Physics - Elasticity Question with Solution | TestHub

Consider an element of length $dx$ at a distance $x$ from the fixed end. The force $F(x)$ acting on this element is the weight of the rod below $x$.$$F(x)={\int }_x^L\rho A(x^{\prime })d{x}^{\prime }=\rho A_0{\int }_x^L{e}^{k{x}^{\prime }}d{x}^{\prime }=\frac{\rho A_0}{k}(e^{kL}-e^{kx})$$ The extension $d(\Delta L)$ of this element is $d(\Delta L)=\frac{F(x)dx}{A(x)Y}$.$$d(\Delta L)=\frac{\frac{\rho A_0}{k}(e^{kL}-e^{kx})}{A_0{e}^{kx}Y}dx=\frac{\rho }{kY}(e^{k(L-x)}-1)dx$$ Total extension $\Delta L={\int }_0^L\frac{\rho }{kY}(e^{k(L-x)}-1)dx$. Let $u=L-x$, so $du=-dx$.$$\Delta L=\frac{\rho }{kY}{\int }_L^0(e^{ku}-1)(-du)=\frac{\rho }{kY}{\int }_0^L(e^{ku}-1)du=\frac{\rho }{kY}{\left[\frac{e^{ku}}{k}-u\right]}_0^L$$ $$\Delta L=\frac{\rho }{kY}\left[\left(\frac{e^{kL}}{k}-L\right)-\left(\frac{1}{k}-0\right)\right]=\frac{\rho }{k^{2}Y}(e^{kL}-1-kL)$