1. Identifying the “wavemaker” of fluid/structure instabilities
Olivier Marquet1 & Lutz Lesshafft2
1 Department of Fundamental and Experimental Aerodynamics
2 Laboratoire d’Hydrodynamique, CNRS-Ecole Polytechnique
24th ICTAM
21-26 August 2016, Montreal, Canada

2. Context 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

3. Model problem: spring-mounted cylinder flow
One spring - cross-stream
𝜅 𝑠 = 𝜅 𝑚
Structural frequency
𝛾 𝑠
Structural damping
𝜌= 𝜌 𝑠 𝜌 𝑓
Density ratio
𝑅𝑒= 𝑈 ∞ 𝐷 𝜈
Reynolds number
0.4< 𝑘 𝑠 <1.2
𝛾 𝑠 =0.01
𝜌= 10 6 , 200, 10
𝑅𝑒=40
( 𝜔 𝑣 =0.73)

4. Stability analysis of the coupled fluid/solid problem
Steady solution – No solid component
𝑞′(𝑥,𝑡)=( 𝑞 𝑓 , 𝑞 𝑠 ) 𝑥 𝑒 𝜆+𝑖 𝜔 𝑡 +𝑐.𝑐.
Fluid/solid
components
Growth rate/frequency
𝐿 𝑓 (𝑅𝑒) 𝐶 𝑓𝑠 𝜌 −1 𝐶 𝑠𝑓 𝐿 𝑠 ( 𝑘 𝑠 , 𝛾 𝑠 ) 𝑞 𝑓 𝑞 𝑠 =(𝜆+𝑖𝜔) 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
Cossu & Morino (JFS, 2000)

5. Diagonal operators: intrinsinc dynamics
Steady solution – No solid component
𝑞′(𝑥,𝑡)=( 𝑞 𝑓 , 𝑞 𝑠 ) 𝑥 𝑒 𝜆+𝑖 𝜔 𝑡 +𝑐.𝑐.
Fluid/solid
components
Growth rate/frequency
Fluid operator
𝐿 𝑓 (𝑅𝑒) 𝐶 𝑓𝑠 𝜌 −1 𝐶 𝑠𝑓 𝐿 𝑠 ( 𝑘 𝑠 , 𝛾 𝑠 ) 𝑞 𝑓 𝑞 𝑠 =(𝜆+𝑖𝜔) 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
Solid operator
(damped harmonic oscillator)
Cossu & Morino (JFS, 2000)

6. Off-diagonal operators: coupling terms
Steady solution – No solid component
𝑞′(𝑥,𝑡)=( 𝑞 𝑓 , 𝑞 𝑠 ) 𝑥 𝑒 𝜆+𝑖 𝜔 𝑡 +𝑐.𝑐.
Fluid/solid
components
Growth rate/frequency
Coupling operator
(boundary condition + non-inertial terms)
𝐿 𝑓 (𝑅𝑒) 𝐶 𝑓𝑠 𝜌 −1 𝐶 𝑠𝑓 𝐿 𝑠 ( 𝑘 𝑠 , 𝛾 𝑠 ) 𝑞 𝑓 𝑞 𝑠 =(𝜆+𝑖𝜔) 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
Coupling operator
Weighted fluid force
Mougin & Magaudet (IJMF, 2002), Jenny et al. (JFM 2004)

7. Stability analysis at large density ratio
𝜌= 10 6
𝑘 𝑠 =0.75
0.4< 𝑘 𝑠 <1.2
Structural Mode ( 𝑘 𝑠 =0.75)
Wake Mode
Methods to identify structural and wake modes
- Vary the structural frequency
- Look at the vertical displacement/velocity (not the fluid component)

8. Stability analysis at smaller density ratio
Effect of decreasing the density ratio WM
𝜌= 10 6
𝜌=200
𝜌=10
Stable
SM is destabilized
WM is destabilized
Stronger interaction between the two branches
The two branches exchange their « nature » for small 𝜌
Coalescence of modes (not seen here)
Zhang et al (JFM 2015), Meliga & Chomaz (JFM 2011)

9. Objective and outlines
- 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 ( 𝜌= 10 6 )
3 - Finite mass ratio ( 𝜌=200 and 𝜌=10 )

10. State of the art for the method
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

11. From operator to eigenvalue decomposition
Operator decomposition
𝐿 𝑞= 𝐿 𝑎 + 𝐿 𝑏 𝑞=𝜎 𝑞
In general, 𝑞 is not an eigenmode of 𝐿 𝑎 or 𝐿 𝑏 , so
𝐿 𝑎 𝑞= 𝜎 𝑎 𝑞+ 𝑟 𝑎
𝐿 𝑏 𝑞= 𝜎 𝑏 𝑞+ 𝑟 𝑏
𝑟 𝑎 ≠0, 𝑟 𝑏 ≠0 but 𝑟 𝑎 =− 𝑟 𝑏 =𝑟
with residuals
Eigenvalue decomposition
𝜎 𝑎 + 𝜎 𝑏 =𝜎
How to compute the eigenvalue contributions 𝜎 𝑎 / 𝜎 𝑏 ?

12. Computing eigenvalue contributions
Expansion of the residual on the set of other eigenmodes 𝑞 𝑘
𝐿 𝑎 𝑞= 𝜎 𝑎 𝑞+ 𝑘 𝑟 𝑘 𝑞 𝑘
𝑟= 𝑘 𝑟 𝑘 𝑞 𝑘
Orthogonal projection on the mode 𝑞
using the adjoint mode 𝑞 +
𝑞 + 𝐻 (𝐿 𝑎 𝑞)= 𝜎 𝑎 ( 𝑞 + 𝐻 𝑞)+ 𝑘 𝑟 𝑘 ( 𝑞 + 𝐻 𝑞 𝑘 ) =1
Bi-orthogonality =0
Normalisation
𝜎 𝑎 = 𝑞 + 𝐻 (𝐿 𝑎 𝑞)
𝜎 𝑏 = 𝑞 + 𝐻 (𝐿 𝑏 𝑞)
Adjoint mode-based decomposition
𝜎= 𝜎 𝑎 + 𝜎 𝑏

13. Why this particular eigenvalue decomposition?
For an identical decomposition of the operator,
other eigenvalue decompositions are possible
Non-orthogonal projection on the mode 𝑞
𝜎 𝑎 = 𝑞 𝐻 (𝐿 𝑎 𝑞)
𝜎 𝑏 = 𝑞 𝐻 (𝐿 𝑏 𝑞)
Direct mode-based decomposition
𝜎= 𝜎 𝑎 + 𝜎 𝑏
But it includes contributions from other eigenmodes
𝐿 𝑎 𝑞= 𝜎 𝑎 𝑞+ 𝑘 𝑟 𝑘 𝑞 𝑘
𝐿 𝑏 𝑞= 𝜎 𝑏 𝑞− 𝑘 𝑟 𝑘 𝑞 𝑘
𝜎 𝑎/𝑏 = 𝜎 𝑎/𝑏 ± 𝑘 𝑟 𝑘 ( 𝑞 𝐻 𝑞 𝑘 ) ≠0
M.Juniper (private communication)

14. Application to the spring-mounted cylinder flow
𝐿 𝑓 𝐶 𝑓𝑠 𝜌 −1 𝐶 𝑠𝑓 𝐿 𝑠 𝑞 𝑓 𝑞 𝑠 =𝜎 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
𝜎= 𝜎 𝑓 + 𝜎 𝑠
Adjoint mode-based decomposition
𝜎 𝑓 = 𝑞 𝑓 +𝐻 (𝐿 𝑓 𝑞 𝑓 + 𝐶 𝑓𝑠 𝑞 𝑠 )
Fluid contribution
𝜎 𝑠 = 𝑞 𝑠 +𝐻 ( 𝐿 𝑠 𝑞 𝑠 + 𝜌 −1 𝐶 𝑠𝑓 𝑞 𝑓 )
Solid contribution
𝜎 𝑓 = 𝐷 𝑞 𝑓 +∗ (𝑥)⋅(𝐿 𝑓 𝑞 𝑓 + 𝐶 𝑓𝑠 𝑞 𝑠 ) 𝑥 𝑑𝑥
Local contributions

15. Infinite mass ratio - Fluid Modes
𝐿 𝑓 𝐶 𝑓𝑠 𝟎 𝐿 𝑠 𝑞 𝑓 𝑞 𝑠 =𝜎 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
𝜌 −1 =0
𝜎= 𝜎 𝑓 + 𝜎 𝑠
Adjoint mode-based decomposition
Fluid contribution
Solid contribution
𝜎 𝑓 = 𝑞 𝑓 +𝐻 (𝐿 𝑓 𝑞 𝑓 + 𝐶 𝑓𝑠 𝑞 𝑠 )
𝜎 𝑠 = 𝑞 𝑠 +𝐻 ( 𝐿 𝑠 𝑞 𝑠 + 𝜌 −1 𝐶 𝑠𝑓 𝑞 𝑓 )
𝜌 −1 =0
𝒒 𝒔 =𝟎
Fluid Modes
𝜎 𝑓 = 𝑞 𝑓 +𝐻 𝐿 𝑓 𝑞 𝑓 =𝜎 OK
𝜎 𝑠 =0

16. Infinite mass ratio - Structural Mode
𝐿 𝑓 𝐶 𝑓𝑠 𝟎 𝐿 𝑠 𝑞 𝑓 𝑞 𝑠 =𝜎 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
𝜌 −1 =0
𝜎= 𝜎 𝑓 + 𝜎 𝑠
Adjoint mode-based decomposition
Fluid contribution
Solid contribution
𝜎 𝑓 = 𝑞 𝑓 +𝐻 (𝐿 𝑓 𝑞 𝑓 + 𝐶 𝑓𝑠 𝑞 𝑠 )
𝜎 𝑠 = 𝑞 𝑠 +𝐻 ( 𝐿 𝑠 𝑞 𝑠 + 𝜌 −1 𝐶 𝑠𝑓 𝑞 𝑓 )
𝜌 −1 =0
𝒒 𝒇 + =𝟎
Structural Mode
𝜎 𝑓 =0 OK
𝜎 𝑠 = 𝑞 𝑠 +𝐻 𝐿 𝑠 𝑞 𝑠 =𝜎

17. Infinite mass ratio – Direct mode-based decomposition
𝐿 𝑓 𝐶 𝑓𝑠 0 𝐿 𝑠 𝑞 𝑓 𝑞 𝑠 =𝜎 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
𝜌 −1 =0
𝜎= 𝜎 𝑓 + 𝜎 𝑠
Direct mode-based decomposition
Fluid contribution
Solid contribution
𝜎 𝑓 = 𝒒 𝒇 𝑯 (𝐿 𝑓 𝑞 𝑓 + 𝐶 𝑓𝑠 𝑞 𝑠 )
𝜎 𝑠 = 𝒒 𝒔 𝑯 ( 𝐿 𝑠 𝑞 𝑠 + 𝜌 −1 𝐶 𝑠𝑓 𝑞 𝑓 )
𝐿 𝑓 𝑞 𝑓 + 𝐶 𝑓𝑠 𝑞 𝑠 =𝜎 𝑞 𝑓
𝜌 −1 =0
Structural Mode
NOT OK
𝜎 𝑓 =𝜎 ( 𝑞 𝑓 𝐻 𝑞 𝑓 )
𝜎 𝑠 =𝜎 ( 𝑞 𝑠 𝐻 𝑞 𝑠 )

18. Infinite mass ratio – Direct mode-based decomposition
𝐿 𝑓 𝐶 𝑓𝑠 0 𝐿 𝑠 𝑞 𝑓 𝑞 𝑠 =𝜎 𝑀 𝑓 0 0 𝐼 𝑞 𝑓 𝑞 𝑠
𝜌 −1 =0
𝜎= 𝜎 𝑓 + 𝜎 𝑠
Direct mode-based decomposition
Fluid contribution
Solid contribution
𝜎 𝑓 = 𝒒 𝒇 𝑯 (𝐿 𝑓 𝑞 𝑓 + 𝐶 𝑓𝑠 𝑞 𝑠 )
𝜎 𝑠 = 𝒒 𝒔 𝑯 ( 𝐿 𝑠 𝑞 𝑠 + 𝜌 −1 𝐶 𝑠𝑓 𝑞 𝑓 )
(𝜎𝐼−𝐿 𝑓 ) 𝑞 𝑓 = 𝐶 𝑓𝑠 𝑞 𝑠
𝜌 −1 =0
Structural Mode
NOT OK
(large fluid response)
𝜎 𝑓 =𝜎 ( 𝑞 𝑓 𝐻 𝑞 𝑓 )
𝜎 𝑠 =𝜎 ( 𝑞 𝑠 𝐻 𝑞 𝑠 )

19. Mass ratio 𝝆=𝟐𝟎𝟎: Structural Mode
Frequency (SM) WM SM
Growth rate (SM)
The frequency is quasi-equal to 𝑘 𝑠
The growth rate gets positive for 𝑘 𝑠 close 𝜔 𝑣

20. Structural Mode: frequency/growth rate decomposition
Fluid
Solid
Growth rate
Solid
Fluid
Very large (resonance) and opposite contributions

21. Spatial distribution of the fluid component
Growth rate
The solid contribution
induces
destabilisation
The fluid contribution
induces
destabilisation
Local fluid contributions
Phase change between and
𝑞 𝑓 +
𝐿 𝑓 𝑞 𝑓
Schmid & Brandt (AMR 2014)

22. Mass ratio 𝝆=𝟏𝟎: Wake Mode
Frequency (WM) WM SM
Growth rate (WM)
For small 𝑘 𝑠 , 𝜔∼𝜔 𝑣 - For large 𝑘 𝑠 , 𝜔∼ 𝑘 𝑠

23. Wake Mode: frequency decomposition
𝜌=10
Fluid
Frequency
Solid
For small 𝑘 𝑠 : 𝜔 𝑓 ∼𝜔 = frequency selection by the fluid
Unstable range : 𝜔 𝑓 ∼ 𝜔 𝑠
For large 𝑘 𝑠 : 𝜔 𝑠 ∼𝜔 = frequency selection by the solid

24. Wake Mode: growth rate decomposition
𝜌=10
Solid
Growth rate
Fluid
Small 𝑘 𝑠
destabilization due to the solid
Large 𝑘 𝑠
destabilization due to the fluid

25. Wake Mode: growth rate decomposition
𝜌=10
Solid
Growth rate
Fluid
Small 𝑘 𝑠
destabilization due to the solid
Large 𝑘 𝑠
destabilization due to the fluid

26. Wake Mode: growth rate decomposition
𝜌=10
Solid
Growth rate
Fluid
Small 𝑘 𝑠
destabilization due to the solid
Large 𝑘 𝑠
destabilization due to the fluid

27. Conclusion & perspectives
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

28. Pure modes (infinite mass ratio)
Direct
Adjoint
𝜎 ∗ 𝑎 𝑓 𝑎 𝑠 = 𝐹 𝐻 0 𝐶 𝑓𝑠 𝐻 𝑆 𝐻 𝑎 𝑓 𝑎 𝑠
𝜎 𝑞 𝑓 𝑞 𝑠 = 𝐹 𝐶 𝑓𝑠 0 𝑆 𝑞 𝑓 𝑞 𝑠
𝜎 𝑓 ∗ 𝑎 𝑓 𝑝 = 𝐹 𝐻 𝑎 𝑓 𝑝
𝜎 𝑞 𝑓 =𝐹 𝑞 𝑓
Fluid
modes
𝑞 𝑠 =0
𝜎 𝑓 ∗ 𝐼− 𝑆 𝐻 𝑎 𝑠 𝑝 = 𝐶 𝑓𝑠 𝐻 𝑎 𝑓 𝑝
𝜎 𝑞 𝑠 =𝑆 𝑞 𝑠
Solid
modes
𝜎 𝑠 ∗ 𝑎 𝑠 𝑝 = 𝑆 𝐻 𝑎 𝑠 𝑝
(𝜎 𝐼−𝐹) 𝑞 𝑓 = 𝐶 𝑓𝑠 𝑞 𝑠
𝑎 𝑓 𝑝 =0
Titre présentation

29. Projection of coupled problem on pure fluid modes
(𝜎 𝐼−𝐹) 𝑞 𝑓 = 𝐶 𝑓𝑠 𝑞 𝑠
(𝜎𝐼 −𝑆) 𝑞 𝑠 = 𝜌 −1 𝐶 𝑠𝑓 𝑞 𝑓
𝑎 𝑓 𝑝 𝐻 𝜎 𝐼 −𝐹 𝑞 𝑓 + 𝑎 𝑠 𝑝 𝐻 𝜎 𝐼−𝑆 𝑞 𝑠 − 𝑎 𝑓 𝑝 𝐻 𝐶 𝑓𝑠 𝑞 𝑠 = 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓
𝜎 ∗ 𝑎 𝑓 𝑝 − 𝐹 𝐻 𝑎 𝑓 𝑝 𝐻 𝑞 𝑓 + 𝜎 ∗ 𝑎 𝑠 𝑝 − 𝑆 𝐻 𝑎 𝑠 𝑝 − 𝐶 𝑓𝑠 𝐻 𝑎 𝑓 𝑝 𝐻 𝑞 𝑠 = 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓
𝜎− 𝜎 𝑓 (𝑎 𝑓 𝑝 𝐻 𝑞 𝑓 )+ 𝜎− 𝜎 𝑓 (𝑎 𝑠 𝑝 𝐻 𝑞 𝑠 )= 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓
𝜎− 𝜎 𝑓 = 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓 (𝑎 𝑓 𝑝 𝐻 𝑞 𝑓 + 𝑎 𝑠 𝑝 𝐻 𝑞 𝑠 )
Titre présentation

30. Projection of coupled problem on pure solid modes
(𝜎 𝐼−𝐹) 𝑞 𝑓 = 𝐶 𝑓𝑠 𝑞 𝑠
(𝜎𝐼 −𝑆) 𝑞 𝑠 = 𝜌 −1 𝐶 𝑠𝑓 𝑞 𝑓
𝑎 𝑓 𝑝 𝐻 𝜎 𝐼 −𝐹 𝑞 𝑓 + 𝑎 𝑠 𝑝 𝐻 𝜎 𝐼−𝑆 𝑞 𝑠 − 𝑎 𝑓 𝑝 𝐻 𝐶 𝑓𝑠 𝑞 𝑠 = 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓
𝜎 ∗ 𝑎 𝑓 𝑝 − 𝐹 𝐻 𝑎 𝑓 𝑝 𝐻 𝑞 𝑓 + 𝜎 ∗ 𝑎 𝑠 𝑝 − 𝑆 𝐻 𝑎 𝑠 𝑝 − 𝐶 𝑓𝑠 𝐻 𝑎 𝑓 𝑝 𝐻 𝑞 𝑠 = 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓
(𝜎− 𝜎 𝑠 )(𝑎 𝑠 𝑝 𝐻 𝑞 𝑠 )= 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓
𝜎− 𝜎 𝑠 = 𝜌 −1 𝑎 𝑠 𝑝 𝐻 𝐶 𝑠𝑓 𝑞 𝑓 𝑎 𝑠 𝑝 𝐻 𝑞 𝑠
Titre présentation

31. Direct-based decomposition of the unstable mode
𝜌=200
Frequency
Fluid
Solid
Growth rate
Solid
Fluid
Titre présentation

32. Free oscillation 𝑅𝑒=40 ; 𝜌=50 𝜔 𝑠 =0.60 𝜔 𝑠 =0.66 𝜔 𝑠 =0.90 𝜔 𝑠 =1.10
No oscillation
𝜔 𝑠 =0.60
𝜔 𝑠 =0.66
Weak oscillation
𝜔 𝑠 =0.90
Strong oscillation
Make a clear distinction between the structural frequency (a parameter) and the vibration frequency of the system (an outcome)
𝜔 𝑠 =1.10
No oscillation

33. Solid displacement – Fluid fields
Multiple solutions