multi degree of freedom systems_1

8/8/2019 Multi Degree of Freedom Systems_1

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DR. MUHD. HAFEEZ BIN ZAINULABIDINFACULTY OF MECHANICAL & MANUFACTURING ENGINEERING

UTHM

NOISE & VIBRATION

MULTI DEGREE OF FREEDOM SYSTEMS


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CHAPTER 2

Multi Degree of Freedom Systems

Figure source: http://cite.iiit.ac.in/vlab/Expr6.htm


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Multi Degree of Freedom Systems

1. Mass Spring System

2. Generalized Coordinates and GeneralizedForces

3. Lagranges Equations

4. General Equations in Matrix Form

5. Eigenvalues and Eigenvectors


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Mass Spring System

a) Newtons 2nd Law Approach

Figure 2(a) and 2(b)


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The application of Newtons 2nd law of motion to

mass mi gives

Equations of motion of the masses m1 and mnwhere i = 1, xo = 0 and i = n, xn+1 = 0.

1...,,3,2;

11

111111

!!

niFxk

xkkxkxcxccxcxm

iii

iiiiiiiiiiiiii

1221212212111Fxkxkkxcxccxm !

nnnnnnnnnnnnn

Fxkkxkxccxcxm ! 1111


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Equations of motion can be expressed in matrix

form as

where [m],[c] and [k] are the mass, damping and

stiffness matrices, respectively.

? A ? A ? A FxkxcxmTTTT !

? A

-

!

nm

m

m

m

m

.

/1///

.

.

.

000

000

000

000

3

2

1


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? A

-

yy

!

1

4

433

3322

221

000

00

000

00

000

nnnccc

c

ccc

cccc

ccc

c

.

.///

//1

.

.

.

? A

-

yy

!

1

4

433

3322

221

000

00

000

00

000

nnnkkk

k

kkk

kkkk

kkk

k

.

.///

//1

.

.

.


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are the displacement, velocity,acceleration and force vectors.

FxxxT

TTT and,,

,

2

1

y

y!

tx

tx

tx

x

n

T

,

2

1

y

y!

tx

tx

tx

x

n

T

,

2

1

y

y

tx

tx

tx

x

n

T

.

2

1

y

y!

tF

tF

tF

F

n

T


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In general form, the mass, damping and stiffness

matrices are

? A

-

!

nnnnn

n

n

n

mmmm

mmmm

mmmm

mmmm

m

.////

.

.

.

321

3333213

2232212

1131211

? A

-

!

nnnnn

n

n

n

cccc

cccccccc

cccc

c

.

////

.

.

.

321

3333213

2232212

1131211

? A

-

!

nnnnn

n

n

n

kkkk

kkkkkkkk

kkkk

k

.

////

.

.

.

321

3333213

2232212

1131211


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b) Influence Coefficients Approach

The equations of motion of a multi degree offreedom system can also be written in terms ofinfluence coefficients.

There are 3 main influence coefficients:

i. Stiffness Influence Coefficients

ii. Flexibility Influence Coefficientsiii. Inertia Influence Coefficients


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Stiffness Influence Coefficients

Defined as the force at point idue to a unitdisplacement at point jwhen all the points otherthan the point jare fixed.

Total force at point i,Fi can be found by summingup the forces due to all displacementsxj j = 1, 2,, n).

or

nixkFn

j jiji

...,,2,1,1

!!!

? AxkF TT

!


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Steps in determining the stiffness influence

coefficients.1. Assume x1 =1 and x2,x3, ,xn = 0. The set of

forces will maintain the system in the assumedconfiguration.

2. Write static equilibrium equations for eachmass.

3. Solve the equations to find the influence

coefficients.4. Repeat Step 1to 3by assuming x2 =1 and so

on.


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Example:

Consider multi degree of freedom mass springsystem shown in Figure 2(c).

Figure 2(c)


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x1=1

Figure 2(d)

0:

:

:

313

2212

11211

!

!

!

km

kkm

kkkm


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x2=1

Figure 2(e)

3323

23222

2121

:

:

0:

kkm

kkkm

kkm

!

!

!


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x3=1

Figure 2(f)

3333

3232

131

:

0:

0:

kkm

kkm

km

!

!

!


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Solution forx1=1

Solution forx2=1

Solution forx3=1

Stiffness matrix

0,, 312212111 !!! kkkkkk

3323222212 ,, kkkkkkk !!!

33332313 ,,0 kkkkk !!!

? A

-

!

33

3322

221

0

0

kk

kkkk

kkk

k


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Flexibility Influence Coefficients

Defined as the deflection at point idue to a unitload at point jwhen all the loads at other pointsother than the point jare zero.

Total deflection at point i,xi can be found bysumming up the contributions of all forces Fj.

or

niFaxn

j

jiji ...,,2,1,1

!! !

? AFaxTT

!


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Steps in determining the flexibility influence

coefficients.1. Assume F1 =1 and F2, F3, , Fn = 0.

2. Write static equilibrium equations for each mass.

3. Solve the equations to find the flexibilitycoefficients.

4. Repeat Step 1to 3by assuming F2 =1 and so on.

OR5. Find inverse [k] if the stiffness matrix is available.


20/46

F1=1

Figure 2(g)

0:

:

1:

213133

21313112122

112121111

!

!

!

aakm

aakaakm

aakakm


21/46

F2=1

Figure 2(h)

0:

1:

:

223233

22323122222

122221211

!

!!

aakm

aakaakm

aakakm


22/46

F3=1

Figure 2(i)

1:

:

:

233333

23333132322

132321311

!

!!

aakm

aakaakm

aakakm


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Solution for F1=1

Solution for F2=1

Solution forF

3=1

Flexibility matrix

131

121

111

1

,

1

,

1

ka

ka

ka !!!

21

32

21

22

1

12

11,

11,

1

kka

kka

ka !!!

321

33

21

23

1

13

111,

11,

1

kkka

kka

ka !!!

? A

-

!

321211

21211

111

111111

11111

111

kkkkkk

kkkkk

kkka


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Inertia Influence Coefficients

Defined as the set of impulses applied at point itoproduce a unit velocity at point jand zero atevery other points.

Total impulse at pointi,

can be found bysumming up the impulses causing the velocities

or

jx

nixmFn

j

jiji

...,,2,1,1

~!!

!

? AxmFT

T!

~

iF~


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Steps in determining the inertia influence

coefficients.1. Assume and .

2. Write equilibrium equations for each mass.

3. Solve the equations to find the inertiacoefficients.

4. Repeat Step 1to 3 by assuming and so

on.

11 !x 0...,,, 32 !nxxx

12 !x


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Generalised Coordinates &

Generalised Forces

Equations of motion of a vibrating system can beformulated in a number of different coordinate

systems. n independent coordinates are necessary to

describe the motion of a system having n d.o.f.

Any set ofn independent coordinates is calledgeneralised coordinates, designated by

q1,q2,,qn.


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Generalised coordinates may be lengths, angles,

or any other set of numbers that define theconfiguration of the system at any time uniquely.

Generalised coordinates also independent of the

conditions of constraint. When external forces act on the system, the

configuration of the system changes.

New configuration of the system can be obtainedby changing the generalised coordinates qjby qj,j =1, 2, , n.


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IfUj denotes the work done in changing the

generalised coordinate qjby the amount qj, thecorresponding generalised forceQj can bedefined as

where Qj will be a force (moment) when qj is alinear (angular) displacement.

njq

UQj

j

j ...,,2,1, !x

!


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Example: Triple Pendulum


30/46

To illustrate the concept of generalized coordinates, consider thetriple pendulum system shown.

The system can be specified by the six coordinates (xj,yj),

j =1, 2, 3.

However the coordinates are not independent but are constrainedby the relations

Since the coordinates (xj,yj) are not independent, they cannot be

called generalized coordinates. The angular displacement are used to specify the locations of

the masses without constraint.

Thus, they form a set of generalized coordinates .

23

2

23

2

23

2

2

2

12

2

12

2

1

2

1

2

1

lyyxx

lyyxx

lyx

!

!

!

jU

jjq U!


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32/46

Example:

The arrangement of the compressor, turbine and generator in athermal power plant is shown in figure above. This arrangementcan be considered as a torsional system whereJi denote the massmoments of inertia of the three components,Mti indicate theexternal moments acting on the components and kti represent the

torsional spring constants of the shaft between the components.Derive the equations of motion of the system using Lagrangesequations by treating the angular displacements of the componentsas generalized coordinates.


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Substituting [1], [2] and [3] into Lagranges equations to obtain

Equations of motion expressed in matrix form

3233333

2331223222

12212111

ttt

ttttt

tttt

MkkJ

MkkkkJ

MkkkJ

!

!!

UUU

UUUUUUU

!

-

-

3

2

1

3

2

1

33

3322

221

3

2

1

3

2

1

0

0

0000

00

t

t

t

tt

tttt

ttt

MM

M

kkkkkk

kkk

J

J

J

UU

U

UU

U


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General Equations in Matrix Form

Equations of motion of a multi d.o.f system can be derived inmatrix form from Lagranges equations

The kinetic and potential energy in matrix form

Rayleigh dissipation function

niFx

V

x

R

x

T

x

T

dt

di

iiii

...,,2,1, !!x

x

x

x

x

x

x

x

? A

? Axkx

xmxT

T

T

TT

T

T

2

1

2

1

!

!

? AxcxTT

T

2

1!


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From the theory of matrices

Substituting into Lagranges equation to obtain

? A ? A ? A

nixm

xmmxxm

x

T

T

i

TTT

i

...,,2,1,2

1

2

1

!!

!!

x

x

T

T

T

TTT

T

T

HHH

nix

Txm

x

T

dt

d

i

T

i

i

...,,2,10,and !!x

x!

x

x T

T

? A ? A ? A

nixk

xkkxxkx

T

i

TTT

i

...,,2,1,

21

21

!!

!!xx

TT

TTT

TTT

HHH

? A ? A ? A

nixc

xccxxcx

R

T

i

TTT

i

...,,2,1,

2

1

2

1

!!

!!x

x

T

T

T

TTT

T

T

HHH

? A ? A ? A FxkxcxmTTT

T !


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Eigenvalues and Eigenvectors

Solution of Eigenvalue Problem

Consider equation of motion in the form of

The equation can also be expressed aswhere

Premultiplying equation by [k]-1to obtain

where [I] is the identity matrix and

For nontrivial solution of

? A ? A? A 02TT

! Xmk [

? A ? A? A 0TT

! XmkP 21

[P !

? A ? A? A 0TT

! XDIP

? A ? A ? AmkD1!

X

T

? A ? A 0!!( DIP


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Example:

Find the natural frequencies and mode shapes of the systemshown in figure above for k1 = k2 = k3 = kand m1 = m2 = m3= m.


39/46

Solution:

The dynamical matrix is

Where the stiffness, flexibility and mass matrix can be obtainedusing influence coefficient approach or Lagranges equation

and

Thus

? A ? A ? A ? A? AmamkD !! 1

? A

-

!

321

221

1111

ka ? A

-

!

100

010

001

mm

? A

-

!

321

220

111

k

mD


40/46

Setting the characteristics determinant equal to zero to obtainthe frequency equation

where

Dividing throughout by

where

? A ? A ? A 0321

221

111

00

00

00

!

-

-

!!(k

mDI

P

P

P

P

P

0165312221

123

!!

EEEEEE EEE

EEE

k

m

k

m2

[

PE !!

2

1

[P !


41/46

The roots of cubic equation are

Once the natural frequencies are known, the mode shapes oreigenvectors can be calculated using equation

where

m

k

k

m

44504.0,19806.0 1

2

1

1!!! [

[

E

m

k

k

m

8025.1,2490.3 3

2

3

3 !!! [[

E

m

k

k

m2471.1,5553.1 2

2

22 !!! [

[E

? A ? A? A 3,2,1,0 !! iXDI iiTT

P

!i

i

i

i

X

X

X

X

3

2

1T


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First Mode: Substituting the value of

From the first two rows of equation

Solutions

and

The mode shape

km0489.51 !P

!

-

-

-

0

0

0

321

221

111

100

010

001

0489.51

3

1

2

1

1

X

X

X

km

km

!

-

!

0

0

0

0489.221

20489.31

110489.4

13

1

2

1

1

X

X

X

11

1

3

1

2

1

1

1

3

1

2

20489.3

0489.4

XXX

XXX

!

!

11

1

2 8019.1 XX ! 1

1

1

3 2470.2 XX !

!

!

2470.2

8019.1

11

1

1

3

1

2

1

1

1X

X

X

X

X

T


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Second Mode: Substituting the value of

From the first two rows of equation

Solutions

and

The mode shape

!

-

-

-

0

0

0

321

221

111

100

010

001

6430.02

3

2

2

2

1

X

X

X

km

km

!

-

0

0

0

3570.221

23570.11

113570.0

23

2

2

2

1

X

X

X

21

2

3

2

2

2

1

2

3

2

2

23570.1

3570.0

XXX

XXX

!

!

21

2

2 4450.0 XX ! 2

1

2

3 8020.0 XX !

!

!

8020.0

4450.0

12

1

2

3

2

2

2

1

2X

X

X

X

X

T

km6430.02 !P


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Mode Shapes


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