Aero631 Multiple DOF

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  • 1. Multiple Degree of Freedom Systems Aero631 Dr. Eng. Mohammad Tawfik
  • 2. Multiple Degrees of Freedom Aero631 Dr. Eng. Mohammad Tawfik
  • 3. Objectives • What is a multiple degree of freedom system? • Obtaining the natural frequencies of a multiple degree of freedom system • Interpreting the meaning of the eigenvectors of a multiple degree of freedom system • Understanding the mechanism of a vibration absorber Aero631 Dr. Eng. Mohammad Tawfik
  • 4. Two Degrees of Freedom Systems Aero631 Dr. Eng. Mohammad Tawfik
  • 5. Two Degrees of Freedom Systems • When the dynamics of the system can be described by only two independent variables, the system is called a two degree of freedom system Aero631 Dr. Eng. Mohammad Tawfik
  • 6. Two Degrees of Freedom Aero631 Dr. Eng. Mohammad Tawfik
  • 7. Free-Body Diagram Aero631 Dr. Eng. Mohammad Tawfik
  • 8. Equations of Motion m11 (t )  k1 x1 (t )  k2 x2 (t )  x1 (t )  x m2 2 (t )  k2 x2 (t )  x1 (t )  x Rearranging: m11 (t )  (k1  k2 ) x1 (t )  k2 x2 (t )  0 x m2 2 (t )  k2 x1 (t )  k2 x2 (t )  0 x Aero631 Dr. Eng. Mohammad Tawfik
  • 9. Initial Conditions • Two coupled, second -order, ordinary differential equations with constant coefficients • Needs 4 constants of integration to solve • Thus 4 initial conditions on positions and velocities x1 (0)  x10 , x1 (0)  x10 , x2 (0)  x20 , x2 (0)  x20     Aero631 Dr. Eng. Mohammad Tawfik
  • 10. In Matrix Form M  Kx  0 x Where: m1 0  k1  k2  k2  M  , K    k  0 m2   2 k2   x (t )   x (t )     (t )  x x(t )   1 , x(t )   1 , (t )   1   x  x2 (t )   x2 (t )  2 (t ) x  x10   x10   With initial conditions: x(0)   , x(0)      x20    x20  Aero631 Dr. Eng. Mohammad Tawfik
  • 11. Recall: For SDOF • The ODE is m(t )  kx(t )  0 x • The proposed t solution: x (t )  ae • Into the ODE you get k t t the characteristic equation:  ae  ae  0 2 m k k • Giving:   2 j m m Aero631 Dr. Eng. Mohammad Tawfik
  • 12. Solving the system • The ODE is M  Kx  0 x • The proposed jt solution: x(t )  ae • Into the ODE you get the characteristic jt jt equation:   Ma e  Kae 2 0 • Giving:   M  K ae 2 jt 0 Aero631 Dr. Eng. Mohammad Tawfik
  • 13. Giving: Either:  a1  a 0 Trivial solution; a2  No motion!  M  K  0 OR: 2 Aero631 Dr. Eng. Mohammad Tawfik
  • 14. Giving:   m1  k1  k 2 2  k2 0  k2   2 m2  k 2 m1m2  (m1k 2  m2 k1  m2 k 2 )  k1k 2  0 4 2 Which can be solved as a quadratic equation in 2. Aero631 Dr. Eng. Mohammad Tawfik
  • 15. NOTE! • For spring mass systems, the resulting roots are always positive, real, and distinct • Which give two couples of distinct roots. 1, 2   2 1 & 3, 4    2 2 Aero631 Dr. Eng. Mohammad Tawfik
  • 16. Example • m1=9 kg,m2=1kg, k1=24 N/m and k2=3 N/m • In Matrix form: 9 0  27  3 0 1   3 3  x  0 x     Aero631 Dr. Eng. Mohammad Tawfik
  • 17. Example (cont’d) jt • The proposed solution: x(t )  ae • Into the ODE you get the characteristic equation: 4-62+8=(2-2)(2-4)=0 • Giving: 2 =2 and 2 =4 Each value of 2 yields an expression for a: Aero631 Dr. Eng. Mohammad Tawfik
  • 18. Calculating the corresponding vectors a1 and a2 A vector equation for each square frequency ( M  K )a1  0 2 1 And: ( M  K )a 2  0 2 2 4 equations in the 4 unknowns (each vector has 2 components, but... Aero631 Dr. Eng. Mohammad Tawfik
  • 19. Computing the vectors a  a11  For  = 2, let a1    2 1 a12  (- M  K )a  0  2 1 27  9(2)  3   a11  0  3     0   3  (2) a12     9a11  3a12  0 and  3a11  a12  0 2 equations, 2 unknowns but DEPENDENT! Aero631 Dr. Eng. Mohammad Tawfik
  • 20. continued a11 1 1   a11  a12 from both equations : a12 3 3 only the direction, not the magnitude can be determined! This is because:   M  K  0. 2 1 The magnitude is arbitrary.Supposeu1 satisfies (12 M  K )a1  0, so does ca1 , c arbitrary : (12 M  K )ca1  0  (12 M  K )ca1  0 Aero631 Dr. Eng. Mohammad Tawfik
  • 21. For the second value of  2:  a21  For  = 4, let a 2    then wehave 2 2 a22  (-12 M  K )a  0  27  9(4)  3   a21  0  3     0   3  (4) a22     1  9a21  3a22  0 or a 21   a22 3 Note that the other equation is the same Aero631 Dr. Eng. Mohammad Tawfik
  • 22. What to do about the magnitude! Several possibilities, here we just fix one element:  13  Choose: a12  1  a1    1 13  Choose: a22  1  a 2    1 Aero631 Dr. Eng. Mohammad Tawfik
  • 23. Thus the solution to the algebraic matrix equation is:  13  1,3   2 , a1    1 13  2, 4  2, a 2    1 Aero631 Dr. Eng. Mohammad Tawfik
  • 24. Return now to the time response: We have four solutions: x(t )  a1e  j1t , a1e j1t , a 2 e  j2t , a 2 e j2t  Since linear we can combine jas:  j t t x(t )  aa1e 1  ba1e 1  ca 2 e  j2t  da 2 e j2t     x(t )  ae  j1t  be j1t a1  ce  j2t  de j2t a 2   A1 sin(1t  1 )a1  A2 sin(2t  2 )a 2 where A1 , A2 , 1 , and 2 are constantsof integration determined by initial conditions Aero631 Dr. Eng. Mohammad Tawfik
  • 25. Physical interpretation of all that math! • Each of the TWO masses is oscillating at TWO natural frequencies 1 and 2 • The relative magnitude of each sine term, and hence of the magnitude of oscillation of m1 and m2 is determined by the value of A1a1 and A2a2 • The vectors a1 and a2 are called mode shapes Aero631 Dr. Eng. Mohammad Tawfik
  • 26. What is a mode shape? • First note that A1,A2, 1 and 2 are determined by the initial conditions • Choose them so that A2 = 1 = 2 =0 • Then:  x1 (t )   a11  x(t )     A1 a  sin 1t  x2 (t )  12  • Thus each mass oscillates at (one) frequency 1 with magnitudes proportional to a1 the1st mode shape Aero631 Dr. Eng. Mohammad Tawfik
  • 27. Multiple Degrees of Freedom Aero631 Dr. Eng. Mohammad Tawfik
  • 28. Things to note • Two degrees of freedom implies two natural frequencies • Each mass oscillates at these two frequencies present in the response • Frequencies are not those of two component systems Aero631 Dr. Eng. Mohammad Tawfik
  • 29. Eigenvalues and Eigenvectors • Can connect the vibration problem with the algebraic eigenvalue problem • This will give us some powerful computational skills • And some powerful theory • All the codes have eigensolvers so these painful calculations can be automated Aero631 Dr. Eng. Mohammad Tawfik
  • 30. Compound Pendulum Aero631 Dr. Eng. Mohammad Tawfik
  • 31. Pendulum Video Aero631 Dr. Eng. Mohammad Tawfik
  • 32. Frequency Response Aero631 Dr. Eng. Mohammad Tawfik
  • 33. Frequency Response • Similar to SDOF systems, the frequency response of a MDOF system is obtained by assuming harmonic excitation. • An analytical relation between all the possible input forces and output displacements may be obtained, called transfer function • For our course, we will pay more attention to the plot of the relation. Aero631 Dr. Eng. Mohammad Tawfik
  • 34. Dynamic Stiffness • The system of equations we obtain for an undamped vibrating system is always in the form M  Kx  f x • For harmonic excitation  harmonic response, we may write   M  K x  f 2 KD x  f Aero631 Dr. Eng. Mohammad Tawfik
  • 35. Dynamic Stiffness • Now, we have a system of algebraic equations that may be solved for the amplitude of vibration of each DOF as a response to given harmonic excitation at a certain frequency! 1 x  KD f Aero631 Dr. Eng. Mohammad Tawfik
  • 36. Example • For the three DOF system given in the sketch, consider all stiffness values to be 2 and m1=2, m2=1, m3=3 Aero631 Dr. Eng. Mohammad Tawfik
  • 37. Example • The equations of motion may be written in the form: 2 0 0  1   4  2 0   x1   f1  x 0 1 0      2 4  2  x    f     x2    2   2  0 0 3  3   0  2 4   x3   f 3    x       M  Kx  F x Aero631 Dr. Eng. Mohammad Tawfik
  • 38. Example • Getting the eigenvalues, and frequencies 5.023 0 0   2.241  0  ,   1.295 1.677 0     0 0.633 0.796  0    Aero631 Dr. Eng. Mohammad Tawfik
  • 39. Getting the Frequency Response  4 2 0   2 0 0 K D    2 4  2   2 0 1 0      0 2 4     0 0 3    f1  0      f 2   1  f  0  3   Aero631 Dr. Eng. Mohammad Tawfik
  • 40. Aero631 Dr. Eng. Mohammad Tawfik
  • 41. Notes: • For all degrees of freedom, as the frequency reaches one of the natural frequencies, the amplitudes grows too much • For some frequencies, and some degrees of freedom, the response becomes VERY small. If the system is designed to tune those frequencies to a certain value, vibration is absorbed: “Vibration absorber” Aero631 Dr. Eng. Mohammad Tawfik
  • 42. Vibration Absorber The first passive damping technique we will learn! Aero631 Dr. Eng. Mohammad Tawfik
  • 43. For a 2-DOF System • For the shown 2-DOF system, the equations of motion may be written as: M  Kx  f x  f1  • Where: f    f2  Aero631 Dr. Eng. Mohammad Tawfik
  • 44. For Harmonic Excitation • We may write the  f1Cos1t  equation for each of M  Kx   x   0  the excitation frequency in the form  0  M  Kx   x   f 2Cos2t  of: • Then we may add both solutions! Aero631 Dr. Eng. Mohammad Tawfik
  • 45. Consider the first force • We may write the 1 equation in the form: M  Kx    f1Cost  x 0 • And the solution in  x1  x   Cost  the form:  x2  • Which will give:  x1      Cost    2 x x 2  x2  Aero631 Dr. Eng. Mohammad Tawfik
  • 46. The equation of motion becomes    2 m1 0  k1  k 2  k 2   x1   f1           0   m2    k 2 2 k 2    x2   0   • Get x1() and find out when does it equal to zero! Aero631 Dr. Eng. Mohammad Tawfik
  • 47. Using the Dynamic Stiffness Matrix • Writing down the dynamic stiffness matrix:   2 m1  K1  K 2  K2   x1   f1          K2   m2  K 2   x2   0  2 Getting the inverse:   2 m2  K 2 K2     x1   K2   m1  K1  K 2  2  f1  x    2      2 m1  K1  K 2   2 m2  K 2  K 2 2 0   Aero631 Dr. Eng. Mohammad Tawfik
  • 48. Obtaining the Solution • Multiply the inverse by the right-hand-side  x1   1      2 m2  K 2 f1   x  m m  4  m K  K   m K  2  K K    2 1 2 2 1 2 1 2 1 2  K 2 f1  • For the first degree of freedom: x1    m  K 2 f1 2 2  0 m1m2  m2 K1  K 2   m1 K 2   K1 K 2 4 2 Aero631 Dr. Eng. Mohammad Tawfik
  • 49. Vibration Absorber • For the first degree of freedom to be stationary, i.e. x1=0 • The excitation frequency have to satisfy: K2  m2 • Note that this frequency is equal to the natural frequency of the auxiliary spring- mass system alone Aero631 Dr. Eng. Mohammad Tawfik
  • 50. Vibration absorber Aero631 Dr. Eng. Mohammad Tawfik
  • 51. Vibration absorber Aero631 Dr. Eng. Mohammad Tawfik
  • 52. Homework #2 • Repeat the example of this lecture using f2=f3=0 and f1=1 AND f1=f2=0 and f3=1 • Plot the response of each mass for each of the excitation functions • Comment on the results in the lights of your understanding of the concept of vibration absorber Aero631 Dr. Eng. Mohammad Tawfik
  • 53. Homework #2 (cont’d) • Use modal decomposition (diagonalization) to obtain the same results. Aero631 Dr. Eng. Mohammad Tawfik