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Why only the second invariant is considered? What about the third invariant?

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    $\begingroup$ Could you add more information to your question. It will help people to give better answers if they have an idea of what you already know and what your possible reasons are. $\endgroup$
    – hazzey
    Oct 5 at 19:46
  • $\begingroup$ "Because the von Mises yield criterion is independent of the first stress invariant, it is applicable for the analysis of plastic deformation for ductile materials such as metals, as onset of yield for these materials does not depend on the hydrostatic component of the stress tensor." Does this answer your question? $\endgroup$
    – r13
    Oct 5 at 19:59
  • $\begingroup$ @r13 What about third invariant? $\endgroup$
    – Bhushan
    Oct 6 at 9:53
  • $\begingroup$ Please see this article. simscale.com/blog/2017/04/von-mises-stress $\endgroup$
    – r13
    Oct 6 at 14:47
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The invariants affect the shape of the yield surface. The von Mises condition assumes that the yield surface remains cylindrical in principal stress space. If you want pressure-dependence (the circular cylinder becomes a circular cone), then you add the first invariant into the mix. If the yield surface varies depending on whether you are in pure triaxial tension or triaxial compression, then you need the third invariant to represent the shape. See, for example, the Willam-Warnke condition. https://en.wikipedia.org/wiki/Willam%E2%80%93Warnke_yield_criterion

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  • $\begingroup$ Since this thread is already opened, I will also try to pitch in my query. Why does the ductile failure depend upon deviatoric stress only and not on hydrostatic stress, while the brittle failure depend upon hydrostatic stress and not on deviatoric stress? $\endgroup$ Oct 5 at 21:20
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    $\begingroup$ @RameezUlHaq: An engineering answer would be that metals don't show much pressure-dependence in experiments, while many ceramics/rocks etc. do. We model what we see. The physics is a bit involved and depends on grain sizes, and deep observations about dislocation development and flow. Keep in mind that all these yield models are gross simplifications of reality, and it's a miracle that they even work. $\endgroup$ Oct 5 at 21:27
  • $\begingroup$ I'm sorry but can you explain why yield surface needs to be cylindrical? While deriving the mathematical formulation the second stress invariant is considered and then we arrive at the final equation which represents the cylinder. And am I correct in saying that if I want to use this criteria for a material whose yield values vary in compression and in tension then we have to use criteria which includes both the invariants - 2nd and 3rd? $\endgroup$
    – Bhushan
    Oct 6 at 10:07
  • $\begingroup$ A cylindrical surface is the simplest model that can capture several experimentally observed phenomena in nominally isotropic metals and alloys. The cylinder has a circular cross-section because of isotropy which means that you don't give preference to any particular eigenvalue of the stress tensor. The second invariant is used because it is a measure of the shear stress (experimentally found to be the most relevant parameter in the plastic flow of pure metals). ... (see next comment) $\endgroup$ Oct 6 at 18:42
  • $\begingroup$ The first invariant also distinguishes between compression and tension. The third invariant is well described at en.wikipedia.org/wiki/Lode_coordinates. $\endgroup$ Oct 6 at 18:43

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