Document version history
===================================


.. csv-table::

    "**Version Aster**", "**Author (s) or Contributor (s),**
    **organization**", "**Description of changes**"
    ":math:`7.4` ", "J.M. PROIX, O. DIARD, T., T. KANIT, T. EDF /R & D", "Initial Release"
    ":math:`8.4` ", "J.EL- GHARIB, J.M. PROIX, J.M., EDF /R & D", "Complements"
    ":math:`9.2` ", "J.EL- GHARIB, J.M. PROIX, J.M., EDF /R & D", "Added behavior MONO_DD_KR."
    ":math:`9.4` ", "J.EL- GHARIB, J.M. PROIX, J.M., EDF /R & D", "Behavior Optimization MONO_DD_KR."
    ":math:`10.2` ", "J.M. PROIX ", "note on the interaction matrix"
    ":math:`10.3` ", "N. RUPIN, J.M. PROIX, EDF /R & D", "Add behavior DD_CFC"
    ":math:`10.5` ", "N. RUPIN, J.M. PROIX, EDF /R & D", "Addition of crystal lattice rotation"
    ":math:`11.1` ", "N. RUPIN, F. LATOURTE, J.M. PROIX, J.M., EDF /R & D", "Adding major deformations"
    ":math:`11.1` ", "N. RUPIN, J.M. PROIX, EDF /R & D", "Files 16373 (convergence criterion) and 17422: correction of POLYCRISTAL internal variables."
    ":math:`11.1` ", "P. DE BONNIERES, J.M. PROIX ", "Sheet 14586: Add behavior DD_FAT"
    ":math:`11.2` ", "J.M. PROIX ", "Sheet 18398: resorption of ALGO_C_PLAN"
    ":math:`11.2` ", "J.M. PROIX ", "Sheet 18692 added to DD_CC"
    ":math:`11.3` ", "J.M. PROIX ", "Form 19021 adding internal variables for DD_CC_IRRA"
    ":math:`12.2` ", "J.M. PROIX ", "Fiches 22563: simplification of the law DD_CC and 22491 coefficient :math:`{\mu }^{\mathit{loca}}` for homogenization,"




**Jacobian expression for integrated elasto-visco-plastic equations**

The system to be solved is of the form:

:math:`R(Y)=R(\Sigma ,\Delta {E}^{\text{vp}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})=\{\begin{array}{}s(\Sigma ,\Delta {E}^{\text{vp}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\\ e(\Sigma ,\Delta {E}^{\text{vp}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\\ {n}_{s}\left\{\begin{array}{}a(\Sigma ,\Delta {E}^{\text{vp}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\\ g(\Sigma ,\Delta {E}^{\text{vp}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\\ p(\Sigma ,\Delta {E}^{\text{vp}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\end{array}\right\}\end{array}`

:math:`=\{\begin{array}{}{\Lambda }^{\text{-1}}\Sigma -({\Lambda }_{\text{-}}^{\text{- 1}}){\Sigma }^{\text{-}}-(\Delta E-\Delta {E}^{\text{th}}-\Delta {E}^{\text{vp}})\\ \Delta {E}^{\text{vp}}-{\sum }_{s}{\mu }_{s}\Delta {\gamma }_{s}\\ {n}_{s}\left\{\begin{array}{}\Delta {\alpha }_{s}-h({\tau }_{s}^{\text{+}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\\ \Delta {\gamma }_{s}-g({\tau }_{s}^{\text{+}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\\ {\mathrm{\Delta p}}_{s}-f({\tau }_{s}^{\text{+}},{\alpha }_{s}^{\text{+}},{\gamma }_{s}^{\text{+}},{p}_{s}^{\text{+}})\end{array}\right\}\end{array}`

:math:`=0\text{avec}{\tau }_{s}^{\text{+}}={\Sigma }^{\text{+}}:{\mu }_{s}`

So let's evaluate the terms of the Jacobian hypermatrix :math:`J` at the moment :math:`t+\Delta t`

:math:`J=\left[\begin{array}{ccccc}\frac{\partial s}{\partial \text{ΔΣ}}& \frac{\partial s}{\partial {\mathrm{\Delta E}}^{\text{vp}}}& \frac{\partial s}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial s}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial s}{\partial {\mathrm{\Delta p}}_{s}}\\ \frac{\partial e}{\partial \text{ΔΣ}}& \frac{\partial e}{\partial {\mathrm{\Delta E}}^{\text{vp}}}& \frac{\partial e}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial e}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial e}{\partial {\mathrm{\Delta p}}_{s}}\\ \frac{\partial a}{\partial \text{ΔΣ}}& \frac{\partial a}{\partial {\mathrm{\Delta E}}^{\text{vp}}}& \frac{\partial a}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial a}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial a}{\partial {\mathrm{\Delta p}}_{s}}\\ \frac{\partial g}{\partial \text{ΔΣ}}& \frac{\partial g}{\partial {\mathrm{\Delta E}}^{\text{vp}}}& \frac{\partial g}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial g}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial g}{\partial {\mathrm{\Delta p}}_{s}}\\ \frac{\partial p}{\partial \text{ΔΣ}}& \frac{\partial p}{\partial {\mathrm{\Delta E}}^{\text{vp}}}& \frac{\partial p}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial p}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial p}{\partial {\mathrm{\Delta p}}_{s}}\end{array}\right]`

With regard to the first row of the matrix, independently of the work-hardening and flow equations, we have:

:math:`\begin{array}{cccc}\frac{\partial s}{\partial \text{ΔΣ}}={D}^{-1}& \frac{\partial s}{\partial {\mathrm{\Delta E}}^{\text{vp}}}=\text{Id}& \frac{\partial s}{\partial {\mathrm{\Delta \alpha }}_{s}}=& \frac{\partial s}{\partial {\mathrm{\Delta \gamma }}_{s}}=\frac{\partial s}{\partial {\mathrm{\Delta p}}_{s}}\end{array}=0`

The second line can also be written independently of the flow and the workings:

:math:`\begin{array}{ccc}\begin{array}{ccc}\frac{\partial e}{\partial \text{ΔΣ}}=0& \frac{\partial e}{\partial {\mathrm{\Delta E}}^{\text{vp}}}=\text{Id}& \frac{\partial e}{\partial {\mathrm{\Delta \alpha }}_{s}}=0\end{array}& \frac{\partial e}{\partial {\mathrm{\Delta \gamma }}_{s}}=-{\mu }_{s}& \frac{\partial e}{\partial {\mathrm{\Delta p}}_{s}}=0\end{array}`

The first column of the rows corresponding to equations (a), (g), and (p) is written:

:math:`\begin{array}{c}\frac{\partial a}{\partial \text{ΔΣ}}=\frac{\partial a}{\partial {\mathrm{\Delta \tau }}_{s}}\frac{{\mathrm{\Delta \tau }}_{s}}{\text{ΔΣ}}\\ \frac{\partial g}{\partial \text{ΔΣ}}=\frac{\partial g}{\partial {\mathrm{\Delta \tau }}_{s}}\frac{{\mathrm{\Delta \tau }}_{s}}{\text{ΔΣ}}\\ \frac{\partial p}{\partial \text{ΔΣ}}=\frac{\partial p}{\partial {\mathrm{\Delta \tau }}_{s}}\frac{{\mathrm{\Delta \tau }}_{s}}{\text{ΔΣ}}\end{array}`

with

:math:`\frac{{\mathrm{\Delta \tau }}_{s}}{\text{ΔΣ}}={({\mu }_{s})}^{T}`

The second column is identically zero (because of equation (e)): the flow and work hardening relationships can only be expressed as a function of :math:`\Delta {\gamma }_{s}` and not of :math:`{\mathrm{\Delta E}}^{\text{vp}}`.

The last block of equations, for its part, depends on the behaviors chosen:

:math:`\begin{array}{ccc}\frac{\partial a}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial a}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial a}{\partial {\mathrm{\Delta p}}_{s}}\\ \frac{\partial g}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial g}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial g}{\partial {\mathrm{\Delta p}}_{s}}\\ \frac{\partial p}{\partial {\mathrm{\Delta \alpha }}_{s}}& \frac{\partial p}{\partial {\mathrm{\Delta \gamma }}_{s}}& \frac{\partial p}{\partial {\mathrm{\Delta p}}_{s}}\end{array}`

**Example**

Let's choose the viscoplastic flow relationship **MONO_VISC1**

:math:`\begin{array}{}(g){\mathrm{\Delta \gamma }}_{s}-{\mathrm{\Delta p}}_{s}\frac{{\tau }_{s}-{\mathrm{c\alpha }}_{s}}{\mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid }=0\\ (p){\mathrm{\Delta p}}_{s}-\mathrm{\Delta t}\text{.}{\langle \frac{\mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid -{R}_{s}({p}_{s})}{k}\rangle }^{n}=0\end{array}`

with isotropic work hardening **MONO_ISOT1 :** :math:`{R}_{s}({p}_{s})={R}_{0}+Q(\sum _{r=1}^{N}{h}_{\text{sr}}(1-{e}^{-{\text{bp}}_{r}}))`,

.. image:: images/Object_304.svg
    :width: 122
    :height: 24


.. _RefImage_Object_304.svg:

and kinematic work hardening defined by **MONO_CINE1**

:math:`(a){\mathrm{\Delta \alpha }}_{s}-{\mathrm{\Delta \gamma }}_{s}+{\mathrm{d\alpha }}_{s}{\mathrm{\Delta p}}_{s}=0`

so:

:math:`\begin{array}{c}\frac{\partial a}{\partial {\mathrm{\Delta \tau }}_{s}}=0\\ \frac{\partial g}{\partial {\mathrm{\Delta \tau }}_{s}}=0\\ \frac{\partial p}{\partial {\mathrm{\Delta \tau }}_{s}}=\frac{-\mathrm{n\Delta t}}{{K}^{n}}{\langle \mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid -{R}_{s}({p}_{s})\rangle }^{n-1}\frac{{\tau }_{s}-\mathrm{c\alpha }}{\mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid }\end{array}`

:math:`\begin{array}{ccc}\frac{\partial a}{\partial {\mathrm{\Delta \alpha }}_{s}}& =& 1+{\mathrm{d\Delta p}}_{s}\\ \frac{\partial g}{\partial {\mathrm{\Delta \alpha }}_{s}}& =& 0\\ \frac{\partial p}{\partial {\mathrm{\Delta \alpha }}_{s}}& =& \frac{\text{nc}\mathrm{\Delta t}}{{K}^{n}}{\langle \mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid -{R}_{s}({p}_{s})\rangle }^{n-1}\frac{{\tau }_{s}-{\mathrm{c\alpha }}_{s}}{\mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid }\end{array}`

:math:`\begin{array}{ccc}\frac{\partial a}{\partial {\mathrm{\Delta \gamma }}_{s}}& =& -1\\ \frac{\partial g}{\partial {\mathrm{\Delta \gamma }}_{s}}& =& 1\\ \frac{\partial p}{\partial {\mathrm{\Delta \gamma }}_{s}}& =& 0\end{array}`

:math:`\begin{array}{}\begin{array}{ccc}\frac{\partial a}{\partial {\mathrm{\Delta p}}_{s}}& =& {\mathrm{d\alpha }}_{s}\\ \frac{\partial g}{\partial {\mathrm{\Delta p}}_{s}}& =& \frac{{\tau }_{s}-{\mathrm{c\alpha }}_{s}}{\mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid }\\ \frac{\partial p}{\partial {\mathrm{\Delta p}}_{s}}& =& 1+\frac{\mathrm{n\Delta t}}{{K}^{n}}{\langle \mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid -{R}_{s}({p}_{s})\rangle }^{n-1}\frac{{\text{dR}}_{s}({p}_{s})}{{\mathrm{d\Delta p}}_{s}}\end{array}\\ \frac{{\text{dR}}_{s}({p}_{s})}{{\mathrm{d\Delta p}}_{s}}={\text{Qbh}}_{\text{ss}}{e}^{-{\text{bp}}_{s}}\end{array}`

and, concerning the interaction between sliding systems, there is only one non-zero term:

:math:`\begin{array}{}\frac{\partial p}{\partial {\mathrm{\Delta p}}_{r}}=1+\frac{\mathrm{n\Delta t}}{{K}^{n}}{\langle \mid {\tau }_{s}-{\mathrm{c\alpha }}_{s}\mid -{R}_{s}({p}_{s})\rangle }^{n-1}\frac{{\text{dR}}_{s}({p}_{s})}{{\mathrm{d\Delta p}}_{r}}\\ \frac{{\text{dR}}_{s}({p}_{s})}{{\mathrm{d\Delta p}}_{r}}={\text{Qbh}}_{\text{sr}}{e}^{-{\text{bp}}_{r}}\end{array}`



**Consistent tangent operator evaluation**

It is a question of finding the coherent tangent operator, i.e. calculated from the solution of :math:`(R(Y)=0)` at the end of the increment. For a small variation of :math:`R`, this time considering :math:`\Delta E` as a variable and not as a parameter, we get:

:math:`\frac{\mathrm{\partial }R}{\mathrm{\partial }\Sigma }\delta \Sigma +\frac{\mathrm{\partial }R}{\mathrm{\partial }\Delta E}\delta \Delta E+\frac{\mathrm{\partial }R}{\mathrm{\partial }\Delta {\gamma }_{s}}\delta \Delta {\gamma }_{s}\mathrm{=}0`

This system can be written as:

:math:`\frac{\mathrm{\partial }R}{\mathrm{\partial }Y}\delta (Y)\mathrm{=}X` with :math:`Y\mathrm{=}\left[\begin{array}{c}\Sigma \\ \Delta {\gamma }_{s}\end{array}\right]` and :math:`X\mathrm{=}\left[\begin{array}{c}\delta \Delta E\\ 0\end{array}\right]`

By writing the Jacobian matrix in the form:

:math:`J\text{.}\delta Y\mathrm{=}\left[\begin{array}{cc}{Y}_{0}& {Y}_{1}\\ {Y}_{2}& {Y}_{3}\end{array}\right]\left[\begin{array}{c}\Sigma \\ \Delta Z\end{array}\right]`

With:

:math:`\begin{array}{}{Y}_{0}={\Lambda }^{-1}\\ \Delta Z=\left\{\Delta {\gamma }_{s}\right\}\times {n}_{s}\end{array}`

The dimensions of the sub-matrices are:

:math:`\begin{array}{}\text{dim}({Y}_{0}={\Lambda }^{-1})=\left[\mathrm{6,6}\right]\\ \text{dim}{Y}_{1}=\left[\mathrm{6,}{n}_{s}\right]\\ \text{dim}{Y}_{2}=\left[{n}_{s}\mathrm{,6}\right]\\ \text{dim}{Y}_{3}=\left[{n}_{s},{n}_{s}\right]\end{array}`

By operating by successive removals and substitutions, the third block of the equation system gives:

:math:`\begin{array}{c}\Delta Z\mathrm{=}\mathrm{-}{({Y}_{3})}^{\mathrm{-}1}{Y}_{2}\Sigma \\ ({Y}_{0}\mathrm{-}{Y}_{1}{({Y}_{3})}^{\mathrm{-}1}{Y}_{2})\Sigma \mathrm{=}\Delta E\end{array}`

the tangent operator sought can therefore be written directly:

:math:`{(\frac{\mathrm{\partial }\Sigma }{\mathrm{\partial }E})}_{t+\Delta t}\mathrm{=}{(\frac{\mathrm{\partial }\Sigma }{\mathrm{\partial }\Delta E})}_{t+\Delta t}\mathrm{=}{({Y}_{0}\mathrm{-}{Y}_{1}{Y}_{3}^{\mathrm{-}1}{Y}_{2})}^{\mathrm{-}1}`