2. Anisotropic Fourier analysis

2.1. General theory

All the fields considered (forces, displacements, deformations, constraints) are expressed in cylindrical coordinates with the following convention on the order of the components:

1	radial component next \(r\)
2	axial component next \(z\)
3	tangential component next \(\theta\)

Example: \(({u}_{r},{u}_{z},{u}_{\theta }),({f}_{r},{f}_{z},{f}_{\theta })\)

The mesh is located in plane \((r,z)\), the symmetry of revolution taking place around the \(\mathrm{Oz}\) axis. Triad \((r,z,\theta )\) is oriented in the anticlockwise direction.

We decompose the displacement \(\mathrm{u}\) (or the load \(f\)) as follows \(\mathrm{u}\mathrm{=}{\mathrm{u}}^{s}+{\mathrm{u}}^{a}\) where \({\mathrm{u}}^{s}\) (resp. \({u}^{a}\)) designates the symmetric (or antisymmetric) part of the Fourier series expansion of \(u\) with respect to the variable \(\theta\).

We get:

\(\begin{array}{}\begin{array}{}{u}_{r}^{s}=\sum _{l=0}^{\infty }{u}_{l}^{s}(r,z)\text{cos}l\theta \\ {u}_{z}^{s}=\sum _{l=0}^{\infty }{v}_{l}^{s}(r,z)\text{cos}l\theta \\ {u}_{\theta }^{s}=\sum _{l=0}^{\infty }{w}_{l}^{s}(r,z)(-\text{sin}l\theta )\end{array}\}\text{partie symétrique}{u}^{s}\\ \begin{array}{}{u}_{r}^{a}=\sum _{l=0}^{\infty }{u}_{l}^{a}(r,z)\text{sin}l\theta \\ {u}_{z}^{a}=\sum _{l=0}^{\infty }{v}_{l}^{a}(r,z)\text{sin}l\theta \\ {u}_{\theta }^{a}=\sum _{l=0}^{\infty }{w}_{l}^{a}(r,z)\text{cos}l\theta \end{array}\}\text{partie antisymétrique}{u}^{a}\end{array}\)

Note the choice of the sign \(–\) for \({u}_{\theta }^{s}\), which simplifies subsequent calculations. If we note \({U}_{l}^{s}=({u}_{l}^{s},{v}_{l}^{s},{w}_{l}^{s})\text{}(\text{resp}\text{.}{U}_{l}^{a})\) the \(l-\text{ième}\) symmetric (resp. antisymmetric) component of the Fourier series expansion of \(u\), we get:

\(u=\sum _{l=0}^{\infty }\left[(\begin{array}{ccc}\text{cos}l\theta & & 0\\ & \text{cos}l\theta & \\ 0& & -\text{sin}l\theta \end{array}){U}_{l}^{s}+(\begin{array}{ccc}\text{sin}l\theta & & 0\\ & \text{sin}l\theta & \\ 0& & \text{cos}l\theta \end{array}){U}_{l}^{a}\right]\) eq 2.1-1

If we designate the linearized deformation vector by \(\varepsilon\), we see that \(\varepsilon\) can be decomposed into the following Fourier series:

\(\varepsilon \mathrm{=}\mathrm{\sum }_{l\mathrm{=}0}^{\mathrm{\infty }}(\begin{array}{cc}\text{cos}l\theta {I}_{4}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& \mathrm{-}\text{sin}l\theta {I}_{2}\end{array}){\varepsilon }_{l}^{s}+(\begin{array}{cc}\text{sin}l\theta {I}_{4}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& \text{cos}l\theta {I}_{2}\end{array}){\theta }_{l}^{a}\) eq 2.1-2

with \(\varepsilon =\left\{{\varepsilon }_{r},{\varepsilon }_{z},{\varepsilon }_{q},{\gamma }_{\text{rz}},{\gamma }_{\mathrm{rq}},{\gamma }_{\mathrm{zq}}\right\}\)

\({\varepsilon }_{l}^{s}={B}_{l}^{s}{U}_{l}^{s}\text{}{\varepsilon }_{l}^{a}={B}_{l}^{a}{U}_{l}^{a}\)

with (see [bib1]):

\({B}_{l}^{s}=\left[\begin{array}{ccc}\frac{\partial }{\partial r}& 0& 0\\ 0& \frac{\partial }{\partial z}& 0\\ \frac{1}{r}& 0& -\frac{l}{r}\\ \frac{\partial }{\partial z}& \frac{\partial }{\partial r}& 0\\ \frac{l}{r}& 0& \frac{\partial }{\partial r}-\frac{1}{r}\\ 0& \frac{l}{r}& \frac{\partial }{\partial z}\end{array}\right]\)

We have \({B}_{l}^{a}={B}_{l}^{s}\text{}\forall l\) (this is due to choosing the symmetric expansion of \(u\) in \((\mathrm{cos},\mathrm{cos},–\mathrm{sin})\) instead of \((\mathrm{cos},\mathrm{cos},\mathrm{sin})\)). From now on, the indices \(a\) and \(s\) will be omitted and the operator used to calculate the deformations corresponding to the harmonic \(l\) will be noted \({B}_{l}\).

2.2. Coupling and decoupling symmetric and antisymmetric modes

Using the previous notations, we have:

\(u=\sum _{l}(\begin{array}{cc}\text{cos}l\theta {I}_{2}& {0}_{\mathrm{2,1}}\\ {0}_{\mathrm{1,2}}& -\text{sin}l\theta \end{array}){u}_{l}^{s}+\sum _{l}(\begin{array}{cc}\text{sin}l\theta {I}_{2}& {0}_{\mathrm{2,1}}\\ {0}_{\mathrm{1,2}}& \text{cos}l\theta \end{array}){u}_{l}^{a}\)

What can be written, by introducing matrices \({M}_{l}^{s}\text{et}{M}_{l}^{a}\):

\(\begin{array}{}u=\sum _{l}({M}_{l}^{s}{U}_{l}^{s}+{M}_{l}^{a}{U}_{l}^{a})\\ {u}_{l}={M}_{l}^{s}{U}_{l}^{s}+{M}_{l}^{a}{U}_{l}^{a}\end{array}\)

We deduce that: \({\varepsilon }_{l}=M{\text{'}}_{l}^{s}{\varepsilon }_{l}^{s}+M{\text{'}}_{l}^{a}{\varepsilon }_{l}^{a}\)

\(\begin{array}{}\begin{array}{cc}\text{avec}& M{\text{'}}_{l}^{s}=(\begin{array}{cc}\text{cos}l\theta {I}_{4}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& -\text{sin}l\theta {I}_{2}\end{array})\\ & M{\text{'}}_{l}^{a}=(\begin{array}{cc}\text{sin}l\theta {I}_{4}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& \text{cos}l\theta {I}_{2}\end{array})\end{array}\\ \end{array}\)

Calculation of deformation energy

\(\begin{array}{}\begin{array}{cc}{W}_{l}& =\underset{0}{\overset{2\pi }{\int }}\underset{s}{\int }{}^{t}{\varepsilon }_{l}D{\varepsilon }_{l}\mathrm{ds}d\theta \text{avec}\mathrm{ds}=\text{rdrdz}\\ & =\underset{0}{\overset{2\pi }{\int }}d\theta \underset{s}{\int }{}^{t}{\varepsilon }_{l}^{s\text{}t}M{\text{'}}_{l}^{s}\text{DM}{\text{'}}_{l}^{s}{\varepsilon }_{l}^{s}\text{ds}+\underset{0}{\overset{2\pi }{\int }}d\theta \underset{s}{\int }{}^{t}{\theta }_{l}^{a\text{}t}M{\text{'}}_{l}^{a}\text{DM}{\text{'}}_{l}^{a}{\varepsilon }_{l}^{a}\text{ds}\\ & +\underset{0}{\overset{2\pi }{\int }}d\theta \underset{s}{\int }{}^{t}{\varepsilon }_{l}^{a\text{}t}M{\text{'}}_{l}^{a}\text{DM}{\text{'}}_{l}^{s}{\varepsilon }_{l}^{s}\text{ds}+\underset{0}{\overset{2\pi }{\int }}d\theta \underset{s}{\int }{}^{t}{\varepsilon }_{l}^{s\text{}t}M{\text{'}}_{l}^{s}\text{DM}{\text{'}}_{l}^{a}{\varepsilon }_{l}^{a}\text{ds}\end{array}\\ \text{Puisque}M{\text{'}}_{l}^{a}\text{DM}{\text{'}}_{l}^{s}=(\begin{array}{cc}\text{sin}l\theta {I}_{4}& 0\\ 0& \text{cos}l\theta {I}_{2}\end{array})(\begin{array}{cc}{D}_{1}& {D}_{3}\\ {}^{t}\text{}{D}_{3}& {D}_{2}\end{array})(\begin{array}{cc}\text{cos}l\theta {I}_{4}& 0\\ 0& -\text{sin}l\theta {I}_{2}\end{array})\\ \text{}M{\text{'}}_{l}^{a}\text{DM}{\text{'}}_{l}^{s}=(\begin{array}{cc}{D}_{1}\text{sin}l\theta \text{cos}l\theta & -{D}_{3}{(\text{sin}l\theta )}^{2}\\ {}^{t}\text{}{D}_{3}{(\text{cos}l\theta )}^{2}& -{D}_{2}\text{sin}l\theta \text{cos}l\theta \end{array})\end{array}\)

and that \(\underset{0}{\overset{2\pi }{\int }}\text{sin}l\theta \text{cos}l\theta d\theta =\mathrm{0,}\text{si}{D}_{3}=0\) so there is no \(({}^{t}\text{}{\varepsilon }_{l}^{a},{\varepsilon }_{l}^{s})\text{ou}({}^{t}\text{}{\varepsilon }_{l}^{s},{\varepsilon }_{l}^{a})\) term in \(W\). There is then no \(({U}^{a},{U}^{s})\text{ou}({U}^{s},{U}^{a})\) coupling.

2.3. Stress calculation

Just as \(\varepsilon ,\sigma\) can be broken down into the following Fourier series:

\(\sigma =\sum _{l}(M{\text{'}}_{l}^{s}{\sigma }_{l}^{s}+M{\text{'}}_{l}^{a}{\sigma }_{l}^{a})\)

From Hooke’s law \(\sigma =D\varepsilon\), we deduce:

\(s=\sum _{l}(\begin{array}{cc}\text{cos}l\theta {D}_{1}& -\text{sin}l\theta {D}_{3}\\ \text{cos}l\theta {D}_{3}^{t}& -\text{sin}l\theta {D}_{2}\end{array}){\varepsilon }_{l}^{s}+(\begin{array}{cc}\text{sin}l\theta {D}_{1}& \text{cos}l\theta {D}_{3}\\ \text{sin}l\theta {D}_{3}^{t}& \text{cos}l\theta {D}_{2}\end{array}){\varepsilon }_{l}^{a}\)

Or, by making the \(M{\text{'}}_{l}^{s}\text{et}M{\text{'}}_{l}^{a}\) matrices appear:

\(\begin{array}{cc}\sigma =& \sum _{l}M{\text{'}}_{l}^{s}\left[(\begin{array}{cc}{D}_{1}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& {D}_{2}\end{array}){\varepsilon }_{l}^{s}+(\begin{array}{cc}{0}_{\mathrm{4,4}}& {D}_{3}\\ -{D}_{3}^{t}& {0}_{\mathrm{2,2}}\end{array}){\varepsilon }_{l}^{a}\right]\\ & +M{\text{'}}_{l}^{a}\left[(\begin{array}{cc}{0}_{\mathrm{4,4}}& -{D}_{3}\\ {D}_{3}^{t}& {0}_{\mathrm{2,2}}\end{array}){\varepsilon }_{l}^{s}+(\begin{array}{cc}{D}_{1}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& {D}_{2}\end{array}){\varepsilon }_{l}^{a}\right]\end{array}\)

By setting \({D}^{s}(\begin{array}{cc}{D}_{1}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& {D}_{2}\end{array})\) and \({D}^{a}\mathrm{=}(\begin{array}{cc}{0}_{\mathrm{4,4}}& {D}_{3}\\ \mathrm{-}{D}_{3}^{t}& {0}_{\mathrm{2,2}}\end{array})\), we deduce the parts symmetric and antisymmetric of the constraint relating to harmonic \(l\):

\(\{\begin{array}{}{\sigma }_{l}^{s}={D}^{s}{\varepsilon }_{l}^{s}+{D}^{a}{\varepsilon }_{l}^{a}={D}^{s}{B}_{l}{u}_{l}^{s}+{D}^{a}{B}_{l}{u}_{l}^{a}\\ {\sigma }_{l}^{a}=-{D}^{a}{\varepsilon }_{l}^{s}+{D}^{s}{\varepsilon }_{l}^{a}=-{D}^{a}{B}_{l}{u}_{l}^{s}+{D}^{s}{B}_{l}{u}_{l}^{a}\end{array}\) eq 2.3-1

Note:

In the case of orthotropy with respect to \(\mathrm{Oz}\) , we have \({D}^{a}=0\) and [éq 2.1-1] is reduced to:

\(\{\begin{array}{}{\sigma }_{l}^{s}={D}^{s}{B}_{l}{u}_{l}^{s}\\ {\sigma }_{l}^{a}={D}^{s}{B}_{l}{u}_{l}^{a}\end{array}\)

That is to say, if the movements are symmetric (or antisymmetric), so are the constraints.