Question For the beam and loading shown in Fig. P10.16, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.

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For the beam and loading shown in Fig. P10.16, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.

Transcribed Image Text: For the beam and loading shown in Fig. P10.16, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.

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, 40WenIntegratie on bothsides N_{x}=\frac{1}{2} \times \frac{\omega_{0} x}{2} \times \frac{x}{3}\text { FI } \frac{d y}{d x}=\frac{w_{0} x^{4}}{24 L}+G \rightarrow M_{x}=\frac{\omega_{0} x}{6 L}We know that \frac{d y}{d x}=0 at x=LE \pm(0)=\frac{w_{0} L^{3}}{24}+G_{1} \Rightarrow G_{1}=-\frac{\omega_{0} L^{3}}{24}In E \Delta \frac{d y}{d x}=\frac{\omega_{0} x^{4}}{24 L}-\frac{w_{0} L^{3}}{24}Integrate on both sidesE I y=\frac{w_{0} x^{5}}{120 L}-\frac{w_{0} L^{3}}{24}+C_{i}We know that deflection =0 at x=L\begin{array}{l}E T(0)=\frac{\omega_{0} L^{5}}{120 L}-\frac{\omega_{0} L^{4}}{24}+C_{2} \\0=\frac{\omega_{0} L^{4}}{120}-\frac{\omega_{0} L^{4}}{24}+c^{2} \\\Rightarrow-\frac{\not \omega_{0} L^{4}}{122_{30}}+c_{2} \\q_{2}=\frac{\omega_{0} L^{4}}{30} \\E+y=\frac{\omega_{0} x^{5}}{120 L}-\frac{\omega_{0} L^{3} x}{24}+\frac{\omega_{0} L^{4}}{30} \\y=\frac{\omega_{0} x^{5}}{E I(120 L)}-\frac{\omega_{0} L^{3} x}{2 A E I}+\frac{\omega_{0} L^{4}}{30 E I} \\\end{array} ...