Anisotropic Fourier analysis
=============================




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:


.. csv-table::

    "1", "radial component next :math:`r`"
    "2", "axial component next :math:`z`"
    "3", "tangential component next :math:`\theta`"


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


.. image:: images/10000ACC000069BB00007BB781EFBDEDAC3DE5CC.svg
    :width: 218
    :height: 255


.. _RefImage_10000ACC000069BB00007BB781EFBDEDAC3DE5CC.svg:

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


.. image:: images/10000B14000069D500005A0565C67827BF00BDBF.svg
    :width: 218
    :height: 255


.. _RefImage_10000B14000069D500005A0565C67827BF00BDBF.svg:

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

We get:

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

:math:`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 :math:`\varepsilon`, we see that :math:`\varepsilon` can be decomposed into the following Fourier series:


.. _RefEquation 2.1-2:

:math:`\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 :math:`\varepsilon =\left\{{\varepsilon }_{r},{\varepsilon }_{z},{\varepsilon }_{q},{\gamma }_{\text{rz}},{\gamma }_{\mathrm{rq}},{\gamma }_{\mathrm{zq}}\right\}`

:math:`{\varepsilon }_{l}^{s}={B}_{l}^{s}{U}_{l}^{s}\text{}{\varepsilon }_{l}^{a}={B}_{l}^{a}{U}_{l}^{a}`

with (see [:ref:`bib1 <bib1>`]):

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




Coupling and decoupling symmetric and antisymmetric modes
----------------------------------------------------------------

Using the previous notations, we have:

:math:`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 :math:`{M}_{l}^{s}\text{et}{M}_{l}^{a}`:

:math:`\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: :math:`{\varepsilon }_{l}=M{\text{'}}_{l}^{s}{\varepsilon }_{l}^{s}+M{\text{'}}_{l}^{a}{\varepsilon }_{l}^{a}`

:math:`\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**

:math:`\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 :math:`\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 :math:`({}^{t}\text{}{\varepsilon }_{l}^{a},{\varepsilon }_{l}^{s})\text{ou}({}^{t}\text{}{\varepsilon }_{l}^{s},{\varepsilon }_{l}^{a})` term in :math:`W`. There is then no :math:`({U}^{a},{U}^{s})\text{ou}({U}^{s},{U}^{a})` coupling.




Stress calculation
----------------------

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

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

From Hooke's law :math:`\sigma =D\varepsilon`, we deduce:

:math:`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 :math:`M{\text{'}}_{l}^{s}\text{et}M{\text{'}}_{l}^{a}` matrices appear:

:math:`\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 :math:`{D}^{s}(\begin{array}{cc}{D}_{1}& {0}_{\mathrm{4,2}}\\ {0}_{\mathrm{2,4}}& {D}_{2}\end{array})` and :math:`{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 :math:`l`:


.. _RefEquation 2.3-1:

:math:`\{\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* :math:`\mathrm{Oz}` *, we have* :math:`{D}^{a}=0` and [:ref:`éq 2.1-1 <éq 2.1-1>`] is reduced to:

:math:`\{\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.*