Jtam.dvi JOURNAL OF THEORETICAL AND APPLIED MECHANICS 49, 2, pp. 343-354, Warsaw 2011 MODAL ANALYSIS OF MULTI-DEGREE-OF-FREEDOM SYSTEMS WITH REPEATED FREQUENCIES – ANALYTICAL APPROACH Rafał Palej Artur Krowiak Cracow University of Technology, Institute of Computing Science, Cracow, Poland e-mail: palej@mech.pk.edu.pl; krowiak@mech.pk.edu.pl The paper deals with the modal analysis of mechanical systems consi- sting of n identical masses connected with springs in such a way that the stiffness matrix has the form of amultiband symmetric matrix. The eigenvalue problem formulated for such systems is characterisedby repe- ated eigenvalues to which linearly independent eigenvectors correspond. The solution to the eigenvalue problem has been found for an arbitrary, finite number of degrees of freedom for the fully coupled systems and the systems in which the masses are exclusively connected with the nearest neighbours.Depending on the property of the eigenvectors, two forms of the solution to the initial value problem have been derived. The gene- ral deliberations are illustrated with an example of the system with 10 degrees of freedom for 5 different degrees of coupling. Key words: modal analysis, conservative systems, repeated eigenvalues 1. Introduction Dynamic analysis of systems with repeated frequencies is interesting not only from the theoretical point of view. There aremany technical problems concer- ning such systems, e.g. the influence of the imperfection on the mode shapes (Wei andPierre, 1998).Theexamples analysed inWang et al. (2005) andPešek (1995) refer to a special type of two degrees of freedom systems possessing do- uble frequency. It seems to be worthwhile to study multi-degree-of-freedom systems with many repeated eigenvalues possessing various multiplicity. One of the simplest systems which possesses natural frequency of multi- plicity 2 consists of three identical masses, connected with one another with 344 R. Palej, A. Krowiak the use of identical springs. Such a system has one zero frequency to which corresponds rigid-body motion and a double frequency to which corresponds harmonicmotion.One should expect that systemswith a larger number of de- grees of freedom, consisting of identical masses and springs, will also possess repeated frequencies. It turns out that depending on the degree of coupling, the number of frequencies and their multiplicity may be different. The eige- nvalue problem of such systems possesses, among others, repeated eigenvalues and mutually orthogonal eigenvectors belonging to them. Since such systems have a regular structure, there is a possibility of deriving analytical formulae for natural frequencies andmodes of vibration for an arbitrary, finite number of degrees of freedom. The paper presents the way of deriving formulae for eigenvalues and eigenvectors of regular systems for various degrees of coupling as well as two forms of analytical solution to the initial value problem. 2. Fully coupled systems Let us consider a mechanical system consisting of n identical elements of mass m, connected – each one with each one – through springs of stiffness k and, additionally, connected with the base through springs of stiffness p. The potential energy U and the kinetic energy E of the system have the form U = 1 2 k n ∑ i=1 n ∑ j=i+1 (qj −qi)2+ 1 2 p n ∑ i=1 q2i E= 1 2 m n ∑ i=1 q̇2i (2.1) where qi and q̇i denote generalized coordinates and velocities, respectively. 2.1. Equations of motion Equations of motion derived by means of Lagrange’s equations have the form Mq̈+Kq=0 (2.2) where the stiffness matrix is the symmetric Toeplitz matrix of the form K=         (n−1)k+p −k · · · −k −k −k (n−1)k+p · · · −k −k ... ... ... ... ... −k −k · · · (n−1)k+p −k −k −k · · · −k (n−1)k+p         n×n (2.3) Modal analysis of multi-degree-of-freedom systems... 345 while the inertiamatrix is a scalarmatrix, containing themass m on themain diagonal. 2.2. The eigenvalue problem Seeking a solution to Eq. (2.2) in the form q=ueαt (2.4) we obtain the algebraic eigenvalue problem in the standard form Au=λu (2.5) where matrix A has the following form A=          (n−1) k m + p m − k m · · · − k m − k m − k m (n−1) k m + p m · · · − k m − k m ... ... ... ... ... − k m − k m · · · (n−1) k m + p m − k m − k m − k m · · · − k m (n−1) k m + p m          n×n (2.6) and the unknown scalar λ is related to exponent α by the formula λ=−α2 (2.7) The determinant of A can be written as (Bernstein, 2005) detA= p m ( n k m + p m )n−1 (2.8) Since the determinant of thematrix is equal to the product of its eigenvalues, one can suppose that eigenvalues of Awill have the form        λ1 = p m λj =n k m + p m j=2,3, . . . ,n (2.9) As can be seen λ1 is a single eigenvalue while λ2 is repeated n− 1 times. Putting the eigenvalue λ1 into Eq. (2.5) we obtain a set of equations with respect to the elements of the eigenvector belonging to λ1 in the form         n−1 −1 · · · −1 −1 −1 n−1 · · · −1 −1 ... ... ... ... ... −1 −1 · · · n−1 −1 −1 −1 · · · −1 n−1                 u1 u2 ... un−1 un         =0 (2.10) 346 R. Palej, A. Krowiak The rank of the coefficientmatrix is equal to n−1 and, as it is easy to notice, the solution to Eq. (2.10) has the form u1 = [1,1, . . . ,1,1] ⊤ (2.11) The eigenvectors corresponding to the eigenvalue λ2 should fulfill the following set of equations         −1 −1 · · · −1 −1 −1 −1 · · · −1 −1 ... ... ... ... ... −1 −1 · · · −1 −1 −1 −1 · · · −1 −1                 u1 u2 ... un−1 un         =0 (2.12) Now, the rank of the coefficient matrix is equal to 1, which means that it is possible to choose freely n− 1 components of the vector u. The simplest choice consists in imposing the values: 1 and −1 on two components of uj, j=2,3 . . . ,nwhileputtingothers as zero.Thevalue −1canbeassigned to the first component,whereas the value 1 shouldhave a changeable position – from the second one to the last one. Thematrix U created from such eigenvectors has the form U=         1 −1 · · · −1 −1 1 1 · · · 0 0 ... ... ... ... ... 1 0 · · · 1 0 1 0 · · · 0 1         n×n (2.13) The eigenvectors uj, j=2,3 . . . ,n constitute a set of n−1 linearly indepen- dent vectors, so the geometrical multiplicity of the eigenvalue λ2 is equal to its algebraic multiplicity, therefore the matrix U diagonalizes the matrix A, i.e. U −1 AU= diag(λ1,λ2, . . . ,λn) (2.14) Relation (2.14) confirms the accuracy of Eqs. (2.9) and (2.13) which describe the eigenvalues and eigenvectors of matrix A. 2.3. Solution of the initial value problem in the base of linearly independent vectors To every (different from zero) eigenvalue λj there correspond, according to Eq. (2.7), two imaginary mutually coupled values of the exponent α Modal analysis of multi-degree-of-freedom systems... 347 αj = √ λj i α ∗ j =− √ λj i i = √ −1 (2.15) to which, in turn, corresponds a solution to Eq. (2.2) in the form qj =(Aje αjt+A∗je α∗ j t )uj = [1 2 (Cj − iSj)e √ λjit+ 1 2 (Cj +iSj)e − √ λjit ] (2.16) = (Cj cos √ λjt+Sj sin √ λj t)uj Due to the fact that the eigenvectors uj, j=1,2, . . . ,n are linearly indepen- dent, the complete solution will have the form q= n ∑ j=1 qj = n ∑ j=1 (Cj cosωjt+Sj sinωjt)uj (2.17) where ωj denote natural frequencies of the system, related to the eigenvalues by the formula ωj = √ λj j=1,2, . . . ,n (2.18) In the case of the lack of the couplingwith thebase (p=0), thefirst eigenvalue equals 0, so Eq. (2.17) will now have the form q=(C1+ tS1)u1+ n ∑ j=2 (Cj cosωjt+Sj sinωjt)uj (2.19) The constants Cj and Sj should be found out from the initial conditions q(0)= q0 q̇(0)=p0 (2.20) Equation (2.20) constitutes a set of 2n linear algebraic equations. 2.4. Solution of the initial value problem in the base of orthogonal vectors The eigenvectors uj, j = 2,3, . . . ,n are orthogonal with respect to the eigenvector u1 but they are not mutually orthogonal. Thus they cannot be orthogonal with respect to the scalar matrix either. For this reason, there is a need to solve the set of Eq. (2.20) to determine the constants Cj and Sj. It is possible to avoid this necessity using Gram-Schmidt orthogonalization 348 R. Palej, A. Krowiak process (Meirovitch, 1997) that renders the independent vectors orthogonal. Next, normalizing theorthogonal vectors tobeorthonormalwith respect to the matrix M, we receive vectors nj, j =1,2, . . . ,n. The elements of matrix N, composed of the vectors nj, are described by formulae ni1 = 1√ mn i=1,2, . . . ,n (2.21) nij =              1√ m −1 √ j(j−1) i< j 1√ m j−1 √ j(j−1) i= j j=2,3, . . . ,n 0 j < i¬n Thematrix N fulfils the following relations N ⊤ MN= I N⊤KN= diag(λ1,λ2, . . . ,λn) (2.22) N −1 AN= diag(λ1,λ2, . . . ,λn) where I denotes the identity matrix. As can be seen, the matrix N, alike matrix U, is the similarity transformation matrix of the matrix A. To uncouple the equations of motion, we should introduce the vector x of natural coordinates by means of the linear transformation q=Nx (2.23) Substituting Eq. (2.23) into Eq. (2.2) and premultiplying the result by N⊤, we obtain an uncoupled set of equations in the form ẍ+Λx=0 (2.24) where Λ= diag(λ1,λ2, . . . ,λn). Making use of linear transformation (2.23), one can easily determine the initial conditions for the natural coordinates in the form x0 =N ⊤ Mq0 v0 =N ⊤ Mp0 (2.25) The solution to Eq. (2.24) for j-th natural coordinate (λj 6=0) can be expres- sed as xj =n ⊤ j M ( q0cosωjt+ p0 ωj sinωjt ) (2.26) Modal analysis of multi-degree-of-freedom systems... 349 whereas the complete solution becomes q=Nx= n ∑ j=1 xjnj = n ∑ j=1 n ⊤ j M ( q0cosωjt+ p0 ωj sinωjt ) nj (2.27) In the case of the lack of the coupling with the base (p= 0), solution (2.27) will have the form q=n⊤1M(q0+ tp0)n1+ n ∑ j=2 n ⊤ j M ( q0cosωjt+ p0 ωj sinωjt ) nj (2.28) Thanks to the use of the orthonormal vectors nj, the solutions to Eq. (2.2) in the form (2.27) or (2.28) do not require any set of algebraic equations to be solved. 3. Systems not fully coupled In the case when each individual element of the system is not coupled with all remaining ones, the stiffness matrix K will have subdiagonals containing zeroes. The number of zero subdiagonals depends on the total number ofmas- ses (n) and the number of masses connected with individual element (s). As- suming the samenumberof neighbouringmasses connectedwith every element on both their sides, the stiffness matrix will have the form of a symmetrical multibandmatrix. For example, for n=8 and s=4 the stiffness matrix has the form K=               4k+p −k −k 0 0 0 −k −k −k 4k+p −k −k 0 0 0 −k −k −k 4k+p −k −k 0 0 0 0 −k −k 4k+p −k −k 0 0 0 0 −k −k 4k+p −k −k 0 0 0 0 −k −k 4k+p −k −k −k 0 0 0 −k −k 4k+p −k −k −k 0 0 0 −k −k 4k+p               (3.1) In such systems, there are repeated eigenvalues as well as single ones. The formulae for eigenvalues and eigenvectors are not as simple as in fully co- upled systems. However, the described procedure for obtaining the analytical solution is still the same. 350 R. Palej, A. Krowiak 4. Systems with the lowest degree of coupling The systems with the lowest degree of coupling consist of identical masses connected exclusivelywith thenearest neighbours (s=2)where thefirstmass is connected with the second one and the last one. The schematic diagram of such a system (not coupled with the base) is presented in Fig.1. Fig. 1. Schematic diagram of the systemwith the lowest degree of coupling The system shown in Fig.1 has its continuous counterpart in the form of an unrestrained prismatic bar with the ends connected with each other by a rigid weightless link. Making use of the dispersion formula, which relates the natural frequency and the length of thewave, it is possible to derive a formula for natural frequencies of the system in the form (Palej and Goik, 2003) ωj =2 √ k m sin π(j−1) n j=1,2, . . . ,n (4.1) The squares of natural frequencies (4.1) determine, according to Eq. (2.18), the eigenvalues of the matrix A= k m             2 −1 0 · · · 0 0 −1 −1 2 −1 · · · 0 0 0 0 −1 2 · · · 0 0 0 ... ... ... ... ... ... ... 0 0 0 · · · 2 −1 0 0 0 0 · · · −1 2 −1 −1 0 0 · · · 0 −1 2             n×n (4.2) In the case when themasses are also coupledwith the base (p 6=0), we obtain the eigenvalue problem in the standard form of the matrix B B=A+ p m I (4.3) Modal analysis of multi-degree-of-freedom systems... 351 The addition of p/m to the diagonal elements of A produces a shift in the eigenvalues by the same constant, so the eigenvalues of thematrix B take the form λj =4 k m sin2 π(j−1) n + p m j=1,2, . . . ,n (4.4) It appears from Eq. (4.4) that the smallest eigenvalue λ1 is single, while the remaining eigenvalues, in the case when n is odd, are double. In the case when n is even, the smallest eigenvalue λ1 as well as the biggest one λn/2+1 are single, while the remaining eigenvalues are double. It should be pointed out that Eqs. (2.9) and (4.4) determine the same eigenvalues for the system consisting of three masses (n = 3, s = 2). The way of coupling in such a system fulfils the requirements characteristic of fully coupled systems and the systems with the lowest degree of coupling. The eigenvectors uj of matrix A form the modal matrix U, which has elements of the form (Palej and Goik, 2003) uij =        cos 2π(j−1)i n j¬ jsep sin 2π(j−1)i n jsep