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


Calculation method used for reference solutions
-----------------------------

The whole of this demonstration can be read in more detail in document :math:`[1]`

The stress tensor is written as:

 :math:`\sigma =(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& {\sigma }_{z}\end{array})`

As a result of loading, we have:

 :math:`\{\begin{array}{c}{\varepsilon }_{z}-{\varepsilon }_{\nu z}=\frac{{\sigma }_{z}}{E}\\ {\varepsilon }_{\theta }-{\varepsilon }_{\nu \theta }=-\frac{\nu }{E}{\sigma }_{z}\\ {\varepsilon }_{r}-{\varepsilon }_{\nu r}=-\frac{\nu }{E}{\sigma }_{z}\end{array}` and :math:`\dot{{\varepsilon }_{\nu }}=\frac{3}{2}g(\sigma \text{*})\frac{{\sigma }^{D}}{\sigma \text{*}}`



therefore

     :math:`\{\begin{array}{c}\dot{{\varepsilon }_{\nu z}}=g({\sigma }_{z})\\ \dot{{\varepsilon }_{\nu \theta }}=-\frac{1}{2}g({\sigma }_{z})\\ \dot{{\varepsilon }_{\nu r}}=-\frac{1}{2}g({\sigma }_{z})\end{array}`

If :math:`t\le {t}_{0}`, we have :math:`{\varepsilon }_{z}=\frac{{\varepsilon }_{0}}{{t}_{0}}t`,

 Be :math:`a=\sqrt{\frac{{\varepsilon }_{0}}{{t}_{0}}}`

 We get :math:`{\varepsilon }_{z}={a}^{2}t`

 Substituting, we find:

     :math:`\dot{{\varepsilon }_{\nu z}}=g(({a}^{2}t-{\varepsilon }_{\nu z})E)`

We set :math:`E=1` and :math:`z={a}^{2}t-{\varepsilon }_{\nu z}`, we get: :math:`\dot{z}={a}^{2}–{z}^{2}`

Integrating with :math:`z(0)=0` we get:



 :math:`z=a\mathrm{tanh}(\mathrm{at})`

For :math:`t\le {t}_{0}`

 :math:`\{\begin{array}{c}{\sigma }_{r}={\sigma }_{\theta }=0\\ {\sigma }_{z}=a\mathrm{tanh}(\mathrm{at})\\ {\varepsilon }_{r}={\varepsilon }_{\theta }=a[(\frac{1}{2}-\nu )\mathrm{tanh}(\mathrm{at})-\frac{1}{2}\mathrm{at}]\\ {\varepsilon }_{z}={a}^{2}t\\ w=\mathrm{ar}[(\frac{1}{2}-\nu )\mathrm{tanh}(\mathrm{at})-\frac{1}{2}\mathrm{at}]\end{array}`

If :math:`t\ge {t}_{0}`

 :math:`{\varepsilon }_{z}={a}^{2}{t}_{0}`

 :math:`\dot{{\varepsilon }_{\nu z}}=g({a}^{2}{t}_{0}-{\varepsilon }_{\nu z})={({a}^{2}{t}_{0}-{\varepsilon }_{\nu z})}^{2}`

What results when integrating:



 :math:`{\varepsilon }_{\nu z}={a}^{2}{t}_{0}–\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}`

So we finally have

 :math:`\{\begin{array}{c}{\sigma }_{r}={\sigma }_{\theta }=0\\ {\sigma }_{z}=\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}\\ {\varepsilon }_{r}={\varepsilon }_{\theta }=(\frac{1}{2}-\nu )\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}–\frac{1}{2}{a}^{2}{t}_{0}\\ {\varepsilon }_{z}={a}^{2}{t}_{0}\\ w=r\left[(\frac{1}{2}-\nu )\frac{1}{\frac{1}{a\mathrm{tanh}({\mathrm{at}}_{0})}+t–{t}_{0}}–\frac{1}{2}{a}^{2}{t}_{0}\right]\end{array}`




Reference quantities
-----------------------


* Move :math:`\mathrm{DX}` to node :math:`B`


* Constraints :math:`\mathrm{SIXX}`, :math:`\mathrm{SIYY}`, and :math:`\mathrm{SIZZ}` at node :math:`B`


Benchmark results
-----------------------


.. csv-table::

    "Size", "Point", "Moments", "Reference"
    ":math:`\mathrm{DX}` "," :math:`B` ", "4"," :math:`-0.2109`"
    ":math:`\mathrm{SIXX}` "," :math:`B` ", "4"," :math:`0.`"
    ":math:`\mathrm{SIYY}` "," :math:`B` ", "4"," :math:`0.2616`"
    ":math:`\mathrm{SIZZ}` "," :math:`B` ", "4"," :math:`0.`"



Uncertainty about the solution
---------------------------

Analytical solution


Bibliographical references
---------------------------

[:ref:`1 <1>`] Ph. De BONNIERES, M. ZIDI: Introduction of viscoplasticity into the Cyrano3 thermomechanics module: Principle, description and validation, Note HI-71/8334.




*