Mechanical Vibrations Chapter 6 Solution Methods for the Eigenvalue Problem

Transcription

1 Mechanical Vibrations Chapter 6 Solution Methods for the Eigenvalue Problem

2 Introduction Equations of dynamic equilibrium eigenvalue problem K x = ω M x The eigensolutions of this problem are written in the following order: 0 ω x 1 (1), ω x (),, ω x n ( n)

3 3 Criteria for selecting the solution method Number of degrees of freedom in the system (n) Class I Class II 1 n n 50 Development of the characteristic equation Jacobi s method, power algorithm Class III Class IV Class V 50 n n 5000 n > 5000 Band character of K and M Inverse iteration method, subspace method, Lanczos method Reduction methods

4 4 Criteria for selecting the solution method Required frequency spectrum Ability to separate close eigenvalues Rate of convergence Computational cost Automatic extraction of rigid-body modes Handling of coupled problems Use within the substructuring context Interested readers may refer to Géradin s book.

5 5 Reduction and substructuring methods The reduction and substructuring methods are often used in industry for two reasons: 1. As only the low frequency range is of interest for mechanical design purposes, it is advantageous to reduce from the start the eigenvalue problem to a smaller dimension.. In the context of large projects, the analysis is divided into several parts (often performed by distinct teams). A separate model is constructed for each part of the system and will be used to reconstruct the whole original model. That is what is called substructuring techniques.

6 6 Industrial examples Stator of a turbojet engine Courtesy of Techspace Aero Automated Transfer Vehicle Courtesy of EADS Space Transportation DOF DOF

7 7 Reduction and substructuring methods Initial problem Reduced problem Kn n M n n Km m Mm m n ~ 10 5, 10 6 m ~ 10, 10 3 Transformation matrix?

8 8 Reduction and substructuring methods The general principle of a method for reducing the size of an eigenvalue problem K x = ω M x consists of building a subspace R of dimension n x m (m < n) so that the solution can be written in the form x = R y ( n 1) ( n m) ( m 1)

9 9 Reduction and substructuring methods Variational problem ( T V ) δ = max max 0 δ 1 x T K x ω x T M x = 0 δ 1 y T K y ω y T M y = 0 K y = ω M y Reduced stiffness and mass matrices K T = R K R and M = R T M R

10 10 Static condensation (Guyan-Irons reduction) The aim of the Guyan s condensation method is to obtain an eigenvalue problem of reduced size without altering too much the low eigenfrequency spectrum of the initial problem. For this purpose, the degrees of freedom are partitioned into n R dynamic (retained) coordinates (with n R << n) and n C condensed coordinates. x xr = xc K RR RC RR RC K = K M CR K CC M CR M CC K M M = The dynamic behaviour of the structure will be described by the retained coordinates only.

11 11 Static condensation (Guyan-Irons reduction) K x = ω M x The equation can be put in the form Kx= F where F= ω M x is the vector of inertia forces. KRR KRC xr FR = K K x F CR CC C C with F C 0 The inertia forces F C may be neglected if the masses affected to the condensed degrees of freedom are equal to zero or negligible. If it is the case, one finds 1 C = CC CR R x K K x

12 1 Static condensation (Guyan-Irons reduction) Thus we can define the transformation matrix R x xr xr I = = 1 = 1 xr = xc KCC KCR xr KCC KCR R x R It follows that the reduced stiffness and mass matrices are given by K RR = R T K R = K RR K RC K 1 CC K CR M RR = R T M R = M RR M RC K 1 CC + K RC K CR K 1 CC K M RC CC K K 1 CC 1 CC M K CR CR

13 13 Static condensation (Guyan-Irons reduction) Remarks The validity of the Guyan s reduction method depends on the extent to which the vector of inertia forces F C is negligible. It can be shown that static condensation always leads to an excess approximation to the eigenvalue spectrum. In computational practice, the reduction matrix R I I = = 1 KCC KCR RCR is computed by solving the static problem (with n R second members) KCC RCR = KCR

14 14 Example: the beam clamped at both ends Two finite element model Three finite element model w Ψ 1 w w 3 Ψ 3 1 Ψ Comparison of the results (no reduction) ω r 4 ml EI Mode n elements DOF 3 elements 4 DOF exact

15 15 Example: the beam clamped at both ends Three finite element model 1 w w 3 Ψ 3 1 Ψ Condensation of the rotational degrees of freedom ω r 4 ml EI Mode n elements DOF 3 elements DOF 3 elements 4 DOF exact

16 16 Example: the beam clamped at both ends Beam clamped at both ends with 100 finite elements (= 396 DOF) Reduction of the rotational degrees of freedom 198 DOF ω r 4 ml EI 14 x ω r 4 ml EI 1 x Close-up Mode n Mode n The relative error on the first 10th modes is less than %!

17 17 Example: the beam clamped at both ends Guyan reduction: 396 DOF 198 DOF 5 DOF ω r 4 ml EI 14 x x exact o Guyan (5 DOF) FE (13 elements, 6 DOF) Mode n 11 x Close-up 10 Close-up

18 18 Example: the beam clamped at both ends Guyan reduction: 396 DOF 198 DOF 5 DOF Relative error % 5 4 o Guyan (5 DOF) FE (13 elements, 6 DOF) Mode n Guyan 5 DOF FE 6 DOF exact Mode n

19 19 Substructuring methods Let us consider a substructure which is connected to the rest of the system by a set of boundary degrees of freedom q boundary DOF q internal DOF q 1 The internal degrees of freedom q 1 are free. The substructure is described by its stiffness and mass matrices K and M

20 0 Concept of mechanical impedance boundary DOF q The dynamic equilibrium equation of the substructure writes ( K ω M) q = g internal DOF q 1 applied force amplitudes and the impedance matrix is defined as ( ω ) ω 1( ω = = ) Z K M H

21 1 Concept of mechanical impedance boundary DOF q internal DOF q 1 Since the internal degrees of freedom q 1 are not loaded, we may write Z Z 11( ω ) Z q1 0 1 = 1( ω ) Z q g ( ω ) ( ω ) External loads and/or boundary reactions From the first equation, we can eliminate the internal degrees of freedom.

22 Concept of mechanical impedance 1 1 = 11 1 q Z Z q So we deduce the relationship ( ) * ω = Z q g where Z * 1 = Z Z 1 Z11 Z1 Reduced impedance matrix Z * One notes that admits as poles the zeros of Z 11 which corresponds to the eigenfrequencies of the subsystem with its boundary degrees of freedom q fixed.

23 3 Concept of mechanical impedance Let us consider the subsystem clamped on its boundary boundary DOF q The eigensolutions of the subsystem ( ω ) K M x = 0 internal DOF q 1 are numbered in the following order 0 ω1 ωn < x x () 1 ( n)

24 4 Concept of mechanical impedance Based on the spectral expansion of, it can be shown that the reduced impedance matrix takes the form Z 1 11 * 1 = Z K K K K 4 ( M ) M1 K11 K1 K1 K11 M1 K1 K11 M11 K11 K1 ω + ω n 1 i= 1 ( ) T ( K ) 1 ωi M1 x ( i) x ( i) K1 ωi M1 4 i ( i ) ω ω ω T where the terms of orders 0, 1 and in ω have been isolated. What do we recognize in this equation?

25 5 Concept of mechanical impedance The first two terms corresponds to a static condensation of the substructure on its boundary (Guyan s reduction method). RR RR K = K K K K M = M M K K K K M + K K M K K Z * = K ω ω RR 4 M n 1 RR Guyan s reduction method ( ) T ( K ) 1 ωi M1 x ( i) x ( i) K1 ωi M1 4 i= 1 ωi ( ωi ω ) T The last term represents a correction term that can be exploited to improve Guyan s reduction method.

26 6 Craig and Bampton s method Z * = K ω ω RR 4 M n 1 RR Guyan s reduction method ( ) T ( K ) 1 ωi M1 x ( i) x ( i) K1 ωi M1 4 i= 1 ωi ( ωi ω ) T The term of order ω 4 represents the contribution of the subsystem eigenmodes in clamped boundary configuration.

27 7 Craig and Bampton s method So the dynamic behaviour of a substructure is fully described by: the static boundary modes resulting from the static condensation, the subsystem eigenmodes in clamped boundary configuration.

28 8 Craig and Bampton s method Accordingly, it means that the following transformation may be applied to the initial degrees of freedom x I 0 xr = 1 KCC KCR ΦI y n R n I boundary DOF intensity parameters (n I = n n R ) where the Guyan s reduction matrix has been complemented by the set of n internal vibration modes obtained by solving ( ω ) K M x = 0

29 9 Craig and Bampton s method In practice, only a certain number m < n I modes are kept: of internal vibration Φ x x m = () 1 ( m) They can be selected according to the intensity of the associated boundary reactions ( ω ) K M x K x RC i RC () i RC () i This yields the final reduction matrix of dimension n x (n R + m) R I 0 = 1 KCC KCR Φm

30 30 Craig and Bampton s method Working the reduced stiffness and mass matrices explicit gives with KRR 0 MRR MRm K = and = M 0 Ωm M mr I 1 RR = RR RC CC CR K K K K K RR = RR RC CC CR RC CC CR + RC CC CC CC CR M M M K K K K M K K M K K ( 1 ) T T mr = m CR CC CC CR = Rm M Φ M M K K M In the finite element context, matrices and constitute a so-called superelement. K M

31 31 Example: the beam clamped at both ends Beam clamped at both ends with 100 finite elements (= 396 DOF) 40 elements (80 DOF) Superelement Guyan s reduction method 80 DOF Craig-Bampton s substructuring method 80 DOF + internal modes 80 DOF + 5 internal modes

32 3 Example: the beam clamped at both ends Relative error % ω r 4 ml EI.5 x Mode n x exact (FE with 396 DOF) o Guyan (80 DOF) Craig-Bampton ( modes) Δ Craig-Bampton (5 modes) Close-up Mode n

33 33 Example (Courtesy of Techspace Aero) Analysis of the radial freeplay Casing DOF Stator DOF Rotor DOF

34 34 After reduction: validation in [0-300 Hz] Analysis of the radial freeplay Casing 81 DOF Craig-Bampton Stator DOF Rotor 85 DOF Craig-Bampton Guyan