lecture_14

Lecture_14 - Lecture 14 1 Matrix mechanics In this lecture we will use matrix algebra to the full extent The matrix version of equations of motion

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Lecture 14 1 Matrix mechanics In this lecture we will use matrix algebra to the full extent. The matrix version of equations of motion for the system shown in figure ?? is ± m 0 0 2 m ²± ¨ x 1 ¨ x 2 ² = ± - 2 k k k - 2 k ²± x 1 x 2 ² Thus, using our definition of the mass matrix and the stiffness matrix, we get M ¨ x = - Kx, or M ¨ x + Kx = 0 . multiplying the above equation by the matrix inverse of M , denoted M - 1 , and remembering that M - 1 M = I , where I is the 2 × 2 identity matrix, we obtain ¨ x = - M - 1 Kx, (1) where x = ( x 1 ,x 2 ) T is the vertical coordinate vector, and M - 1 is the matrix inverse of the mass matrix M . Recall, that the matrix inverse of a 2 × 2 matrix A = ± a b c d ² is obtained by A - 1 = 1 det ( A ) ± d - b - c a ² = 1 ad - bc ± d - b - c a ² Thus, the inverse of the mass matrix reads M - 1 = 1 det ( M ) ± 2 m 0 0 m ² = 1 2 m 2 ± 2 m 0 0 m ² = ± 1 /m 0 0 1 / (2 m ) ² Note that the equation (1) looks very much like our canonical equation of motion for a single degree of freedom system, ¨ x = - ω 2 n x 1
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placed by the matrix product M - 1 K . In fact, if there was no coupling spring, k 12 = 0, then we would have two decoupled equations: ± m 0 0 2 m ²± ¨ x 1 ¨ x 2 ² = ± - k 0 0 - k ²± x 1 x 2 ² Wouldn’t it be great if we had a way of taking the coupled equations and finding new coordinates in which equations of motion decouple like that? In fact, it is possible, due to same matrix magic we will perform next 1.1 Normal mode coordinates We’d like to find coordinates y = ( y 1 ,y 2 ) such that ¨ y = - Dy, where y = ( y 1 ,y 2 ) T is the vertical ”new coordinate” vector, D is a diagonal, positive matrix (i.e. all the entries of D that are not on the main diagonal are zero, and entries on the main diagonal are either zero or positive): D = ³ ( ω 1 ) 2 0 0 ( ω 2 ) 2 ´ , and ω 1 2 are the frequencies for this 2 DOF system. This can be achieved by setting y = V - 1 x where V = ± ( φ 1 ) 1 ( φ 2 ) 1 ( φ 1 ) 2 ( φ 2 ) 2 ² , and φ 1 = (( φ 1 ) 1 , ( φ 1 ) 2 ) T 2 = (( φ 2 ) 1 , ( φ 2 ) 2 ) T are the column-vector normal modes that we discussed in the last lecture! To show this, note that ¨ y = V - 1 ¨ x = V - 1 ( - M - 1 K ) x = - V - 1 ( M - 1 K ) V y From linear algebra we know that, if we set the columns of V to be the eigenvectors of M - 1 K , then V - 1 ( - M - 1 K ) V = D a diagonal matrix, with eigenvalues of M - 1
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This note was uploaded on 03/23/2009 for the course ME 163 taught by Professor Mezic,i during the Winter '08 term at UCSB.

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Lecture_14 - Lecture 14 1 Matrix mechanics In this lecture we will use matrix algebra to the full extent The matrix version of equations of motion

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