3. Calculation of the stiffness matrix

3.1. General case

Let \(\mathrm{u}\) and \(\varepsilon\) be any two kinematically admissible fields. Applying the principle of virtual work to volume element \(v\), we can write:

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

After decomposition into a Fourier series and integration with respect to \(\theta\), we obtain, for any \({\varepsilon }_{l}^{s},{\varepsilon }_{l}^{a},{u}_{l}^{s},{u}_{l}^{a}\) AC fields and for any harmonic \(l\):

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

Or, using [éq 2.3-1] and setting:

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

The following system of equations is obtained:

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

where \({}^{t}\text{}{K}_{l}^{\text{as}}=-{K}_{l}^{\text{as}}\) we see that if \({D}_{a}\ne 0\), the decoupling of modes into symmetric and antisymmetric harmonics is no longer possible. On the other hand, if \({D}_{a}=0\) (orthotropy with respect to \(\mathrm{Oz}\)) then \({K}_{l}^{\text{as}}=0\) and [éq 3-1] are reduced to:

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

By taking as displacement vectors (resp. force) corresponding to harmonic \(l\) the vectors:

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

We then have:

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

3.2. Calculation of \({K}_{l}^{g}\) in the isotropic case

So in this case we have \({K}_{l}^{\text{as}}=0\). The calculation of \({K}_{l}={}_{{s}_{l}}\text{}\int {}^{t}\text{}{B}_{l}{D}^{s}{B}_{l}{\text{ds}}_{l}\) is detailed below

In the isotropic case, we have:

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

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

We can write:

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

← form facts → ← derivatives of form facts →

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

By designating the shape functions of the element in question by \({\left\{{W}_{J}\right\}}_{J=1\text{à ˆn}}\), we have:

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

Note \((P)=({P}_{1},\dots ,{P}_{N})\) where \(N\) is the number of nodes in the element.

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

\({K}_{l}\) is symmetric and made up of blocks \({({K}_{l})}^{I,J}3\times 3\):

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

The calculation of blocks \({({K}_{l})}^{I,J}\) is explained below:

\(\begin{array}{}{\text{tB}}_{l}^{\text{'}}{\mathrm{DB}}_{l}^{\text{'}}=\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}^{\text{'}}{\mathrm{DB}}_{l}^{\text{'}}{P}_{J}=({K}_{\text{ij}}^{l})\begin{array}{c}I,J\\ 3\times 3\end{array}=\left[\begin{array}{ccc}{K}_{\text{11}}^{I,J}& {K}_{\text{12}}^{I,J}& {K}_{\text{13}}^{I,J}\\ {K}_{\text{21}}^{I,J}& {K}_{\text{22}}^{I,J}& {K}_{\text{23}}^{I,J}\\ {K}_{\text{31}}^{I,J}& {K}_{\text{32}}^{I,J}& {K}_{\text{33}}^{I,J}\end{array}\right]\text{avec}\\ \{\begin{array}{}{K}_{\text{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}_{\text{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}_{\text{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}_{\text{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}_{\text{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}_{\text{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}_{\text{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}_{\text{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}_{\text{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}\)

Blocks \({K}^{I,J}\) are not symmetric except for \(I=J\) (on the diagonal of \(K\)). We actually notice that blocks \({K}^{I,J}\) can be written for any harmonic (\(l=0\) included).

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

where blocks \({\mathrm{K0}}^{I,J}\) are independent of harmonic \(l\).