Reference solutions
======================


Calculation method used for reference solutions
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The whole of this demonstration can be read in more detail in the [:ref:`bib1 <bib1>`] document.

**Contactless phase**

We want to find the value of :math:`{p}_{1}(t)` to be applied to the inner wall of the pellet for which the contact takes place.

For the pellet, we find:

:math:`\sigma \mathrm{=}(\begin{array}{ccc}\gamma (1\mathrm{-}\frac{{r}_{2}^{2}}{{r}^{2}})& 0& 0\\ 0& \gamma (1+\frac{{r}_{2}^{2}}{{r}^{2}})& 0\\ 0& 0& 2\nu \gamma \end{array})` where :math:`\gamma \mathrm{=}\frac{{p}_{1}(t){r}_{1}^{2}}{{r}_{2}^{2}\mathrm{-}{r}_{1}^{2}}`

:math:`{\varepsilon }_{\theta }\mathrm{=}\frac{1+\nu }{E}\gamma \left[1\mathrm{-}2\nu +\frac{{r}_{2}^{2}}{{r}^{2}}\right]\mathrm{=}\frac{w}{r}`.

The contact condition is written as: :math:`w({r}_{3})\mathrm{-}w({r}_{2})\mathrm{=}0`, we have :math:`{r}_{3}\mathrm{-}{r}_{2}\mathrm{=}{r}_{2}\frac{2(1+\nu )\gamma }{E}(1\mathrm{-}\nu )`

Where :math:`\gamma \mathrm{=}(\frac{{r}_{3}}{{r}_{2}}\mathrm{-}1)\frac{E}{2{(1\mathrm{-}\nu )}^{2}}`

:math:`{p}_{1}\text{lim}\mathrm{=}(\frac{{r}_{3}}{{r}_{2}}\mathrm{-}1)\frac{E({r}_{2}^{2}\mathrm{-}{r}_{1}^{2})}{2{r}_{1}^{2}{(1\mathrm{-}\nu )}^{2}}`.

**Contact phase**

We want the sheath to behave the same way as in the ssna104a test from moment :math:`t\mathrm{=}0`.

When there is contact, we have:

:math:`{w}_{P}({r}_{2})\mathrm{=}{W}_{G}({r}_{3})+{r}_{3}\mathrm{-}{r}_{2}`,

so by retrieving the value of the movements in the ssna104 test, we must obtain:

:math:`{w}_{P}({r}_{2})\mathrm{=}{r}_{3}\mathrm{-}{r}_{2}+\frac{{p}_{0}{r}_{3}^{3}}{{r}_{4}^{2}\mathrm{-}{r}_{3}^{2}}\left\{\frac{1}{E}((1+\nu )\frac{{r}_{4}^{2}}{{r}_{3}^{2}}+\frac{1\mathrm{-}2\nu }{2}(3\mathrm{-}(1\mathrm{-}2\nu ){e}^{\mathrm{-}\mathit{Ekt}}))+\frac{3}{2}k\frac{{r}_{4}^{2}}{{r}_{3}^{2}}t\right\}`.

The stress field of the pellet is given by

:math:`\sigma \mathrm{=}(\begin{array}{ccc}{\gamma }_{1}(1\mathrm{-}\frac{{r}_{2}^{2}}{{r}^{2}})\mathrm{-}{\gamma }_{0}(1\mathrm{-}\frac{{r}_{2}^{2}}{{r}^{2}})& 0& 0\\ 0& {\gamma }_{1}(1+\frac{{r}_{2}^{2}}{{r}^{2}})\mathrm{-}{\gamma }_{0}(1+\frac{{r}_{2}^{2}}{{r}^{2}})& 0\\ 0& 0& {\sigma }_{Z}\end{array})`

with :math:`{\gamma }_{1}\mathrm{=}\frac{{p}_{1}{r}_{1}^{2}}{{r}_{2}^{2}\mathrm{-}{r}_{1}^{2}}` and :math:`{\gamma }_{0}\mathrm{=}\frac{{p}_{0}{r}_{1}^{2}}{{r}_{2}^{2}\mathrm{-}{r}_{1}^{2}}`.

Like :math:`{\varepsilon }_{Z}\mathrm{=}\frac{1+\nu }{E}{\sigma }_{Z}\mathrm{-}\frac{\nu }{E}(2({\gamma }_{1}\mathrm{-}{\gamma }_{0})+{\sigma }_{Z})\mathrm{=}0`, we find: :math:`{\sigma }_{Z}\mathrm{=}2\nu ({\gamma }_{1}\mathrm{-}{\gamma }_{0})`.

So we have :math:`{\varepsilon }_{\theta }\mathrm{=}\frac{1+\nu }{E}{\sigma }_{\theta }\mathrm{-}\frac{\nu }{E}({\sigma }_{r}+{\sigma }_{\theta }+{\sigma }_{Z})\mathrm{=}\frac{1+\nu }{E}\left[(1\mathrm{-}2\nu )({\gamma }_{1}\mathrm{-}{\gamma }_{0})+{\gamma }_{1}\frac{{r}_{2}^{2}}{{r}^{2}}\mathrm{-}{\gamma }_{0}\frac{{r}_{1}^{2}}{{r}^{2}}\right]\mathrm{=}\frac{w}{r}`

:math:`{w}_{P}({r}_{2})\mathrm{=}\frac{1+\nu }{E}{r}_{2}\left[2(1\mathrm{-}\nu ){\gamma }_{1}\mathrm{-}{\gamma }_{0}(1\mathrm{-}2\nu +\frac{{r}_{1}^{2}}{{r}_{2}^{2}})\right]`, we find :math:`{p}_{1}(t)` given by the formula a little above.


Benchmark results
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Move :math:`\mathit{DX}` on node :math:`B`


Uncertainty about the solution
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:math:`\text{0\%}`: analytical solution


Bibliographical references
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* Ph. Of BONNIERES, two analytical solutions to axisymmetric problems in linear viscoelasticity and with unilateral contact, Note HI-71/8301