A reciprocity relationship for random diffuse fields in acoustics and structures with application to the analysis of uncertain systems.

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A reciprocity relationship for random diffuse fields in acoustics and structures with application to the analysis of uncertain systems Robin Langley

Example 1: Acoustic loads on satellites Arianne 5 Launch Acoustic test facility at ESTEC

Satellite computational model Plane wave Pressure p Structural model { } p = α exp ikrsinθ cosφ + ikrsinθ sinφ + ikr cosθ + iωt j k Diffuse acoustic field jk j k j k j Pressure modelled as a summation of plane waves at each frequency This is very time consuming - is there a more efficient approach?

Example 2: High frequency structural vibrations Thin panels very many modes, response can be viewed as a diffuse wave field Stiff components detailed FE model needed

Structure computational model FE model displacement dofs SEA model diffuse wave fields How do we couple these very different types of model? Again we need to compute the forces arising from the diffuse waves.

The diffuse field reciprocity relation Finite element model has dof x Pressure p produces force vector f Can we avoid summing waves to compute f? The diffuse field reciprocity relation states: *T 4E S ff = E ff = Im D ωπ n { } E is the energy of the diffuse field n is the modal density of the diffuse field (for acoustics E/n is specified by the noise level) D is the dynamic stiffness matrix for radiation into an infinite acoustic system readily computable

Short proof of the relation (structural component) q B = boundary freedoms D D q g = D D q 0 11 12 B T 12 22 I q I = interior freedoms Express q I in terms of blocked modes q I = Φa where a are the modal amplitudes D11 C12 qb g = T C12 Λ a 0 Λ = + 2 2 kk ωk (1 iη) ω Boundary dynamic stiffness Force on boundary due to interior waves { D C Λ C } q = g or Dq = g 1 T 11 12 12 f = C a ff = C aa C B *T *T T 12 12 12 B

Ensemble averaging Assume that: -the natural frequencies are random -The modal amplitudes a correspond to a diffuse field, i.e. they are statistically independent and each mode stores equal energy [ ] DD= D= D C C Λ Λ CC = D = C C C C 2 T T ηω 1 T k[ C1212 C12 ij ] 1 T *T ij 11 12 12 Dij D 11 11 12 12 12 12 E[Im( Dij)] ) 11, ij E[ 122 12 2 ]( πn/ 2 N 2 2 ω) 2 2 k ( kωω k (1 + ) iη) ( ω k ω + ηω ) ff = C aa C E[ ff ] = E[ C C ](2 E/ Nω ) *T *T T *T T 2 12 12 12 12 Thus: *T 4E E ff = E[Im { D} ] ωπ n The proof is completed by noting that E[ D]= D Smith, P. JASA 34, 640-647, 1962 Shorter, P.J. and Langley, R.S. JASA 117, 85-95, 2005

Application 1: diffuse acoustic loading D is the radiation dynamic stiffness of the structure. This can be found by using the boundary element method. The diffuse field reciprocity relation *T 4E E ff = E[Im { D} ] ωπ n replaces the need for a direct summation { } p = α exp ikrsinθ cosφ + ikrsinθ sinφ + ikr cosθ + iωt j k jk j k j k j

Example acoustic transmission, ensemble of random cavities

Application 2: the hybrid FE-SEA method Thin panels very many modes, response can be viewed as a diffuse wave field. SEA degree of freedom E k Stiff components detailed FE model needed FE degrees of freedom q

Hybrid Equations the response of subsystem k Direct Field Reverberant Field + Boundary freedoms q Boundary motions generate a direct field of ingoing waves. Associated boundary forces written as: D ( k ) dir q Reflections produce the reverberant field. Associated boundary forces written as: ( ) k f rev

Hybrid Equations continued So the system equations are written as: D q = f + f, D = D + D ( k) ( k) tot rev tot d dir k k D ( k ) dir ( k ) frev can be found by considering wave radiation into a semi-infinite system is accounted for by using a diffuse field reciprocity result: 4E S E f f = Im D { } ( k), rev ( k) ( k)* T k ( k) ff rev rev dir ωπ nk

Final form of the Hybrid Equations ωη ( + η ) E + ωη n ( E / n E / n ) = P ext j d, j j jk j j j k k in, j k 4E S = D S + Im{ D k } D 1 k ( ) 1*T qq tot ff dir tot k ωπ nk where: ωη ωη P ( j) 1 1*T { D }( ) = ( ω /2) Im D S D ext in, j dir, rs tot ff tot rs n j k { D, } { } rs ( ) = (2 / π ) Im D Im D D ( ) 1 ( ) 1*T jk j dir rs tot dir tot rs 2 j = Im{ D } Im D { D } D ( 1 ( ) 1*T ) d, j d, rs tot dir tot π n j rs rs rs

Example application of the hybrid method Test box frame: welded beam sections 1 x1 x1/8 5 plates of different thickness Courtesy of General Motors Corporation

Example application of the hybrid FE-SEA method -50 Frame 1-50 Frame 2 2 H 2 (db ref. 1 m 2 /N 2 ) -100-150 -200 20 40 60 80 100-100 -150-200 20 40 60 80 100 1 1 2 3-50 Plate 1-50 Plate 2-50 Plate 3 H 2 (db ref. 1 m 2 /N 2 ) -100-150 -100-150 -100-150 -200 20 40 60 80 100 Frequency (Hz) -200 20 40 60 80 100 Frequency (Hz) -200 20 40 60 80 100 Frequency (Hz)

Further Development: Variance Prediction in the Hybrid Method Research example: two plates coupled through 3 points Plate 1 is deterministic (long wavelength, ~ 9 modes below 100 Hz) Plate 2 is randomized by masses and edge springs (short wavelength, ~ 148 modes below 100 Hz) Harmonic point force on plate 1 1 2

Ensemble mean & relative variance of plate 2 energy response -20 Energy of plate 2 10 1 Relative variance of energy -30-40 10 0 T 2 (db ref. 1 J) -50-60 Var[T 2 ] / E[T 2 ] 2 10-1 -70 10-2 -80-90 0 20 40 60 80 100 Frequency (Hz) 10-3 0 20 40 60 80 100 Frequency (Hz)

0 Auto-spectra of velocity response 10 1 Relative variance of veloc ity auto-spectra Ensemble mean & relative variance auto-spectra response at 3 points on plate 1 Sq 1 q 1 (db ref. 1 m 2 /N 2 ) Sq 1 q 1 (db ref. 1 m 2 /N 2 ) -20-40 -60-80 -100-120 0 20 40 60 80 100-20 -40-60 -80-100 0 20 40 60 80 100 Var[Sq 1 q 1 ] / E[Sq 1 q 1 ] 2 Var[Sq 1 q 1 ] / E[Sq 1 q 1 ] 2 10 0 10-1 10-2 10-3 10 1 10 0 10-1 10-2 10-3 0 20 40 60 80 100 0 20 40 60 80 100-20 10 1 Sq 1 q 1 (db ref. 1 m 2 /N 2 ) -40-60 -80-100 -120 0 20 40 60 80 100 Frequency (Hz) Var[Sq 1 q 1 ] / E[Sq 1 q 1 ] 2 10 0 10-1 10-2 10-3 0 20 40 60 80 100 Frequency (Hz)

Postcript For systems that are driven purely at the boundary, it is unlikely that the excitation will produce a perfect diffuse field. In this case it has recently been found that the reciprocity relationship must be extended: ff 4E 2 = + + πωn D π m S S { } ˆˆ qm ˆˆ *T E[ ] Im( ) 2 Re( ) ( ) ff This extension clears up an apparent anomaly between the reciprocity relationship and recent results concerning the variance of the response of a random system. ff Langley, R.S. JASA submitted, 2006