_{B}}{E I}\end{aligned}due to Reaction at C\begin{aligned}\Delta_{B} \text { due to } R_{C} & =\Delta \text { due to } R_{B}(M c \\\Delta_{B_{C}} & =\Delta_{C_{B}}=\frac{180 R_{c}}{E I} \\\Delta_{c_{C}} & =\frac{R_{c} \times(12)^{3}}{3 E I}=\frac{576 R_{c}}{E I}\end{aligned}\Delta_{B}=0\Delta_{B_{30}}+\Delta_{B 60}+\Delta_{B_{B}}+\Delta_{B_{C}}=0\begin{array}{l}\left(-\frac{675}{E I}\right)+\left(\frac{-2160}{E I}\right)+\frac{72 R_{B}}{E I}+\frac{180 R_{C}}{E I}=0 \\72 R_{B}+180 R_{C}=2835 \\8 R_{B}+20 R_{C}=315-0 \\\Delta_{C}=0 \\\Delta_{C 30}+\Delta_{C_{60}}+\Delta_{C_{B}}+\Delta_{C C}=0 \\\frac{-1485}{E I}-\frac{4050}{E I}+\frac{180 R_{B}}{E I}+\frac{576 R_{C}}{E I}=0 \\180 R_{B}+576 R_{C}=5535 \\20 R_{B}+64 R_{C}=615\end{array}Solue (1) k (2), we get\begin{array}{l}R_{B}=70.178 \mathrm{kN} \\R_{C}=-12.321 \mathrm{kN}\end{array}\begin{array}{l} \sum f_{y}=0 \\R_{A}=30-70.178+12.321 \\=-27.857\end{array}\sum M=0\begin{aligned}M_{A} & =30 \times 3+60-70.178 \times 6-(-12.321 \times 12) \\& =+123.216 \mathrm{kN-m}(C)\end{aligned}(clockuise\begin{aligned}\theta_{c} & =\frac{30(3)^{2}}{2 E I}+\frac{60 \times 9}{E I}+\frac{12.321 \times 12^{2}}{2 E I}-\frac{70.178 \times 6^{2}}{2 E I} \\& =\frac{298 \cdot 908}{E I}\end{aligned}\begin{array}{l}\Delta_{E}=\left(\frac{30 \times 3^{3}}{3 E I}+\frac{30 \times 3^{2}}{2 E I} \times 6\right) \\ -\left(\frac{70.178 \times 6^{3}}{3 E I}+\frac{70.178 \times 6^{2}}{2 E 1} \times 3\right) \\ +\left(\frac{60 \times 9^{2}}{2 E I}\right)+\left(\frac{12.321 \times 9^{3}}{3 E I}+\frac{12.321 \times 9^{2}}{2 E I} \times 3\right) \\ =\frac{-841.423}{E I} \text { (upword) }(\uparrow) \\\end{array} Please do rate the answer and comment for any clarification. ...