JEE Main & Advanced Physics Elasticity Relation Between Volumetric Strain, Lateral Strain and Poisson's Ratio

If a long bar have a length L and radius r then volume \[V=\pi {{r}^{2}}L\]           

Differentiating both the sides \[dV=\pi {{r}^{2}}dL+\pi 2rL\,dr\]           

Dividing both the sides by volume of bar  \[\frac{dV}{V}=\frac{\pi {{r}^{2}}dL}{\pi {{r}^{2}}L}+\frac{\pi 2rL\,dr}{\pi {{r}^{2}}L}\]\[=\frac{dL}{L}+2\frac{dr}{r}\]           

\[\Rightarrow \]  Volumetric strain = longitudinal strain + 2(lateral strain)                           

\[\Rightarrow \]\[\frac{dV}{V}=\frac{dL}{L}+2\sigma \frac{dL}{L}\]\[=(1+2\sigma )\frac{dL}{L}\]                                                                                        \[\left[ \text{As}\,\,\sigma =\frac{dr/r}{dL/L}\,\,\,\,\Rightarrow \,\,\,\frac{dr}{r}=\sigma \frac{dL}{L} \right]\]

or \[\sigma =\frac{1}{2}\left[ \frac{dV}{AdL}-1 \right]\]                                                                                         

[where A = cross-section of bar]

(i) If a material having \[\sigma =-0.5\] then \[\frac{dV}{V}=[1+2\sigma ]\frac{dL}{L}=0\]

\[\therefore \] Volume = constant  or  \[K=\infty \] i.e. the material is incompressible.

(ii) If a material having \[\sigma =0,\] then lateral strain is zero i.e. when a substance is stretched its length increases without any decrease in diameter e.g. cork. In this case change in volume is maximum.

(iii) Theoretical value of Poisson?s ratio \[-1<\sigma <0.5\].

(iv) Practical value of Poisson?s ratio \[0<\sigma <0.5\]