Part 1: Discrete systems

Transcription

1 Part 1: Discrete systems Introduction Single degree of freedom oscillator Convolution integral Beat phenomenon Multiple p degree of freedom discrete systems Eigenvalue problem Modal coordinates Damping Anti resonances 1

2 Why suppress vibrations? Failure Building response to earthquakes (excessive strain) Wind on bridges (flutter instability) Fatigue Comfort Car suspensions Noise in helicopters Wind-induced sway in buildings Operation of precision devices DVD readers Wafer steppers Telescopes & interferometers 2

3 How? Vibration damping: Reduce the resonance peaks Vibration isolation: Prevent propagation of disturbances to sensitive payloads 3

4 Active damping in civil engineering structures TMD: Tuned Mass Damper = DVA: Dynamic Vibration Absorber AMD: Active Mass Damper 4

5 Single degree of freedom (d.o.f.) oscillator Free body diagram Free response: Characteristic equation: Eigenvalues: Solution? 5

6 6

7 (A,B, A1, B1 depend on initial conditions) Impulse response Spring and damping forces Have finite amplitudes 7

8 Impulse response for various damping ratios 8

9 Convolution Integral Linear system For a causal system: 9

10 Harmonic response 1. Undamped oscillator Dynamic amplification 10

11 Harmonic response 2. Damped oscillator Dynamic amplification 11

12 Bode plots Quality factor 12

13 Nyquist plot 13

14 Frequency Response Function (FRF) Harmonic excitation: FRF: The FRF is the Fourier transform of the impulse response 14

15 Fourier transform Convolution integral (linear systems): Parseval theorem: = energy spectrum of f(t) 15

16 Transient response (Beat) Undamped oscillator starting from rest: [ ] Modulating function At resonance 16

17 Transient response Beat Steady-state amplitude: The beat is a transient Phenomenon! 17

18 State Space form (system of first order differential equations) Oscillator: State variables: Alternative choice Of state variables: 18

19 Problem 1: Find the natural frequency of the single story building 19

20 Problem 2: write the equation of motion of the hinge rigid bar 20

21 Multiple degree of freedom systems In matrix form: 21

22 Mass matrix Stiffness matrix Damping matrix Symmetric & semi positive definite [ ] 22

23 Eigenvalue problem Free response of the conservative system (C=0) A non trivial solution exists if Eigenvalue problem The eigenvalues s are solutions of Because M and K are symmetric and semi-positive definite, the eigenvalues are purely imaginary: Natural frequency Mode shape Two-mass system: 23

24 Example: Two-mass system: Mode shapes: Natural frequencies: 24

25 Orthogonality of the mode shapes Upon permuting i and j, Subtracting: The mode shapes corresponding to distinct natural frequencies are orthogonal with respect to M and K Modal mass (or generalized mass) [Can be selected freely] Rayleigh quotient: 25

26 Orthogonality relationships in matrix form: with Notes: (1) Multiple natural frequencies: If several modes have the same natural frequency, they form a subspace and any vector in this subspace is also solution of the eigenvalue problem. (2) Rigid body modes: They have no strain energy: They also satisfy eigenvalue problem with which means that they are solutions of the 26

27 Free response from initial conditions 2n constants to determine from the initial conditions. Using the orthogonality conditions, 27

28 If there are rigid body modes ( i =0) Rigid body modes Flexible modes 28

29 Problem: write the mass and stiffness matrices for the structures 29

30 Kinetic energy: Strain energy: 30

31 Building with n identical floors Natural frequencies: Mode shapes: 31

32 Orthogonality relationships Modal coordinates x Assumption of modal damping: Set of decoupled equations of single d.o.f. oscillators: Mode i: Work of the external forces on mode i 32

33 Modal truncation Mode i: The modes within the bandwidth of f respond dynamically; y Those outside the bandwidth respond in a quasi-static manner. 33

34 Modal truncation If The response may be split into two groups of modes: Responding dynamically Responding in a quasi-static manner 34

35 Damping 35

36 36

37 Passive damping of very lightly damped Structures ( ) with shunted PZT patches 37

38 Rayleigh damping and are free parameters that Can be selected to match the Damping of two modes. 38

39 Dynamic flexibility matrix Harmonic response of: [ ] Modal expansion of G( ): 39

40 Modal truncationti [m<<n] Dynamic flexibility matrix Dynamic amplification of mode i 40

41 Modal truncation Residual modes Dynamic part restricted to the low frequency modes 41

42 Anti-resonances Diagonal terms of the dynamic flexibility matrix: collocated If the system is undamped,g kk is purely real: All the residues are positive and G kk is a monotonously increasing function of G kk ( ) = 0 42

43 G kk ( ) = 0 1. There is always one anti-resonance frequency between two resonance frequencies. 2. The anti-resonances depend d on the location of the collocated actuator/sensor t pair. 43

44 Poles of a single degree of freedom oscillator 44

45 Pole-zero pattern of a structure t with collocated actuator/sensor t pair With damping 45

46 Alternative form of the open-loop transfer function of collocated systems: undamped damped 46

47 Nyquist & Bode plots of collocated systems 47

48 Anti-resonances and constrained system 48

49 Lagrangian dynamics Principle of virtual work D Alembert principle Hamilton s s principle Lagrange s equations Examples First integrals of Lagrange s equations Green strain tensor Geometric strain energy (prestress) Buckling 1

50 Lagrangian Dynamics Newton ( ) introduced the equation of dynamics in vector form Hamilton ( ) wrote them in variational form, which ismoregeneral Because it can be extended to distributed and electromechanical systems. Generalized coordinates q i : set of coordinates describing the kinematics of the system. If minimum, they are independent. If not, they are connected by kinematic constraints. 2

51 Principle of virtualwork (example) Find the relationship between f and w at the static equilibrium The static equilibrium problem is transformed into kinematics: Virtual displacements: Principle of virtual work: =0 3

52 D Alembert s principle (Extension of the principle of virtual work to dynamics) The inertia forces are added: The virtual work of the effective forces on the virtual displacements Compatible with the constraints is zero. D Alembert s principle is most general, but it is difficult to apply because it refers to vector quantities expressed in inertial frame; it cannot be transformed directly in generalized coordinates. This achieved with Hamilton s principle. 4

53 Hamilton s Principle x i does not measure the displacements on the true path, but the separation between the true path and a perturbed one at a given time. Lagrangian: The actual path is that which cancels the variational indicator VI V.I. with respect To all arbitrary variations of the path between t 1 and t 2, compatible with the Kinematic constraints, and such that 5

54 Example: Hamilton: is eliminated by integrating by parts (differential equation of the pendulum) 6

55 Lagrange s equation Hamilton s principle contains only scalar work and energy quantities. Does not refer to any specific coordinate system and the system configuration may be expressed In terms of generalized coordinates: We assume an explicit dependency on time which is important for gyroscopic systems. Kinetic energy: General forms of T and V: 7

56 Lagrange s equation Example 1: Example 2: 8

57 Vibration of a linear discrete non gyroscopic system Lagrange Dissipation function All the forces non already included in D Viscous damping: 9

58 Example 3: Pendulum with a sliding mass Gravity Spring elastic energy Note: Try to obtain these results by writing the absolute acceleration in moving frame and applying Newton s law. Which way is easier? 10

59 11

60 Example 4: Pendulum with a sliding disk Example 3: The rod is a uniform bar of length l and mass M Two additional terms: Kinetic energy of the rod: [ ] Potential energy of the rod: 12

61 Example 5: Rotating pendulum T 2 T 0 13

62 Example 6: Rotating spring mass system (constant rotation speed) T 2 T 0 Lagrange: The system becomes unstable when 14

63 Example 7: Two axis oscillator with anisotropic stiffness (constantrotation speed) Generalized coordinates: position (x,y) in the rotating frame. Absolute velocity in rotating frame: Kinetic energy: Potential energy: [T 1 is responsible for the gyroscopic effects] Non conservative forces: or Note: This model is representative of a «Jeffcott rotor» with anisotropic shaft 15

64 Coupling between x and y Anti symmetric matrix of gyroscopic forces, which couples the motion in the two directions Modified potential: 16

65 In the particular case: (isotropic i & undamped) d) [ ] To study the stability, we assume a solution Characteristic equation: The solutions are purely imaginary: Campbell diagram 17

66 Anisotropic stiffness, undamped Characteristic equation: This term is negative for: The system is unstable (Routh Hurwitz) 18

67 Vibrating angular rate sensor 19

68 Vibrating angular rate sensor (2) Assuming: Harmonic excitation: 20

69 «First integrals» of the Lagrange equations: 1. Jacobi integral If the system is conservative and if the Lagrangian does not depend explicitly on time Total time derivative of L: Lagrange equation: 21

70 Euler theorem on homogeneous functions: If T n is homogeneous of degree n in some variables q i, It satisfies the identity: Since Jacobi integral or Painlevé integral If T=T 2, it reduces to the Conservation of the total energy: 22

71 2. Ignorable coordinate If the Lagrangian does not depend explicitly on some coordinate q s, the coordinate is ignorable: Lagrange equation: The generalized momentum associated with an ignorable coordinate is conserved. 23

72 Example: the sphericalpendulum p is ignorable: [Conservation of the angular momentum about Oz] 24

73 Green Strain Tensor Green Tensor: Quadratic part 25

74 Global rigid body rotation: 26

75 27

76 Geometric stiffness Tension prestresses tend to rigidify the system and increase the natural frequencies Compression prestresses tend to soften the system and decrease the natural frequencies For a linear elastic material: Constitutive equations: Strain energy density: Alternative form of the Constitutive equations: 28

77 Geometric strain energy due to prestress [Additional stress and strain From prestressed state] It can be shown that the strain energy can be written: It is impossible to account for the strain energy associated with the prestress if the linear strain tensor is used Geometric strain energy due to prestress (it may be >0 or <0 depending on the prestress) If >0, it rigidify the system and increases the resonances 29

78 Discrete system with prestresses For a discrete sytem, V g takes the form of a quadratic function of the generalized coordinates: Where K g is the geometric stiffness (no longer positive definite) The Lagrangian of the system is: Leading to the equation of motion: The natural frequencies are solutions of the eigenvalue problem: Buckling occurs when the smallest natural frequency is reduced to 0 30

79 Buckling If a loading produces a prestress and a geometric stiffness matrix Then, a proportional loading produces a prestress And a geometric stiffness matrix The solution of the eigenvalue problem gives the critical buckling amplification lf factor for the loaddistribution df definedby And gives the buckling mode. 31

80 Part 3 Continuous beams, bars and string Rayleigh Ritz method Beam with prestresses Rayleigh quotient 1

81 Vibration of beams (Euler Bernoulli) Vertical equilibrium: Rotational equilibrium: 2

82 Kinematic assumptions: The fibers are in a uniaxial state of stress and strain: (No axial loading) Bending moment: Partial differential equation 3

83 Alternative derivation from Hamilton s principle Strain Energy: Kinetic Energy: Virtual work: Hamilton: (The configuration is fixed at t 1 and t 2 ) 4

84 Finally: Boundary Conditions: At x=0, x=l PDE 5

85 Beam with axial prestress Vertical equilibrium: Moment about P: 6

86 Beam with axial prestress (from Hamilton s principle) One must add the Geometric energy of prestress: Green strain: Hamilton: PDE: Boundary conditions: 7

87 Free vibration of a uniform beam A solution of the form exists if: Introducing: Eigenvalue Problem: Non dimensional Frequency: Characteristic equation: General solution: 8

88 Decoupling the boundary conditions We define: Alternative form of the general solution: Decoupled!! 9

89 Example 1: Simply supported beam Solutions: (natural frequency) (mode shape) 10

90 Simply supported beam (modes) 11

91 Example 2: Free free beam 12

92 1. Rigid body modes: Double root at Boundary conditions: 2. Flexible modes: 13

93 For any given value of µ, 14

94 Orthogonality relationships Mode i satisfies: and 15

95 Orthogonality relationships (2) Modal mass Rayleigh quotient: Compare to similar results for discrete systems: 16

96 Modal decomposition (integrating by parts) (using the orthogonality relations) (work of p on mode k) 17

97 [Previous discussion About discrete systems] Modal truncation Mode i: The modes within the bandwidth of f respond dynamically; Those outside the bandwidth respond in a quasi static manner. 18

98 [Previous discussion About discrete systems] Modal truncation If The response may be split into two groups of modes: Responding dynamically Responding in a quasi static manner 19

99 Vibration of a string Can you derive this from Hamilton s principle?? Free vibration: Assuming: General solution: Boundary conditions at x=0 and x=l: 20

100 Axial vibration of a bar Free vibration: Speed of sound General solution: 21

101 Boundary conditions: fixed at x=0, free at x=l 22

102 Rayleigh Ritz method (global assumed mode method) Transform s PDE to ODE by making global assumptions on the displacement field Shape functions (assumed modes): Linearly independent, Complete, Satisfying the boundary conditions Example: simply supported beam: Note: In this particular example, the assumed modes are orthogonal functions, which makes the coefficients of the expansion independent. 23

103 Axial vibration of a bar (Rayleigh Ritz) Strain energy: The strain is uniform in the cross section Stiffness matrix: Positive definite 24

104 Similarly, the kinetic energy Mass matrix: Positive semi definite 25

105 Virtual displacements: (work of the distributed loads On the shape function) 26

106 Planar vibration of a beam (Rayleigh Ritz) Euler Bernoulli assumption: Strain energy: Rayleigh Ritz Assumption: Stiffness matrix 27

107 Similarly, the kinetic energy: Mass matrix: Virtual work of external forces: Generalized force: 28

108 With distributed damping: Damping matrix: 29

109 Beam with axial preload Geometric energy due to pre stress: Quadratic part of The Green strain Pre stress Geometric stiffness matrix: (not positive definite, depending on N) 30

110 Simply supported beam with uniform axial load P Natural frequency? N(x)= P Assumed mode: Equation of motion: P cr = Critical buckling load Analytical solution 31

111 Rayleigh quotient 1. Continuous beams The mode shapes and the natural frequencies satisfy Rayleigh quotient: where v(x) is any displacement Compatible with the kinematics Expanding v(x) in terms Ofthe mode shapes, 32

112 Rayleigh quotient 2.Discretesystems t The mode shapes and the natural frequencies satisfy: Rayleigh quotient: where x is any vector of generalized displacements compatible with the kinematics. Principle of stationarity: The Rayleigh quotient is stationary in the vicinity of the natural frequencies and an error of the first order on the mode shape produces anerror of the second order on the natural frequency. Consider: normalized such that 33

113 It follows that: and If the error is expanded in the mode shapes: is called the Hessian matrix 34

114 Recursive search for eigenvectors based on the Rayleigh quotient: The eigenvalue and the eigenvector of order k are such that How to project a vector in a space orthogonal to a set of modes? 1. The coefficients of the expansion are obtained from the orthogonality condition: 2. One removes the component along mode k: The projection matrix A k projects an arbitrary vector in the subspace orthogonal to mode k 35

115 Return to Problem 1: Find the natural frequency of the single story building Based on the exactstatic solution: 36

116 Single storey building with gravity loads Assumption: (one column) >24!! Geometric stiffness: Total stiffness: Natural frequency: 37

117 Finite Elements: Bar element Plane truss Beam element Beam with geometric stiffness Guyan reduction Craig Bampton reduction 1

118 Rayleigh Ritz Ritz vs. Finite Elements «local assumed modes» On the contrary to the Rayleigh Ritz method, the shape functions within an element are selected once and for all for every type of element. 2

119 Bar element Kinematic assumptions (within an element): (uniform strain within the element) Strain energy: 3

120 Generalized coordinates: Strain energy: Stiffness matrix: Kinetic energy: Mass matrix: 4

121 Summary: This model is sufficient for the analysis of the axial vibration of a bar However, if the bar is partof a truss structure, one mustalso consider the kinetic energy associated with the transverse motion. 5

122 Generalized coordinates: 6

123 Truss structure (assembly) 6 nodes with 2 d.o.f. each: (u i, v i ) Global l coordinates: Using the topology ofthe structure, the local coordinates of every element are related to the global coordinates: y 7

124 The total strain energy and kinetic energy are expressed in global cooordinates Total strain energy = sum of the strain energy of all the elements Global stiffness matrix: Total kinetic energy = sum of the kinetic energy of all the elements Global mass matrix: 8

125 Beam element (Euler Bernoulli) 1. Kinematics Shape functions: 9

126 Euler Bernoulli beam: In the element: «consistent» mass matrix (based on the same shape functions as the stiffness matrix) 10

127 «Lumped» mass matrix Theinertia associated with the rotation is neglected, and one half of the total mass is lumped at both ends of the element 11

128 Beam structure (assemby) Assembled stiffnessmatrix: 12

129 Assembled mass matrix: 13

130 Boundary conditions: Partition of the coordinates: where 14

131 After enforcing the boundary conditions: 15

132 Eigenvalue problem: Reduced Eigenvalue: In the theory of beams, The reduced frequency was defined Larger but quite close First mode: > (analytical result) >> (the approximation is poor) 1. The FE method overestimates the natural frequencies (Rayleigh quotient) 2. Good accuracy of m natural frequencies requires N>>m degrees of freedom 3. High frequency modes depend on the dicretization (no physical meaning). 16

133 Convergence of the F.E. model Consistent mass model Lumped mass model Number of elements 17

134 Geometric stiffness of a planar beam element Geometric energy due to an axial load N(x): Positive in traction 18

135 Example: Single storey building with gravity loads Each column may be modeled by a single finite element After enforcing the boundary conditions: Geometric stiffness: N= mg/2 Total stiffness : Previous result obtained with the Rayleigh Ritz method: (section 5.6.1) 19

136 Previous result obtained by the Rayleigh Ritz method (section 5.6.1) Single storey building with gravity loads Assumption: (one column) >24!! Geometric stiffness: Total stiffness: Natural frequency: 20

137 Guyan Reduction 1. The size of FE models is governed by the representation of the stiffness. 2. Automatic mesh generators tend to produce very big models (N>10 5 dof) d.o.f.). 3. Guyan s idea ( 60): quasi static condensation before solving the eigenvalue problem. The d.o.f. are separated in two groups: Masters : x 1 Slaves : x 2 (will be eliminated) Case 1: The slaves have no inertia and have no external forces applied: Involves only the master d.o.f. There isno approximation i in thiscase 21

138 Guyan s assumption: The quasi static relationship between masters and slaves applies in all cases Coordinate Transformation: Kinetic energy: Strain energy: Reduced mass and siffness matrices : Virtual work of External forces: Equation of motion after reduction: 22

139 Guidelines for selecting the master d.o.f. 1. The d.o.f. without inertia and external forces applied may be condensed without affecting the accuracy 2. The translation d.o.f. carry more information than the rotation d.o.f. 3. The master d.o.f. should be selected in order to maximize the first natural frequency i of the constrained system (x 1 blocked) [the error is an increasing function of the ratio: ( i i ) 2 ] 4. The frequency i of the first constrained mode should be far above the frequency band where the model is expected to be accurate. 23

140 Example 1: Clamped beam modelled with a single finite element Reduced eigenvalue : Eigenvalue problem: F.E. (2 dof) Analytical 24

141 (second row of the stiffness matrix) Static deflection Mass and stiffness After reduction: Constrained system: 25

142 First row of the stiffness matrix) Static deflection Constrained sytem: Poor quality! 26

143 Example 2: Comparison of various Guyan reductions 27

144 Craig Bampton reduction: X 1 contains all the d.o.f. of ofinterest Step 1: Guyan reduction: Step 2: Consider the constrained system (x 1 =0): Let 2 be a set of modes normalized according to The solution is enriched by adding a set of fixed boundary modes: 28

145 Finally, one gets the reduced equation: which h may beused with an increasingi number of constrained modes. 29

146 Example: dynamics of a segmented mirror E ELT Telescope Primary mirror One segment 30

147 31

148 32

149 Seismic Response 1

150 2

151 Hamilton: X 0 is arbitrary y satisfies the clamped condition at the base The kinetic energy depends On the absolute velocities The strain energy depends on the relative displacements =1 for a point force eliminated by Integrating by parts 3

152 Equations of motion: With viscous damping and assuming f=0 Structure clamped at the base: Orthogonality conditions: Assumption of modal damping: 4

153 Equations of motion in modal coordinates: Modal participation factor of mode i Modal participation vector Rigid body velocity: Total mass Of the structure 5

154 Seismic response Modal response: Modal participation factor Acceleration inthe structure: 6

155 Fig.2.11 Natural frequencies: Mode shapes: 7

156 Modal participation factor: Figure.7.2 8

157 Amplification of the response within the structure Precision devices should always be placed onlower floors!!! 9

158 Reaction force f 0 and Dynamic mass: f 0 is expressed in terms of the inertia forces Dynamic mass: 10

159 Alternative representation: f 0 is expressed in terms of the elastic forces (assuming no damping) For =0: Effective modal mass of mode i 11

160 Effective Modal mass Figure

161 Truncation to m modes: Quasi static contribution of the high frequency modes Statically correct (no error on the static mass) 13

162 Vibration alleviation with a Dynamic Vibration Absorber (DVA) =m a /m T =0.01 Mass ratio= m a /m 1 m 1? DVA parameters: 14

163 Dynamic Vibration Absorber (DVA) Equal peak design: (Den Hartog) 15

164 Equal peak design 16

165 Effect of the DVA on the reaction force 17

166 Response Spectrum (pseudo velocity) 18

167 19

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171 23

172 Rotor Dynamics: Jeffcott rotor Jeffcott rotor with damping; stability Gyroscopic effects Campbell diagram Rigid rotor on elastic supports Anisotropic shaft and supports 1

173 Jeffcott rotor (1919) Perfectly balanced: The point mass P coincide with the elastic center C Unbalanced: P has a small Excentricity with respect to C Jeffcott rotor: the stiffness k is isotropic the disk remains aligned along z In the fixed reference frame Oxyz, the coordinates of the elastic center C are taken as generalized coordinates: (x c,y c ) 2

174 Lagrange equations: External forces in fixed frame (if any) (1) Unbalanced response (F x =F y =0): A particular solution of the form: If Response= synchronous Whirl with amplitude: (critical velocity) 3

175 (critical velocity) Subcritical region: Supercritical region: (response 180 out of phase with the excitation) Self centering at high speed Laval (1889) was the first to run a steam turbine at a supercritical velocity, Demonstrating the weakness of Rankine s model. 4

176 Rankine s model Example 6: Rotating spring mass system (constant rotation speed) T 2 T 0 Lagrange: The system becomes unstable when 5

177 Complex coordinates: with (Non rotating external forces) With these notations, the unbalanced response reads: Free whirl: Free motion in complex coordinates: Forward whirl Backward whirl 6

178 7

179 Jeffcott rotor with viscous damping Two types of damping forces: 1. Stationary damping, associated with the non rotating parts (always stabilizing) 2. Rotating damping, associated with the motion of the rotor (destabilizing at supercritical velocities) (1) Stationary damping forces: Rotating damping forces: (velocity in Moving frame) Rotating damping forces Expressed in fixed frame: 8

180 Rotating damping forces expressed in fixed frame Skew symmetric Contribution to The stiffness matrix Or, using the complex coordinates 9

181 Stability analysis: Free whirling with damping The characteristic equation depends on the spin velocity of the rotor: Stable if: Routh Hurwitz (after some algebra!..): the system becomes unstable for At supercritical velocities, the stability is controlled by the ratio between the stationary and the rotating damping 10

182 Gyroscopic effects The Jeffcott model ignores the angular momentum of the disk. It cannot account for the gyroscopic effects which result from the interaction between the spin of the disk and the bending of the shaft. Rayleigh Ritz approximation: The transverse displacements in inertial frame are assumed of the form: The function f(z) is normalized in such way that f(a)=1. 11

183 Strain energy (assuming a uniform beam): with (same form as for the Jeffcott rotor) 12

184 Kinetic energy: Since the disk is axisymmetric, two rotations are neededto transform the inertial frame {x,y,z} into a moving frame where the inertia tensor is diagonal and constant: In this referential, the disk rotates about z 2 at and the inertia tensor Is diagonal: 13

185 The rotation velocity of the disk is: These contributions Belong to the same frame (moving frame) Must be transformed into the moving frame: The absolute rotation velocity is expressed in moving frame: Rotational kinetic energy: Responsible for the gyroscopic effects 14

186 Total kinetic energy: translation rotation Kinematic Assumptions: with (Generalized mass) (Gyroscopic constant) If there is a small excentricity of the center of mass, one must add (same as for Jeffcott rotor) 15

187 Dynamics with gyroscopiceffects (in fixed frame) Lagrange: External forces in the fixed reference frame In complex coordinates Reduces to the Jeffcott rotor in the angular momentum is neglected, 16

188 Free whirl, Campbell diagram Free motion in complex coordinates: Solution Campbell diagram Forward whirl Backward whirl 17

189 At point A: Critical velocity: 18

190 Unbalanced response Particular solution: At the critical velocity, the unbalanced response is tuned onthe forward whirl 19

191 Response to asynchronous force The response is obtained by superposition: 20

192 Rigid rotor on elastic support Potential energy: Kinetic energy: In what follows, we assume a=b 21

193 Lagrange equations: for a=b, two decoupled sets of equations: With complex cordinates: Cylindrical mode Conical mode 22

194 Conical mode: The eigenvalues are solutions of With the notations: Forward whirl (s=j 1 ): Backward whirl (s= j 2 ): 2 23

195 Critical velocity : No critical velocity for a disk (I z > I x ) 24

196 Rigid rotor on elastic support: Campbell diagram 25

197 Jeffcott rotor with anisotropic shaft (the equations are written in moving frame where the elastic properties are constant) In moving frame: 26

198 (From chapter on Lagrange s equation) Example 7: Two axis oscillator with anisotropic stiffness (constantrotation speed) Generalized coordinates: position (x,y) in the rotating frame. Absolute velocity in rotating frame: Kinetic energy: Potential energy: [T 1 is responsible for the gyroscopic effects] Non conservative forces: or Note: This model is representative of a «Jeffcott rotor» with anisotropic shaft 27

199 Stationary and rotating Damping forces: In moving frame: Lagrange: Stability of the anisotropic shaft: (without damping) 28

200 Stability of the anisotropic shaft: Characteristic equation: Unstable if: 29

201 Stability in presence of rotating damping: Characteristic equation: Unstable for all spin velocities above the critical speeds 30