QUARK.ACOUSTICS copybilla/ACOUSTICS.GUIDE.pdf · ON PISTON FACE A piston moves in a tube filled...

1

2

3

iv

v

vi

vii

viii

1.1

1.2

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ACOUSTICSPRODUCTION OF SOUND

IINNTTRROODDUUCCTTIIOONN

William J. Anderson

PROPAGATION OF SOUNDRECEPTION OF SOUND

PRESSURE ON PISTON FACELINEAR PISTON THEORY

ACOUSTICS VS. HYDRODYNAMICSACOUSTICS VS. WING THEORY

ACOUSTICS VS. PISTON THEORY

ACOUSTIC WAVESMEASUREMENT OF SOUND

IMPEDANCE

COMPARISON OF FLUID REGIMES

1.3

ACOUSTICSGreatest researcher in acoustics: Lord Rayleigh(John William Strutt), 1842-1919.

"Theory of Sound" was published in 1877.

Acoustics is divided into:

Production

Propagation

Reception

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2

1.4

1.5

1.6

1.7

where the reference pressure is typically:

Sound Intensity (I): The rate at which soundenergy is transmitted through a unit area.For a harmonic, plane wave, on a plane perpendicular to the wave path:

I = p2

ρ0c

watts/m2

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Sound Pressure Level (SPL): A logarithmicratio of pressures:

SPL = 20 log10

ppref

10

rms dB

pref

= 2 ×10− 5

N/m 2 in air

pref

= N/m2 in liquids1 ×10− 6

20 µPa

1 µPa

( )

( )

1.8

Sound Energy Density (D): The sound energycontained per unit volume of medium:

where γ = 1.4 for diatomic gases (air). This sound energy density is for a reverberant soundfield (see next definition).

D = p2

ρ0c2 = p2

γp0

Total Radiated Power: The radiated poweremanating from a source. Can be measuredin a reverberant (hard-walled) room.

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11

1.9

1.10

1.11

PRESSUREON PISTON FACE

A piston moves in a tube filled with air:

Assume 1-D fluid motion.

Find the pressure on the piston face (x=0).

The boundary condition at the left end is:

∇2p(x, t) − 1c 2

∂ 2p(x , t)

∂ t 2= 0

v (0,t) = v0(t)

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∞

A. PHYSICAL PROBLEM

18

1.12

1.13

1.14

1.15

˙

Does the fluid provide stiffness, added mass and damping?

Do the fluid terms depend on ω and V?

Regime; assumptions M C K

Hydrodynamics

Acoustics

Subsonic wing theory

Linear piston theory

M

M(ω)

M

_

__

K(ω)

K(ω,V)

K(ω,V)

C(ω,V)

C(ω,V)

c = ∞ V = 0

V = 0

vorticesMach<< 1

2 ≤ Mach ≤ 5

θ m[ ]+ M[ ]( ) ˙ u + c[ ]+ C[ ]( ) ˙ u

˙ θ

+ k[ ]+ K[ ]( ) u

θ

=F(t)

M(t)

radiation

26

1.16

1.17

1.18

2.1

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William J. Anderson

TTHHEEOORRYY OOFF AACCOOUUSSTTIICCSS

PREVIEW

ADIABATIC PROCESS

1-D EQUATIONS

3-D WAVE EQUATION

SPEED OF ACOUSTIC DISTURBANCE

FEM VS BEM

PROBLEM SESSION

SPEED OF SOUND IN BAROTROPIC FLUIDS

2.2

2.3

ρ+dρc+dv

pressure distribution

xp+dp

p

4

SPEED OF AN ACOUSTIC DISTURBANCE

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p+dpc

p

ρ(undisturbed fluid)

A 1-D pressure disturbance dp travels at constant speed c to the left. For a small (isentropic) disturbance, show that c is dp/dρ. Sketch the flow as seen by an observer moving with the disturbance.

(1,2)

2.4

Continuity of mass through the pressure pulse:

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5

ρcA = (ρ + dρ)(c + dv) A

ρdv + cdρ = 0

dv = cρ dρ−

c c+dv

ρ+dρρ

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ΣFx = (accelerationx)(mass)

c+dvc

dx

mass = (ρ+ dρ/2) Adx

accelerationx = dvdt

= dvdx /(c + dv/2)

The control volume has:

=A −p ( p+ dp) A (ρ + dρ /2)(Adx) dvdx

(c + dv /2)

Adp =− ρAcdv + 0(dv2)

dp = −ρcdv

p p+dp

(momentum)

(continuity)

2.5

2.6

2.7

2.8

2.9

2.10

2.11

2.12

2.13

2.14

2.15

Problem 1. Solutions of Wave Equation

Solution Substitute each proposed solution into the wave equation.

Show that these pressure perturbations are solutions of the wave equation:

p(x, t) = A sin(ωt)cos(2 π x

L)

p(x, t) = Af (x + at)

p(x, t) = eiω tei 2πx /L

A)

B)

C) A

28

PROBLEM SESSION

2.16

2.17

2.18

3.1

3.2

VELOCITY POTENTIAL

∂ρ∂ t

+ ρ0∇ • v = 0

ρ0

∂v

∂ t+ ∇p = 0

Continuity:

Momentum:

p = c2ρAdiabatic change (linearized):

c 2 = ∂ p

∂ρ 0Speed of sound:

Summarize the 3-D equations:

2

'

'

'

' '

3.3

3.4

fcn(t)ρ0∂∂ tΦ ='+ p

We will assume that this integration constantis zero, hence: ρ

0∂∂ tΦ= −'p

This holds in 3-D, as well.

∂ρ '

∂ t+ ρ0∇ • v = 0

Return to the 3-D continuity equation:

∂ p '

∂ t+ ρ0∇ • = 0

c 2( ) ∇Φ

( + ρ0 = 0c

∇ Φ∂∂ t 2

)ρ

0 ∂∂ tΦ−

2

Eliminate density by the adiabaticargument:

Use the unsteady pressure perturbation:

5

'

3.5

∇ 2Φ + k2Φ( )w dV = 0

V

∫The quantity w(x,y,z) is a "weight function."

Apply Green's 1st identity to obtain:

∫V

∂Φ∂nS

∫ w dS − ∇Φ ∇ wdV + k 2Φ wdVV∫ = 0

WEIGHTED RESIDUALMETHOD

Set the error in the Helmholtz equation to zero, in an averaged way:

7

•

This equation drives error in Φ to zero in an averaged way.

Φ

8

∇Φ ∇ w − k 2Φ w( ) dV = γ w dSSU

∫V

∫ •

On the other hand, one can directly cause the error in Φ to be zero on a boundary where Φ is specified:

SU

Sp

pressure specified velocity

specified:

∂∂n

= γΦ = ˜ Φ

Hence the residual error equation need not contain a term for integrating over Sp:

3.6

3.7

∂12

The derivatives on Φ are related to derivativesof the shape functions Nj:

∂∂xk

Φ = ΦjNj

∂ x kj =1

N

∑

We use the shape functions as weighting functions (Galerkin's method):

∇(V∫ Φj

j =1

N∑ Nj ∇ N − k2

i )( ) • Φjj=1

N∑ Nj dV = γ dS

SU

∫ Nj

(i = 1,2,...N)

Ni

∇Φ ∇ N − k2Φ N( ) dV = γ N dSSU

∫V

∫ (i = 1,2,...N)i i i

Φ jj =1

N

∑ Nj

•

3.8

Rearrange:

j=1

N∑ ∇

VNi •∇ Nj dV Φj − k2

j=1

N∑

VNi dVNj Φj

= γ dSS

U

Nj

(i = 1,2,...N)k[ ] Φ − k2

m[ ] Φ = f where

∇V∫ N j•∇ Ni dVkij =

V∫ Nj dVNimij =

γ dSS

U

∫ Njfj =

These are general equations for any acousticelement shape. Consider a special shape next.

13

(L)

(L3)

(L3/T)(Volume flow/time. Positive out of the node)

∫ ∫ ∫

hdA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAN (x,y)

3

1

x

y

1 2

3

N1(x, y) = 1− xb

− y

d

N2 (x, y) = xb

N3(x, y) = yd

The shape functions are:


k11

=

∂N1∂x

∂N1∂y

T

A∫

∂N1∂x

∂N1∂y

=−1b

−1d

T

A∫ dA

−1b

−1d

15

h

3.9

hdA

This is a "constant velocity" element.

k11

= ( 1

b2+ 1

d 2)

A

∫ = ( 1b2 + 1

d2)hbd2

= (d

2b+ b

2d)

k12

=

∂N1∂x

∂N1∂y

T

A∫ hdV

∂N2∂x

∂N2∂y

=

T

hbd2

−1b

−1d

1b

0

= hd2b

−

k13

=

∂N1∂x

∂N1∂y

T

A∫ hdA

∂N3∂x

∂N3

∂y

=

T

hbd2

−1b

−1d

1d

0= hb

2d−

16

h

3.10

k23

=

∂N2∂x

∂N2∂y

T

V∫

∂N3

∂x

∂N3∂y

=

T

1d

0=

1b

0

0

k33

=

∂N3∂x

∂N3∂y

T

V∫

∂N3∂x

∂N3∂y

=

T

1d

0=

0

1d

hb2d

k22

=

∂N2∂x

∂N2∂y

T

A∫ hdA

∂N2∂x

∂N2∂y

=

T

hbd2

1b

1b

0= hd

2b0

17

hdA

hdA

hbd2

hbd2

k[ ] =

( d2b

+ b2d

) − d2b

− b2d

d2b

d2b 0

− b2d 0

b2d

The acoustic stiffness matrix is:

V∫ Nj dVNimij =

Acoustic mass is:

A typical term is

m23

=0

b

∫ x

b0

d−dx/b

∫ y

dhdydx =

0

b∫ x

b

y 2

2d0

d− dx/b

dx

=0

b

∫ xb

d2(1 − x/b)2

2 ddx ∫= hd

2b 0

b( x − 2 x 2/b + x 3/b2) dx

−

18

h

h

h

3.11

= hd2b

( b2

2−2 b

2

3+ b

2

4)m

23= hbd

24= hd

2bx 2

2− 2 x 3

3b+ x4

4 b20

b

m[ ]=hbd2

16

112

112

1

12

1

6

1

12112

112

16

The mass matrix becomes:

The equivalent nodal fluxes are:

f1

= Γb

0∫ N

1(x,0)hdx

= ( )(1 − x /b)dx = Γh x − x 2

2b

0

b

∫0

b

= Γh(b − b2

) = Γhb

(Note that the totalmass adds up to thetriangle volume.)

19

h Γ2

f2

= Γb

0∫ N2(x,0)hdx

= ( x /b )dx = Γh x2

2b

0

b

∫0

b

= (b2

) = Γbh

f3

= Γb

0∫ N3( x,0)dx = ( )( )dx

0

b

∫ = 0

f =110

The velocity flux:

1 2

3

20

h h

(Γ) Γh2

Γ

Γbh2 Γbh

2

Γbh2

0

3.12

NUMERICAL EXAMPLEFind the normal modes and frequencies of a right triangle element with 10 m sides. The walls are rigid andthe fluid is air(2):

k[ ] =

( d2b

+ b2d

) − d2b

− b2d

d2b

d2b 0

− b2d 0

b2d

− =1

1−1

2−1

212

12

−2

12

−

ρ = 1.2 kg/m3

c = 333 m/sec

21

hh

1 2

3

10 m

10 m

(0,d)

(b,0)

0

0

The determinant of the acoustic stiffnessmatrix is:

Det k[ ] = 14

− 18

− 18

= 0

therefore, there are constant pressure ("rigid body") modes possible.

The finite element equation of motion is:

1 −12

−12

−12

12

0

−12

0 12

− hk2b

2

2

16

112

112

112

16

112

112

112

16

Φ1

Φ2

Φ3

=0

0

0

Multiply both sides by 2; define λ = k2b2/12.

22

h

3.13

2 −1 −1−1 1 0−1 0 1

− λ

2 1 11 2 11 1 2

Φ1Φ2Φ3

=000

2−2λ −1−λ −1−λ−1−λ 1−2λ −λ

− λ −λ 1−2λ

Φ1Φ2Φ3

=000

Combine terms:

λ1 = 0 λ2,3

= 4 ± 16 − 4 × 3

2= 1.0 , 3.0

Solutions are:

2−2λ −1−λ −1−λ−1−λ 1−2λ −λ−1−λ −λ 1−2λ

Det = 02λ(3 − 4λ + λ2)=

Eigenvalue solution by determinant method

23

−1

For the second mode, λ = 1.0:

The 1st Eqn. is:

−2Φ2 − 2Φ3 = 0

−2Φ1 − Φ2 − Φ3 = 0

−2Φ1 − Φ2 + Φ2 = 0

Φ3 = −Φ2

Φ1 = 0

2−2λ −1− λ −1−λ−1−λ 1−2λ −λ−1−λ −λ 1−2λ

Φ1Φ2Φ3

000

0 −2 −2

−2 −1 −1−2 −1 −1

Φ1Φ2Φ3

=2

Φ1Φ2Φ

=

01

−1

3 2

≡

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x

yΦSet Φ2 = 1:

26

The 2nd Eqn. is:

3.14

Eigenvector solution

Show that the zero frequency solution corre-sponds to a uniform perturbation pressure.For λ = 0:

−Φ1 + Φ2= 0

− Φ1 + Φ3 = 0

Eqn. 2 gives

Eqn. 3 gives

Set Φ1 = 1.0:

Φ2 = Φ1

Φ3

= Φ1

2 −1 −1−1 1 0−1 0 1

Φ1

Φ2

Φ3

=

000

1

Φ1Φ2Φ3

1

=111


x

yΦ

25

3.15

For the third mode, λ = 3.0:

2−2λ −1− λ −1−λ−1−λ 1−2λ −λ−1−λ −λ 1−2λ

Φ1Φ2Φ3

000

4 −4 −4

− 4 −5 −3− 4 −3 −5

Φ1

Φ2

Φ3

=−

3

Subtract Eqn. 2 from Eqn. 1:

Subtract 5/4 of Eqn. 1 from Eqn. 3:

Set Φ1 = 1.0:

Φ2 − Φ3 = 0

Φ1 + 2Φ2 = 0

Φ1Φ2Φ

=

1−0.5−0.5

3 3

≡

5Φ1 + 5Φ2 + 5Φ3 − 4Φ1 − 3Φ2 − 5Φ3 = 0

Φ2 = − 0.5Φ1

Φ3 = Φ2

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x

yΦ

27

3.16

The second mode is a standing wave withlongest wavelength along the hypotenuse:

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x

yΦΦ

y

L

Our boundary condition at the nodes is zerovelocity (rigid boundary), so we have a paradoxwhere we have a discontinuous velocity field:

The velocity field is imagined to continue outside the element (a mental experiment).

y

v

29

AAAAAAAA

AAAAAAAAA

cc Φ

AAAAAAAAAAAAAAAAAA

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We satisfy boundary conditions on the field var-iable, but have a discontinuity in the derivative of the field variable. (This also happens in the structural constant-strain triangle).

Do a traveling wave mental experiment. If the wave travels a distance L in a time ∆t,reinforcement of the wave (resonance) occurs.

At time 0:

At time ∆t: L

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AAAAAAAA

AAAAAAAAAAAAAAAAAAAt time 2∆t:

30

3.17

3.18


kij = AdNi

dx0

L

∫ dN j

dxdx

The general expression for acoustic stiffness is:

For our 1-D element:

−k

11= A

d(1− x / L)

dx0

L

∫ d(1− x / L)

dxdx = A

0

L

∫ 1L

dx = AL

1L

−

k12 = k21 = A−1L

0

L

∫ 1L

dx = − AL

k22

= A1

L0

L

∫ 1

Ldx = A

L34

3.19

The acoustic stiffness is:

k[ ] = AL

1 −1

−1 1

V∫ Nj dVNimij =

The general form for acoustic mass is:

For our 1-D line element:

0∫ Nj dxNimij =

L

A

m11

= A (1 − x /L)0

L

∫ 2dx = A (1 − 2 x /L + x

2/L2)0

L

∫ dx

= A( x − x 2/L + x 3/3L2

)0

L = A( L − L + L/3 ) = AL/ 3

35

m12

= m21

= A (1− x/L)0

L

∫ ( x /L) dx = A ( x/L − x2/L2)

0

L

∫ dx

= A( x2/2 L + x3/3L

2)0

L= A( L / 2 − L / 3) = AL/6

m22

= A0

L

∫ (x /L)2dx = A (x

3/3L2)

0

L

= AL/ 3

m[ ] = AL

1

3

1

61

6

1

3

The equation of motion for a single element:

=AL

13

16

16

13

−A

L

1 −1

−1 1

k

2Φ1

Φ2

0

0( )

36

3.20

Solution

Problem 2. Acoustic modes of closed tubeFind the acoustic modes of a closed tube oflength L using one line element. The exact answer for the second acoustic frequency is f2 = c/2L.

Given:

λ = k2L2Set

=AL

1

3

1

616

13

−A

L

1 −1

−1 1

k

2Φ

1

Φ2

0

0( )

(1− λ /3 ) (−1− λ /6)

(−1− λ /6) (1− λ /3)

Φ1

Φ2

= 0

0

LΦ(x)

37

3.21

Find the eigenvectors. For λ = 0:

(1 − λ /3 ) (−1− λ /6)

(−1− λ /6) (1 − λ /3)

Φ1

Φ

= 0

0

2

1 −1−1 1

Φ1

Φ2

= 0

0

Either equation gives Φ1 = Φ2 and hence:

Φ 1

=1

1

The second mode for λ = 12 gives:

−3 −3

−3 −3

Φ1

Φ2

= 0

0

Φ

2= 1

1

−

x

x

Φ(x)

Φ( x)

39

3.22

Solution

Problem 3. Convergence of line elementsHow many line elements are needed to get 1 % accuracy for the second frequency of a closed tube 10 m long?

Carry out 2 to 9 element solutions.For 2 elements:

A

L

1 −1 0

−1 2 −1

0 −1 1

− k

2AL

1 / 3 1 / 6 0

1 / 6 2 / 3 1 /6

0 1 / 6 1 / 3

Φ1

Φ2

Φ3

=

0

0

0

The characteristic equation is found algebraic-ally or by use of a symbolic language. 41

10 mL

Φ

3.23

3.24

4.1

4.2

1) Create elements in MSC/NASTRAN format.2) Use COMET/Vision to create a data set. 3) Execute in COMET/Acoustics.

D. SOLUTION PROCEDURE

Side view of the passenger compartment:

4

4.3

4.4

4.5

4.6

4.7

4.8

4.9

4.10

5.1

A. FUNDAMENTAL SOLUTIONTHEORY

Vibrating bodies radiate sound:

The body is mathematically modeled by a collection of point pressures on its surface. The radiating point is at r and the sound is observed at a "data recovery point" at point rdr. The data recovery point could

also be on the surface or inside the body.

r rdr

2

5.2

5.3

Integrate Eqn. 3 over the entire acousticdomain (either the internal cavity or outer radiation region):

Ψ r , r dr

( )∇2p r ( ) − p r ( )∇

2Ψ r ,r

dr( )( )

V

∫ dV = δ r − r dr

( )p r ( )V∫ dV

=

p(r dr), for r dr12

p(r dr), for r dr on a smooth boundary of V

0, for r dr outside V

inside V

5AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA


cavityregion

radiationregion

∞∞

Green's theorem converts the volume integral into a surface integral:

Ψ( r ,r dr

)∇ 2p (r ) − p(r )∇2Ψ(r , r dr

)( ) dVV∫

where:S = finite boundary of the acoustic domainΣ = boundary of the acoustic domain at infinity, for radiation problems = partial derivative with respect to the unit normal on the surface

∂∂n

C. GREEN'S THEOREM

6

= Ψ(r ,r

dr)

∂p(r )

∂n−

∂Ψ(r , r dr

)

∂np (r )

S+Σ∫ dS

5.4

Acoustic pressure p and acoustic velocity v are used as variables. They are defined on the exterior (or the interior) side of the boundary. Radiation problem: The integral over the surface Σ at infinity is equal to zero due to the Sommerfeld radiation condition. The integral is only over the bounded surface S. Interior problem: The integration surface consists only of S.

where: C= 1, for inside the acoustic domain = 0, for outside the acoustic domain = 0.5 for on a smooth boundary

Ψ(r ,r dr

)∂p

∂n(r )− ∂Ψ

∂n(r ,r

dr) p(r )

S∫ dS = Cp(r

dr)

r drr drr dr 7

(Helmholtz integral eqn.)

Consider an interior problem. To get the pressure at any point (blue dot), one must

integrate over all wetted solid surfaces makingup the enclosure (cutaway view).

r dr

8

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

5.5

5.6

D. DISCRETIZATION Isoparametric elements use the same shape functions to interpolate the field variable as are used to interpolate the boundaries:

11

ξ2

ξ1

ξ2

ξ1

5.7

E. ACOUSTIC EQUATIONS In the direct method, only one boundary vari-able (p or v) or a relationship between the two is specified in the boundary conditions. To compute unknown boundary variables, a data recovery point is located on every one of the nodes. This leads to N equations:

13

I

Ψ(r ,r j )

∂p∂n

(r ) − ∂Ψ∂n

(r ,r j ) p(r )

Si

∫i=1

∑ dSi= Cp(r j )

= the data recovery point is on the boundary S at the jth node rj

r j r dr

( j= 1,2,... N)

i = element numberI = total number of elementsN = total number of nodes

5.8

The determinant of the Jacobian accounts for the area change to curvilinear coordinates, and

are values on the mth node of the ith element

p im ∂p im

∂nand

15

A change is made to curvilinear coordinates:

∂pim

∂nΨ r (ξ1

,ξ2),r j

( ) Nm(ξ

1,ξ

2) J(ξ

1,ξ

2)

Si

∫m=1

M

∑i=1

I

∑ dSi(ξ

1,ξ

2)

− pim ∂Ψ∂n

r (ξ1,ξ

2),r

j( )N

m(ξ

1,ξ

2) J(ξ

1,ξ

2)

Si

∫m=1

M

∑i =1

I

∑ dSi(ξ

1,ξ

2)

= C p(r j ) ( j= 1,2,... N)

5.9

AAAAAAAAAAAAAAAAAAAA

G. MULTI-ZONE CONCEPTConsider the internal acoustic problem for a container with stratified fluid regions, such as a fuel tank with half fuel, half air. The method can include several internal zones, but only one infinite exterior zone.

Consider two fluid zones, SJ and SK. They have

a common boundary SC. This interface SC be-

tween the fluids must now be meshed as well as all other "wetted" solid surfaces.

fluidinterface

SC

SJ

SK


17

H

JHC

J − GCJ

0

0 HC

KG

C

KH

K

pJ

pC

vC

pK

= GJ

0

0 GK

v

J

vK

H JHCJ[ ] pJ

pC

J

= GJG

CJ[ ] vJ

vCJ

HK

HCK[ ] p K

pCK

= GKGC

K[ ] vK

vCK

The governing equations in each region relatep and v at solid and fluid boundaries:

The regions use a common set of pk and vk

variables, and can be assembled:c c

The assembled equations are banded. 18

5.10


The zoning method is required to study the effects of different acoustic fluids. It can also be used as a trick to reduce bandwidth in interior problems for long, tubelike enclosures.


Without zoning:

With zoning:

The zoned formulation has half the bandwidth.

19

H. DATA RECOVERY

Once the boundary variables have been found, one can compute acoustic results at any data recovery point within the acoustic domain.

Ψ(r ,r dr

)∂p

∂n(r ) − ∂Ψ

∂n(r ,r

dr) p(r )

S∫ dS = Cp(r

dr)

Such data recovery is typically done on viewingsurfaces such as planes, cylinders and spheres.

20

5.11

= π r

− i k

4eΨ (r)

r The fundamental solution:

is substituted into the Helmholtz equation:

∂ 2Ψ∂r2 + 2

r

∂Ψ∂r

+ k2Ψ= 0

∂∂ r

−ike−ikr

4πr+ − e

−ikr

4π r 2

+ 2

r−ike

−ikr

4π r+ − e

−ikr

4π r2

+ k

2 e−ikr

4π r= 0

(−ik)2e−ikr

r − − ike−ikr

r2− −ike−ikr

r2−−2e−ikr

r3

+ 2r

− ike− ikr

r + − e−ikr

r 2

+ k

2 e−ikr

r = 0

22

5.12

5.13

6.1

6.2

4as seen in Comet/Vision.

The ends are identified as rigid surfaces:

6.3

6.4

6.5

6.6

6.7

6.8

6.9

6.10

6.11

6.12

7.1

7.2

7.3

Integrate Eqn. 3 over the acoustic domain V (either the internal cavity or outer region):

Ψ r , r dr

( )∇2p r ( ) − p r ( )∇

2Ψ r ,r

dr( )( )

V

∫ dV = δ r − r dr

( )p r ( )V∫ dV

=

p(r dr), for r dr12

p(r dr), for r dr on a smooth boundary of V

0, for r dr outside V

inside V

5AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA


radiationregion

∞∞

(4)

cavityregion

7.4

= δ r − r dr

( )p r ( )V∫ dV

= δ r −r dr

( )p r ( )V∫ dV

Ψ r , r dr

( )∇2p r ( ) − p r ( )∇

2Ψ r ,r

dr( )( )

V

∫ dV

1

Ψ r , r dr

( )∇2p r ( ) − p r ( )∇

2Ψ r ,r

dr( )( )

V

∫ dV

2

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Use the symmetric form of Green's theorem:

Remember that Ψ is known in V1 and V2.

8

1

2

Use Helmholtz' integral equation with surface acoustic variables, on each side of the model.

7.5

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

p2

p1

n1

n2= δ r − r

dr( )p r ( )

V∫ dV

= δ r −r dr

( )p r ( )V∫ dV

Ψ(r , r dr

)∂p

∂n(r )− ∂Ψ

∂n(r ,r

dr) p (r )

S∫ dS+

1

1

1

Ψ r , r dr

( ) ∇p r ( ) − p r ( ) ∇Ψ r ,r dr

( )( )V∫ dV

2

− ∇ ∇• •

Ψ(r ,r dr

)∂p

∂n(r ) − ∂Ψ

∂n(r ,r

dr) p (r )

S∫ dS+

2 2 2

Ψ r , r dr

( ) ∇p r ( ) − p r ( ) ∇Ψ r ,r dr

( )( )V

∫ dV1

− ∇ ∇• •0

0

1

2

1

2

9

Add the two equations, (S1 = S2 = S, n1 = − n2):

Ψ(r ,r dr

)∂p

∂n(r ) − ∂Ψ

∂n(r ,r

dr) p (r )

S∫ dS

1

Ψ(r ,r dr

)∂p

∂n

(r ) − ∂Ψ∂n

(r ,r dr

) p (r )

S∫ dS+

2 2

= δ r − r dr

( )p r ( )V∫ dV δ r −r

dr( )p r ( )

V∫ dV+

21

1

2 2

2

2

=

p(r dr)

p(r dr)

(0+1)

for r dr

for r dr on S

for r dr inside V2

inside V1

p(r dr)

12+ 1

2( )

(1+0) = p(r dr)

r

S∫ 1 2Ψ(r r

dr)

∂p∂n

(r ) − ∂p∂n

(r )

,−−∂Ψ∂n

(r ,r dr) p r )

1 (

p ( )2

dS2 2 2

V1

V2

S

10

− )( )(−

7.6

This equation allows one to compute acoustic response at any data recovery point once the boundary variables are known. 11

p(r dr

) = ∆p(r a)

∂Ψ r dr

, r a

( )∂n

a

− Ψ r dr

,r a

( )∆dp r a( )

s∫ dSa

unit normal uniformly directed over the entire model towards domain 1 or 2 (choice of domain does not matter)

where: a = point on the boundary S, (a ∈ S)

n ˆ =

Use the notation for the differences in pressure and gradient, and realize that either normal n1

or n2 could have been used for both integrals:

(5)

E. DISCRETIZATION An isoparametric approach is used, wherethe same shape functions are used to interp-olate the field variable as the boundaries.

12

ξ2

ξ1

ξ2

ξ1

7.7

7.8


+ ∆pS

U

∫ (r a) ∂Ψ∂na

(r b,r a) dSUa

+ ∆p(r a)SI

∫ ∂Ψ∂na

(r b,r a) + iβna kΨ(r b,r a) dSIa

p(r b ) =− Ψ(r bSP

∫ ,r a)∆dp(r a)dSPa

15

r dr

≡ r b b ∈ S

P( )

The equation for the case of a data recovery point on a pressure-specified surface is:

(6)

where: = specific normal admittance.βna

SU

SI

SP

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

−iρω vn(r

b) = − ∆ dp(r

a)

∂Ψ(r b,r

a)

∂nb

dSPa

SP

∫

+ ∆pS

U

∫ (r a)

∂2Ψ(r b,r

a)

∂nb∂n

a

dSUa

+ ∆p(r a)

SI

∫ ∂2 Ψ(r b,r

a)

∂nb∂n

a

+ i βna

kΨ(r

b,r

a)

∂nb

dS

Ia

16

The equation for the case of a data recovery point on a velocity-specified surface is:

r dr

≡ r b b ∈ S

U( )

(7)

SU

SI

SP

7.9


−∆dp(r a

) ∂Ψ(r b,r

a)

∂nb

dSPaSP

∫

+ ∆ p(r a)

∂2Ψ(r b, r

a)

∂nb∂n

aSU

∫ dSU

+ ∆ p(r a)

SI

∫ ∂2Ψ(r b, r

a)

∂nb∂n

a

+i βnak

∂Ψ(r b,r

a)

∂nb

dSIa

+ ( iβnbk) ∆dp(r

a) Ψ(r

b,r

a)

SP

∫ dSPa−

+ ∆p (r

a)

SU

∫ ∂Ψ(r b,r

a)

∂na

dSUa

+ ∆ p(r a)

∂Ψ(r b,r

a)

∂na

+ iβnakΨ(r

b,r

a)

dSIa

SI

∫

= 0

17

r dr

≡ r b b ∈ SI( )

The equation for the case of a data recovery point on an impedance-specified surface is:

(8)

SU

SI

SP

7.10

∫

∫

Ψ

Ψ

Ψ

∂)

)

)

a

,

U

∂

,

n n

SP

F = ∆dp (r b)p (r

b) dS

PbS

P

∫ + i ρ ω vn(r

b)∆ p (r

b) dS

UbSU

∫

+ 12

Ψ(r b, r

a)∆ dp (r

b)∆ dp(r

a) dS

PadS

PbS

P

∫

+ ∆p (r b)∆p (r

a)∂2 (r

b,r

a)

∂na∂nb

dSUa

dSUb

SU

∫S∫

+ ∆p(r a)∆p (r

b)

∂2 (r b,r

a)

∂na nb

+ i β

nak ∂Ψ(r b , r a)

∂nbSI

∫SI

+i βnbk

∂ (r b

r a

)

∂na

−βnbβnak2 Ψ(r

b,r

a) dS

IadS

Ib

− ∆p(r a

)SU

∫SP

∫ ∆ dp (r b)

∂Ψ(r b

r a

∂na

dSUa

dSPb

− ∆p (r a)∆dp (r

b)

∂Ψ(r b, r

a∂n

a

+ i βnakΨ(r

b, r

a)

SI

∫SP

∫ dSIa

dSPb

+ ∆p(r b)∆p(r

a) ∂2Ψ(r

b,r

∂na∂ b

+ i βnak

Ψ(r b,r

a)

∂ b

S

I

∫SU

∫ dSIb

dSUa

12

12

19

7.11

7.12

I. DATA RECOVERY Once the boundary variables have been found, one can compute acoustic results at any data recovery point within the acoustic domain.

Such data recovery is typically done on viewingsurfaces such as planes, cylinders and spheres.

24

pp(r dr

) = ∆ (r a )∂Ψ r dr , r a( )

∂na

− Ψ r dr,r a( )∆ dp r a( )

s

∫ dSa (9)

7.13

OVERVIEW

2

If coupling between structure and acoustic field is strong, one does a "coupled analysis." The acoustic field excites the structure and the structure excites the acoustic field.

The structural vibration and acoustic responseare solved simultaneously. Transmission loss problems can also be solved simultaneously.

7

A. COUPLED EQUATIONSA coupled structure/acoustic problem is:

where:

= structural degrees of freedom (d.o.f.) = acoustic d.o.f. on the boundary element model = external load applied on structure = external acoustic load = coupling matrices

UstUa

FstaF

[C1]

−ω2[M] +[K] [ ]

− ρ ω2[ ]T

[A ]

UstUa

= F

Fst

a

(1)

THEORY

[C2],

C2

C1

9

B. MODAL BASISThe structure is modeled by its normal modes:

Ust = [Φ]η

η = modal d.o.f. [Φ] = mass normalized modal matrix.

where:

Change variables in Eqn. 1:

−ω2[M] +[K] [C1]

− ρω2[ ]T [A ]

Ua

= F

Fst

a

[Φ]η

Absorb the modal matrix [Φ]:

−ω2[M]+ [K] [ ]

− ρ ω 2[ ]T [A ]

=

F

Fst

a

Ua

η[Φ][Φ]( )

C2

C1

C2

10Premultiply the first equation by [Φ]T:

−ω2 [M ] + [K ] [ ]

− ρω 2[ ]T [A ]

Ua

η[Φ][Φ]( )T[Φ] [Φ]

T[Φ] [Φ]

T

= F

Fst a

[Φ]T

Generalized masses are unity:

[M[Φ]T

[Φ] = [ I]]

[K[Φ]T

[Φ] =] [ωi2 ]

[ωi2]=

ω12 0 • 0

0 ω22 • 0

• • • •0 0 • ωn

2

Generalized stiffnesses are natural frequencies:

C2

C1

C2[A] + ρω2[ ]T[Φ] [ωi2] − [I] ω 2[ ]

−1[Φ]T[ ][ ]Ua

= Fa + ρω2[ ]T[Φ ] [ωi2] − [ I]ω 2[ ]

−1[Φ]TFst

11

− ω 2[I ] + [ω i2] [ Φ]T[ ]

− ρω 2[ ]T[Φ ] [ A]

η

Ua

=[Φ]T Fst

Fa

Introduce generalized masses and stiffnesses:

Structural modal d.o.f. are found from the upper part and substituted in the lower part:

This gives the acoustic variables. One can thenfind the structural variables from above.

C1

C2

C1

C2

12

Incompatibility between structural finite element mesh and acoustic boundary element mesh is resolved.

The structural model is divided into areas thatdo or do not interface with the acoustic medium. Likewise the acoustic mesh is divided into regions that join or do not join with the structure.


AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Modal damping can be added.

16

Import the fuselage model from SDRC I-DEAS:

front view

side view

E. SOLUTION

Create and positionthe right-wing source:

Add left-wingsource.

20

Import the structural interface mesh:

Enter the acoustic material properties(default values): (343 m/s) of sound, density (1.21 kg/m3), and ref pressure (20E-6 Pa).

c = 343 m/s (speed of sound)ρ = 1.21 kg/m3 (density)

p = 20E-6 Pa (reference acoustic pressure)

21

∂u∂h

−ω 2[ M] + i ω [B ] + [K ]

The discrete structural equation of motion is:

Take a derivative w. r. t. a structural parameter:

Solve for the sensitivity vector:

−ω2 ∂[M ]∂h

+ iω ∂[B ]∂h

+ ∂ [ K ]∂h

u

= 0−ω2 ∂[M ]∂h

+ iω ∂[B ]∂h

+ ∂ [K ]∂h

u

−ω 2[ M ] + i ω [ B ] + [K ]

∂u∂h

= − ×−1

4(3)

−ω 2[ M ] + iω [ B ] + [K ]

u = (2)

+

f

where: [M] = mass matrix [B] = damping matrix [K] = stiffness matrix u = structural dynamic response

Structural dynamic sensitivities depend on mass, damping, stiffness, and their derivatives.These sensitivities also depend on the appliedload and baseline structural dynamic response.

5Solve for the sensitivity vector:

−ω2 ∂[M ]∂h

+ iω ∂[B ]∂h

+ ∂ [K ]∂h

u

−ω 2[ M ] + i ω [ B ] + [K ]

∂u∂h

= − ×−1

where: Ω = domain of the element. A =area of the element.vni = element normal velocity.

n =unit normal.

The structural/acoustic sensitivity is a complex number and it needs to be interpreted appropriately. The baseline acoustic response (pressure) at a data recovery point can be written as pdr = pdrR +

ipdrI.

∂vni∂h =

∂v ∂h n dΩ

Ω∫

A

•

11

(10)

obtained from MSC/NASTRAN

5) Assign the material properties of the air. 6) Set the analysis parameters. 7) Write a COMET/ACOUSTICS data deck. 8) Add the data recovery node. 9) Run COMET/ACOUSTICS sensitivities.10) Post-process acoustic sensitivities relative to surface velocities.11) Combine structural and acoustic sensitivities with merge_sens.exe program.12) Review results in tabular form.13) Run acoustic response analyses for baseline and modified plates.14) Post-process acoustic response of baseline and modified plates.

17

NUMERICALACOUSTICS

MULTIMEDIA STUDY GUIDE

Professor William J. Anderson

The University of Michigan

Produced by:Automated Analysis Corporation

2805 S. Industrial Ann Arbor, MI 48104

This is the second in a series of Personal ProfessorTM

courses produced by Automated Analysis Corporation.

The lectures use full motion video, and LivePenTM

annotation which allows the instructor to sketch on thefigures. The video procedures for recording were devel-oped by Online Training, Inc., which has a patent pend-ing for the synchronized annotation.

The course consists of 11 lectures, originally developedas a part of an MS level course at The University ofMichigan. Viewers can proceed at their own pace andcan back up and replay any portions. Keyword searchhelps viewers find important technical topics, theoremsand people.

Professor William J. Anderson received his Ph. D. fromCaltech in 1963. He has been a Professor at The

University of Michigan since1965. He is best known forhis teaching of finite ele-ment theory to thousandsof students over the pasttwenty-six years. Bill con-sults for Caterpillar, Ford,GM, Chrysler and othercompanies. He is a founderof Automated AnalysisCorporation and OnlineTraining, Inc., of Ann Arbor,Mi. He has authored eight

FEA tutorial books and fifty journal and conferencepapers.

Guest speakers include Nick Vlahopoulos and ThomasTecco from Automated Analysis Corporation.

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