Isotropic elastic modulus relation ($E=3B(1-2\nu )=2G_s(1+\nu )$)

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 iswhere $E$ 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 giveFiniteness 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 $B=500/21\mathrm{GPa}\simeq 23.8\mathrm{GPa}$ and $G_s=62.5\mathrm{GPa}$, both positive. A negative Poisson's ratio is compatible with stability. The limit $\nu \to 1/2$ at fixed $E$ sends the bulk modulus to infinity and represents an incompressible limit, rather than a finite positive-modulus point.