For the beam and loading shown in Fig. P10.15, 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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soludion:distance of x from poin A as follows:\begin{array}{l}\frac{g}{x}=\frac{w_{0}}{2} \\9=\frac{w_{0}}{2} x\end{array}distance x from point A for the sedelion A B :\begin{aligned}V_{A B} & =-\frac{1}{2} x \times q \\& \frac{W_{0}}{2} x \text { for } q . \\V_{A B} & =-\frac{1}{2} x \times \frac{W_{0}}{L} x=-\frac{W_{0}}{2 L} x^{2}\end{aligned}distance x from peint A for the scction A :m_{A D}=-\frac{w_{0}}{2 L} x^{2} \times \frac{x}{3}=-\frac{w_{0}}{6 L} x^{3}elastic curve for the section A B as follows: EI \frac{d^{2} v}{d x^{2}}=M_{A B}\begin{aligned}& -W_{0} x^{3} \text { for } m_{A B} . \\E L \frac{d^{2} v}{d x^{2}} & =-\frac{w_{0} x^{3}}{6 L}\end{aligned}E I \frac{d v}{d x}=\left(\frac{w_{0} x^{4}}{24 L}\right)+C_{1}Obtain the deflection equation:\text { EIV }=-\left(\frac{w_{0} x^{5}}{220 L}\right)+c_{1} x+c_{2} \text {. }At x=L, \frac{d v}{d x}=0these values in equation\begin{array}{l}\text { El } \frac{d V}{d x}=\frac{w_{0} x^{4}}{24 L}+c_{1} \\E I(0)=-\frac{w_{0}(L)^{4}}{24 L}+c_{1}=c_{1}=\frac{w_{0}(L)^{3}}{24}\end{array}At V=0 at x=L :V, L for x and \frac{W_{0}(L)^{3}}{24} for clin equation\begin{array}{l}E I V=-\left(\begin{array}{l}w_{0} x^{5} \\120 L\end{array}\right)+c_{1} x+c_{2} \\E I(0)=-\left(\frac{w_{0}\left(L^{5}\right)}{120 L}\right)+\left(\frac{w_{0}\left(l^{3}\right.}{24}\right)(l)+c_{2} \\C_{2}=\frac{w_{0} L^{4}}{120}-\frac{w_{0} L^{4}}{24}--\frac{w_{0} L^{4}}{30}\end{array}\frac{W_{0}(L)^{3}}{24} for c_{1} and -\frac{W_{0} L^{4}}{30} for c_{1} inV=-\frac{W_{0}}{\text { T2OLEI }}\left[x^{5}-5 L^{4} x+4 L^{5}\right]Therefar, the required equation of the elartic curve for the section A B is \left.-\frac{w_{0}}{120 L E I}\left[x^{2}-5 L\right)^{2} c+4 L\right]of for x in the elastic crrve equation.\begin{aligned}V_{A} & =-\frac{W_{0}}{120 L E I}\left[(0)^{5}-5 L^{4}(0)+4 L^{5}\right] \\& =-\frac{W_{0} L^{4}}{30 E I}\end{aligned}The given beam at point A using equation\text { EI } \frac{d v}{d x}=-\frac{w_{0} x^{4}}{24 L}+c_{1}\frac{W_{0}(L)^{3}}{24} for C_{1} and 0 for x i.\left(\frac{d v}{d x}\right)_{A}=-\frac{W_{0}(0)^{4}}{24 L E I}+\frac{W_{0} L^{3}}{24 L E I}=\frac{W_{0} L^{3}}{24 E I} ...