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.

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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