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!


This is a tricky issue because the parameters used in Hill's criterion are determined on the basis of tests with respect to a particular set of orthogonal world coordinates (not necessarily Cartesian). Therefore, you should express the stress in that coordinate system before you compute the value.

As long as you are working in an orthogonal curvilinear coordinate system, the easiest way of dealing with the issue is to transform your stress components at each point so that they are with respect to the experimental coordinates system.

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.

(Update: The previous axisymmetric expression was wrong and has been corrected.)

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  • $\begingroup$ 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? $\endgroup$ – user2822693 Mar 13 '14 at 9:29

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