Modulus of Elasticity
What is Modulus of Elasticity?
Modulus of elasticity is the measure of stiffness or rigidity of a material. This decides the extent of deformation the material will undergo when a stress within elastic limit acts on it. Modulus of elasticity is the ratio of stress developed inside the body to the
strain responsible for it. We represent it by the symbol \(E\).
Its SI unit is Pascal same as that of pressure.
Modulus of Elasticity for Longitudinal strain
Young’s Modulus of Elasticity\[\text{Young’s modulus}=\frac{\text{longitudinal stress}}{\text{longitudinal strain}}\]
\[Y=\frac{\sigma }{\epsilon } = \frac{FL}{A \Delta L} \]
Modulus of Elasticity for Shear strain
\(\text{Shear strain}=\tan \theta = \frac{\Delta L}{L}\)
For small strain, \(\tan \theta = \theta\)
Modules of elasticity for shear deformation is modulus of rigidity. We represent it by G.
\[\text{Modulus of Rigidity}=\frac{\text{shear stress}}{\text{shear strain}}\]
\[G=\frac{\tau}{\gamma} = \frac{F}{A \theta} \]
Modulus of Elasticity for Volume strain
\(\text{Volume strain}= – \frac{\Delta V}{V}\), where negative sign indicates that the volume decreases with increase in hydraulic pressure.
For small strain, \(\tan \theta = \theta\)
Modules of elasticity for hydraulic deformation is Bulk Modulus. We represent it by B.
\[\text{Bulk Modulus }=\frac{\text{hydraulic stress}}{\text{volume strain}}\]
\[B=-\frac{PV}{\Delta V} \]
Compressibility
Compressibility is defined as the reciprocal of bulk modulus.
\[\text{Compressibility}=\frac{\text{1}}{\text{Bulk Modulus }}\]
SI unit of Compressibility is \(N^{-1}m^2\)
Dimension is \([M^{-1}L^{1}T^{2}]\)
Poisson’s Ratio
Longitudinal strain = change in length/ original length
\[\epsilon=\frac{\Delta L}{L_0}\]
Lateral strain = change in diameter/ original diameter
\[\epsilon_{l}=\frac{\Delta d}{d_0}\]
Poisson’s ratio, \(\nu\) = -(Lateral strain/ Longitudinal strain)
\[\nu=\frac{\frac{\Delta d}{d_0}}{\frac{\Delta L}{L_0}}\]
\[\nu=\frac{\Delta d}{\Delta L }\frac{L_0}{d_0}\]
Elastic potential energy in a stretched wire
Let, a wire of length L is stretched by applying a force F. This elongates the wire by l. If at any instant, the elongation was x, then the longitudinal stress opposing the elongation will be \(\sigma =Y\dfrac{x}{L}\) where Y = Young’s modulus
\[\dfrac{F}{A}=Y\dfrac{x}{L}\]\[F=\dfrac{YAx}{L}\]
The work done by the restoring force in causing a small elongation \(dx\)
\[d\omega =\vec {F}\cdot \vec{dx}\]\[=\dfrac{YAx}{L}\cdot dx\cos 180^{0}\] \[=-\dfrac{YAx}{L}\cdot dx\]
Total work done for the entire elongation,\(l\)
\[\int dW= \int ^{l}_{0}-\dfrac{YAx}{L}dx\] \[W= -\dfrac{YA}{L} \int ^{l}_{0}xdx\] \[W=-\dfrac{YA}{L}\left[ \dfrac{x^{2}}{2}\right] _{0}^{l}\] \[W=-\dfrac{YA}{L} \dfrac{l^{2}}{2}\] \[W=-\frac{1}{2}Y\frac{l}{L}Al\]
\[W=- \dfrac{1}{2}\sigma \epsilon V\] where \(Y\frac{l}{L} = \sigma\), \(\frac{l}{L} = \epsilon\) and \(AL = V\)
Change in potential energy \(\Delta U\) =- work done \[\Delta U=\dfrac{1}{2}\sigma \epsilon V\]
Change in potential energy per unit volume, \(= \frac{\Delta U}{V} =\dfrac{1}{2}\sigma \epsilon \)
