2. Benchmark solution#

Here we develop an analytical solution for the problem presented above. This solution will be developed on the hypothesis of small deformations by considering that the materials of the crowns are isotropic, governed by a linear elastic law without temperature variation.

Because of the symmetries of the problem, the on-the-go solution of the problem has the following generic form:

\(u={u}_{r}(r,z)\mathrm{.}\underline{{e}_{r}}+{u}_{z}(r,z)\mathrm{.}\underline{{e}_{z}}\)

2.1. Case 1: plane deformations#

Using the symmetries of the problem and the assumption of invariance according to z of the plane constraints, the solution of the problem takes the following form:

\({u}_{r}={u}_{r}(r)\)

\({u}_{\theta }=0\) eq 2.1

\({u}_{z}=0\)

Using the Lamé-Navier equation:

\(\begin{array}{c}(\lambda +\mu )\underline{\mathrm{grad}}(d(\underline{u}))+\mu \Delta \underline{u}+\underline{\mathrm{fd}}=\underline{0}\end{array}\) eq 2.2

where \(\underline{\mathrm{fd}}=\underline{0}\) is the zero volume efforts here, and the Laplacian formula:

\(\begin{array}{c}\Delta \underline{u}\mathrm{=}\underline{\mathit{grad}}(\mathit{div}(\underline{u}))+\underline{\mathit{rot}}\underline{\mathit{rot}}(\underline{u})\end{array}\) eq 2.3

We can write eq 2.2 in the form:

\(\begin{array}{c}(\lambda +2\mu )\underline{\mathit{grad}}(\mathit{div}(\underline{u}))+\mu \underline{\mathit{rot}}\underline{\mathit{rot}}(\underline{u})+\underline{\mathit{fd}}\mathrm{=}\underline{0}\end{array}\) eq 2.4

or again using \(\underline{\mathrm{rot}}(\underline{u})=\overrightarrow{0}\text{et}\underline{\mathrm{fd}}=\overrightarrow{0}\text{et}u={u}_{r}(r)\mathrm{.}\underline{{e}_{r}}\):

\(\begin{array}{c}d(\underline{u})=\frac{d}{\mathrm{dr}}{u}_{r}(r)+\frac{1}{r}{u}_{r}(r)\\ \underline{\mathrm{grad}}(d\underline{u})=\frac{d}{{d}_{r}}[\frac{1}{r}\frac{d}{{d}_{r}}(r{u}_{r}(r))]\mathrm{.}{\underline{e}}_{r}\\ \text{soit encore}(\lambda +2\mu )\frac{d}{{d}_{r}}[\frac{1}{r}\frac{d}{{d}_{r}}(r{u}_{r}(r))]=0\end{array}\) eq 2.5

by integrating the equation, the following form of the displacement field is obtained for the rings (exterior and interior):

\({u}_{r}={C}_{i}r+\frac{{D}_{i}}{r}\text{}{u}_{\theta }=0\text{}{u}_{z}=0\) eq 2.6

To determine \({C}_{i}\) and \({D}_{i}\), we still have to impose the limit conditions in terms of pressure and movement. To do this, it is first necessary to calculate the deformations and then the stresses associated with the field of displacement.

Deformations are the symmetric part of the displacement gradient. We get:

\(\begin{array}{c}{\epsilon }_{\mathrm{rr}}={C}_{i}-\frac{{D}_{i}}{\mathrm{r²}}\\ {\epsilon }_{\theta \theta }={C}_{i}+\frac{{D}_{i}}{\mathrm{r²}}\\ {\epsilon }_{\mathrm{zz}}={\epsilon }_{r\theta }={\epsilon }_{\mathrm{rz}}={\epsilon }_{\theta z}=0\end{array}\) eq 2.7

Applying Hooke’s law:

\(\underline{\underline{\sigma }}=\lambda \mathrm{tr}(\underline{\underline{\epsilon }})\underline{\underline{1}}+2\mu \underline{\underline{\epsilon }}\) eq 2.8

the following general form is obtained for the constraints:

\(\begin{array}{c}{\sigma }_{\mathrm{rr}}=\frac{E}{1+\nu }(\frac{{C}_{i}}{1-2\nu }-\frac{{D}_{i}}{\mathrm{r²}})\\ {\sigma }_{\theta \theta }=\frac{E}{1+\nu }(\frac{{C}_{i}}{1-2\nu }+\frac{{D}_{i}}{\mathrm{r²}})\\ {\sigma }_{\mathrm{zz}}=\frac{2\nu {\mathrm{EC}}_{i}}{(1+\nu )(1-2\nu )}\\ {\sigma }_{r\theta }={\sigma }_{\mathrm{rz}}={\sigma }_{\theta z}=\underline{0}\end{array}\) eq 2.9

By posing:

\(\begin{array}{c}{A}_{i}\mathrm{=}\frac{{E}_{i}}{(1+{\nu }_{i})(1\mathrm{-}2{\nu }_{i})}{C}_{i}\text{}{B}_{i}\mathrm{=}\frac{{E}_{i}}{1+{\nu }_{i}}{D}_{i}\end{array}\) eq 2.10

non-zero constraints become:

\(\begin{array}{c}{\sigma }_{\mathrm{rr}}={A}_{i}-\frac{{B}_{i}}{\mathrm{r²}}\\ {\sigma }_{\theta \theta }={A}_{i}+\frac{{B}_{i}}{\mathrm{r²}}\\ {\sigma }_{\mathrm{zz}}=2\nu {A}_{i}\end{array}\) eq 2.11

All we have to do is calculate the values of \({A}_{i}\) and \({B}_{i}\) for each of the crowns. Note \({\lambda }_{n}\) the contact pressure between the two rings such that:

\(\begin{array}{}{\underline{\underline{\sigma }}}_{\mathrm{1rr}}({R}_{2})\mathrm{.}(-{\underline{e}}_{r})={\lambda }_{n}{\underline{e}}_{r}\\ {\underline{\underline{\sigma }}}_{\mathrm{2rr}}({R}_{2})\mathrm{.}{\underline{e}}_{r}=-{\lambda }_{n}{\underline{e}}_{r}\end{array}\) eq 2.12

with the boundary conditions:

\(\begin{array}{}{\underline{\underline{\sigma }}}_{\mathrm{1rr}}({R}_{1})\mathrm{.}{\underline{e}}_{r}=-\mathrm{p.}{\underline{e}}_{r}\\ {\underline{\underline{\sigma }}}_{\mathrm{2rr}}({R}_{3})\mathrm{.}(-{\underline{e}}_{r})=\underline{0}\end{array}\) eq 2.13

The condition of continuity on the movement at the interface between the two rings in contact also gives:

\(\begin{array}{}{u}_{r;1}(\mathrm{R2})={u}_{r;2}(\mathrm{R2})\end{array}\) eq 2.14

So we have 5 equations for the 5 unknowns \(\begin{array}{}{A}_{1},{B}_{1},{A}_{2},{B}_{\mathrm{2,}}{\lambda }_{n}\end{array}\).

The system of the first 4 equations allows us to obtain:

\(\begin{array}{c}{A}_{1}\mathrm{=}\frac{\mathrm{-}p{R}_{1}^{2}+{\lambda }_{n}{R}_{2}^{2}}{{R}_{1}^{2}\mathrm{-}{R}_{2}^{2}};{B}_{1}\mathrm{=}(\mathrm{-}p+{\lambda }_{n})\frac{{R}_{1}^{2}{R}_{2}^{2}}{{R}_{1}^{2}\mathrm{-}{R}_{2}^{2}}\\ {A}_{2}\mathrm{=}\mathrm{-}{\lambda }_{n}\frac{{R}_{2}^{2}}{{R}_{2}^{2}\mathrm{-}{R}_{3}^{2}};{B}_{2}\mathrm{=}\mathrm{-}{\lambda }_{n}\frac{{R}_{2}^{2}{R}_{3}^{2}}{{R}_{2}^{2}\mathrm{-}{R}_{3}^{2}}\end{array}\) eq 2.15

and the equation of continuity on the displacement finally makes it possible to have the contact pressure:

\({\lambda }_{n}\mathrm{=}\frac{2p{R}_{1}^{2}(1\mathrm{-}{\nu }_{1})}{{R}_{1}^{2}+{R}_{2}^{2}(1\mathrm{-}2{\nu }_{1})+\frac{{E}_{1}}{{E}_{2}}\frac{1+{\nu }_{2}}{1+{\nu }_{1}}\frac{{R}_{1}^{2}\mathrm{-}{R}_{2}^{2}}{{R}_{2}^{2}\mathrm{-}{R}_{3}^{2}}({R}_{2}^{2}(1\mathrm{-}2{\nu }_{2})+{R}_{3}^{2})}\) eq 2.16

2.2. Case 2: plane constraints#

First, we assume that there are no constraints in the direction perpendicular to the plane of the crowns (\({\underline{\underline{\sigma }}}_{i}\mathrm{.}{\underline{e}}_{z}\mathrm{=}\underline{0}\)). The symmetries of the problem lead us to a field of constraints that can be written in the form:

\(\underline{\underline{{\sigma }_{i}}}(r,z)={\sigma }_{\mathrm{rr};i}{\underline{e}}_{r}\otimes {\underline{e}}_{r}+{\sigma }_{\theta \theta ;i}{\underline{e}}_{\theta }\otimes {\underline{e}}_{\theta }+{\sigma }_{r\theta ;i}{\underline{e}}_{r}\otimes {\underline{e}}_{\theta }\)

where the index \(i\) is 1 for the outer ring and 2 for the inner ring. In the absence of volume forces and considering the quasistatic problem (we neglect the effects of acceleration), we have:

\(\mathit{div}\underline{\underline{\sigma }}\mathrm{=}\underline{0}\)

Using the on-the-go solution of the generic problem:

\(u={u}_{r}(r,z)\mathrm{.}\underline{{e}_{r}}+{u}_{z}(r,z)\mathrm{.}\underline{{e}_{z}}\)

The deformation field is then written as:

\(\begin{array}{}{\epsilon }_{\mathrm{rr}}=\frac{\partial {u}_{r}(r,z)}{\partial r}\\ {\epsilon }_{\theta \theta }=\frac{{u}_{r}}{r}\\ {\epsilon }_{\mathrm{zz}}=\frac{\partial {u}_{z}(r,z)}{\partial z}\\ {\epsilon }_{\theta z}={\epsilon }_{r\theta }=0\\ {\epsilon }_{\mathrm{rz}}=\frac{1}{2}(\frac{\partial {u}_{r}}{\partial z}+\frac{\partial {u}_{z}}{\partial r})\end{array}\) eq 2.17

Like:

\(\begin{array}{}d\underline{\underline{\sigma }}=\underline{0}\text{devient}\frac{\partial {\sigma }_{\mathrm{rr}}}{\partial r}+\frac{{\sigma }_{\mathrm{rr}}-{\sigma }_{\theta \theta }}{r}=0\\ \text{avec}\\ {\sigma }_{\mathrm{rr}}=\frac{E}{1+\nu }({\epsilon }_{\mathrm{rr}}+\frac{\nu }{1-2\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }+{\epsilon }_{\mathrm{zz}}))\\ {\sigma }_{\theta \theta }=\frac{E}{1+\nu }({\epsilon }_{\theta \theta }+\frac{\nu }{1-2\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }+{\epsilon }_{\mathrm{zz}}))\\ {\sigma }_{\mathrm{zz}}=\frac{E}{1+\nu }({\epsilon }_{\mathrm{zz}}+\frac{\nu }{1-2\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }+{\epsilon }_{\mathrm{zz}}))\\ {\sigma }_{\mathrm{zz}}={\sigma }_{z\theta }={\sigma }_{\mathrm{zr}}=0\\ \text{impliquant}\\ {\epsilon }_{\mathrm{zr}}=0=\frac{\partial {u}_{r}}{\partial z}+\frac{\partial {u}_{z}}{\partial r}\end{array}\) eq 2.18

The fact that \({\sigma }_{\mathrm{zz}}=0\) gives us:

\(\begin{array}{}\frac{E}{1+\nu }({\epsilon }_{\mathrm{zz}}+\frac{\nu }{1-2\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }+{\epsilon }_{\mathrm{zz}}))=0\\ {\epsilon }_{\mathrm{zz}}(1-\nu )=-\nu ({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta })\\ {\epsilon }_{\mathrm{zz}}=\frac{-\nu }{(1-\nu )}({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta })\end{array}\) eq 2.19

which involves:

\(\begin{array}{}{\sigma }_{\mathrm{rr}}-{\sigma }_{\theta \theta }=\frac{E}{1+\nu }({\epsilon }_{\mathrm{rr}}-{\epsilon }_{\theta \theta })\\ \mathrm{tr}\underline{\underline{\epsilon }}=\frac{\nu -1}{\nu }{\epsilon }_{\mathrm{zz}}+{\epsilon }_{\mathrm{zz}}\\ \mathrm{tr}\underline{\underline{\epsilon }}=\frac{2\nu -1}{\nu }{\epsilon }_{\mathrm{zz}}\end{array}\) eq 2.20

and allows you to write:

\(\begin{array}{}\mathrm{tr}\underline{\underline{\epsilon }}=\frac{1-2\nu }{1-\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta })\end{array}\) eq 2.21

We then replace \(\mathrm{tr}\underline{\underline{\epsilon }}\) with its value in \({\sigma }_{\mathrm{rr}}\) and \({\sigma }_{\theta \theta }\) to obtain:

\(\begin{array}{}{\sigma }_{\mathrm{rr}}=\frac{E}{1+\nu }({\epsilon }_{\mathrm{rr}}+\frac{\nu }{1-2\nu }\frac{1-2\nu }{1-\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }))\\ {\sigma }_{\theta \theta }=\frac{E}{1+\nu }({\epsilon }_{\theta \theta }+\frac{\nu }{1-2\nu }\frac{1-2\nu }{1-\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }))\\ \text{}\\ {\sigma }_{\mathrm{rr}}=\frac{E}{1+\nu }({\epsilon }_{\mathrm{rr}}+\frac{\nu }{1-\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }))\\ {\sigma }_{\theta \theta }=\frac{E}{1+\nu }({\epsilon }_{\theta \theta }+\frac{\nu }{1-\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }))\end{array}\) eq 2.22

The equilibrium equation becomes as follows:

\(\begin{array}{c}\frac{\mathrm{\partial }{\sigma }_{\mathit{rr}}}{\mathrm{\partial }r}+\frac{{\sigma }_{\mathit{rr}}\mathrm{-}{\sigma }_{\theta \theta }}{r}\mathrm{=}0\text{}\to \text{}\frac{\mathrm{\partial }{\epsilon }_{\mathit{rr}}}{\mathrm{\partial }r}+\frac{\nu }{1\mathrm{-}\nu }(\frac{\mathrm{\partial }}{\mathrm{\partial }r}({\epsilon }_{\mathit{rr}}+{\epsilon }_{\theta \theta }))+\frac{{\epsilon }_{\mathit{rr}}\mathrm{-}{\epsilon }_{\theta \theta }}{r}\mathrm{=}0\end{array}\) eq 2.23

As we have:

\(\begin{array}{}{\epsilon }_{\mathrm{rr}}=\frac{\partial {u}_{r}(r,z)}{\partial r}\\ {\epsilon }_{\theta \theta }=\frac{{u}_{r}}{r}\end{array}\) eq 2.24

the substitution of the stresses by the deformations in the equilibrium equation makes it possible to finally write:

\(\begin{array}{}\frac{\partial \mathrm{²}{u}_{r}(r,z)}{\partial \mathrm{r²}}+\frac{\nu }{1-\nu }(\frac{\partial \mathrm{²}{u}_{r}(r,z)}{\partial \mathrm{²}r}+\frac{\partial {u}_{r}(r,z)}{\partial r})+\frac{1}{r}(\frac{\partial {u}_{r}(r,z)}{\partial r}-\frac{{u}_{r}(r,z)}{r})=0\\ \frac{\partial \mathrm{²}{u}_{r}(r,z)}{\partial \mathrm{r²}}+\frac{\nu }{1-\nu }(\frac{\partial \mathrm{²}{u}_{r}(r,z)}{\partial \mathrm{r²}}+\frac{1}{r}\frac{\partial {u}_{r}(r,z)}{\partial r}-\frac{1}{\mathrm{r²}}{u}_{r}(r,z))+\frac{1}{r}(\frac{\partial {u}_{r}(r,z)}{\partial r}-\frac{{u}_{r}(r,z)}{r})\\ \frac{\partial \mathrm{²}{u}_{r}(r,z)}{\partial \mathrm{r²}}+\frac{1}{r}\frac{\partial {u}_{r}(r,z)}{\partial r}-\frac{1}{\mathrm{r²}}{u}_{r}(r,z)=0\\ \frac{\partial }{\partial r}[\frac{1}{r}\frac{\partial }{\partial r}({\mathrm{r.u}}_{r}(r,z))]=0\end{array}\) eq 2.25

By successive integrations of eq 2.25 we obtain the following form of field \({u}_{r}(r,z)\):

\(\begin{array}{}\frac{1}{r}\frac{\partial }{\partial r}({\mathrm{r.u}}_{r})=f(z)\\ \frac{\partial }{\partial r}({\mathrm{r.u}}_{r})=\mathrm{r.f}(z)\\ {\mathrm{r.u}}_{r}=\frac{\mathrm{r²}}{2}f(z)+g(z)\\ {u}_{r}=\frac{r}{2}f(z)+\frac{g(z)}{r}\end{array}\) eq 2.26

Either:

\(\begin{array}{}{u}_{r}(r,z)=C(z)r+\frac{D(z)}{r}\end{array}\) eq 2.27

As we have:

\(\begin{array}{}\frac{\partial {u}_{z}}{\partial r}+\frac{\partial {u}_{r}}{\partial z}=0\\ {\epsilon }_{\mathrm{rr}}=\frac{{u}_{r}(r,z)}{\partial r}=C(z)-\frac{D(z)}{\mathrm{r²}}\\ {\epsilon }_{\theta \theta }=\frac{{u}_{r}}{r}=C(z)+\frac{D(z)}{\mathrm{r²}}\\ {\epsilon }_{\mathrm{zz}}=\frac{\partial {u}_{z}}{\partial z}=\frac{\nu }{1-\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta })\\ {\epsilon }_{\mathrm{zz}}=\frac{-2\nu }{1-\nu }C(z)\end{array}\) eq 2.28

We thus end up with:

\(\begin{array}{}\frac{\partial {u}_{z}}{\partial z}=\frac{-2\nu }{1-\nu }{C}_{1}z+{C}_{2}\end{array}\) eq 2.29

So we can write by integrating \({\epsilon }_{\mathrm{zz}}\) than \(\begin{array}{}{u}_{z}(r,z)=f(z)+g(r)\end{array}\). Using the first relationship of eq 2.28 we then obtain:

\(\begin{array}{}\frac{\partial {u}_{z}}{\partial r}=g\text{'}(r)=\frac{-\partial {u}_{r}}{\partial z}\\ g\text{'}(r)=-C\text{'}(z)r-\frac{D\text{'}(z)}{r}\\ C\text{'}(z)=\mathrm{cte}\to C(z)={C}_{1}z+{C}_{2}\\ D\text{'}(z)=\mathrm{cte}\to D(z)={D}_{1}z+{D}_{2}\\ g(r)=-{C}_{1}\frac{\mathrm{r²}}{2}-{D}_{1}\mathrm{ln}(r)+{C}_{0}\end{array}\) eq 2.30

We thus end up with:

\(\begin{array}{}\frac{\partial {u}_{z}}{\partial r}=g\text{'}(r)\end{array}\) eq 2.31

By integrating the two partial differential equations eq 2.29 and eq 2.31 we obtain for each of the two rings i=1 exterior and i=2 interior:

\(\begin{array}{}{u}_{z}=\frac{-2\nu }{1-\nu }({C}_{1}\frac{\mathrm{z²}}{2}+{C}_{2}z)-{C}_{1}\frac{\mathrm{r²}}{2}-{D}_{1}\mathrm{ln}(r)+{C}_{0}\\ {u}_{z}(r\mathrm{,0})=g(r)+f(0)=0\to {C}_{0}={C}_{1}={D}_{1}=0\\ {u}_{z}(r,z)=\frac{-2\nu }{1-\nu }{C}_{2}z\end{array}\) eq 2.32

From where we get:

\(\begin{array}{}{\sigma }_{\mathrm{rr}}=\frac{E}{1+\nu }({\epsilon }_{\mathrm{rr}}+\frac{\nu }{1-\nu }({\epsilon }_{\mathrm{rr}}+{\epsilon }_{\theta \theta }))\\ {\sigma }_{\mathrm{rr}}=\frac{E}{1+\nu }(\frac{\partial {u}_{r}(r,z)}{\partial r}+\frac{\nu }{1-\nu }(\frac{\partial {u}_{r}(r,z)}{\partial r}+\frac{{u}_{r}}{r}))\\ {\sigma }_{\mathrm{rr}}=\frac{E}{1+\nu }(\frac{1+\nu }{1-\nu }{C}_{i}+\frac{{D}_{i}}{\mathrm{r²}})\end{array}\) eq 2.33

By posing:

\({A}_{i}=\frac{E}{1+\nu }\frac{1+\nu }{1-\nu }{C}_{i}\text{}{B}_{i}=\frac{E}{1+\nu }{D}_{i}\) eq 2.34

the constraints field is written as:

\(\begin{array}{c}{\sigma }_{\mathit{rr};i}(r)\mathrm{=}{A}_{i}\mathrm{-}\frac{{B}_{i}}{{r}^{2}}\\ {\sigma }_{\theta \theta ;i}(r)\mathrm{=}{A}_{i}+\frac{{B}_{i}}{{r}^{2}}\\ {\underline{\underline{\sigma }}}_{i}(r)\mathrm{=}{A}_{i}\underline{\underline{1}}\mathrm{-}\frac{{B}_{i}}{{r}^{2}}({\underline{e}}_{r}\mathrm{\otimes }{\underline{e}}_{r}\mathrm{-}{\underline{e}}_{\theta }\mathrm{\otimes }{\underline{e}}_{\theta })\end{array}\) eq 2.35

All we have to do is calculate the values of \({A}_{i}\) and \({B}_{i}\) for each of the crowns. Note \({\lambda }_{n}\) the contact pressure between the two rings such that:

\(\begin{array}{c}{\underline{\underline{\sigma }}}_{1}({R}_{2})\mathrm{.}(\mathrm{-}{\underline{e}}_{r})\mathrm{=}{\lambda }_{n}{\underline{e}}_{r}\\ {\underline{\underline{\sigma }}}_{2}({R}_{2})\mathrm{.}{\underline{e}}_{r}\mathrm{=}\mathrm{-}{\lambda }_{n}{\underline{e}}_{r}\end{array}\) eq 2.36

with the boundary conditions:

\(\begin{array}{c}{\underline{\underline{\sigma }}}_{1}({R}_{1})\mathrm{.}{\underline{e}}_{r}\mathrm{=}\mathrm{-}\mathit{p.}{\underline{e}}_{r}\\ {\underline{\underline{\sigma }}}_{2}({R}_{3})\mathrm{.}(\mathrm{-}{\underline{e}}_{r})\mathrm{=}\underline{0}\end{array}\) eq 2.37

we have:

In addition, the law of behavior of materials makes it possible to write:

\(\begin{array}{c}{\underline{\underline{\epsilon }}}_{i}\mathrm{=}\frac{1+{\nu }_{i}}{{E}_{i}}{\underline{\underline{\sigma }}}_{i}\mathrm{-}\frac{{\nu }_{i}}{{E}_{i}}\mathit{Tr}({\underline{\underline{\sigma }}}_{i})\underline{\underline{1}}\mathrm{=}\frac{1}{{E}_{i}}\mathrm{[}{A}_{i}(1\mathrm{-}{\nu }_{i})\underline{\underline{1}}\mathrm{-}{B}_{i}(1+{\nu }_{i})\frac{1}{{r}^{2}}({\underline{e}}_{r}\mathrm{\otimes }{\underline{e}}_{r}\mathrm{-}{\underline{e}}_{\theta }\mathrm{\otimes }{\underline{e}}_{\theta })\mathrm{]}\\ \text{}\mathrm{=}\frac{\mathrm{\partial }{u}_{r;i}}{\mathrm{\partial }r}{\underline{e}}_{r}\mathrm{\otimes }{\underline{e}}_{r}+\frac{1}{2}(\frac{1}{r}\frac{\mathrm{\partial }{u}_{r;i}}{\mathrm{\partial }\theta }+\frac{\mathrm{\partial }{u}_{\theta ;i}}{\mathrm{\partial }r}\mathrm{-}\frac{{u}_{\theta ;i}}{r})({\underline{e}}_{r}\mathrm{\otimes }{\underline{e}}_{\theta }+{\underline{e}}_{\theta }\mathrm{\otimes }{\underline{e}}_{r})+(\frac{1}{r}\frac{\mathrm{\partial }{u}_{\theta ;i}}{\mathrm{\partial }\theta }+\frac{{u}_{r;i}}{r}){\underline{e}}_{\theta }\mathrm{\otimes }{\underline{e}}_{\theta }\end{array}\) eq 2.39

which makes it possible to obtain:

\({\underline{u}}_{i}\mathrm{=}\frac{1}{{E}_{i}}\mathrm{[}{A}_{i}(1\mathrm{-}{\nu }_{i})r+{B}_{i}(1+{\nu }_{i})\frac{1}{r}\mathrm{]}{\underline{e}}_{r}\mathrm{=}{f}_{\mathit{CP}}(r){\underline{e}}_{r}\) eq 2.40

The function \(f(r)\) of the radial displacement is given as a function of the properties of the materials and the pressure \(p\). In the case of plane constraints (MODELISATION = “C_ PLAN “) we have:

\(\begin{array}{c}f({R}_{1})\mathrm{=}\frac{1}{{E}_{1}}({A}_{1}(1\mathrm{-}{\nu }_{1}){R}_{1}+\frac{{B}_{1}(1+{\nu }_{1})}{{R}_{1}})\\ f({R}_{3})\mathrm{=}\frac{1}{{E}_{2}}({A}_{2}(1\mathrm{-}{\nu }_{2}){R}_{3}+\frac{{B}_{2}(1+{\nu }_{2})}{{R}_{3}})\end{array}\) eq 2.41

To obtain the value of \({\lambda }_{n}\), we impose the continuity of the displacement vector in \(r\mathrm{=}{R}_{2}\):

\(\begin{array}{c}{\underline{u}}_{1}({R}_{2})\mathrm{=}{\underline{u}}_{2}({R}_{2})\\ \frac{1}{{E}_{1}}\mathrm{[}{A}_{1}(1\mathrm{-}{\nu }_{1}){R}_{2}+{B}_{1}(1+{\nu }_{1})\frac{1}{{R}_{2}}\mathrm{]}{\underline{e}}_{r}\mathrm{=}\frac{1}{{E}_{2}}\mathrm{[}{A}_{2}(1\mathrm{-}{\nu }_{2}){R}_{2}+{B}_{2}(1+{\nu }_{2})\frac{1}{{R}_{2}}\mathrm{]}{\underline{e}}_{r}\\ \frac{1}{{E}_{1}}\mathrm{[}\frac{\mathrm{-}p{R}_{1}^{2}+{\lambda }_{n}{R}_{2}^{2}}{{R}_{1}^{2}\mathrm{-}{R}_{2}^{2}}(1\mathrm{-}{\nu }_{1}){R}_{2}+(\mathrm{-}p+{\lambda }_{n})\frac{{R}_{1}^{2}{R}_{2}^{2}}{{R}_{1}^{2}\mathrm{-}{R}_{2}^{2}}(1+{\nu }_{1})\frac{1}{{R}_{2}}\mathrm{]}{\underline{e}}_{r}\\ \text{}\mathrm{=}\frac{1}{{E}_{2}}\mathrm{[}\mathrm{-}{\lambda }_{n}\frac{{R}_{2}^{2}}{{R}_{2}^{2}\mathrm{-}{R}_{3}^{2}}(1\mathrm{-}{\nu }_{2}){R}_{2}\mathrm{-}{\lambda }_{n}\frac{{R}_{2}^{2}{R}_{3}^{2}}{{R}_{2}^{2}\mathrm{-}{R}_{3}^{2}}(1+{\nu }_{2})\frac{1}{{R}_{2}}\mathrm{]}{\underline{e}}_{r}\end{array}\) eq 2.42

or again:

\({\lambda }_{n}\mathrm{=}\frac{2p{R}_{1}^{2}}{{R}_{1}^{2}(1+{\nu }_{1})+{R}_{2}^{2}(1\mathrm{-}{\nu }_{1})+\frac{{E}_{1}}{{E}_{2}}\frac{{R}_{1}^{2}\mathrm{-}{R}_{2}^{2}}{{R}_{2}^{2}\mathrm{-}{R}_{3}^{2}}({R}_{2}^{2}(1\mathrm{-}{\nu }_{2})+{R}_{3}^{2}(1+{\nu }_{2}))}\) eq 2.43 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~

2.3. note#

Once we have calculated the value of the displacements and of the contact pressure for the case under plane stresses, we can easily calculate these values in plane deformations by replacing the values of the Young’s modulus and the Poisson’s ratio:

\({E}_{\mathit{DP}}\mathrm{=}\frac{{E}_{\mathit{CP}}}{1\mathrm{-}{\nu }_{\mathit{CP}}^{2}};{\nu }_{\mathit{DP}}\mathrm{=}\frac{{\nu }_{\mathit{CP}}}{1\mathrm{-}{\nu }_{\mathit{CP}}}\) eq 2.44

Where \({E}_{\mathit{CP}}\) and \({\nu }_{\mathit{CP}}\) take the values \({E}_{i}\) and \({\nu }_{i}\) from §2.1.2.

The values of the displacements are thus obtained:

\({\underline{u}}_{i}\mathrm{=}\frac{1+{\nu }_{i}}{{E}_{i}}\mathrm{[}{A}_{i}(1\mathrm{-}2{\nu }_{i})r+\frac{{B}_{i}}{r}\mathrm{]}{\underline{e}}_{r}\mathrm{=}{f}_{\mathit{DP}}(r){\underline{e}}_{r}\) eq 2.45

and contact pressure:

Note that the values of \({A}_{i}\) and \({B}_{i}\) remain the same.

2.4. Calculation of the L2 norm of contact pressure#

Let \({\Gamma }_{c}\) be the contact surface. Contact pressure standard \({L}^{2}\) is defined by:

\({\parallel \lambda \parallel }_{{L}^{2}}^{2}={\int }_{{\Gamma }_{c}}{\lambda }^{2}\mathit{dS}\mathrm{.}\)

In the present case, \({\Gamma }_{c}\) l is the circle with center \(O\) and radius \({R}_{2}\) and the contact pressure is uniform.

So we have:

\({\parallel \lambda \parallel }_{{L}^{2}}^{2}={\lambda }^{2}\mid {\Gamma }_{c}\mid =2\pi {\lambda }^{2}{R}_{2}\mathrm{.}\)

And finally:

\({\parallel \lambda \parallel }_{{L}^{2}}=\mid \lambda \mid \sqrt{2\pi {R}_{2}}\mathrm{.}\)

2.5. Tested values#

The contact pressure is tested at the interface between the 2 rings. For an interface node, the analytical solution is obtained by equations 1.3 and 1.5 in plane deformations and plane stresses respectively.

The value of the pressure applied to the edge at each time step is given by the formula:

\(p(t)\mathrm{=}{p}_{0}{10}^{\frac{t}{10}\mathrm{-}\mathrm{1,1}},t\mathrm{\in }\left\{n\mathrm{\in }\mathrm{\mathbb{Z}},1\le n\le 21\right\},{p}_{0}\mathrm{=}\mathrm{1,0}\mathit{MPa}\) eq 2.47

For \(t\mathrm{=}1\), the pressure is \(\mathrm{0,1}\mathit{MPa}\) and for \(t\mathrm{=}21\) it is equal to \(\mathrm{10MPa}\). Note that the simulations were done by neglecting the effects of acceleration (command STAT_NON_LINE), which implies that the time steps have arbitrary units.