= Isotropic elastic modulus relation
{title2=$E=3B(1-2\nu)=2G_s(1+\nu)$}
Under uniform compressive <pressure> $p$, isotropic <Hooke's law> gives each normal <strain> as $-p(1-2\nu)/E$. The <volume> <strain> is three times this, so the <bulk modulus> is
$$
B=\frac{E}{3(1-2\nu)},
$$
where $E$ is <Young's modulus> and $\nu$ is <Poisson's ratio>. The <shear modulus> is $G_s=E/[2(1+\nu)]$.
For stable three-dimensional <isotropic> <linear elasticity>, both $B$ and $G_s$ must be positive. With finite $E>0$, these two positivity requirements give
$$
-1<\nu<\frac12.
$$
Finiteness of $B$ alone does not establish the upper inequality: its positive sign is essential. The bounds come from positive <strain energy density> for both <volumetric strain> and <deviatoric strain>, rather than a prohibition on negative <Poisson's ratio>.
For example, $E=100\,\mathrm{GPa}$ and $\nu=-1/5$ give $B=500/21\,\mathrm{GPa}\simeq23.8\,\mathrm{GPa}$ and $G_s=62.5\,\mathrm{GPa}$, both positive. A negative <Poisson's ratio> is compatible with stability. The limit $\nu\to1/2$ at fixed $E$ sends the <bulk modulus> to infinity and represents an incompressible limit, rather than a finite positive-modulus point.
Back to article page
