7. Eigenmodes and Lagrange parameters

7.1. Introduction

This paragraph is intended to answer the following two questions:

Q1:	What is the reduced system of eigenvalues (and vectors) to solve when a mechanical model is subject to linear homogenous kinematic constraints?
Q2:	What is the equivalent or previous dualized model (with Lagrange parameters)?

7.2. Mechanical problem to be solved

We assume a mechanical system that is already discretized by finite elements.

Nodal unknowns are noted \(U=\left\{{u}_{i}\right\}(i=\mathrm{1,}n)\).

Nodal displacements are not all independent: there are \(p(\text{<}n)\) homogeneous linear relationships between these displacements: \({B}_{j}(U)=0(j=\mathrm{1,}p)\).

These linear relationships are independent of each other, i.e. the rank of the matrix \(B\) containing the coefficients of the \(p\) linear relationships is \(p\).

Let \(K\) be the stiffness matrix of the mechanical system without constraints.

Let \(M\) be the mass matrix of the mechanical system without constraints.

What is the eigenvalue system to solve to find the eigenmodes of the constrained structure?

7.3. Reduced system

Note that if we write linear kinematic relationships in the form:

\(\mathrm{BU}=0\) eq 7.3-1

where:	\(B\) is a \(p\times n\) matrix
	\(U\) is the vector of the nodal unknowns \(\in {ℝ}^{n}\)

so:

\(B\dot{U}=0\) eq 7.3-2

and the relationship is also valid for speeds.

Also, if \(B\) is of rank \(p\), then there is a square submatrix of \(B\) of rank \(p\). Let’s denote this sub-matrix \({B}_{1}\).

Let’s then make a partition of the unknowns of \(U\) in \({U}_{1}\) and \({U}_{2}\) such as:

\(\begin{array}{}\mathrm{BU}=0\text{}\iff \text{}\left[\begin{array}{cc}{B}_{1}& {B}_{2}\end{array}\right]\left[\begin{array}{c}{U}_{1}\\ {U}_{2}\end{array}\right]=0\\ \text{}{U}_{1}\in {ℝ}^{p}\\ \text{}{U}_{2}\in {ℝ}^{n-p}\\ \text{}{B}_{1}=\text{matrice}p\times p\\ \text{}{B}_{2}=\text{matrice}p\times (n-p)\end{array}\)

Linear relationships can then be written as:

\({B}_{1}{U}_{1}+{B}_{2}{U}_{2}=0\)

This allows the unknowns \({U}_{1}\) to be expressed as a function of \({U}_{2}\) since \({B}_{1}\) is invertible.

\({U}_{1}=-{B}_{1}^{-1}{B}_{2}{U}_{2}\) eq 7.3-3

Reduced stiffness matrix:

The elastic deformation energy of the unstressed discretized structure is \({W}_{\text{def}}=\frac{1}{2}{U}^{T}\mathrm{KU}\). If we partition matrix \(K\) in the same way we partitioned \(U\), we get:

\(K=\left[\begin{array}{cc}{K}_{1}& {K}_{\text{12}}\\ {K}_{\text{12}}^{T}& {K}_{2}\end{array}\right]\)

so:

\({\mathrm{2W}}_{\text{def}}={U}_{1}^{T}{K}_{1}{U}_{1}+{U}_{2}^{T}{K}_{2}{U}_{2}+{U}_{2}^{T}{K}_{\text{12}}{U}_{1}+{U}_{1}^{T}{K}_{\text{12}}^{T}{U}_{2}\)

Let’s then introduce linear constraints [éq 7.3-3]:

\(\begin{array}{}{\mathrm{2W}}_{\text{def}}={U}_{2}^{T}{B}_{2}^{T}{B}_{1}^{-T}{K}_{1}{B}_{1}^{-1}{B}_{2}{U}_{2}+{U}_{2}^{T}{K}_{2}{U}_{2}\\ \text{}-{U}_{2}^{T}{K}_{\text{12}}{B}_{1}^{-1}{B}_{2}{U}_{2}-{U}_{2}^{T}{B}_{2}^{T}{B}_{1}^{-T}{K}_{\text{12}}^{T}{U}_{2}\\ \text{}={U}_{2}^{T}{\tilde{K}}_{2}{U}_{2}\end{array}\)

with:

\({\tilde{K}}_{2}={K}_{2}+{B}_{2}^{T}{B}_{1}^{-T}{K}_{1}{B}_{1}^{-1}{B}_{2}-{K}_{\text{12}}{B}_{1}^{-1}{B}_{2}-{B}_{2}^{T}{B}_{1}^{-T}{K}_{\text{12}}^{T}\) eq 7.3-4

We can therefore see that we have expressed the deformation energy of the reduced structure as a bilinear form of \({U}_{2}\). The nodal unknowns of \({U}_{1}\) have been removed. Nodal unknowns \({U}_{2}\) are no longer constrainted.

Reduced mass matrix:

Let’s adopt the same partition for mass matrix \(M\). We can write the relationship [éq 7.3-2]:

\({B}_{1}{\dot{U}}_{1}+{B}_{2}{\dot{U}}_{2}=0\)

The same calculation as before then leads us to:

\({\mathrm{2W}}_{\text{cin}}={\dot{U}}_{2}^{T}{\tilde{M}}_{2}{\dot{U}}_{2}\)

with:

\({\tilde{M}}_{2}={M}_{2}+{B}_{2}^{T}{B}_{1}^{-T}{M}_{1}{B}_{1}^{-1}{B}_{2}-{M}_{\text{12}}{B}_{1}^{-1}{B}_{2}-{B}_{2}^{T}{B}_{1}^{-T}{M}_{\text{12}}^{T}\) eq 7.3-5

Conclusion:

The system to be solved in order to find the natural modes (and frequencies) of the constrained structure is:

Find \((n\mathrm{-}p)({\mathrm{X}}_{\mathrm{i}}\text{,}{\omega }_{\mathrm{i}}^{2})\mathrm{\in }{\mathrm{ℝ}}^{n\mathrm{-}p}\mathrm{\times }\mathrm{ℝ}\) such as \(({\tilde{\mathrm{K}}}_{2}\mathrm{-}{\omega }_{\mathrm{i}}^{2}{\tilde{\mathrm{M}}}_{2}){\mathrm{X}}_{\mathrm{i}}\mathrm{=}0\) with \({\tilde{K}}_{2}\) and \({\tilde{M}}_{2}\) defined by [éq 7.3-4] and [éq 7.3-5].

Application to blocked degrees of freedom:

In this case:

\(\{\begin{array}{}{B}_{1}=I\\ {B}_{2}=0\end{array}\)

from where: \({\tilde{K}}_{2}={K}_{2}\) and \({\tilde{M}}_{2}={M}_{2}\)

that is, all you have to do is « forget » in \(K\) and \(M\) the rows and columns corresponding to the blocked degrees of freedom.

7.4. Dualized system

We saw in [§5] that taking into account Lagrange coefficients (double) in a \(A\) matrix led to the \(A\) matrix:

\(A\text{'}=\left[\begin{array}{ccc}A& \beta {B}^{T}& \beta {B}^{T}\\ \beta B& -\alpha I& +\alpha I\\ \beta B& \alpha I& -\alpha I\end{array}\right]\)

where:

• \(B\) is the matrix of kinematic conditions: \(\mathrm{BU}=0\),

• \(\alpha \in {ℝ}^{+}\text{arbitraire}\alpha \ne 0\),

• \(\beta \in ℝ\text{arbitraire}\beta \ne 0\),

Let’s apply the dualization of boundary conditions to matrices \(K\) and \(M\), by partitioning the degrees of freedom in \({X}_{1}\), \({X}_{2}\) as in [§7.3]. We get the following eigenvalue problem:

\((S)\left\{\left[\begin{array}{cccc}{K}_{1}& {K}_{\text{12}}& {\beta }_{k}{B}_{1}^{T}& {\beta }_{k}{B}_{1}^{T}\\ {K}_{\text{12}}& {K}_{2}& {\beta }_{k}{B}_{2}^{T}& {\beta }_{k}{B}_{2}^{T}\\ {\beta }_{k}{B}_{1}& {\beta }_{k}{B}_{2}& -{\alpha }_{k}I& +{\alpha }_{k}I\\ {\beta }_{k}{B}_{1}& {\beta }_{k}{B}_{2}& {\alpha }_{k}I& -{\alpha }_{k}I\end{array}\right]-{\omega }^{2}\left[\begin{array}{cccc}{M}_{1}& {M}_{\text{12}}& {\beta }_{m}{B}_{1}^{T}& {\beta }_{m}{B}_{1}^{T}\\ {M}_{\text{12}}& {M}_{2}& {\beta }_{m}{B}_{2}^{T}& {\beta }_{m}{B}_{2}^{T}\\ {\beta }_{m}{B}_{1}& {\beta }_{m}{B}_{2}& -{\alpha }_{m}I& {\alpha }_{m}I\\ {\beta }_{m}{B}_{1}& {\beta }_{m}{B}_{2}& {\alpha }_{m}I& -{\alpha }_{m}I\end{array}\right]\right\}X=0\)

for a natural pulsation \(\omega\) and an eigenvector \(X\) of this system, we can write:

\({\stackrel{ˉ}{K}}_{1}{X}_{1}+{\stackrel{ˉ}{K}}_{\text{12}}{X}_{2}+\beta {B}_{1}^{T}({\lambda }_{1}+{\lambda }_{2})=0\) eq 7.4-1

\({\stackrel{ˉ}{K}}_{\text{12}}{X}_{1}+{\stackrel{ˉ}{K}}_{2}{X}_{2}+\beta {B}_{2}^{T}({\lambda }_{1}+{\lambda }_{2})=0\) eq 7.4-2

\(\beta ({B}_{1}{X}_{1}+{B}_{2}{X}_{2})-\alpha ({\lambda }_{1}-{\lambda }_{2})=0\) eq 7.4-3

\(\beta ({B}_{1}{X}_{1}+{B}_{2}{X}_{2})+\alpha ({\lambda }_{1}-{\lambda }_{2})=0\) eq 7.4-4

with: \({\stackrel{ˉ}{K}}_{i}={K}_{i}-{\omega }^{2}{M}_{i}\text{;}\beta ={\beta }_{k}-{\omega }^{2}{\beta }_{m}\text{;}\alpha ={\alpha }_{k}-{\omega }^{2}{\alpha }_{m}\)

System \((S)\) is of order \((n+2p)\) if \(\{\begin{array}{cc}n& \text{= nombre de ddls physiques}\\ p& \text{= nombre de relations cinŽmatiques}\end{array}\).

The characteristic polynomial in \({\omega }^{2}\) is a priori of degree \(n+2p\). Its highest degree term is: \(\prod _{i=1}^{n}(-{m}_{i})\cdot {(+{\alpha }_{m})}^{2p}\) if \({m}_{i}\) is the \({i}^{\mathit{ème}}\) diagonal term of \(M\).

we therefore see**that if* \({\alpha }_{m}\ne 0\), the term with the highest degree is \(\ne 0\) (because the \({m}_{i}\) are \(>0\)) and therefore the dualized system \((S)\) has the most eigenvalues than the reduced system: \((n-p)\). The two systems are therefore not equivalent. This is what we see on the example of [§7.5],

• let’s choose \({\alpha }_{n}={\beta }_{n}=0\):

[éq 7.4-3] and [éq 7.4-4] \(\mathrm{\Rightarrow }\)

\(\{\begin{array}{}{\lambda }_{1}={\lambda }_{2}\\ {X}_{1}=-{B}_{1}^{-1}{B}_{2}{X}_{2}\end{array}\)

[éq 7.4-1] \(\mathrm{\Rightarrow }\)

\({\lambda }_{1}={\lambda }_{2}=\frac{-1}{\mathrm{2\beta }}{B}_{1}^{-T}(-{\stackrel{ˉ}{K}}_{1}{B}_{1}^{-1}{B}_{2}+{\stackrel{ˉ}{K}}_{\text{12}}){X}_{2}\)

[éq 7.4-2] \(\mathrm{\Rightarrow }\)

\(-{\stackrel{ˉ}{K}}_{\text{12}}{B}_{1}^{-1}{B}_{2}{X}_{2}+{\stackrel{ˉ}{K}}_{2}{X}_{2}-{B}_{2}^{T}{B}_{1}^{-T}(-{\stackrel{ˉ}{K}}_{1}{B}_{1}^{-1}{B}_{2}+{\stackrel{ˉ}{K}}_{\text{12}}){X}_{2}=0\)

\(\iff ({\tilde{K}}_{2}-{\omega }^{2}{\tilde{M}}_{2}){X}_{2}=0\)

with:

\(\{\begin{array}{}{\tilde{K}}_{2}=-{K}_{\text{12}}^{T}{B}_{1}^{-1}{B}_{2}+{K}_{2}+{B}_{2}^{T}{B}_{1}^{-T}{K}_{1}{B}_{1}^{-1}{B}_{2}-{B}_{2}^{T}{B}_{1}^{-T}{K}_{\text{12}}\\ {\tilde{M}}_{2}=-{M}_{\text{12}}^{T}{B}_{1}^{-1}{B}_{2}+{M}_{2}+{B}_{2}^{T}{B}_{1}^{-T}{M}_{1}{B}_{1}^{-1}{B}_{2}-{B}_{2}^{T}{B}_{1}^{-T}{M}_{\text{12}}\end{array}\)

It can be seen that the definitions of \({\tilde{K}}_{2}\) and \({\tilde{M}}_{2}\) are identical to those of equations [éq7.3‑4] and [éq7.3-5].

We can therefore see that any eigenvector \(X\) of the dualized system is also eigenvector of the reduced system (with the same natural pulsation) if it is projected onto space \({U}_{2}\).

Conversely, any eigenvector \({X}_{2}\) of the reduced problem can be extended into an eigenvector of the dualized system \({X}^{T}=\left[{X}_{1}^{T},{X}_{2}^{T},{\lambda }_{1}^{T},{\lambda }_{2}^{T}\right]\).

with:

\(\{\begin{array}{}{X}_{1}=-{B}_{1}^{-1}{B}_{2}{X}_{2}\\ {\lambda }_{1}=-\frac{1}{\mathrm{2\beta }}{B}_{1}^{-T}(-{\stackrel{ˉ}{K}}_{1}{B}_{1}^{-1}{B}_{2}+{\stackrel{ˉ}{K}}_{\text{12}}){X}_{2}\\ {\lambda }_{2}=-\frac{1}{\mathrm{2\beta }}{B}_{1}^{-T}(-{\stackrel{ˉ}{K}}_{1}{B}_{1}^{-1}{B}_{2}+{\stackrel{ˉ}{K}}_{\text{12}}){X}_{2}\end{array}\)

The two systems are therefore equivalent, they have the same eigenmodes and the same eigenvalues.

The dualized system, although larger in size than the reduced system, does not have more eigenvalues than the reduced system (the dimension of the eigenspace is the same).

Conclusion:

The dualized system is equivalent to the reduced system as soon as you choose \({\alpha }_{m}={\beta }_{m}=0\) , that is, if you take the dualized stiffness matrix but don’t change the mass matrix. That’s what’s done in Aster.

7.5. example

Let the system be:

\(K=\left[\begin{array}{cc}k& -k\\ -k& k\end{array}\right]\text{}M=\left[\begin{array}{cc}m& 0\\ 0& m\end{array}\right]\)

we add the \(\alpha {u}_{1}+\beta {u}_{2}=0(\alpha \ne 0)\) constraint; that is \(\gamma =\frac{\beta }{\alpha }\).

The reduced system is then:

• \({K}_{1}=k;{K}_{2}=k;{K}_{\text{12}}\text{=-}k;{M}_{1}=m;{M}_{2}=m;{M}_{\text{12}}=0\)

• \({B}_{1}=\alpha ;{B}_{2}=\beta\)

\(\Rightarrow {\tilde{K}}_{2}=k{(1+\gamma )}^{2};{\tilde{M}}_{2}=m(1+{\gamma }^{2})\)

\(\Rightarrow {\omega }^{2}=\frac{k}{m}\frac{{(1+\gamma )}^{2}}{1+{\gamma }^{2}};{X}_{2}=1\)

We can see that the eigenvalue \({\omega }^{2}\) depends on the \(\gamma \mathrm{=}\frac{\beta }{\alpha }\) ratio.

If \(\gamma =0\text{,}{\omega }^{2}=\frac{k}{m}\)

If \(\gamma =\mathrm{1,}{\omega }^{2}=\frac{2k}{m}\)

If \(g\to \mathrm{\infty },{\omega }^{2}\to \frac{k}{m}\)

Let’s choose \((\alpha =\beta =1)\) to simplify and write the dualized system:

\(\left\{\left[\begin{array}{cccc}k& -k& {\beta }_{k}& {\beta }_{k}\\ -k& k& {\beta }_{k}& {\beta }_{k}\\ {\beta }_{k}& {\beta }_{k}& -{\alpha }_{k}& {\alpha }_{k}\\ {\beta }_{k}& {\beta }_{k}& {\alpha }_{k}& -{\alpha }_{k}\end{array}\right]-\lambda \left[\begin{array}{cccc}m& 0& {\beta }_{m}& {\beta }_{m}\\ 0& m& {\beta }_{m}& {\beta }_{m}\\ {\beta }_{m}& {\beta }_{m}& -{\alpha }_{m}& {\alpha }_{m}\\ {\beta }_{m}& {\beta }_{m}& {\alpha }_{m}& -{\alpha }_{m}\end{array}\right]\right\}\left\{\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {\lambda }_{1}\\ {\lambda }_{2}\end{array}\right\}=0\)

the eigenvalues of this system are:

\({\omega }^{2}=\left\{\frac{{\beta }_{k}}{{\beta }_{m}},\frac{{\beta }_{k}}{{\beta }_{m}},\frac{{\beta }_{k}}{{\beta }_{m}},\frac{2k}{m}\right\}\)

We note that we find the real eigenvalue (the fourth), but that we find 3 parasitic eigenvalues due to the non-nullity of \({\alpha }_{m}\) and \({\beta }_{m}\).

If we choose \({\alpha }_{m}={\beta }_{m}=0\), the calculation shows that the characteristic polynomial is of degree 1 and that its only solution is:

\(\{\begin{array}{}{\omega }^{2}=\frac{2k}{m}\\ X=\text{{}-1\text{,}+1\text{}}\end{array}\)

which is the solution sought.

7.6. Conclusions

• If \(\mathrm{K}\) and \(M\) are the stiffness and mass matrices of an unstressed system.

• If linear constraints can be written as:

\(\left[{\mathrm{B}}_{1}{\mathrm{B}}_{2}\right]\left[\begin{array}{c}{\mathrm{U}}_{1}\\ {\mathrm{U}}_{2}\end{array}\right]\mathrm{=}0\) with \({\mathrm{B}}_{1}\) reversible

• Then the natural modes of the constrained structure are those of the reduced system:

\(({\tilde{K}}_{2}-{\omega }^{2}{\tilde{M}}_{2}){X}_{2}=0\)

with:

\(\{\begin{array}{}{\tilde{K}}_{2}=-{K}_{\text{12}}^{T}{B}_{1}^{-1}{B}_{2}+{K}_{2}+{B}_{2}^{T}{B}_{1}^{-T}{K}_{1}{B}_{1}^{-1}{B}_{2}-{B}_{2}^{T}{B}_{1}^{-T}{K}_{\text{12}}\\ {\tilde{M}}_{2}=-{M}_{\text{12}}^{T}{B}_{1}^{-1}{B}_{2}+{M}_{2}+{B}_{2}^{T}{B}_{1}^{-T}{M}_{1}{B}_{1}^{-1}{B}_{2}-{B}_{2}^{T}{B}_{1}^{-T}{M}_{\text{12}}\end{array}\)

• The dual system (Lagrange doubles) which is written as:

\((\tilde{\tilde{\mathrm{K}}}\mathrm{-}{\omega }^{2}\tilde{\tilde{\mathrm{M}}})\tilde{\tilde{\mathrm{X}}}\mathrm{=}0\)

with:

\(\begin{array}{c}{\tilde{\tilde{\mathrm{X}}}}^{T}\mathrm{=}\left[\begin{array}{cccc}{\mathrm{X}}_{1}& {\mathrm{X}}_{\text{12}}& {\lambda }_{1}& {\lambda }_{2}\end{array}\right]\\ \tilde{\tilde{\mathrm{K}}}\mathrm{=}\left[\begin{array}{cccc}{\mathrm{K}}_{1}& {\mathrm{K}}_{\text{12}}& {\mathrm{B}}_{1}^{\mathrm{T}}& {\mathrm{B}}_{1}^{\mathrm{T}}\\ {\mathrm{K}}_{\text{12}}^{\mathrm{T}}& {\mathrm{K}}_{2}& {\mathrm{B}}_{2}^{\mathrm{T}}& {\mathrm{B}}_{2}^{\mathrm{T}}\\ {\mathrm{B}}_{1}& {\mathrm{B}}_{2}& \mathrm{-}\mathrm{I}& \mathrm{I}\\ {\mathrm{B}}_{1}& {\mathrm{B}}_{2}& \mathrm{I}& \mathrm{-}\mathrm{I}\end{array}\right]\\ \tilde{\tilde{\mathrm{M}}}\mathrm{=}\left[\begin{array}{cccc}{\mathrm{M}}_{1}& {\mathrm{M}}_{\text{12}}& 0& 0\\ {\mathrm{M}}_{\text{12}}^{\mathrm{T}}& {\mathrm{M}}_{2}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\end{array}\)

then has the same solutions as the reduced system.