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


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

The whole of this demonstration can be read in more detail in the [:ref:`bib1 <bib1>`] document.

In the case of a linear isotropic viscoelastic material, the behavior over time can be described using two functions :math:`I(t)` and :math:`K(t)` in such a way that the deformations and the stresses can be written as:

:math:`\varepsilon (t)\mathrm{=}(I+K)\mathrm{\ast }\frac{d\sigma (t)}{d\tau }\mathrm{-}K\mathrm{\ast }\frac{d(\mathit{Tr}(\sigma (t)))}{d\tau }{I}_{3}`

where :math:`{I}_{3}` refers to the identity matrix of rank 3

and :math:`\mathrm{\ast }` the convolution product: :math:`(f\mathrm{\ast }g)(t)\mathrm{=}{\mathrm{\int }}_{0}^{t}f(t\mathrm{-}\tau )g(\tau )d\tau`

We find :math:`I(t)\mathrm{=}\frac{1}{E}+\mathit{kt}`, :math:`K(t)\mathrm{=}\frac{\nu }{E}+\frac{1}{2}\mathit{kt}`

We impose the pressure :math:`{P}_{0}` at the moment :math:`t\mathrm{=}0`, the internal pressure is equal to :math:`p(t)\mathrm{=}H(t){P}_{0}` or :math:`H(t)\mathrm{=}\mathrm{\{}\begin{array}{c}0\mathit{si}t\mathrm{-}\tau <0\\ 1\mathit{si}t\mathrm{-}\tau \mathrm{\ge }0\end{array}` with in this case :math:`\tau \mathrm{=}0`

We use the Laplace Carson transform :math:`{f}^{\text{+}}(n)\mathrm{=}L(f(t))\mathrm{=}n{\mathrm{\int }}_{0}^{\mathrm{\infty }}f(t){e}^{\mathrm{-}\mathit{nt}}\mathit{dt}`

From where :math:`{p}^{+}\mathrm{=}{P}_{0}`

The solution of the equivalent elastic problem is:

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

We determine :math:`{\sigma }_{Z}^{\text{+}}` by the condition on :math:`{\varepsilon }_{Z}^{\text{+}}` given by the boundary conditions:

:math:`{\varepsilon }_{Z}^{\text{+}}\mathrm{=}0\mathrm{=}({I}^{\text{+}}+{K}^{\text{+}}){\sigma }_{Z}^{\text{+}}\mathrm{-}{K}^{\text{+}}(2\gamma +{\sigma }_{Z}^{\text{+}})\mathrm{=}{I}^{\text{+}}{\sigma }_{Z}^{\text{+}}\mathrm{-}2{K}^{\text{+}}\gamma`

Hence :math:`{\sigma }_{Z}^{\text{+}}\mathrm{=}\gamma (1+\frac{(2\nu \mathrm{-}1)p}{p+\mathit{Ek}})`.

We find by the inverse Laplace transform :math:`{\sigma }_{z}(t)\mathrm{=}\gamma (1\mathrm{-}(1\mathrm{-}2\nu ){e}^{\mathrm{-}\mathit{Eht}})`, similarly by applying the inverse Laplace transform on :math:`{\sigma }_{r}` and :math:`{\sigma }_{\theta }`, we find

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

From this we deduce:

:math:`\dot{{\varepsilon }_{V}}\mathrm{=}(\begin{array}{ccc}\frac{3}{2}k\gamma (\frac{1\mathrm{-}2\nu }{3}{e}^{\mathrm{-}\mathit{Ekt}}\mathrm{-}\frac{{r}_{1}^{2}}{{r}^{2}})& 0& 0\\ 0& \frac{3}{2}k\gamma (\frac{1\mathrm{-}2\nu }{3}{e}^{\mathrm{-}\mathit{Ekt}}\mathrm{-}\frac{{r}_{1}^{2}}{{r}^{2}})& 0\\ 0& 0& \mathrm{-}k\gamma ((1\mathrm{-}2\nu ){e}^{\mathrm{-}\mathit{Eht}})\end{array})`

and integrating with :math:`{\varepsilon }_{V}(0)\mathrm{=}0`;

:math:`\dot{{\varepsilon }_{V}}\mathrm{=}(\begin{array}{ccc}\frac{3}{2}\gamma (\frac{1\mathrm{-}2\nu }{3}{e}^{\mathrm{-}\mathit{Ekt}}\mathrm{-}k\frac{{r}_{1}^{2}}{{r}^{2}}t)& 0& 0\\ 0& \frac{3}{2}\gamma (\frac{1\mathrm{-}2\nu }{3}{e}^{\mathrm{-}\mathit{Ekt}}\mathrm{-}k\frac{{r}_{1}^{2}}{{r}^{2}}t)& 0\\ 0& 0& \mathrm{-}\gamma \frac{(1\mathrm{-}2\nu )}{E}(1\mathrm{-}{e}^{\mathrm{-}\mathit{Eht}})\end{array})`.

The radial displacement is deduced

:math:`w(r,t)\mathrm{=}r\gamma \left[\frac{1}{E}\left[(1+\nu )\frac{{r}_{1}^{2}}{{r}^{2}}+\frac{1\mathrm{-}2\nu }{2}(3\mathrm{-}(1\mathrm{-}2\nu ){e}^{\mathrm{-}\mathit{Ekt}})\right]+\frac{3}{2}k\frac{{r}_{1}^{2}}{{r}^{2}}t\right]`


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

Move :math:`\mathit{DX}` on node :math:`B` and constraints :math:`\mathit{SIXX}`, :math:`\mathrm{SIYY}`, and :math:`\mathrm{SIZZ}` in :math:`B`


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

:math:`\text{0\%}`: analytical solution


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

Ph. Of BONNIERES: Two analytical solutions to axisymmetric problems in linear viscoelasticity and with unilateral contact, Note HI-71/8301