2. Benchmark solution

Here we develop an analytical solution for the problem presented above. This solution will be developed under 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:

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

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

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

Using the Lamé-Navier equation:

\(\begin{array}{c}(\lambda +\mu )\underline{\mathit{grad}}(\nabla \mathrm{.}(\underline{u}))+\mu \Delta \underline{u}+\underline{\mathit{fd}}=\underline{0}\end{array}\) 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 field of displacement is obtained for solid 1:

\({u}_{r}={C}_{1}r+\frac{{D}_{1}}{r}\text{}{u}_{\theta }=0\text{}{u}_{z}=0\) eq 2.6

and for solid 2 the following form of the displacement fields:

\({u}_{r}={C}_{2}r\text{}{u}_{\theta }=0\text{}{u}_{z}=0\) eq 2.7

To determine the constants \({C}_{1},{D}_{1},{C}_{2}\), all we have to do is impose the pressure and displacement limit conditions. 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. For solid 1, we obtain:

\(\begin{array}{c}{ϵ}_{\mathit{rr}}={C}_{1}-\frac{{D}_{1}}{\mathit{r²}}\\ {ϵ}_{\theta \theta }={C}_{1}+\frac{{D}_{1}}{\mathit{r²}}\\ {ϵ}_{\mathit{zz}}={ϵ}_{r\theta }={ϵ}_{\mathit{rz}}={ϵ}_{\theta z}=0\end{array}\) eq 2.8

And for solid 2:

\(\begin{array}{c}{ϵ}_{\mathit{rr}}={C}_{1}\\ {ϵ}_{\theta \theta }={C}_{1}\\ {ϵ}_{\mathit{zz}}={ϵ}_{r\theta }={ϵ}_{\mathit{rz}}={ϵ}_{\theta z}=0\end{array}\) eq 2.9

Applying Hooke’s law:

\(\underline{\underline{\sigma }}=\lambda \mathit{tr}(\underline{\underline{ϵ}})\underline{\underline{1}}+2\mu \underline{\underline{ϵ}}\) eq 2.10

the following general form is obtained for the stresses for solid 1:

\(\begin{array}{c}{\sigma }_{\mathit{rr}}=\frac{{E}_{1}}{1+{\nu }_{1}}(\frac{{C}_{1}}{1-2{\nu }_{1}}-\frac{{D}_{1}}{\mathit{r²}})\\ {\sigma }_{\theta \theta }=\frac{{E}_{1}}{1+{\nu }_{1}}(\frac{{C}_{1}}{1-2{\nu }_{1}}+\frac{{D}_{1}}{\mathit{r²}})\\ {\sigma }_{\mathit{zz}}=\frac{2{\nu }_{1}{E}_{1}{C}_{1}}{(1+{\nu }_{1})(1-2{\nu }_{1})}\\ {\sigma }_{r\theta }={\sigma }_{\mathit{rz}}={\sigma }_{\theta z}=\underline{0}\end{array}\) eq 2.11

and the following general form for the stresses for solid 2:

\(\begin{array}{c}{\sigma }_{\mathit{rr}}=\frac{{E}_{2}}{1+{\nu }_{2}}(\frac{{C}_{2}}{1-2{\nu }_{2}})\\ {\sigma }_{\theta \theta }=\frac{{E}_{2}}{1+{\nu }_{2}}(\frac{{C}_{2}}{1-2{\nu }_{2}})\\ {\sigma }_{\mathit{zz}}=\frac{2{\nu }_{2}{E}_{2}{C}_{2}}{(1+{\nu }_{2})(1-2{\nu }_{2})}\\ {\sigma }_{r\theta }={\sigma }_{\mathit{rz}}={\sigma }_{\theta z}=\underline{0}\end{array}\) eq 2.12

We ask:

\(\begin{array}{c}{A}_{1}=\frac{{E}_{1}}{(1+{\nu }_{1})(1-2{\nu }_{1})}{C}_{1}\text{}{B}_{1}=\frac{{E}_{1}}{1+{\nu }_{1}}{D}_{1}\text{}{A}_{2}=\frac{{E}_{2}}{(1+{\nu }_{2})(1-2{\nu }_{2})}{C}_{2}\end{array}\) eq 2.13

Now all we have to do is calculate the values for \({A}_{\mathrm{1,}}{B}_{\mathrm{1,}}{A}_{2}\). 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 2.14

with the boundary conditions:

\(\begin{array}{c}{\underline{\underline{\sigma }}}_{\mathrm{1rr}}({R}_{1})\mathrm{.}{\underline{e}}_{r}=-\mathit{p.}{\underline{e}}_{r}\end{array}\) eq 2.15

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 2.16

So we have 4 equations for the 4 unknowns \(\begin{array}{c}{A}_{1},{B}_{1},{A}_{\mathrm{2,}}{\lambda }_{n}\end{array}\).

The system of the first 3 equations allows us to obtain:

\(\begin{array}{c}{A}_{1}=\frac{-p{R}_{1}^{2}+{\lambda }_{n}{R}_{2}^{2}}{{R}_{1}^{2}-{R}_{2}^{2}};{B}_{1}=(-p+{\lambda }_{n})\frac{{R}_{1}^{2}{R}_{2}^{2}}{{R}_{1}^{2}-{R}_{2}^{2}}\\ {A}_{2}=-{\lambda }_{n}\end{array}\) eq 2.17

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

\({\lambda }_{n}=\frac{2p{R}_{1}^{2}(1-{\nu }_{1})}{{R}_{1}^{2}+{R}_{2}^{2}(1-2{\nu }_{1})+\frac{{E}_{1}}{{E}_{2}}\frac{1+{\nu }_{2}}{1+{\nu }_{1}}(1-2{\nu }_{2})({R}_{1}^{2}-{R}_{2}^{2})}\) eq 2.18

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 2.19

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 2.20

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 2.21

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:

\(\chi (r,\theta )=(A{r}^{2}+B{r}^{4}+\frac{C}{{r}^{2}}+D)\mathrm{cos}(2\theta )\) eq 2.22

From there, we can express the non-zero constraints in the polar coordinate system for solid 1:

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

And for solid 2:

\(\begin{array}{c}{\sigma }_{\mathit{rr}}^{2}=(-{\mathrm{2A}}_{2})\mathrm{cos}(2\theta )\\ {\sigma }_{\theta \theta }^{2}=({\mathrm{2A}}_{2}+{\mathrm{12B}}_{2}{r}^{2})\mathrm{cos}(2\theta )\\ {\sigma }_{r\theta }^{2}=2({A}_{2}+{\mathrm{3B}}_{2}{r}^{2})\mathrm{sin}(2\theta )\\ {\sigma }_{\mathit{zz}}^{2}=\nu ({\sigma }_{\mathit{rr}}+{\sigma }_{\theta \theta })\end{array}\) eq 2.24

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 2.25

We will use the expression deformations to express the displacements 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 2.26

By integrating these relationships and using the symmetries of the problem, we can express the displacements for solid 1:

\(\begin{array}{c}{u}_{r}^{1}=\frac{1+{\nu }_{1}}{{E}_{1}}[(-{\mathrm{2A}}_{1}r+2\frac{{C}_{1}}{{r}^{3}}+4\frac{{D}_{1}}{r})-{\nu }_{1}({\mathrm{4B}}_{1}{r}^{3}+4\frac{{D}_{1}}{r})]\mathrm{cos}(2\theta )\\ {u}_{\theta }^{1}=\frac{1+{\nu }_{1}}{{E}_{1}}[({\mathrm{2A}}_{1}r+{\mathrm{6B}}_{1}{r}^{3}+2\frac{{C}_{1}}{{r}^{3}}-2\frac{{D}_{1}}{r})-{\nu }_{1}({\mathrm{4B}}_{1}{r}^{3}-4\frac{{D}_{1}}{r})]\mathrm{sin}(2\theta )\end{array}\) eq 2.27

And for solid 2:

\(\begin{array}{c}{u}_{r}^{2}=\frac{1+{\nu }_{2}}{{E}_{2}}[(-{\mathrm{2A}}_{2}r)-{\nu }_{2}({\mathrm{4B}}_{2}{r}^{3})]\mathrm{cos}(2\theta )\\ {u}_{\theta }^{2}=\frac{1+{\nu }_{2}}{{E}_{2}}[({\mathrm{2A}}_{2}r+{\mathrm{6B}}_{2}{r}^{3})-{\nu }_{1}({\mathrm{4B}}_{1}{r}^{3})]\mathrm{sin}(2\theta )\end{array}\) eq 2.28

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

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

The boundary conditions are:

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

So we have 6 equations for the 6 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 asking: \({f}_{1}={(\frac{{R}_{2}}{{R}_{1}})}^{2}\)

The system of 6 equations allows us to have:

\(\begin{array}{c}{A}_{1}=\frac{{\alpha }_{1}(2{f}_{1}^{2}+{f}_{1}+1)-\lambda ({f}_{1}^{3}+{f}_{1}^{2}+2{f}_{1})}{2{(1-{f}_{1})}^{3}};{B}_{1}=\frac{-1}{{R}_{2}^{2}}\frac{{\alpha }_{1}(3{f}_{1}^{2}+{f}_{1})-\lambda ({f}_{1}^{3}+3{f}_{1}^{2})}{6{(1-{f}_{1})}^{3}};\\ {C}_{1}={R}_{2}^{4}\frac{{\alpha }_{1}({f}_{1}+3)-\lambda ({\mathrm{3f}}_{1}+1)}{6{(1-{f}_{1})}^{3}};{D}_{1}=-{R}_{2}^{2}\frac{{\alpha }_{1}({f}_{1}^{2}+{f}_{1}+2)-\lambda (2{f}_{1}^{2}+{f}_{1}+1)}{2{(1-{f}_{1})}^{3}}\\ {A}_{2}=\frac{\lambda }{2};{B}_{2}=\frac{-\lambda }{{\mathrm{6R}}_{2}^{2}}\end{array}\) eq 2.29

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 2.30

contact pressure can be expressed analytically:

\(\lambda =\frac{{\mathit{coef}}_{1}}{{\mathit{coef}}_{2}+{\mathit{coef}}_{3}}{\alpha }_{1}\) eq 2.31

such as:

\(\begin{array}{c}{\mathit{coef}}_{1}=\frac{1+{\nu }_{1}}{6{E}_{1}{(1-{f}_{1})}^{3}}[(-12{f}_{1}^{2}-8{f}_{1}-12)+{\nu }_{1}(12{f}_{1}^{2}+8{f}_{1}+12)]\\ {\mathit{coef}}_{2}=\frac{1+{\nu }_{1}}{6{E}_{1}{(1-{f}_{1})}^{3}}[(-3{f}_{1}^{3}-15{f}_{1}^{2}-9{f}_{1}-5)+{\nu }_{1}(2{f}_{1}^{3}+18{f}_{1}^{2}+6{f}_{1}+6)]\\ {\mathit{coef}}_{3}=\frac{1+{\nu }_{2}}{6{E}_{2}}(-3+2{\nu }_{2})\end{array}\) eq 2.32

2.3. Tested values

The contact pressure at the interface between the two solids, as well as the displacements along X and Y: \({u}_{x},{u}_{y}\), are tested 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 \(\theta =\mathrm{arctan}(\frac{Y}{X})\).

The values the min and max values of the movements and of the contact pressure will be tested. For each modeling, two calculations will be carried out: a first calculation with contact and a second calculation without contact. The values will be tested for each calculation. We will call « calcul_1 » the calculation with contact and « calcul_2 » the calculation without contact.