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rotordynamics lecture3
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Bernhard Bettig
Mechanical Design Research LabMechanical Engineering - Engineering Mechanics Dept.
Michigan Technological University
Web site: http://www.me.mtu.edu/~mdrl
Rotordynamics: Unit 3- Modeling Coupling
Misalignments
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• In real rotordynamic systems, the vibration response does not approach zero as the rotation speed approaches zero.
• This is because of coupling misalignments.
Motivation for using Coupling Element
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Motivation for using Coupling Element
• Coupling misalignments must be added to conventional rotordynamic analysis to do accurate simulations.
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Coupling Element
• The coupling element we will be looking at is implemented using the “penalty function method”.
• It links two nodes at the coupling location using a “stiff” spring.
• It has 8 degrees of freedom (4 per node):
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Important Variables
• The displacements and rotations at Node 1: u1, v1, θ1, φ1.
• The displacements and rotations at Node 2: u2, v2, θ2, φ2.
• The coupling misalignments (magnitude e and phase τ; lateral cl and angular ca): ecl, τcl, eca, τca
• The coupling stiffnesses: kx = ky = kcl, kφ = kθ = kca
• The shaft rotation speed: Ω
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Important Assumptions
• No displacements in the Z direction.
• Linearization: u1, v1, θ1, φ1, u2, v2, θ2, and φ2 are small.
• Penalty Function Method:
if (k + δ1) u1 – (k + δ2) u2 = (k + δ3) e,
and δ1, δ2 , δ3 << k,
then u1 – u2 ≈ e.
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The Finite Element Matrices
• The coupling element is represented similar to a spring, however, the amount of coupling misalignment is given in the RHS harmonic force vector.
( ) ( )tt sc Ω+Ω= sincos FFku
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Finite Element Matrices
• The stiffness matrix is symmetric.
−−
−−
=
2
2
2
2
1
1
1
1
0.
00
000
000
0000
00000
000000
θ
φ
θ
φ
v
u
v
u
k
k
k
k
kk
kk
kk
kk
ca
cl
ca
cl
caca
clcl
caca
clcl
Sym
ku
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Finite Element Matrices
• The coupling misalignment appears similar to an imbalance.
( )( )( )
( )( )( )
( )( )
( )
( )( )( )( )( )( )
( )( )
( )t
ek
ek
ek
ek
ek
ek
ek
ek
t
ek
ek
ek
ek
ek
ek
ek
ek
cacaca
clclcl
cacaca
clclcl
cacaca
clclcl
cacaca
clclcl
cacaca
clclcl
cacaca
clclcl
cacaca
clclcl
cacaca
clclcl
Ω
−−−−
+Ω
−
−
−
−
= sin
cos
cos
sin
sin
cos
cos
sin
sin
cos
sin
sin
cos
cos
sin
sin
cos
cos
ττττττ
ττ
τττ
τττττ
ku
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Choosing Arbitrary Large k Values
Choosing too small a value will result in incorrect answers (the coupling is too “soft”, too “springy”).
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Choosing Arbitrary Large k Values
Choosing too large a value results in numerical instabilities. This occurs because some terms in the finite element equations are much larger than others.
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Calculating Realistic k Values
• If we consider deformations occurring at the coupling flanges, we can calculate realistic k values.
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Calculating Realistic k Values
• We need to consider displacements that are (a) lateral, (b) angular
(a) (b)
kcl (uouter – uinner) = Fx kca (φouter – φinner) = Mφ
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Lateral Stiffness kx Calculation
• Assume Airy stress function: Φ = C1 r ψ sinψ
where:
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Lateral Stiffness kx Calculation
• Solving:
E is the modulus of elasticity
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Lateral Stiffness kφ Calculation
• Assume Mφ is causing a shear stress in the shaded region (parabolic on h).
• Assume boundary conditions on vertical displacement w:
• Solving:
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Considering both Flanges of Coupling
• To include the effect of both flanges (one from each connected shaft) the stiffnesses must be added in parallel:
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Sample Questions
1. What are the equivalent spring stiffnesses for a coupling consisting of two similar steel flanges, each having:
• diameter 250 mm,• thickness of 40 mm, • shaft diameter 150 mm,• bolt hole circle diameter 200 mm, and• bolt hole diameter 20 mm.
2. What are the finite element matrix entries if there are coupling misalignments:
• Lateral – 2 mm at 30 degrees, and• Angular – 0.001 rad at 45 degrees.
3. What vibration amplitude would appear just below the coupling if a stiff journal bearing held the shaft just above the coupling?
