MDOF Systems

8/10/2019 MDOF Systems

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CHAPTER-8

MULTI DEGREE OF FREEDOM SYSTEMS

Topics covered:

Eigen values and Eigen vectors

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. Eigen values and Eigen vectors

In vibration problems Eigen values are the natural frequencies and the Eigen vectors

are modal vectors. This method is a basic tool, which can be used to analyse any

vibratory problems.

Example-1

Obtain natural frequencies and modal vectors and mode shapes of the system shown

in Fig.1, using eigen value method.

The equations of motions of the system can be obtained by Newton’s or Lagrange’s

method.

The governing equations of motion of the system shown in Fig.1 are:

m1

K1

m2

K2

x1

x2

m3

K3

x3

Fig.1 Linear vibratory system



VTU e-learning Course ME65 Mechanical Vibrations

Dr. S. K. Kudari, Professor Sessions: 10,11,12,13,14,15,16 &17

Deptt. Mech. Engg.,

B. V. B. College of Engineering and Technology, Hubli - 580031.

2

0xK)xK(Kxm 2212111 (1)

0xK)xK(KxKxm 332321222 (2)

0xKxKxm 332333 (3)

The above equations can be written in matrix form as:

0

00

x

xx

KK0

K)K(KK0K)K(K

x

xx

m00

0m000m

3

2

1

33

3322

221

3

2

1

3

2

1

(4)

0xKxM (5)

where, M -Mass/inertia matrix

K - Stiffness matrix

x -generalized displacement vector

x -generalized acceleration vector

Eqn.(5) can be written as:

0xKMx 1 (6)

0xDx (7)

where, D - is referred as Dynamic matrix

M

M AdjM

1 (8)

For harmonic analysis with frequency ,

xωx 2 (9)

substitute (8) in (7) 0xωxD 2 (10)

λω2 substitute in above Eqn.

0xλxD (11)

where, I -Identity matrix

Eigen values can be obtained by,

0D-Iλ (12)

For simplification, let us consider,

mmmmandKKKK 321321

The equations in matrix form changes to:

0

0

0

x

x

x

KK0

K2KK

0K2K

x

x

x

m00

0m0

00m

3

2

1

3

2

1

(13)

KMDhaveWe 1





Deptt. Mech. Engg.,


3

M

M AdjM

1

3mM

100010

001

m

1

m000m0

00m

m

1M

2

2

2

3

1

KK-0

K-2KK-

0K-2K

10

010

001

m

1D

11-0

1-21-

01-2

m

KD

Eigen valuesEigen values can be obtained by,

0D-Iλ

0

11-0

1-21-

01-2

m

K-

100

010

001

λ

0

K/m)(λK/m-0

K/m-2K/m)(λK/m-

0K/m-2K/m)-(λ

Solving the above eqn.,

0m

K

m

K6λ

m

K5λλ

3

3

2

223 (14)

Solve the above Eqn. to get three values of ,

m

K0.198ωλ 2

11 rad/sm

K0.44ω1

m

K1.55ωλ 2

22 rad/sm

K1.24ω2

mK3.24ωλ 2

33 rad/smK1.80ω3

Eigen vectors

For obtaining Eigen vectors consider the equation





Deptt. Mech. Engg.,


4

0xD-Iλ

0sinω AD-Iλ t (15)

0 AD-Iλ (16)

where, A - eigen vector or modal vector

First modal vector,

0 AD-Iλ 11 (17)

0

0

0

A

A

A

11-0

1-21-

01-2

m

K-

100

010

001

m

0.198K

31

21

11

..

0

0

0

A

A

A

..

0.80210

11.8021

011.802

31

21

11

Solving the above Eqn,

0 A1.802A 2111 (18)

0 A1.802A A 312111 (19)

00.802A A 3121 (20)

the first eigen vector,

31

21

11

1

A

A

A

A

The above eqn in terms of amplitude ratio can be written as:

11

31

11

21

11

1

A A

A A

A

A

From Eqns. (18),(19) and (20), we get,

1.802

A

A

11

21

2.247 A

A

11

31





Deptt. Mech. Engg.,


5

2.247

1.802

1

A

A

A A

A

A

11

31

11

21

11

1 (21)

It is the modal vector in terms of amplitude ratios.

Second modal vector,

0 AD-Iλ 22

0

0

0

A

A

A

11-0

1-21-

01-2

m

K-

100

010

001

m

1.55K

32

22

12

..

0

0

0

A

A

A

..

0.5510

10.4451

010.445

32

22

12

0 A0.445A 2212 (22)

0 A0.445A A 322212 (23)

00.55A A 3222 (24)

32

22

12

2

A

A

A

A

In terms of amplitude ratio,

12

32

12

22

12

2

A A

A A

A

A


0.445 A

A

12

22

-0.80 A

A

12

32





Deptt. Mech. Engg.,


6

0.80-

0.445

1

A

A

A A

A

A

12

32

12

22

12

2 (25)

Third modal vector

0 AD-Iλ 33

0

0

0

A

A

A

11-0

1-21-

01-2

m

K-

100

010

001

m

3.247K

33

23

13

..

0

0

0

A

A

A

..

2.24710

11.2471

011.247

33

23

13

Solving the above Eqn.

0 A1.247A 2313 (26)

02.247A A 3323 (27)

0 A1.247A A 332313 (28)

33

23

13

3

A

A

A

A

13

33

13

23

13

3

A A

A A

A

A


-1.247 A

A

13

23

0.55 A

A13

33





Deptt. Mech. Engg.,


7

0.55

1.247-

1

A

A

A A

A

A

13

33

13

23

13

3 (29)

Modal vectors:

2.247

1.802

1

A 1

0.80-

0.445

1

A 2

0.55

1.247-1

A 3

Modal matrix:

0.550.802.247

1.2470.4451.802

111

(30)

Mode Shapes

m

K1

m

K2

x1

x2

m

K3

x3 Mode-I Mode-II Mode-III

Fig.2 Mode shapes of the system





Deptt. Mech. Engg.,


8

2. 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 = 1K

The elements corresponds to inverse mass matrix are referred as flexibility

mass/inertia co-efficients.

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

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

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, k 11, ……..k 33 are referred as stiffness influence coefficients

k 11-stiffness influence coefficient at point 1 due to a unit deflection at point 1

k 21- stiffness influence coefficient at point 2 due to a unit deflection at point 1

k 31- stiffness influence coefficient at point 3 due to a unit deflection at point 1





Deptt. Mech. Engg.,


9

Example-1.

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

I-step:

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

equations. We get,

2111 KKk

221 Kk 0k31

II-step:

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

equations. We get,

0k13

323 -Kk

333 Kk

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.3 Stiffness influence coefficients of the system





Deptt. Mech. Engg.,


10

333231

322221

131211

kkk

kkk

kkk

K

33

3322

221

KK-0K-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, 11, ……..33 are referred as stiffness influence coefficients

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

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

31- 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.

I-step:

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

equations. We get,

21

3222K

1

K

1αα





Deptt. Mech. Engg.,


11

III-step:

Apply 1 unit Force at point 3 as shown in Fig.4(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

For simplification, let us consider : KKKK 321

K1

K1ααααα

1

1331122111

K

2

K

1

K

1ααα 233222

K

3

K

1

K

1

K

1α33

(b) (c)

Fig.4 Flexibility influence coefficients of the system

x2=

22

m1

K1

m2

K2

x1=12

m3

K3

x3=32

F1=0

F2=1

F3=0

m1

K1

m2

K2

x1=13

x2=23

m3

K3

x3=33

F1=0

F2=1

F3=0

(a)

m1

K1

m2

K2

m3

K3

x3=31

F1=0

F2=1

F3=0

x1=11

x2=21





Deptt. Mech. Engg.,


12

333231

322221

131211

ααα

ααα

ααα

α

321221

111

K

1

α

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.5.

I-step:

Apply 1 unit Force at point 1 as shown in Fig.6 and write the force equilibrium

equations. We get,

lθsinT 3mgm)mg(mθcosT

3mg

1θtan

θsinθtansmall,isθ

l

αθsin 11

θsinlα11

m

m

l

l

m

l

Fig.5 Pendulum system

m

m

l

l

m

l

F=1

T

11

Fig.6 Flexibility influence

co-efficents





Deptt. Mech. Engg.,


13

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





Deptt. Mech. Engg.,


14

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 (31)

0MωK 2

0MKIω

1 1

2

0MαIω

12

(32)

For n DOF systems,

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

(33)

It is the polynomial equation of nth degree in (1/ 2). Let the roots of above Eqn. are:

2

n

2

2

2

1 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

22

n

2

2

2

1

n

2

2

n

22

2

22

1

2

(34)





Deptt. Mech. Engg.,


15

Comparing Eqn.(33) and Eqn. (34), we get,

2

n

2

2

2

1 ω

1......

ω

1

ω

1 nnn222111 mα...mαmα (35)

In mechanical systems higher natural frequencies are much larger than the

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

2

1ω

1 nnn222111 mα...mαmα (36)

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 (37)

The Dunkerley’s formula can be written as:

2

nn

2

2n

2

1n

2

1 ω1......

ω1

ω1

ω1 (38)

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

considering each mass separately.

The above formula given by Eqn. (38) 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.7

using Dunkerley’s method.

m

K

m

K

x1

x2

m

K

x3






Deptt. Mech. Engg.,


17

Obtain the influence co-efficents:

EI

1.944x10α

-3

11

EI

9x10α

-3

22

2

nω

1 222111 mαmα

rad/s1.245ωn





Deptt. Mech. Engg.,


18

(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.9. 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.

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

n

1i

iimax gym2

1V (39)

n

1i

2

iimax ym2

1T (40)

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

Considering the system vibrates with SHM,i

2

i yωy (41)

substitute Eqn. (41) in (40)

n

1i

2

ii

2

max ym2

ωT (42)

According to Rayleigh’s method,

maxmax TV (43)

substitute Eqn. (39) and (42) in (43)

n

1i

2

ii

n

1i ii2

ym

gym

ω (44)

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

1 2 3

m1 m2 m3

A B

y1y2

y3

Fig.9 A rotor system.





Deptt. Mech. Engg.,


19

gmαgmαgmαy 3132121111 (45)

gmαgmαgmαy 3232221212 (46)

gmαgmαgmαy 3332321313 (47)

Eqn.(44) is the Rayleigh’s formula, which is used to estimate frequency of transverse

vibrations of a vibratory systems.

Examples: 1

Estimate the approximate fundamental natural frequency of the system shown in

Fig.10 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:

gmαgmαgmαy 3132121111

2000

5g

2K

5mg122

2K

mgy1



2m

2K

2m

K

x1

x2

m

K

x3






Deptt. Mech. Engg.,


20

2000

11g

2K

11mg362

2K

mgy2



2000

13g

2K

13mg

5622K

mg

y3 Rayleigh’s formula is:

n

1i

2

ii

n

1i

ii2

ym

gym

ω

2

222

2

2

g2000

132

2000

112

2000

52

g2000

132x

2000

112x

2000

52x

ω

rad/s12.41ω

Examples: 2

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

Fig.11 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.

1 2

m1 m2

B

160 80 180

A

Fig.11 A rotor system.

bx

l

W

Fig.12 A simply supported beam





Deptt. Mech. Engg.,


21

For a simply supported beam shown in Fig.12, the deflection of beam at distance x

from left is given by:

b)(l xfor bxl6EIl

Wbxy 222 (48)

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

deflections due to each load acting separately.

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

1iii

2

ym

gymω

22

2

20x0.8240x0.803

20x0.8240x0.8039.81ω

rad/s1541.9ωn





Deptt. Mech. Engg.,


22

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 1< 2<……….< n

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

as:

0xKxM (49)

where,

φωtsin Ax (50)

substitute Eqn.(50) in (51)

0 AK AMω2 (51)

For principal modes of oscillations, for rth

mode,

0 AK AMω r r

2

r

r 2

r

r Aω

AMK11

r 2

r

r Aω

AD1

(52)

where, D is referred as Dynamic matrix.

Eqn.(52) converges to first natural frequency and first modal vector.

The Equation,

r 2r r Aω AKM 1

r

2

r r 1 Aω AD (53)

where, 1D is referred as inverse dynamic matrix.

Eqn.(53) converges to last natural frequency and last modal vector.

In above Eqns (52) and (53) 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





Deptt. Mech. Engg.,


23

2K

3ααα 233222

2K

5α33

333231

322221

131211

ααα

ααα

ααα

α

531

331

111

2K

1Kα

1

First natural frequency and modal vector

r 2r

r Aω

AMK11

r 2r

r Aω

AD 1

Obtain Dynamic matrix MKD1

562

362

122

2K

m

100

020

002

531

331

111

2K

mD

Use basic Eqn to obtain first frequency

12r

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 newvector as assumed vector in next iteration.





Deptt. Mech. Engg.,


24

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





Deptt. Mech. Engg.,


25

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.13. Vectors banda are orthogonal to each

other if and only if

0ba T (54)

0b

baa

2

1

21

0b

b

10

01aa

2

1

21

0bIa T (55)

where, I is Identity matrix.

From Eqn.(55), Vectors banda 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 writtenas:

0xKxM

φωtsin Ax

0 AK AMω2 11

x1

x2

2

1

b

bb

2

1

a

aa

Fig.13 Vector representation graphically





Deptt. Mech. Engg.,


27

Therefore the trial vector is:

3

2

32

3

2

1

V

V

βVαV

V

V

V

3

2

1

V

VV

100

010βα0

1VS (66)

where S is referred as Sweeping matrix and 1V is the trial vector.

New dynamics matrix is:

SDDs

22

2

1s Aω

VD1

(67)

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





Deptt. Mech. Engg.,


28

2.612(1)

2(2.61)

Am

Amα

11

22

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

22

2

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-

K0.5m

15.77

0.50-

10.64-

Km

8.14

1

9.71-

1.890.390

0.110.390

1.111.610

Km

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

13.781-

8.98-

K

2.08m

28.672.08-

18.68-

K

m

15.381-

8.67-

1.890.3900.110.390

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





Deptt. Mech. Engg.,


29

Sixth Iteration

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

K1087m

ω12

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.





Deptt. Mech. Engg.,


30

(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,

1

2

11 xωmF for Mass 1 (68)

2

2

22

xωmF for Mass 2 (69)

3

2

33 xωmF for Mass 3 (70)

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

formula,

333322311

233222211

133122111

3

2

1

αFαFαF

αFαFαF

αFαFαF

x

x

x

(71)

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 (72)

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.





Deptt. Mech. Engg.,


31

First iteration

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

2

1

2

11 2mωxωmF 2

2

2

22 2mωxωmF 2

3

2

33 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 ,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





Deptt. Mech. Engg.,


32

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

Second iteration

1. Initial vector of system

2

u =

2.6

2.2

1

x

x

x

3

2

1


2

1

2

11 2mωxωmF 2

2

2

22 mω4xωmF 4.2

3

2

33 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

mωx2222

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






Deptt. Mech. Engg.,


33

Third iteration

1. Initial vector of system 3u =

3.13

2.55

1

x

x

x

3

2

1


2

1

2

11 2mωxωmF

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.

12K

10.23mω2

m

K0.44ω rad/s

2

3

2

33 3.13mωxωmF

2

2

2

22 5.1mωxωmF





Deptt. Mech. Engg.,


35


2

1

2

11 4mωxωmF 2

2

2

22 2mωxωmF 2

3

2

33 mωxωmF

3. New deflection vector using flexibility influence coefficients,

1331221111 αFαFαFx

3K

7mω

3K

mω

3K

2mω

3K

4mωx

2222

1

2332222112 αFαFαFx

3K

16mω

3K

4mω

3K

8mω

3K

4mωx

2222

2

3333223113 αFαFαFx

3K19mω

3K7mω

3K8mω

3K4mωx

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


Second iteration

1. Initial vector of system 2u =

2.71

2.28

1

x

x

x

3

2

1


2

1

2

11 4mωxωmF 2

2

2

22 4.56mωxωmF 2

3

2

33 2.71mωxωmF





Deptt. Mech. Engg.,


36

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


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

mK0.52ω

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.





Deptt. Mech. Engg.,


38

In Eqn. (83) and i both are unknowns. Using this Eqn. one can obtain natural

frequencies and modal vectors by assuming a trial frequency and amplitude 1 so

that the above Eqn is satisfied.

Steps involved

1. Assume magnitude of a trial frequency 2. Assume amplitude of first disc/mass (for simplicity assume 1=1

3. Calculate the amplitude of second disc/mass 2 from first Eqn. of motion

0)φ(φKφJω 21111

2

1

11

2

12K

φJωφφ (84)

4. Similarly calculate the amplitude of third disc/mass 3 from second Eqn. of motion.

0)φ(φK)φ(φKφJω 32212122

2

0)φ(φK)φK

φJω(φKφJω 3221

1

112

1122

2

0)φ(φKφJωφJω 32211

2

22

2

22

2

11

2

322 φJωφJω)φ(φK

2

22

2

11

2

23K

φJωφJω-φφ

(85)

The Eqn (85) can be written as:

2

2

1i

2

ii

23K

ωφJ

-φφ

(86)

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

is:

n

n

1i

2

ii

1-nnK

ωφJ

-φφ

(87)

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

0φJωn

1i

ii2

(88)

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

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

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





Deptt. Mech. Engg.,


39

Table-1. Holzar’s Table

1 2 3 4 5 6 7 8

S No J J2 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 from lower value to a higher value in proper steps.

φ2Jω φ2JωK

1

1 K1 2 3K2

J3J2J1

Fig.16 A torsional semi-definite system





Deptt. Mech. Engg.,


40

Table-2. Holzar’s Table for Example-1

1 2 3 4 5 6 7 8

S No J J2 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

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

φ2Jω φ2JωK

1





Deptt. Mech. Engg.,


41

Table.3 Iteration summary table

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

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

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

rad/s1.71ω

rad/s1ωrad/s0ω

3

2

1

φ2Jω

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,

Fig.17. Holzar’s plot of Table-3





Deptt. Mech. Engg.,


43

Table-4. Holzar’s Table for Example-2

1 2 3 4 5 6 7 8

S No J J2 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

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

φ2Jω φ2JωK

1





Deptt. Mech. Engg.,


44

1 2 3 4 5 6 7 8

S No J J2 K

VII-iteration

1.75

1 3 1 9.18 9.18 1 9.182 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

Table.5 Iteration summary table

4

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 16.332.5 -23.09

The values in above table are plotted in Fig.19.

φ2Jω φ2JωK

1



VTU e-learning Course ME65 Mechanical Vibrations 45

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

rad/s2.30ω

rad/s1.15ω

rad/s0.35ω

3

2

1

0.0 0.5 1.0 1.5 2.0 2.5

-20

-10

0

10

20

D i s p l a c e m e n t ,

4

Frequency,

1ω 2ω 3ω

Fig.17. Holzar’s plot of Table-5

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