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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
VTU e-learning Course ME65 Mechanical Vibrations
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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.
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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
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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
VTU e-learning Course ME65 Mechanical Vibrations
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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,
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[ ]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)
VTU e-learning Course ME65 Mechanical Vibrations
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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
VTU e-learning Course ME65 Mechanical Vibrations
Prof. S. K. Kudari, Principal APS College of Engineering, Bengaluru-28.
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≅
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.
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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 ++=
gmαgmαgmαy 3232221212 ++=
gmαgmαgmαy 3332321313 ++=
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.
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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:
gmαgmαgmαy 3132121111 ++=
( )2000
5g
2K
5mg122
2K
mgy1 ==++=
Deflection at point 2 is:
gmαgmαgmαy 3232221212 ++=
( )2000
11g
2K
11mg362
2K
mgy2 ==++=
Deflection at point 3 is:
gmαgmαgmαy 3332321313 ++=
( )2000
13g
2K
13mg562
2K
mgy3 ==++=
Rayleigh’s formula is:
2m
2K
2m
K
x1
x2
m
K
x3
Fig.8 Linear vibratory system
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∑
∑
=
==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
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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 =
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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 ===
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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.
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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
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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
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[ ]{ } [ ]{ }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:
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( )
+
=
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 −=
−=
−=
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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
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=
=
−
−−
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.
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(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
22 xωmF = for Mass 2
32
33 xωmF = for Mass 3
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
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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 ++=′
Substitute for F’s and α,s
2K
11mω
2K
3mω
2K
6mω
K
mωx
2222
2 =++=′
3333223113 αFαFαFx ++=′
Substitute for F’s and α,s
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
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1. Initial vector of system { }2u =
=
′
′
′
2.6
2.2
1
x
x
x
3
2
1
2. Find out inertia forces of system at each mass point
21
211 2mωxωmF =′=′
22
222 mω4xωmF 4.=′=′
23
233 2.6mωxωmF =′=′
3. New deflection vector,
1331221111 αFαFαFx ′+′+′=′′
Substitute for F’s and α,s
2K
9mω
2K
2.6mω
2K
4.4mω
K
mωx
2222
1 =++=′′
2332222112 αFαFαFx ′+′+′=′′
Substitute for F’s and α,s
2K
23mω
2K
7.8mω
2K
13.2mω
K
mωx
2222
2 =++=′′
3333223113 αFαFαFx ′+′+′=′′
Substitute for F’s and α,s
2K
28.2mω
2K
13mω
2K
13.2mω
K
mωx
2222
3 =++=′′
4. New deflection vector is:
=
′′
′′
′′
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
1. Initial vector of system { }3u =
=
′′
′′
′′
3.13
2.55
1
x
x
x
3
2
1
2. Find out inertia forces of system at each mass point
21
211 2mωxωmF =′′=′′
2
22
22 5.1mωxωmF =′′=′′
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3. new deflection vector,
1331221111 αFαFαFx ′′+′′+′′=′′′
Substitute for F’s and α,s
2K
10.23mω
2K
3.13mω
2K
5.1mω
K
mωx
2222
1 =++=′′′
2332222112 αFαFαFx ′′+′′+′′=′′′
Substitute for F’s and α,s
2K
26.69mω
2K
9.39mω
2K
15.3mω
K
mωx
2222
2 =++=′′′
3333223113 αFαFαFx ′′+′′+′′=′′′
Substitute for F’s and α,s
2K
28.2mω
2K
16.5mω
2K
15.3mω
K
mωx
2222
3 =++=′′′
4. New deflection vector is:
=
′′′
′′′
′′′
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 =′′=′′
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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
2. Find out inertia forces of system at each mass point
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
Fig.12 Linear vibratory system
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2332222112 αFαFαFx ++=′
3K
16mω
3K
4mω
3K
8mω
3K
4mωx
2222
2 =++=′
3333223113 αFαFαFx ++=′
3K
19mω
3K
7mω
3K
8mω
3K
4mωx
2222
3 =++=′
4. New deflection vector is:
=
′
′
′
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
1. Initial vector of system { }2u =
=
′
′
′
2.71
2.28
1
x
x
x
3
2
1
2. Find out inertia forces of system at each mass point
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 =++=′′
4. New deflection vector is:
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=
′′
′′
′′
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.
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(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
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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
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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
∑ φ2Jω ∑ φ2JωK
1
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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
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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
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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
∑ φ2Jω ∑ φ2JωK
1
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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
∑ φ2Jω ∑ φ2JωK
1
VTU e-learning Course ME65 Mechanical Vibrations
Prof. S. K. Kudari, Principal APS College of Engineering, Bengaluru-28.
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