# Relation between Volumetric Strain, Lateral Strain and Poisson’s Ratio

## Relation between Volumetric Strain, Lateral Strain and Poisson’s Ratio

If the deforming force produces a change in volume along the strain produced in the body is called Volumetric Strain, the ratio of the change in radius or diameter to the original radius or diameter is called Lateral Strain, the ratio of lateral strain to longitudinal strain is called Poisson’s Ratio.

If a long bar has a length L and radius r then Volume, (V) = πr²L.

Differentiating Volume on both the sides, we get:

dV = πr² dL + π 2rL dr

Dividing both the sides by volume of bar, we get:

$$\frac{dV}{V}=\frac{\pi {{r}^{2}}\,dL}{\pi {{r}^{2}}L}+\frac{\pi \,2r\,L\,dr}{\pi {{r}^{2}}L}=\frac{dL}{L}+2\frac{dr}{r}$$.

Now, Volumetric Strain = Longitudinal Strain + 2 (Lateral Strain)

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

$$\sigma =\frac{1}{2}\left( \frac{dV}{AdL}-1 \right)$$; Where, A = Cross Section of bar

(i) If a material having σ = – 0.5 then

$$\frac{dV}{V}=\left( 1+2\sigma \right)\frac{dL}{L}=0$$.

Therefore, Volume = Constant, i.e., the material is incompressible.

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

(iii) Theoretical value of Poisson’s ratio, -1 < σ < 0.5

(iv) Practical value of Poisson’s ratio, 0 < σ < 0.5.