Note3_98F - Eigenvalues and Eigenvectors : Basis for Modal...

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Unformatted text preview: Eigenvalues and Eigenvectors : Basis for Modal Analysis September 22, 1998 In the first model problem, we have consider vibration of an elastic string: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 , , , , , , 2 2 2 2 = = = = = +- = L t u t u x u x u x v x t u L in f ku x u T t u m Here a distributed elastic foundation is assumed to be connected on the elastic string spanned by a tensile force T , and u and v are the initial displacement and velocity at the initial time t = 0. Applying the weighted residual method with the finite number of trial functions and test functions ( 29 ( 29 { } x x i j y f , , a discrete problem ( 29 ( 29 2 2 , v u u u f Ku u M = = = + dt d dt d where ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 = = = = = = + = + = = = L i i i L i i i L i i i L i j i j i j i j ij L i j i j ij dx x x v v v dx x x u u u dx x x t f f f dx x x k dx d dx d T k dx d dx d T k dx x x m m m , , , , , , , , y y y y y y y f y f y f y f y f y f We shall now consider a homogeneous problem that the right hand side, the applied distributed force f is zero: Ku u M = + 2 2 dt d and we shall consider a stationary problem with harmonic motion such that (29 1 ,- = = i e t t i x u w where is a frequency of the harmonic motion in time, and x is independent of time t . Substitution of the harmonic motion into the equation of motion, we have 2 , w l l = = +- Kx Mx that is 2 , w l l = = Mx Kx . When m = n in the discrete problem, this is called a generalized eigenvalue problem , where is an eigenvalue and x is an associated eigenvector of K and M . If we can decompose the mass matrix into the following form T LL M = where L is non-singular in the sense that its inverse L-1 exists, then x L Mx L x L KL L x LL Mx Kx T T T T l l l l = = = =--- 1 1 Putting x L y T = we can convert the generalized eigenvalue problem to the form of the standard eigenvalue problem ( 29 y Ay y y L K L l l = =-- , 1 1 T . In this case, if M and K are symmetric, then A is also symmetric. That is, the transformation x L y T = of the eigenvector, yields a symmetric eigenvalue problem. However, M-1 K need not be symmetric even for both symmetric M and K , and then the eigenvalue problem x Kx M l =- 1 must be solved only by a un-symmetric eigenvalue solver. Suppose that we have n number of eigenvalues, 1 ,...., n , and n number of linearly independent eigenvectors, x 1 ,...., x n , of the generalized eigenvalue problem: n i i i i ,....., 2 , 1 , = = Mx Kx l . Applying the Gram-Schmidt orthogonalization process with respect to a given matrix M of n number of linearly independent vectors z 1 ,....., z n : ( 29 ( 29 ( 29 n i i i i i i j j i j i i ,...., 2 , , , , , , 1 1 1 1 1 1 = =- = = - = z M z z x x Mz x z z Mz z z x n number of linearly independent eigenvectors x 1 ,...., x n , can be orthonomalized with respect to the matrix M ( i.e. ( 29 ) , ij j i d = Mx x , and then they can span the n dimensional space...
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Note3_98F - Eigenvalues and Eigenvectors : Basis for Modal...

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