Elasticity
6.0 Elongation of Rod Under its Self Weight
6.0 Elongation of Rod Under its Self Weight
Let a rod having mass $m$, area of cross-section $A$ and length $L$. $$\lambda = \frac{m}{l},{\text{ mass per unit length}}$$
Take a small element of $dx$ length at a distance $x$ from $A$ as shown in Fig. E. 10.
Let the extension produced in small element of $dx$ length be $\Delta x$. This extension is produced due to the weight of $CB$.
Length of $CB$ = $l-x$
Mass of $CB$ = $\lambda (l - x)$
Weight of $CB$ = $\lambda (l - x)g$
From the FBD of the small segment $dx$,
$$\begin{equation} \begin{aligned} Stress = \frac{{\lambda \left( {l - x} \right)g}}{A} \\ Strain = \frac{{\Delta x}}{{dx}} \\\end{aligned} \end{equation} $$
So from Hooke’s law,
$$\begin{equation} \begin{aligned} Strain = \frac{{Stress}}{Y} \\ \frac{{\Delta x}}{{dx}} = \frac{{\lambda \left( {l - x} \right)g}}{{AY}} \\ \Delta x = \frac{{\lambda g}}{{AY}}\left( {l - x} \right)dx \\\end{aligned} \end{equation} $$
Integration both sides with proper limits we get,
$$\begin{equation} \begin{aligned} \mathop \smallint \limits_0^{\Delta l} \Delta x = \frac{{\lambda g}}{{AY}}\mathop \smallint \limits_0^l \left( {l - x} \right)dx \\ \Delta l = \frac{{\lambda g}}{{AY}}\left[ {lx - \frac{{{x^2}}}{2}} \right]_0^l \\ \Delta l = \frac{{\lambda g{l^2}}}{{2AY}} \\ \Delta l = \frac{{mgl}}{{2AY}}\left( {As,\lambda = \frac{m}{l}} \right) \\\end{aligned} \end{equation} $$
Example 3. A steel bar of mass $m$, cross-section area $A$ and length $l$ is hanged from the ceiling of the wall. A block of mass $M$ is hanged from the free end of the bar by a massless string. Find out the elongation in the bar as shown in the Fig. E. 11. (Young’s modulus of steel = $Y$).
Solution: $$\lambda = \frac{m}{l},{\text{ mass per unit length of bar}}$$
Take a small element of $dx$ length at a distance $x$ from $A$ as shown in Fig. E. 12.
Let the extension produced in small element of $dx$ length be $\Delta x$. This extension is produced due to the weight of $CB$ and the weight of the block of mass $M$.
Length of $CB$ = $l-x$
Mass of $CB$ = $\lambda (l - x)$
Weight of $CB$ = $\lambda (l - x)g$
Weight of block = $Mg$
From the FBD of the small segment $dx$,
$$\begin{equation} \begin{aligned} Stress = \frac{{\left[ {\lambda \left( {l - x} \right) + M} \right]g}}{A} \\ Strain = \frac{{\Delta x}}{{dx}} \\\end{aligned} \end{equation} $$
So from Hooke’s law,
$$\begin{equation} \begin{aligned} Strain = \frac{{Stress}}{Y} \\ \frac{{\Delta x}}{{dx}} = \frac{{\left[ {\lambda \left( {l - x} \right) + M} \right]g}}{{AY}} \\ \Delta x = \frac{g}{{AY}}\left[ {\lambda \left( {l - x} \right) + M} \right]dx \\\end{aligned} \end{equation} $$
Integration both sides with proper limits we get,
$$\begin{equation} \begin{aligned} \mathop \smallint \limits_0^{\Delta l} \Delta x = \frac{g}{{AY}}\mathop \smallint \limits_0^l \left[ {\lambda \left( {l - x} \right) + M} \right]dx \\ \Delta l = \frac{g}{{AY}}\left[ {\lambda \left( {lx - \frac{{{x^2}}}{2}} \right) + Mx} \right]_0^l \\ \Delta l = \frac{g}{{AY}}\left( {\lambda \frac{{{l^2}}}{2} + Ml} \right) \\ \Delta l = \frac{g}{{AY}}\left( {\frac{{ml}}{2} + Ml} \right),\left( {As,\lambda = \frac{m}{l}} \right) \\ \Delta l = \left( {M + \frac{m}{2}} \right)\frac{{lg}}{{AY}} \\\end{aligned} \end{equation} $$
