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Mar 3 '17 at 10:36 comment added 1QuickQuestion Thank you for the detailed answer and clarification. One last thing to clear up. Must the equation for $\sigma_{n+1}$ have to be solved iteratively with the consistency condition $f(\sigma_{n+1})=0$? I can't figure out how to solve for $\dot{\lambda}$ within the yield function for $n+1$ with $\sqrt{\frac{3\sigma_{n+a}^TT\sigma_{n+1}}{2}}=\sigma_y$.
Mar 2 '17 at 5:26 history edited Biswajit Banerjee CC BY-SA 3.0
Updated explanation.
Mar 2 '17 at 4:47 history edited Biswajit Banerjee CC BY-SA 3.0
Corrected T to Tbar.
Mar 2 '17 at 4:44 comment added Biswajit Banerjee That should be $\bar{T}$ in the first relation and not $T$. I have corrected the answer. I'll try to write out the return algo when I get the time.
Mar 1 '17 at 10:12 comment added 1QuickQuestion Can you confirm that $T \sigma = [s_{11}\ s_{22} \ 2s_{12}]^T$ or have I made a mistake? Secondly, the yield function you have given is interchangeable with $f = \sqrt{\sigma_{11}^2-\sigma_{11}\sigma_{22}+\sigma_{22}^2+3\sigma_{12}^2} - \sigma_Y$ (in plane stress) which I have no problem with. My current issue is how to correctly compute the return in plane stress as in 3D as simple scaling factor (scalar value) can be used as show here. In the link given $\alpha = \sigma_Y / \sigma_{VM}$.
Mar 1 '17 at 2:43 history answered Biswajit Banerjee CC BY-SA 3.0