Anisotropic Hill Strain

PolyUMod Theory for the Anisotropic Hill Strain

The PolyUMod© library uses the following equation to calculate an effective strain for anisotropic materials. This equation is based on the work by Rodney Hill:

\( \varepsilon_{hill} = \displaystyle\left\{ \frac{2}{9} \left[ F(\varepsilon_{22} – \varepsilon_{33})^2 + G(\varepsilon_{33} – \varepsilon_{11}) + H(\varepsilon_{11}-\varepsilon_{22})^2\right] + \frac{4}{3} \left[ L\varepsilon_{23}^2 + M \varepsilon_{31}^2 + N \varepsilon_{12}^2 \right] \right\}^{1/2}, \)

where \(\varepsilon_{ij}\) are the true strain components.

  • If F=G=H=L=M=N=1, then the response becomes isotropic.
  • If the strain state is incompressible and uniaxial then \( \varepsilon_{hill} = \varepsilon\).
  • If the only non-zero strain component is \(\varepsilon_{12}\) then \(\varepsilon_{hill}=2 \varepsilon_{12}/\sqrt{3}\).

For incompressible uniaxial loading the anisotropic Hill strain equation can be simplified to:

  • For tension in the 1-direction: \(\varepsilon_{hill} = \varepsilon \cdot \sqrt{(G+H)/2}\).
  • For tension in the 2-direction: \(\varepsilon_{hill} = \varepsilon \cdot \sqrt{(F+H)/2}\).
  • For tension in the 3-direction: \(\varepsilon_{hill} = \varepsilon \cdot \sqrt{(F+G)/2}\).


  • If the anisotropic Hill strain should be 10% higher in the 1-direction, then set F=0.79, G = H = 1.21.
  • If the anisotropic Hill strain should be 10% higher in the 2-direction, then set G=0.79, F = H = 1.21.
  • If the Hill strain should be 10% higher in the 3-direction, then set H=0.79, F = G = 1.21.

Example: PolyUMod TNV Model

In this example I’m using MCalibration with a PolyUMod TNV model with a Hill strain failure model. I set F=0.79, G=H=1.21, and as expected, the failure occurs at a small strain in the 1-direction since the Hill strain is higher in that direction.

Anisotropic Hill strain prediction

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