Relation Between Volumetric Strain, Lateral Strain And Poisson’s Ratio

Solved Example Based on Relation Between Volumetric Strain, Lateral Strain And Poisson’s Ratio

Example 1: If $\mathrm{Y},\mathrm{K}$ and $\eta$ are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.

1) ${}^{\eta }=\frac{3YK}{9K+Y}N/m^2$
2) $Y=\frac{9K\eta }{2\eta +3K}N/m^2$
3) $K=\frac{Y\eta }{9\eta -3Y}N/m^2$
4) $Y=\frac{9K\eta }{3K-\eta }N/m^2$

Solution:

Let $\alpha =$ longitudinal strain per unit stress $=\frac{\mathrm{\Delta }l}{l}$
And $\beta =$ lateral strain per unit stress. $=\frac{\mathrm{\Delta }d}{d}$
Then Poisson's ratio, $\sigma =\frac{\text{lateral strain}}{\text{longitudinal strain}}=\frac{\beta }{\alpha }$
Young's modulus, $Y=\frac{\text{longitudinal stress}}{\text{longitudinal strain}}=\frac{1}{\frac{\text{longitudinal stress}}{\text{longitudinal strain}}}=\frac{1}{\alpha }$
And Bulk modulus, $K=\frac{\text{hydrostatic strain}}{\text{volumetric strain}}=\frac{1}{3(\alpha -2\beta )}$

$K=\frac{1}{3(\alpha -2\beta )}=\frac{1}{3\alpha (1-2\frac{\beta }{\alpha })}$

Since $\sigma =\frac{\beta }{\alpha }$ and $Y=\frac{1}{\alpha }$, so by substituting these values in above equation, we get
$K=\frac{Y}{3(1-2\sigma )}\cdots \cdots \cdot (1)$

Now the options are given in terms of $\mathrm{Y},\mathrm{K}$ and modulus of rigidity $(\eta )$.
For this we can use this formula
$Y=2\eta (1+\sigma )\cdots \cdots \cdot (2)$

From equation( 1)
$\begin{array}{rl}& 1-2\sigma =\frac{Y}{3K}\\ & \Rightarrow 2\sigma =1-\frac{Y}{3K}\\ & \Rightarrow \sigma =\frac{1}{2}(1-\frac{Y}{3K})\cdots \cdots (3)\end{array}$

Now from equation (2)
$\sigma =\frac{Y}{2\eta }-1\cdots \dots$

By comparing equations (3) and (4), we have
$\begin{array}{rl}& \frac{1}{2}(1-\frac{Y}{3K})=\frac{Y}{2\eta }-1\\ & \Rightarrow \frac{3}{2}-\frac{Y}{6K}=\frac{Y}{2\eta }\\ & \Rightarrow 3-\frac{Y}{3K}=\frac{Y}{\eta }\end{array}$

n further solving, we get
$\begin{array}{rl}& 3=\frac{Y}{\eta }+\frac{Y}{3K}\\ & \Rightarrow \frac{Y}{3K}=3-\frac{Y}{\eta }\\ & \Rightarrow \frac{Y}{3K}=\frac{3\eta -Y}{\eta }\\ & \Rightarrow K=\frac{Y\eta }{9\eta -3Y}\end{array}$

Hence, the answer is option (3).

Example 2: A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of the area floats on the surface of the liquid, covering the entire cross-section of the cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $\left(\frac{dr}{r}\right)$ is:

1) $\frac{mg}{Ka}$
2) $\frac{Ka}{mg}$
3) $\frac{Ka}{3mg}$
4) $\frac{-mg}{3Ka}$

Solution:

Bulk Modulus -

The ratio of normal stress to volumetric strain.

$\begin{array}{rl}& K=\frac{f/A}{-\mathrm{\Delta }v/v}=\frac{-Fv}{A\mathrm{\Delta }v}\\ & K=\frac{-Pv}{\mathrm{\Delta }v}\end{array}$

v = Original volume

$\mathrm{\Delta }v=\text{Change in volume}$

P = Increase in pressure

$\text{-ve(sign) shows volume}(\mathrm{\Delta }v)\text{decrease.}$

- wherein

$\begin{array}{ll}\mathrm{\Delta }P=\frac{mg}{a}& V=\frac{4\pi }{3}{r}^3\\ K=-\frac{\mathrm{\Delta }P}{\left(\frac{\mathrm{\Delta }V}{V}\right)}& \therefore \frac{dV}{V}=3\cdot \frac{dr}{r}\\ K=\frac{-\mathrm{\Delta }P}{3\cdot \left(\frac{dr}{r}\right)}& \\ \text{or}\frac{dr}{r}=\frac{-mg}{3Ka}& \end{array}$

Hence, the answer is option (4).

Example 3: A material has Poisson's ratio of 0.5. If a uniform rod of it suffers a longitudinal strain of $2\times {10}^{-3}$. The percentage change in its volume is:

1) $5\mathrm{\%}$
2) $10\mathrm{\%}$
3) $12\mathrm{\%}$
4) $0\mathrm{\%}$

Solution:

Use,

$\begin{array}{rl}\frac{\mathrm{\Delta }V}{V}\times 100& =\frac{\mathrm{\Delta }l}{l}(1-2\sigma )\\ & =\frac{\mathrm{\Delta }l}{l}(1-2\times 0.5)\\ & =\frac{\mathrm{\Delta }l}{l}(1-1)=0\mathrm{\%}\end{array}$

Hence, the answer is option (4).

Example 4: Choose the correct relationship between the Poisson ratio $(\sigma )$and bull modulus (k) of rigidity $(\eta )$ of a given solid object :

1) $\sigma =\frac{3K+2\eta }{6K+2\eta }$
2) $\sigma =\frac{3K-2\eta }{6K+2\eta }$
3) $\sigma =\frac{6K+2\eta }{3K-2\eta }$
4) $\sigma =\frac{6K-2\eta }{3K-2\eta }$

Solution:

$\begin{array}{ll}Y=2\eta [1+\sigma ]& \\ \text{and}& Y=3K[1-2\sigma ]\\ \text{Now}& 2\eta (1+\sigma )=3K(1-2\sigma )\\ & 2\eta \sigma +2\eta =3K-6K\sigma \\ & (2n+6K)\sigma =3K-2n\\ & \sigma =\frac{3K-2n}{2\eta +6K}\end{array}$

Hence, the answer is option (2).

Example 5: A solid sphere of radius R made of material of buck modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid when a mass m is placed on the piston to compress the liquid, the fractional change in the radius of the sphere $\frac{\mathrm{\Delta }R}{R}$ is _____.

1) $\frac{mg}{AK}$
2) $\frac{mg}{3AK}$
3) $\frac{mg}{A}$
4) $\frac{3mg}{AK}$

Solution:

$\mathrm{\Delta }P=\frac{mg}{A}$

Volumetric strain is
$\begin{array}{rl}\frac{\mathrm{\Delta }V}{V}& =\frac{\mathrm{\Delta }P}{K}\\ 3\frac{\mathrm{\Delta }R}{R}& =\frac{mg/A}{K}\\ \frac{\mathrm{\Delta }R}{R}& =\frac{mg}{3KA}\end{array}$

Hence, the answer is option (2).

Summary

Volumetric strain is a measure of volume changes of a material under stress. Lateral strain is a measure of deformation perpendicular to the applied force. The ratio of lateral to axial strains is Poisson's ratio. All of these concepts are connected by the fact that Poisson's ratio makes predictions for the lateral and volumetric changes a material will undergo upon stretching or compressing.