美国麻省理工大学：《结构力学》（英文版） Unit 22 Vibration of Multi Degree-Of- Freedom Systems

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

MT-1620 al.2002 For each eigenvalue, the homogeneous solution is e sina. t coSO. t homogeneous Still an undetermined constant in each case(An ) which can be determined from the initial conditions Each homogeneous solution physically represents a possible free vibration mode Arrange natural frequencies from lowest (o, to highest (o By superposition, any combinations of these is a valid solution Example: Two mass system(from Unit 19) Figure 22.1 Representation of dual spring-mass system Paul A Lagace @2001 Unit 22-6

MIT - 16.20 Fall, 2002 For each eigenvalue, the homogeneous solution is: r r r q i hom = φi () e iω r t = C1 φi () sinωr t + C2 φi () cosωr t ~ ~ ~ ~ homogeneous Still an undetermined constant in each case (An) which can be determined from the Initial Conditions • Each homogeneous solution physically represents a possible free vibration mode • Arrange natural frequencies from lowest (ω1) to highest (ωn) • By superposition, any combinations of these is a valid solution Example: Two mass system (from Unit 19) Figure 22.1 Representation of dual spring-mass system Paul A. Lagace © 2001 Unit 22 - 6

MT-1620 al.2002 The governing equation was m. 0 +)-k]91F 0m22 Thus, from equation(22-5) (k+k)-02m-k2 k k2-0 This gives: (k+k)-a m, k2 0m,一 k2=0 This leads to a quadratic equation in (2. Solving gives two roots(o 1 2 and 022)and the natural frequencies are o, and O2 Find the associated eigenvectors in terms of A2(i.e normalized by a2) Paul A Lagace @2001 Unit 22-7

q q MIT - 16.20 Fall, 2002 The governing equation was: m1 0  ˙˙1  (k1 + k2 ) −k2  q1  F1   0 m2  ˙˙2      +    =    −k2 k2  q2  F2  Thus, from equation (22-5): (k1 + k2 ) − ω2 m1 −k2 = 0 2 −k2 k2 − ω m2 This gives: [(k1 + k2 ) − ω2 m1 ][k2 − ω2 m2 ] − k22 = 0 This leads to a quadratic equation in ω2. Solving gives two roots (ω 12 and ω22) and the natural frequencies are ω1 and ω2 Find the associated eigenvectors in terms of A2 (i.e. normalized by A2) Paul A. Lagace © 2001 Unit 22 - 7

MT-1620 a.2002 Go back to equation (22-4)and divide through by A2 421「41 k Normalized constant k for o. mode k+k2-0n2m1 Thus the eigenvectors are k 2 k k2-02m 2)|k+k2-m2mn1 For the case of Initial Conditions of 0, the cos term goes away and are left with sin w. t Physically the modes are Paul A. Lagace @2001 Unit 22-8

        MIT - 16.20 Fall, 2002 Go back to equation (22-4) and divide through by A2: (k1 + k2 ) − ωr 2 m1 −k2   A1      = 0  −k2 k2 − ωr 2 m2   1  Normalized constant k ⇒ A1 = k1 + k2 2 − ωr 2 m1 for ωr mode Thus the eigenvectors are:  k2   k2  1 k1 + k2 − ω12 m1  φ() k1 + k2 − ω22 2 m1  φi () =   i =   ~ ~      1   1  For the case of Initial Conditions of 0, the cos term goes away and are left with… r q t() = φ() sin ωr t ~ ~ Physically the modes are: Paul A. Lagace © 2001 Unit 22 - 8

MT-1620 al.2002 Figure 22.2 Representation of modes of spring-mass system (lowest frequency) M L (higher frequency) masses move in opposite direction Paul A Lagace @2001 Unit 22-9

φ φ ω MIT - 16.20 Fall, 2002 Figure 22.2 Representation of modes of spring-mass system φ(1) ω1 (lowest frequency) masses move in same direction φ(2) frequency) ω2 (higher masses move in opposite direction Paul A. Lagace © 2001 Unit 22 - 9

MT-1620 al.2002 General Rules for discrete systems Can find various modes(without amplitudes) by considering combinations of positive and negative(relative)motion However, be careful of (-1)factor across entire mode For example in two degree- of-freedom case +十 same mode same mode The more reversals" in direction, the higher the mode(and the frequency It is harder to excite higher modes This can be better illustrated by considering the vibration of a beam. So look at Representation of a Beam as a Discrete Mass System Paul A Lagace @2001 Unit 22-10

MIT - 16.20 Fall, 2002 General Rules for discrete systems: • Can find various modes (without amplitudes) by considering combinations of positive and negative (relative) motion. However, be careful of (-1) factor across entire mode. For example, in two degree-of-freedom case + + + - same mode same mode - - - + • The more “reversals” in direction, the higher the mode (and the frequency) • It is harder to excite higher modes This can be better illustrated by considering the vibration of a beam. So look at: Representation of a Beam as a Discrete Mass System Paul A. Lagace © 2001 Unit 22 - 10