Finite element analysis of rotating structures

Finite element analysis of rotating structures Dr. Louis Komzsik Chief Numerical Analyst Siemens PLM Software

Why do rotor dynamics with FEM? Very complex structures with millions of degrees of freedom Wide spread industrial applications Aviation industry: Auto industry: Propellers Aero engines Starters Drive trains Energy industry: Power generator turbines Windmills

Tutorial topics Theoretical foundation of rotor dynamics Critical speeds and stability analysis Dynamic response solutions A demonstration example An industrial case study Application recommendations

A rotating mass point z z e 3 k e 2 Ω j y y r r x m r 3 = i= 1 3 = i= 1 r e i r r ( t) e ( t) i i Ω = Ω xi + Ω y j + Ω z k i i e 1 x r ( t) = r ( Ωt) i i e ( t) = e ( Ωt) i i

Velocities and accelerations Velocity dr dr de de = e + r = v e + r = v dt dt dt dt 3 3 3 3 i i i i i i i i i= 1 i= 1 i= 1 i= 1 Acceleration Special derivatives Acceleration dv dv de de d e = + + + dt dt dt dt dt 3 3 3 3 2 i i i i ei vi vi ri 2 i = 1 i = 1 i = 1 i = 1 de a = a + v + r dei ( t) = Ω e dt 3 3 2 i i 2 i i 2 i= 1 dt i= 1 dt i 2 d ei 2 d e dω de = ei + Ω dt dt dt dω a = a + 2 Ω v + r + Ω ( Ω r) dt i

Equilibrium of forces Inertial force Fin = ma Force in rotating system F = ma Centrifugal force F = mω ( Ω r) cf Coriolis and Eotvos force Euler force Force equilibrium in rotating system F = 2mΩ v C dω FE = m r dt F = Fin Fcf FC FE

Rotating mass particle of a flexible structure Ω = 0i + 0 j + Ωk z = z i j e 1 y y r r Ωt x x u m ux u = uy u z cos Ωt sin Ωt 0 H = sin t cos t 0 Ω Ω 0 0 1 r = [ H] r + [ H] u i j k Ω e1 = 0 0 Ω = cos Ωt sin Ωt 0 deɺ 1 dt

Analysis of flexible rotating structure Fundamental problem of calculus of variations Euler-Lagrange differential equation solution t 1 I = f ( uɺ, u, t) dt = extremum t 0 d f f = 0 dt uɺ u Hamilton s principle of conservative systems t t 1 0 Ldt = extremum The Lagrangian Lagrange s equation of motion L = E ( uɺ, u, t) E ( u) k d E E dt u ɺ u u E k k p + = p 0

Kinetic energy of mass particle Kinetic energy Ek 1 T = mrɺ rɺ 2 Mass particle velocity Rotation matrix derivative Auxiliary matrices Mass particle kinetic energy rɺ = Hɺ ( r + u) + Huɺ sin Ωt cos Ωt 0 Hɺ = Ω cos t sin t 0 Ω Ω = ΩH 0 0 0 1 0 0 0 1 0 T T J = H H = 0 1 0, P = H H = 1 0 0 0 0 0 0 0 0 m Ek = Ω r Jr + Ω r Ju + Ω u Ju + Ω uɺ P r + Ω u Puɺ + uɺ Iuɺ 2 2 T 2 T 2 T T T T T ( 2 2 2 )

Kinetic energy based equation of motion Displacement derivative Velocity and time derivative Lagrange s equation u d dt E k uɺ = m Ω Jr + Ω Ju + ΩPuɺ E k 2 2 T ( ) T = m( Ω P uɺ + Iuɺɺ ) d Ek Ek dt u = ɺ u m Ω P uɺ + Iuɺɺ Ω Jr Ω Ju + ΩP uɺ T 2 2 T ( ) Equation of motion miuɺɺ + mωp uɺ Ω mju Ω mjr = T 2 2 2 0 2 [ ] ɺɺ [ ] ɺ [ ] m u + 2 Ω c u Ω z u = { f c }

Non-conservative systems Full Lagrange s equation d E E dd dw dt uɺ u u duɺ du E k k p + + = 0 Elastic potential energy Dissipative energy E p D D W m u E p Work potential W

Elastic structure equilibrium equation Full Lagrange s equation Potential energy Stiffness matrix Geometric stiffness Damping matrices External forces d E k E E k p dd dw + + = ; i = 1,2,... n dt uɺ i ui ui duɺ i dui de p, E T { } [ B] [ E][ B]{ u} dv [ K ]{ u} d u = = de { } T ( σ ε ) E = 1 2 ( dv + σ ε dv ) = E + E T p 0 0 p, G p, E [ B '][ S ][ B' ]{ u} dv [ K ]{ u} p, G 0 d u = = dd d u = + { ɺ} dw d u = { } [ D]{ u ɺ } [ K ]{ u} { F} B G

Equation of motion in rotating system Second order, full, ordinary, non-homogeneous system M uɺɺ t D C uɺ t K Z K K u t F t 2 ( ) + ( + 2 Ω ) ( ) + ( Ω ( G ) + Ω B ) ( ) = ( ) Standard structural matrices (symmetric, pos. def.) M, D, K Geometric stiffness (centrifugal stiffening, spd) Centrifugal softening (symmetric, positive definite) Circulatory matrix (skew-symmetric) Gyroscopic (Coriolis) matrix (skew-symmetric) Active load vector K G Z K B C F

Centrifugal softening concept Particle at the end of flexible rod (only axial displacement) K r m u F c Small displacement (linear) solution Ku = Fc = mrω 2 Large displacement (nonlinear) solution 2 Ku = m( r + u) Ω Large displacement adjustment ( K Ω m) u = mrω 2 2 Static equilibrium with centrifugal softening 2 ( K Ω Z) u = Fc

Centrifugal stiffening concept F b Particle at the end of flexible beam (bending deformation) r u K b F c Moment equilibrium Force equilibrium Linear formulation Centrifugal stiffening matrix Fb r Fcu = K bur F r Fc ( K b + ) u = F r c Fb u = K bu F r c 2 mrω = = mω = Ω r b 2 2 K G

Equation of motion in fixed system Equations of motion M uɺɺ ( t) + ( D + Ω C) uɺ ( t) + ( K + Ω K ) u( t) = F ( t) No centrifugal (softening or stiffening) effect present Change in coefficient and content of gyroscopic matrix No Coriolis, only gyroscopic terms related to rotational DOF B Rotating system solution may be converted to fixed for visualization

θ z σ Coupling flexible rotating and zɶ = Bβ z stationary structures yɶ y r y ψ r x ϕ Ωt xɶ x θ w v ψ u ϕ ρ Aα r m u ϕ { ρ} = v ;{ α} = ψ w θ ϕ σ x β = ψ σ = σ y θ σ z { } ;{ } { u} σ β = ρ α cos Ωt sin Ωt 0 H = sin Ωt cos Ωt 0 0 0 1 { } { } { } r = σ + [ B]{ β} + [ H]{ r} + [ H] ρ + [ H][ A]{ α}

Time dependent coupled matrices Mass matrix M( Ω t) = M + M sin( Ω t) + M cos( Ω t) + M sin(2 Ω t) + M cos(2 Ωt ) 0 1 2 3 4 Coriolis matrix C( Ω t) = C + C sin( Ω t) + C cos( Ω t) + C sin(2 Ω t) + C cos(2 Ωt) 0 1 2 3 4 Centrifugal matrix Z( Ω t) = Z + Z sin( Ω t) + Z cos( Ω t) + Z sin(2 Ω t) + Z cos(2 Ωt) 0 1 2 3 4

Frequency domain solutions Harmonic solution iωt u( t) = e u( ω) ( ) i ω t F t = e F ( ω ) Frequency domain problem 2 ( ω M + iωd + K) u( ω) = F( ω) Time dependent matrices M ( Ω t ) = M + M ( Ω t) D ( Ω t ) = D + 2 Ω C ( Ω t ) K Ω t = K Ω Z Ω t K + Ω K 2 ( ) ( ( ) G ) B Free vibrations problem λ M + λd + K ϕ = λ = α + iω 2 ( ) 0;

Next topic Theoretical foundation of rotor dynamics Critical speeds and stability analysis Dynamic response solutions A demonstration example An industrial case study Application recommendations

Critical speed analysis Frequency of a mode (in RPM) f ( Ω ) = ω( Ω) / (2 π ) The critical speeds are found by intersecting the mode lines with the np lines where frequency is the nth multiple of rotor speed f ( Ω ) = np In rotating system intersection with - 0P indicates forward whirl - 2P indicates backward whirl 0 P : f = 0 2 P : f = 2* Ω In fixed system intersection with - 1P line indicates both whirls = negative slope - backward = positive slope forward 1 P : f = Ω Zero slope is indicative of a linear motion in both systems

Whirl directions Complex eigenvector ϕ( Ω ) = ϕ + ϕ i Re Im The mode shape whirl is w = ϕ ϕ = w i + w j + w k Re Im x y z Forward Backward Linear motion w z > w z < 0 0 w < ε z

Stability analysis Complex eigenvalue real part α ( Ω) The damping ratio Modal convention by 2 g ( Ω ) = 2 α ω The mode is in the Stable region Unstable region g ( Ω ) < 0 g ( Ω ) > 0

Undamped rotating mass particle Equation of motion in time domain problem m 0 0 m k 0 m 0 0 m m 0 0 k 0 m 2 { uɺɺ } + 2Ω { uɺ } + Ω { u} = { p} y Frequency domain free vibrations k m x m 0 0 m k 0 m 0 2 0 0 m + Ω m 0 + Ω = 0 k 0 m λ 2 λ 2 ϕ { } { } z k Eigenvalue solution is purely imaginary λ ( Ω ) = 0 ± ( Ω ± 1)i

Critical speeds of undamped rotating mass particle Rotating system Campbell diagram Critical speeds - 0P at 1 Hz : forward whirl - 2P at 1 Hz: backward whirl Stability: Unconditional since g = 0

Damped rotating mass particle problem Equation of motion in time domain m 0 d 0 0 m k 0 m 0 0 m 0 d m 0 0 k 0 m 2 { uɺɺ } + + 2Ω { uɺ } + Ω { u} = { p} y Frequency domain free vibrations + + = 0 m 2Ωm d 0 k Ω m 2 2 m 0 d 2Ωm k Ω m 0 λ λ ϕ 2 Eigenvalue solution has real part λ ( Ω ) = ± α ( Ω ) ± ω ( Ω )i { } { 0} z d m k x

Damped rotating mass particle stability Damping ratio plot Mode shapes Forward whirl: 1 unstable above 1 Hz Backward whirl: 2 stable throughout

Next topic Theoretical foundation of rotor dynamics Critical speeds and stability analysis Dynamic response solutions A demonstration example An industrial case study Application recommendations

Dynamic response analyses Frequency response Synchronous analysis: excitation frequency is a function of rotor speed load is scaling with speed: simulate mass unbalance Asynchronous analysis: constant rotor speed, excitation frequency is changing load does not scale with speed: simulate gravity load Transient response Synchronous analysis: excitation time function is in synchrony with rotor speed simulate the start-up or wind-down process Asynchronous analysis: excitation time function is changing for constant rotor speed simulate behavior at the operational point of the rotor

Damped rotating mass particle frequency response Forward rotating force load exciting forward whirl: mode 1 Asynchronous analysis: dotted Synchronous analysis: solid Peaks increase due to the decreasing damping of mode 1 Fixed reference system result

Damped rotating mass particle frequency response Backward rotating force load exciting backward whirl: mode 2 Asynchronous and synchronous analysis both dotted Peaks decrease due to the increasing damping of mode 2 Fixed reference system result

Damped rotating mass particle transient response Response to a sweep load Sine excitation in x-direction Cosine excitation in y-direction Synchronous analysis Resonance at critical speed: 1 Hz Fixed reference system results

Damped rotating mass particle transient response Response to a sweep load Sine excitation in x-direction Cosine excitation in y-direction Asynchronous analysis Resonance at critical speed: 1 Hz Fixed reference system results

Next topic Theoretical foundation of rotor dynamics Critical speeds and stability analysis Dynamic response solutions A demonstration example An industrial case study Application recommendations

A demonstration example Long cylindrical shaft Heavy central hub Spring and damper supports Ω Rotor speed: Ω = 0 24000 RPM Solutions sought: Critical speeds and mode shapes Mass unbalance and start-up

Critical speed analysis Modes 1,2 Symmetric bending forward-backward whirl Mode 3 - Torsion mode Mode 4 Asymmetric bending backward whirl Mode 5 - Asymmetric bending forward whirl Mode 6 Axial mode Mode 5 Forward Whirl Mode 4 Backward Whirl 1P Mode 6 Axial Critical speeds: Independent: 1,460 RPM Backward: 8,420 RPM Forward: 17,800 RPM Mode 3 Torsion Mode 1, 2 Forward and backward 1460 RPM Critical Speed Backward & Forward 8420 RPM Critical Speed Backward 17800 RPM Critical Speed Forward

0 RPM Symmetric bending @24 Hz

24,000 RPM Symmetric forward whirl

24,000 RPM Symmetric backward whirl

0 RPM Asymmetric bending @179 Hz

24,000 RPM Asymmetric forward whirl

24,000 RPM Asymmetric backward whirl

Mass unbalance simulation Force load on central hub Lateral and vertical direction Phased with positive rotation F me π π t 2 2 x( Ω ) = 4 Ω sin(2 Ω ) F me π π t π 2 2 y ( Ω ) = 4 Ω sin(2 Ω + / 2) Unbalance: mass = m kg eccentricity = e mm Critical Speed 1460 RPM (24 Hz) Critical Speed 17800 RPM (296 Hz) Frequency response analysis 0-500 RPM Result: Forward whirl modes excited Due to force positive rotation Rotating Rotor Non-Rotating Rotor

Rotor start-up simulation Force load and eccentricity as before t=.3 sec Critical Speed at 24 Hz t = 3.0 sec Critical Speed at 296 Hz Transient response analysis 0-4 sec Result: Forward whirl modes excited Short dwell time

Next topic Theoretical foundation of rotor dynamics Critical speeds and stability analysis Dynamic response solutions A demonstration example An industrial case study Application recommendations

Industrial case study Turbine wheel with 30 blades Range: 0 25,000 RPM Number of steps: 100 Number of nodes: 45,350 Number of elements: 24,199 Degrees of freedom: 135,035

Critical speed analysis 16,969 RPM Critical Speed Backward whirl 1,389 RPM Critical Speed Backward whirl 4,664 RPM Critical Speed Backward whirl 4,085 RPM Critical Speed Forward whirl

Rotor disk tilting modes Mode shape at 20,000 RPM Real Imaginary

Synchronous circumferential bending Real Imaginary Mode shape at 20,000 RPM

Asynchronous circumferential bending Real Mode shape at 20,000 RPM Imaginary

Shaft symmetric bending Real Mode shape at 20,000 RPM Imaginary

Next topic Theoretical foundation of rotor dynamics Critical speeds and stability analysis Dynamic response solutions A demonstration example An industrial case study Application recommendations

Coupled solution strategies The fully coupled equations of motion between the support and rotor contain periodic terms Hence evaluation of the solution at multiple azimuth angles is required and makes the solution very expensive Coupled solution strategies: execute at a discrete azimuth angle for a rotor speed sweep with specified speed discretization execute a full circle azimuth sweep with specified angular discretization at a given fixed rotor speed

Reference system selection The fixed (non-rotating) reference system is an inertial system while the rotating coordinate system is not The equations of motion are different in the two reference systems, resulting in different but compatible solutions for certain models The content of some of the participating matrices is also different between the two reference systems Rotating reference frame is the preferred analysis solution for general finite element models with large radius components It is also possible to solve the equation of motion in one reference frame and convert the solution to the other one Presenting results in the fixed reference system is more intuitive for the engineer

Modeling approach Models built from solid elements must be analyzed in the rotating system, since the nodal rotations are not present in the model, but the rotational matrices in the fixed system contain only terms related to the nodal rotations This can be overcome by covering solid model surfaces with thin shells, but this approach has challenges in establishing the proper characteristics Models built from shell elements may be analyzed in both systems, but rotational system is still preferred, because the shell normal DOFs are either not present or approximated

The effect of rotor symmetry Symmetric rotors Turbine engines I x = I y Unsymmetric rotors Automobile crankshaft I x I y Fixed reference system formulation Rotating reference system formulation C = f ( I ); Z = 0 z C = f ( I, I, I ); Z = f ( I, I, I ) z x y z x y Symmetric rotor may be analyzed with fixed zero azimuth angle based time independent coupling terms Unsymmetric rotor must be analyzed with azimuth angle based time dependent coupling terms

The effect of support symmetry Symmetric support Classical bearings via spring and dampers k = k ; b = b xy yx xy yx Example: tower of a vertical axis wind turbine Unsymmetric support Fluid film (journal) bearings k k ; b b xy yx xy yx Example: support of horizontal axis turbines Symmetric support may be analyzed with fixed zero azimuth angle based time independent coupling terms Unsymmetric support must be analyzed with azimuth angle based time dependent coupling terms

Advanced applications of rotordynamics Simulate the rotational loads occurring during aircraft maneuvers Simulate the blade loss and rubbing phenomena in aero engines with nonlinear rotor dynamic solutions Simulate the dynamic behavior of wind turbines by real-time coupling with unsteady airflow solution Simulate the operation of fossil turbines by co-simulating with heat-fluid flow solution

References

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