Benchmark solutions
======================


Calculation method
-----------------

In simple traction and with the hypothesis of small deformations, the tensile stress

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    :width: 36
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.. _RefImage_Object_4.svg:

as well as the plastic multiplier

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Just now

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are given in the case under consideration by:


* yes

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    .. _RefImage_Object_7.svg:

: :math:`\mathrm{\sigma }(u)=E\frac{l(u)-{L}_{0}}{{L}_{0}}`, :math:`\dot{p}(u)=0`, :math:`l({t}_{1}^{p})={L}_{0}(1+\frac{{\mathrm{\sigma }}_{Y}(-50°C)}{E})`


* yes

    .. image:: images/Object_9.svg
        :width: 36
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    .. _RefImage_Object_9.svg:

: :math:`\mathrm{\sigma }(u)={E}_{t}\frac{l(u)-{L}_{0}}{{L}_{0}}+\frac{E-{E}_{t}}{E}{\mathrm{\sigma }}_{Y}(-50°C)`, :math:`\dot{p}(u)=(1-\frac{{E}_{t}}{E})\frac{\dot{l}(u)}{{L}_{0}}`


* yes

    .. image:: images/Object_11.svg
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    .. _RefImage_Object_11.svg:

: :math:`\mathrm{\sigma }(u)=\mathrm{\sigma }(u=10)-E\frac{l(u=10)-l(u)}{{L}_{0}}`, :math:`\dot{p}(u)=0`


* yes

    .. image:: images/Object_13.svg
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    .. _RefImage_Object_13.svg:

: :math:`\mathrm{\sigma }(u)=\mathrm{\sigma }(u=20)`, :math:`\dot{p}(u)=0`


* yes

    .. image:: images/Object_15.svg
        :width: 36
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    .. _RefImage_Object_15.svg:

: :math:`\mathrm{\sigma }(u)=\mathrm{\sigma }(u=20)+{E}_{t}\frac{l(u)-l(u=20)}{{L}_{0}}`, :math:`\dot{p}(u)=(1-\frac{{E}_{t}}{E})\frac{\dot{l}(u)}{{L}_{0}}`


Weibull
-------

The cumulative probability of breakage

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Just now

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is given by (cf. POST_ELEM and POST_BEREMIN):

:math:`{P}_{f}(t)=1-\mathrm{exp}(-\sum _{\mathit{dV}}{[\begin{array}{c}\mathit{max}\\ {t}^{p}\le u\le t\end{array}(\frac{{\mathrm{\sigma }}_{I}(u)}{{\mathrm{\sigma }}_{u}(\mathrm{\theta }(u))})]}^{m}\frac{\mathit{dV}}{{V}_{0}})`.

The summation relates to the volumes of matter.

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laminated (from now on

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),

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and

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designating the maximum principal stress and the temperature in each of these volumes at the various times

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. Here, the volume

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The reference value is equal to :math:`50\mu {m}^{3}`. The Weibull module

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is equal to 24 while the cleavage stress

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depends on the temperature according to:


.. csv-table::

    "**Temperature** :math:`\mathrm{[}°C\mathrm{]}` ", "**—50**", "**—100**", "**—150**", "**—150**"
    ":math:`{\sigma }_{u}\mathrm{[}\mathit{MPa}\mathrm{]}` ", "2800", "2700", "2600"


The cumulative probability of breakage varies according to (

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.. _RefImage_Object_28.svg:

) according to:

:math:`{P}_{f}(t)=1-\mathrm{exp}(-{[\begin{array}{c}\mathit{max}\\ {t}^{p}\le u\le t\end{array}(\frac{\mathrm{\sigma }(u)}{{\mathrm{\sigma }}_{u}(\mathrm{\theta }(u))})]}^{m}\frac{V}{{V}_{0}})`.


Rice and Tracey
--------------

In simple traction, the natural logarithm of the cavity growth rate at the moment

.. image:: images/Object_30.svg
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is given by (cf. POST_ELEM):

:math:`\mathrm{log}(\frac{R(t)}{{R}_{0}})=0.283\times \mathrm{exp}(0.5)\times \underset{0}{\overset{t}{\int }}\dot{p}(u)\mathit{du}`


Reference quantities and results
-----------------------------------


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and

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.. _RefImage_Object_33.svg:

for the following couples (temperature, travel = :math:`(l\mathrm{-}{l}_{0})`): :math:`(–\mathrm{50,0}°C,\mathrm{20,35}\mathit{mm})`;

:math:`(–\mathrm{50,0}°C,\mathrm{20,30}\mathit{mm})`; :math:`(–\mathrm{150,0}°C,\mathrm{20,30}\mathit{mm})` and :math:`(–\mathrm{150,0}°C,\mathrm{32,53}\mathit{mm})`.


Uncertainties about the solution
----------------------------

Analytical solution.