Question For ACB beam with elasticity module E, inertia I; a- Find the elastic curve equation by singularity method? b- Find the reaction forces at point A. c- Find the reaction force at point B. pls 30 min IL_SB:O_MAKİNE MÜHEND (2/3) 2W B C -1/2 1.12-. Soru : Elastisite modulo E, ataleti I olan ACB kirişi için; a- elastik eğri denklemini singularite yöntemi ile bulunuz? b- A noktasındaki reaksiyon kuvvetlerini bulunuz. c- B noktasındaki reaksiyon kuvvetini bulunuz. W

\&\Rightarrow Draming Free Body Diagram of Lean\begin{array}{l}\Rightarrow \quad F_{y}=0 \Rightarrow R_{B}+R_{A}+\omega \frac{L}{2}=\frac{2}{3} \omega+\frac{1}{2} 2 \omega \times L \\R_{B}+R_{A}+\frac{\omega L}{2}=\frac{2}{3} \omega+\omega L \\R A+R B=\frac{2}{3} \omega+\frac{\omega L}{2} \\\sum M_{A}=0 . \Rightarrow \frac{R_{B} \frac{B L}{2}}{}+M_{A}+\omega \frac{L}{2}\left(\frac{L}{4}\right)=\frac{2}{3} \omega\left(\frac{L}{2}\right)+\frac{1}{2} L \omega L \times \frac{7}{6} L \\\frac{3 R_{B} L}{2}+M A+\frac{\omega L^{2}}{8}=\frac{\omega L}{3}+\frac{7}{6} \omega L^{2} \\\frac{3 R B L}{2}+M A=\frac{\omega L}{3}-\frac{2}{24} \omega L^{2} \\\end{array}Using singularity functin\frac{d v}{d x}=-\frac{2 \omega}{L}\left\langle x-\frac{1}{2}\right\rangle+\omega\langle x\rangleIntergretuig\begin{array}{l}V(x)-V(0)=\frac{-2 \omega}{L}\left\langle\frac{\langle x-L / 2\rangle^{2}}{2}+\omega\langle x\rangle\right. \\\therefore N(x)=-\frac{2 \omega}{L} \frac{\langle x-L / 2\rangle^{2}}{2}+\omega\langle x\rangle+R_{A}\end{array}\Rightarrow \frac{d M}{d x}=v(x)Integreting fuam 0 to x\begin{array}{l}\Rightarrow M(x)-M(0)=\frac{-2 \omega}{L} \frac{\left\langle x-\frac{1}{2}\right\rangle^{3}}{6}+\frac{\omega\langle x\rangle^{2}}{2}+R A\langle x\rangle \\M(u)=\frac{-\frac{2 \omega}{L}\left\langle x-\frac{L}{2}\right\rangle^{3}+\frac{\omega\langle x\rangle^{2}}{2}+R A\langle x\rangle-M A}{6}\end{array}\begin{array}{l} E I \frac{d^{2} y}{d x^{2}}=0 M(x) \\\therefore \frac{d^{2} y}{d x^{2}}=\frac{1}{E I}\left[\frac{-\frac{2 \omega}{L}\left\langle x-\frac{L}{2}\right\rangle^{3}}{6}+\frac{\omega\langle x\rangle^{2}}{2}+R A\langle x\rangle-M A\right] \\\frac{d y}{d x}=\frac{1}{E I}\left[\frac{-\frac{2 \omega}{L}\left\langle x-\frac{L}{2}\right\rangle^{4}}{24}+\frac{\omega\langle x\rangle^{3}}{6}+\frac{R A\langle x\rangle^{2}}{2}-M_{A}\langle x\rangle+4\right]\end{array}At x=0, \frac{d y}{d x} or slape of curve is 0 , therefore 9=0y=\frac{1}{E I}\left[\frac{-\frac{2 \omega}{L}\left\langle x-\frac{L}{2}\right\rangle^{5}}{120}+\frac{\omega\langle x\rangle^{4}}{24}+\frac{R A\langle x\rangle^{3}}{6}-\frac{H A\langle x\rangle^{2}}{2}+c_{2}\right]At x=0, y deflection is 0 , therefore c_{2}=0y(x)=\frac{1}{E I}\left[\frac{-\frac{2 \omega}{L}\left\langle x-\frac{L}{2}\right\rangle^{5}}{120}+\frac{\omega\langle x\rangle^{4}}{24}+\frac{R_{A}\langle x\rangle^{3}}{6}-\frac{M A^{\langle}\langle x\rangle^{2}}{2}\right]\Rightarrow At x=\frac{3 L}{2}, y=0, putting x=\frac{31}{2} in equation of y(a)\Rightarrow 0=\frac{1}{E I}\left[\frac{-2 \omega L^{4}}{120}+\frac{81 \omega L^{4}}{384}+\frac{27 R A L^{3}}{48}-\frac{M A 9 L^{2}}{8}\right]0=\frac{-32 \omega L^{2}+405 \omega L^{2}+1080 R A L-2160 \mathrm{MA}}{1920}373 \omega L^{2}+1080 R A L-2160 M A=0Using equetion (1), (2) and (3)\begin{array}{l}\Rightarrow \text { Fran (B) }-R_{A}=\frac{\left(2160 M A-373 \omega L^{\prime}\right)}{1080 L} \\\text { Fram (2) }-R B=\left[\frac{\omega L}{3}-\frac{25}{24} \omega L^{\circ}-M A\right] \frac{L}{3 L}\end{array}Putting in equotion (1)\begin{array}{l}\text { RA } \frac{720 \omega}{1080}+\frac{38187 \omega L}{1080} \\\end{array}\begin{array}{l}R_{B}=\left\{\frac{\omega L}{3}-\frac{25}{24} \omega L^{2}-\left[\frac{12}{36} \omega L+\frac{1663}{1440} \omega L^{2}\right]\right\} \frac{2}{3 L} \\ R_{B}=\left[\frac{2 \omega}{9}-\frac{25}{36} \omega L-\frac{2 / \omega}{9}-\frac{1663 \omega L}{2160}\right] \\ R_{B}=\frac{-3163 \omega L}{2160} \\ y(u)=\frac{1}{E I}\left[\frac{-2 \omega}{L}\left\langle x-\frac{L}{2}\right\rangle+\frac{\omega\langle x\rangle^{4}}{120}+\frac{720 \omega}{24}+\frac{38187 \omega L}{1080}\langle x\rangle\right. \\ \left.-\frac{\omega L}{3}+\frac{1663 \omega L^{2}\langle x\rangle^{2}}{1440}\right] \\\end{array} ...