# How does Abaqus calculate Hill's function for non-rectangular coordinate systems?

Within the manual, the effective/von Mises stress or Hill's potential for anisotropic bodies is calculated in Abaqus in cartesian rectangular coordinates as

$\sigma_{eff}=\sqrt{I_{1}^{2}-3I_{2}} \\ f({\sigma})_{Hill}=\sqrt{\big( F(\sigma_{22}-\sigma_{33})^{2}+ G(\sigma_{33}-\sigma_{11})^{2}+H(\sigma_{11}-\sigma_{22})^{2}+\\ 2L\sigma_{23}^{2}+2M\sigma_{31}^2+2N\sigma_{12}^{2}\big) }$

where $I_{1}$ and $I_{2}$ are the first and second invariants of the stress tensor and F,G,H,L,M,N are constants you define.

Whereas it is clear that in non-rectangular systems the effective stress is calculated in the coordinate system you work in, this is not clear when one applies Hill's potential. I am trying to implement anisotropy with axial symmetry/axisymmetric model, but Hill's potential function is given in terms of rectangular coordinates.

Does Abaqus carry out the transformation from rectangular to polar in non rectangular coordinate systems automatically? Do I have to supply the values that correspond to a transformed coordinate system? Or are do the ${1,2,3}$ coordinates correspond to ${r,z,\theta}$ in Hill's potential function?

Any help appreciated, Thank you!

In particular, you are limited to transverse isotropy if the model is axisymmetric and Hill's criterion takes the form $$f(\boldsymbol{\sigma}) = \sqrt{F\,[\sigma_{rr}^2 + \sigma_{\theta\theta}^2 + 2\sigma_{zz}^2 - 2(\sigma_{rr}+\sigma_{\theta\theta})\sigma_{zz}] + H\,(\sigma_{rr}-\sigma_{\theta\theta})^2 + 2\,L\,\sigma_{rz}^2 }$$ where $1\rightarrow r, 2 \rightarrow \theta, 3 \rightarrow z$. In some axisymmetric tests, the load is applied in the form of pressure in the $r-\theta$ plane and we have $\sigma_{rr} = \sigma_{\theta\theta}$. For that situation, $$f(\boldsymbol{\sigma}) = \sqrt{2F\,(\sigma_{rr} - \sigma_{zz})^2 + 2\,L\,\sigma_{rz}^2 } \,.$$ A good description of the treatment of axisymmetric problems in FEA can be found here.
• Thank you for you response! I cannot see how it would take this form. You seem to assume that $\sigma_{rr}-\sigma_{\theta\theta}=\sigma_{zz}-\sigma_{\theta\theta}$. Also, since it is axisymmetric, then $\sigma_{r\theta}=\sigma_{z\theta}=0$ meaning $L=M=0$ leaving N as the only non-zero constant. Am I wrong? How did you arrive to this form? – user2822693 Mar 13 '14 at 9:29