7.0 Bulk Modulus of Elasticity

  Elasticity

Within the elastic limit, it is defined as the ratio of normal stress to the volumetric strain.

$$B = \frac{{normal\ stress}}{{volumetric\ strain}}$$

Consider a spherical solid body of volume $V$ and surface area $A$. In order to compress the body, let a force $F$ be applied normally on the entire surface. And its volume decreases by $\Delta V$ as shown in the Fig. E. 13.

Consider a spherical solid body of volume V and surface area A. In order to compress the body, let a force F be applied normally on the entire surface. And its volume decreases by ?V as shown in the Fig. E. 13.

Therefore, $$Volumetric\ Strain = - \frac{{\Delta V}}{V}$$

Note: Negative (-ve) sign shows that the volume is decreasing when the external force is applied.

$$Normal\ Stress = \frac{F}{A}$$

So, from the definition of bulk modulus of elasticity, $$B = \frac{{\frac{F}{A}}}{{ - \frac{{\Delta V}}{V}}} = - \frac{{FV}}{{A\Delta V}}$$

Pressure ($P$) on the surface of the body is, $$P = \frac{F}{A}$$ So, $$B = - \frac{P}{{\frac{{\Delta V}}{V}}}$$

Note:

• A solid have all the three moduli of elasticity $Y$, $B$ and $\eta $. But in case of liquid or gas only $B$ can be defined because the liquid or gas cannot be framed into a wire and no shear force and be applied on them.

• For a liquid or gas, $$B = \left( {\frac{{ - P}}{{dV/V}}} \right)$$ So, instead of $P$ as we are more interested in change in pressure $dP$.

• When a pressure ($dP$) is applied on a substance its density is changed. The change in density can be calculated as under: $$\begin{equation} \begin{aligned} \rho = \frac{{mass}}{{volume}} \\ \rho \propto \frac{1}{{volume}}\left( {mass = constant} \right) \\\end{aligned} \end{equation} $$ Therefore, $$\begin{equation} \begin{aligned} \frac{{\rho '}}{\rho } = \frac{V}{{V'}} = \frac{V}{{V + dV}} \\ \rho ' = \rho \left( {\frac{V}{{V + dV}}} \right) \\\end{aligned} \end{equation} $$ Dividing numerator and denominator by $V$ we get, $$\rho ' = \rho \left( {\frac{1}{{1 + \frac{{dV}}{V}}}} \right)$$ We know that, $$B = \left( {\frac{{ - dP}}{{dV/V}}} \right)or\frac{{dV}}{V} = - \frac{{dP}}{B}$$ So, $$\rho ' = \rho \left( {\frac{1}{{1 - \frac{{dP}}{B}}}} \right)$$

• Relation between compressibility and bulk modulus of elasticity. $$Compressibility = \frac{1}{B}$$

Example 4. Find the depth of lake at which density of water is $1\% $ greater than at the surface. Given compressibility as $50 \times {10^{ - 6}}/atm.$.

Solution: We know that, $$B = \frac{1}{{Compressibility}} = \frac{1}{{50 \times {{10}^{ - 6}}}}atm$$ Also, $$1atm = 1.01 \times {10^5}N{m^{ - 2}}$$ So, $$B = \frac{{1.01 \times {{10}^{5}}}}{{50 \times {{10}^{ - 6}}}}N{m^{ - 2}}$$ Let the density at the surface be $\rho $ and that at depth $h$ be $c$.

As from question, $$\rho ' = 1.01{\text{}}\rho {\text{}}$$

As from the Fig. E. 14. The pressure difference between the surface and depth $h$ be $dP$. $$dP = \rho gh$$

Relation between density and bulk modulus of elasticity is, $$\rho ' = \rho \left( {\frac{1}{{1 - \frac{{dP}}{B}}}} \right)$$ Therefore, $$\begin{equation} \begin{aligned} 1.01 = \left( {\frac{1}{{1 - \frac{{dP}}{B}}}} \right) \\ \frac{{101}}{{100}} = \left( {\frac{1}{{1 - \frac{{dP}}{B}}}} \right) \\ \frac{{100}}{{101}} = 1 - \frac{{dP}}{B} \\ \frac{1}{{101}} = \frac{{dP}}{B} \\ \frac{1}{{101}} = \frac{{\rho gh}}{B} \\ h = \frac{B}{{101\rho g}} = \frac{{\frac{{1.01 \times {{10}^{5}}}}{{50 \times {{10}^{ - 6}}}}}}{{101 \times 1000 \times 10}} = \frac{{1.01 \times {{10}^5}}}{{50 \times {{10}^{ - 6}} \times 101 \times 1000 \times 10}} \\ h = 2kms \\\end{aligned} \end{equation} $$