MT-1620 al.2002 Unit 22 Vibration of multi Degree-Of- Freedom Systems Paul A Lagace Ph D Professor of aeronautics Astronautics and engineering systems Paul A Lagace @2001
MIT - 16.20 Fall, 2002 Unit 22 Vibration of Multi Degree-Of- Freedom Systems Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001
MT-1620 al.2002 Previously saw (in Unit 19)that a multi degree-of-freedom system has the same basic form of the governing equation as a single degree-of-freedom system The difference is that it is a matrix equation q+kg 221) matrix So apply the same solution technique as for a single degree-of-freedom system. Thus, first deal with Free Vibration Do this by again setting forces to zero mq+ kq= o (222) Paul A Lagace @2001 Unit 22-2
MIT - 16.20 Fall, 2002 Previously saw (in Unit 19) that a multi degree-of-freedom system has the same basic form of the governing equation as a single degree-of-freedom system. The difference is that it is a matrix equation: m q˙˙ + k q = F (22-1) ~ ~ ~~ ~ ~ = matrix So apply the same solution technique as for a single degree-of-freedom system. Thus, first deal with… Free Vibration Do this by again setting forces to zero: F = 0 ~ ~ m q˙˙ + k q = 0 (22-2) ~ ~ ~ ~ ~ Paul A. Lagace © 2001 Unit 22 - 2
Again assume a solution which has harmonic motion. It now has mulc all 2 MT-1620 components where o are the natural frequencies of the system and」 a is a vector of constants = J4 Substituting the assumed solution into the matrix set of governing equations →-02mAem+kAe01=0 o be true for all cases k-02m (224 This is a standard eigenvalue problem Either ( trivial solution) Paul A Lagace @2001 Unit 22-3
i t ω ω MIT - 16.20 Fall, 2002 Again assume a solution which has harmonic motion. It now has multiple components: ω q t() = Ae (22-3) ~ ~ where ω are the natural frequencies of the system and: M A is a vector of constants = ~ Ai M Substituting the assumed solution into the matrix set of governing equations: it ⇒ −ω2 mA e + kA e it = 0 ~~ ~ ~ ~ To be true for all cases: [ k − ω2 m ] A = 0 (22-4) ~ ~ ~ ~ This is a standard eigenvalue problem. Either: A = 0 (trivial solution) ~ or Paul A. Lagace © 2001 Unit 22 - 3
MT-1620 al.2002 The determinant 0 (225) There will be n eigenvalues for an n degree-of-freedom system In this case eigenvalue natural frequency n degree-of-freedom system has n natural frequencies Corresponding to each eigenvalue(natural frequency), there is an Eigenvector-- Natural Mode Place natural frequency o into equation(22-4) Since determinant =0, there is one dependent equation, so one cannot solve explicitly for A. However, one can solve for the relative values of the components of A in terms of (normalized by) one component Paul A Lagace @2001 Unit 22-4
MIT - 16.20 Fall, 2002 The determinant: k − ω2 m = 0 (22-5) ~ ~ There will be n eigenvalues for an n degree-of-freedom system. In this case: eigenvalue = natural frequency ⇒ n degree-of-freedom system has n natural frequencies Corresponding to each eigenvalue (natural frequency), there is an… Eigenvector -- Natural Mode • Place natural frequency ωr into equation (22-4): [ k − ωr 2 m ] A = ~0 ~ ~ ~ • Since determinant = 0, there is one dependent equation, so one cannot solve explicitly for A. However, one can solve for the ~ relative values of the components of A in terms of (normalized ~ by) one component Paul A. Lagace © 2001 Unit 22 - 4
MT-1620 al.2002 Say divide through by An a. m Solve for A, /An for each Call the eigenvector /A, =9/ 4 Indicates solution Do this for each eigenvalue frequency associated mode Paul A Lagace @2001 Unit 22-5
MIT - 16.20 Fall, 2002 • Say divide through by An: M A A i n [~ k − ωi 2 m ] = ~ 0 ~ M 1 • Solve for Ai / An for each ωr M A A i n • Call the eigenvector r Indicates solution = φi () M ~ for ωr 1 • Do this for each eigenvalue frequency: ω1, ω2 ……. ωn 1 2 φi ( ) associated mode: φi () φi () n ~ ~ ~ Paul A. Lagace © 2001 Unit 22 - 5