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

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

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)

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)

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)

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)

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)

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)

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 )

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

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 ?

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

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)

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

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

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

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 )

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 )

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

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

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)

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

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

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

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

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

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

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

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

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

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

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

Solid displacement – Fluid fields Multiple solutions