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S. Ilanko 2007
3.1 Equations of Motion using Newtons 2nd
Law & Calculation of Natural Frequencies
Even in a free country there are laws that its people must abide by. In vibration, the motion
along any degree of freedom is also subject to a law, which may be expressed as Newtons
second law of motion. We were able to write an equation of motion for a single degree of
freedom system by summing all the actions (forces or moments) along the direction of motion
in which the system is free to move, and equating the sum of the actions to the product of
acceleration and an inertial resistance factor (this factor being mass for a translational degree
of freedom, or moment of inertia for a rotational degree of freedom). A discrete system with n
degrees of freedom will have n equations of motion. This means more work for us. In
vibration, freedom seems to come with a price! These equations may be obtained by applying
Newtons second law of motion to the discrete masses along each of the n independent
translational/rotational co-ordinates. This is not the only way to obtain the equations of
motion. In another section, we will study an alternative method based on an energy principle
to derive these equations. For now, let us consider the application of Newtons second law of
motion to find the natural frequencies and modes, using a very simple 2-DOF spring-mass
system (throughout this text, the term mass will be used to indicate a particle possessing
mass) as shown in Figure 3.1.1, as an illustration. The 2-DOF system is only an illustration
and we will present the steps using matrix notation in a general way so that one can see how
this could be done for any discrete system.
State 2 (at time tduring vibration)
State 1 (equilibrium, springs unstressed)
T1 T2
T2
u1 u2
m2
k1 k2
Figure 3.1.1 A 2 DOF system
m1
3. Systems with Several Degrees of Freedom
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S. Ilanko 2007 50
Let the dynamic displacement of the masses be u1, u2 and their amplitudes be 1, 2. Sincex,y
are used for representing the static co-ordinates of continuous systems, the dynamic
displacements will be denoted by u,v, and to be consistent we will use the same notation in
this book.
For determining the natural modes and frequencies, the motion may be assumed to be simple
harmonic.
Ifu1 =1 sin(t+) and u2 = 2 sin(t+)
then the accelerations are given by
( ) 12
1
2
1 sin utuu =+= and 22
2 uu = .
Applying Newtons second law to m1 along u1 (see Figure 3.1.2), 1112 umTT =
Using the constitutive equations for the springs, ( ) 12
11111122 umumukuuk ==
Rearranging this we get, (k1+k2)u1 -k2u2 = m12u1 ...(3.1.1)
This is the first equation of motion.
Similarly for the second mass, k2(u2-u1) = m22u2 ...(3.1.2)
These equations of motion may be given in the following matrix form,
...(3.1.3)
where the elements of the generalized stiffness and mass matrices are as follows:
+=
22
221 )(][
kk
kkkK and
=
2
1
0
0][
m
mM ...(3.1.4a,b)
This is a standard eigenvalue problem, and for any linear, nDOF discrete system, one can
derive n equations of motion in the form of (3.1.3). That is the resulting matrix equation will
contain an n n stiffness matrix and an n n mass matrix. This eigenvalue equation may be
solved using iterative methods, or by finding the roots of a determinantal frequency equation
giving n natural frequencies and modes. For a 2-DOF system, explicit expressions for the
natural frequencies can be obtained as follows.
[K]{u}=2[M]{u}
m11u
T1 T2
Figure 3.1.2 Freebody diagramforparticle m1
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From equation (3.1.1), 22121
21
)(u
mkk
ku
+= ...(3.1.4c)
Substituting this into equation (3.1.2) gives
2
2
222
121
2222
)( umumkk
kkuk =+
0)(
2
2
22
121
222 =
+ um
mkk
kkk
This is true if either 02 =u
or 0)(
2
22
121
222 =
+
m
mkk
kkk
However, if 02=
u then from equation (3.1.4c), 01=
u and the system will not vibrate. This is
a trivial solution. For non-trivial solution, 0)(
2
22
121
222 =
+
m
mkk
kkk
This is the frequency equation. The roots of this frequency equation are the two natural
frequencies. Substituting the natural frequencies into either equation (3.1.1) or (3.1.2) gives
the ratio of the displacements (u1/u 2).
As an example, let k1 = 100N/m, k2 =200 N/m, m1 = 0.2 kg and m2 = 0.3 kg. This yields the
following values for the natural frequencies (eigenvalues) and modes (eigenvectors):
First mode: 1= 12.91 rad/sec, (1/ 2)1 = 0.75 ..(3.1.5a,b)
Second mode: 2= 44.72 rad/sec, (1/ 2)2 = -2.0 ..(3.1.6a,b)
The system can vibrate freely at any of the two frequencies without any external dynamic
force. If it vibrates purely in one of the above modes at the corresponding frequency, it is
then called a principal vibration. The actual dynamic displacement of a system depends on
initial conditions, and is generally a combination of all its natural modes as described in the
next section.
For an n-DOF system equation (3.1.3) may be solved using iteration. The following procedure
may also be used, but it is often not convenient.
Rewriting equation (3.1.3),
Let [Kd] =[K]-2[M],
then [Kd] {u}={0}.
For non-trivial solution, |Kd|=0This is the frequency equation.
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From here onwards, the following matrix notation will also be used to represent the principal
modes as a matrix called modal matrix:
..(3.1.7)
In this matrix the ith
row represents the ith
degree of freedom, and thejth
column represents the
jth
mode. i.e. jia , is the displacement (or rotation) corresponding to the ith
degree of freedom
in thejth
principal mode. The modes are normalized either by setting the maximum amplitude
to unity, or by setting energy terms to unity. In the above example, by setting the amplitudes
to unity, the modal matrix may be written as:
0.51.00
1.000.75..(3.1.8)
The normalized modes are referred to as normal modes. This is not the only way to normalise
the modal matrix. Another more useful normalisation method will be discussed later.
Derivation of the Equations of Motion using Newtons Second Law
A freebody diagram of each particle of rigid body must be sketched. In addition to the forces
and moments acting on each freebody, the accelerations and the relevant mass or moment of
inertia must also be shown.
Then Newtons second law should be applied either in a translational direction or a rotational
direction.
The simple harmonic relationship may be used to express the accelerations in terms of the
frequency and displacement.
The resulting equations are arranged in a matrix form to obtain the eigenvalue equation.
The derivation of the equations of motion by applying Newtons second law of motion for
many practical problems can be very difficult and a more convenient method based on the
potential and kinetic energies of the system will be discussed in another chapter.
n,n
i,j
,,
,,
a
aaa
aa
a2212
2111
][matrixmodalThe =
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S. Ilanko 2007 53
mb,Ib
mb,Ib
Example 1
Two rigid bars are hinged at points O and
P, and connected by means of an elastic
spring of stiffness k as shown in Figure
3.1.3. They also carry four concentrated
masses at their ends. The bars are
identical having a mass mb (which is
equal to 0.2m) and a moment of inertia of
Ib (which is 0.2mL2 /12) about their
centroidal axis perpendicular to the planeof oscillation. This system is in static
equilibrium under gravity field g.
P
O
g
0.2L
0.3L
0.5L
0.2L
0.7L
0.1L
k
m
m
2 m
2.5 m
Figure 3.1.3
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Example 2
Figure 3.1.4 is a schematic representation of a power transmission system afterthe sudden
application of a brake on the left end of the top shaft.
The relevant moments of inertia of the gears and the rotor are:
I2 = 1.2 kgm2;
I3 = 0.15 kgm2; and
I4 = 3 kgm2
(a.)Obtain the equations of motion for the torsional vibration of the above system
neglecting the mass of the shafts (treating the system as discrete).
(b.)Show that the natural frequencies and modes of the above system are as follows:
1 = 18.34 rad/s, and for the first mode, 2/4 = 0.374
2 = 115.0 rad/s, and for the second mode, 2/4 = 4.458
I2
k1=6 kNm/rad
I3I4k2= 4 kNm/rad
2r
r
Figure 3.1.4
4 (t)
2 (t)
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3.2 Response of a 2-DOF Undamped System
Transient Response
In the preceding example, if the masses were initially displaced by 5 mm from their
equilibrium state, and then released, the response may be calculated as follows:
Initial conditions:
u1 (0) = 5 mm ; u2 (0) = 5 mm.
Also 0)0(1 =u ; 0)0(2 =u
Let ( ) ( )22
2111
2
1sin
50
001sin
001
750 t+
.-
.Gt+
.
.G
u
u
+
=
Substituting the first two initial conditions we get,
)+sin(00.5-
1.00G)+sin(0
1.00
0.75G
5
52211
+
=
..(3.2.1)
Using the last two initial conditions,
)+cos(00.5-
1.00G)+cos(0
1.00
0.75G
0
022111 2
+
=
..(3.2.2)
The above equation is satisfied if1 = 2 =/2. Substituting these into equation (3.2.1) gives:
+
=
50
001
001
750
5
521
.-
.G
.
.G
Solving the two simultaneous equations gives: G1= 5.45 and G
2= 0.90
Hence the response is:
( ) ( )27244sin450
91.029112sin
45.5
09.4
2
1/
+/
=
t+..-
t+.u
u
i.e. ( ) ( )t.t.u
u7244cos
45.0
91.09112cos
45.5
09.4
2
1
+
=
..(3.2.3)
This is the required dynamic response. To develop a more convenient general procedure forobtaining the dynamic displacement, it is necessary to consider an important relationship
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S. Ilanko 2007 56
between the natural modes called the orthogonality relationship. This is done in section 3.3
and a more convenient procedure for calculating the transient response of a discrete system is
described in section 3.4.
Steady State Response Due to Simple Harmonic Force
Let us consider the vibration of a 2-DOF spring-mass system subject to a harmonic excitation
as shown in Figure 3.2.1.
Applying Newtons second law of motion to the two particles gives the following equations:
11121 sin umTTtF =+
222 umT =
Substituting the constitutive equations for the springs into the above equations and
rearranging leads to the following equations of motion:
tFukumukk =++ sin)( 12211121
0222212 =++ umukuk
By inspection, it may be seen that the forms ptuu sin11 = and ptuu sin22 = satisfy the above
equations of motion. Substitution gives:
tFukupmkk =+ sin)( 12212
221 ..(3.2.4a)
State 2 (at time tduring vibration)
State 1 (equilibrium, springs unstressed)
T1 T2
T2
u1
m2
k1 k2
Figure 3.2.1 A 2 DOF system Subject to a SH Force
m1
F1 sin t
F2 sin t
T2
F1 sin t
T1 T2
12
FBD ofm1FBD ofm2
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0)( 22
2212 =+ umkuk ..(3.2.4b)
From equation (3.2.4b),)( 222
122
=
mk
uku ..(3.2.4c)
Substituting this into equation (3.2.4a) gives:
( )tFu
mk
kmkmkk=
+sin
)(
))((112
22
2
2
2
22
2
221
( )222
22
2
221
1
2
221
))((
sin)(
kmkmkk
tFmku
+
= ..(3.2.4d)
Substituting this into equation (3.2.4c) gives:
( )222
22
2
221
122
))((
sin
kmkmkk
tFku
+
= ..(3.2.4d)
It may be noted that if 0222 = mk then 01 =u
This is significant and can be used to control the vibration due to harmonic excitation.
An undamped vibration absorber
For example, if a SDOF is subject to a harmonic excitation as shown in Figure 3.2.2, then
adding another SDOF consisting of a mass m2 and stiffness k2 such that 02
22 = mk will
ensure that the original system will remain undisturbed by the excitation as all the excitation
force will be absorbed by the subsystem.
Consider the attached spring-mass system as a separate system. The natural frequency of this
system is given by 22/mk . This means that if the attached system has a natural frequency
equal to the excitation frequency of the main system then the main system will not vibrate.
This is called an undamped vibration absorber or a detuner.
u1
m2
Figure 3.2.2 An undamped vibration absorber
m1
F1 sin t
Original SDOF system Detuned subsystem
k1 k2m2
u1
m1
F1 sin t
k2k1
u2
Added subsystem
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3.3 Orthogonality of Modes.
In the preceding example, it may be seen that m1. 1,1a . 2,1a + m2. 1,2a . 2,2a = 0.
For any discrete system with n degrees of freedom, we will show that there is a similarrelationship between any two different natural modes. We will initially restrict our arguments
to systems that have a diagonal mass matrix. If all displacements share the same vector
direction then we will prove that
...........(3.3.1a)
The above relationship is called the orthogonality property of the natural modes. In a situation
where the displacements are in different directions, the above equation still holds, if care is
taken to ensure that a mass that has two or three degrees of freedom, appears twice or three
times in the expression, the number of such appearances being equal to the number of degrees
of freedom. That is to say that the subscript i strictly refers to a degree of freedom and not a
mass. For example if a particle having mass m0 is associated with the third and fourth degrees
of freedom, then both m1 and m2 should be set to m0. This is explained through a 2-DOF
spring-mass system at the end of this section.
In general the mass terms mi may include moments of inertia i in which case the
displacement co-ordinates ai,j would be angle of rotations along the degrees of freedom.
It is convenient to express the orthogonality relationship in matrix form:
...........(3.3.1b)
where [M] is a diagonal matrix which is sometimes referred to as the generalised mass
matrix. In this book, we will simply call it the diagonalised inertia matrix. The elements of
this matrix may be associated with either rotational or translational motion, and inertia matrix
is a more suitable term.
,0
1
==
i,ki,ji aamn
i
forjk
[a]T[M][a] = [M]
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For a n DOF system, there are n(n-1) number of orthogonality relationships which are
represented by the equations from (3.3.1b) that give rise to the off-diagonal terms in [M].
The modes may be normalised by making this an identity matrix.
For example, to normalise the first mode in the previous example,
m1 ( 1,1a )2
+ m2 ( 1,2a )2
=1 .............(3.3.2)
But 1,1a / 1,2a = 0.75/1.0 .............(3.3.3)
Solving these two equations gives: 1,2a = 1.557 and 1,1a = 1.168
Similarly the second mode may be normalized to obtain:
2,1a = 1.907 and 2,2a = -0.953
Hence the alternative to normalised modes in equation (3.1.8) is:
0.9531.557
1.9071.168.............(3.3.4)
This would give an identity matrix on the right hand side of equation (3.3.1b). This type of
normalisation is often used in response calculations as explained later.
Proof of Orthogonality:
Consider the free vibration of an n degree of freedom
system in its jth
mode. The freebody diagram of a
typical mass with forces and acceleration along the ith
degree of freedom is shown in Figure 3.3.1.
The net elastic restoring force on the mass is
Fi,j = - mij2ai,j
By d'Alambert's principle, if each of the
masses were subject to a static force Fi,j = mi
j2ai,j then the resulting displacements of the
masses will be given by ai,j. Now consider
the forces associated with the kth
mode of
vibration. Similarly we can say that forces
Fi,k = mi k2ai,k will result in displacements
Fi,j
Fi,k
ai,j ai,k
Force
Displacement
Figure 3.3.2
Acceleration
-j2ai,j
mi
Fi,j
Figure 3.3.1 Freebody
Diagram of a typical mass
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ai,k.
From Betty-Maxwell's reciprocal theorem, for linear elastic structures, the work done by the
forces Fi,j while undergoing displacements ai,kis equal to the work done by the second set of
forces Fi,k over displacements ai,j. This can be explained by equating the work done by the
two sets of forces applied in different order.
First let us consider the work done on mass mi by forces Fi,jwhile causing displacement ai,j.
This is given by the dotted triangular area under the force-displacement curve shown in
Figure 3.3.2 and is equal to jiji aF ,,2
1
Now let us apply forces Fikwhich cause displacements ai,k. The work done by the force Fik on
mass mi is given by the hatched triangular area and is equal to kiki aF ,,2
1. However as
displacements ai,k. are being caused by Fik the force Fi,j that is already acting on the system
would also do work equal to Fi,j ai,k. shown by the shaded rectangular area in Figure 3.3.2. The
net work done on the ith mass is therefore jiji aF ,,
2
1+ kiki aF ,,
2
1+ Fi,j ai,k.
The total work done on the system is given by summing the work done for all masses. This is
=
++
n
i
kijikikijiji aFaFaF1
,,,,,,2
1
2
1
If the order of application of the forces were reversed (that is forces Fi,k is applied first
followed by Fi,k) then the net work done is =
++
n
i
jikikikijiji aFaFaF1
,,,,,,2
1
2
1
For a linear elastic structure, the order of application of force has no effect on its behaviour.
Therefore the total work done may be equated.
=
++
n
i
kijikikijiji aFaFaF1
,,,,,,2
1
2
1=
=
++
n
i
jikikikijiji aFaFaF1
,,,,,,2
1
2
1
This is a statement of the reciprocal theorem.
Cancelling the common terms, we get ==
=
n
i
jiki
n
i
kiji aFaF1
,,
1
,,
Using dAlemberts principle,
( ) =n
i kijijiaam
1 ,,
2 = ( ) =n
i jikikiaam
1 ,,
2
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Rearranging gives: ( )( )221 ,, kj
n
i kijiiaam = = 0
Ifjk, then ( ) =n
i kijiiaam
1 ,,= 0.
This means that if two modes have different natural frequencies then the sum of the product
of the masses and the displacements/rotations corresponding to those modes will be zero. The
modes are said to be orthogonal. In cases where the co-ordinates ai,j and aj,k are in different
directions, the work done is obtained by taking the dot product of the force vector (and
displacement vector. An alternative matrix approach for a general proof of orthogonality is
given below:
When a discrete system vibrates in thejth mode, equation (3.1.3) reduces to:
[K]{a}j = (j)2[M]{a}j
where {a}j is thejth
mode which is thejth
column of the modal matrix.
Pre-multiplying both sides of this equation by T}{ ka gives:
T}{ ka [K]{a}j = (j)2 T}{ ka [M]{a}j
Similarly pre-multiplying the equations of motion corresponding to the kth
mode by the
transpose of thejth
mode gives:
T}{ ja [K]{a}k= (k)2 T}{ ja [M]{a}k
Since the stiffness matrix [K] is symmetrical the left hand side of the last two equations are
equal. Therefore we can equate the right hand side of these equations to obtain:
(j)2 T
}{ ka [M]{a}j = (k)2 T
}{ ja [M]{a}k ..(3.3.5)
Since the mass matrix is also symmetricalT
}{ ka [M]{a}j =T}{ ja [M]{a}k
Substituting this into equation (3.3.5) gives:
(j)2 T}{ ja [M]{a}k= (k)
2 T}{ ja [M]{a}k
If the frequencies jand kare not equal, for the above equation to be true,
..(3.3.6)T}{ ja [M]{a}k= 0
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This is a matrix representation of the orthogonality property of natural modes. Sincej and k
are arbitrary, a series of the above equations for any combination ofj and k (jk) may be
written.
Ifj = k, then
..(3.3.7a)
Therefore we can write a matrix equation of the form (3.3.1b) which we will restate now.
[a]T[M][a] = [M] ....(3.3.1b)
where [M] is a diagonal matrix.
The elements of the diagonal matrix are defined by equation (3.3.7a) and in systems where
the original mass matrix is diagonal, they may also be expressed in the following form:
..(3.3.7b)
For an n-DOF system, it is possible to write n equation of the form (3.3.7). This gives us
another way to normalise the modes. We can make [M]an identity matrix [I] by scaling the
modes such that T}{ ja [M]{a}j=1
Such a normalized mode is called the normal mode, and may be denoted by [N]
..(3.3.8)
Orthogonality with respect to stiffness matrix
We will now show that a similar orthogonal relationship with respect to stiffness also exists.
Applying equation (3.1.3) to a system that is vibrating in its kth
natural mode we get
[K]{a}k= (k)2
[M]{a}k
[N]T[M][N] = [I]
2,, )( ri
n
i
irr amM =
T}{ ja [M]{a}j = jjM , 0
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Substituting this into equation (3.3.6) gives
T}{ ja 21
k[K]{a}k= 0
This gives T}{j
a [K]{a}k= 0 ..(3.3.9)
It also follows that ifj = k, then T}{ ja [K]{a}j = jjK , = (j)2 jjM , ..(3.3.10)
and [a]T[K][a] = [ K], where jjK , = (j)
2 jjM , ..(3.3.11)
In terms of the normal modes,
[N]T[K][N] = [ K] ..(3.3.12)
where [ K] is a diagonal matrix whose elements contain the eigenvalues.
That is2
, jjjK = ..(3.3.13)
These relationships will be very useful in the calculation of dynamic displacements of
vibratory systems as explained in the following sections.
A geometrical and vectorial interpretation of orthogonality
To illustrate the situation where one needs to be careful in applying the orthogonality
condition to a system where a mass may be associated with more than one degree of
freedom, let us consider a spring mass system that is free to vibrate inx-y plane.
This system consists of a particle of mass m, restrained by two springs, one oriented along the
y axis and another at angle to the y axis as shown in Figure 3.3.3. Let the stiffness of the
springs in the y and the inclined directions be k1 and k2 and the force in these springs be F1
Figure 3.3.3
mv
k1
k2
u
F1
F2u
v
m
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S. Ilanko 2007 64
and F2 respectively. Let the independent coordinates be u and v, these being measured in thex
andy directions respectively.
For small displacements, the elongation of the springs 1 and 2 would be v and (u sin +v cos
) respectively.From these we can write the constitutive equations in terms of the displacements:
vkF 11 = and ( ) cossin22 vukF +=
Writing Newtons second law inx andy directions, we get:
umF =sin2 = um2
and vmFF =+ cos21 = vm2
Substituting the constitutive equations into the above we get,
[ ]
=
v
uM
v
uK
2][ (3.3.14)
where[ ]
+=
2
212
2
2
2
coscossin
cossinsin
kkk
kkK and [ ]
=
10
01mM
For convenience, let kkk == 21
[ ]
+=
2
2
cos1cossin
cossinsin
kK
For non-trivial solution, | K- M2 | = 0 gives the following roots for 2
( )( )mk/cos121 = and ( )( )mk/cos12
2 +=
The corresponding modes may be found by evaluation ( )vu / which gives the tangent of
direction of motion of the mass. From the first row of equation (3.3.14),
v
u=
22sin
cossin
mk
k
which for the fits and second natural frequencies give the following
results:)cos1(sin
cossin2
1
=
kk
k
v
u=
2coscos
cossin
=
cos1
sin
and)sin1(sin
cossin2
2
+=
kk
k
v
u=
2coscos
cossin
)cos1(
sin
+=
Multiplying these tangents give 1 showing that these directions are perpendicular to each
other. This is a situation where one can see that two modes are geometrically orthogonal
(perpendicular). The dot product of two non-zero vectors would be zero only if they are
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perpendicular, and for a situation involving vibration of one particle possessing mass in
possibly more than one translational sense, then the modal directions must be perpendicular to
each other. It should be mentioned here that this statement is only true in the absence of
rotational motion. The orthogonality statement for a rigid body that is capable of rotational
motion would include a term associated with its moment of inertia and two rotational
displacement values corresponding to two modes, and in such a situation the geometric
interpretation of orthogonality of modes does not apply.
Let us write down the modal matrix for this problem in terms of its vector components in the
x andy directions. i.e. [ ]
=
21
21
vv
uua =
+
11
cos1
sin
cos1
sin
The mass matrix is [ ]
=
m
mM
0
0
[ ] [ ][ ]=aMa T
+
1cos1
sin
1cos1
sin
m
+
11cos1
sin
cos1
sin
= m( )
( )
+
+
+
1cos1
sin0
01cos1
sin
2
2
2
2
= [ ]M - a diagonal matrix.
Note that although there is only one particle there are two mass terms, each referring to the
inertial resistance of the particle against displacements in two directions. We will now look at
an interesting example where the vibration of a single mass in two orthogonal directions is
affected by different system properties.
Lateral and axial vibration of a spring-mass system
Consider the vibration of the spring mass system shown in Figure 3.3.4. The springs have
stiffness coefficients k1,k2 and are under a static tensile force TS. Let us say that the mass and
springs lie on a smooth horizontal table, and that we are interested in the natural frequencies
and modes of this system when it vibrates in the horizontal plane.
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Figure 3.3.54. Spring Mass System Under Static Force
Let the static tension in the springs be TS and the dynamic tensions be TD1, TD2. Now,
Newtons second law may be applied in u, v directions:
In the u direction,
( ) ( ) umTTTT DSDS =++ 1122 coscos
For small displacements, cos 1, cos 2 1, giving umTT DD = 12
Using the spring properties, this becomes umekek = 1122
where e1,e2 are the elongation of the springs 1,2 respectively. For small displacements, it can
be shown that e2= -u and e1 = u. Equation of motion thus reduces to:
( ) umukk =+ 21
But uu 2=
Therefore ((k1+k2) -m2
)u = 0.
The non-trivial solution of this frequency equation ism
kk 21 += (3.3.15)
It is interesting to note that this natural frequency does not depend on TDand corresponds to a
non-trivial solution ofu. This means that there is a natural mode of vibration that is entirely
axial, and the corresponding frequency is independent of the static force in the axial direction.
Now let us obtain the equation of motion in the v direction.
TS TS
TS+TD1
TS+TD2
TS+TD1TS+TD2
Freebody Diagram
m1
L1 L2
k2k1
At equilibrium state
During vibration
v
v
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( ) ( ) vmTTTT DSDS =++ 2211 sinsin
For small displacements, sin 1v/L1 and sin 2v/L2
Also the dynamic forces TD1,TD2 are negligible compared to TS. Using the above into the
equation of motion gives:
vmdt
vdm
LLvT
S
2
2
2
21
11==
+
For non-trivial solution of v,
m
LLTS
+
=21
11
..(3.3.16)
This is a pure lateral mode, and its frequency is not dependent on the spring properties.
However, it is a function of the axial force (geometric stiffness) and will be zero in the
absence of any axial tension. The frequency increases with axial tension. This is true for all
mechanical systems. That is, the natural frequency in lateral vibration depends on the axial
force, and increases with tension. A good illustration of this is the well-known fact that the
pitch of string instruments increases with tension.
It is not possible to give a geometric interpretation of the orthogonality principle in general.
The main application of the orthogonality property of the modes is in response calculations.
3.4 General procedure for finding the transient response of an undamped system
Having established the orthogonality relationship, we can now set up a general procedure for
solving an initial value problem. Let us suppose, we have the frequencies and modes of an n
degree of freedom system that is subject to prescribed initial conditions. i.e. { }u and { }u at t
= 0 are known.
Due to the orthogonality of modes with respect to mass,
0=n
i
s,ir,ii aam for rs (3.4.1)
where, n is the number of degrees of freedom.
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This is to say that the rth
mode and sth
mode are orthogonal with respect to mass. The modes
may be arranged in a matrix called modal matrix, which gives the jth
mode in thejth
column
and the rows correspond to mass (the term 'mass' is used here generically to include mass and
moment of inertia) numbers. The modal matrix may be denoted by [a], and it is not unique
for any given system.
For the two degree of freedom system considered in this chapter, any of the following may be
used as the modal matrix, as long as the relative ratios of the displacement (translation or
rotation) of various masses for a given mode remain unchanged thus defining a unique shape
or pattern for each mode.
14
23or
14
23or
5.01
175.0could be used as the modal matrix.
In terms of the modal matrix, the orthogonality property leads to a matrix manipulation
technique which can be used to simplify vibration analysis.
Recalling from equation (3.3.1b) the diagonalised inertia matrix ][M is obtained from
[a]T[M][a] = ][M (3.3.1b)
The rth
diagonal element of ][M is given by equation (3.3.7)
2
,, )( ri
n
i
irr amM = (3.3.7b)
Solving an initial Value Problem:
If an undamped vibratory system is given an initial displacement or velocity in the form of
one of its natural modes and then released, it will continue to vibrate in that particular mode.
For any other initial condition, its response will consist of more than one mode. The
contribution from each mode to the vibration of the system may be expressed in the following
form:
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(3.4.2)
where {u} gives the dynamic displacement, [a] is the modal matrix and {p} is a vector
consisting of the contribution from each mode in the form of a scaled time function. These
contribution factors {p} arecalled the principal co-ordinates.
i.e.pr= Arsinrt + Brcosrt (3.4.3a)
We can also obtain the time derivative
rp = rArcosrt - rBrsinrt (3.4.3b)
Use of initial conditions:
Note that at t= 0,pr= Br (3.4.3c)
and rp = rAr (3.4.3d)
The actual displacement of the ith
mass would be given by
ui= =
l
r
rri pa1
, (3.4.4)
where l is the number of modes.
Pre-multiplying both sides of equation (3.4.2) by [a]T[M] gives:
[a]T[M]{u} = [a]
T[M][a] {p} (3.4.5)
Substituting equation (3.3.1b) into this gives:
(3.4.6)
Since ][M is diagonal, these equations are decoupled, and for any given initial conditions,
{p} may be easily determined as follows:
If the original mass matrix is also diagonal, then expanding the rth row of equation (3.4.6), and
substituting equation (3.3.7b) into the resulting equation, we get,
[a]T[M]{u} = ][M {p}
{u} = [a]{p}
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==
n
i
irii uam1
, pr2
)a(m r,i
n
i
i (3.4.7)
Therefore pr= /1
,
=
n
i
irii uam
2)a(m r,i
n
i
i (3.4.8)
Since only the initial values ofui would be known, the components ofpr(Ar andBr) need to
be determined using the initial conditions and equations (3.4.3c) and (3.4.3d). This gives:
(3.4.9a)
(3.4.9b)
If the original mass matrix contains off-diagonal elements then the matrix equation (3.4.6)
must be used to obtain the principal coordinates.
Once the principal co-ordinates are found, the displacement is given by equation (3.4.2). So
equations (3.4.9a), (3.4.9b), (3.4.3a) and (3.4.2) give the response of a general multi-degree of
freedom system subject to any initial conditions, without requiring any simultaneous
equations solver. Another advantage of this approach is that one does not need the knowledge
of all modes of a system to estimate the response. In practice, the use of the first few modes
(equation (3.4.2)) may be sufficient. That is one could use a narrow rectangular matrix
consisting of as many modes as necessary to reach the desired accuracy. However, all masses
must be included to get correct results. (i.e. the number of modes l may be smaller than the
degree of freedom nf, but the number of masses (n) must be at least equal to nf, and in some
cases it may exceed nf. The latter occurs when there are several masses whose relative motion
is constrained for example as in gears. Using a different modal matrix would give different
{p} but the final response will be the same.
Example 1:
Let us solve the example in Section 3.2 using the general approach.
Br= /)0(1
,
=
n
i
irii uam
2)a(m r,i
n
i
i
Ar= /)0(1
,
=
n
i
irii uam
2)a(m r,i
n
i
ir
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Let [a] =
5.01
175.0
[M] =
3.00
02.0
Firs we can express initial conditions in terms of principal coordinates.
{u} = [a]{p}
[a]T[M]{u} = [a]
T[M][a]{p}= ][M {p}
5.01
175.0
3.00
02.0{u}=
5.01
175.0
3.00
02.0
5.01
175.0{p}
15.02.0
3.015.0{u}=
275.00
04125.0{p}
This is true at any time. Therefore at t=0,
15.02.0
3.015.0{u(0)} =
275.00
04125.0{p(0)}
But we have {u(0)} =
5
5mm
Substituting this into the previous equation gives:
275.0
4125.0{p(0)}=
25.0
25.2mm giving
=909.0
45.5)}0({p
Similarly,
15.02.0
3.015.0{ )0(u } =
275.00
04125.0{ )0(p }
and { )0(u } ={ }0
gives { )0(p }={ }0
From equations (3.4.3c,d) we get
==909.0
45.5)}0({}{ pB mm and { }0}
)0({}{ ==
pA
The frequencies are given in equations (3.1.5,6).
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}]{[}{ pau = =
5.01
175.0
)72.44cos(909.0
)91.12cos(45.5
t
tmm
=
+
)72.44(cos45.0)91.12(cos.455)72.44(cos91.0)91.12(cos09.4
tttt mm
These results could have been obtained by substituting the elements of the modal matrix and
mass matrix into equations (3.4.9a,b), and then using equation (3.4.2), but we have gone
through the problem step-by-step. Using a normal mode [N] instead of [a] would have given
the same result for the response.
[N]T[M]{u} = [N]
T[M][N]{p}=[I]{p}
{p}=[N]T[M]{u}
{p(0)}= [N]T[M] {u(0)}
From equation (3.3.4) [N] =
0.9531.557
1.9071.168
{p(0)}=
0.9531.907
1.5571.168
3.00
02.0
5
5=
2859.03814.0
4671.02336.0
5
5=
4775.0
5035.3
Since { )0(u } ={ }0 , { )0(p }={ }0 giving { }0})0(
{}{ ==
pA
== )}0({}{ pB
4775.0
5035.3mm
The principal coordinates are different now, but we will get the same displacement for the
masses.
}]{[}{ qPu = =
0.9531.557
1.9071.168
)72.44cos(.47750
)91.12cos(.50353
t
tmm
=
+
)72.44(cos45.0)91.12(cos.455
)72.44(cos91.0)91.12(cos09.4
tt
ttmm
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Compared to the way in which this was solved in Section 3.2, we should note that using the
modal matrix and principal coordinates, we did not have to invert a matrix nor solve a set of
simultaneous equations. The use of orthogonality relationship has enabled us to solve the
problem using matrix multiplications and additions only.
Example 2
The response of a four dof undamped spring-mass system with the following properties is
required.
Mass
(kg)
Mode Number Mode 1 Mode 2 Mode 3 Mode 4
Nat. frequencies (rad/s) 0.9031 2.6312 4.8944 6.9064
Displacements
1.0 u1 0.2390 0.5342 1.0000 1.0000
1.2 u2 0.5445 1.0000 0.7364 -0.8466
2.5 u3 0.7470 0.9340 -0.5198 0.1913
3.1 u4 1.0000 -0.8148 0.0809 -0.0138
Initially the system is given the following displacements and then released:
Show that the dynamic displacement u1 is given by:
u1= 0.119 cos .9031t- 0.020 cos 2.6312t+ 0.002 cos 4.8944 t+ 0.003 cos 6.9064 tmm
Solution: See Transient response.ppt
3.5 Free vibration analysis of damped systems
The analysis will be confined to viscous damping only. The differential equation of motion
will be of the form:
(3.5.1)[ ]{ } [ ]{ } [ ]{ } { }0=++ uKuCuM
=
53.0
34.0
23.0
1.0
)0(
)0(
)0(
)0(
4
3
2
1
u
u
u
u
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[ ] [ ] [ ]KMC +=
This may be obtained by applying Newtons law of motion or using an energy method (see
Chapter 6).
In terms of principal coordinates,
{ } [ ]{ }pau = (3.5.2a)
Therefore, { } [ ]{ }pau = and { } [ ]{ }pau = (3.5.2b, c)
Substituting these into equation (3.5.1) and pre-multiplying each term by [ ]Ta gives:
[ ][ ]{ } [ ] [ ]{ } [ ][ ]{ } { }0][][][ =++ paKapaCapaMa TTT (3.5.3)
Usually, [ ] [ ]aCa T ][ is not a diagonal matrix, and the above equations are therefore coupled. A
practical approach to overcome this problem is to introduce the concept of proportional
damping in which the damping matrix [ ]C is assumed proportional to either the mass matrix
or the stiffness matrix, or a linear combination of the two.
i.e. Let (3.5.4)
Then [ ] [ ]aCa T ][ = [ ][ ] [ ][ ]aKaaMa TT ][][ + = [ ]rrr M ,2 )( + (3.5.5)
Here we are using the notation that rrr M ,2 )( + is a diagonal matrix whose rth diagonal
element is given by (rrr M ,
2 )+
Substituting equation (3.5.5) into equation (3.5.1), we get the following decoupled equations:
[ ]{ }+pM rr , [ ]{ } [ ]{ } { }02 ,2
, =+ pMpM rrrrrrr (3.5.6)
where )(2 2rrr += (3.5.7)
This gives the uncoupled equations
(3.5.8)022
=++ rrrrrr ppp , r= 1,2,3l,
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which are of the same form as the damped, single degree of freedom system for which the
solution is
(3.5.9)
{ }u may be found using equation (3.5.2a)
Usually it is no significance whether modal damping rrrr M ,2 is based upon [ ]C being
proportional to [ ]M or [ ]K , or a combination of both. From a practical point of view,
r and r are interpreted as properties inherent in the system. The frequency r can be
calculated or measured experimentally, and r is usually obtained by experiment or assumed
from experience. A typical value for mechanical or civil structures is r = 0.05 or less.
3.6 Forced Vibration
Response of a multi-DOF system due to an excitation (dynamic force/s) involves the solution
of a system of simultaneous non-homogeneous differential equations. Let us first study a
simple example and then develop a general procedure.
Example: A 2 DOF undamped system subject to simple harmonic excitation
Figure 3.6.1 Forced Vibration of a 2 D.O.F. System
d2u1/dt
2
F1
State 1 (equilibrium, springs unstressed)
State 2 (at time tduring vibration)T1
T2
T2
u1 u2
m2
k1k2
m1
F1+ T2T1
m1
d2u2/dt
2
T2
m2
F2
F2
( )tBtAep rrrrrrtr rr 22 1sin1cos +=
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The system shown in Figure 3.6.1 is the same spring-mass system in Figure 3.1.1, but is
subjected to two dynamic forces F1, F2. Application of Newtons second law to the masses
yields the following equations, which are similar to equations (3.1.1) and (3.1.2) except for
the addition of the forces F1 and F2.
Applying Newtons second law to mass m1, 11121 umTTF =+
which after rearranging gives: 11112 FumTT =++
Using the constitutive relationships we get:
( ) 11122121 Fumukukk =++ ...(3.6.1)
Similarly for the second mass we get,
2222212 Fumukuk =+ ...(3.6.2)
These equations of motion may be given in the following matrix form,
[ ]{ } [ ]{ } { }FuMuK =+ ...(3.6.3)
The solution for {u} consists of a particular integral {up} and a complimentary function {uc}.
ie. {u} = {up} + {uc} ...(3.6.4)
the complimentary function is the solution to the free vibration problem, and depends on
initial conditions. It can be found as described earlier. The particular integral is a function of
the applied force vector {F}. Inclusion of damping (see next section) leads to acomplimentary function that decays with time, and this function is then called the transient
response. In all practical problems, there will be some damping and with time the response
becomes dominated by the particular integral. For this reason it is called the steady state
response. The remainder of this section deals only with the steady state response. Therefore
for convenience we will use {u} instead of {up}
General steady state response due to harmonic excitation
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If the applied forces are simple harmonic then the following simple method gives the
response.
Let {F} = {Q } sin(t+) ..(3.6.5)
where is the frequency of excitation.Then the particular integral will take the form {u} = {} sin (t+) ..(3.6.6)
Substituting the above equations into equation (3.6.3) and cancelling sin (t+) from both
sides give:
[K]{}-2[M]{} = {Q} ...(3.6.7)
Since p and {Q} are known {p} may be calculated by solving the following equation:
[Kd] {} = {Q} ...(3.6.8)
where [Kd] is a dynamic stiffness matrix given by [Kd] = [K] -
2
[M] ...(3.6.9)
Let us consider the variation of the amplitudes of vibration with frequency. This is called the
response. For the above example let the excitation forces be given by:
{ }
=0
N)(cos tFF
[ ]
+=
2
222
2
2
121
-mk-k
-kmkkKd
So we now need to solve
=
+
0
2
1
2
222
2
2
121 F
u
u
-mk-k
-kmkk
By elimination or substitution we get: FK
mku
d
)(2
221
=
and FK
ku
d
)( 22
=
The results for this problem are given in Figure 3.6.2. in a non-dimensional form. The
displacement amplitudes are non-dimensionalised by dividing them by the static deflection
corresponding to = 0.
It is important to note that replacing in [Kd] to would make |Kd| =0 the frequency
equation for the 2-DOF system. This means that |Kd| =0 if 1= or 2= . This implies that
the response of both masses would tend to infinity as the frequency of the forcing function
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approaches any one of its two natural frequencies, as to be expected since the system is
undamped. What is more interesting to observe is that if2
22 mk = then 1=0. This means
that if the excitation force on m1 had a frequency equal to the natural frequency of a single
DOF spring-mass system consisting ofk2 and m2, then the first mass would be stationary. Thisis the basis of the undamped vibration absorber discussed earlier. We will discuss this in
more detail at another section, but let us turn our attention to the problem of solving the
equations of motion. Although we were able to find the solution by elimination or
substitution, we can use a more convenient procedure that only requires matrix additions and
multiplications as has been done for the free vibration problem.
We would consider the effect of one harmonic excitation at a time and finally superimpose
the results. For a simple harmonic excitation of the form {Fsin (t+)} we can express the
displacement vector {p} in the following form.
{p} = {b sin(t+)}={ p }sin(t+) ...........(3.6.10)
Then {} = [a]{ q }and { p }= [a]-1
{} ...........(3.6.11a,b)
Substituting this into equation (3.6.7) gives: [K][a]{ p } - 2[M][a]{ p } = {Q} ...(3.6.12)
Multiplying by [a]T
gives: [a]T[K][a]{ p } - 2[a]T[M][a]{ p }=[a]T{Q} ......(3.6.13)
Due to orthogonality, the product terms [a]T[K][a] and [a]
T[M][a] are both diagonal matrices.
This means the solution to equation (3.6.13) may be obtained without any further matrix
algebra. We must note here that the modal matrix is not unique and can be scaled. To make
the analysis more convenient, let us consider scaling the modal matrix in the following way.
For the free vibrational analysis, {Q} = {0}, and = results in
[a]T[K][a]{ p } -2[a]T[M][ a]{ p } = {0} (3.6.14)
Since the natural modes are orthogonal with respect to the mass matrix, the modes may be
normalized by setting [N]T[M][N] = [I]. (3.6.15)
Putting this into equation (3.6.14) gives:
[N]T[K][N]{ q } = [ K], (3.6.16)
where [ K] is a diagonal matrix containing the square of natural frequencies, i.e. iiK, = i2
Substituting equations (3.6.15) and (3.6.16) into equation (3.6.13) gives:
[ K]{ p }-2[I]{ p }= [a]T {Q}
These equations are decoupled, and elements of { p } are therefore given by:
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( )22
...2,1
j,i
=
=
i
nj
j
i
Qa
p
(3.6.17)
{} can then be found from equation (3.6.11)
In our example, Q1 = )sin( tF and Q2 =0
Therefore( )221,i )sin(
=
i
i
tFap
22,111,11 papau += = F
aa
)()( 222
2
2,1
22
1
2
1,1
+
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3.7 Forced Damped Vibration
In general a discrete system may possess damping properties and may be subject to any set of
dynamic forces. Assuming the damping to be proportional, we will now reduce the problem to
a set of decoupled differential equations which may be solved in a way similar to that of a
single DOF system. Once the solution to each decoupled equation is found, the system
response may be found by superposition.
The equation of motion is:
(3.7.1)
Substituting { } [ ]{ }pau = into the above equation and premultiplying by [a]T gives
{ } { } { } { }FapaKapaCapaMa TTTT ][]][[][]][[][]][[][ =++
Going through the same algebraic manipulations as in Section 3.4, the following decoupled
equations is obtained:
[ ]{ }+pM rr , [ ]{ } [ ]{ } { }FapMpM rrrrrrrT
,
2
, ][2 =+
where [M] is a diagonal matrix.
The rth
equation of this matrix is:
(3.7.2)
If the modal matrix is normalised by setting the diagonal terms of[ ]M to unity, then equation
(3.7.2) reduces to:
{ } { } { } { }FuKuCuM =++ ][][][
rr
n
i iri
rrrrrrM
Fappp
,
1 ,22 ==++
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(3.7.3)
pr may be found by solving this decoupled equation in the same way as a single DOF system
and onceprare determined, u may be found by using
{u} = [a]{p} (3.7.4)
This may also be written as
(3.7.5)
These derivations are applicable, even for continuous systems if they are numerically
discretised as explained in Chapter 6.
It is worth noting that in finding the actual response it is not always possible, nor necessary to
consider all modes. For example, in a 100 DOF structural system (n=100), if it is possible to
get a good estimate of the response by considering only the first 10 terms. Hence the
summation in equation (3.6.22) may be terminated after 10 terms instead of 100 terms.
==++n
i irirrrrrrFappp
1 ,
22
=
=
n
r
rrjk pau1
,
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3.8. Lagranges Equations of Motion
Application of Newtons second law in finding equations of motion often requires careful
handling of the sign convention for forces, displacements etc. A more convenient technique,
particularly useful in complicated systems, is often used to obtain equations of motion based
on scalar energy terms. This is due to Lagrange, and the general form of the equation is:
k
kkkk
V
q
D
q
T)
q
T(
dt
d=
+
+
Where T is the kinetic energy of the system, which usually takes the form
Vis the potential energy, which normally takes the form: .......2
1 2ii
i
xkV =
D is the dissipation function due to factors such as damping and is of the form,
2
2
1ii
i
xcD =
Qk are the generalised forces corresponding to the generalised co-ordinates qk, and t is the
time variable, k positive integer which is less than or equal to the number of degrees offreedom n (1 k n). The generalised co-ordinates qk must be independent. It should be
noted here, that unlike in the Rayleighs method, V, T, etc. are not the maximum values of
potential, kinetic energies. They are time dependent functions.
Derivation of Lagranges Equation of Motion
According to the principle of virtual work, for a system of masses that is in static equilibrium,
0= ii
i rF ,
where iFis the net force corresponding to the ith
mass, and ir is a small virtual displacement
along the position vector rk that is compatible with the system constraints.
dAlemberts Principle:
From Newtons second law of motion, for a system vibrating freely,
.rmF iii 0=
22
2
1
2
1ii
i
ii
iIT.....or.........xmT
==
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If the system is subject to forces of magnitude iirm then it would be in equilibrium, and the
resulting displacements would be equal in magnitude to the vibratory displacements. This
means we may treat the vibratory system as an equivalent static problem, by adding negative
inertia forces. This is dAlemberts principle. This enables us to obtain an equivalentstatement of virtual work theorem for a vibratory system in the following form:
( ) =i
iiii .rrmF 0 (3.9.1)
Let ri be described by a set of independent generalised co-ordinates qk.
ri = ri ( q1, q2, q3, . qn )
Taking its first variation,
kk k
i
i
rr
= ..(3.8.2)
Substituting this into equation (3.8.1) gives:
( ) =
i
k
i k
iiii .q
q
rrmF 0
Since the generalised co-ordinates qkare independent,
( ) =
i k
iiii .
q
rrmF 0 .(3.8.3)
First let us transform the terms associated with mass as a function of the kinetic energy.
Dividing both sides of equation (3.8.2) by t and taking the limit ast 0gives:
=
k
k
k
ii q
q
rr (3.8.4)
Differentiating the above equation with respect to kq gives:k
i
k
i
q
r
q
r
=
.(3.8.5)
Therefore,
=
k
i
k
i
q
r
dt
d
q
r
dt
d
j
j k
i
j
r
q
= (using the chain rule)
j
j j
k
i
r
q
= (interchanging the order of operations)
k
i
ik q
rr
q
=
=
(using equation (3.8.4))
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S. Ilanko 2007 84
i.e.
=
k
i
k
i
q
r
q
r
dt
d
..(3.8.6)
Therefore,
=
k
ii
k
ii
q
rr
q
r
dt
dr
Multiplying the equation by mi and taking all terms to the left-hand side and adding
k
iii
q
rrm
gives:
k
iii
k
iii
k
iii
k
iii
q
rrm
q
rrm
q
r
dt
drm
q
rrm
=
+
Summing over i results in:
=
+
i k
iii
i k
iii
i k
iii
k
iii
q
rrm
q
rrm
q
r
dt
drm
q
rrm
(3.8.7)
The kinetic energy is of the form:
=
i
iirmT2
2
1 (3.8.8)
Thereforek
ii
i
i
k q
rrm
q
T
=
Differentiating this with respect to tgives:
=
k
ii
i
i
k q
rr
dt
dm
q
T
dt
d
+
=
ki
ik
i
ii
i q
r
dt
d
rq
r
rm
(3.8.8a)
Also from equation (3.8.8)
=
i k
iii
k q
rrm
q
T (3.8.8b)
Substituting equations (3.8.8a) and (3.8.8b) into equation (3.8.7) gives:
=
i k
iii
kkq
rrm
q
T
q
T
dt
d
and substituting equation (3.8.5) for the last term on the right-hand side of the above equation
gives:
=
i k
iii
kkq
rrm
q
T
q
T
dt
d
...(3.8.9)
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This forms a part of the virtual work equation (3.8.3). The remaining terms
i k
ii
q
rF come
from the work done by restoring forces, frictional forces and any external forces and it may be
shown that
+
=
kki k
ii
qD
qV
qrF
+Qk ..(3.8.10)
Substituting equations (3.8.9) and (3.8.10) into equation (3.8.3) gives:
k
kkkk
V
q
D
q
T)
q
T(
dt
d=
+
+
..(3.8.11)
3.9 Application of Lagranges Equations of Motion:
3.9.1. A single degree of freedom system
This is a single degree of freedom system.
Generalised co-ordinate qk= x.
Generalised force Qk= F0 sin(pt)
Kinetic Energy 2
2
1xmT =
Potential Energy 2
2
1kxV=
Dissipation function 221 xcD =
Using the above
kxx
V=
xmx
T
=
( ) xmxmdt
d
x
T
dt
d
==
)(
F0 sin(pt)
k
c
m
x
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0=
x
T
xcx
D
=
Substituting the above equations into Lagranges formula yields the following equation of
motion:
)sin(0 ptFxmxckx =++
For free vibration, F0 = 0 which gives
0=++ xmxckx
And for undamped free vibration,
0=+ xmkx
3.9.2. Simple Pendulum
(g/l)solution,trivial-nonFor
0l-Therefore,
-motion,harmonicsimpleFor
0
0
0)(
formula,sLagrange'Using
Therefore
sin,0as,vibrationamplitudesmallFor
)(sin
)cos1(
0
0
)(
212
)(2
1
,1
system.freedomofdegreesingleaisThis
2
2
2
2
2
2
1
=
=+
=
=+
=+
=
+
+
=
=
=
=
=
=
=
==
g
gl
mglml
VDTT
dt
d
mglV
mglV
mglV
D
T
mlT
dt
d
mlT
lmT
qn
l
m
mg
Length = l
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3.9.3. Two Degree of Freedom System
This is a two degree of freedom system. i.e. n = 2
Let q1 =x1 and q2 =x2
Since there are no forces, Q1 = Q2 = 0.
( )
12
2
2121221211
2
2
22
2
1
11
1
11
1
2
122
2
11
22
2)(
0
)(
0
)(
0
)(2
1
xkxkx
V
xkxkkxkxkxkx
V
x
T
xmx
T
dt
dx
T
xmx
T
dt
d
xmx
T
D
xxkxkV
=
+=+=
=
=
=
=
=
=
+=
Using Lagranges formula:
00)(and
0)(0)(
222212
2222
1122121
1111
=++=
+
+
=++=
+
+
xmxkxkx
V
x
D
x
T
x
T
dt
d
xmxkxkkx
V
x
D
x
T
x
T
dt
d
These are the equations of motion.
k1
m1
x1
k2
m2
x2
( )2222
11
2
2
1
2
1xmxmxmT ii
i
+==
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3.9.4. Torsional vibration of a Geared System
Consider the torsional vibration of
the geared shaft system shown in
the figure. It may be assumed
that the mass of the shafts is
negligible, and the system may be
treated as a discrete system.
Although there are four rotors,
since the rotation of the gears in contact are related by the gear ratio, there are only three
degrees of freedom. Therefore, in choosing the generalised co-ordinates, the rotation of only
one of the two gears should be taken. In this example, let us choose the rotations of rotors 1,2
and 4 as the generalised co-ordinates.
{qk} = {1, 2, 4}
3 is related to 2
sincer22= - r33
or3
223 where,
r
r== (1)
The potential energy is given by:
2
34
2
22
12
1
1 )(2
1)(
2
1
+
=
L
GJ
L
GJV
Using equation (1) this may be rewritten
2
24
2
22
12
1
1 )(2
1)(
2
1 +
+
=
L
GJ
L
GJV .(2)
2442332222112
1
2
1
2
1
2
1 IIIIT +++=
Substituting equation (1) into the above gives:
2
44
2
2
2
3
2
22
2
112
1
2
1
2
1
2
1 IIIIT +++= ..(3)
From equation (2), we have,
)( 211
1
1
=
L
GJV(3a)
1 2
3 4
r3
r2
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)()( 242
212
1
1
2
+
+
=
L
GJ
L
GJV(4a)
and )( 242
2
4
+
=
L
GJV(5a)
From equation (3), we have,
( )1111
1
II
dt
dT
dt
d==
..(3b)
( )23
2
22
2
322
2
)(
IIII
dt
dT
dt
d+=+=
(4b)
and ( )4444
4
IIdt
dT
dt
d==
..(5b)
also, 0421
=
=
=
TTT(3c), (4c), (5c)
Substituting equations (3a), (3b) and (3c) into Lagranges general form of equation of motion
gives:
0)( 11211
1=+
I
L
GJ..(3d)
Similarly, using equations (4a), (4b) and (4c) we can get
0)()()( 232
224
2
2
12
1
1=+++
+
II
L
GJ
L
GJ.(4d)
And from equations (5a,b,c) we have,
0)( 44242
2=++
I
L
GJ..(5d)
Equations (3d), (4d) and (5d) are the Lagranges equations of motion for the geared torsional
vibratory system. This problem may be used to illustrate the importance of ensuring that the
generalised co-ordinates are independent. If we had taken four generalised co-ordinates 1,
2, 3, and 4, then equation (4d) would have become
0)( 22121
1=+
I
L
GJ
which is incorrect. The best way to choose the correct generalised co-ordinates is to see if it
is possible to displace the system along each of the generalised co-ordinates without violating
the constraints of the system and its supports. For example, it is not possible to give a rotation
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of one of the gears without rotating the other, as the constraint equation (1) cannot be
violated. Therefore, only one of the gear rotations (2, 3) may be regarded as a generalised
co-ordinate.
3.9.5. Compound Pendulum on a Roller
The system shown here consists of a
compound pendulum hanging from a
moving platform which rests on two
rollers and is connected to a rigid wall
by means of a spring of stiffness k. It
has three degrees of freedom (n = 3).
It is convenient to take the
displacement of the platform (x) andthe rotations of the two strings (1)
and (2) as the three generalised co-ordinates.
i.e Let q1= x, q2= 1, q3= 2For small amplitude oscillations,
222
)cos1()cos1(
mofanslationverticaltr
2
)cos1(
mofontranslativertical
)(mofvelocityand
mofvelocity
2
22
2
11
2
22
2
11
2211
2
2
11
11
1
22112
111
LLLL
LL
L
L
LLx
Lx
+=+
+=
=
+=
=
( )
( )
2
00
2
22
2
112
2
111
2
11321
2
22
2
11232
2
11121
2
111
2
1mofenergyKinetic
)1....()(2
1energypotentialNet
2:mongravitytodueenergyPotential
2:mongravitytodueenergyPotential
2
1
:springthetodueEnergyPotential
xm
LLgmgLmxkVVV
LLgmV
LgmV
xkV
=
+++=++=
+=
=
=
k
x
L1
L2
m1
m2
1
2
g
Rollers have
mass m, radius
a and radius of
gyration r
m0
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( )
( ) )2....()()(2
22
1EnergyKineticNet
)(2
1mmassofEnergyKineticand
)(2
1mmassofEnergyKinetic
22
12rollersbothofenergyKinetic
2
22112
2
111
2
222
0
2
221122
2
1111
2
22
++
++=
=
=
+=
LLxmLxma
xramxm
LLxm
Lxm
a
xram
From equation (1) we can obtain the following expressions, required in Lagranges formula:
222
2
1121
1
)(
gLmV
gLmmV
kxx
V
=
+=
=
Similarly, from equation (2) we can obtain the following expressions:
motion.ofequationssLagrange'theareThese
0
0)()()(
0)()1(2
:motionofequationsfollowingthegivesformulasLagrange'intosexpressionabovethengSubstituti
0havealsoWe
)(
)()()(
)()1(4
)(
2
2
22121222222
2212121211211121
2221121212
2
0
21
2
2
22121222
2
22121
2
121121
1
2221121212
2
0
=++
=+++++
=+
++
+++
=
=
=
++=
++++=
+
++
++=
LmLLmxLmgLm
LLmLmmxLmmgLmm
LmLmmxmma
rmmkx
TT
x
T
LmLLmxLmT
dt
d
LLmLmmxLmm
T
dt
d
LmLmmxmma
rmm
x
T
dt
d
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3.9.6. Two-Link Robot Mechanism
In the first five examples, the kinetic energy T, was a function of the time derivative of the
generalised co-ordinates, but it was independent of the generalised co-
ordinates )0/( =kdqdT . In this example the kinetic energy is a function of generalised co-
ordinates as well as their time derivatives.
Potential Energy due to gravity:
+= cos
2coscos
2
212
111
LLgm
LgmV .(1)
Potential Energy due to the springs: )2.(..........)(2
1
2
1 22
2
11 += kkV
Net potential energy 21 VVV += ....(3)
Kinetic Energy of the first arm: ,22
1
2
12
11
2
11
+=
LmIT .(4)
Where 1I is the moment of inertia of the first arm about its centroidal axis and is given by:
12
2
111
LmI = (4a)
Similarly, kinetic energy of the second arm is:
The two arm robot mechanism shown in
the figure consists of two rigid arms having
masses m1, m2 and lengths L1,L2
respectively. They are hinged together at
B, and one arm is hinged to a rigid frame at
A. Arm AB is connected to the support by
a coil spring of stiffness k1, and the two
arms are also connected by a coil spring of
stiffness k2 at the hinge B. The system is
under a uniform gravity field g. In this
problem we need to obtain the equations of
motion, for large amplitude rotations.
k1
k2
m1,L1
m2,L2
g
B
A
C
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++
++=
2
21
2
212
2
22 )sin(2
)sin()cos(2
)cos(2
1
2
1
LL
dt
dLL
dt
dmIT . (5)
where
12
2
222
LmI = (5a)
After differentiating with respect to t, equation (5) becomes:
++
++=
2
21
2
212
2
22 )cos(2
)cos(()sin(2
)sin((2
1
2
1
LL
LLmIT
Expanding this, and simplifying yields:
( ) ( )
+
++=
)cos(LL2
L
Lm2
1
I2
1
T 21
2
22
12
2
22 (5b)
Total kinetic energy is T = T1 + T2..(6)
Using equations (1), (2) and (3), we get
sin)2
()(
sinsin2
)(
1211
221
1211
221
gLmLm
kkk
gLmgLm
kkkV
+++=
+++=
..(7a)
From equations (4), (5) and (6)
)sin(2
1212
=
LLm
T.(7b), and
)cos(
2
1)
4( 212
2
12
2
111 +++=
LLmLm
LmI
T..(7c)
Differentiating equation (7c) with respect to t gives:
)cos(2
1)4(
212
2
12
2
111 +++=
LLmLm
Lm
IT
dt
d
))(sin(2
1212
+ LLm .(7d)
Substituting equations (7a)..(7d) into Lagranges formula yields:
)cos(2
1)
4(
)sin(2
1sin)
2()(
212
2
12
2
111
2121211
221
++++
++++
LLmLmLm
I
LLmgLmLm
kkk
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0))(sin(2
1212 =+
LLm .(8)
Similarly, we can operate with respect to .
)(sin2
1222 +=
kgLm
V
(9a)
)sin(2
1212
=
LLm
T(9b) and
++=
)cos(
221
2222
L
LLmI
T(9c)
Differentiating equation (9c) with respect to tgives:
+++=
))(sin()cos(
2211
2222
LLLLm
IT
dt
d9(d)
Substituting equations (9a)..(9d) into Lagranges formula yields:
0))(sin()cos(22
)sin(2
1sin
2
1)(
11222
2
212222
=
+++
+
LLLLm
I
LLmgLmk
.(10)
3.9.3.8. Radial Vibration of a Rotating Spring-Mass System
Another example in which the kinetic energy is a function
of a generalised co-ordinate is a rotating spring-mass
system. Consider the radial vibration of the spring-mass
system that is rotating at a constant angular speed of
radians per second, as shown in the figure. The length of
the unstretched spring is L. The extension of the spring
due to the centripetal acceleration during the steady state
rotation is us. A further radial vibratory motion is denoted
by u.
The potential energy of the spring during vibration is given by:
L us u
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2)(2
1uukV s += ,
and the kinetic energy of the mass is given by:
( )( )22
)(2
1
uuuLmT s
+++=
( ) ( )
.......(2)0)(
.....(1)0i.e.
h.must vanisequationtheofpartstwothe,anyforvalidisequationabovetheSince
0)(
:getwe,separatelywithassociatedtermstheGrouping
0)()(
equation,sLagrange'Using
)(
)(
2
2
22
2
2
=+
=+
=+++
=++++=
+
=
++=
+=
ss
ss
ss
s
s
uLmku
umumku
u
uLmkuumumku
u
uuLmumuukuT
uT
dtd
uV
umu
T
dt
d
uuLmu
T
uuku
V
Equation (1) is the Lagranges equation for the vibratory motion, and may be used to obtain
the natural frequency, while equation (1) gives the steady state radial displacement us.
From equation (2),
=
12
m
k
Lus (3)
And substituting the simple harmonic relationship uu 2= into equation (1) gives:
0)( 22 =+ muku .
For non-trivial solution,
=
2
m
k
..(4)
It should be noted here that (k/m) represents the square of the natural frequency of the non-
rotating system (s) and hence equation (4) may also be written:
)( 22 = s ..(4a)
It may be noted that when s, the natural frequency of the rotating system would
approach zero, thus indicating a state of critical equilibrium. This is also evident from
equation (3) as the steady state displacement us would also be infinite as
2
(k/m). Thistype of instability occurs in shafts also, when the speed of rotation reaches the natural
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frequencies of the shaft in flexural vibration. This is called whirling of shafts, and the speeds
at which whirling occurs are referred to as the whirling speeds.
3.9.8 Two DOF System with Translation and Rotation
2
2
2
1 ))(2/())(2/( GGGG bykaykV ++=
( )22)2/1( IymT G += 2
2
2
1 ))(2/())(2/( GGGG bycaycD ++=
)()( 21 GGGGG
bykayky
V ++=
;
)()( 21 GGGGG
bybkayakV
++=
G
G
ymy
T
dt
d
=
;
G
G
IT
dt
d
=
)()( 21 GGGGG
bycaycy
D
++=
Figure 3.9.8
Gy
k2k1
m,I
G
G
a b
Gy
k2k1
m, I
G
G
a b
c2c2
c1 c1
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)()( 21 GGGGG
bybcayacD
++=
Equations of Motion:
0)()()()( 2121 =++++++ GGGGGGGGG ymbycaycbykayk
Rearranging gives:
0)()()()( 12211221 =++++++ GGGGG ymacbcyccakbkykk (1)
0)()()()( 2121 =+++++ GGGGGGGGG Ibybcayacbybkayak
Rearranging gives:
0)()()()( 222
112
2
2
2
112 =++++++ GGGGG Ibcacyacbcbkakyakbk (2)
If the system is undamped, these reduce to:
0)()( 122
21 =++ GG akbkymkk
0)()( 2222
112 =++ GG Ibkakyakbk
This may be written as:
=
+
+
0
0
)()(
)()(22
2
2
112
12
2
21
G
Gy
Ibkakakbk
akbkmkk
For N.T.S., 0)()(
)()(
222
2112
12
2
21=
+
+
Ibkakakbk
akbkmkk
i.e. 0)())(( 21222
2
2
1
2
21 =++ akbkIbkakmkk is the frequency equation the roots
of which give the first and second natural frequencies.
Note: If )( 12 akbk =0 then the system is decoupled!
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m
l
l/2
Figure 3.8.2.9
k
mg
c
3.9.9 Inverse Pendulum
cos222
12
lmg
lkV +
=
22
2
222
62122
1
mllm
mlT =
+=
2
22
1
=
lcD
Applying Lagranges formula
0)( =
+
+
kkkk q
V
q
D
q
T
q
T
dt
d
,
the equation of motion is:
0sin244
03
222
=++ mglklclml
For small amplitude motion, as sin this reduces to:
02443
222
=
++
mglklclml
Note, if lmgk /2 the system is unstable!
Substituting lmgk /10= and dividing by l2
gives the following simplified equation.
0243
=++ lmgcm
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From the formula supplied, lgm
lmgn /6
)3/(
)/2(== = 2.45 rad/s
Damping ratio 031.0
)3/(28
/2.0)2/(
2===
lgm
lgmkmc
2031.016 =d = 2.45 rad/s
t
ddnetBtAt
+= )cossin()(
2020)0( == B radians
t
ddd
t
ddnnn etBtAetBtAt
++= )sincos()cossin()(
dn
dn
BA
AB
/
0)()(0)0(
=
=+=
= 0.62 radians
tettt
076.0)45.2cos2045.2sin62.0()( +=
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3.10 The Energy Terms in the Rayleigh-Ritz Method and Lagranges Equations
It is important to know the differences between the energy terms used in the Rayleigh-Ritz
method and the energy terms in Lagranges equations of motion. In the Rayleigh-Ritz
method, which is based on the principle of conservation of energy, the maximum kinetic
energy associated with vibration and the corresponding maximum total potential energy terms
are used. In Lagranges equations, the energy terms used are time dependent functions.
Another difference comes in the consideration of some potential energy terms due to external
forces. This may be illustrated by considering the static deflection of a spring mass system.
Static case may be treated as a special dynamic case with the applied force in the form Q0
cos(pt) where the excitation frequency p = 0. Both Lagranges equations of motion and the
Rayleigh-Ritz procedure are applicable in static equilibrium problems too.
In this single degree of freedom system, the only generalised co-ordinate (qk) isx.
For Lagranges equation of motion, the potential energy may be taken as
V= (1/2) kx2
and the kinetic energy
T= (1/2)m(dx/dt)2.
The generalised force Q1 = {Q0 cos(pt)}.
Applying Lagranges equation we get the following equation of motion:
)cos(0 ptQxmkx =+
For the steady state solution, we may usex=Xcosptwhich gives
kx-mp2x = Q0 cos(pt)
And since p=0 this reduces to kX= Q0 andX= Q0/kas expected.
In using the Rayleigh-Ritz procedure for static analysis, the total potential energy should
include the potential of the applied force.
Vm = (1/2) kX2
- Q0X
Tm = 0
Rayleigh-Ritz equation is Vm/X=0 gives kX-Q0 =0.
ThereforeX=Q0/k
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What is important here is that the total potential energy used in the Rayleigh-Ritz method
includes the potential of the applied load where as in Lagranges equation the applied load
comes in as a generalised force.
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mb, Ib
mb, Ib
Multi DOF Systems: Problems
1. A rigid rod of length l2 and mass m is hinged to another
rigid but light (of negligible mass) rod of length l1 which
is hinged to a fixed point O as shown in Figure 1. Thesystem is under gravity field g.
a) Derive the linearized equations of motion of this
system giving the mass and stiffness matrices.
b) Ifl2 = 6 l1 = 6 l, calculate the natural frequencies
and eigenvectors (modes) of the system.
Note: The moment of inertia of a rod of mass m and
length l, about a centroidal axis perpendicular to its
axis is given by12
2mlI=
2. Two rigid bars are hinged at points O and P, and connected by means of an elastic spring of
stiffness k as shown in Figure 2.
They also carry four concentrated
masses at their ends. The bars are
identical having a mass mb
(which is equal to 0.2m) and a
moment of inertia of Ib (which is
0.2mL2
/12) about their centroidalaxis perpendicular to the plane of
oscillation. This system is in
static equilibrium under gravity
field g.
Figure 1
1
2
l1
l2, m
g
O
P
O
g
0.2L
0.3L
0.5L
0.2L
0.7L
0.1L
k
m
m
2 m
2.5 m
Figure 2
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3. (a) Derive the equations of motion for the
torsional vibration of the system in
Figure 3a, neglecting the mass of the
shafts (treating the system as discrete).
Do not solve the equations. Note r3 =
r2.(b)If 41 II = = 32 II = = Iand 21 kk = = k,
write down the expressions for any
two natural frequencies of this system.
(c)
Would your answers to the above
questions be any different, if the
system had a geometric arrangement
as shown in Figure 3b.
4. Figure 4 is a schematic
representation of a
power transmission
system after the
sudden application of
a brake on the left end
of the top shaft.The relevant moments
of inertia of the gears
and the rotor are:
I2 = 1.2 kgm2;
I3 = 0.15 kgm2; and
I4 = 3 kgm2
(a.)Obtain the equations of motion for the torsional vibration of the above system
neglecting the mass of the shafts (treating the system as discrete).
(b.)Show that the natural frequencies and modes of the above system are as follows:
1 = 18.34 rad/s, and for the first mode, 2/ 4 = 0.374
2 = 115.0 rad/s, and for the second mode, 2/ 4 = 4.458
2 1
r3 = r2
r2
43k2
k1
I2I1
I4
I3
Figure 3a
1
2
r2
k1
I2 I1
r3 = r2
34k2
I3
I4Figure 3b
I2
k1=6 kNm/rad
I3I4k2= 4 kNm/rad
r
Figure 4
4 (t)
2 (t)
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(c.)Obtain an expression for the subsequent torsional oscillation, 2 (t) of I2 for the
following initial conditions:
2(0) = 4(0) = 0 and rad/s120-)0(andrad/s60)0( 42 ==
5. The natural frequencies and some values of a modal matrix for the three-degree of freedom
spring-mass system shown in Figure 5 are given below.
Modal matrix:
11?
?0?
111
The natural frequencies are:
82.20
00.10
0
rad/sec.
(a)Complete the missing values in the modal matrix.
(b)Obtain an expression for the displacement of the second mass as a function of time,
for the following initial conditions:
Displacement at t = 0, are
)0(u
)0(u
)0(u
3
2
1
= mm
7
2
5
and
zero initial velocity. i.e.
)0(u
)0(u
)0(u
3
2
1
=
0
0
0
(c)Comment on how these results could have been obtained more conveniently by
making use of symmetry.
6. The two degree of freedom system shown in Figure 6 has the
following natural frequencies and modal matrix:
=1 5.41 rad/s, =2 13.07 rad/s; [a] =
462.01913.0
1913.0462.0
The first and second degrees of freedom correspond to
translations in x and y directions respectively.
(a) If the mass matrix [M] is
40
04kg, find [ ]M .
(b) If the 4 kg mass is given an initial displacement of 5 mm in
thex direction, and then released, find an expression for the subsequent displacement
of the mass in they direction. i.e.{ }
=0
5)0(u mm, and { } { }0)0( =u , ?)()(2 == tytu
3 kg5 kg 5 kg
u1 u2 u3
Figure 5
Figure 6
m= 4 kg
y
k
kx
45
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7. Derive the equations of
motion for the system
shown in Figure 7. This
consists of two rigid
bodies of mass m0,
connected by a spring ofstiffness k, and
supported on rollers
resting on a non-slip
surface. One of the
bodies is partly
restrained by a light
elastic cable of lengthL.
The cable is under a
tension T and at
equilibrium state it is
perpendicular to thedirection of motion of
the bodies. All
displacements may be assumed to be small.
8. In the 2-DOF oscillator shown in Figure 8, the spring is unextended for 021 == . For
the small vibration studied here, the spring may be assumed to remain horizontal during
the motion. The stiffness coefficient of the
spring is k.
(a.)Give the expressions for the kinetic
energy Tand the potential energy Vin
terms of 2121 ,,, .
(b.)From these energy expressions, find
the elements of the mass and stiffness
matrices (M and K) describing the
linearised system.
(c.)For, mmm == 21 and lll == 21 , find
the natural frequencies and the modes (intermediate steps may be skipped). Hint: You
may make use of the symmetry of the system.
k
L
T
m0m0
x2
Figure 7
All rollers have
mass m, radius
a and radius of
gyration r
m1
1
l1
m2
l2
s2
Figure 8
k
