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According to efunda, enter image description here

But if the Poisson's ratio is 0.5 for incompressible materials, the stress tensor (vector here) cannot be determined for an arbitrary strain, which does not make any sense, since it is possible for an incompressible material to have a state of stress at a point for a given strain.

What is wrong here?

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2 Answers 2

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The basic point that for an incompressible material, the compliance matrix is not invertible. To see this, recall that schematically $$\text{strain}=\text{(compliance matrix)}\cdot\text{(stress)}\,.$$

However, incompressible means that (following your quetsion, I assume an isotropic material here)

  • there is no volume strain, $\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}=0$, and
  • correspondingly, a purely isotropic stress produces no strain. In your notation, this means $$\text{(compliance matrix)}\cdot\begin{pmatrix}1\\1\\1\\0\\0\\0\end{pmatrix}=0\,.$$

Hence, you should isolate the isotropic component (where any arbitrary stress will produce zero strain) and consider deviatoric loads separately.

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Incompressibility doesn't follow Hooke's law. What we should use here is $\lambda\,\mathrm{Tr}(\varepsilon) = \text{constant}$, where $\lambda$ approaches infinity. So approximately, $\mathrm{Tr}(\varepsilon) = 0$.

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