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.

Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

, 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} ...