structural sound and vibrations

Vocabulary:

discrete system: mass is concentrated in isolated points

flexible solid: mass is distributed continuously

structure: solid satisfies certain geometrical assumptions (e.g. thin-walled). Beams, plates, shells.

Structural Sound and Vibrations

HAW/M+P, Ihlenburg, CompA Vibrations: Flexible Bodies 1

plates, shells.

structural vibrations: free and forced vibrations of (bounded) structures.

structural sound: waves propagating through (unbounded) structures.

A bar is vibrating in its axial direction. The axial displacements are denoted U(x,t).Consider an arbitrary section of length dx

m = dx

x,UE,A,

A=FdA dx = + F

dx

1. Longitudinal Waves in Bars


m = dx

dxUAUmm....

==F

A=

F A dxdx

+

= +

F

= E , dxdu

= 2

2

2

2

t

UEx

U

=

22

2

2

t

UEx

U

=

-i( , ) ( ) e tU x t u x =

2 ( ) ( ) 0u x u x

E

+ =

( ) i -i1 2L Lk x k xu x C e C e= + ( ) ( )i -i1 2( , ) L Lk x t k x tU x t C e C e += +

The infinite bar admits longitudinal waves, traveling right and left, with the

Speed of Sound and Wave Length


The infinite bar admits longitudinal waves, traveling right and left, with thespeed of sound,

corresponding to the longitudinal wave number . Waves in an infinite bar can propagate at all frequencies.

LE

c

=

LL

kc

=

2LL

c Ef

pi

= =

Longitudinal Modes

0)()0( == luu

l

If the bar is of finite length, the displacement must satisfy boundary conditions, e.g. (0) = (l) = 0 for the free bar.

, 0,1, 2,n E npi = =

Free vibrations in finite bars are possible only at certain discrete eigenfrequencies. The corresponding shapes of vibration are called eigenmodes.


, 0,1, 2,nn E

nlpi

= =

( ) cosnn

u x xlpi

=

2. Bending Waves in Beams

Replace in beam equation EIw(x)=q(x) the static load by the inertia force Aw

x

E,I,

( ),W x t


4 2

4 2( , ) ( , ) 0W x t W x tEI Ax t

+ =

( ) ( ), i tW x t w x e =4

24 0

d w Aw

dx EI

+ = ( )i t kxW e

4B

AkEI

=

Speed of Sound and Wave Length

412B

EIA

pi

=

Remark: the speed of sound depends on the driving frequency. Such waves arecalled dispersive.

4BBk EI

cA

= =


nlpi

Bending Modes (Standing Waves)Beams of finite length admit free vibrations only at certain frequencies, depending on the beam length and the boundary conditions at the ends. Free vibrations can be interpreted also as standing waves, which result from the interference of traveling waves and their reflections.


pi

2pi3pi4pi

AEI

nn 2=

Nodes of eigenmode

The eigenfrequencies and modes can be analytically computed for various boundary conditions; see table in the appendix.

hx

y,z W

( )3

20,

12 1EhD W hW D

+ = =

i t=

3. Bending Waves in Plates


( )( , , ) ( , ) i tW x y t w x y e =

4 0D W W =4 2 h

D

=

4BD

ch

=

Bending waves can propagate through infinite plates at arbitrary frequencies. The speed of sound and the wavelength are

412B

Dh

pi

=

Plates with finite dimensions can vibrate freely at their eigenfrequencies. Each eigenfrequency corresponds to a standing wave (bending mode).

a

h

b

Standing Waves (Bending Modes)


The eigenfrequencies and modes are known analytically only for some specialcases. For instance, the eigenfrequencies of a rectangular plate with simplysupported edges are

2 22

mn

m n Da b h

pi

= +

cf. Gross, Hauger, Schnell, TM4 5. Auflage S. 247

, 1, 2,m n =

where m,n indicate the number of sine-half-waves in the mode shape.

The FEM is a discretisation method. FE models of structural vibrations are linear matrix-vector equations, similar to the vibrations of discrete systems.

Example: E,A,, l

le le

4. Modal analysis with FEM


=

eAl

e

EAc

l=

e

EAc

l=

2 0 = K M u

2eAl 2eAl

2 2

2 2 0U Ux E t

=

4 2

4 2( , ) ( , ) 0W x t W x tEI Ax t

+ =

0D W h W + =

Bar:

Beam:

Plate:

Analytical and Computational (FEM) Models of Structural Vibrations


0D W h W + =

Work of stresses on strains (strain energy) =

potential energyWork of inertia forces on displacements =

kinetic energy

FE models of free structural vibrations

2 0 = K M u

Stiffness matrix Mass matrix

Plate:

Example: simply supported beam

FEM

Exact


Clamped/ clamped

Example ctd.


Longitudinal stiffness of beams >> bending stiffness large eigenfrequencies

Example ctd. (Longitudinal vibrations)


, 0,1,2,3,...nn E

nlpi

= =

In principle, any construction from flexible material represents a distributed-mass system that will vibrate under dynamic loads. The loads can be:- harmonic or periodic, - transient, - stochasticOnly harmonic loads are considered in this course.

5. Forced vibrations (Frequency Response Analysis)


x( ),F x t

( )4 2

4 2( , ) ( , )

,

W x t W x tEI A F x tx t

+ =

1. Separation ansatz:

2. Decomposition of load

( , ) ( ) , ( , ) ( )i t i tW x t w x e F x t f x e = =

( ) ( )f x f w x=

5.1. Analytical Solution


into the eigenfunctions

3. Decomposition of unknown displacement ( ) ( )1

k kk

w x y w x

=

=

( ) ( )1

k kk

f x f w x=

=

4. Solution of modal equations ( ) 2 21 ik

kk

fyA

=

+

x( ) ( )0, cosF x t x t =

0x x=

1. Fourier decomposition of force:

( ) ( ) 00

2sin

l

k kxf f x w x dx k

l lpi

= =

( ) 2 sin2

sin

k kw x xlxk

l l

pi

=

=

Normed by:

l

Modal Basis

Example: Local Stiffness (1/4)


( ) ( )1

k kk

w x y w x

=

=

0 l l

2. Solution of modal equations: ( )20

1l

kw x dx =

( )( )

0

42

2sin

1 ik

xkl ly

EI kAA l

pi

pi

=

+

3. Modal Superposition:

4. Local Stiffness: 0x x=

( ) ( )( )

0

42

sin sin2

1 ik k

x xk kl ly w x

Al EI kA l

pi pi

pi

=

+

Local Stiffness (2/4)

3. Modal Superposition:

where

( ) ( )1

k kk

w x y w x

=

=


( ) ( ) ( )( )

2 0

0 0 41 1 2

sin2,

1 ik k

k k

xklw x y w x

m EI kA l

pi

pi

= =

= =

+

0, 0 = =5. Verification: Static Stiffness:

( ) ( ) ( )

1

4 4 42 2 2

1 1 12 3 51 i 1 i 1 i

m

EI EI EIA l A l A l

pi pi pi

+ + + + + +

0, 0 = = 1

41 1 1Al EI pi

+ + +

Local Stiffness (3/4)x

0 2x l=


0, 0 = =

x=1:2:1000;0.5*pi^4/sum(1./x.^4)

48.0000

44 4 4 41 1 1

2 1 3 5Al EI

Al

pi

+ + +

Mechanical Eng. Table:

100

101

102

103Local Stiffness of Beam @ le/2, material damping 0.02

1/ ( )w

Local Stiffness (4/4)

EigenMode 7


0 10 20 30 40 5010-3

10-2

10-1

10

( ) 1f

EigenMode 1

EigenMode 3

EigenMode 5Static stiffness

Continuous Structure Discrete System

FEM

5.2. Frequency response with FEM


- direct or

- modal reduction

( ) ( )2i + K C M u = fComputational model: Solution:

Example: Plate Vibrations, Direct solution

PARAMETERS FOR SPARSE DECOMPOSITION OF DATA BLOCK SCRATCH ( TYPE=RDP ) FOLLOWMATRIX SIZE = 36966 ROWS NUMBER OF NONZEROES = 386249 TERMS

Size of system matrix = # of unsupported DOF ( = 6*(# of free Nodes) for shell models)

( ) ( )1


12:40:24 Analysis started.12:40:24 Geometry access/verification to CAD part initiated (if needed).12:40:24 Geometry access/verification to CAD part successfully completed (if needed).12:40:24 Finite element model generation started.12:40:25 Finite element model generated 12000 degrees of freedom.12:40:25 Finite element model generation successfully completed.12:40:25 Application of Loads and Boundary Conditions to the finite element model started.12:40:25 Application of Loads and Boundary Conditions successfully completed.12:40:25 Solution of the system equations for frequency response started.12:40:25 Solution of the system equations for frequency response successfully completed.12:40:25 Frequency response analysis completed.

( ) ( )12i + u = K C M fExtract from solver protocol:

Numerical Results

Forced Vibration @200Hz

LF:

MF:


Eigenmode @ 74Hz


Eigenmode @ 50Hz

Harmonic force


HF:

Forced Vibration @5kHz

Forced Vibration @1kHz


2000 1200 2mm,=0.1 (0.2 HF)

MAT1 1 2.1+11 .3 8000.

Example ctd./ Modal Reduction (S111)

( ) ( )2i K C M u = fFind: Solution of in a frequency band:0 min max

1. Modal Analysis: Solve eigenvalue problem 2 0i i K M x =

1

1.5


3. Solution and Back-Transform:

1max1.5n

2. Modal Superposition: 1 1 2 2 n ny y y= + + =u x x x Xy

2iT T + = X K C M Xy X f

K f1

,

= =y K f u Xy

Modal Reduction (Details)

Recall:

2iT T + = X K C M Xy X f

2, ,

T Tn n= =X KX X M X I

1

1

n

=


:TX CX

diagonal: modal damping, Rayleigh damping, material damping fully populated: local damping (G on MAT1, PBUSH, )

Example: Plate Vibrations

1,

= =y K f u Xy

2 0i i K M x =

( ) ( )2i + K C M u = f

2iT T = X K C M Xy X f

K f

1

2

3


09:58:52 Finite element model generation started.09:58:52 Finite element model generated 7349 degrees of freedom.09:58:52 Finite element model generation successfully completed.09:58:52 Application of Loads and Boundary Conditions started.09:58:52 Application of Loads and Boundary Conditions successfully completed.09:58:53 Solution of the system equations for normal modes started.09:58:57 Solution of the system equations for normal modes successfully completed.09:58:57 Solution of the system equations for frequency response started.09:58:57 Solution of the system equations for frequency response successfully completed.09:58:57 Frequency response analysis completed.

Extract from solver protocol:

12

3

Appendix: Longituninal vibrations of bars

Ec =


Quelle: Inman S. 322

Ec

=

Bending vibrations of beams (1/2)


Quelle: Inman S. 3352

n n

EIA

=n s. Folgeseite

Bending vibrations of beams (2/2)


The table entries are obtained from the general ansatz

xAxAxAxAxw coshsinhcossin)( 4321 +++=and the boundary conditions of the particular load case.

nL4.7300

nL1.875

AEI

nn 2=

Eigenmodes of beams

clamped/ clamped clamped (cantilever)


4.7300

7.8532

10.996

14.1372

1.875

4.6941

7.8547

10.995

pi2

12:5 + nn pi

212

:5 nn

The modal effective mass (MEM) of a vibration mode with respect to a rigid body mode is defined as

It can be seen from the definition that the MEM are non-negative and their unit is kg (assumingthat the modes are non-dimensional). Essentially the MEM represent mass-weighted scalarproducts of vibration modes and rigid body modes,

ix

, 1, , , 1, ,6.L i N j= = =x Mr

jr

( )2i jij

i i

m =x Mrx Mx

Appendix: Modal Effective Mass


where N is the number of vibration modes. The rigid body modes consist of uniform unittranslations/ rotations in the direction of/ around each coordinate axis. The scalar product is takenonly over the free nodes of the FE model.

, 1, , , 1, ,6.ij i jL i N j= = =x Mr

Example: Vertical rigid-body mode of simply supported beam.

The scalar product indicates how close a vibrational mode is to a rigid body mode. If the scalar product is one the modes are exactly aligned, if it is zero they are orthogonal.

Origin of Method, References

Modal effective masses are frequently calculated in spacecraft design. Spacecraft structures undergo large accelerated motions during launch and landing, and hence it is of critical interest to establish which deformational modes are likely to be excited by rigid body motions.

In practice, the MEM are also used to determine whether a mode is global or local.

The following simple tests illustrate the method. The results will show that a small MEM does not necessarily indicate local modes.


Tom Irvine, EFFECTIVE MODAL MASS & MODAL PARTICIPATION FACTORS, http://www.vibrationdata.com/tutorials2/ModalMass.pdf (28.01.2010)

Wijker, Jacob Job, Spacecraft structures, Springer Verlag Berlin, 2008 (Chapter 16: Modal effective mass)

A. Paolozzi and I. Peroni, A PROCEDURE FOR THE DETERMINATION OF EFFECTIVE MASS SENSITIVITIES IN A GENERAL TRIDIMENSIONAL STRUCTURE, Computers & Structures Vol. 62, No. 6. pp. 1013-1024. 1997

RK, Using Modal Effective Mass to Determine Modes for Frequency Response Analysis, http://feadomain.com/e107_plugins/content/content.php?content.76 (28.01.2010)

FE model: four elements, rectangular cross-section, both ends simply supported. Mass:

30.8m, =8000kgml =x

y

z

0.02m 0.04m

800*20*40*8 9 5.12 3tm e e= =

Test 1: Simply Supported beam


Result: Nine modes. Computation with lumped masses rotational dof are condensed out, 3 trans dof per node.

1,2 (y and z)3,4 (y and z)

5,6 (y and z)7-9: Longitudianal and

torsional modes.

EFFECTIVE MASS MATRIX *** ***

* 3.840000E-03 7.595561E-12 -3.844592E-12 0.000000E+00 -6.286070E-09 -2.332034E-09 ** 7.595561E-12 3.840000E-03 1.766097E-15 0.000000E+00 -8.835683E-13 1.536000E+00 ** -3.844592E-12 1.766097E-15 3.840000E-03 0.000000E+00 -1.536000E+00 -8.579544E-13 ** 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 ** -6.286070E-09 -8.835683E-13 -1.536000E+00 0.000000E+00 7.168000E+02 4.088132E-10 ** -2.332034E-09 1.536000E+00 -8.579544E-13 0.000000E+00 4.088132E-10 7.168000E+02 **** ***

A-SET RIGID BODY MASS MATRIX*** ***

3 3.84 3t4

m e=

Rigid body mass matrix


*** ***

* 3.840000E-03 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 3.840000E-03 0.000000E+00 0.000000E+00 0.000000E+00 1.536000E+00 ** 0.000000E+00 0.000000E+00 3.840000E-03 0.000000E+00 -1.536000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 -1.536000E+00 0.000000E+00 7.168000E+02 0.000000E+00 ** 0.000000E+00 1.536000E+00 0.000000E+00 0.000000E+00 0.000000E+00 7.168000E+02 **** ***

Observation:

rigid body translations in y, z direction are coupled with rotations.

effective mass identical with rigid body mass

four elements each interior node gets of total mass, free boundary nodes get 1/8.

Since boundary nodes are fixed only of total mass is included in rigid body mass matrix.

MODAL EFFECTIVE MASS FRACTION(FOR TRANSLATIONAL DEGREES OF FREEDOM)

MODE FREQUENCY T1 T2 T3NO. FRACTION SUM FRACTION SUM FRACTION SUM

1 7.077026E+01 9.314766E-31 9.314766E-31 9.714045E-01 9.714045E-01 2.673639E-32 2.673639E-322 1.411843E+02 1.945421E-32 9.509307E-31 1.114075E-34 9.714045E-01 9.714045E-01 9.714045E-013 2.802581E+02 1.935277E-25 1.935287E-25 8.024709E-29 9.714045E-01 6.461121E-29 9.714045E-014 5.541207E+02 1.947883E-22 1.949818E-22 3.381378E-25 9.714045E-01 2.764477E-25 9.714045E-015 5.913062E+02 1.512678E-22 3.462496E-22 2.859548E-02 1.000000E+00 5.836389E-26 9.714045E-016 1.148360E+03 8.641317E-22 1.210381E-21 8.764064E-24 1.000000E+00 2.859548E-02 1.000000E+007 3.045298E+03 9.714045E-01 9.714045E-01 5.222461E-18 1.000000E+00 1.346433E-18 1.000000E+008 5.626977E+03 3.037006E-15 9.714045E-01 1.207543E-17 1.000000E+00 3.084168E-18 1.000000E+009 7.352000E+03 2.859547E-02 1.000000E+00 2.592362E-18 1.000000E+00 6.609832E-19 1.000000E+00

Effective modal mass fraction

Observation:


Observation:

MEM is zero for symmetric mode shapes 3,4.

97%

0%

3%

Same Beam model, four elements, left end clamped.

x

y

z

Result: 12 modes. Computation with lumped masses rotational dof are condensed out, 3 trans dof per node.

Test 2: Cantilever beam


Result: 12 modes. Computation with lumped masses rotational dof are condensed out, 3 trans dof per node.

1,2 (y and z)

3,4 (y and z)

68% 22%

Mode shapes and modal effective mass fraction..

5,7 (y and z)6,8(y and z)

7% 3%

MODAL EFFECTIVE MASS FRACTION(FOR TRANSLATIONAL DEGREES OF FREEDOM)


(FOR TRANSLATIONAL DEGREES OF FREEDOM)

MODE FREQUENCY T1 T2 T3NO. FRACTION SUM FRACTION SUM FRACTION SUM

1 2.452855E+01 8.438346E-39 8.438346E-39 6.846376E-01 6.846376E-01 1.080871E-31 1.080871E-312 4.900471E+01 4.317162E-40 8.870062E-39 9.489781E-32 6.846376E-01 6.851764E-01 6.851764E-013 1.438673E+02 8.686789E-41 8.956930E-39 2.171252E-01 9.017628E-01 1.157240E-34 6.851764E-014 2.856255E+02 1.992160E-33 1.992169E-33 1.382351E-31 9.017628E-01 2.179096E-01 9.030859E-015 3.793786E+02 6.398226E-32 6.597443E-32 7.258439E-02 9.743472E-01 6.754913E-29 9.030859E-016 6.570369E+02 8.113327E-30 8.179302E-30 2.565278E-02 1.000000E+00 1.016997E-26 9.030859E-017 7.441080E+02 6.852732E-30 1.503203E-29 7.966597E-28 1.000000E+00 7.224850E-02 9.753344E-018 1.266119E+03 5.962455E-27 5.977486E-27 1.014358E-26 1.000000E+00 2.466557E-02 1.000000E+009 1.552479E+03 9.026479E-01 9.026479E-01 2.163141E-29 1.000000E+00 2.481903E-28 1.000000E+0010 4.421087E+03 7.999389E-02 9.826418E-01 3.634619E-29 1.000000E+00 2.992522E-28 1.000000E+0011 6.616625E+03 1.594510E-02 9.985870E-01 2.315860E-28 1.000000E+00 8.756785E-28 1.000000E+0012 7.804841E+03 1.413076E-03 1.000000E+00 8.412071E-27 1.000000E+00 4.169802E-27 1.000000E+00

A-SET RIGID BODY MASS MATRIX*** ***

* 4.480000E-03 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 4.480000E-03 0.000000E+00 0.000000E+00 0.000000E+00 2.048000E+00 ** 0.000000E+00 0.000000E+00 4.480000E-03 0.000000E+00 -2.048000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 -2.048000E+00 0.000000E+00 1.126400E+03 0.000000E+00 ** 0.000000E+00 2.048000E+00 0.000000E+00 0.000000E+00 0.000000E+00 1.126400E+03 **** ***

78

m

Rigid body mass matrix (clamped beam)


*** ***

The MEM

indicate the dominant direction of global modes. allow to separate lateral bending modes from axial modes. can be zero for global modes of symmetric models.

Conclusions


structural sound and vibrations

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