Incremental shape
==================

We are placing ourselves here in the context of finite increases. The index — refers to a component at the beginning of a loading step and the absence of an index a component at the end of a loading step. The :math:`\Delta` operator refers to the increase of a component.

In pure mechanics or when the law is used in modeling THM in effective constraints, the equations translating elastic behavior are written:

:math:`\begin{array}{}s\text{=}{s}^{\text{-}}\text{+}2\mu (\Delta e\text{-}\Delta {e}^{p})\text{=}{s}^{e}\text{-}2\mu \Delta {e}^{p}\text{}\begin{array}{cc}\text{où}& {s}^{e}\text{=}{s}^{\text{-}}\text{+}2\mu \Delta e\end{array}\\ {I}_{1}\text{=}{I}_{{1}^{\text{-}}}\text{+}\mathrm{3K}(\Delta {\varepsilon }_{\nu }\text{-}\Delta {\varepsilon }_{\nu }^{p})\text{=}{I}_{1}^{e}\text{-}\mathrm{3K}\Delta {\varepsilon }_{\nu }^{p}\text{}\begin{array}{cc}\text{où}& {I}_{1}^{e}\text{=}{I}_{1}^{\text{-}}+\mathrm{3K}\Delta {\varepsilon }_{\nu }\end{array}\end{array}`

When the law is used in total stress modeling THM, the stress tensor and the equations translating elastic behavior are written as follows:

:math:`\begin{array}{}\sigma \text{=}H(\varepsilon \text{-}{\varepsilon }^{p})\text{+}{\sigma }_{p}I\\ {\sigma }^{e}\text{=}{\sigma }^{\text{-}}\text{+}H\Delta \varepsilon +\Delta {\sigma }_{p}I\text{=}\sigma {\text{'}}^{\text{-}}\text{+}H\Delta \varepsilon \text{+}{\sigma }_{p}I\\ s\text{=}{s}^{e}\text{-}2\mu \Delta {e}^{p}\text{où}{s}^{e}\text{=}{s}^{\text{-}}\text{+}2\mu \Delta e\\ {I}_{1}\text{=}{I}_{1}^{e}\text{-}\mathrm{3K}\Delta {\varepsilon }_{v}^{p}\text{où}{I}_{1}^{e}\text{=}{I}_{1}^{\text{-}}\text{+}\mathrm{3K}\Delta {\varepsilon }_{v}\text{+}3\Delta {\sigma }_{p}\end{array}`

:math:`\Delta {\sigma }_{p}\text{=}\text{-}b(S\Delta {p}_{\text{lq}}\text{+}(1\text{-}S)\Delta {p}_{\text{gz}})\text{=}b(S\Delta {p}_{c}\text{-}\Delta {p}_{\text{gz}})`

In addition, the flow rule is written as:

:math:`d{\varepsilon }_{\text{ij}}^{p}\text{=}d\lambda \text{.}\frac{\partial G}{\partial {\sigma }_{\text{ij}}}(\sigma ,\gamma )\text{=}d\lambda (\eta (\gamma ){\delta }_{\text{ij}}\text{+}\sqrt{\frac{3}{2}}\frac{{s}_{\text{ij}}}{{s}_{\text{II}}})`

Like :math:`{e}^{p}\text{=}{\varepsilon }^{p}\text{-}\frac{\text{tr}({\varepsilon }^{p})}{3}I`, :math:`\Delta {\varepsilon }^{p}\text{=}\Delta {e}^{p}\text{+}\frac{\Delta {\varepsilon }_{\nu }^{p}}{3}I`. From this we deduce that :math:`\{\begin{array}{c}\Delta {e}^{p}\text{=}\frac{3}{2}\frac{s}{{\sigma }_{\text{eq}}}\Delta \lambda \\ \Delta {\varepsilon }_{\nu }^{p}\text{=}3\eta (\gamma )\Delta \lambda \end{array}`, and therefore, like :math:`{\sigma }_{\text{eq}}^{e}s\text{=}{\sigma }_{\text{eq}}{s}^{e}`, the following relationships are identical for both aspects of the law:

:math:`\{\begin{array}{c}s\text{=}{s}^{e}\text{-}3\mu \frac{s}{{\sigma }_{\text{eq}}}\Delta \lambda \text{=}{s}^{e}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\\ \begin{array}{}{I}_{1}\text{=}{I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda \text{}\\ {\sigma }_{\text{eq}}\text{=}{\sigma }_{\text{eq}}^{e}\text{-}3\mu \Delta \lambda \text{}\end{array}\end{array}`

Finally, :math:`\Delta {\gamma }^{p}\text{=}\sqrt{\frac{2}{3}\Delta {e}^{p}:\Delta {e}^{p}}\text{=}\sqrt{\frac{2}{3}\Delta {\lambda }^{2}(\frac{3}{2}\frac{s}{{\sigma }_{\text{eq}}}):(\frac{3}{2}\frac{s}{{\sigma }_{\text{eq}}})}\text{=}\Delta \lambda`. So, taking up the result obtained in paragraph :ref:`3.2.3 <Ref100116782>`, we see that :math:`\Delta \gamma \text{=}\Delta \lambda [:ref:`\eta (\gamma )\pm 1 <\eta (\gamma )\pm 1>`] `and like :math:`\mathrm{\Delta \gamma },\mathrm{\Delta \lambda }\ge 0`, we necessarily have :math:`\Delta \gamma \text{=}\Delta \lambda [:ref:`\eta (\gamma )+1 <\eta (\gamma )+1>`]\ ge 0\ text { . } `

*Note* **: The calculation of** :math:`\mathrm{\Delta \lambda }` **is the same in all formulations.**


.. _Ref100129806:




Calculation of :math:`\Delta \lambda`
--------------------------------



We denote :math:`P` the transition matrix such as:

:math:`\tilde{P}\text{.}{s}^{e}\text{.}P\text{=}{\stackrel{ˉ}{s}}^{e}\text{avec}{\stackrel{ˉ}{s}}^{e}\text{=}\text{diag}({s}_{1}^{e},{s}_{2}^{e},{s}_{3}^{e})`, or

:math:`\tilde{P}\text{.}{\sigma }^{e}\text{.}P\text{=}{\stackrel{ˉ}{\sigma }}^{e}\text{avec}{\stackrel{ˉ}{\sigma }}^{e}=\text{diag}({\sigma }_{1}^{e},{\sigma }_{2}^{e},{\sigma }_{3}^{e})\text{et}{\sigma }_{i}^{e}={s}_{i}^{e}+\frac{{I}_{1}^{e}}{3}`

So:

:math:`\begin{array}{}\tilde{P}\text{.}\sigma \text{.}P\text{=}\tilde{P}\text{.}(s\text{+}\frac{{I}_{1}}{3}I)\text{.}P\text{=}\tilde{P}\text{.}({s}^{e}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )I)\text{.}P\\ \text{}\text{=}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}}){\stackrel{ˉ}{s}}^{e}\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )I\\ \text{}=\text{diag}({\sigma }_{1},{\sigma }_{2},{\sigma }_{3})\\ \text{avec}{\sigma }_{i}\text{=}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}}){s}_{i}^{e}\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )\text{=}{\sigma }_{i}^{e}\text{-}3\Delta \lambda (\frac{\mu {s}_{i}^{e}}{{\sigma }_{\text{eq}}^{e}}\text{+}K\eta (\gamma ))\end{array}`

So, if :math:`(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})>0`, that is :math:`{s}_{\text{II}}>0`, the :math:`{\sigma }_{i}` are ordered like the :math:`{s}_{i}^{e}`. If :math:`(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\text{=}0`, that is :math:`{s}_{\text{II}}\text{=}0`, the :math:`{\sigma }_{i}` are all equal to :math:`\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta \Delta \lambda )\text{=}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{3K}\eta \frac{{\sigma }_{\text{eq}}^{e}}{\mu })`. We can thus write the criterion :math:`F` as a function of :math:`\Delta \lambda \text{ou}\Delta \gamma` only, the :math:`{s}_{i}^{e}` being given and ordered, by taking:

:math:`\begin{array}{}{\sigma }_{3}\text{-}{\sigma }_{1}\text{=}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})({s}_{3}^{e}\text{-}{s}_{1}^{e})\\ {\sigma }_{3}\text{=}{s}_{3}^{e}(1\text{-}3\mu \frac{\Delta \lambda }{{\sigma }_{\text{eq}}^{e}})\text{+}\frac{1}{3}({I}_{1}^{e}\text{-}\mathrm{9K}\eta (\gamma )\Delta \lambda )\end{array}`

We then obtain a non-linear function :math:`F(\Delta \lambda )` solved by a Newton algorithm, given by:


.. _Ref100130223:


.. _eq1:

:math:`\begin{array}{}F(\Delta \gamma )\text{=}0\text{=}({s}_{3}^{e}\text{-}{s}_{1}^{e})\left[1\text{-}\frac{3\mu }{{\sigma }_{\text{eq}}^{e}}h(\Delta \gamma )\Delta \gamma \right]\text{-}b(\Delta \gamma )\left[1\text{-}\frac{1}{{\sigma }_{3}^{b\text{-}d}}({s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma )\right]\\ \text{-}{(S(\Delta \gamma ){\sigma }_{c}^{2}(\Delta \gamma )\text{-}m(\Delta \gamma ){\sigma }_{c}(\Delta \gamma )\left[{s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma \right])}^{\frac{1}{2}}\end{array}` **Eq 3.1**

 where we noted:

:math:`\begin{array}{}\\ \begin{array}{cc}h(\Delta \gamma )\text{=}\frac{1}{\eta +1}\text{=}\frac{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}{3(1\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma ))}& \text{et}g(\Delta \gamma )\text{=}\frac{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}{3(1\text{+}\text{sin}\phi ({\gamma }^{\text{-}}+\Delta \gamma ))}(\frac{\mathrm{6K}\text{sin}\phi ({\gamma }^{\text{-}}+\Delta \gamma )}{3\text{+}\text{sin}\phi ({\gamma }^{\text{-}}\text{+}\Delta \gamma )}\text{+}\frac{3\mu {s}_{3}^{e}}{{\sigma }_{\text{eq}}^{e}})\end{array}\end{array}`.

In practice, we will apply Newton's algorithm to the function

:math:`\begin{array}{}\stackrel{ˉ}{F}(\Delta \gamma )\text{=}{\left[({\sigma }_{3}\text{-}{\sigma }_{1})\text{-}b(1\text{-}\frac{{\sigma }_{3}}{{\sigma }_{3}^{b\text{-}d}})\right]}^{2}\text{-}(S{\sigma }_{c}^{2}\text{-}m{\sigma }_{c}{\sigma }_{3})\\ \text{=}{\left[({s}_{3}^{e}\text{-}{s}_{1}^{e})\left[1\text{-}\frac{3\mu }{{\sigma }_{\text{eq}}^{e}}h(\Delta \gamma )\Delta \gamma \right]\text{-}b(\Delta \gamma )\left[1\text{-}\frac{1}{{\sigma }_{3}^{b\text{-}d}}({s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma )\right]\right]}^{2}\\ \text{-}S(\Delta \gamma ){\sigma }_{c}^{2}(\Delta \gamma )\text{+}m(\Delta \gamma ){\sigma }_{c}(\Delta \gamma )\left[{s}_{3}^{e}\text{+}\frac{{I}_{1}^{e}}{3}\text{-}g(\Delta \gamma )\Delta \gamma \right]\end{array}`

in order to overcome the difficulties associated with the sign of the element under the root during iterations. The derivative of :math:`\stackrel{ˉ}{F}` with respect to :math:`\Delta \gamma` is given in Appendix 2.


.. _Ref101847972:




Existence of the solution
------------------------

The principle of analytical resolution consists in determining point :math:`({I}_{1},s)` as the projection of point :math:`({I}_{1}^{e},{s}^{e})` on the charge surface with respect to the plastic flow potential:

:math:`{s}_{\text{II}}\ge 0`

.. image:: images/100001920000319C00003839321796D10A9F159E.svg
    :width: 19
    :height: 24


.. _RefImage_100001920000319C00003839321796D10A9F159E.svg:

:math:`\mathrm{\Delta \lambda }\le \frac{{\mathrm{\sigma }}_{\text{eq}}^{e}}{\mathrm{3\mu }}` :math:`\mathrm{\Delta \lambda }>\frac{{\mathrm{\sigma }}_{\text{eq}}^{e}}{\mathrm{3\mu }}` :math:`\frac{\partial F}{\partial \mathrm{\sigma }}=\frac{\partial F}{\partial \stackrel{ˉ}{\mathrm{\sigma }}}\frac{\partial \stackrel{ˉ}{\mathrm{\sigma }}}{\partial \mathrm{\sigma }}`

:math:`\stackrel{ˉ}{\mathrm{\sigma }}`

No solution

:math:`\stackrel{ˉ}{\mathrm{\sigma }}`

However, the solution must respect condition :math:`{s}_{\text{II}}>0`, that is to say :math:`F(\stackrel{ˉ}{\sigma },\gamma )=({\stackrel{ˉ}{\sigma }}_{3}-{\stackrel{ˉ}{\sigma }}_{1})-\sqrt{-{\stackrel{ˉ}{\sigma }}_{3}\text{.}m{\sigma }_{c}+S{\sigma }_{c}^{2}}-b\text{.}(1-\frac{{\stackrel{ˉ}{\sigma }}_{3}}{{\sigma }_{3}^{b-d}})`: we can therefore see that there is an area in which the problem does not admit a solution, which corresponds to :math:`\frac{\partial F}{\partial {\stackrel{ˉ}{\mathrm{\sigma }}}_{i}}={\mathrm{\delta }}_{\mathrm{i3}}-{\mathrm{\delta }}_{\mathrm{i1}}+\frac{1}{2}{\mathrm{\delta }}_{\mathrm{i3}}{\mathrm{\sigma }}_{c}m{\left[-{\stackrel{ˉ}{\mathrm{\sigma }}}_{3}\text{.}{\mathrm{m\sigma }}_{c}+{\mathrm{S\sigma }}_{c}^{2}\right]}^{-\frac{1}{2}}+b\frac{{\mathrm{\delta }}_{\mathrm{i3}}}{{\mathrm{\sigma }}_{3}^{b-d}}`.