NFnMS_5

1Determination of Natural Frequency and Mode

Shapes

Eigenvalue and Eigenvector

Standard Matrix Iteration Method

Dunkerlys Formula

Rayleighs Method

Holzers Method

2Eigenvalue and Eigenvector

An eigenvector describes the movement of each mass and the direction of each mass relative to each others.

3What is the eigenvector for the 3 mass system ?

4For the 1st mode of vibration shown there are 3 natural frequencies and 3 mode shapes. The 1st mode shape is characterized by all the mass moving in phase to each other.

5

3

2

1

x

x

x

3

2

1

x

x

xEigenvalue 1

2

or

mode 1

6mode 2

3

2

1

x

x

x

3

2

1

x

x

x

3

2

1

x

x

x

2 > 1

Eigen value 12

or

or

7

3

2

1

x

x

x

3

2

1

x

x

x

3 > 2 >

mode 3

Eigenvalue 32 or

1

11

(7)

Standard Matrix Iteration Method

20

21

Mass matrix

In the mass matrix of a MDOF system such as in the

example , the leading diagonal represent each masses as a

matrix elements , the rest is zero.

m00

02 / m0

00m

[ m ] =

3

2

1

x

x

x

3

2

1

x

x

x

characteristic equation of a MDOF system

2 [ m ] [ x ] = [ k ] [ x ]

= 2

24

Inverse Matrix.

35

Dunkerleys Formula

It gives the approx value of the fundamental freq of a

composite system.

Consider the following general n DOF system:

For a lumped mass system with diagonal mass

matrix, the equation becomes:

01or 02

2 maImk

0

0...0

0

0

0...0

...

...

...

10...0

0

10

0...01

1 2

1

21

22221

11211

2

nnnnn

n

n

m

m

m

aaa

aaa

aaa

36

i.e.

Expanding:

0

1...

...1

...1

22211

22222121

12121112

nnnnn

nn

nn

mamama

mamama

mamama

(E.1) 0...1

)...

...(

1...

1

2

211,,1212112

11,1313311212211

1

22221112

n

nnnnnn

nnnnnn

n

nnn

n

mmaammaa

mmaammaammaa

mamama

37

Let the roots of this equation be 1/12, 1/22,, 1/n2. Thus

Equating coefficients of (1/2)n-1 in (E.1) and (E.2):

In most cases,

(E.2) 0...11

...11111

...1111

1

222

2

2

1

2222

2

22

1

2

n

n

n

n

n22211122

2

2

1

m...mm1

...11

nn

n

aaa

n2,3,...,i ,11

2

1

2

i

38

Thus

Can also be written as

where in=(1/aiimi)1/2=(kii/mi)1/2

formula) s'(Dunkerley ...1

2221112mamama nn

i

22

2

2

1

2

1...

111

nnnni

39

Example 1

Estimate the fundamental natural

frequency of a simply supported beam

carrying 3 identical equally spaced

masses, as shown below.

40

Solution

m1=m2=m3=m

EI

la

EI

laa

3

22

3

331148

1,

256

3

31

33

2

1

75375.4

04427.0256

3

48

1

256

31

ml

EI

EI

ml

EI

ml

41

Example 2

Estimate the fundamental frequency of the

beam shown using Dunkerleys formula for the following data: m1=m3=5m, m2=m.

49

Rayleighs Method

Based on Rayleighs Principle

Kinetic and potential energies of an n DOF

discrete system:

Assume harmonic motion to be

where is the mode shape and is the natural frequency

xkxVxmxT TT 2

1 ,

2

1

tXx cos

X

50

Maximum KE:

Maximum PE:

For a conservative system, Tmax=Vmax

XkXV T

2

1max

2max2

1XmXT T

quotient sRayleigh' 2 XRXmX

XkXT

T

51

Example

Estimate the fundamental

frequency of vibration of the

system as shown. Assume

that m1=m2=m3=m,

k1=k2=k3=k, and the mode

shape is

3

2

1

X

52

Stiffness matrix

Mass matrix

Substitute the assumed mode shape into

110

121

012

kk

100

010

001

mm

m

k

m

k

m

k

XR 4629.02143.0

3

2

1

100

010

001

321

3

2

1

110

121

012

321

2

53

Fundamental Frequency of Beams and Shafts

Static deflection curve is used to approximate the

dynamic deflective curve.

Consider a shaft carrying several masses as shown

below.

54

Potential energy of the system is strain energy of the

deflected shaft, which is the work done by the static

loads.

For free vibration, max kinetic energy due to the masses

is

Equating Vmax and Tmax,

2211max2

1gwmgwmV

2222112

max2

wmwmT

2

22

2

11

2211

wmwm

wmwmg

55

Example

Estimate the fundamental frequency of the lateral vibration of a shaft carrying 3 rotors (masses), as shown below with m1=20kg, m2=50kg, m3=40kg, l1=1m, l2=3m, l3=4m and l4=2m. The shaft is made of steel with a solid circular cross section of diameter 10cm.

56

The deflection of the beam shown below due to

a static load P is given by

lxalxxaEIl

xlPa

axxblEIl

Pbx

xw

;26

0 ;6

22

222

57

Deflection due to m1

At the location of m1:



EIEI

w74.529

181100106

1981.920'1

EIEI

w06.1236

4102161106

6181.920'2

EIEI

w3.621

8102641106

2181.920'3

58





EIEI

w15.3090

136100106

1681.950''1

EIEI

w6.9417

1636100106

4681.950''2

EIEI

w5232

81026416106

2481.950''3

59





EIEI

w6.1242

14100106

1281.940'''1

EIEI

w6.4185

164100106

4281.940'''2

EIEI

w48.3348

644100106

8281.940'''3

60

Total deflections of m1,m2 and m3:

EIwwww

EIwwww

EIwwww

78.9201

26.14839

49.4862

'''

3

''

3

'

33

'''

2

''

2

'

22

'''

1

''

1

'

11

61

Substituting into

For the shaft, E=2.07x1011 N/m2 and I=(0.1)4/64 m4

Hence =28.4482 rad/s

,2

22

2

11

2211

wmwm

wmwmg

EIEI

028222.078.92014026.148395049.486220

78.92014026.148395049.48622081.9222

62

Holzers Method

A trial-and-error scheme to find natural

frequencies of systems

A trial frequency is first assumed, and a

solution is found when the constraints are

satisfied.

Requires several trials

The method also gives mode shapes

63

Holzers Method: Torsional Systems

Consider the undamped torsional semidefinite system

shown below.

Equations of motion

0

0

0

23233

32212122

12111

t

tt

t

kJ

kkJ

kJ

64

Since the motion is harmonic, i=icos(t+)

Summing these equations gives

This states that the sum of the inertia torques of the

system must be zero.

The trial freq must satisfy this requirement.

232332

21121122

2

21111

2

t

tt

t

kJ

kkJ

kJ

03

1

2 i

iiJ

65

1 is arbitrarily chosen as 1.

Substitute these values into to see

whether the constraints are satisfied.

If not, repeat the process with a new trial value of .

These equations can be generalized for a n-disc

system as follows:

22112

2

23

1

11

2

12 ,

JJkk

J

tt

03

1

2 i

iiJ

01

2

n

i

iiJ niJk

i

k

kk

ti

ii ,,3,2 ,1

11

2

1

66

The graph below plots the torque Mt applied at the last disc against the chosen .

The natural frequencies are the at which Mt=0.

The amplitudes i (i=1,2,,n) are the mode shapes of the system

67

Example

The arrangement of the compressor, turbine and

generator in a thermal power plant is shown

below. Find the natural frequencies and mode

shapes of the system.

69

Mt3 is the torque to the right

of the generator, which must

be zero at the natural

frequencies.

Closely-spaced trial values

of are used in the vicinity of Mt3=0 to obtain accurate

values of the 1st two flexible

mode shapes, as shown.

70

Holzers Method: Spring-Mass Systems

Holzers method is also applicable to vibration analysis of spring-mass systems.

Equations of motion:

For harmonic motion, xi(t)=Xicost where Xi is the amplitude of mass mi. Thus

0

0

32212122

21111

xxkxxkxm

xxkxm

32211

2

32212122

2

21111

2

XXkXmXXkXXkXm

XXkXm

71

The resultant force applied to the last (nth) mass can be

computed as follows:

Repeat for several other trial frequencies .

Plot a graph of F vs . The natural frequencies are those that give F=0.

niXmk

XX

XmXmk

XX

k

XmXX

i

k

kk

i

ii ,...,3,2 ,1

11

2

1

2211

2

2

23

1

11

2

12

n

i

iiXmF1

2

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