Machdyn Chapt 5

7/30/2019 Machdyn Chapt 5

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5 Vibration of Linear Multiple-Degree-of-

Freedom Systems

F (t)1

F (t)2

F (t)3

x2

k1

c1

x1

x3

k2

k3

m1

m3

m2

=0

=0

x1x1x1x1x1

F (t)1m1

FC1

FK1FK3

FK2

FK3

FK2 m2

F (t)3m3

F (t)2

x1x1x1 x2

x3

Fig. 5.1: Multi-Degree-of-Freedom System with Free Body Diagram

5.1 Equation of Motion

The equation of motion can be derived by using the principles we have learned such as

Newtons/Eulers laws or Lagranges equation of motion. For a general linear system mdof

system we found that we can write in matrix form

FxNKxGCxM =++++ &&& (5.1.1)

with the matrices

M: Mass matrix (symmetric)T

MM =

C: Damping matrix (symmetric)T

CC=

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K: Stiffness matrix (symmetric)T

KK=

G : Gyroskopic matrix (skew-symmetric)T

GG =

N: Matrix of non-conservative forces (skew-symmetric)T

NN =

F: External forces

Note:A general matrixA can be decomposed into the symmetric part and the skew-symmetric part

by the following manipulation:

( ) ( )

4444 34444 21

4342143421

part

symmetricskew

T

symmetric

TAAAAA

++=2

1

2

1

In the standard case that we have no gyroscopic forces and no non-conservative displacement

dependent forces but only inertial forces, damping forces and elastic forces the last equation

reduces to

FxKxCxM =++ &&& (5.1.2)

Example:

The system shown in fig. 5.1, where the masses can slide without friction ( = 0), has thefollowing equation of motion.

=

++

+

+

)(

)(

)(

0

0

000

000

00

00

00

00

3

2

1

3

2

1

33

22

32321

3

2

11

3

2

1

3

2

1

tF

tF

tF

x

x

x

kk

kk

kkkkk

x

x

xc

x

x

x

m

m

m

&

&

&

&&

&&

&&

(5.1.3)

5.2 Influence of the Weight Forces and Static Equilibrium

The static equilibrium displacements are calculated by ( 0== statstat xx &&& ):

statstat FxK = (5.2.1)

which in the case of the example shown in fig. 5.2:

==

gm

gmFxK statstat

2

1

The dynamic problem for this example is

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( )

{

mequilibriustatictheabout

vibrationthedescribingmotiontheofpart

dynstat

forcesstatic

stat

xxx

tFFxKxM

+=

+=+321

&&

(5.2.2)

g

k2

k1

m1

m1

m2

m2

k1

k2

2

1

Fig. 5.2: Static equilibrium position of a two dof system

From the last equation also follows that

dyndyn xxxx &&&&&& ==

so that

( )tFFxxKxM statstatdyndyn +=++ )(&& (5.2.3)

and after rearrangement

( )tFxKFxKxM statstatdyndyn +=+=

44 344 21&&

0

(5.2.4)

)(tFxKxM dyndyn =+&& (5.2.5)

As can be seen the static forces and static displacements can be eliminated and the equation of

motion describes the dynamic process about the static equilibrium position.

Note:

In cases where the weight forces influences the dynamic behavior a simple elimination of the

static forces and displacements is not possible. In the example of an inverted pendulum shown

in fig. 5.3 the restoring moment is mglsin, where lis the length of the pendulum.

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k k

mg

m

Fig. 5.3: Case where the static force also influences the dynamics

5.3 Ground Excitation

Fig. 5.4 shows a mdof system

k2

k1

m1

m2

c1

x2

x1

xo

f2

f1

Fig. 5.4: 2dof System with excitation by ground motionx0

Without ground motionx0 = 0 the equation of motion is

=

+

+

+

2

1

2

1

22

221

2

11

2

1

2

1

00

0

0

0

f

f

x

x

kk

kkk

x

xc

x

x

m

m

&

&

&&

&&

(5.3.1)

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Now, if we include the ground motion, the differences (x1-x0) and the relative velocity

d(x1-x0)/dtdetermine the elastic and the damping force, respectively at the lower mass. This

can be expressed by addingx0 to the last equation in the following manner

)(0)(000

0

0

0

0

1

0

1

2

1

2

1

22

221

2

11

2

1

2

1

tx

c

tx

k

f

f

x

x

kk

kkk

x

xc

x

x

m

m

&&

&

&&

&&

+

+

=

+

+

+

(5.3.2)

The dynamic forcef0 of the vibrating system on the foundation is

(5.3.3)( ) ( 0110110 xxcxxkf && += )

Fig. 5.5: Example for ground motion excitation of a building structure (earthquake excitation)

Fig. 5.6: Excitation of a vehicle by rough surface

5.4 Free Undamped Vibrations of the Multiple-Degree-of-Freedom

System

5.4.1 Eigensolution, Natural Frequencies and Mode Shapes of the System

The equation of motion of the undamped system is

0=+ xKxM && (5.4.1)

To find the solution of the homogeneous differential equation, we make the harmonic solution

approach as in the sdof case. However, now we have to consider a distribution of the

individual amplitudes for each coordinate. This is done by introducing (the unknown) vector:

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ti

ti

ex

ex

=

=

&&(5.4.2)

Putting this into eqn. (5.4.1) yields

) 0 = MK (5.4.3)

This is a homogeneous equation with unknown scalarand vector . If we set we see

that this is a general matrix eigenvalue problem

2=1:

) 0= MK (5.4.4)

where is the eigenvalue and is the eigenvector. Because the dimension of the matrices isf

byfwe get f pairs of eigenvalues and eigenvectors:

fiii ...,2,1......... == (5.4.5)

i is the i-th natural circular frequency and

ithe i-th eigenvector which has the physical meaning of a vibration mode shape

The solution of the characteristic equation

0det = MK (5.4.6)

yields the eigenvalues and natural circular frequencies , respectively. The natural

frequencies are:

2=

2

iif = (5.4.7)

The natural frequencies are the resonant frequencies of the structure.

The eigenvectors can be normalized arbitrarily, because they only represent a vibration mode

shape, no absolute values. Commonly used normalizations are

1) Normalizei

so that 1=i

2) Normalizei

so that the maximum component is 1.

3) Normalizei

so that the modal mass (the generalized mass) is 1.

Generalized mass or modal mass:i

T

iiMM = (5.4.8)

1The well-known special eigenvalue problem has the form 0)( = xIA , where I is the identity matrix, x

the eigenvector and the eigenvalue.

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Generalized stiffness:i

T

iiKK = (5.4.9)

1:

=

==

i

Ti

Mif

KK iii(5.4.10)

The so-calledRayleigh ratio is

ii

T

iM

K

Tii = (5.4.11)

It allows the calculation of the frequency if the vectors are already known.

5.4.2 Modal Matrix, Orthogonality of the Mode Shape Vectors

If we order the natural frequencies so that

f ...321

and put the corresponding eigenvectors columnwise in a matrix, the so-called modal matrix,

we get

Modal Matrix: [ ]

==

ffff

f

f

f

...

............

...

...

,...,,

21

22221

11211

21(5.4.12)

The first subscript of the matrix elements denotes the no. of the vector component while the

second subscript characterizes the number of the eigenvector.

The eigenvectors are linearly independent and moreover they are orthogonal. This can be

shown by a pairi andj

( ) 02 =ii

MK and ( ) 02 =jj

MK (5.4.13)

Premultplying by the transposed eigenvector with indexj and i respectively:

02 =ii

Tj MK and 0

2 =jj

Ti MK (5.4.14)

If we take the transpose of the second equation:

02 = iTjTTj MK (5.4.15)

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and consider the symmetry of the matrices:T

MM = and TKK= and subtract this equation

( ) 02 =ij

Tj MK from the first equation (5.4.14) we get

( ) ( )022 =

i

T

jijM (5.4.16)

which means that if the eigenvalues are distinct ji for ji the second scalar product

expression must be equal to zero:

0=i

Tj M (5.4.17)

That means that the two distinct eigenvectors ji are orthogonal with respect to the massmatrix. For all combinations we can write:

SymbolKroneckerji

ji

for

for

MK

MM

ij

iiiji

T

j

iiji

T

j

==

=

=.............

0

1

(5.4.18)

or with the modal matrix:

{ }

{ }

==

==

ff

iiT

f

iT

M

M

M

MdiagK

M

M

M

MdiagM

...0

0

...0

0

22

11

2

1

(5.4.19)

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Example: Mode shapes and natural frequencies of a two storey structure


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5.4.3 Free Vibrations, Initial Conditions

The free motion of the undamped systemx(t)is a superposition of the modes vibrating with

the corresponding natural frequency:

( ) (=

+=f

iisiiciitAtAtx

1

sincos ) (5.4.20)

Each mode is weighted by a coefficientAci andAsi which depend on the initial displacement

shape and the velocities. In order to get these coefficients, we premultiply (5.4.20) by the

transposedj-th eigenvector:

( ) ( ) ( )tAtAMtAtAMtxM jsjjcjM

j

T

j

jifr

f

iisiicii

T

j

T

j

j

sincossincos

0.

1

+=+=

=

=

43421444444 3444444 21

(5.4.21)

All but one of the summation terms are equal to zero due to the orthogonality conditions.

With the initial conditions fort= 0 we can derive the coefficients:

( )

cjjT

jAMxM

xtx

t

=

==

=

0

00

0

j

T

jcj

M

xMA

0= (5.4.22)

( ) ( )=

+=f

i

isiiciiitAtAtx

1

cossin &

( )

sjjjT

jAMxM

vtx

t

=

==

=

0

00

0

&

& jj

T

jsj

M

vMA

0= (5.4.23)

which we have to calculate for modesj .

5.4.4 Rigid Body Modes

x1 x2

m1 m2k

Fig. 5.7: A two-dof oscillator which can perform rigid body motion

As learned earlier the constraints reduce the dofs of the rigid body motion. If the number of

constraints is not sufficient to suppress rigid body motion the system has also zero

eigenvalues. The number of zero eigenvalues corresponds directly to the number of rigid body

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modes. In the example shown in Fig. 5.7 the two masses which are connected with a spring

can move with a fixed distance as a rigid system. This mode is the rigid body mode, while the

vibration of the two masses is a deformation mode.

The equation of motion of this system is

=

+

0

0

0

0

2

1

2

1

2

1

x

x

kk

kk

x

x

m

m

&&

&&

The corresponding eigenvalue problem is

=

0

0

2

1

2

1

mkk

kmk

The eigenvalues follow from the determinant which is set equal to zero:

[ ] 0))((det 221 == kmkmk L

( ) 021212 =+ kmkmmm

Obviously, this quadratic equation has the solution

0211 == and

21

21222

mm

mmk +==

The corresponding (unnormalized) eigenvectors are

=

1

1

1

which is the rigid body mode: both masses have the same displacement, no potential energy is

stored in the spring and hence no vibration occurs. The second eigenvector, the deformation

mode is

=

2

12

1

m

m

which is a vibration of the two masses. Other examples for systems with rigid body modes are

shown in the following figures.

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Fig. 5.8: Examples for systems with torsional and transverse bending motion with rigid body

motion

Fig. 5.9: Flying airplane (Airbus A318) as a system with 6 rigid body modes and deformation

modes

Fig. 5.10: Commercial communication satellite system (EADS) with 6 rigid body modes and

deformation modes

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5.5 Forced Vibrations of the Undamped Oscillator under Harmonic

Excitation

k1

k2

m2

m1

k2

x1

x2f (t)2

f (t)1

2 2

Fig. 5.11: Example for a system under forced excitation

The equation of motion for this type of system is

( )tFxKxM =+&& (5.5.1)

For a harmonic excitation we can make an exponential approach to solve the problem as we

did with the sdof system

{

ti

vectorAmplitudecomplex

eFtF )( = (5.5.2)

We make a complex harmonic approach for the displacements with as the excitation

frequency:

tieXx = (5.5.3)

The acceleration vector is the second derivativetieXx = && (5.5.4)

Putting both into the equation of motion and eliminating the exp-function yields

( ) FXMK = (5.5.5)

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which is a complex linear equation system that can be solved by hand for a small number of

dofs or numerically. The formal solution is

( ) FMKX 1= (5.5.6)

which can be solved if determinant of the coefficient matrix : 0det MK

If the excitation frequency coincides with one of the natural frequencies i we getresonance of the system with infinitely large amplitudes (in the undamped case)

Resonance: ( ) iMK == 0det

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Machdyn Chapt 5