Hill Stress

General Hill Stress Theory

The Hill stress is an anisotropic scalar (effective) stress that can be calculated from:

\( \sigma_{hill} = \sqrt{ F (\sigma_{22}-\sigma_{33})^2 + G (\sigma_{33}-\sigma_{11})^2 + H (\sigma_{11}-\sigma_{22})^2 + 2 L \sigma_{23}^2 + 2M \sigma_{31}^2 + 2N\sigma_{12}^2 } \).

  • If \(F=G=H=1/2\) and \(L=M=N=3/2\) then the Hill stress becomes equal to the Mises stress.

The von Mises stress is a scalar (effective) stress that can be calculated from:

\( \sigma_{vm} = \sqrt{ \displaystyle\frac{1}{2}(\sigma_{22} – \sigma_{33})^2 + \frac{1}{2}(\sigma_{33}-\sigma_{11})^2 + \frac{1}{2} (\sigma_{11}-\sigma_{22})^2 + 3 \sigma_{23}^2 + 3 \sigma_{31}^2 + 3 \sigma_{12}^2 }, \)

which can also be written:

\( \sigma_{vm} = \displaystyle\sqrt{\frac{3}{2} s_{ij} s_{ij}} \).

  • \(s_{ij}\) is the deviatoric stress.
  • If \( \sigma_{11} \) is the only non-zero component then the Mises stress becomes: \( \sigma_{vm} = \sigma_{11}\).
  • If \(\sigma_{12}\) is the only non-zero component then the Mises stress becomes: \(\sigma_{vm} = \sqrt{3}\, \sigma_{12}\).

PolyUMod Theory

The PolyUMod© library uses the following isotropic scalar effective stress  to drive viscoplastic flow (\(s_{ij}\) is the deviatoric stress given by \(\mathbf{s}\equiv\text{dev}[\boldsymbol{\sigma}]\) ):

\( \tau = ||\mathbf{s}||_F = \displaystyle \sqrt{\mathbf{s}:\mathbf{s}}= \sqrt{\text{tr}[\mathbf{s} \mathbf{s}]} = \sqrt{ s_{ij} s_{ij}} \),

which is the same as the Mises stress multiplied by \(\sqrt{2/3}\).

For anisotropic flow, the PolyUMod library uses he following equation for the Hill stress:

\( \tau_{hill} = \displaystyle\sqrt{ \frac{F}{2} (\sigma_{22}-\sigma_{33})^2 + \frac{G}{2} (\sigma_{33}-\sigma_{11})^2 + \frac{H}{2} (\sigma_{11}-\sigma_{22})^2 + L \sigma_{23}^2 + M \sigma_{31}^2 + N\sigma_{12}^2 } \).

  • If \(F=G=H=1\) and \(L=M=N=3\), then the Hill stress becomes equal to the Mises stress.
  • If \(F=G=H=2/3\) and \(L=M=N=2\), then the Hill stress becomes equal to the effective stress that is used by PolyUMod for isotropic materials.
  • If \(\sigma_{11}\) is the only non-zero component then the Hill stress becomes \( \tau_{hill} = \sqrt{(G+H)/2}\, \sigma_{11}\).
  • If \(\sigma_{22}\) is the only non-zero component then the Hill stress becomes \( \tau_{hill} = \sqrt{(F+H)/2}\, \sigma_{22}\).
  • If \(\sigma_{33}\) is the only non-zero component then the Hill stress becomes \( \tau_{hill} = \sqrt{(F+G)/2}\, \sigma_{33}\).
  • If \(\sigma_{12}\) is the only non-zero component then the Hill stress becomes \( \tau_{hill} = \sqrt{N}\, \sigma_{12}\).
  • If \(\sigma_{23}\) is the only non-zero component then the Hill stress becomes \( \tau_{hill} = \sqrt{L}\, \sigma_{23}\).
  • If \(\sigma_{31}\) is the only non-zero component then the Hill stress becomes \( \tau_{hill} = \sqrt{M}\, \sigma_{31}\).
  • If a material is almost isotropic, but has a slightly higher yield stress in the 1-dir, then select: \(F=1+x\), \(G=H=1-x\), \(L=M=N=3\), where \(x > 0\), or alternatively set: \(F=2/3+x\), \(G=H=2/3-x\), \(L=M=N=2\).
  • If a material is almost isotropic, but has a slightly higher yield stress in the 2-dir, then select: \(G=2/3+x\), \(F=H=2/3-x\), \(L=M=N=2\).
  • If a material is almost isotropic, but has a slightly higher yield stress in the 3-dir, then select: \(H=2/3+x\), \(F=G=2/3-x\), \(L=M=N=2\).

Example: Comparison of Isotropic and Anisotropic PolyUMod TNV Model

Stress-strain prediction (generated using MCalibration©) from the TNV model with isotropic flow.

Hill Stress Example

Stress-strain prediction from the TNV model with anisotropic flow. Note that F=G=H=2/3. With these values we get the same stress-strain prediction as in the isotropic case.

If I change to F=0.747 and G=H=0.587, the yield stress in the 1-direction is slightly increased, and the yield stress in the 2- and 3-directions are left unchanged.

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