3. Calcul de la matrice de rigidité

3.1. Cas général

Soient \(\mathrm{u}\) et \(\varepsilon\) deux champs cinématiquement admissibles quelconques. En appliquant le principe des travaux virtuels à l’élément de volume \(v\) , on peut écrire:

\(\underset{v}{\int}({}^{t}\text{}\delta \varepsilon .\mathrm{s})\text{dv}=\underset{v}{\int}({}^{t}\text{}\delta \mathrm{u}.f)\text{dv}\)

Après décomposition en série de Fourier et intégration par rapport à \(\theta\) , on obtient, pour des champs \({\varepsilon}_{l}^{s},{\varepsilon}_{l}^{a},{u}_{l}^{s},{u}_{l}^{a}\) C.A. quelconques et pour toute harmonique \(l\) :

\(\underset{{s}_{l}}{\int}({}^{t}\text{}\delta {\varepsilon}_{l}^{s}.{\sigma}_{l}^{s}+{}^{t}\text{}\delta {\varepsilon}_{l}^{a}.{\sigma}_{l}^{a}){\text{ds}}_{l}=\underset{{s}_{l}}{\int}({}^{t}\text{}\delta {u}_{l}^{s}.{f}_{l}^{s}+{}^{t}\text{}\delta {u}_{l}^{a}.{f}_{l}^{a}){\text{ds}}_{l}\)

Soit, en utilisant [éq 2.3-1] et en posant:

\(\begin{array}{}{K}_{l}^{s}=\underset{{s}_{l}}{\int}{}^{t}\text{}{B}_{l}{D}^{s}{B}_{l}{\text{ds}}_{l}\\ {K}_{l}^{a}=\underset{{s}_{l}}{\int}{}^{t}\text{}{B}_{l}{D}^{s}{B}_{l}{\text{ds}}_{l}={K}_{l}^{s}={K}_{l}\\ {K}_{l}^{\text{as}}=\underset{{s}_{l}}{\int}{}^{t}\text{}{B}_{l}{D}^{a}{B}_{l}{\text{ds}}_{l}\end{array}\)

On obtient le système d’équations suivant:

\(\lbrace \begin{array}{}{K}_{l}{u}_{l}^{s}+{K}_{l}^{\text{as}}{u}_{l}^{a}={f}_{l}^{s}\\ {}^{t}\text{}{K}_{l}^{\text{as}}{u}_{l}^{s}+{K}_{l}{u}_{l}^{a}={f}_{l}^{a}\end{array}\) éq 3-1

où \({}^{t}\text{}{K}_{l}^{\text{as}}=-{K}_{l}^{\text{as}}\) on voit que si \({D}_{a}\ne 0\) , le découplage des modes en harmoniques symétriques et antisymétriques n’est plus possible. Par contre, si \({D}_{a}=0\) (orthotropie par rapport à \(\mathrm{Oz}\) ) alors \({K}_{l}^{\text{as}}=0\) et [éq 3-1] se réduit à:

\(\lbrace \begin{array}{}{K}_{l}{u}_{l}^{s}={f}_{l}^{s}\\ {K}_{l}{u}_{l}^{a}={f}_{l}^{a}\end{array}\)

En prenant pour vecteurs déplacement (resp. force) correspondant à l’harmonique \(l\) les vecteurs:

\(\begin{array}{}{u}_{l}={\left\lbrace {u}_{r}^{s},{u}_{z}^{s},{u}_{\theta}^{s},{u}_{r}^{a},{u}_{z}^{a},{u}_{\theta}^{a}\right\rbrace }_{l}\\ {f}_{l}={\left\lbrace {f}_{r}^{s},{f}_{z}^{s},{f}_{\theta}^{s},{f}_{r}^{a},{f}_{z}^{a},{f}_{\theta}^{a}\right\rbrace }_{l}\end{array}\)

On a alors:

\({K}_{l}^{g}{u}_{l}={f}_{l}\text{avec}{K}_{l}^{g}=(\begin{array}{cc}{K}_{l}& {K}_{l}^{\text{as}}\\ {}^{t}\text{}{K}_{l}^{\text{as}}& {K}_{l}\end{array})\)

3.2. Calcul de :math:`{K}_{l}^{g}`dans le cas isotrope

Dans ce cas on a donc \({K}_{l}^{\text{as}}=0\) . On détaille dans la suite le calcul de \({K}_{l}={}_{{s}_{l}}\text{}\int{}^{t}\text{}{B}_{l}{D}^{s}{B}_{l}{\text{ds}}_{l}\)

Dans la cas isotrope, on a:

\(D={D}^{s}=\left[\begin{array}{cc}\begin{array}{cccc}\mathrm{D1}& \mathrm{D2}& \mathrm{D2}& 0\\ \mathrm{D2}& \mathrm{D1}& \mathrm{D2}& 0\\ \mathrm{D2}& \mathrm{D2}& \mathrm{D1}& 0\\ 0& 0& 0& \mathrm{D3}\end{array}& 0\\ 0& \begin{array}{cc}\mathrm{D3}& 0\\ 0& \mathrm{D3}\end{array}\end{array}\right]\)

\(\begin{array}{cc}\text{avec}& \mathrm{D1}=\frac{E(1-\nu )}{(1+\nu )(1-2\nu )}\\ & \mathrm{D2}=\frac{E\nu }{(1+\nu )(1-2u)}\\ & \mathrm{D3}=\frac{E}{2(1+\nu )}\end{array}\)

On peut écrire:

\(\left\lbrace \begin{array}{}{\varepsilon}_{r}\\ {\varepsilon}_{z}\\ {\varepsilon}_{\theta}\\ {\gamma}_{\text{rz}}\\ {\gamma}_{r\theta }\\ {\gamma}_{z\theta }\end{array}\right\rbrace ={B}_{l}\left\lbrace \begin{array}{}{u}_{r}^{s}\\ {u}_{z}^{s}\\ {u}_{\theta}^{s}\end{array}\right\rbrace ={B}_{l}^{'}{}^{t}\text{}\left\lbrace \frac{{u}_{r}}{r},\frac{{u}_{z}}{r},\frac{{u}_{\theta}}{r},\frac{\partial {u}_{r}}{\partial r},\frac{\partial {u}_{z}}{\partial r},\frac{\partial {u}_{\theta}}{\partial r},\frac{\partial {u}_{r}}{\partial z},\frac{\partial {u}_{z}}{\partial z},\frac{\partial {u}_{\theta}}{\partial z}\right\rbrace\)

← fcts de forme → ← dérivées des fcts de forme →

\(\text{avec}B{'}_{l}=\left[\begin{array}{ccccccccc}0& 0& 0\text{}& 1& 0& 0\text{}& 0& 0& 0\\ 0& 0& 0\text{}& 0& 0& 0\text{}& 0& 1& 0\\ 1& 0& -l\text{}& 0& 0& 0\text{}& 0& 0& 0\\ 0& 0& 0\text{}& 0& 1& 0\text{}& 1& 0& 0\\ l& 0& -1\text{}& 0& 0& 1\text{}& 0& 0& 0\\ 0& l& 0\text{}& 0& 0& 0\text{}& 0& 0& 1\end{array}\right]\)

En désignant par \({\left\lbrace {W}_{J}\right\rbrace }_{J=1\text{à ˆn}}\) les fonctions de forme de l’élément considéré, on a:

\(U=\left[\begin{array}{}\frac{{u}_{r}}{r}\\ \frac{{u}_{z}}{r}\\ \frac{{u}_{\theta}}{r}\\ \frac{\partial {u}_{r}}{\partial r}\\ \frac{\partial {u}_{z}}{\partial r}\\ \frac{\partial {u}_{\theta}}{\partial r}\\ \frac{\partial {u}_{r}}{\partial z}\\ \frac{\partial {u}_{z}}{\partial z}\\ \frac{\partial {u}_{\theta}}{\partial z}\end{array}\right]=\left[\begin{array}{ccccc}\cdots & \frac{{W}_{J}}{r}& \stackrel{\text{noeud}J}{\stackrel{}{0}}& 0& \cdots \\ \cdots & 0& \frac{{W}_{J}}{r}& 0& \cdots \\ \cdots & 0& 0& \frac{{W}_{J}}{r}& \cdots \\ \cdots & \frac{\partial {W}_{J}}{\partial r}& 0& 0& \cdots \\ \cdots & 0& \frac{\partial {W}_{J}}{\partial r}& 0& \cdots \\ \cdots & 0& 0& \frac{\partial {W}_{J}}{\partial r}& \cdots \\ \cdots & \frac{\partial {W}_{J}}{\partial z}& 0& 0& \cdots \\ \cdots & 0& \frac{\partial {W}_{J}}{\partial z}& 0& \cdots \\ \cdots & 0& \underset{\text{bloc}{P}_{J}}{\underset{\underbrace{}}{0}}& \frac{\partial {W}_{J}}{\partial z}& \cdots \end{array}\right]\left\lbrace \begin{array}{}\cdot \\ \cdot \\ \cdot \\ {u}_{r}(J)\\ {u}_{z}(J)\\ {u}_{\theta}(J)\\ \cdot \\ \cdot \\ \cdot \end{array}\right\rbrace\)

On note \((P)=({P}_{1},\dots ,{P}_{N})\) où \(N\) est le nombre de nœuds de l’élément.

Alors \({K}_{l}={\int}_{{s}_{l}}{}^{t}\text{}P{}^{t}\text{}{B}_{l}^{'}{\mathrm{DB}}_{l}^{'}P{\text{ds}}_{l}\)

\({K}_{l}\) est symétrique et formée de blocs \({({K}_{l})}^{I,J}3\times 3\) :

\({({K}_{l})}^{I,J}={\int}_{{s}_{l}}{}^{t}\text{}{P}_{I}{}^{t}\text{}{B}_{l}^{'}{\mathrm{DB}}_{l}^{'}{P}_{J}{\text{ds}}_{l}\)

Le calcul des blocs \({({K}_{l})}^{I,J}\) est explicité ci-dessous:

\(\begin{array}{}{\text{tB}}_{l}^{'}{\mathrm{DB}}_{l}^{'}=\left[\begin{array}{ccccccccc}\mathrm{D1}+{1}^{2}\mathrm{D3}& 0& -l(\mathrm{D1}+\mathrm{D3})& \mathrm{D2}& 0& \text{lD}3& 0& \mathrm{D2}& 0\\ 0& {l}^{2}\mathrm{D3}& 0& 0& 0& 0& 0& 0& \text{lD}3\\ -l(\mathrm{D1}+\mathrm{D3})& 0& {l}^{2}\mathrm{D1}+\mathrm{D3}& -\text{lD}2& 0& -\mathrm{D3}& 0& -\text{lD}2& 0\\ \mathrm{D2}& 0& -\text{lD}2& \mathrm{D1}& 0& 0& 0& \mathrm{D2}& 0\\ 0& 0& 0& 0& \mathrm{D3}& 0& \mathrm{D3}& 0& 0\\ \text{lD}3& 0& -\mathrm{D3}& 0& 0& \mathrm{D3}& 0& 0& 0\\ 0& 0& 0& 0& \mathrm{D3}& 0& \mathrm{D3}& 0& 0\\ \mathrm{D2}& 0& -\text{lD}2& \mathrm{D2}& 0& 0& 0& \mathrm{D1}& 0\\ 0& \text{lD}3& 0& 0& 0& 0& 0& 0& \mathrm{D3}\end{array}\right]\\ {}^{t}\text{}{P}_{I}{}^{t}\text{}{B}_{l}^{'}{\mathrm{DB}}_{l}^{'}{P}_{J}=({K}_{ij}^{l})\begin{array}{c}I,J\\ 3\times 3\end{array}=\left[\begin{array}{ccc}{K}_{11}^{I,J}& {K}_{12}^{I,J}& {K}_{13}^{I,J}\\ {K}_{21}^{I,J}& {K}_{22}^{I,J}& {K}_{23}^{I,J}\\ {K}_{31}^{I,J}& {K}_{32}^{I,J}& {K}_{33}^{I,J}\end{array}\right]\text{avec}\\ \lbrace \begin{array}{}{K}_{11}^{I,J}=(\frac{\mathrm{D1}+{l}^{2}\mathrm{D3}}{{r}^{2}}){W}_{I}{W}_{J}+\mathrm{D1}\frac{\partial {W}_{I}}{\partial r}\frac{\partial {W}_{J}}{\partial r}+\mathrm{D3}\frac{\partial {W}_{I}}{\partial z}\frac{\partial {W}_{J}}{\partial z}+\frac{\mathrm{D2}}{r}({W}_{I}\frac{\partial {W}_{J}}{\partial r}+{W}_{J}\frac{\partial {W}_{I}}{\partial r})\\ {K}_{22}^{I,J}=(\frac{{l}^{2}\mathrm{D3}}{{r}^{2}}){W}_{I}{W}_{J}+\mathrm{D3}\frac{\partial {W}_{I}}{\partial r}\frac{\partial {W}_{J}}{\partial r}+\mathrm{D1}\frac{\partial {W}_{I}}{\partial z}\frac{\partial {W}_{J}}{\partial z}\\ {K}_{33}^{I,J}=(\frac{{l}^{2}\mathrm{D1}+\mathrm{D3}}{{r}^{2}}){W}_{I}{W}_{J}+\mathrm{D3}(\frac{\partial {W}_{I}}{\partial r}\frac{\partial {W}_{J}}{\partial r}+\frac{\partial {W}_{I}}{\partial z}\frac{\partial {W}_{J}}{\partial z})-\frac{\mathrm{D3}}{r}({W}_{I}\frac{\partial {W}_{J}}{\partial r}+\frac{\partial {W}_{I}}{\partial r}{W}_{J})\\ {K}_{12}^{I,J}=\mathrm{D2}\frac{\partial {W}_{I}}{\partial r}\frac{\partial {W}_{J}}{\partial z}+\mathrm{D3}\frac{\partial {W}_{I}}{\partial z}\frac{\partial {W}_{J}}{\partial r}+\frac{\mathrm{D2}}{r}{W}_{I}\frac{\partial {W}_{J}}{\partial z}\\ {K}_{21}^{I,J}=\mathrm{D3}\frac{\partial {W}_{I}}{\partial r}\frac{\partial {W}_{J}}{\partial z}+\mathrm{D2}\frac{\partial {W}_{I}}{\partial z}\frac{\partial {W}_{J}}{\partial r}+\frac{\mathrm{D2}}{r}{W}_{J}\frac{\partial {W}_{I}}{\partial z}\\ {K}_{13}^{I,J}=-\frac{l}{{r}^{2}}(\mathrm{D1}+\mathrm{D3}){W}_{I}{W}_{J}-\frac{l}{r}{\mathrm{D2W}}_{J}\frac{\partial {W}_{I}}{\partial r}+\frac{l}{r}{\mathrm{D3W}}_{I}\frac{\partial {W}_{J}}{\partial r}\\ {K}_{31}^{I,J}=-\frac{l}{{r}^{2}}(\mathrm{D1}+\mathrm{D3}){W}_{I}{W}_{J}-\frac{l}{r}{\mathrm{D2W}}_{I}\frac{\partial {W}_{J}}{\partial r}+\frac{l}{r}{\mathrm{D3W}}_{J}\frac{\partial {W}_{I}}{\partial r}\\ {K}_{23}^{I,J}=-\frac{l}{r}\mathrm{D2}\frac{\partial {W}_{I}}{\partial z}{W}_{J}+\frac{l}{r}{\mathrm{D3W}}_{I}\frac{\partial {W}_{J}}{\partial z}\\ {K}_{32}^{I,J}=-\frac{l}{r}{\mathrm{D2W}}_{I}\frac{\partial {W}_{J}}{\partial z}+\frac{l}{r}\mathrm{D3}\frac{\partial {W}_{I}}{\partial z}{W}_{J}\end{array}\end{array}\)

Les blocs \({K}^{I,J}\) ne sont pas symétriques sauf pour \(I=J\) (sur la diagonale de \(K\) ). On remarque en fait que les blocs \({K}^{I,J}\) peuvent s’écrire pour tout harmonique (\(l=0\) compris).

\(\lbrace \begin{array}{}{K}_{11}^{I,J}={\mathrm{K0}}_{11}^{I,J}+{l}^{2}\frac{\mathrm{D3}}{{r}^{2}}{W}_{I}{W}_{J}\\ {K}_{22}^{I,J}={\mathrm{K0}}_{22}^{I,J}+{l}^{2}\frac{\mathrm{D3}}{{r}^{2}}{W}_{I}{W}_{J}\\ {K}_{33}^{I,J}={\mathrm{K0}}_{33}^{I,J}+{l}^{2}\frac{\mathrm{D1}}{{r}^{2}}{W}_{I}{W}_{J}\\ {K}_{12}^{I,J}={\mathrm{K0}}_{12}^{I,J}\\ {K}_{21}^{I,J}={\mathrm{K0}}_{21}^{I,J}\\ {K}_{13}^{I,J}=-{\mathrm{lK0}}_{13}^{I,J}\\ {K}_{31}^{I,J}=-{\mathrm{lK0}}_{31}^{I,J}\\ {K}_{23}^{I,J}=-{\mathrm{lK0}}_{23}^{I,J}\\ {K}_{32}^{I,J}=-{\mathrm{lK0}}_{32}^{I,J}\end{array}\)

où les blocs \({\mathrm{K0}}^{I,J}\) sont indépendants de l’harmonique \(l\) .