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Identifying the “wavemaker” of fluid/structure instabilities
Olivier Marquet1 & Lutz Lesshafft2
1 Department of Fundamental and Experimental Aerodynamics2 Laboratoire d’Hydrodynamique, CNRS-Ecole Polytechnique
24th ICTAM
21-26 August 2016, Montreal, Canada
Context
2
Flow-induced structural vibrations
Civil engineering Aeronautics Offshore-marine industry
Stability analysis of the fluid/structure problem
A tool to predict the onset of vibrations
Model problem: spring-mounted cylinder flow
3
�� = ����
Reynolds number
= �⁄Structural frequency
�� = 40� = 0.010.4 < � < 1.2 � = 10�, 200, 10(�� = 0.73)
�Structural damping
� = � ���Density ratio
One spring - cross-stream
Stability analysis of the coupled fluid/solid problem
4
�(��) !��"#!� (� , �)
$�$ = (% + '�) (� 00 )
$�$
$′(+, ,) = ($�, $) + � -./0 1 + 2. 2.Fluid/solid
components
Growth rate/frequency
Cossu & Morino (JFS, 2000)
Steady solution – No solid component
Diagonal operators: intrinsinc dynamics
5
�(��) !��"#!� (� , �)
$�$ = (% + '�) (� 00 )
$�$
$′(+, ,) = ($�, $) + � -./0 1 + 2. 2.Fluid/solid
components
Growth rate/frequency
Cossu & Morino (JFS, 2000)
Steady solution – No solid component
Solid operator
(damped harmonic oscillator)
Fluid operator
Off-diagonal operators: coupling terms
6
�(��) !��"#!� (� , �)
$�$ = (% + '�) (� 00 )
$�$
$′(+, ,) = ($�, $) + � -./0 1 + 2. 2.Fluid/solid
components
Growth rate/frequency
Mougin & Magaudet (IJMF, 2002), Jenny et al. (JFM 2004)
Steady solution – No solid component
Coupling operator
(boundary condition + non-inertial terms)
Coupling operator
Weighted fluid force
Stability analysis at large density ratio
7
� = 0.75 0.4 < � < 1.2Structural Mode (� = 0.75)
Wake Mode
� = 10�
Methods to identify structural and wake modes
- Vary the structural frequency
- Look at the vertical displacement/velocity (not the fluid component)
Stability analysis at smaller density ratio
8
SM WMEffect of decreasing the density ratio
� = 200� = 10� � = 10
- Stronger interaction between the two branches
- The two branches exchange their « nature » for small �- Coalescence of modes (not seen here)
SM is destabilized WM is destabilizedStable
Zhang et al (JFM 2015), Meliga & Chomaz (JFM 2011)
Objective and outlines
9
Objective:
- Identify the « wavemaker » of a fluid/solid eigenmode
- Quantify the respective contributions of fluid and solid
dynamics to the eigenvalue of a coupled eigenmode
Outlines:
1 - Presentation of the operator/eigenvalue decomposition
2 - The infinite mass ratio limit ( � = 10�)3 - Finite mass ratio ( � = 200and � = 10)
State of the art for the method
10
• Energetic approach of eigenmodes (Mittal et al, JFM 2016)
- Transfer of energy from the fluid to the solid / Growth rate
- Does not identify the « wavemaker » region in the fluid
• Wavemaker analysis (Giannetti & Luchini, JFM 2008, …)
- Structural sensitivity analysis of the eigenvalue problem.
- Largest eigenvalue variation induced by any perturbation of the
operator ?
- Output of this analysis is an inequality. We would like an identify !
• Operator/Eigenvalue decomposition
From operator to eigenvalue decomposition
11
$ = 4 + 5 $ = 6$Operator decomposition
In general, $ is not an eigenmode of 4 or 5 , so
4$ = 64$ + 74 5$ = 65$ + 75
Eigenvalue decomposition
with residuals 74 ≠ 0, 75 ≠ 0 but 74 = −75 = 7
64 + 65 = 6
How to compute the eigenvalue contributions 64/65 ?
Computing eigenvalue contributions
12
Expansion of the residual on the set of other eigenmodes $;
Orthogonal projection on the mode $using the adjoint mode $.
$.<( 4$) = 64($.<$) +=7;($.<$;);= 1 Bi-orthogonality= 0Normalisation
4$ = 64$ +=7;$;;
7 = =7;$;;
64 = $.<( 4$) 65 = $.<( 5$)
Adjoint mode-based decomposition
6 = 64 + 65
Why this particular eigenvalue decomposition?
13
For an identical decomposition of the operator,
other eigenvalue decompositions are possible
6>4/5 = 64/5 ±=7;($<$;);
≠ 0
Non-orthogonal projection on the mode $
6>4 = $<( 4$) 6>5 = $<( 5$)
Direct mode-based decomposition
6 = 6>4 + 6>5
4$ = 64$ +=7;$;;
But it includes contributions from other eigenmodes
M.Juniper (private communication)
5$ = 65$ −=7;$;;
Application to the spring-mounted cylinder flow
14
6 = 6� + 6Adjoint mode-based decomposition
6� = $�.<( �$� + !�$)Fluid contribution
6 = $.<( $ + �"#!� $�)Solid contribution
� !��"#!�
$�$ = 6 (� 00 )
$�$
6� = @ $�.∗(+) ⋅ ( �$� + !�$) + C+D
Local contributions
Infinite mass ratio - Fluid Modes
15
6 = 6� + 6Adjoint mode-based decomposition
6� = $�.<( �$� + !�$)Fluid contribution
6 = $.<( $ + �"#!� $�)Solid contribution
Fluid ModesEF = G
6� = $�.< �$� = 6
� !�G
$�$ = 6 (� 00 )
$�$�"# = 0
�"# = 0
6 = 0OK
Infinite mass ratio - Structural Mode
16
6 = 6� + 6Adjoint mode-based decomposition
6� = $�.<( �$� + !�$)Fluid contribution
6 = $.<( $ + �"#!� $�)Solid contribution
Structural Mode
6� = 0
� !�G
$�$ = 6 (� 00 )
$�$�"# = 0
�"# = 0
6 = $.< $ = 6OK
EH. = G
Infinite mass ratio – Direct mode-based decomposition
17
6 = 6>� + 6>Direct mode-based decomposition
6>� = EHI( �$� + !�$)Fluid contribution
6> = EFI( $ + �"#!� $�)Solid contribution
Structural Mode
6>� = 6($�<$�)
� !�0
$�$ = 6 (� 00 )
$�$�"# = 0
�"# = 0
6> = 6($<$)NOT OK
�$� + !�$ = 6$�
Infinite mass ratio – Direct mode-based decomposition
18
6 = 6>� + 6>Direct mode-based decomposition
6>� = EHI( �$� + !�$)Fluid contribution
6> = EFI( $ + �"#!� $�)Solid contribution
Structural Mode
6>� = 6($�<$�)
� !�0
$�$ = 6 (� 00 )
$�$�"# = 0
�"# = 0
6> = 6($<$)NOT OK(large fluid response)
(6) − �)$� = !�$
Mass ratio J = KGG: Structural Mode
19
SMWM
Growth rate (SM)
Frequency (SM)
• The frequency is quasi-equal to �• The growth rate gets positive for �close ��
Structural Mode: frequency/growth rate decomposition
20
Frequency
Fluid
Solid
FluidSolid
Growth rate
Very large (resonance) and opposite contributions
Spatial distribution of the fluid component
21
Growth rate
Local fluid contributions
The solid contribution
induces
destabilisation
The fluid contribution
induces
destabilisation
Phase change between and $�. �$�Schmid & Brandt (AMR 2014)
Mass ratio J = LG: Wake Mode
22
SMWMFrequency (WM)
For small �, � ∼ �� - For large �, � ∼ �
Growth rate (WM)
Wake Mode: frequency decomposition
23
Frequency SolidFluid� = 10
For small �: �� ∼ � = frequency selection by the fluid
Unstable range : �� ∼ �
For large �: � ∼ � = frequency selection by the solid
Wake Mode: growth rate decomposition
24
Growth rate Solid Fluid� = 10
Small �destabilization due to the solid
Large �destabilization due to the fluid
Wake Mode: growth rate decomposition
25
Growth rate Solid Fluid� = 10
Small �destabilization due to the solid
Large �destabilization due to the fluid
Wake Mode: growth rate decomposition
26
Growth rate Solid Fluid� = 10
Small �destabilization due to the solid
Large �destabilization due to the fluid
Conclusion & perspectives
27
• The adjoint-based eigenvalue decomposition enables to discuss
- the frequency selection/ the destabilization of coupled modes
- The localization of this process in the fluid
• Comparison with structural sensitivity (not shown here)
Identification of the same spatial regions
• Use this decomposition in more complex fluid/structure problem
Cylinder with a flexible splitter plate (Jean-Lou Pfister)
Thank you to
Pure modes (infinite mass ratio)
28 Titre présentation
6 $�$ = O !�0 P
$�$
Fluid
modes
6∗ Q�Q = O< 0!�< P<
Q�Q
Direct Adjoint
Solid
modes
6$� = O$�$ = 0
6$ = P$(6) − O)$� = !�$
6�∗Q�R = O<Q�R
6�∗) − P< QR = !�<Q�R
6∗QR = P<QRQ�R = 0
Projection of coupled problem on pure fluid modes
29 Titre présentation
(6) − O)$� = !�$(6) − P)$ = �"#!� $�
Q�R< 6) − O $� + QR< 6) − P $ − Q�R<!� $= �"#QR<!�$�6∗Q�R − O<Q�R
<$� + 6∗QR − P<QR − !�<Q�R<$ = �"#QR<!�$�
6 − 6� (Q�R<$�) + 6 − 6� (QR<$) = �"#QR<!� $�
6 − 6� = �"#QR<!�$�(Q�R<$� + QR<$)
Projection of coupled problem on pure solid modes
30 Titre présentation
(6) − O)$� = !�$(6) − P)$ = �"#!� $�
Q�R< 6) − O $� + QR< 6) − P $ − Q�R<!� $= �"#QR<!�$�6∗Q�R − O<Q�R
<$� + 6∗QR − P<QR − !�<Q�R<$ = �"#QR<!�$�
(6 − 6)(QR<$) = �"#QR<!�$�
6 − 6 = �"#QR<!� $�QR<$
Direct-based decomposition of the unstable mode
31 Titre présentation
Frequency
Fluid
Solid
Growth rate
� = 200
Solid
Fluid
Free oscillation
�� = 40; � = 50
No oscillation� = 0.60
� = 0.66 Weak oscillation
� = 0.90 Strong oscillation
� = 1.10 No oscillation
Solid displacement – Fluid fields
Multiple solutions