2. Reference solution

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

The on-the-go solution to the problem has the following generic form:

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

The problem will be solved within the framework of the hypothesis of plane deformations. As our load is written in the form \(p={\alpha }_{0}+{\alpha }_{1}\mathrm{.}\mathrm{cos}(2\theta )\), we will decouple the solution of the problem into a part where the pressure is uniform \(p={\alpha }_{0}\), and a part where the pressure is variable \(p={\alpha }_{1}\mathrm{.}\mathrm{cos}(2\theta )\).

2.1. Uniform pressure

Using the symmetries of the problem and the assumption of z-invariance of plane deformations, the solution of the problem takes the following form:

\(\begin{array}{c}{u}_{r}={u}_{r}(r)\\ {u}_{\theta }=0\\ {u}_{z}=0\end{array}\) eq 2.1

Using the Lamé-Navier equation:

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

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

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

We can write eq 2.2 in the form:

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

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

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

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

Applying Hooke’s law:

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

the following general form is obtained for the constraints:

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

By posing:

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

non-zero constraints become:

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

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

with boundary conditions:

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

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

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

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

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

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

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

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

2.2. Variable pressure

Using the hypothesis of z-invariance of plane deformations, the solution of the problem takes the following form:

\(\begin{array}{c}{u}_{r}={u}_{r}(r,\theta )\\ {u}_{\theta }={u}_{\theta }(r,\theta )\\ {u}_{z}=0\end{array}\) eq 2.17

Throughout the following, the parameters specific to each solid will be noted by an index i, with i=1.2.

In the absence of volume forces, we will use a form of the Airy function proposed by Michell [1], in polar coordinates:

\(\begin{array}{c}\chi (r,\theta )={A}_{01}{r}^{2}+{A}_{02}{r}^{2}\mathrm{log}(r)+{A}_{03}\mathrm{log}(r)+{A}_{04}\theta \\ +({A}_{11}{r}^{3}+{A}_{12}\mathit{rlog}(r)+{A}_{13}{r}^{-1})\mathrm{cos}(\theta )+{A}_{14}r\theta \mathrm{sin}(\theta )\\ +({B}_{11}{r}^{3}+{B}_{12}r\mathrm{log}(r)+{B}_{13}{r}^{-1})\mathrm{sin}(\theta )+{B}_{14}r\theta \mathrm{cos}(\theta )\\ +\sum _{n=2}^{\infty }({A}_{\mathit{n1}}{r}^{n+2}+{A}_{\mathit{n2}}{r}^{-n+2}+{A}_{\mathit{n3}}{r}^{n}+{A}_{\mathit{n4}}{r}^{-n})\mathrm{cos}(n\theta )\\ +\sum _{n=2}^{\infty }({B}_{\mathit{n1}}{r}^{n+2}+{B}_{\mathit{n2}}{r}^{-n+2}+{B}_{\mathit{n3}}{r}^{n}+{B}_{\mathit{n4}}{r}^{-n})\mathrm{sin}(n\theta )\end{array}\) eq 2.18

The terms of the non-zero Cauchy stress tensor are expressed in terms of the Airy function as follows:

\(\begin{array}{c}{\sigma }_{\mathit{rr}}=\frac{1}{r}\frac{\partial \chi }{\partial r}+\frac{1}{{r}^{2}}\frac{{\partial }^{2}\chi }{\partial {\theta }^{2}}\\ {\sigma }_{\theta \theta }=\frac{{\partial }^{2}\chi }{\partial {r}^{2}}\\ {\sigma }_{r\theta }=\frac{-\partial }{\partial r}(\frac{1}{r}\frac{\partial \chi }{\partial \theta })\\ {\sigma }_{\mathit{zz}}=\nu ({\sigma }_{\mathit{rr}}+{\sigma }_{\theta \theta })\end{array}\) eq 2.19

As our pressure \(p={\alpha }_{1}\mathrm{.}\mathrm{cos}(2\theta )\) varies in \(\mathrm{cos}(2\theta )\), we will only take the part varying in \(\mathrm{cos}(2\theta )\) in the Airy function. The Airy function will then be written as:

\({\chi }_{i}(r,\theta )=({A}_{i}{r}^{2}+{B}_{i}{r}^{4}+\frac{{C}_{i}}{{r}^{2}}+{D}_{i})\mathrm{cos}(2\theta )\) eq 2.20

From there, we can express the non-zero constraints in the polar coordinate system:

\(\begin{array}{c}{\sigma }_{\mathit{rr}}^{i}=(-{\mathrm{2A}}_{i}-6\frac{{C}_{i}}{{r}^{4}}-4\frac{{D}_{i}}{{r}^{2}})\mathrm{cos}(2\theta )\\ {\sigma }_{\theta \theta }^{i}=({\mathrm{2A}}_{i}+{\mathrm{12B}}_{i}{r}^{2}+6\frac{{C}_{i}}{{r}^{4}})\mathrm{cos}(2\theta )\\ {\sigma }_{r\theta }^{i}=2({A}_{i}+{\mathrm{3B}}_{i}{r}^{2}-3\frac{{C}_{i}}{{r}^{4}}-\frac{{D}_{i}}{{r}^{2}})\mathrm{sin}(2\theta )\\ {\sigma }_{\mathit{zz}}^{i}=\nu ({\sigma }_{\mathit{rr}}+{\sigma }_{\theta \theta })\end{array}\) eq 2.21

and the terms of the strain tensor using Hooke’s law:

\(\underline{\underline{\epsilon }}=\frac{1}{E}((1+\nu )\underline{\underline{\sigma }}-\nu \mathit{tr}(\underline{\underline{\sigma }})\underline{\underline{I}})\) eq 2.22

The non-zero terms of the strain tensor are then expressed in terms of the constants of the problem:

\(\begin{array}{c}{\epsilon }_{\mathit{rr}}^{i}=\frac{1+{\nu }_{i}}{{E}_{i}}[(-{\mathrm{2A}}_{i}-6\frac{{C}_{i}}{{r}^{4}}-4\frac{{D}_{i}}{{r}^{2}})-{\nu }_{i}({\mathrm{12B}}_{i}{r}^{2}-4\frac{{D}_{i}}{{r}^{2}})]\mathrm{cos}(2\theta )\\ {\epsilon }_{\theta \theta }^{i}=\frac{1+{\nu }_{i}}{{E}_{i}}[({\mathrm{2A}}_{i}+{\mathrm{12B}}_{i}{r}^{2}+6\frac{{C}_{i}}{{r}^{4}})-{\nu }_{i}({\mathrm{12B}}_{i}{r}^{2}-4\frac{{D}_{i}}{{r}^{2}})]\mathrm{cos}(2\theta )\\ {\epsilon }_{r\theta }^{i}=\frac{1+{\nu }_{i}}{{E}_{i}}[2({A}_{i}+{\mathrm{3B}}_{i}{r}^{2}-3\frac{{C}_{i}}{{r}^{4}}-\frac{{D}_{i}}{{r}^{2}})]\mathrm{sin}(2\theta )\end{array}\) eq 2.23

These fields will be used to express the movements in the polar coordinate system.

We have:

\(\begin{array}{c}\frac{\partial {u}_{r}}{\partial r}={\epsilon }_{\mathit{rr}}\\ \frac{\partial {u}_{\theta }}{\partial \theta }=r{\epsilon }_{\theta \theta }-{u}_{r}\\ \frac{1}{2}(\frac{\partial {u}_{\theta }}{\partial r}+\frac{1}{r}\frac{\partial {u}_{r}}{\partial \theta }-\frac{{u}_{\theta }}{r})={\epsilon }_{r\theta }\end{array}\) eq 2.24

By integrating these relationships, we can express displacements:

\(\begin{array}{c}{u}_{r}^{i}=\frac{1+{\nu }_{i}}{{E}_{i}}[(-{\mathrm{2A}}_{i}r+2\frac{{C}_{i}}{{r}^{3}}+4\frac{{D}_{i}}{r})-{\nu }_{i}({\mathrm{4B}}_{i}{r}^{3}+4\frac{{D}_{i}}{r})]\mathrm{cos}(2\theta )\\ {u}_{\theta }^{i}=\frac{1+{\nu }_{i}}{{E}_{i}}[({\mathrm{2A}}_{i}r+{\mathrm{6B}}_{i}{r}^{3}+2\frac{{C}_{i}}{{r}^{3}}-2\frac{{D}_{i}}{r})-{\nu }_{i}({\mathrm{4B}}_{i}{r}^{3}-4\frac{{D}_{i}}{r})]\mathrm{sin}(2\theta )\end{array}\) eq 2.25

Now that we’ve expressed all of our fields in terms of constants \({A}_{i},{B}_{i},{C}_{i},{D}_{i}(i=\mathrm{1,2})\), we need to calculate them based on geometric characteristics and loading.

Note \(\lambda\) the contact pressure between the two rings.

The boundary conditions are:

\(\begin{array}{c}{\sigma }_{\mathit{rr}}^{1}({R}_{1})=-\alpha \mathrm{cos}(2\theta )\mathrm{:}\text{pression externe appliquée}\\ {\sigma }_{r\theta }^{1}({R}_{1})=0\mathrm{:}\text{pression tangentielle nulle sur le bord extérieur de la couronne 1}\\ {\sigma }_{\mathit{rr}}^{2}({R}_{2})=\lambda \mathrm{:}\text{pression de contact appliquée par la couronne 1 sur la couronne 2}\\ {\sigma }_{\mathit{rr}}^{1}({R}_{2})=\lambda \mathrm{:}\text{pression de contact appliquée par la couronne 2 sur la couronne 1}\\ {\sigma }_{r\theta }^{1}({R}_{2})=0\mathrm{:}\text{pas de frottement entre les deux couronnes}\\ {\sigma }_{r\theta }^{2}({R}_{2})=0\mathrm{:}\text{pas de frottement entre les deux couronnes}\\ {\sigma }_{\mathit{rr}}^{2}({R}_{3})=0\mathrm{:}\text{bord extérieur de la couronne 2 libre}\\ {\sigma }_{r\theta }^{2}({R}_{3})=0\mathrm{:}\text{bord extérieur de la couronne 2 libre}\end{array}\)

So we have 8 equations for the 8 unknowns: \({A}_{\mathrm{1,}}{B}_{\mathrm{1,}}{C}_{\mathrm{1,}}{D}_{\mathrm{1,}}{A}_{\mathrm{2,}}{B}_{\mathrm{2,}}{C}_{\mathrm{2,}}{D}_{2}\).

By posing:

\({f}_{1}={(\frac{{R}_{2}}{{R}_{1}})}^{2}\text{et}{f}_{2}={(\frac{{R}_{3}}{{R}_{2}})}^{2}\)

The system of 8 equations allows us to have:

\(\begin{array}{c}{A}_{1}=\frac{{\alpha }_{1}(2{f}_{1}^{2}+{f}_{1}+1)-\lambda ({f}_{1}^{3}+{f}_{1}^{2}+2{f}_{1})}{2{(1-{f}_{1})}^{3}};{B}_{1}=\frac{-1}{{R}_{2}^{2}}\frac{{\alpha }_{1}(3{f}_{1}^{2}+{f}_{1})-\lambda ({f}_{1}^{3}+3{f}_{1}^{2})}{6{(1-{f}_{1})}^{3}};\\ {C}_{1}={R}_{2}^{4}\frac{{\alpha }_{1}({f}_{1}+3)-\lambda ({\mathrm{3f}}_{1}+1)}{6{(1-{f}_{1})}^{3}};{D}_{1}=-{R}_{2}^{2}\frac{{\alpha }_{1}({f}_{1}^{2}+{f}_{1}+2)-\lambda (2{f}_{1}^{2}+{f}_{1}+1)}{2{(1-{f}_{1})}^{3}}\\ {A}_{2}=\frac{\lambda (2{f}_{2}^{2}+{f}_{2}+1)}{2{(1-{f}_{2})}^{3}};{B}_{2}=\frac{-1}{{R}_{3}^{2}}\frac{\lambda (3{f}_{2}^{2}+{f}_{2})}{6{(1-{f}_{2})}^{3}};\\ {C}_{2}={R}_{3}^{4}\frac{\lambda ({f}_{2}+3)}{6{(1-{f}_{2})}^{3}};{D}_{2}=-{R}_{3}^{2}\frac{\lambda ({f}_{2}^{2}+{f}_{2}+2)}{2{(1-{f}_{2})}^{3}}\end{array}\) eq 2.26

Contact pressure can be expressed analytically

Using the continuity of the radial displacement at the contact interface:

\({u}_{r}^{1}({R}_{2})={u}_{r}^{2}({R}_{2})\) eq 2.27

contact pressure can be expressed analytically:

\(\lambda =\frac{{\mathit{coef}}_{1}}{{\mathit{coef}}_{2}+{\mathit{coef}}_{3}}{\alpha }_{1}\) eq 2.28

such as:

\(\begin{array}{c}{\mathit{coef}}_{1}=\frac{1+{\nu }_{1}}{6{E}_{1}{(1-{f}_{1})}^{3}}[(-12{f}_{1}^{2}-8{f}_{1}-12)-{\nu }_{1}(-12{f}_{1}^{2}-8{f}_{1}-12)]\\ {\mathit{coef}}_{2}=\frac{1+{\nu }_{1}}{6{E}_{1}{(1-{f}_{1})}^{3}}[(-3{f}_{1}^{3}-15{f}_{1}^{2}-9{f}_{1}-5)-{\nu }_{1}(-2{f}_{1}^{3}-18{f}_{1}^{2}-6{f}_{1}-6)]\\ {\mathit{coef}}_{3}=\frac{1+{\nu }_{2}}{6{E}_{2}{(1-{f}_{2})}^{3}}[(-5{f}_{2}^{3}-9{f}_{2}^{2}-15{f}_{2}-3)-{\nu }_{2}(-6{f}_{2}^{3}-6{f}_{2}^{2}-18{f}_{2}-2)]\end{array}\) eq 2.29

2.3. Tested values

We test the contact pressure at the interface between the two rings, at the level of the slave surface, as well as the movements along X and Y: \({u}_{x},{u}_{y}\) (and along Z also for 3D models), in plane deformations.

The value of the pressure applied to the outer edge of the crown in \(r={R}_{1}\) is expressed in the form: \(p(\theta )={10}^{7}+{10}^{5}\mathrm{cos}(2\theta )(\mathit{Pa})\), with \(\theta =\mathrm{arctan}(\frac{Y}{X})\).

We will test the values of the points located on radius \(r={R}_{2}\), on the side of crown 2, at different polar angles: \(\theta \in \text{{}\mathrm{0,}\frac{\pi }{4},\frac{\pi }{2},\frac{3\pi }{4},\pi ,\frac{5\pi }{4},\frac{3\pi }{2},\frac{7\pi }{4}\text{}}\) for models A, B, C, D.

For the E, G and H models, a min-max test will be done on three groups of nodes. Each group of nodes contains three nodes with equal X and Y coordinates, which makes it possible to test the invariance of the result obtained according to Z. The reference value used in the test is the analytical minimum value (respectively maximum) of the component tested for each node group. With respect to the F model, only three nodes with equal X and Y coordinates were tested, at the center of the structure, in order to avoid edge effects due to the fact that the meshes of the master and slave contact surfaces are not compatible.

2.3.1. Bibliography

[1] J.R. Barber, « Elasticity, » Kluwer Academic Publishers, 1982