Math 2.0

Mechanics of deformable solids Definition

Poisson's ratio

Builds on Lamé parametersStrain tensorCauchy stress tensor

Definition

Let a homogeneous isotropic linear elastic material have Lamé parameters λ\lambda and μ\mu, with μ>0\mu > 0 and 3λ+2μ>03\lambda + 2\mu > 0, so that λ+μ>0\lambda + \mu > 0 as well. Consider the state of uniaxial stress σ=s e1⊗e1\sigma = s \, e_1 \otimes e_1 with s≠0s \ne 0, where σ\sigma is the Cauchy stress tensor, and let ε\varepsilon be the strain that produces it. The Poisson's ratio of the material is ν=−ε22ε11=λ2(λ+μ).\nu = -\frac{\varepsilon_{22}}{\varepsilon_{11}} = \frac{\lambda}{2(\lambda + \mu)}.

Remarks

Example: for σ=s e1⊗e1\sigma = s \, e_1 \otimes e_1 one finds ε11=s(λ+μ)μ(3λ+2μ)\varepsilon_{11} = \frac{s(\lambda + \mu)}{\mu(3\lambda + 2\mu)} and ε22=ε33=−sλ2μ(3λ+2μ)\varepsilon_{22} = \varepsilon_{33} = -\frac{s \lambda}{2\mu(3\lambda + 2\mu)}, and the quotient gives the formula. For λ=120\lambda = 120 and μ=80\mu = 80 (in GPa, steel) it gives ν=120/400=0.3\nu = 120/400 = 0.3.

Non-example: in uniaxial strain the sides are held fixed, ε22=0\varepsilon_{22} = 0, and the quotient −ε22/ε11=0-\varepsilon_{22}/\varepsilon_{11} = 0 is not Poisson's ratio; the lateral faces must be free of stress.

Positivity of the elasticity tensor, μ>0\mu > 0 and 3λ+2μ>03\lambda + 2\mu > 0, is equivalent to E>0E > 0 and −1<ν<1/2-1 < \nu < 1/2. The value ν=1/2\nu = 1/2 is the incompressible limit, and materials with ν<0\nu < 0 are called auxetic. The volume change in uniaxial tension is tr⁡ε=(1−2ν)ε11\tr \varepsilon = (1 - 2\nu) \varepsilon_{11}.

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  1. Cauchy stress tensor concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Poisson%27s_ratio
About this page
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MSC74B05
DefinesPoisson's ratio
UsesLamé parameters, Strain tensor, Cauchy stress tensor
Changed3 November 2026

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