Connect stress and strain in an isotropic solid

A small symmetric strain tensor decomposes into a spherical part proportional to the identity and a trace-free deviatoric part. The trace describes fractional volume change to first order, while the deviator describes shape change at fixed volume to that order. Isotropy requires the same response after rotating the entire loading experiment relative to the material. Linearity then permits one coefficient for the volume response and another for the shape response; it does not make all stress components equal.

In Lamé notation the law is $\boldsymbol\sigma=\lambda\operatorname{tr}(\boldsymbol\varepsilon)\mathbf I+2\mu\boldsymbol\varepsilon$. Here $\mu$ is the shear modulus and both parameters have pressure units. This relation assumes an unstressed reference at fixed temperature and omits thermal expansion, plastic strain, and viscosity. A crystal can remain linearly elastic but be anisotropic, requiring more independent constants. A material can also be isotropic yet nonlinear at larger strain, so the two assumptions must be tested separately.

In a long bar loaded only axially with free lateral surfaces, the ideal interior has uniaxial stress. The axial strain is $\sigma_{xx}/E$, and the two lateral strains are $-\nu\sigma_{xx}/E$. Young's modulus $E$ and Poisson ratio $\nu$ summarize that particular response. Poisson contraction is a consequence of the three-dimensional law and the zero lateral-stress conditions. It is not an extra applied lateral compressive force.

If a fixture instead suppresses both lateral strains, the material develops lateral stresses. For prescribed strain $\varepsilon_{xx}=e$ with the other normal strains zero, axial stress is $(\lambda+2\mu)e$ and each lateral normal stress is $\lambda e$. That axial coefficient generally exceeds E for ordinary positive Poisson ratio. Calling both experiments simply tension conceals the constraint that changes the answer. Similarly, plane stress and plane strain are different reductions of the same three-dimensional constitutive law.

Elastic moduli measure stress divided by dimensionless strain and therefore have units of pascals. Bulk and shear moduli answer different experiments, so a very large bulk modulus does not imply an equally large resistance to shear. Nearly incompressible elastomers are an important example: volume change is costly while shape change remains relatively easy. In the formal incompressible limit the volume constraint introduces a pressure-like unknown, so blindly assigning an infinite number to a finite formula is a poor computational strategy.

For positive elastic energy in an unconstrained three-dimensional isotropic solid, the conditions are $\mu>0$ and $K>0$. They imply $E>0$ and $-1<\nu<1/2$. They do not require $\lambda>0$: an isotropic auxetic solid may have negative Poisson ratio while remaining stable. These statements concern infinitesimal material energy near a stress-free state. They do not guarantee that a compressed column cannot buckle, that a flaw cannot grow, or that the response stays reversible at arbitrarily large loads.