02 May 2025
Elasticity and Elastic Constants
An external force changes the relative positions of the particles of a solid. When the force is removed, interatomic forces tend to restore the original configuration. A deformation is elastic if the body recovers its original dimensions after the load is removed.
Stress, strain and Hooke’s law
For a rod of original length $L$ and cross-sectional area $A$, an axial tensile force $F$ produces an extension $\Delta L$. The normal stress and longitudinal strain are
\[\sigma=\frac{F}{A}, \qquad \epsilon_l=\frac{\Delta L}{L}.\]Within the proportional range, Hooke’s law states that stress is proportional to strain:
\[\boxed{\sigma=Y\epsilon_l},\]where $Y$ is Young’s modulus. Stress has dimension $[ML^{-1}T^{-2}]$ and strain is dimensionless, so every elastic modulus has the dimension of pressure.
On the stress-strain diagram, $O$ to $P$ is the straight-line or proportional region, with slope
\[Y=\frac{d\sigma}{d\epsilon_l}.\]Up to the elastic limit $E$, unloading restores the original dimensions. Beyond $E$, permanent strain remains. Plastic flow begins near the yield point $Y_p$; the engineering stress reaches its greatest value at the ultimate point $U$ and the specimen finally breaks at $B$. Only the initial straight segment obeys the simple equation $\sigma=Y\epsilon_l$.
The editable source of the diagram is available as a TikZ file.
Elastic moduli
Three elementary deformations define the usual elastic moduli.
Young’s modulus. Under axial loading,
\[Y=\frac{\text{normal stress}}{\text{longitudinal strain}} =\frac{F/A}{\Delta L/L}.\]Bulk modulus. If a uniform external pressure $p$ changes the volume from $V$ to $V+\Delta V$,
\[K=-\frac{p}{\Delta V/V}.\]The minus sign makes $K$ positive because compression gives $\Delta V<0$.
Shear modulus or modulus of rigidity. If tangential stress $\tau$ produces engineering shear strain $\gamma_s$, then
\[G=\frac{\tau}{\gamma_s}.\]For a small shear angle $\phi$, $\gamma_s=\tan\phi\simeq\phi$.
Poisson’s ratio
When a rod is stretched, it normally contracts in a transverse direction. If a transverse dimension changes from $d$ to $d+\Delta d$, then
\[\epsilon_t=\frac{\Delta d}{d}, \qquad \boxed{\nu=-\frac{\epsilon_t}{\epsilon_l}}.\]Under tension, $\epsilon_l>0$ and $\epsilon_t<0$, so $\nu$ is positive for ordinary materials.
Relations among the elastic constants
For a homogeneous isotropic solid in the linear range, only two elastic constants are independent. For a mechanically stable ordinary isotropic solid, $K>0$ and $G>0$, which imply
\[\boxed{-1<\nu<\frac12}.\]The upper limit corresponds to the incompressible limit $K\to\infty$; it is not reached by a material with finite bulk modulus.
Relation between $Y$, $K$ and $\nu$
Apply equal compressive stresses $-p$ along the three mutually perpendicular axes. Along $x$, the direct strain is $-p/Y$. Each of the stresses along $y$ and $z$ produces a lateral strain $+\nu p/Y$ along $x$. By superposition,
\[\epsilon_x=-\frac{p}{Y}+\frac{\nu p}{Y}+\frac{\nu p}{Y} =-\frac{p}{Y}(1-2\nu).\]The three normal strains are equal. To first order,
\[\frac{\Delta V}{V}=\epsilon_x+\epsilon_y+\epsilon_z =-\frac{3p}{Y}(1-2\nu).\]Using $K=-p/(\Delta V/V)$ gives
\[\boxed{Y=3K(1-2\nu)}.\]Relation between $Y$, $G$ and $\nu$
A state of pure shear stress $\tau$ has principal stresses $+\tau$ and $-\tau$ along directions at $45^\circ$ to the sheared faces. The corresponding principal strains are
\[\epsilon_+=\frac{\tau-\nu(-\tau)}{Y} =\frac{(1+\nu)\tau}{Y},\] \[\epsilon_-= \frac{-\tau-\nu\tau}{Y} =-\frac{(1+\nu)\tau}{Y}.\]For this small deformation the engineering shear strain is their difference:
\[\gamma_s=\epsilon_+-\epsilon_- =\frac{2(1+\nu)\tau}{Y}.\]Since $G=\tau/\gamma_s$,
\[\boxed{Y=2G(1+\nu)}.\]Equivalent forms
Eliminating $\nu$ between the two relations gives
\[\boxed{Y=\frac{9KG}{3K+G}}, \qquad \boxed{\nu=\frac{3K-2G}{2(3K+G)}}.\]The requested expressions for Poisson’s ratio in terms of other elastic constants may also be written as
\[\boxed{ \nu=\frac{Y}{2G}-1 =\frac{3K-Y}{6K} =\frac{3K-2G}{2(3K+G)} }.\]The relations are independently checked in the Maxima worksheet; every displayed residual is zero.
Discussion