2. Reference solutions

2.1. Calculation method used for reference solutions

The whole of this demonstration can be read in more detail in the [bib1] document.

Contactless phase

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

For the pellet, we find:

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

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

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

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

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

Contact phase

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

When there is contact, we have:

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

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

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

The stress field of the pellet is given by

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

with \({\gamma }_{1}\mathrm{=}\frac{{p}_{1}{r}_{1}^{2}}{{r}_{2}^{2}\mathrm{-}{r}_{1}^{2}}\) and \({\gamma }_{0}\mathrm{=}\frac{{p}_{0}{r}_{1}^{2}}{{r}_{2}^{2}\mathrm{-}{r}_{1}^{2}}\).

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

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

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

2.2. Benchmark results

Move \(\mathit{DX}\) on node \(B\)

2.3. Uncertainty about the solution

\(\text{0\%}\): analytical solution

2.4. Bibliographical references

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