Multi degree of freedom systems

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.
[ ]{ } [ ]{ } [ ]
0
x
K
x
M =
+
&
&
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
MULTI DEGREE OF FREEDOM SYSTEMS
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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.
{ } [ ]{ }
x
K
F =
[ ]










=
33
32
31
32
22
21
13
12
11
k
k
k
k
k
k
k
k
k
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,
2
1
11 K
K
k +
=
2
21 K
k −
=
0
k31 =
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
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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,
2
12 -K
k =
3
2
22 K
K
k +
=
3
31 -K
k =
III-step:
Apply 1 unit deflection at point 3 as shown in Fig.1(c) and write the force equilibrium
equations. We get,
0
k13 =
3
23 -K
k =
3
33 K
k =
[ ]










=
33
32
31
32
22
21
13
12
11
k
k
k
k
k
k
k
k
k
K
[ ]
( )
( )










+
+
=
3
3
3
3
2
2
2
2
1
K
K
-
0
K
-
K
K
K
-
0
K
-
K
K
K
From stiffness coefficients K matrix can be obtained without writing Eqns. of motion.
Flexibility influence co-efficents.
{ } [ ]{ }
x
K
F =
{ } [ ] { }
F
K
x
1
−
=
{ } [ ]{ }
F
α
x =
where, [ ]
α - Matrix of Flexibility influence co-efficents given by
[ ]










=
33
32
31
32
22
21
13
12
11
α
α
α
α
α
α
α
α
α
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
31
21
11
K
1
α
α
α =
=
=
II-step:
Apply 1 unit Force at point 2 as shown in Fig.2(b) and write the force equilibrium
equations. We get,
2
1
32
22
K
1
K
1
α
α +
=
=
III-step:
Apply 1 unit Force at point 3 as shown in Fig.2(c) and write the force equilibrium
equations. We get,
2
1
23
K
1
K
1
α +
=
3
2
1
33
K
1
K
1
K
1
α +
+
=
Therefore,
1
13
31
12
21
11
K
1
α
α
α
α
α ==
=
=
=
=
2
1
23
32
22
K
1
K
1
α
α
α +
=
=
=
3
2
1
33
K
1
K
1
K
1
α +
+
=
(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 : K
K
K
K 3
2
1 =
=
=
K
1
K
1
α
α
α
α
α
1
13
31
12
21
11 =
==
=
=
=
=
K
2
K
1
K
1
α
α
α 23
32
22 =
+
=
=
=
K
3
K
1
K
1
K
1
α33 =
+
+
=
[ ]










=
33
32
31
32
22
21
13
12
11
α
α
α
α
α
α
α
α
α
α
[ ]










=
3
2
1
2
2
1
1
1
1
K
1
α
[ ] [ ] 1
K
α
−
=
In Vibration analysis if there is need of [ ] 1
K
−
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
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VTU e-learning Course ME65 Mechanical Vibrations 6
l
θ
sin
T =
3mg
m)
m
g(m
θ
cos
T =
+
+
=
3mg
1
θ
tan =
θ
sin
θ
tan
small,
is
θ =
l
α
θ
sin 11
=
θ
sin
l
α11 =
3mg
l
α11 =
Similarly apply 1 unit force at point 2 and next at point 3 to obtain,
5mg
l
α22 =
the influence coefficients are:
5mg
l
α
α
α
α
α 13
31
12
21
11 ==
=
=
=
=
6mg
11l
α
α
α 23
32
22 =
=
=
6mg
11l
α33 =
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,
[ ] [ ] [ ]
0
K
M
ω2
=
+
−
[ ] [ ] [ ]
0
M
ω
K 2
=
+
−
[] [ ] [ ] [ ]
0
M
K
I
ω
1 1
2
=
+
−
−
[] [ ][ ] [ ]
0
M
α
I
ω
1
2
=
+
−
For n DOF systems,
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7
[ ]
0
m
.
0
0
.
0
.
m
0
0
.
0
m
α
.
α
α
.
α
.
α
α
α
.
α
α
1
.
0
0
.
0
.
1
0
0
.
0
1
ω
1
n
2
1
nn
n2
n1
2n
22
21
1n
12
11
2
=
























+












−
[ ]
0
m
α
ω
1
.
m
α
m
α
.
m
α
.
m
α
ω
1
0
m
α
.
m
α
m
α
ω
1
n
nn
2
2
n2
1
n1
n
2n
2
22
2
n
1n
2
12
1
11
2
=


























+
−






+
−






+
−
.
.
.
Solve the determinant
( )
( ) [ ]
0
...
m
α
...
m
m
α
α
m
m
α
α
ω
1
m
α
...
m
α
m
α
ω
1
n
nn
3
1
33
11
2
1
22
11
1
n
2
n
nn
2
22
1
11
n
2
=
+
+
+
+






+
+
+
−






−
(1)
It is the polynomial equation of nth degree in (1/ω2
). 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
1
n
2
2
n
2
2
2
1
n
2
2
n
2
2
2
2
2
1
2
=
−
















+
+
+
−








=








−








−








−
−
(2)
Comparing Eqn.(1) and Eqn. (2), we get,
=








+
+
+ 2
n
2
2
2
1 ω
1
......
ω
1
ω
1
( )
n
nn
2
22
1
11 m
α
...
m
α
m
α +
+
+
In mechanical systems higher natural frequencies are much larger than the
fundamental (first) natural frequencies. Approximately, the first natural frequency is:
≅








2
1
ω
1
( )
n
nn
2
22
1
11 m
α
...
m
α
m
α +
+
+
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
1
11
1n
m
K
m
α
1
ω =
=
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)
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where, .....
,
ω
,
ω 2n
1n 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:
≅








2
1
ω
1
( )
n
nn
2
22
1
11 m
α
...
m
α
m
α +
+
+ 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
1
α
α
α
α
α 13
31
12
21
11 =
=
=
=
=
K
2
α
α
α 23
32
22 =
=
=
K
3
α33 =
Substitute all influence coefficients in the Dunkerley’s formula.
≅








2
1
ω
1
( )
n
nn
2
22
1
11 m
α
...
m
α
m
α +
+
+
m
K
m
K
x1
x2
m
K
x3
Fig.5 Linear vibratory system
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9
≅








2
1
ω
1
K
6m
K
3m
K
2m
K
m
=






+
+
K/m
0.40
ω1= 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
1.944x10
α
-3
11 =
EI
9x10
α
-3
22 =
≅








2
n
ω
1
( )
2
22
1
11 m
α
m
α +
rad/s
1.245
ωn =
(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
1
i
i
i
max gy
m
2
1
V (4)
∑
=
=
n
1
i
2
i
i
max y
m
2
1
T &
where, mi- masses of the system, yi –displacements at mass points.
Considering the system vibrates with SHM,
i
2
i y
ω
y =
&
From above equations
∑
=
=
n
1
i
2
i
i
2
max y
m
2
ω
T (5)
According to Rayleigh’s method,
max
max T
V = (6)
substitute Eqn. (4) and (5) in (6)
∑
∑
=
=
= n
1
i
2
i
i
n
1
i
i
i
2
y
m
gy
m
ω (7)
The deflections at point 1, 2 and 3 can be found by.
g
m
α
g
m
α
g
m
α
y 3
13
2
12
1
11
1 +
+
=
g
m
α
g
m
α
g
m
α
y 3
23
2
22
1
21
2 +
+
=
g
m
α
g
m
α
g
m
α
y 3
33
2
32
1
31
3 +
+
=
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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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
1
α
α
α
α
α 13
31
12
21
11 =
=
=
=
=
2K
3
α
α
α 23
32
22 =
=
=
2K
5
α33 =
Deflection at point 1 is:
g
m
α
g
m
α
g
m
α
y 3
13
2
12
1
11
1 +
+
=
( )
2000
5g
2K
5mg
1
2
2
2K
mg
y1 =
=
+
+
=
g
m
α
g
m
α
g
m
α
y 3
23
2
22
1
21
2 +
+
=
( )
2000
11g
2K
11mg
3
6
2
2K
mg
y2 =
=
+
+
=
g
m
α
g
m
α
g
m
α
y 3
33
2
32
1
31
3 +
+
=
( )
2000
13g
2K
13mg
5
6
2
2K
mg
y3 =
=
+
+
=
Rayleigh’s formula is:
2m
2K
2m
K
x1
x2
m
K
x3
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12
∑
∑
=
=
= n
1
i
2
i
i
n
1
i
i
i
2
y
m
gy
m
ω
2
2
2
2
2
2
g
2000
13
2
2000
11
2
2000
5
2
g
2000
13
2x
2000
11
2x
2000
5
2x
ω






+






+












+
+
=
rad/s
12.41
ω =
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
x
for
b
x
l
6EIl
Wbx
y 2
2
2
−
≤
−
−
=
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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13
Deflections due to 20 kg mass
( ) ( ) EI
0.265
0.18
0.16
0.42
6EI0.42
x0.18x0.16
9.81x20
y 2
2
2
'
1 =
−
−
=
( ) ( ) EI
0.29
0.18
0.24
0.42
6EIx0.42
x0.18x0.24
9.81x20
y 2
2
2
'
2 =
−
−
=
Deflections due to 40 kg mass
( ) ( ) EI
0.538
0.16
0.26
0.42
6EIx0.42
x0.16x0.26
9.81x40
y 2
2
2
'
'
1 =
−
−
=
( ) ( ) EI
0.53
0.16
0.18
0.42
6EIx0.42
x0.16x0.18
9.81x40
y 2
2
2
'
'
2 =
−
−
=
The deflection at point 1 is:
EI
0.803
y
y
y '
'
1
'
1
1 =
+
=
The deflection at point 2 is:
EI
0.82
y
y
y '
'
2
'
2
2 =
+
=
∑
∑
=
=
= n
1
i
2
i
i
n
1
i
i
i
2
y
m
gy
m
ω
( )
( ) ( )
2
2
2
20x0.82
40x0.803
20x0.82
40x0.803
9.81
ω
+
+
=
rad/s
1541.9
ωn =
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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:
[ ]{ } [ ]{ } [ ]
0
x
K
x
M =
+
&
&
where,
{ } { } ( )
φ
ωt
sin
A
x +
= (8)
substitute Eqn.(8) in (9)
[ ]{ } [ ]{ } [ ]
0
A
K
A
M
ω2
=
+
− (9)
For principal modes of oscillations, for rth
mode,
[ ]{ } [ ]{ } [ ]
0
A
K
A
M
ω r
r
2
r =
+
−
[ ] [ ]{ } { }r
2
r
r A
ω
A
M
K
1
1
=
−
[ ]{ } { }r
2
r
r A
ω
A
D
1
= (10)
where, [ ]
D is referred as Dynamic matrix.
Eqn.(10) converges to first natural frequency and first modal vector.
The Equation,
[ ] [ ]{ } { }r
2
r
r A
ω
A
K
M =
−1
[ ]{ } { }r
2
r
r
1 A
ω
A
D = (11)
where, [ ]
1
D 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
1
α
α
α
α
α 13
31
12
21
11 =
=
=
=
=
2K
3
α
α
α 23
32
22 =
=
=
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2K
5
α33 =
[ ]










=
33
32
31
32
22
21
13
12
11
α
α
α
α
α
α
α
α
α
α
[ ] [ ]










=
=
−
5
3
1
3
3
1
1
1
1
2K
1
K
α
1
First natural frequency and modal vector
[ ] [ ]{ } { }r
2
r
r A
ω
A
M
K
1
1
=
−
[ ]{ } { }r
2
r
r A
ω
A
D
1
=
Obtain Dynamic matrix [ ] [ ] [ ]
M
K
D
1
−
=
[ ]










=




















=
5
6
2
3
6
2
1
2
2
2K
m
1
0
0
0
2
0
0
0
2
5
3
1
3
3
1
1
1
1
2K
m
D
Use basic Eqn to obtain first frequency
[ ]{ } { }1
2
r
1 A
ω
A
D
1
=
Assume trial vector and substitute in the above Eqn.
Assumed vector is: { }










=
1
1
1
u 1
First Iteration
[ ]{ } =
1
u
D




















1
1
1
5
6
2
3
6
2
1
2
2
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
[ ]{ } =
2
u
D




















2.6
2.2
1
5
6
2
3
6
2
1
2
2
2K
m










=
3.13
2.55
1
K
4.5m
Third Iteration
[ ]{ } =
3
u
D




















3.133
2.555
1
5
6
2
3
6
2
1
2
2
2K
m










=
3.22
2.61
1
K
5.12m
Fourth Iteration
[ ]{ } =
4
u
D




















3.22
2.61
1
5
6
2
3
6
2
1
2
2
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.
[ ]{ } { }1
2
1
1 A
ω
A
D
1
=
K
5.22m
ω
1
2
1
=
m
K
5.22
1
ω2
1 =
m
K
0.437
ω1 = Rad/s
Modal vector is:
{ }










=
3.23
2.61
1
A 1
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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
{ } { } 0
b
a
T
=
{ } 0
b
b
a
a
2
1
2
1 =






{ } 0
b
b
1
0
0
1
a
a
2
1
2
1 =












{ } []{ } 0
b
I
a
T
= (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:
[ ]{ } [ ]{ } [ ]
0
x
K
x
M =
+
&
&
{ } { } ( )
φ
ωt
sin
A
x +
=
[ ]{ } [ ]{ } [ ]
0
A
K
A
M
ω2
=
+
− 1
1
x1
x2
{ }






=
2
1
b
b
b
{ }






=
2
1
a
a
a
Fig.11 Vector representation graphically
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18
[ ]{ } [ ]{ }1
1 A
K
A
M
ω2
=
If system has two frequencies ω1 and ω2
[ ]{ } [ ]{ }1
1 A
K
A
M
ω2
1 = (13)
[ ]{ } [ ]{ }2
2 A
K
A
M
ω2
2 = (14)
Multiply Eqn.(13) by { }T
2
A and Eqn.(14) by { }T
1
A
{ } [ ]{ } { } [ ]{ }1
T
2
1
T
2
2
1 A
K
A
A
M
A
ω = (15)
{ } [ ]{ } { } [ ]{ }2
T
1
2
T
1
2
1 A
K
A
A
M
A
ω = (16)
Eqn.(15)-(16)
{ } [ ]{ } 0
A
M
A 2
T
1 =
Above equation is a condition for mass orthogonality.
{ } [ ]{ } 0
A
K
A 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 { } 1
1 ω
and
A to obtain { } 2
2 ω
and
A choose a trial vector { }1
V
orthogonal to { }1
A ,which gives constraint Eqn.:
{ } [ ]{ } 0
A
M
V 1
T
1 =
{ } 0
A
A
A
m
0
0
0
m
0
0
0
m
V
V
V
3
2
1
3
2
1
3
2
1 =




















( ) ( ) ( )
{ } 0
A
m
V
A
m
V
A
m
V 3
3
3
2
2
2
1
1
1 =
+
+
( ) ( ) ( )
{ } 0
V
A
m
V
A
m
V
A
m 3
3
3
2
2
2
1
1
1 =
+
+
3
2
1 βV
αV
V +
=
where α and β are constants








−
=
1
1
2
2
A
m
A
m
α








−
=
1
1
3
3
A
m
A
m
β
Therefore the trial vector is:
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19
( )









 +
=










3
2
3
2
3
2
1
V
V
βV
αV
V
V
V




















=
3
2
1
V
V
V
1
0
0
0
1
0
β
α
0
[ ]{ }1
V
S
=
where [ ]
S
= is referred as Sweeping matrix and { }1
V is the trial vector.
New dynamics matrix is:
[ ] [ ][ ]
S
D
Ds +
[ ]{ } { }2
2
2
1
s A
ω
V
D
1
=
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
[ ]










=
1
0
0
0
2
0
0
0
2
M is the mass matrix
Find sweeping matrix
[ ]










=
1
0
0
0
1
0
β
α
0
S
2.61
2(1)
2(2.61)
A
m
A
m
α
1
1
2
2
−
=








−
=








−
=
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1.615
2(1)
1(3.23)
A
m
A
m
β
1
1
3
3
−
=








−
=








−
=
Sweeping matrix is:
[ ]










=
1
0
0
0
1
0
1.615
-
2.61
-
0
S
New Dynamics matrix is:
[ ] [ ][ ]
S
D
Ds +
[ ]










−
−
−
=




















=
1.89
0.39
0
0.11
0.39
0
1.11
1.61
0
K
m
1
0
0
0
1
0
1.615
-
2.61
-
0
5
3
1
3
6
2
1
2
2
2K
m
Ds
First Iteration
[ ]{ } { }2
2
2
1
s A
ω
V
D
1
=










=










=




















−
−
−
8.14
1
9.71
-
K
0.28m
2.28
0.28
2.27
-
K
m
1
1
1
1.89
0.39
0
0.11
0.39
0
1.11
1.61
0
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
-
1.89
0.39
0
0.11
0.39
0
1.11
1.61
0
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
-
1.89
0.39
0
0.11
0.39
0
1.11
1.61
0
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
-
1.89
0.39
0
0.11
0.39
0
1.11
1.61
0
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
-
1.89
0.39
0
0.11
0.39
0
1.11
1.61
0
K
m
Sixth Iteration
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21










=










=




















−
−
−
13.43
1
-
7.08
-
K
1.87m
25.12
1.87
-
13.24
-
K
m
13.5
1
-
7.2
-
1.89
0.39
0
0.11
0.39
0
1.11
1.61
0
K
m
K
1087m
ω
1
2
2
=
m
K
1.87
1
ω2
1 =
m
K
0.73
ω1 =
Modal vector
{ }










=
1.89
0.14
-
1
-
A 2
Similar manner the next frequency and modal vectors can be obtained.
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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,
1
2
1
1 x
ω
m
F = for Mass 1
2
2
2
2 x
ω
m
F = for Mass 2
3
2
3
3 x
ω
m
F = for Mass 3
3. Find new deflection vector using flexibility influence coefficients, using the
formula,










+
+
+
+
+
+
=










′
′
′
33
3
32
2
31
1
23
3
22
2
21
1
13
3
12
2
11
1
3
2
1
α
F
α
F
α
F
α
F
α
F
α
F
α
F
α
F
α
F
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,
13
3
12
2
11
1
1 α
F
α
F
α
F
x +
+
=
=
′ 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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23
1. Assume a modal vector of system { }1
u =










=










1
1
1
x
x
x
3
2
1
2. Find out inertia forces of system at each mass point
2
1
2
1
1 2mω
x
ω
m
F =
=
2
2
2
2
2 2mω
x
ω
m
F =
=
2
3
2
3
3 mω
x
ω
m
F =
=
3. Find new deflection vector using flexibility influence coefficients
Obtain flexibility influence coefficients of the system:
2K
1
α
α
α
α
α 13
31
12
21
11 =
=
=
=
=
2K
3
α
α
α 23
32
22 =
=
=
2K
5
α33 =
13
3
12
2
11
1
1 α
F
α
F
α
F
x +
+
=
′
Substitute for F’s and α,s
2K
5mω
2K
mω
K
mω
K
mω
x
2
2
2
2
1 =
+
+
=
′
23
3
22
2
21
1
2 α
F
α
F
α
F
x +
+
=
′
2K
11mω
2K
3mω
2K
6mω
K
mω
x
2
2
2
2
2 =
+
+
=
′
33
3
32
2
31
1
3 α
F
α
F
α
F
x +
+
=
′
2K
13mω
2K
5mω
2K
6mω
K
mω
x
2
2
2
2
3 =
+
+
=
′
4. New deflection vector is:










=










′
′
′
13
11
5
2K
mω
x
x
x 2
3
2
1










=










′
′
′
2.6
2.2
1
2K
5mω
x
x
x 2
3
2
1
={ }2
u
The new deflection vector{ } { }1
2 u
u ≠ . Iterate again using new deflection vector { }2
u
Second iteration
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24
1. Initial vector of system { }2
u =










=










′
′
′
2.6
2.2
1
x
x
x
3
2
1
2
1
2
1
1 2mω
x
ω
m
F =
′
=
′
2
2
2
2
2 mω
4
x
ω
m
F 4
.
=
′
=
′
2
3
2
3
3 2.6mω
x
ω
m
F =
′
=
′
3. New deflection vector,
13
3
12
2
11
1
1 α
F
α
F
α
F
x ′
+
′
+
′
=
′
′
2K
9mω
2K
2.6mω
2K
4.4mω
K
mω
x
2
2
2
2
1 =
+
+
=
′
′
23
3
22
2
21
1
2 α
F
α
F
α
F
x ′
+
′
+
′
=
′
′
2K
23mω
2K
7.8mω
2K
13.2mω
K
mω
x
2
2
2
2
2 =
+
+
=
′
′
33
3
32
2
31
1
3 α
F
α
F
α
F
x ′
+
′
+
′
=
′
′
2K
28.2mω
2K
13mω
2K
13.2mω
K
mω
x
2
2
2
2
3 =
+
+
=
′
′










=










′
′
′
′
′
′
28.2
23
9
2K
mω
x
x
x 2
3
2
1










=










′
′
′
′
′
′
3.13
2.55
1
2K
9mω
x
x
x 2
3
2
1
={ }3
u
3 u
u ≠ . Iterate again using new deflection vector{ }3
u
Third iteration
u =










=










′
′
′
′
′
′
3.13
2.55
1
x
x
x
3
2
1
2
1
2
1
1 2mω
x
ω
m
F =
′
′
=
′
′
2
2
2
2
2 5.1mω
x
ω
m
F =
′
′
=
′
′
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25
3. new deflection vector,
13
3
12
2
11
1
1 α
F
α
F
α
F
x ′
′
+
′
′
+
′
′
=
′
′
′
2K
10.23mω
2K
3.13mω
2K
5.1mω
K
mω
x
2
2
2
2
1 =
+
+
=
′
′
′
23
3
22
2
21
1
2 α
F
α
F
α
F
x ′
′
+
′
′
+
′
′
=
′
′
′
2K
26.69mω
2K
9.39mω
2K
15.3mω
K
mω
x
2
2
2
2
2 =
+
+
=
′
′
′
33
3
32
2
31
1
3 α
F
α
F
α
F
x ′
′
+
′
′
+
′
′
=
′
′
′
2K
28.2mω
2K
16.5mω
2K
15.3mω
K
mω
x
2
2
2
2
3 =
+
+
=
′
′
′










=










′
′
′
′
′
′
′
′
′
33.8
26.69
10.23
2K
mω
x
x
x 2
3
2
1










=










′
′
′
′
′
′
′
′
′
3.30
2.60
1
2K
10.23mω
x
x
x 2
3
2
1
={ }4
u
The new deflection vector { } { }3
4 u
u ≅ stop Iterating
Fundamental natural frequency can be obtained by.
1
=
2K
10.23mω2
m
K
0.44
ω = rad/s
Modal vector is:
{ }










=
3.30
2.60
1
A 1
2
3
2
3
3 3.13mω
x
ω
m
F =
′
′
=
′
′
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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
1
α
α
α
α
α 13
31
12
21
11 =
=
=
=
=
3K
4
α
α
α 23
32
22 =
=
=
3K
7
α33 =
First iteration
1. Assume a modal vector of system { }1
u =










=










1
1
1
x
x
x
3
2
1
2
1
2
1
1 4mω
x
ω
m
F =
=
2
2
2
2
2 2mω
x
ω
m
F =
=
2
3
2
3
3 mω
x
ω
m
F =
=
3. New deflection vector using flexibility influence coefficients,
13
3
12
2
11
1
1 α
F
α
F
α
F
x +
+
=
′
3K
7mω
3K
mω
3K
2mω
3K
4mω
x
2
2
2
2
1 =
+
+
=
′
4m
3K
2m
K
x1
x2
m
K
x3
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27
23
3
22
2
21
1
2 α
F
α
F
α
F
x +
+
=
′
3K
16mω
3K
4mω
3K
8mω
3K
4mω
x
2
2
2
2
2 =
+
+
=
′
33
3
32
2
31
1
3 α
F
α
F
α
F
x +
+
=
′
3K
19mω
3K
7mω
3K
8mω
3K
4mω
x
2
2
2
2
3 =
+
+
=
′










=










′
′
′
19
16
7
3K
mω
x
x
x 2
3
2
1










=










′
′
′
2.71
2.28
1
3K
7mω
x
x
x 2
3
2
1
={ }2
u
2 u
u ≠ . Iterate again using new deflection vector { }2
u
Second iteration
u =










=










′
′
′
2.71
2.28
1
x
x
x
3
2
1
2
1
2
1
1 4mω
x
ω
m
F =
′
=
′
2
2
2
2
2 4.56mω
x
ω
m
F =
′
=
′
2
3
2
3
3 2.71mω
x
ω
m
F =
′
=
′
3. New deflection vector
13
3
12
2
11
1
1 α
F
α
F
α
F
x ′
+
′
+
′
=
′
′
3K
11.27mω
3K
2.71mω
3K
4.56mω
3K
4mω
x
2
2
2
2
1 =
+
+
=
′
′
23
3
22
2
21
1
2 α
F
α
F
α
F
x ′
+
′
+
′
=
′
′
3K
33.08mω
3K
10.84mω
3K
18.24mω
3K
4mω
x
2
2
2
2
2 =
+
+
=
′
′
33
3
32
2
31
1
3 α
F
α
F
α
F
x ′
+
′
+
′
=
′
′
3K
41.21mω
3K
18.97mω
3K
18.24mω
3K
4mω
x
2
2
2
2
3 =
+
+
=
′
′
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









=










′
′
′
′
′
′
41.21
33.08
11.27
3K
mω
x
x
x 2
3
2
1










=










′
′
′
′
′
′
3.65
2.93
1
K
3.75mω
x
x
x 2
3
2
1
={ }3
u
Stop Iterating as it is asked to carry only two iterations. The Fundamental natural
frequency can be calculated by,
1
2K
3.75mω2
=
m
K
0.52
ω =
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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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:
0
θ
(θ
K
θ
J 2
1
1
1
1 =
−
+ )
&
&
0
θ
(θ
K
θ
(θ
K
θ
J 3
2
2
1
2
1
2
2 =
−
+
−
+ )
)
&
&
0
θ
(θ
K
θ
(θ
K
θ
J 4
3
3
2
3
2
3
3 =
−
+
−
+ )
)
&
&
0
θ
(θ
K
θ
J 3
4
3
4
4 =
−
+ )
&
&
The Motion is harmonic,
( )
ωt
sin
φ
θ i
i = (17)
where i=1,2,3,4
Substitute above Eqn.(17) in Eqns. of motion, we get,
)
φ
(φ
K
φ
J
ω 2
1
1
1
1
2
−
= (18)
)
φ
(φ
K
)
φ
(φ
K
φ
J
ω 3
2
2
1
2
1
2
2
2
−
+
−
=
)
φ
(φ
K
)
φ
(φ
K
φ
J
ω 4
3
3
2
3
2
3
3
2
−
+
−
=
)
φ
(φ
K
φ
J
ω 3
4
3
4
4
2
−
+ (19)
Add above Eqns. (18) to (19), we get
0
φ
J
ω
4
1
i
i
i
2
=
∑
=
For n dof system the above Eqn changes to,
0
φ
J
ω
n
1
i
i
i
2
=
∑
=
(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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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
(φ
K
φ
J
ω 2
1
1
1
1
2
=
−
=
1
1
1
2
1
2
K
φ
J
ω
φ
φ −
=
4. Similarly calculate the amplitude of third disc/mass φ3 from second Eqn. of motion.
0
)
φ
(φ
K
)
φ
(φ
K
φ
J
ω 3
2
2
1
2
1
2
2
2
=
−
+
−
=
0
)
φ
(φ
K
)
φ
K
φ
J
ω
(φ
K
φ
J
ω 3
2
2
1
1
1
1
2
1
1
2
2
2
=
−
+
−
−
=
0
)
φ
(φ
K
φ
J
ω
φ
J
ω 3
2
2
1
1
2
2
2
2
=
−
+
−
=
2
2
2
1
1
2
3
2
2 φ
J
ω
φ
J
ω
)
φ
(φ
K +
=
−
2
2
2
2
1
1
2
2
3
K
φ
J
ω
φ
J
ω
-
φ
φ
+
= (21)
The Eqn (21) can be written as:
2
2
1
i
2
i
i
2
3
K
ω
φ
J
-
φ
φ
∑
=
=
5. Similarly calculate the amplitude of nth disc/mass φn from (n-1)th Eqn. of motion
is:
n
n
1
i
2
i
i
1
-
n
n
K
ω
φ
J
-
φ
φ
∑
=
=
6. Substitute all computed φi values in basic constraint Eqn.
0
φ
J
ω
n
1
i
i
i
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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31
1 2 3 4 5 6 7 8
ω
S No J φ Jω2
φ 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 Jω2
φ 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
∑ φ
2
Jω ∑ φ
2
Jω
K
1
θ1 K1 θ2 θ3
K2
J3
J2
J1
Fig.14 A torsional semi-definite system
∑ φ
2
Jω ∑ φ
2
Jω
K
1
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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
∑ φ
2
Jω
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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
1.71
ω
rad/s
1
ω
rad/s
0
ω
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
∑ φ
2
Jω
Frequency, ω
Fig.15. Holzar’s plot of Table-3
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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 Jω2
φ 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
∑ φ
2
Jω ∑ φ
2
Jω
K
1
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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 Jω2
φ 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    
∑ φ
2
Jω ∑ φ
2
Jω
K
1
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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
2.30
ω
rad/s
1.15
ω
rad/s
0.35
ω
3
2
1
=
=
=
0.0 0.5 1.0 1.5 2.0 2.5
-20
-10
0
10
20
Displacement,
φ
4
Frequency,ω
1
ω
2
ω 3
ω
Fig.17. Holzar’s plot of Table-5
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