ae 2403 vibrations and elements of aeroelasticity n.
Skip this Video
Loading SlideShow in 5 Seconds..
Download Presentation

play fullscreen
1 / 51
Download Presentation
Download Presentation


- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript


  2. Fundamentals of Linear Vibrations Single Degree-of-Freedom Systems Two Degree-of-Freedom Systems Multi-DOF Systems Continuous Systems

  3. Single Degree-of-Freedom Systems • A spring-mass system • General solution for any simple oscillator • General approach • Examples • Equivalent springs • Spring in series and in parallel • Examples • Energy Methods • Strain energy & kinetic energy • Work-energy statement • Conservation of energy and example

  4. A spring-mass system General solution for any simple oscillator: Governing equation of motion: where:

  5. Any simple oscillator General approach: • Select coordinate system • Apply small displacement • Draw FBD • Apply Newton’s Laws:

  6. + Simple oscillator – Example 1

  7. + Simple oscillator – Example 2

  8. + Simple oscillator – Example 3

  9. + Simple oscillator – Example 4

  10. Springs in series: same force - flexibilities add Springs in parallel: same displacement - stiffnesses add Equivalent springs

  11. Equivalent springs – Example 1

  12. + Equivalent springs – Example 2 Consider: ka2 > Wln2 is positive - vibration is stable ka2 = Wl statics - stays in stable equilibrium ka2 < Wl unstable - collapses

  13. + Equivalent springs – Example 3 We cannot definen since we have sin term If  < < 1, sin   :

  14. Strain energy U: energy in spring = work done Kinetic energy T: Energy methods Conservation of energy: work done = energy stored

  15. Work done = Change in kinetic energy Conservation of energy for conservative systems Work-Energy principles E = total energy = T + U = constant

  16. Energy methods – Example Work-energy principles have many uses, but one of the most useful is to derive the equations of motion. Conservation of energy: E = const. Same as vector mechanics

  17. Two Degree-of-Freedom Systems • Model problem • Matrix form of governing equation • Special case: Undamped free vibrations • Examples • Transformation of coordinates • Inertially & elastically coupled/uncoupled • General approach: Modal equations • Example • Response to harmonic forces • Model equation • Special case: Undamped system

  18. Two-DOF model problem Matrix form of governing equation: where: [M] = mass matrix; [C] = damping matrix; [K] = stiffness matrix; {P} = force vector Note: Matrices have positive diagonals and are symmetric.

  19. Undamped free vibrations Assumed general solutions: Zero damping matrix [C] and force vector {P} Characteristic equation: Characteristic polynomial (for det[ ]=0): Eigenvalues (characteristic values):

  20. Undamped free vibrations Eigenvalues and frequencies: Special case when k1=k2=k and m1=m2=m Two mode shapes (relative participation of each mass in the motion): The two eigenvectors are orthogonal: Eigenvector (1) = Eigenvector (2) =

  21. Undamped free vibrations (UFV) Single-DOF: For any set of initial conditions: • We know {A}(1) and {A}(2), 1 and 2 • Must find C1, C2, 1, and 2 – Need 4 I.C.’s For two-DOF:

  22. UFV – Example 1 Given: No phase angle since initial velocity is 0: From the initial displacement:

  23. UFV – Example 2 Now both modes are involved: From the given initial displacement: Solve for C1 and C2: Hence, or Note: More contribution from mode 1

  24. Transformation of coordinates UFV model problem: “inertially uncoupled” Introduce a new pair of coordinates that represents spring stretch: “elastically coupled” z1(t) = x1(t) = stretch of spring 1 z2(t) = x2(t) - x1(t) = stretch of spring 2 or x1(t) = z1(t) x2(t) = z1(t) + z2(t) Substituting maintains symmetry: “inertially coupled” “elastically uncoupled”

  25. We have found that we can select coordinates so that: Inertially coupled, elastically uncoupled, or Inertially uncoupled, elastically coupled. Big question: Can we select coordinates so that both are uncoupled? Notes in natural coordinates: Transformation of coordinates The eigenvectors are orthogonal w.r.t [M]: The modal vectors are orthogonal w.r.t [K]: Algebraic eigenvalue problem:

  26. Transformation of coordinates Governing equation: General approach for solution Let or We were calling “A” - Change to u to match Meirovitch Substitution: Modal equations: Known solutions Solve for these using initial conditions then substitute into (**).

  27. Transformation - Example Model problem with: 1) Solve eigenvalue problem: 2) Transformation: So As we had before. More general procedure: “Modal analysis” – do a bit later.

  28. Response to harmonic forces Model equation: {F} not function of time [M], [C], and [K] are full but symmetric. Assume: Substituting gives: Hence:

  29. Special case: Undamped system Entries of impedance matrix [Z]: Zero damping matrix [C] Substituting for X1 and X2: For our model problem (k1=k2=k and m1=m2=m), let F2 =0: Notes: 1) Denominator originally (-)(-) = (+). As it passes through w1, changes sign. 2) The plots give both amplitude and phase angle (either 0o or 180o)

  30. Multi-DOF Systems • Model Equation • Notes on matrices • Undamped free vibration: the eigenvalue problem • Normalization of modal matrix [U] • General solution procedure • Initial conditions • Applied harmonic force

  31. Model equation: Notes on matrices: Multi-DOF model equation Multi-DOF systems are so similar to two-DOF. Vector mechanics (Newton or D’ Alembert) Hamilton's principles Lagrange's equations We derive using: • They are square and symmetric. • [M] is positive definite (since T is always positive) • [K] is positive semi-definite: • all positive eigenvalues, except for some potentially 0-eigenvalues which occur during a rigid-body motion. • If restrained/tied down  positive-definite. All positive.

  32. UFV: the eigenvalue problem Equation of motion: Matrix eigenvalue problem in terms of the generalized D.O.F. qi Substitution of leads to For more than 2x2, we usually solve using computational techniques. Total motion for any problem is a linear combination of the natural modes contained in {u} (i.e. the eigenvectors).

  33. Do this a row at a time to form [U]. This is a common technique for us to use after we have solved the eigenvalue problem. Normalization of modal matrix [U] We know that:  Let the 1st entry be 1 So far, we pick our eigenvectors to look like: Instead, let us try to pick so that: Then: and

  34. Consider the cases of: Initial excitation Harmonic applied force Arbitrary applied force General solution procedure For all 3 problems: • Form [K]{u} = w2 [M]{u} (nxn system) Solve for all w2 and {u} [U]. • Normalize the eigenvectors w.r.t. mass matrix (optional).

  35. Initial conditions General solution for any D.O.F.: 2n constants that we need to determine by 2n conditions Alternative: modal analysis Displacement vectors: UFV model equation: n modal equations:  Need initial conditions on h, not q.

  36. Initial conditions - Modal analysis Using displacement vectors: As a result, initial conditions: is: Since the solution of hence we can easily solve for And then solve

  37. Applied harmonic force Equation of motion: Driving force {Q} = {Qo}cos(wt) Substitution of leads to Hence, then

  38. Continuous Systems • The axial bar • Displacement field • Energy approach • Equation of motion • Examples • General solution - Free vibration • Initial conditions • Applied force • Motion of the base • Ritz method – Free vibration • Approximate solution • One-term Ritz approximation • Two-term Ritz approximation

  39. Displacement field: u(x, y, z) = u(x, t) v(x, y, z) = 0 w(x, y, z) = 0 The axial bar Main objectives: • Use Hamilton’s Principle to derive the equations of motion. • Use HP to construct variational methods of solution. A = cross-sectional area = uniform E = modulus of elasticity (MOE) u = axial displacement r = mass per volume

  40. Energy approach For the axial bar: Hamilton’s principle:

  41. Axial bar - Equation of motion Hamilton’s principle leads to: If area A = constant  Since x and t are independent, must have both sides equal to a constant. Separation of variables: Hence

  42. Fixed-free bar – General solution Free vibration: = wave speed EBC: NBC: General solution: EBC  NBC  For any time dependent problem:

  43. Fixed-free bar – Free vibration For free vibration: General solution: Hence are the frequencies (eigenvalues) are the eigenfunctions

  44. Fixed-free bar – Initial conditions Give entire bar an initial stretch. Release and compute u(x, t). Initial conditions: Initial velocity: Initial displacement: or Hence

  45. Fixed-free bar – Applied force Now, B.C’s: From we assume: Substituting: B.C. at x = 0: B.C. at x = L: or Hence

  46. Fixed-free bar – Motion of the base From Using our approach from before: B.C. at x = 0: B.C. at x = L: Hence Resonance at:

  47. Ritz method – Free vibration Start with Hamilton’s principle after I.B.P. in time: • Seek an approximate solution to u(x, t): • In time: harmonic function cos(wt) (w = wn) • In space: X(x) = a1f1(x) where: a1 = constant to be determined f1(x) = known function of position f1(x) must satisfy the following: • Satisfy the homogeneous form of the EBC. u(0) = 0 in this case. • Be sufficiently differentiable as required by HP.

  48. One-term Ritz approximation 1 Substituting: Hence Ritz estimate is higher than the exact Only get one frequency If we pick a different basis/trial/approximation function f1, we would get a different result.

  49. One-term Ritz approximation 2 Substituting: Hence Both mode shape and natural frequency are exact. But all other functions we pick will never give us a frequency lower than the exact.

  50. Two-term Ritz approximation In matrix form: where: