Mechanical Vibrations (ME 65) CHAPTER-8 MULTI · PDF fileMechanical Vibrations (ME 65) CHAPTER-8 MULTI DEGREE OF FREEDOM SYSTEMS Topics covered: Influence co-efficents

Mechanical Vibrations (ME 65)

CHAPTER-8

MULTI DEGREE OF FREEDOM SYSTEMS

Topics covered:

Influence co-efficents

Approximate methods

(i) Dunkerley’s method

(ii) Rayleigh’s method

Influence co-efficients

Numerical methods

(i) Matrix iteration method

(ii) Stodola’s method

(iii) Holzar’s method

1. Influence co-efficents

It is the influence of unit displacement at one point on the forces at various points of a

multi-DOF system.

OR

It is the influence of unit Force at one point on the displacements at various points of

a multi-DOF system.

The equations of motion of a multi-degree freedom system can be written in terms of

influence co-efficients. A set of influence co-efficents can be associated with each of

matrices involved in the equations of motion.

[ ]{ } [ ]{ } [ ]0xKxM =+&&

For a simple linear spring the force necessary to cause unit elongation is referred as

stiffness of spring. For a multi-DOF system one can express the relationship between

displacement at a point and forces acting at various other points of the system by

using influence co-efficents referred as stiffness influence coefficents

The equations of motion of a multi-degree freedom system can be written in terms of

inverse of stiffness matrix referred as flexibility influence co-efficients.

Matrix of flexibility influence co-efficients = [ ] 1−K

The elements corresponds to inverse mass matrix are referred as flexibility

mass/inertia co-efficients.

Matrix of flexibility mass/inertia co-efficients =[ ] 1−M

The flexibility influence co-efficients are popular as these coefficents give elements

of inverse of stiffness matrix. The flexibility mass/inertia co-efficients give elements

of inverse of mass matrix

VTU e-learning Course ME65 Mechanical Vibrations

Prof. S. K. Kudari, Principal APS College of Engineering, Bengaluru-28.

2

Stiffness influence co-efficents.

For a multi-DOF system one can express the relationship between displacement at a

point and forces acting at various other points of the system by using influence co-

efficents referred as stiffness influence coefficents.

{ } [ ]{ }xKF =

[ ]

=

333231

322221

131211

kkk

kkk

kkk

K

wher, k11, ……..k33 are referred as stiffness influence coefficients

k11-stiffness influence coefficient at point 1 due to a unit deflection at point 1

k21- stiffness influence coefficient at point 2 due to a unit deflection at point 1

k31- stiffness influence coefficient at point 3 due to a unit deflection at point 1

Example-1.

Obtain the stiffness coefficients of the system shown in Fig.1.

I-step:

Apply 1 unit deflection at point 1 as shown in Fig.1(a) and write the force equilibrium

equations. We get,

2111 KKk +=

221 Kk −=

0k31 =

m1

K1

m2

K2

x1=1 Unit

x2=0

m3

K3

x3=0

k11

k21

k31

x2=1 Unit

m1

K1

m2

K2

x1=0

m3

K3

x3=0

K12

k22

k32

m1

K1

m2

K2

x1=0

x2=0

m3

K3

x3=1 Unit

k13

k23

k33

(a) (b) (c)

Fig.1 Stiffness influence coefficients of the system



3

II-step:

Apply 1 unit deflection at point 2 as shown in Fig.1(b) and write the force equilibrium

equations. We get,

212 -Kk =

3222 KKk +=

331 -Kk =

III-step:

Apply 1 unit deflection at point 3 as shown in Fig.1(c) and write the force equilibrium

equations. We get,

0k13 =

323 -Kk =

333 Kk =

[ ]

=

333231

322221

131211

kkk

kkk

kkk

K

[ ]( )

( )

+

+

=

33

3322

221

KK-0

K-KKK-

0K-KK

K

From stiffness coefficients K matrix can be obtained without writing Eqns. of motion.

Flexibility influence co-efficents.

{ } [ ]{ }xKF =

{ } [ ] { }FKx1−

=

{ } [ ]{ }Fαx =

where, [ ]α - Matrix of Flexibility influence co-efficents given by

[ ]

=

333231

322221

131211

ααα

ααα

ααα

α

wher, α11, ……..α33 are referred as stiffness influence coefficients

α11-flexibility influence coefficient at point 1 due to a unit force at point 1

α21- flexibility influence coefficient at point 2 due to a unit force at point 1

α31- flexibility influence coefficient at point 3 due to a unit force at point 1

Example-2.

Obtain the flexibility coefficients of the system shown in Fig.2.



4

I-step:

Apply 1 unit Force at point 1 as shown in Fig.2(a) and write the force equilibrium

equations. We get,

1

312111K

1ααα ===

II-step:

Apply 1 unit Force at point 2 as shown in Fig.2(b) and write the force equilibrium

equations. We get,

21

3222K

1

K

1αα +==

III-step:

Apply 1 unit Force at point 3 as shown in Fig.2(c) and write the force equilibrium

equations. We get,

21

23K

1

K

1α +=

321

33K

1

K

1

K

1α ++=

Therefore,

1

1331122111K

1ααααα ======

21

233222K

1

K

1ααα +===

321

33K

1

K

1

K

1α ++=

(b) (c)

Fig.2 Flexibility influence coefficients of the system

x2=α22

m1

K1

m2

K2

x1=α12

m3

K3

x3=α32

F1=0

F2=1

F3=0

m1

K1

m2

K2

x1=α13

x2=α23

m3

K3

x3=α33

F1=0

F2=1

F3=0

(a)

m1

K1

m2

K2

m3

K3

x3=α31

F1=0

F2=1

F3=0

x1=α11

x2=α21



5

For simplification, let us consider : KKKK 321 ===

K

1

K

1ααααα

1

1331122111 =======

K

2

K

1

K

1ααα 233222 =+===

K

3

K

1

K

1

K

1α33 =++=

[ ]

=

333231

322221

131211

ααα

ααα

ααα

α

[ ]

=

321

221

111

K

1α

[ ] [ ] 1Kα

−=

In Vibration analysis if there is need of [ ] 1K

−one can use flexibility co-efficent matrix.

Example-3

Obtain of the Flexibility influence co-efficents of the pendulum system shown in the

Fig.3.

I-step:

Apply 1 unit Force at point 1 as shown in Fig.4 and write the force equilibrium

equations. We get,

m

m

l

l

m

l

Fig.3 Pendulum system

m

m

l

l

m

l

F=1

T θ

α11

Fig.4 Flexibility influence

co-efficents



6

lθ sin T =

3mgm)mg(mθ cos T =++=

3mg

1θ tan =

θ sinθ tan small, is θ =

l

αθ sin 11=

θ sin lα11 =

3mg

lα11 =

Similarly apply 1 unit force at point 2 and next at point 3 to obtain,

5mg

lα22 =

the influence coefficients are:

5mg

lααααα 1331122111 ======

6mg

11lααα 233222 ===

6mg

11lα33 =

Approximate methods

In many engineering problems it is required to quickly estimate the first

(fundamental) natural frequency. Approximate methods like Dunkerley’s method,

Rayleigh’s method are used in such cases.

(i) Dunkerley’s method

Dunkerley’s formula can be determined by frequency equation,

[ ] [ ] [ ]0KMω2 =+−

[ ] [ ] [ ]0MωK 2 =+−

[ ] [ ] [ ] [ ]0MKIω

1 1

2=+−

−

[ ] [ ][ ] [ ]0MαIω

12

=+−

For n DOF systems,



7

[ ]0

m.00

.

0.m0

0.0m

α.αα

.

α.αα

α.αα

1.00

.

0.10

0.01

ω

1

n

2

1

nnn2n1

2n2221

1n1211

2=

+

−

[ ]0

mαω

1.mαmα

.

mα.mαω

10

mα.mαmαω

1

nnn22n21n1

n2n2222

n1n2121112

=

+−

+−

+−

...

Solve the determinant

( )

( ) [ ]0...mα...mmααmmαα

ω

1mα...mαmα

ω

1

nnn313311212211

1n

2nnn222111

n

2

=++++

+++−

−

(1)

It is the polynomial equation of nth degree in (1/ω2). Let the roots of above Eqn. are:

2n

22

21 ω

1...... ,

ω

1 ,

ω

1

0...ω

1

ω

1......

ω

1

ω

1

ω

1

ω

1

ω

1...... ,

ω

1

ω

1 ,

ω

1

ω

1

1n

22n

22

21

n

2

2n

222

221

2

=−

+++−

=

−

−

−

− (2)

Comparing Eqn.(1) and Eqn. (2), we get,

=

+++

2n

22

21 ω

1......

ω

1

ω

1 ( )nnn222111 mα...mαmα +++

In mechanical systems higher natural frequencies are much larger than the

fundamental (first) natural frequencies. Approximately, the first natural frequency is:

≅

21ω

1( )nnn222111 mα...mαmα +++

The above formula is referred as Dunkerley’s formula, which can be used to estimate

first natural frequency of a system approximately.

The natural frequency of the system considering only mass m1 is:

1

1

111

1nm

K

mα

1ω ==

The Dunkerley’s formula can be written as:

2

nn

2

2n

2

1n

2

1 ω

1......

ω

1

ω

1

ω

1+++≅ (3)



8

where, ..... ,ω ,ω 2n1n are natural frequency of single degree of freedom system

considering each mass separately.

The above formula given by Eqn. (3) can be used for any mechanical/structural

system to obtain first natural frequency

Examples: 1

Obtain the approximate fundamental natural frequency of the system shown in Fig.5

using Dunkerley’s method.

Dunkerley’s formula is:

≅

21ω

1 ( )nnn222111 mα...mαmα +++ OR

2

nn

2

2n

2

1n

2

1 ω

1......

ω

1

ω

1

ω

1+++≅

Any one of the above formula can be used to find fundamental natural frequency

approximately.

Find influence flexibility coefficients.

K

1ααααα 1331122111 =====

K

2ααα 233222 ===

K

3α33 =

Substitute all influence coefficients in the Dunkerley’s formula.

≅

21ω

1 ( )nnn222111 mα...mαmα +++

m

K

m

K

x1

x2

m

K

x3

Fig.5 Linear vibratory system



9

≅

21ω

1

K

6m

K

3m

K

2m

K

m=

++

K/m0.40ω1= rad/s

Examples: 2

Find the lowest natural frequency of the system shown in Figure by Dunkerley’s

method. Take m1=100 kg, m2=50 kg

VTU Exam July/Aug 2006 for 20 Marks

Obtain the influence co-efficents:

EI

1.944x10α

-3

11 =

EI

9x10α

-3

22 =

≅

2nω

1 ( )222111 mαmα +

rad/s 1.245ωn =

(ii) Rayleigh’s method

It is an approximate method of finding fundamental natural frequency of a system

using energy principle. This principle is largely used for structural applications.

Principle of Rayleigh’s method

Consider a rotor system as shown in Fig.7. Let, m1, m2 and m3 are masses of rotors on

shaft supported by two bearings at A and B and y1, y2 and y3 are static deflection of

shaft at points 1, 2 and 3.

1 2

180 120

m1 m2

Fig.6 A cantilever rotor system.



10

For the given system maximum potential energy and kinetic energies are:

∑=

=n

1iiimax gym

2

1V (4)

∑=

=n

1i

2iimax ym

2

1T &

where, mi- masses of the system, yi –displacements at mass points.

Considering the system vibrates with SHM,

i2

i yωy =&

From above equations

∑=

=n

1i

2ii

2

max ym2

ωT (5)

According to Rayleigh’s method,

maxmax TV = (6)

substitute Eqn. (4) and (5) in (6)

∑

∑

=

==n

1i

2

ii

n

1i

ii2

ym

gym

ω (7)

The deflections at point 1, 2 and 3 can be found by.

gmαgmαgmαy 3132121111 ++=



Eqn.(7) is the Rayleigh’s formula, which is used to estimate frequency of transverse

vibrations of a vibratory systems.

1 2 3

m1 m2 m3

A B

y1 y2

y3

Fig.7 A rotor system.



11

Examples: 1

Estimate the approximate fundamental natural frequency of the system shown in Fig.8

using Rayleigh’s method. Take: m=1kg and K=1000 N/m.

Obtain influence coefficients,

2K

1ααααα 1331122111 =====

2K

3ααα 233222 ===

2K

5α33 =

Deflection at point 1 is:


( )2000

5g

2K

5mg122

2K

mgy1 ==++=



( )2000

11g

2K

11mg362

2K

mgy2 ==++=



( )2000

13g

2K

13mg562

2K

mgy3 ==++=

Rayleigh’s formula is:

2m

2K

2m

K

x1

x2

m

K

x3




12

∑

∑

=

==n

1i

2ii

n

1i

ii2

ym

gym

ω

2

222

2

2

g2000

132

2000

112

2000

52

g2000

132x

2000

112x

2000

52x

ω

+

+

++

=

rad/s 12.41ω =

Examples: 2

Find the lowest natural frequency of transverse vibrations of the system shown in

Fig.9 by Rayleigh’s method.

E=196 GPa, I=10-6 m4, m1=40 kg, m2=20 kg

VTU Exam July/Aug 2005 for 20 Marks

Step-1:

Find deflections at point of loading from strength of materials principle.

For a simply supported beam shown in Fig.10, the deflection of beam at distance x

from left is given by:

( ) b)(l xfor bxl6EIl

Wbxy 222 −≤−−=

For the given problem deflection at loads can be obtained by superposition of

deflections due to each load acting separately.

1 2

m1 m2

B

160 80 180

A

Fig.9 A rotor system.

b x

l

W

Fig.10 A simply supported beam



13

Deflections due to 20 kg mass

( ) ( ) EI

0.2650.180.160.42

6EI0.42

x0.18x0.169.81x20y 222'

1 =−−=

( ) ( ) EI

0.290.180.240.42

6EIx0.42

x0.18x0.249.81x20y 222'

2 =−−=

Deflections due to 40 kg mass

( ) ( ) EI

0.5380.160.260.42

6EIx0.42

x0.16x0.269.81x40y 222''

1 =−−=

( ) ( ) EI

0.530.160.180.42

6EIx0.42

x0.16x0.189.81x40y 222''

2 =−−=

The deflection at point 1 is:

EI

0.803yyy ''

1'11 =+=

The deflection at point 2 is:

EI

0.82yyy ''

2'22 =+=

∑

∑

=

==n

1i

2

ii

n

1i

ii2

ym

gym

ω

( )( ) ( )22

2

20x0.8240x0.803

20x0.8240x0.8039.81ω

+

+=

rad/s 1541.9ωn =



14

Numerical methods

(i) Matrix iteration method

Using this method one can obtain natural frequencies and modal vectors of a vibratory

system having multi-degree freedom.

It is required to have ω1< ω2<……….< ωn

Eqns. of motion of a vibratory system (having n DOF) in matrix form can be written

as:

[ ]{ } [ ]{ } [ ]0xKxM =+&&

where,

{ } { } ( )φωtsinAx += (8)

substitute Eqn.(8) in (9)

[ ]{ } [ ]{ } [ ]0AKAMω2 =+− (9)

For principal modes of oscillations, for rth

mode,

[ ]{ } [ ]{ } [ ]0AKAMω rr2r =+−

[ ] [ ]{ } { }r2

r

r Aω

AMK11

=−

[ ]{ } { }r2

r

r Aω

AD1

= (10)

where, [ ]D is referred as Dynamic matrix.

Eqn.(10) converges to first natural frequency and first modal vector.

The Equation,

[ ] [ ]{ } { }r

2

rr AωAKM =−1

[ ]{ } { }r2rr1 AωAD = (11)

where, [ ]1D is referred as inverse dynamic matrix.

Eqn.(11) converges to last natural frequency and last modal vector.

In above Eqns (10) and (11) by assuming trial modal vector and iterating till the Eqn

is satisfied, one can estimate natural frequency of a system.

Examples: 1

Find first natural frequency and modal vector of the system shown in the Fig.10 using

matrix iteration method. Use flexibility influence co-efficients.

Find influence coefficients.

2K

1ααααα 1331122111 =====

2K

3ααα 233222 ===



15

2K

5α33 =

[ ]

=

333231

322221

131211

ααα

ααα

ααα

α

[ ] [ ]

==−

531

331

111

2K

1Kα

1

First natural frequency and modal vector

[ ] [ ]{ } { }r2

r

r Aω

AMK11

=−

[ ]{ } { }r2

r

r Aω

AD1

=

Obtain Dynamic matrix [ ] [ ] [ ]MKD1−

=

[ ]

=

=

562

362

122

2K

m

100

020

002

531

331

111

2K

mD

Use basic Eqn to obtain first frequency

[ ]{ } { }12

r

1 Aω

AD1

=

Assume trial vector and substitute in the above Eqn.

Assumed vector is: { }

=

1

1

1

u 1

First Iteration

[ ]{ } =1uD

1

1

1

562

362

122

2K

m

=

2.6

2.2

1

2K

5m

As the new vector is not matching with the assumed one, iterate again using the new

vector as assumed vector in next iteration.



16

Second Iteration

[ ]{ } =2uD

2.6

2.2

1

562

362

122

2K

m

=

3.13

2.55

1

K

4.5m

Third Iteration

[ ]{ } =3uD

3.133

2.555

1

562

362

122

2K

m

=

3.22

2.61

1

K

5.12m

Fourth Iteration

[ ]{ } =4uD

3.22

2.61

1

562

362

122

2K

m

=

3.23

2.61

1

K

5.22m

As the vectors are matching stop iterating. The new vector is the modal vector.

To obtain the natural frequency,

[ ]

3.22

2.61

1

D

=

3.23

2.61

1

K

5.22m

Compare above Eqn with with basic Eqn.

[ ]{ } { }12

1

1 Aω

AD1

=

K

5.22m

ω

12

1

=

m

K

5.22

1ω2

1 =

m

K0.437ω1 = Rad/s

Modal vector is:

{ }

=

3.23

2.61

1

A 1



17

Method of obtaining natural frequencies in between first and last one

(Sweeping Technique)

For understanding it is required to to clearly understand Orthogonality principle of

modal vectors.

Orthogonality principle of modal vectors

Consider two vectors shown in Fig.11. Vectors { } { }b and a are orthogonal to each

other if and only if

{ } { } 0baT

=

{ } 0b

baa

2

1

21 =

{ } 0b

b

10

01aa

2

1

21 =

{ } [ ]{ } 0bIaT

= (12)

where, [ ]I is Identity matrix.

From Eqn.(12), Vectors { } { }b and a are orthogonal to each other with respect to

identity matrix.

Application of orthogonality principle in vibration analysis

Eqns. of motion of a vibratory system (having n DOF) in matrix form can be written

as:

[ ]{ } [ ]{ } [ ]0xKxM =+&&

{ } { } ( )φωtsinAx +=

[ ]{ } [ ]{ } [ ]0AKAMω2 =+− 11

x1

x2

{ }

=2

1

b

bb

{ }

=2

1

a

aa

Fig.11 Vector representation graphically



18

[ ]{ } [ ]{ }11 AKAMω2 =

If system has two frequencies ω1 and ω2

[ ]{ } [ ]{ }11 AKAMω21 = (13)

[ ]{ } [ ]{ }22 AKAMω22 = (14)

Multiply Eqn.(13) by { }T

2A and Eqn.(14) by { }T

1A

{ } [ ]{ } { } [ ]{ }1

T

21

T

2

21 AKAAMAω = (15)

{ } [ ]{ } { } [ ]{ }2

T

12

T

1

21 AKAAMAω = (16)

Eqn.(15)-(16)

{ } [ ]{ } 0AMA 2

T

1 =

Above equation is a condition for mass orthogonality.

{ } [ ]{ } 0AKA 2

T

1 =

Above equation is a condition for stiffness orthogonality.

By knowing the first modal vector one can easily obtain the second modal vector

based on either mass or stiffness orthogonality. This principle is used in the matrix

iteration method to obtain the second modal vector and second natural frequency.

This technique is referred as Sweeping technique

Sweeping technique

After obtaining { } 11 ω and A to obtain { } 22 ω and A choose a trial vector { }1V

orthogonal to { }1A ,which gives constraint Eqn.:

{ } [ ]{ } 0AMV 1

T

1 =

{ } 0

A

A

A

m00

0m0

00m

VVV

3

2

1

3

2

1

321 =

( ) ( ) ( ){ } 0AmVAmVAmV 333222111 =++

( ) ( ) ( ){ } 0VAmVAmVAm 333222111 =++

321 βVαVV +=

where α and β are constants

−=

11

22

Am

Amα

−=

11

33

Am

Amβ

Therefore the trial vector is:



19

( )

+

=

3

2

32

3

2

1

V

V

βVαV

V

V

V

=

3

2

1

V

V

V

100

010

βα0

[ ]{ }1VS=

where [ ]S= is referred as Sweeping matrix and { }1V is the trial vector.

New dynamics matrix is:

[ ] [ ][ ]SDDs +

[ ]{ } { }222

1s Aω

VD1

=

The above Eqn. Converges to second natural frequency and second modal vector.

This method of obtaining frequency and modal vectors between first and the last one

is referred as sweeping technique.

Examples: 2

For the Example problem 1, Find second natural frequency and modal vector of the

system shown in the Fig.10 using matrix iteration method and Sweeping technique.

Use flexibility influence co-efficients.

For this example already the first frequency and modal vectors are obtained by matrix

iteration method in Example 1. In this stage only how to obtain second frequency is

demonstrated.

First Modal vector obtained in Example 1 is:

{ }

=

=

3.23

2.61

1

A

A

A

A

3

2

1

1

[ ]

=

100

020

002

M is the mass matrix

Find sweeping matrix

[ ]

=

100

010

βα0

S

2.612(1)

2(2.61)

Am

Amα

11

22 −=

−=

−=



20

1.6152(1)

1(3.23)

Am

Amβ

11

33 −=

−=

−=

Sweeping matrix is:

[ ]

=

100

010

1.615-2.61-0

S

New Dynamics matrix is:

[ ] [ ][ ]SDDs +

[ ]

−

−−

=

=

1.890.390

0.110.390

1.111.610

K

m

100

010

1.615-2.61-0

531

362

122

2K

mDs

First Iteration

[ ]{ } { }222

1s Aω

VD1

=

=

=

−

−−

8.14

1

9.71-

K

0.28m

2.28

0.28

2.27-

K

m

1

1

1

1.890.390

0.110.390

1.111.610

K

m

Second Iteration

=

=

−

−−

31.54

1-

21.28-

K

0.5m

15.77

0.50-

10.64-

K

m

8.14

1

9.71-

1.890.390

0.110.390

1.111.610

K

m

Third Iteration

=

=

−

−−

15.38

1-

8.67-

K

3.85m

59.52

3.85-

33.39-

K

m

31.54

1-

21.28-

1.890.390

0.110.390

1.111.610

K

m

Fourth Iteration

=

=

−

−−

13.78

1-

8.98-

K

2.08m

28.67

2.08-

18.68-

K

m

15.38

1-

8.67-

1.890.390

0.110.390

1.111.610

K

m

Fifth Iteration

=

=

−

−−

13.5

1

7.2-

K

1.90m

25.65

1.90-

13.68-

K

m

13.78

1

8.98-

1.890.390

0.110.390

1.111.610

K

m

Sixth Iteration



21

=

=

−

−−

13.43

1-

7.08-

K

1.87m

25.12

1.87-

13.24-

K

m

13.5

1-

7.2-

1.890.390

0.110.390

1.111.610

K

m

K

1087m

ω

12

2

=

m

K

1.87

1ω2

1 =

m

K0.73ω1 =

Modal vector

{ }

=

1.89

0.14-

1-

A 2

Similar manner the next frequency and modal vectors can be obtained.



22

(ii) Stodola’s method

It is a numerical method, which is used to find the fundamental natural frequency and

modal vector of a vibratory system having multi-degree freedom. The method is

based on finding inertia forces and deflections at various points of interest using

flexibility influence coefficents.

Principle / steps

1. Assume a modal vector of system. For example for 3 dof systems:

=

1

1

1

x

x

x

3

2

1

2. Find out inertia forces of system at each mass point,

12

11 xωmF = for Mass 1

22


32


3. Find new deflection vector using flexibility influence coefficients, using the

formula,

++

++

++

=

′

′

′

333322311

233222211

133122111

3

2

1

αFαFαF

αFαFαF

αFαFαF

x

x

x

4. If assumed modal vector is equal to modal vector obtained in step 3, then solution

is converged. Natural frequency can be obtained from above equation, i.e

If

′

′

′

≅

3

2

1

3

2

x

x

x

x

x

x1

Stop iterating.

Find natural frequency by first equation,

1331221111 αFαFαFx ++==′ 1

5. If assumed modal vector is not equal to modal vector obtained in step 3, then

consider obtained deflection vector as new vector and iterate till convergence.

Example-1

Find the fundamental natural frequency and modal vector of a vibratory system shown

in Fig.10 using Stodola’s method.

First iteration



23

1. Assume a modal vector of system { }1u =

=

1

1

1

x

x

x

3

2

1

2. Find out inertia forces of system at each mass point

21

211 2mωxωmF ==

22

222 2mωxωmF ==

23

233 mωxωmF ==

3. Find new deflection vector using flexibility influence coefficients

Obtain flexibility influence coefficients of the system:

2K

1ααααα 1331122111 =====

2K

3ααα 233222 ===

2K

5α33 =

1331221111 αFαFαFx ++=′

Substitute for F’s and α,s

2K

5mω

2K

mω

K

mω

K

mωx

2222

1 =++=′

2332222112 αFαFαFx ++=′


2K

11mω

2K

3mω

2K

6mω

K

mωx

2222

2 =++=′

3333223113 αFαFαFx ++=′


2K

13mω

2K

5mω

2K

6mω

K

mωx

2222

3 =++=′

4. New deflection vector is:

=

′

′

′

13

11

5

2K

mω

x

x

x2

3

2

1

=

′

′

′

2.6

2.2

1

2K

5mω

x

x

x2

3

2

1

={ }2u

The new deflection vector{ } { }12 uu ≠ . Iterate again using new deflection vector { }2u

Second iteration



24

1. Initial vector of system { }2u =

=

′

′

′

2.6

2.2

1

x

x

x

3

2

1


21

211 2mωxωmF =′=′

22

222 mω4xωmF 4.=′=′

23

233 2.6mωxωmF =′=′

3. New deflection vector,

1331221111 αFαFαFx ′+′+′=′′


2K

9mω

2K

2.6mω

2K

4.4mω

K

mωx

2222

1 =++=′′

2332222112 αFαFαFx ′+′+′=′′


2K

23mω

2K

7.8mω

2K

13.2mω

K

mωx

2222

2 =++=′′

3333223113 αFαFαFx ′+′+′=′′


2K

28.2mω

2K

13mω

2K

13.2mω

K

mωx

2222

3 =++=′′


=

′′

′′

′′

28.2

23

9

2K

mω

x

x

x2

3

2

1

=

′′

′′

′′

3.13

2.55

1

2K

9mω

x

x

x2

3

2

1

={ }3u

The new deflection vector{ } { }23 uu ≠ . Iterate again using new deflection vector{ }3u

Third iteration


=

′′

′′

′′

3.13

2.55

1

x

x

x

3

2

1


21

211 2mωxωmF =′′=′′

2

22

22 5.1mωxωmF =′′=′′



25

3. new deflection vector,

1331221111 αFαFαFx ′′+′′+′′=′′′


2K

10.23mω

2K

3.13mω

2K

5.1mω

K

mωx

2222

1 =++=′′′

2332222112 αFαFαFx ′′+′′+′′=′′′


2K

26.69mω

2K

9.39mω

2K

15.3mω

K

mωx

2222

2 =++=′′′

3333223113 αFαFαFx ′′+′′+′′=′′′


2K

28.2mω

2K

16.5mω

2K

15.3mω

K

mωx

2222

3 =++=′′′


=

′′′

′′′

′′′

33.8

26.69

10.23

2K

mω

x

x

x2

3

2

1

=

′′′

′′′

′′′

3.30

2.60

1

2K

10.23mω

x

x

x2

3

2

1

={ }4u

The new deflection vector { } { }34 uu ≅ stop Iterating

Fundamental natural frequency can be obtained by.

1=2K

10.23mω2

m

K0.44ω = rad/s

Modal vector is:

{ }

=

3.30

2.60

1

A 1

23

233 3.13mωxωmF =′′=′′



26

Example-2

For the system shown in Fig.12 find the lowest natural frequency by Stodola’s method

(carryout two iterations)

July/Aug 2005 VTU for 10 marks

Obtain flexibility influence coefficients,

3K

1ααααα 1331122111 =====

3K

4ααα 233222 ===

3K

7α33 =

First iteration

1. Assume a modal vector of system { }1u =

=

1

1

1

x

x

x

3

2

1


21

211 4mωxωmF ==

22

222 2mωxωmF ==

23

233 mωxωmF ==

3. New deflection vector using flexibility influence coefficients,

1331221111 αFαFαFx ++=′

3K

7mω

3K

mω

3K

2mω

3K

4mωx

2222

1 =++=′

4m

3K

2m

K

x1

x2

m

K

x3




27

2332222112 αFαFαFx ++=′

3K

16mω

3K

4mω

3K

8mω

3K

4mωx

2222

2 =++=′

3333223113 αFαFαFx ++=′

3K

19mω

3K

7mω

3K

8mω

3K

4mωx

2222

3 =++=′


=

′

′

′

19

16

7

3K

mω

x

x

x2

3

2

1

=

′

′

′

2.71

2.28

1

3K

7mω

x

x

x2

3

2

1

={ }2u

The new deflection vector{ } { }12 uu ≠ . Iterate again using new deflection vector { }2u

Second iteration


=

′

′

′

2.71

2.28

1

x

x

x

3

2

1


21

211 4mωxωmF =′=′

22

222 4.56mωxωmF =′=′

23

233 2.71mωxωmF =′=′

3. New deflection vector

1331221111 αFαFαFx ′+′+′=′′

3K

11.27mω

3K

2.71mω

3K

4.56mω

3K

4mωx

2222

1 =++=′′

2332222112 αFαFαFx ′+′+′=′′

3K

33.08mω

3K

10.84mω

3K

18.24mω

3K

4mωx

2222

2 =++=′′

3333223113 αFαFαFx ′+′+′=′′

3K

41.21mω

3K

18.97mω

3K

18.24mω

3K

4mωx

2222

3 =++=′′




28

=

′′

′′

′′

41.21

33.08

11.27

3K

mω

x

x

x2

3

2

1

=

′′

′′

′′

3.65

2.93

1

K

3.75mω

x

x

x2

3

2

1

={ }3u

Stop Iterating as it is asked to carry only two iterations. The Fundamental natural

frequency can be calculated by,

12K

3.75mω2

=

m

K0.52ω =

Modal vector,

{ }

=

3.65

2.93

1

A 1

Disadvantage of Stodola’s method

Main drawback of Stodola’s method is that the method can be used to find only

fundamental natural frequency and modal vector of vibratory systems. This method is

not popular because of this reason.



29

(iii) Holzar’s method

It is an iterative method, used to find the natural frequencies and modal vector of a

vibratory system having multi-degree freedom.

Principle

Consider a multi dof semi-definite torsional semi-definite system as shown in Fig.13.

The Eqns. of motions of the system are:

0θ(θKθJ 21111 =−+ )&&

0θ(θKθ(θKθJ 32212122 =−+−+ ))&&

0θ(θKθ(θKθJ 43323233 =−+−+ ))&&

0θ(θKθJ 34344 =−+ )&&

The Motion is harmonic,

( )ωtsinφθ ii = (17)

where i=1,2,3,4

Substitute above Eqn.(17) in Eqns. of motion, we get,

)φ(φKφJω 211112 −= (18)

)φ(φK)φ(φKφJω 322121222 −+−=

)φ(φK)φ(φKφJω 433232332 −+−=

)φ(φKφJω 343442 −+ (19)

Add above Eqns. (18) to (19), we get

0φJω4

1i

ii2 =∑

=

For n dof system the above Eqn changes to,

0φJωn

1i

ii2 =∑

=

(20)

The above equation indicates that sum of inertia torques (torsional systems) or inertia

forces (linear systems) is equal to zero for semi-definite systems.

θ1 K1 θ2 θ3 K2 K3 θ4

J4 J3 J2 J1

Fig.13 A torsional semi-definite system



30

In Eqn. (20) ω and φi both are unknowns. Using this Eqn. one can obtain natural

frequencies and modal vectors by assuming a trial frequency ω and amplitude φ1 so

that the above Eqn is satisfied.

Steps involved

1. Assume magnitude of a trial frequency ω

2. Assume amplitude of first disc/mass (for simplicity assume φ1=1

3. Calculate the amplitude of second disc/mass φ2 from first Eqn. of motion

0)φ(φKφJω 211112 =−=

1

11

2

12K

φJωφφ −=

4. Similarly calculate the amplitude of third disc/mass φ3 from second Eqn. of motion.

0)φ(φK)φ(φKφJω 322121222 =−+−=

0)φ(φK)φK

φJω(φKφJω 3221

1

11

2

1122

2 =−+−−=

0)φ(φKφJωφJω 322112

222 =−+−=

222

112

322 φJωφJω)φ(φK +=−

2

22

2

11

2

23K

φJωφJω-φφ

+= (21)

The Eqn (21) can be written as:

2

2

1i

2

ii

23K

ωφJ

-φφ∑

==

5. Similarly calculate the amplitude of nth disc/mass φn from (n-1)th Eqn. of motion

is:

n

n

1i

2

ii

1-nnK

ωφJ

-φφ∑

==

6. Substitute all computed φi values in basic constraint Eqn.

0φJωn

1iii

2 =∑=

7. If the above Eqn. is satisfied, then assumed ω is the natural frequency, if the Eqn is

not satisfied, then assume another magnitude of ω and follow the same steps.

For ease of computations, Prepare the following table, this facilitates the calculations.

Table-1. Holzar’s Table



31

1 2 3 4 5 6 7 8

ω

S No J φ Jω2φ

K

Example-1

For the system shown in the Fig.16, obtain natural frequencies using Holzar’s method.

Make a table as given by Table-1, for iterations, follow the steps discussed earlier.

Assume ω from lower value to a higher value in proper steps.

Table-2. Holzar’s Table for Example-1

1 2 3 4 5 6 7 8

ω

S No J φ Jω2φ

K

I-iteration

0.25

1 1 1 0.0625 0.0625 1 0.0625

2 1 0.9375 0.0585 0.121 1 0.121

3 1 0.816 0.051 0.172

II-iteration

0.50

1 1 1 0.25 0.25 1 0.25

2 1 0.75 0.19 0.44 1 0.44

3 1 0.31 0.07 0.51

III-iteration

0.75

1 1 1 0.56 0.56 1 0.56

2 1 0.44 0.24 0.80 1 0.80

3 1 -0.36 -0.20 0.60

IV-iteration

1.00 1 1 1 1 1 1 1

2 1 0 0 1 1 1

∑ φ2Jω ∑ φ2JωK

1

θ1 K1 θ2 θ3 K2

J3 J2 J1

Fig.14 A torsional semi-definite system


1



32

3 1 -1 -1 0

V-iteration

1.25

1 1 1 1.56 1.56 1 1.56

2 1 -0.56 -0.87 0.69 1 0.69

3 1 -1.25 -1.95 -1.26

VI-iteration

1.50

1 1 1 2.25 2.25 1 2.25

2 1 -1.25 -2.82 -0.57 1 -0.57

3 1 -0.68 -1.53 -2.10

VII-iteration

1.75

1 1 1 3.06 3.06 1 3.06

2 1 -2.06 -6.30 -3.24 1 -3.24

3 1 1.18 3.60 0.36

Table.3 Iteration summary table

ω

0 0

0.25 0.17

0.5 0.51

0.75 0.6

1 0

1.25 -1.26

1.5 -2.1

1.75 0.36

∑ φ2Jω



33

The values in above table are plotted in Fig.15.

From the above Graph, the values of natural frequencies are:

rad/s 1.71ω

rad/s 1ω

rad/s 0ω

3

2

1

=

=

=

Definite systems

The procedure discussed earlier is valid for semi-definite systems. If a system is

definite the basic equation Eqn. (20) is not valid. It is well-known that for definite

systems, deflection at fixed point is always ZERO. This principle is used to obtain the

natural frequencies of the system by iterative process. The Example-2 demonstrates

the method.

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

∑ φ2Jω

Frequency, ω

Fig.15. Holzar’s plot of Table-3



34

Example-2

For the system shown in the figure estimate natural frequencies using Holzar’s

method.

July/Aug 2005 VTU for 20 marks

Make a table as given by Table-1, for iterations, follow the steps discussed earlier.

Assume ω from lower value to a higher value in proper steps.

Table-4. Holzar’s Table for Example-2

1 2 3 4 5 6 7 8

ω

S No J φ Jω2φ

K

I-iteration

0.25

1 3 1 0.1875 0.1875 1 0.1875

2 2 0.8125 0.1015 0.289 2 0.1445

3 1 0.6679 0.0417 0.330 3 0.110

4 0. 557

II-iteration

0.50

1 3 1 0.75 0.75 1 0.75

2 2 0.25 0.125 0.875 2 0.437

3 1 -0.187 -0.046 0.828 3 0.27

4 -0.463

III-iteration

0.75

1 3 1 1.687 1.687 1 1.687

2 2 -0.687 -0.772 0.914 2 0.457

3 1 -1.144 -0.643 0.270 3 0.090

4 -1.234

IV-iteration

1.00 1 3 1 3 3 1 3

2 2 -2 -4 -1 2 -0.5

J 2J

3K 2K K

3J

Fig.16 A torsional system


1



35

3 1 -1.5 -1.5 -2.5 3 -0.833

4 -0.667

V-iteration

1.25

1 3 1 4.687 4.687 1 4.687

2 2 -3.687 -11.521 -6.825 2 -3.412

3 1 -0.274 -0.154 -6.979 3 -2.326

4 2.172

VI-iteration

1.50

1 3 1 6.75 6.75 1 6.75

2 2 -5.75 -25.875 -19.125 2 -9.562

3 1 3.31 8.572 -10.552 3 -3.517

4 7.327

1 2 3 4 5 6 7 8

ω

S No J φ Jω2φ

K

VII-iteration

1.75

1 3 1 9.18 9.18 1 9.18

2 2 -8.18 -50.06 -40.88 2 -20.44

3 1 12.260 37.515 -3.364 3 -1.121

4 13.38

VIII-iteration

2.0

1 3 1 12 12 1 12

2 2 -11 -88 -76 2 -38

3 1 -27 108 32 3 10.66

4 16.33

IX-iteration

2.5

1 3 1 18.75 18.75 1 18.75

2 2 -17.75 -221.87 -203.12 2 -101.56

3 1 83.81 523.82 320.70 3 106.90

4 -23.09


1



36

Table.5 Iteration summary table

ω φ4

0 0

0.25 0.557

0.5 -0.463

0.75 -1.234

1 -0.667

1.25 2.172

1.5 7.372

1.75 13.38

2 16.33

2.5 -23.09

The values in above table are plotted in Fig.17.

From the above Graph, the values of natural frequencies are:

rad/s 2.30ω

rad/s 1.15ω

rad/s 0.35ω

3

2

1

=

=

=

0.0 0.5 1.0 1.5 2.0 2.5

-20

-10

0

10

20

Dis

pla

ce

me

nt,

φ4

Frequency,ω

1ω2ω 3ω

Fig.17. Holzar’s plot of Table-5

Documents

Mechanical Vibrations (ME 65) CHAPTER-8 MULTI · PDF fileMechanical Vibrations (ME 65) CHAPTER-8 MULTI DEGREE OF FREEDOM SYSTEMS Topics covered: Influence co-efficents