Question

A bar made of material whose Young’s modulus is equal to$E$and Poisson’s ratio to $\mu $ is subjected to the hydrostatic pressure $p$. Find the relationship between the compressibility $\beta $and elastic constants $E$ and $\mu $
$\eqalign{
  & {\text{A}}{\text{. }}\beta = \left( {1 - 2\mu } \right)\dfrac{3}{E} \cr
  & {\text{B}}{\text{. }}\beta = \left( {1 - 3\mu } \right)\dfrac{2}{E} \cr
  & {\text{C}}{\text{. }}\beta = \left( {1 + 2\mu } \right)\dfrac{E}{3} \cr
  & {\text{D}}{\text{. }}\mu = \left( {1 - 2\beta } \right)\dfrac{3}{E} \cr} $

Answer 1

Hint: Young’s modulus is the ratio of longitudinal stress over strain, and poisson's ratio is the ratio of lateral increment or decrement of the material. Use these two to form an equation for compression on the whole bar and then on each side. Later equate it to the expression of strain to get the required answer.

Formula used:
The Strain acting on the bar, $ - \dfrac{{\Delta V}}{V}$
The compression on the bar, $\dfrac{p}{E}$

Complete step by step answer:
Consider a cube having unit length before any pressure is applied to it. Later an equal amount of pressure is applied to each of its faces.
The pressure acting on the opposite faces will produce tensile stress which in turn would result in longitudinal compression and lateral strain.

Given:
The Young’s modulus of the material of bar, $Y = E$
The Poisson’s ratio of the material of bar $ = \mu $
The hydrostatic pressure acting on the bar $ = p$
So, the compression will be given by: $\dfrac{p}{E}$
Similarly, the lateral extension will be given by: $\mu \dfrac{p}{E}$

All these forces acting on the bar will result in compression of the bar. This compression on each side of the cube will be:

$\dfrac{p}{E}\left( {1 - 2\mu } \right)$
But because this compression is responsible for the volumetric change in the bar, thus we have:
$\dfrac{{\Delta V}}{V} = - \dfrac{{3p}}{E}\left( {1 - 2\mu } \right) \cdots \cdots \cdots \left( 1 \right){\text{ }}\left[ {\because {\text{by symmetry }}\dfrac{{\Delta V}}{V} = 3\dfrac{{\Delta l}}{l}} \right]$
Now, Then the strain on its volume will be give by:
$ - \dfrac{{\Delta V}}{V} = \dfrac{p}{k} \cdots \cdots \cdots \left( 2 \right)$
where k is the bulk modulus of elasticity.
The equation (1) and (2) represent the same physical quantities, so equating both we get:

$\eqalign{
  & \dfrac{p}{k} = \dfrac{{3p}}{E}\left( {1 - 2\mu } \right) \cr
  & \Rightarrow E = 3k\left( {1 - 2\mu } \right) \cdots \; \cdots \cdots \left( 3 \right) \cr} $
But we know that, $k = \dfrac{1}{\beta }$
So, equation (3) becomes:

$\eqalign{
  & E = \dfrac{3}{\beta }\left( {1 - 2\mu } \right) \cr
  & \therefore \beta = \left( {1 - 2\mu } \right)\dfrac{3}{E} \cr} $
Thus to keep both $E{\text{ and }}\beta $ positive we must have $\mu \leqslant \dfrac{1}{2}$.
Therefore, the correct option is A. i.e., the relationship between the compressibility $\beta $ and elastic constants $E$ and $\mu $ is $\beta = \left( {1 - 2\mu } \right)\dfrac{3}{E}$.

Note:
Materials known as auxetic materials display a negative Poisson’s ratio. This means that when they are subjected to positive strain in a longitudinal axis, the transverse strain in these materials will actually be positive i.e. there would be an increase in the cross-sectional area.