Benchmark solution
=====================

The solution to the problem posed can be determined analytically.

We note:


* :math:`{\varepsilon }_{0}` the deformation applied in the :math:`z` direction,


* :math:`{\varepsilon }_{1}`, :math:`{\varepsilon }_{2}`, and :math:`{\varepsilon }_{3}` the main deformations


First loading step: simple compression
-------------------------------------------------


* The deformation tensor is equal to:

    .. image:: images/Object_6.svg
        :width: 135
        :height: 74


    .. _RefImage_Object_6.svg:

with :math:`{\varepsilon }_{0}<0`


* The equivalent deformation is therefore equal to:


    .. image:: images/Object_7.svg
        :width: 135
        :height: 74


    .. _RefImage_Object_7.svg:


* As soon as

    .. image:: images/Object_8.svg
        :width: 135
        :height: 74


    .. _RefImage_Object_8.svg:

, there is an evolution of the damage which is equal to:


.. image:: images/Object_10.svg
    :width: 135
    :height: 74


.. _RefImage_Object_10.svg:


* Finally, constraint :math:`{\sigma }_{\mathit{zz}}` is equal to:


.. image:: images/Object_11.svg
    :width: 135
    :height: 74


.. _RefImage_Object_11.svg:


Second loading stage: thermal expansion in plane deformations
--------------------------------------------------------------------------


* The tensor of the total deformations is equal to:


.. image:: images/Object_12.svg
    :width: 135
    :height: 74


.. _RefImage_Object_12.svg:

with fixed :math:`{\varepsilon }_{0}<0`


* The elastic deformation equal

    .. image:: images/Object_13.svg
        :width: 135
        :height: 74


    .. _RefImage_Object_13.svg:

, the equivalent deformation is equal to:


.. image:: images/Object_14.svg
    :width: 135
    :height: 74


.. _RefImage_Object_14.svg:


* The damage is worth:


.. image:: images/Object_16.svg
    :width: 135
    :height: 74


.. _RefImage_Object_16.svg:


* Finally, constraint :math:`{\sigma }_{\mathit{zz}}` is equal to:


.. image:: images/Object_17.svg
    :width: 135
    :height: 74


.. _RefImage_Object_17.svg:

**Note:**


*
    * *In a given state, the material parameters used are those defined at the maximum temperature seen by the material and not at the current temperature.*


    * *The assessment of damage* :math:`D` *involves the concept of maximum reached during the history of the load; the solution is therefore not completely analytical but involves discretization. In the case where there is no thermal influence, simply take* :math:`\stackrel{~}{\mathrm{\epsilon }}` *equivalent to the maximum equivalent deformation reached. When the thermal aspect is taken into account, heating can contribute to "reducing" or "delaying" damage at a given deformation; this is the case with the evolution of* :math:`{B}_{c}` *retained. In this case, it is in fact necessary to discretize the load quite finely to have the right damage value* :math:`D` *(which in fact has a maximum in our case) .*