Dynamics of machinery(mgu)

B.TECH. DEGREE COURSE
SCHEME AND SYLLABUS
(2002-03 ADMISSION ONWARDS)
MAHATMA GANDHI UNIVERSITY
KOTTAYAM, KERALA
DYNAMICS OF MACHINERY
Module 1
Balancing: - Balancing of rotating masses, static balancing and dynamic balancing,
Balancing of several masses rotating in same plane, Balancing of several masses
rotating in several planes, Balancing machines.
Balancing of reciprocating masses: - The effect of inertia force of the reciprocating
mass on the engine. Partial primary balance. Partial balancing of locomotive,
Hammer blow, Variation of tractive effort, Swaying couple. Coupled locomotives,
Balancing of multi cylinder inline engines, v-engines, Radial engines, Direct and
Reverse cranks
Module 2
Vibrations: - Definitions, simple harmonic motion. Single degree freedom systems:
Undamped free vibrations: - Equations of motion Natural frequency, Energy
method, Equilibrium methods, Rayleigh’s methods, Equivalent stiffness of spring
combinations.
Damped free vibrations: - Viscous damping, Free vibrations with viscous damping,
over-damped system, critically damped system, under-damped system, Logarithmic
decrement, viscous dampers, coulomb damping.
Forced Vibrations: - Forced harmonic excitation Rotating unbalance, Reciprocating
unbalance. Energy dissipated by damping, vibration isolation and Transmissibility.
Vibration measuring instruments.
Module 3
Two degree freedom systems: - Principal modes of vibration, Rectilinear and
angular modes, systems with damping, vibration absorbers, centrifugal pendulum
damper, dry friction damper, untuned viscous damper.
Multi-degree of freedom system: - Free vibrations, equations of motion, Influence
coefficients method, lumped mass and distributed mass systems, Stodola method,
Dunkerly’s method, Holzer’s method, Matrix iteration method.

Torsional Vibrations: - Torsionally equivalent shaft, torsional vibration of two-
rotor, three-rotor, and geared systems.
Module 4
Critical speeds of shafts: - Critical speed of a light shaft having a single disc
without damping. Critical speeds of a light cantilever shaft with a large heavy disc at
its end.
Transient vibration: - Laplace transformation, response to an impulsive input,
response to a step input, response to a pulse input, phase plane method, shock
spectrum.
Non-linear vibrations: - Phase plane, undamped free vibration with non-linear
spring forces, hard spring, soft spring, Perturbation method, Forced vibration with
nonlinear forces, Duffings equation, self excited vibrations.
Module 5
Noise control: - Sound propagation, decibels, acceptance noise levels, Air columns,
Doppler effect, acoustic measurements, microphones and loud speakers, Recording
and reproduction of sound, fourier’s theorem and musical scale, Acoustics of
buildings, Acoustic impedence filters and human ear.

MODULE 1
Introduction
The high speed of engines and other machines is a common phenomenon
now-a-days. It is, therefore, very essential that all the rotating and reciprocating parts
should be completely balanced as far as possible. If these parts are not properly
balanced, the dynamic forces are set up. These forces not only increase the loads on
bearings and stresses in the various members, but also produce unpleasant and even
dangerous vibrations. In this chapter we shall discuss the balancing of unbalanced
forces caused by rotating masses, in order to minimize pressure on the main bearings
when an engine is running.
BALANCING OF ROTATING MASSES
Static and dynamic unbalance: A rotor can in general have two types of unbalance
viz., “static” and “dynamic”. It is of course to be appreciated that practical systems
will all have dynamic unbalance only and considering it as static unbalance is a
“good-enough” approximation for some cases.
Fig 1.1 A thin Rotor Disc - Illustration of Static Unbalance
If the rotor is thin enough (longitudinally) as shown in Fig. 1.1 the unbalance force
can be assumed to be confined to one plane (the plane of the disc). Such a case is
known as “static” unbalance. Such a system when mounted on a knife-edge as shown
in Fig. 1 will always come to rest in one position only – where the centre of gravity
comes vertically below the knife-edge point. Thus in order to “balance out”, all we
need to do is to attach an appropriate “balancing mass” exactly 1800
opposite to this
position.

Fig 1.2 Thin Rotor on a knife edge - Illustration of Static Unbalance
Thus we first mount the disc on a knife edge and allow it to freely oscillate. Mark the
position when it comes to rest. Choose a radial location (1800
opposite to this
position) where we can conveniently attach a balancing mass. By trial and error the
balancing mass can be found out. When perfectly balanced, the disc will exhibit no
particular preferred position of rest. Also when the disc is driven to rotate by a motor
etc., there will be no centrifugal forces felt on the system (for example, at the
bearings). Thus the condition for static balance is simply that the effective centre of
gravity lie on the axis.
Fig 1.3 A Case of Dynamic Unbalance
Consider the rotor shown in Fig. 1.3. It is easily observed that mass distribution
cannot be approximately confined to just one plane. So unbalance masses and hence
unbalance forces are in general present all along the length of the rotor. Such a case
is known as “dynamic unbalance”.
The fundamental difference between static and dynamic unbalance needs to be
clearly appreciated.
When a rotor as shown in Fig.1. 3 is mounted on a knife edge and allowed to
oscillate freely, it too may come to rest in one particular position all the time – the
position corresponding to the resultant unbalance mass (centre of gravity) vertically
below the knife edge. We could, like earlier, mount an appropriate balance mass
exactly 1800
opposite to this position. It would then have no preferred position of rest
when mounted on a knife-edge. Thus effective center of gravity lies on the axis.

Fig1.4 Example of unbalance masses leading to unbalance force that for a resultant
couple because of axial.
However, when mounted in bearings and driven by a motor etc., it could still wobble
due to the unbalanced moments of these forces as shown in Fig1.4. This becomes
apparent only when the rotor is driven to rotate and hence the name “dynamic
unbalance”. Thus it is not, in general, sufficient to do just static balance but
achieving good dynamic balance is more difficult. We will discuss one important
method of achieving dynamic balance in the next lecture.
Two-plane balancing technique
Consider the turbo-machine rotor that was discussed earlier wherein each stage
contains several blades around the circumference of a disk. Eventhough typically
each stage is balanced in itself to the extent possible, it has a likely net unbalance.
When the rotor is set to spin, it will cause dynamic forces and moments on the
bearings that support the shaft. Therefore it is of interest to achieve “good balance”
of this shaft so that the fluctuating forces on the bearings are reduced. Conceptually
our strategy can be simply stated as follows:
Step 1: Consider the shaft supported on its bearings. For each unbalance mass, there
will be a centrifugal force set-up when the rotor spins at some speed . This would
cause some reactions at the supports. Estimate these support reactions that would
come onto the bearings.
Step 2: Estimate the balancing mass that needs to be placed in the plane of bearings,
to counter this reaction force due to unbalance mass.
Repeat steps 1 and 2 for each unbalance mass in the system and each time add the
balancing masses obtained in step 2 vectorially to determine the resultant balancing
mass required.
Let us now understand the details of the technique mentioned earlier. Firstly we
choose to place “balancing or correcting” masses on the shaft (rotating along with the
shaft) to counter-act the unbalance forces. We understand that this is to be done on

the rotor on site, perhaps during a maintenance period. From the point of view of
accessibility, we therefore choose the balancing masses to be kept near the bearings.
Figure 1.5 Two plane balancing technique
The calculations proceed as shown in Fig1.5. For an unbalance mass mi situated at an
angular location in a plane at an axial distance from the left end bearing and
rotating at a radius as shown in the figure, the unbalance force is . It is
resolved into X and Y components as shown in the figure. These forces are
represented by EQUIVALENT FORCES in the balancing planes ( shown in blue
). These forces can be readily calculated (based on calculations similar to
those involved in finding support reactions for a simply supported beam). In order to
counterbalance this force, we need to place a balancing mass at a radius in the
balancing plane such that it creates an equal and opposite force.
Now we need to repeat the calculations for ALL the unbalance masses mi (i =
1,2,3,…..) and find the resultant equivalent force in the balancing plane as shown in
blue in Fig. 2.3.1. This resultant force is balanced out by placing a suitable balancing
mass creating an equal and opposite force (shown in red).
Since all the masses are rotating at the same speed along with the shaft, we can
drop in our calculations – i.e., a rotor balanced at one speed will remain balanced
at all speeds or in other words, our technique of balancing is independent of speed.
We will review this towards the end of the lecture.
While these calculations can be done in any manner perceived to be convenient, a
tabular form is commonly employed to organize the computations. While doing this,
it is also common practice to include the two balancing masses in the balancing
planes as indicated in the table.

Table 1.1 Tabular form of organizing the computations for two-plane balancing
technique
Sr. No
Cos(
)
Sin(
)
Cos(
)
Sin(
)
1
2
3
....
…..
Balancing
Plane 1
0
Balancing
Plane 2
L
TOTAL FORCES 0 0 0 0
It is observed in Table 1 that the balancing masses and their locations (radial as well
as angular) are unknowns while the location of the balancing plane itself is treated as
a known (any accessible location near the bearings etc). The resultant total forces and
moments must sum up to ZERO and therefore we have four equations but six
unknowns. Thus any two of the six unknowns can be freely chosen and the other four
determined from the computations given in the table. This method of balancing is
known as the “two-plane balancing technique” since balancing masses are kept in
two planes.
Balancing of reciprocating masses:
Figure 1.6 Slider-crank Mechanism of IC Engine

A typical crank-slider mechanism as used in an IC Engine is shown in Fig 1.6 It
essentially consists of four different parts viz., frame(i.e., cylinder) ,crank
,connecting rod and reciprocating piston. The frame is supposedly stationary; crank
is undergoing purely rotary motion while the piston undergoes to-and-fro rectilinear
motion. The connecting rod undergoes complex motion – its one end is connected to
the crank (undergoing pure rotation) and the other end is connected to the piston
(undergoing pure translation).
We know that the inertia forces are given by mass times acceleration and we shall
now estimate the inertia forces (shaking forces and moments) due to the moving
parts on the frame (cylinder block).
CONNECTING ROD
One end of the connecting rod is circling while the other end is reciprocating and any
point in between moves in an ellipse. It is conceivable that we derive a general
expression for the acceleration of any point on the connecting rod and hence estimate
the inertia forces due to an elemental mass associated with that point. Integration
over the whole length of the connecting rod yields the total inertia force due to the
entire connecting rod. Instead we try to arrive at a simplified model of the connecting
rod by replacing it with a “dynamically equivalent link” as shown in Fig 1.7
Figure 1.7 Dynamically Equivalent link for a connecting rod .
In order that the two links are dynamically equivalent, it is necessary that:
Total mass be the same for both the links
Distribution of the mass be also same i.e., location of CG must be same and the
mass moment of inertia also must be same.
Thus we can write three conditions:

For convenience we would like the equivalent link lumped masses to be located at
the big and small end of the original connecting rod and if its center of mass (G)
location is to remain same as that of original rod, distances AG and GB are fixed.
Given the mass m and mass moment of inertia of the original connecting rod, the
problem of finding dynamically equivalent link is to determine , and .
An approximate equivalent link can be found by simply ignoring and treating
just the two lumped masses and connected by a mass-less link as the
equivalent of original connecting rod. In such a case we take:
= (GB)/L
= (AG)/L
Thus the connecting rod is replaced by two masses at either end (pin joints A and B)
of the original rod. rotates along with the crank while purely translates along
with the piston. It is for this reason that we proposed use of crank's effective rotating
mass located at pin A, which can now be simply added up to part of connecting rod's
mass.
On the shop floor , can be immediately determined by mounting the existing
connecting rod on two weighing balances located at A and B respectively. The
readings of the two balance give and directly
Dynamic Model of a single cylinder IC Engine Mechanism
Figure 1.8 Dynamic Model of slider-crank Mechanism
Based on our discussion thus far, we can arrive at a simplified model of the crank-
slider mechanism for the purpose of our dynamic analysis as shown in Fig. 1.8 Thus
we have either purely rotating masses or purely translating masses and these are
given by:

where the first term in the rotating masses is due to the effective crank mass at pin A
and the second term is due to the part of equivalent connecting rod mass located at
pin A. Similarly the first term in reciprocating masses is due to the mass of the piston
and the second is due to the part of equivalent connecting rod mass located at pin B.
There are inertia forces due to .
The inertia forces due to can be nullified by placing appropriate balancing
masses. Thus the effective force transmitted to the frame due to rotating masses can
ideally be made zero.
Figure 1.9Counter balancing of rotating masses
Figure 1.10 Opposed position configuration
However it is not so straight forward to make the unbalanced forces due to
reciprocating masses vanish completely. As given in Equation and depicted in Fig.
1.9 there are components of the force which are at the rotational speed and those at
twice this speed. It is conceivable to use a configuration as shown in Fig.1.10 to
completely balance out these forces but the mechanism becomes too bulky. Thus a
single cylinder engine is inherently unbalanced. .
Partial Balancing of Locomotives
The locomotives, usually, have two cylinders with cranks placed at right
angles to each other in order to have uniformity in turning moment diagram. The two
cylinder locomotives may be classified as :

1. Inside cylinder locomotives ; and
2. Outside cylinder locomotives.
In the inside cylinder locomotives, the two cylinders are placed in between
the planes of two driving wheels whereas in the outside cylinder locomotives, the
two cylinders are placed outside the driving wheels, one on each side of the driving
wheel. The locomotives may be
(a) Single or uncoupled locomotives ; and (b) Coupled locomotives.
A single or uncoupled locomotive is one, in which the effort is transmitted
to one pair of the wheels only ; whereas in coupled locomotives, the driving wheels
are connected to the leading and trailing wheel by an outside coupling rod.
Effect of Partial Balancing of Reciprocating Parts of Two Cylinder Locomotives
We have discussed in the previous article that the reciprocating parts are
only partially lanced. Due to this partial balancing of the reciprocating parts, there is
an unbalanced primary force along the line of stroke and also an unbalanced primary
force perpendicular to the line of stroke. The effect of an unbalanced primary force
along the line of stroke is to produce;
1. Variation in tractive force along the line of stroke ; and 2. Swaying
couple.
The effect of an unbalanced primary force perpendicular to the line of stroke
is to produce variation in pressure on the rails, which results in hammering action on
the rails. The maximum magnitude of the unbalanced force along the perpendicular
to the line of stroke is known as a jammer blow. We shall now discuss the effects of
an unbalanced primary force in the following articles.
Variation of Tractive Force
The resultant unbalanced force due to the two cylinders, along the line of
stroke, is known as tractive force. Let the crank for the first cylinder be inclined at an
angle  with the line of stroke. Since the crank for the second cylinder is at right
angle to the first crank, therefore the angle of inclination for the second crank will be
(90° +  ).

Let m = Mass of the reciprocating parts per cylinder, and
c = Fraction of the reciprocating parts to be balanced.
We know that unbalanced force along the line of stroke for cylinder 1
  2
1 c m. .rcos   
Similarly, unbalanced force along the line of stroke for cylinder 2,
   2 o
1 c m. .rcos 90     
 As per definition, the tractive force,
TF = Resultant unbalanced force along the line of stroke
  2
1 c m. .rcos   
   2 o
1 c m. .rcos 90    
   2
1 c m. .r cos sin     
The tractive force is maximum or minimum when (cos  -sin  ) is
maximum or minimum. For (cos  -sin  ) to be maximum or minimum,
 
d
cos sin 0
d
   

or sin cos 0     or sin cos   
 tan 1   or =  135o
or 315o
Thus, the tractive force is maximum or minnimum when  = 135o
or 315o.
 Maximum and minimum value of the tractive force or the variation in
tractive force
     2 o o 2
1 c m. .r cos135 sin135 2 1 c m. .r        
Swaying Couple
The unbalanced forces along the line of stroke for the two cylinders
constitute a couple.This couple has swaying effect about a vertical axis, and tends to

sway the engine alternate in clockwise and anticlockwise directions. Hence the
couple is known as swaying couple.
Let a = Distance between the centre lines of the two cylinders.
 Swaying couple
  2 a
1 c m. .rcos
2
   
   2 o a
1 c m. .rcos 90
2
    
   2 a
1 c m. .r cos sin
2
      
The swaying couple is maximum or minimum when (cos  + sin  ) to be
maximum or minimum.
 
d
cos sin 0
d
   

or sin cos 0     or sin cos    
 tan 1  or =  45o
or 225o
Thus, the swaying couple is maximum or minimum when  = 45o
or 225o.
 Maximum and minimum value of the swaying couple
     2 o o 2a a
1 c m. .r cos45 sin45 1 c m. .r
2 2
         
Note: In order to reduce the magnitude of the swaying couple, revolving
balancing masses are introduced. But, as discussed in the previous article, the
revolving balancing masses cause unbalanced forces to act at right angles to the line
of stroke. These forces vary the downward pressure of the wheels on the rails and
cause oscillation of the locomotive in a vertical plane about a horizontal axis. Since a
swaying couple is more. harmful than an oscillating couple, therefore a value of ‘c’
from 2/3 to 3/4, in two-cylinder locomotives with two pairs of coupled wheels, is
usually used. But in large four cylinder locomotives with three or more pairs of
coupled wheels, the value of ‘c’ is taken as 2/5.

Hammer Blow
We have already discussed that the maximum magnitude of the unbalanced
force along the perpendicular to the line of stroke is known as hammer blow.
We know that the unbalanced force along the perpendicular to the line of
stroke due to the balancing mass B, at a radius b, in order to balance reciprocating
parts only is B. 2
 .b sin  . This force will be maximum when sin  is unity, i.e.
when  = 90° or 270°.
 Hammer blow = B. 2
 .b (Substituting sin  = 1)
The effect of hammer blow is to cause the variation in pressure between the
wheel and the rail.
Let P be the downward pressure on the rails (or static wheel load).
 Net pressure between the wheel and the rail
2
P B. .b  
If  2
P B. .b  is negative, then the wheel will be lifted from the rails.
Therefore the limiting condition in order that the wheel does not lift from the rails is
given by
2
P B. .b 
and the permissible value of the angular speed,
P
Bb
 
Balancing of Coupled Locomotives
The uncoupled locomotives as discussed in the previous article, are obsolete
now-a-days. In a coupled locomotive, the driving wheels are connected to the leading
and trailing wheels by an outside coupling rod. By such an arrangement, a greater
portion of die engine mass is utilised by tractive purposes. In coupled locomotives,

the coupling rod cranks are placed diametrically opposite to the adjacent main cranks
(i.e. driving cranks). The coupling rods together with cranks and pins may be .treated
as rotating masses and completely balanced by masses in the respective wheels. Thus
in a coupled engine, the rotating and reciprocating masses must be treated separately
and the balanced masses for the two systems are suitably combined in the wheel.
It may be noted that the variation of pressure between the wheel and the rail
(i.e., hammer blow) may be reduced by equal distribution of balanced mass (B)
between the driving, leading and trailing wheels respectively.
Balancing of Primary Forces of Multi-cylinder In-line Engines
The multi-cylinder engines with the cylinder centre lines in the same plane
and on the same side of the centre line of the crankshaft, are known as In-line
engines. The following two conditions must be satisfied in order to give the primary
balance of the reciprocating parts of a multi-cylinder engine:
Figure 1.11 Typical Inline engine
1. The algebraic sum of the primary forces must be equal to zero. In
other words, the primary force polygon must *close; and
2. The algebraic sum of the couples about any point in the plane of the
primary forces must be equal to zero. In other words, the primary
couple polygon must close.
We have already discussed, that the primary unbalanced force due to the

reciprocating masses is equal to the component, parallel to the line of stroke, of the
centrifugal force produced by the equal mass placed at the crankpin and revolving
with it. Therefore, in order to give the primary balance of the reciprocating parts of a
multi-cylinder engine, it is convenient to imagine the reciprocating masses to be
transferred to their respective crankpins and to treat the problem as one of revolving
masses.
Notes: 1. For a two cylinder engine with cranks 180o
, condition (1) may be
satisfied, but this will result in an unbalanced couple. Thus the above method of
primary balancing cannot be applied in this case.
2. For a three cylinder engine with cranks at 120o
and if the reciprocating
masses per cylinder are same, then condition (1) will be satisfied because the forces
may be represented by the sides of an equilateral triangle. However, by taking a
reference plan through one of the cylinder centre lines, two couples with non-parallel
axes will remain and these cannot vanish vectorially. Hence the above method of
balancing fails in this case also.
3. For a four cylinder engine, similar reasoning will show that complete
primary balance is possible and it follows that
‘For a multi-cylinder engine, the primary forces may be completely
balanced by suitably arranging the crank angles, provided that the number of cranks
are not less than four'.
Balancing of Secondary Forces of Multi-cylinder In-line Engines
When the connecting rod is not too long (i.e. when the obliquity of the
connecting rod is considered), then the secondary disturbing force due to the
reciprocating mass arises.
We have the secondary force,
2
S
cos2
F m. .r
n

  
This expression may be written as
 2
S
r
F m. 2 cos2
4n
    

As in case of primary forces, the secondary forces may be considered to be
equivalent to be component, parallel to the line of stroke, of the centrifugal force
produced by an equal mass placed at the imaginary crank of length r/4n and
revolving at twice the speed of the actual crank.Thus, in multi-cylinder in-line
engines, each imaginary secondary crank with a mass attached to the crankpin s
inclined to the line of stroke at twice the angle of the actual crank. The values of the
secondary forces and couples nay be obtained by considering the revolving mass.
This is done in the similar way as discussed for primary forces, the following two
conditions must be satisfied in order to give a complete secondary balance of an
engine :
1. The algebraic sum of the secondary forces must be equal to zero. In other
words, die secondary force polygon must close, and
2. The algebraic sum of the couples about any point in the plane of the
secondary forces must be equal to zero. In other words, the secondary
couple polygon must close.
Balancing of Radial Engines (Direct and Reverse Cranks Method)
The method of direct and reverse cranks is used in balancing of radial or V-
engines, in which the connecting rods are connected to a common crank. Since the
plane of rotation of the various cranks (in radial or V-engines) is same, therefore
there is no unbalanced primary or secondary couple.
Fig. 1.12 Typical Radial Engine (Not to scale)
Consider a reciprocating engine mechanism. Let the crank known as the
direct crank) rotates uniformly at  radians per second in a clockwise direction. Let

at any instant the crank makes an angle  with the line of stroke. The indirect or
reverse crank is the image of the direct crank when seen through the mirror placed at
the line of stroke. A little consideration will show that when the direct crank revolves
in a clockwise direction, the reverse crank will revolve in the anticlockwise direction.
We shall now discuss the primary and secondary forces due to the mass of the
reciprocating parts.
Considering the primary forces
We have already discussed that primary force is 2
m. .rcos  . This force is
equal to the component of the centrifugal force along the line of stroke, produced by
a mass placed at the crank pin. Now let us suppose that the mass of the reciprocating
parts is divided into two parts, each equal to m/2.
It is assumed that m/2 is fixed at the direct crank (termed as primary direct
crank) pin and m/2 at the reverse crank (termed as primary reverse crank) pin . We
know that the centrifugal force acting on the primary direct and reverse crank
2m
.r
2
 
Component of the centrifugal force acting on the primary direct crank
2m
.rcos
2
  
and, the component of the centrifugal force acting on the primary reverse crank
2m
.rcos
2
  
Total component of the centrifugal force along the line of stroke
2 2m
2 .rcos m. .rcos
2
       = Primary force, FP
Hence, for primary effects, the mass m of the reciprocating parts may be
replaced by two masses each of magnitude m/2.

MODULE 2
Vibration (Oscillation)
Any motion which repeats itself after an interval of time is called vibration.
Eg: Swinging of simple pendulum.
Causes of vibration
 Unbalanced forces in the machine.
 External excitations applied on the system.
 Elastic nature of the system
 Winds, Earthquakes etc.
Effect of Vibration
Produces unwanted noise, high stresses, wear, poor reliability and premature
failure of one or more of the parts.
Inspite of these harmful effects, it is used in musical instruments, vibrating
conveyors etc.
Elimination of Vibrations
Using shock absorbers
Using vibration absorbers
Resting the machinery on proper type of isolation.
Definitions
Frequency
Number of cycles per unit time.
Natural Frequency (fn)
Frequency of free vibration of the system.
Expressed in Hz or rad/sec.
Amplitude
The maximum displacement of a vibrating body from its equilibrium
position.

Resonance
When the frequency of external excitation is equal to the natural frequency
of a vibrating body, the amplitude of vibration becomes excessively large. This
concept is known as resonance.
Periodic Motion
A motion which repeats itself after equal intervals of time.
Time Period
Time taken to complete one cycle.
Fundamental mode of vibration
The fundamental mode of vibration of a system is the mode having the lowest natural
frequency.
Degree of Freedom
The minimum number of independent co-ordinates required to specify the
motion of system at any instant is known as degrees of freedom.
It is equal to the number of independent displacements that are possible.
This number varies from zero to infinity.
Zero degree of freedom
The body at rest is said to have zero degree of freedom.
Single Degree of freedom
Here there is only one independent co-ordinate to specify the configuration.
Eg: A mass supported by a spring.
Two degree of freedom
There are two independent co-ordinates to specify the configuration.
Eg: Springs supported Rigid mass. (It can move in the direction of springs
and also have angular motion in one plane)
Multi degrees of freedom
A cantilever beam has inifinite degrees of freedom.

Types of Vibration
1. Free (Natural) Vibration
eg: Simple pendulum.
After disturbing the system the external excitation is removed, then the
system vibrates on its own. This type of vibration is known as free vibrations.
3 types:-
a) Longitudinal vibrations.
When the particles of the shaft or disc move parallel to the axis of the shaft,
then the vibrations are known as longitudinal vibrations. In this case the shaft is
elongated and shortened alternately and thus the tensile and compressive stresses are
induced alternately in the shaft.
b) Transverse vibrations
When the particles of the shaft move approximately perpendicular to the
axis of the shaft, then the vibrations are known as transverse vibrations. In this case
the shaft is straight and bent alternately and bending stresses are induced in the shaft.
c) Torsional vibrations
When the particles of the shaft move in a circle about the axis of the shaft,
then the vibrations are known as torsional vibrations.
In this case the shaft is twisted and untwisted alternately and torsional shear
stresses are induced in the shaft to.
2. Forced Vibration
Eg: Machine tools, Electric bells.
The vibration which is under the influences of external force is called forced
vibration.
The external force applied to the body is periodic disturbing force created by
unbalance.
The vibrations have the same frequency as the applied force. Due to the
application of external forces the amplitude of these vibrations is maintained almost
constant.

3. Damped vibration
When there is a reduction in amplitude over every cycle of vibration, the
motion is said to be damped vibration.
That is if the vibrators system has a damper. The motion of the system will
be opposed by it and the energy of the system will be dissipated in friction.
4. Undamped vibration.
There is no damper. There is no loss of energy due to friction.
5. Deterministic vibration
If in the vibratory system the amount of external excitation is known in
magnitude it is deterministic vibration.
6. Random vibration
Non deterministic vibrations
7. Steady state vibrations
In ideal systems, the free vibrations continue indefinitely as there is no
damping. Such vibration is termed as steady state vibration.
8. Transient vibrations
In real systems, the amplitude of vibration decays continuously because of
damping and vanishes finally. Such vibration is real system is called transient
vibration.
9. Linear vibration
A vibratory system basically consists of there elements:
Mass
Spring
Pamper
Fig.
- If in a vibratory system mass, spring and damper behave in a linear
manner, the vibrations caused are known as linear vibrations.
- Linear vibrations are governed by linear differential equations.
- They follow the law for superposition.

10. Non linear vibrations
- if any of the basic components of a vibratory system behaves non linearly,
the vibration is called non-linear vibration.
- it does not follow the law of superposition.
Linear vibration becomes non linear for very large amplitude of vibration.
Forced vibrations are also known as excitations.
The excitation may be:
a) Periodic
b) Impulsive
c) Random
Vibrations because of impulsive forces are called transient.
Earth quake is because of random forces.
- External force keeps the system vibrating.
This force is called external excitation.
Harmonic motion :
 Simplest form of periodic motion is harmonic motion and it is called simple
harmonic motion (SHM). It can be expressed as
where A is the amplitude of motion, t is the time instant and T is the period of
motion.
 Harmonic motion is often represented by projection on line of a point that is
moving on a circle at constant speed.

Figure 2.1: The Simple Harmonic Motion
From Figure 2.1 , we have
where x is the displacement and is the circular frequency in rad/sec.
where T is the period (sec) and f is the frequency (cycle/sec) of the harmonic motion.
 The SHM repeats itself in radians.
 Displacement can be expressed as
So that the velocity and acceleration can be written as
FREE VIBRATIONS
 In absence of damping, the system can be considered as conservative and
principle of conservation of energy offers another approach to the calculation
of the natural frequency.
 The effect of damping is mainly evident in diminishing of the vibration
amplitude at or near the resonance

Undamped Free Vibration
A spring mass system as shown in Figure 2.2 is considered. For simplicity at present
the damping is not considered.
Figure 2.2
The direction of x in the downward direction is positive. Also velocity, ,
acceleration, , and force, F, are positive in the downward direction as shown in
Figure 2.2. From Figure 2.2(d) on application of Newton's second law, we have
or
From Figure 2.2(b), we have (i.e. spring force due to static deflection is
equal to weight of the suspended mass), so the above equation becomes
The choice of the static equilibrium position as reference for x axis datum has
eliminated the force due to the gravity. Equation can be written as
or

where is the natural frequency (in rads/sec).This Equation satisfies the simple
harmonic motion condition.
The undamped free vibration executes the simple harmonic motion as shown in
Figure 2.3.
Figure 2.3: Simple harmonic motion
Since sine & cosine functions repeat after 2 radians (i.e. Frequency Time
period = 2 ), we have
The time period (in second) can be written as
The natural frequency (in rads/sec or Hertz) can be written as
From Figure 2.2(b), we have

On substitution we get
Here T , f , are dependent upon mass & stiffness of the system, which are
properties of the system.
Above analysis is valid for all kind of SDOF system including beam or torsional
members. For torsional vibrations the mass may be replaced by the mass moment of
inertia and stiffness by stiffness of torsional spring. For stepped shaft an equivalent
stiffness can be taken or for distributed mass an equivalent lumped mass can be
taken.
The undamped free response can also be written as
where A & B are constants to be determined from initial conditions.
Equivalent Stiffness of Series and Parallel Springs :
For this system having springs connected in series or parallel, this equation is still
valid with the equivalent stiffness as shown in Figures 2.4 and 2.5.
Figure 2.4
Figure 2.5

Energy method :
In a conservative system (i.e. with no damping) the total energy is constant, and
differential equation of motion can also be established by the principle of
conservation of energy.
 For the free vibration of undamped system: Energy=(partly kinetic energy +
partly potential energy).
 Kinetic energy T is stored in mass by virtue of its velocity.
Potential energy U is stored in the form of strain energy in elastic
deformation or work done in a force field such as gravity, magnetic field etc.
Our interest is to find natural frequency of the system, writing this equation for two
positions
where, 1 & 2 represents two instants of time.
Let 1 represents a static equilibrium position (choosing this as the reference point of
potential energy, here U1=0 ) and 2 represents the position corresponding to
maximum displacement of mass and at this position velocity of mass will be zero and
hence T2 = 0.
Damped Free Vibration
Vibration systems may encounter damping of following types:
1. Internal molecular friction.
2. Sliding friction
3. Fluid resistance
Generally mathematical model of such damping is quite complicated and not suitable
for vibration analysis.
Simplified mathematical model (such as viscous damping or dash-pot) have been
developed which leads to simplified formulation.
A mathematical model of damping in which force is proportional to displacement
i.e., Fd = cx is not possible because with cyclic motion this model will encounter an
area of magnitude equal to zero as shown inFigure 2.1(a). So dissipation of energy is
not possible with this model.

The damping force (non-linearly related with displacement) versus displacement
curve will enclose an area, it is referred as the hysteresis loop (Figure 2.1(b)), that is
proportional to the energy lost per cycle.
(a): Linear relation (b): Non-linear relation
Figure 2.6: Variation of damping force vs displacement
Viscously damped free vibration :
Viscous damping force is expressed as,
c is the constant of proportionality and it is called damping co-efficient.
Figure 2.7 shows spring-damper-mass system with free body diagram.

From free body diagram, we have
(1)
Let us assume a solution of equation(1) of the following form
(2)
where s is a constant (can be a complex number) and t is time.
So that and , on substituting in equation (1), we get,
From the condition that equation (2) is a solution for all values of t , above equation
gives a characteristic equation (Frequency equation) as
(3)
Equation (3) has the following form
solution of which is given as
Solution of equation (3) can be written as
(4)
Hence the general solution of equation (1) from equations (2) and (4) is given by the
equation
(5)

where A and B are integration constants to be determined from initial conditions.
Substituting equation (4) into equation (5).
(6)
The term outside the bracket in RHS is an exponentially decaying function. The term
can have three cases.
(i) : exponents in equation (6) will be real numbers.
 No oscillation is possible as shown in Figure 2.8.
 This is an overdamped system (Figure 2.8).
Figure 2.8: Overdamped system
ii) : exponents in equation (6) are imaginary numbers :
 we can write
Hence the equation (6) takes the following form

Let and , equation (6) can be written as
(7)
where
(iii) Critical case between oscillatory and non-oscillatory motion :
Damping corresponding to this case is called critical damping, cc
(8)
Any damping can be expressed in terms of the critical damping by a non-dimensional
number called the damping ratio
(9)
Response corresponding to the critical damping case is shown in Figure 2.9 for
various initial conditions.
Figure 2.9: Critical damping

Equation of motion for damped system can be expressed in terms of and as
(10)
This form of equation is useful in identification of natural frequency and damping of
system.
It is useful in modal summation of MDOF system also.
The roots of characteristic equation (10) can be written as
(11)
with
Depending upon value of damping ratio we can have the following cases
, overdamped condition
, underdamped condition
, critical damping
, undamped system
1) Oscillatory motion : [ , underdamped case]
General solution equation (1) becomes:
(12)
(3.30)

(13)
where and and , where C &
D and X, are arbitrary constants to be determined from initial conditions, x (0) and
(0).
From equation (13), we have
On application of initial conditions, we get
x(0)=C
and
; which gives
Hence, equation (13), becomes
(14)
Equation (14) indicates that the frequency of damped system is equal to,
(15)

It should be noted that for small ( which is the case of most engineering systems)
2) Non-oscillatory motion : ( over damped case)
Two roots remain real with one increases and another decreases.
The general solution becomes
(16)
so that
On application of initial conditions, we have
x(0)=A + B
and
or
which gives
and

3) Critically damped systems :
We obtained two roots
Two terms in solution combines to give one constant
From equation (14) for critically damped case (when ), we have
(17)
Hence the general solution will be
(18)
so that,
On application of initial conditions, we get
x(0)=A
and

The necessary and sufficient conditions for crossing once can be obtained as
(19)
or
is a necessary condition for crossing time axis once but sufficient conditions
is given by equation (19) as shown in Figure 2.9.
Logarithmic Decrement :
Rate of decay of free vibration is a measure of damping present in a system. Greater
is the decay, larger will be the damping.
Damped (free) vibration, general equation of the response is given as
Defining a term logarithmic decrement which is defined as the natural logarithm
of the ratio of any two successive amplitudes as shown Figure 2.11.
since
Td = damped period, where = damped natural frequency
We have damped period , we get logarithmic
decrement as
Since , the above equation reduces to

Experimental determination of natural frequency and damping ratio :
rad/sec, Td can be obtained from displacement-time free vibration
oscillations.
, where x1and x2 are two consecute amplitudes in the free vibration
displacement-time curve.
Figure 2.10
The above illustration shows for two successive amplitude. But in case, the
amplitude are recorded after "n" cycles, the formula is modified as
Taking log,
Forced Harmonic Vibration:
Steady State Response due to Harmonic Oscillation :
Consider a spring-mass-damper system as shown in figure 2.11. The equation of
motion of this system subjected to a harmonic force can be given by

where, m , k and c are the mass, spring stiffness and damping coefficient of the
system, F is the amplitude of the force, w is the excitation frequency or driving
frequency.
Figure 2.11 Harmonically excited system
Figure 2.12: Force polygon
The steady state response of the system can be determined by solving the above
equation in many different ways. Here a simpler graphical method is used which will
give physical understanding to this dynamic problem. From solution of differential
equations it is known that the steady state solution (particular integral) will be of the
form
As each term of equation (4.1) represents a forcing term viz., first, second and third
terms, represent the inertia force, spring force, and the damping forces. The term in
the right hand side of equation is the applied force. One may draw a close polygon as
shown in figure 2.12 considering the equilibrium of the system under the action of
these forces. Considering a reference line these forces can be presented as follows.
 Spring force = (This force will make an angle with
the reference line, represented by line OA).
 Damping force = (This force will be perpendicular to
the spring force, represented by line AB).

 Inertia force = (this force is perpendicular to the
damping force and is in opposite direction with the spring force and is
represented by line BC) .
 Applied force = which can be drawn at an angle with respect to
the reference line and is represented by line OC.
From equation the resultant of the spring force, damping force and the inertia force will
be the applied force, which is clearly shown in figure 2.12
It may be noted that till now, we don't know about the magnitude of X and which can
be easily computed from Figure 2. Drawing a line CD parallel to AB, from the triangle
OCD of Figure 2,
or
As the ratio is the static deflection of the spring, is known as the
magnification factor or amplitude ratio of the system
Figure 2.13 shows the magnification factor frequency ratio and phase angle
frequency ratio plot.

Following observation can be made from these plots.
 For undamped system ( i.e. ) the magnification factor tends to infinity
when the frequency of external excitation equals natural frequency of the
system .
 But for underdamped systems the maximum amplitude of excitation has a
definite value and it occurs at a frequency
 For frequency of external excitation very less than the natural frequency of
the system, with increase in frequency ratio, the dynamic deflection ( X )
dominates the static deflection , the magnification factor increases till it
reaches a maximum value at resonant frequency .
 For , the magnification factor decreases and for very high value of
frequency ratio ( say )
 One may observe that with increase in damping ratio, the resonant response
amplitude decreases.
 Irrespective of value of , at , the phase angle .
 For , phase angle .
 For, , phase angle approaches for very low value of .
Figure 2.13 : (a) Magnification factor ~ frequency ratio for different values of
damping ratio.

Figure 2.13 : (b) Phase angle ~frequency ratio for different values of damping ratio.
For a underdamped system the total response of the system which is the combination
of transient response and steady state response can be given by
The parameter will depend on the initial conditions.
It may be noted that as , the first part of equation tends to zero and second part
remains.
Rotating Unbalance:
One may find many rotating systems in industrial applications. The unbalanced force
in such a system can be represented by a mass m with eccentricity e , which is
rotating with angular velocity as shown in Figure 4.1.
Figure 2.14: Vibrating system with rotating unbalance

Figure 2.15 Freebody diagram of the system
Let x be the displacement of the nonrotating mass (M-m) from the static equilibrium
position, then the displacement of the rotating mass m is
From the freebody diagram of the system shown in figure 2.15, the equation of
motion is
or
This equation is same as equation (1) where F is replaced by . So from the
force polygon as shown in figure 2.16
or
or

Figure 2.16: Force polygon
or
and
So the complete solution becomes
Figure 2.17 : plot for system with rotating unbalance

Figure 2.18 : Phase angle ~ frequency ratio plot for system with rotating unbalance
From figure following observations may be made for a rotating unbalanced system.
 For very low value of frequency ratio (say ) ), the response of the
system is very small.
 For frequency ratio between 0.5 and1, there is a sharp increase in system
response with increase in frequency of excitation of the system.
 At frequency ratio equal to 1, the phase angle is 900.
 Maximum response amplitude occurs at a frequency slightly greater than
.
 With increase in damping, the response of the system decreases.
 For higher value of (say >2), the response amplitude approaches and
phase angle approaches 1800
Vibration Isolation & Transmissibility:
In many industrial applications, one may find the vibrating machine transmit forces
to ground which in turn vibrate the neighbouring machines. So in that contest it is
necessary to calculate how much force is transmitted to ground from the machine or
from the ground to the machine.
Figure 2.19 : A vibrating system

Figure.2.19 shows a system subjected to a force and vibrating with
.
This force will be transmitted to the ground only by the spring and damper.
Force transmitted to the ground
It is known that for a disturbing force , the amplitude of resulting
oscillation
Substituting these equations and defining the transmissibility TR as the ratio of the
force transmitted Force to the disturbing force one obtains
Comparing equations for support motion, it can be noted that
When damping is negligible

to be used always greater than
Replacing
To reduce the amplitude X of the isolated mass m without changing TR, m is often
mounted on a large mass M. The stiffness K must then be increased to keep ratio
K/(m+M) constant. The amplitude X is, however reduced, because K appears in the
denominator of the expression
Figure 2.20: Transmissibility ~frequency ratio plot
Figure 2.20 shows the variation transmissibility with frequency ratio and it can be
noted that vibration will be isolated when the system operates at a frequency ratio
higher than

Equivalent Viscous Damping :
In the previous sections, it is assumed that the energy dissipation takes place due to
viscous type of damping where the damping force is proportional to velocity. But
there are systems where the damping takes place in many other ways. For example,
one may take surface to surface contact in vibrating systems and take Coulomb
friction into account. Also in many cases energy is dissipated in joints also, which is
a form of structural damping. In these cases one may still use the derived equations
by considering an equivalent viscous damping. This can be achieved by equating the
energy dissipated in the original and the equivalent system.
The primary influence of damping on the oscillatory systems is that of limiting the
amplitude at resonance. Damping has little influence on the response in the
frequency regions away from resonance. In case of viscous damping, the amplitude
at resonance is
For other type of damping, no such simple expression exists. It is possible to
however, to approximate the resonant amplitude by substituting an equivalent
damping Ceq in the foregoing equation. The equivalent damping Ceq is found by
equating the energy dissipated by the viscous damping to that of the nonviscous
damping with assumed harmonic motion.
Where must be evaluated from the particular type of damping.
Structural Damping :
When materials are cyclically stressed, energy is dissipated internally within the
material itself. Experiments by several investigators indicate that for most structural
metals such as steel and aluminum, the energy dissipated per cycle is independent of
the frequency over a wide frequency range and proportional to the square of the
amplitude of vibration. Internal damping fitting this classification is called solid
damping or structural damping. With the energy dissipation per cycle proportional to
the square of the vibration amplitude, the loss coefficient is a constant and the shape
of the hysteresis curve remains unchanged with amplitude and independent of the
strain rate. Energy dissipated by structural damping can be written as

Where is a constant with units of force displacement.
By the concept of equivalent viscous damping
or
Coulomb Damping :
Coulomb damping is mechanical damping that absorbs energy by sliding friction, as
opposed to viscous damping, which absorbs energy in fluid, or viscous, friction.
Sliding friction is a constant value regardless of displacement or velocity. Damping
of large complex structures with non-welded joints, such as airplane wings, exhibit
coulomb damping.
Work done per cycle by the Coulomb force
For calculating equivalent viscous damping
From the above equation equivalent viscous damping is found
Summary
Some important features of steady state response for harmonically excited systems
are as follows-
 The steady state response is always of the form . Where it
is having same frequency as of forcing. X is amplitude of the response, which
is strongly dependent on the frequency of excitation, and on the properties of
the spring—mass system.

 There is a phase lag between the forcing and the system response, which
depends on the frequency of excitation and the properties of the spring-mass
system.
 The steady state response of a forced, damped, spring mass system is
independent of initial conditions.
In this chapter response due to rotating unbalance, support motion, whirling of shaft
and equivalent damping are also discussed.
Magnification Factor (Dynamic magnifier) or Amplitude Ratio
The ratio of the maximum displacement of the forced vibration (xmax) to the
static deflection under the static force F0 (xo) is known as Magnification factor.
Denoted by M.F.
i.e, M.F = max
0
x
x
We have 0
max
22 2
n n
F /s
x
1 2

     
      
      
0
max
22 2
n n
x
x
1 2

     
      
      

22 2
n n
1
M.F
1 2

     
      
      
From this equation, it is clear that the magnification factor depends upon
-The ratio of circular frequencies
n


-The damping factor (  )

Energy Dissipated by damping
The energy lost per cycle due to a damping force Fd is calculated by
d d xw F .d 
where damping force, Fd =
dx
C.
dt
Energy dissipated/cycle, 2
dW c x  
Vibration Measuring Instruments
The instruments which are used to measure the displacement, velocity or
acceleration of a vibrating body are called vibration measuring instruments.
Widely used → vibrometers (low frequency transducer)
→Accelerometers (high frequency transducer)
Example: Accelerometers
a) Bonded strain gauge accelerometer
i) Cantilever beam type accelerometer
ii) Solid cylinder accelerometer
b) Piezoelectric accelerometer
c) Servo accelerometer (force balance accelerometer)
Piezo electric accelerometers
Certain crystals exhibit the property that they generate a charge across their
faces when a stress is applied to them. This property is made use of in piezoelectric
accelerometer.
The change generated to the device is given by q = f.d.
When f → applied force
d → Piezoelectric constant
When the device is subjected to acceleration the mass exerts a variable force
on the Piezoelectric disc, which is proportional to the acceleration. The charge
developed across the disc is in turn proportional to the acceleration of the mass.

MODULE 3
TWO-DEGREE-OF-FREEDOM-SYSTEMS
A single degree of freedom system has only one natural frequency and
requires only one independent co-ordinate to define the system completely.
But a Two Degree freedom system has two natural frequencies and the free
vibration of any point in the system, in general, is a combination of two harmonies of
these two natural frequencies respectively. Under certain conditions, any point in the
system may execute harmonic vibrations, at any of the two natural frequencies, and
these are known as the principal moles of vibrations.
In Two Degree freedom systems, there are two independent co-ordinates to
specify the configuration.
 The vibrating systems, which require two coordinates to describe its motion,
are called two-degrees-of -freedom systems.
 These coordinates are called generalized coordinates when they are
independent of each other and equal in number to the degrees of freedom of
the system.
 Unlike single degree of freedom system, where only one co-ordinate and
hence one equation of motion is required to express the vibration of the
system, in two-dof systems minimum two co-ordinates and hence two
equations of motion are required to represent the motion of the system. For a
conservative natural system, these equations can be written by using mass
and stiffness matrices.
 One may find a number of generalized co-ordinate systems to represent the
motion of the same system. While using these co-ordinates the mass and
stiffness matrices may be coupled or uncoupled. When the mass matrix is
coupled, the system is said to be dynamically coupled and when the stiffness
matrix is coupled, the system is known to be The set of co-ordinates for
which both the mass and stiffness matrix are uncoupled, are known as
principal co-ordinates. In this case both the system equations are independent
and individually they can be solved as that of a single-dof system.
 A two-dof system differs from the single dof system in that it has two natural
frequencies, and for each of the natural frequencies there corresponds a
natural state of vibration with a displacement configuration known as the
normal mode. Mathematical terms associated with these quantities are
eigenvalues and eigenvectors.
 Normal mode vibrations are free vibrations that depend only on the mass and
stiffness of the system and how they are distributed. A normal mode

oscillation is defined as one in which each mass of the system undergoes
harmonic motion of same frequency and passes the equilibrium position
simultaneously.
 The study of two-dof- systems is important because one may extend the same
concepts used in these cases to more than 2-dof- systems. Also in these cases
one can easily obtain an analytical or closed-form solutions. But for more
degrees of freedom systems numerical analysis using computer is required to
find natural frequencies (eigenvalues) and mode shapes (eigenvectors).
Derivation of Equation of Motion
Few examples of two-degree-of-freedom systems ::
Figure 3.1(a) shows two masses m1 and m2 with three springs having spring stiffness
k1, k2 and k3 free to move on the horizontal surface. Let x1 and x2 be the displacement
of mass respectively.
Figure 3.1(a)
As described in the previous lectures one may easily derive the equation of motion
by using d'Alembert principle or the energy principle (Lagrange principle or
Hamilton 's principle)
Figure 3.1(b): Free body diagrams
Using d'Alembert principle for mass m1 from the free body diagram shown in figure
3.1(b)

and similarly for mass m2
Derivation of Equation of Motion and Coordinate Coupling
Noting , the above two equations in matrix form can be written as
Now depending on the position of point C, few cases can are studied below.
Case 1 : Considering , i.e., point C and G coincides, the equation of motion can
be written as
Figure 3.2
So in this case the system is statically coupled and if , this coupling
disappears, and we obtained uncoupled x and vibrations.
Case 2 : If, , the equation of motion becomes
Hence in this case the system is dynamically coupled but statically uncoupled.
Case 3: If we choose , i.e. point C coincide with the left end, the equation of
motion will become

Here the system is both statically and dynamically coupled.
Normal Mode Vibration
Again considering the problem of the spring-mass system in figure 6.1.1 with
, , , the equation of motion can be written as
We define a normal mode oscillation as one in which each mass undergoes harmonic
motion of the same frequency, passing simultaneously through the equilibrium
position. For such motion, we let
Hence,
or, in matrix form
Hence for nonzero values of and (i.e., for non-trivial response)
Now substituting , equation yields
Hence, and

So, the natural frequecies of the system are and
Now it may be observed that for these frequencies, as both the equations are not
independent, one can not get unique value of and . So one should find a
normalized value. One may normalize the response by finding the ratio of to .
From the first equation. the normalized value can be given by
and from the second equation the normalized value can be given by
Now, substituting in equation the same values, as both these
equations are linearly dependent. Here,
and similarly for
It may be noted
 If one of the amplitudes is chosen to be 1 or any number, we say that
amplitudes ratio is normalized to that number.
 The normalized amplitude ratios are called the normal modes and designated
by .
The two normal modes of this problem are:

In the 1st
normal mode, the two masses move in the same direction and are said to be
in phase and in the 2nd
mode the two masses move in the opposite direction and are
said to be out of phase. Also in the first mode when the second mass moves unit
distance, the first mass moves 0.731 units in the same direction and in the second
mode, when the second mass moves unit distance; the first mass moves 2.73 units in
opposite direction.
VIBRATION ABSORBER
Tuned Vibration Absorber
Consider a vibrating system of mass , stiffness , subjected to a force .
As studied in case of forced vibration of single-degree of freedom system, the system
will have a steady state response given by
(1)
which will be maximum when Now to absorb this vibration, one may add a
secondary spring and mass system as shown in figure .

The equation of motion for this system can be given by
(2)
As we know for steady state vibration, the system will vibrate with a frequency of
the external excitation; we can assume the solution to be
(3)
Substituting Equation (3) in equation (2) one may write
(4)
Or, (5)
Using Cramer's rule one may write
(6)
(7)
where

Now
Here are the roots of the characteristic equation . One may note
that these roots are the normal mode frequency for this two-degrees of freedom
system. These free-vibration frequencies can be given by
From equation (6), it is clear that,
Hence, if a system called the primary system with a stiffness mass is subjected
to an exciting force or base motion to vibrate, it is possible to completely eliminate
the vibration of the primary system by suitably designing an attached spring-mass
system (secondary system) with stiffness and mass such that the natural
frequency of the secondary system coincide with the exciting frequency. .
This is the principle of dynamic vibration absorber.
From equation (1) it may be noted that the primary system will have resonance when
the natural frequency of the primary system coincide with that of the excitation
frequency.
Hence to reduce the vibration at resonance of the primary system one should design
the secondary system such that the natural frequency of both the components
coincides.

For this condition
Substituting and , the above equation reduces to
or,
For, ,
and

To keep the displacement of secondary mass small, the stiffness of the secondary
spring should be very large. To have this the secondary mass should also be large
which is not desirable from practical point of view.
Hence a compromise is usually made between the amplitude and the mass ratio. The
mass ratio is usually kept between 0.05 and 0.25.
Resonant frequency of the vibration absorber
Centrifugal Pendulum Vibration Absorber
The centrifugal pendulum vibration absorber was devised and patented in France
about 1935 and at the same time it was independently conceived and put into practice
by E. S. Taylor. Its purpose was to overcome serious torsional vibration problem
inherent in geared radial aircraft-engine �propeller system. Later it was modified
and incorporated into automobile IC engines in order to reduce the torsional
vibrations of the crankshaft. This was done by integrating the absorber mass with
crankshaft counter balance mass.
The tuned vibration absorber is only effective when the frequency of external
excitation equals to the natural frequency of the secondary spring and mass system.
But in many cases, for example in case of an automobile engine, the exciting torques
are proportional to the rotational speed �n' which may vary over a wide range. For
the absorber to be effective, its natural frequency must also be proportional to the
speed. The characteristics of the centrifugal pendulum are ideally suited for this
purpose.
Placing the coordinates through point O', parallel and normal to r, the line r rotates
with angular velocity ( ).

The acceleration of mass
Since the moment about is zero,
Assuming to be small, , so
If we assume the motion of the wheel to be a steady rotation plus a small
sinusoidal oscillation of frequency , one may write
Hence the natural frequency of the pendulum is
and its steady-state solution is
t may be noted that the same pendulum in a gravity field would have a natural
frequency of . So it may be noted that for the centrifugal pendulum the gravity
field is replaced by the centrifugal field .

Torque exerted by the pendulum on the wheel
With the component of equal to zero, the pendulum force is a tension along ,
given by times the component of .
Now assuming small angle of rotation
Now substituting
Hence the effective inertia can be written as
which can be at its natural frequency. This possesses some difficulties in the
design of the pendulum. For example to suppress a disturbing torque of frequency
equal to four times the natural speed n , the pendulum must meet the requirement
. Hence, as the length of the pendulum becomes very
small it will be difficult to design it. To avoid this one may go for Chilton bifilar
design.
MULTIDEGREE OF FREEDOM SYSTEMS
It must be appreciated that any real life system is actually a continuous or distributed
parameter system (i.e. infinitely many d.o.f). Hence to derive its equation of motion
we need to consider a small (i.e., differential) element and draw the free body
diagram and apply Newton 's second Law. The resulting equations of motion are
partial differential equations and more complex than the simple ordinary differential
equations we have been dealing with so far. Thus we are interested in modeling the
real life system using lumped parameter models and ordinary differential equations.

The accuracy of such models (i.e. how well they can model the behavior of the
infinitely many d.o.f. real life system) improves as we increase the number of d.o.f.
Thus we would like to develop mult-d.o.f lumped parameter models which still yield
ordinary differential equations of motion – as many equations as the d.o.f . We would
discuss these aspects in this lecture.
Derivation of Equations of Motion
Fig 3.3 Typical multi-d.o.f. system
Consider a typical multi-d.o.f system as shown in Fig. 3.3. As mentioned earlier our
procedure to determine the equations of motion remains the same irrespective of the
number of d.o.f of the system and is recalled to be: Step 1 : Consider the system in a
displaced Configuration Step 2 : Draw Free Body diagrams and Step 3 : Use
Newton 's second Law to write the equation of motion From the free body diagrams
shown in Fig. 3.3, we get the equations of motion as follows:
Rewriting the equations of motion in matrix notation, we get:
Or in compact form,

There are “n” equations of motion for an “n” d.o.f system. Correspondingly the mass
and stiffness matrices ([M] and [K] respectively) are square matrices of size (n x n).
If we consider free vibrations and search for harmonic oscillations,
. Substituting these in equation, we get,
ie
For a non-trivial solution to exist, we have the condition that the determinant of the
coefficient matrix must vanish. Thus, we can write,
In principle this (n x n) determinant can be expanded by row or column method and
we can write the characteristic equation (or frequency equation) in terms of ,
solution of which yields the “n” natural frequencies of the “n” d.o.f. system just as
we did for the two d.o.f system case.
We can substitute the values of in eqn and derive a relation between the
amplitudes of various masses yielding us the corresponding normal mode shape.
Typical mode shapes are schematically depicted in Figure for a d.o.f system.
Dunkerly's method of finding natural frequency of multidegree of freedom
system
We observed in the previous lecture that determination of all the natural frequencies
of a typical multi d.o.f. system is quite complex. Several approximate methods such
as Dunkerly's method enable us to get a reasonably good estimate of the fundamental
frequency of a multi d.o.f. system. Basic idea of Dunkerly's method
Fig 3.4 A typical multi d.o.f. system
Consider a typical multi d.o.f. system as shown in fig 3.4 Dunkerly's
approximation to the fundamental frequency of this system can be obatined in two
steps:

Step1: Calculate natural
frequency of all the modified
systems shown in Fig 3.5
These modified systems are
obatined by considering one
mass/inertia at a time. Let
these frequencies be
Step2: Dunkerly's estimate of
fundamental frequency is
now given as:
Fig 3.5 Modified system considered in Dunkerly's
Method
Fig 3.6 A Typical two d.o.f. example
Consider a typical two d.o.f. system as shown in Fig 3.6 and the equations of
motion are given as:
For harmonic vibration, we can write:
Thus,
Inverting the stiffness matrix and re-writing the equations

The equation characteristic can be readily obtained by expanding the determinant as
follows:
As this is a two d.o.f. system, it is expected to have two natural frequencies viz
and ., Thus we can write Equation as:
Comparing coefficients of like terms on both sides, we have:
It would appear that these two equations can be solved exactly for and . While this is
true for this simpleexample, we can't practically implement such a scheme for an n-
d.o.f system, as it would mean similar computational effort as solving the original
problem itself. However, we could get an approximate estimate for the fundamental
frequency. If >> , then we can approximately write
Let us now study the meaning of and . It is easily verified that
These can be readily verified to be the reciprocal of the equivalent stiffness values
for the modified systems.

Thus, we can write:
Holzer's method of finding natural frequency of a multi-degree of freedom
system
Holzer's Method.
This method is an iterative method and can be used to determine any number of
frequencies for a multi-d.o.f system. Consider a typical multi-rotor system as shown
in Fig. 3.7
Fig 3.7 Typical multi-rotor system
The equations of motion for free vibration can be readily written as follows:
For harmonic vibration, we assume
Thus:

Summing up all the equations of motion, we get:
This is a condition to be satisfied by the natural frequency of the freely vibrating
system.
Holzer's method consists of the following iterative steps:
Step 1: Assume a trial frequency
Step 2: Assume the first generalized coordinate say
Step 3: Compute the other d.o.f. using the equations of motion as follows:
Step 4: Sum up and verify if this equation is satisfied to the prescribed degree of
accuracy.
If Yes, the trial frequency is a natural frequency of the system. If not, redo the steps
with a different trial frequency.
In order to reduce the computations, therefore one needs to start with a good trial
frequency and have a good method of choosing the next trial frequency to converge
fast. Two trial frequencies are found by trial and error such that is a
small positive and negative number respectively than the mean of these two trial
frequencies(i.e. bisection method) will give a good estimate of for which
.
Holzer's method can be readily programmed for computer based calculations.
TORSIONAL VIBRATIONS
When the particles of a shaft or disc move in a circle about the axis of a
shaft, then the vibrations are known as torsional vibrations. In this case the shaft is
twisted and untwisted alternately and torsional shear stresses are introduced in the
shaft.

Torsional vibrations may result in shafts from following forcings:
 Inertia forces of reciprocating mechanisms (such as pistons in Internal
Combustion engines)
 Impulsive loads occurring during a normal machine cycle (e.g. during
operations of a punch press)
 Shock loads applied to electrical machineries (such as a generator line fault
followed by fault removal and automatic closure)
 Torques related to gear tooth meshing frequencies, turbine blade passing
frequencies, etc.
For machines having massive rotors and flexible shafts (where system natural
frequencies of torsional vibrations may be close to, or within, the source frequency
range during normal operation) torsional vibrations constitute a potential design
problem area.
In such cases designers should ensure the accurate prediction of machine torsional
frequencies and frequencies of any of the torsional load fluctuations should not
coincide with torsional natural frequencies.
Hence, determination of torsional natural frequencies of a dynamic system is very
important.
Simple systems with a single disc mass: Consider a rotor system as shown in
Figure 4.1. The shaft is considered as massless and it provides the stiffness only. The
disc is considered as rigid and it has no flexibility. If we give a small initial
disturbance to the disc in the torsional mode and allow it to oscillate its own, it will
execute free vibrations. The oscillation will be simple harmonic motion (SHM) with
a unique frequency, which is called natural frequency of the rotor system.
From the theory of torsion of shaft, we have

where, kt is the torsional stiffness of shaft, Ip is the rotor polar mass moment of
inertia, J is the shaft polar second moment of area, l is the length of the shaft and q is
the angular displacement of the rotor.
From free body diagram of the disc
External torque on the disc
or
The free (or natural) vibration has the simple harmonic motion (SHM).
For the simple harmonic motion of the disc, we have
so that
where is the amplitude of the torsional vibration and is natural frequency of the
torsional vibration.
or
Hence, the torsional natural frequency is given by square root of the ratio of torsional
stiffness to the polar mass moment of inertia.
A two-disc torsional system
In a two-discs torsional system as shown in Figure 4.3, whole of the rotor is free to
rotate i.e. the shaft being mounted on frictionless bearings.

From free body diagrams of discs in Figure 4.4, we can write
External torque on disc 1 and External torque on disc 2
and
or and
For SHM,
and
where is the torsional natural frequency.
and
which can be written in the matrix form, as

with and
This equation is a homogeneous equation and the non-trial solution is obtained by
taking determinant of the matrix [k] as
|k| = 0
which gives frequency equation of the following form
which can be simplified as
Roots of equation are given as
and
From equation corresponding to first natural frequency for = 0 , we get
From equation it can be concluded that, the first root of equation represents the case
when both discs simply rolls together in phase with each other as shown in Figure
4.5 i.e. the rigid body mode, which is of a little practical significance.
From equation (4.9), for , we get
On substituting for in the above equation we get

which can be simplified as
or
which gives relative amplitudes of two discs as
The second mode represents the case when both masses vibrate in anti-phase with
one another. Figure 4.6 shows the second mode shape of two-rotor system, showing
two discs vibrating in opposite directions.
From the second mode shape, i.e. from Figure 4.6 and noting equation we have

Since both masses are always moving in the opposite direction, so there must be a
point on the shaft where the torsional vibration is not taking place i.e. a torsional
node. The location of the node may be established by treating each end of the real
rotor system as a separate single-disc cantilever system as shown in Figure 4.5. The
torsional node being treated as the point where the shaft is rigidly fixed.
Since the natural frequency of the system is known and the frequency of oscillation
of each of the single-disc system must be same, hence we write
and
Lengths l1 and l2 then can be obtained by, noting equation as
and
which must satisfy
Torsional Systems with a Stepped Shaft

Figure 4.7(a) shows a two-disc stepped shaft. The polar mass moment of inertia of
the shaft is negligible as compared to discs. In such cases the actual shaft should be
replaced by an unstepped equivalent shaft for the purpose of the analysis as shown in
Figure 4.7(b). The equivalent shaft diameter may be same as the smallest diameter of
the real shaft.
The equivalent shaft must have the same torsional stiffness as the real shaft, since for
the present case torsional springs are connected in series. The equivalent torsional
spring can be written as
we have
which gives
where
where , and are equivalent lengths of shaft segments having equivalent shaft
diameter d3 and le is the total equivalent length of unstepped shaft having diameter d3
as shown in Figure 4.7(b).
From Figure 4.7(b) and noting equations in the equivalent shaft the node
location can be obtained as
with and
From equations the node position a & b can be obtained in the equivalent shaft
length.
Now the node location in real shaft system can be obtained as follows:

we have
Since above equation is for shaft segment in which node is assumed to be present, we
can write
and
It can be combined as
So once a & b are obtained the location of the node in the actual shaft can be
obtained i.e. the final location of the node on the shaft in real system is given in the
same proportion along the length of shaft in the equivalent system in which the node
occurs.

MODULE 4
WHIRLING OF SHAFTS – CRITICAL SPEED
Consider a typical shaft, carrying a rotor (disk) mounted between two bearings as
shown in Figure.Let us assume that the overall mass of the shaft is negligible
compared to that of the rotor (disk) and hence we can consider it as a simple
torsional spring. The rotor (disk) section has a geometric centre i.e., the centre of the
circular cross-section and the mass centre due to the material distribution. These two
may or may not coincide in general, leading to eccentricity. The eccentricity could be
due to internal material defects, manufacturing errors etc. As the shaft rotates, the
eccentricity implies that the mass of the rotor rotating with some eccentricity will
cause in-plane centrifugal force. Due to the flexibility of the shaft, the shaft will be
pulled away from its central line as indicated in the figure. Let us assume that the air-
friction damping force is negligible. The centrifugal force for a given speed is thus
balanced by the internal resistance force in the shaft-spring and the system comes to
an equilibrium position with the shaft in a bent configuration as indicated in the
figure. Thus the shaft is rotating about its own axis and the plane containing the bent
shaft and the line of bearings rotates about an axis coinciding with the line of
bearings. We consider here only the case, wherein these two rotational speeds are
identical, called the synchronous whirl.

Centrifugal force
shaft resistance force
wherein, the shaft stiffness k is the lateral stiffness of a shaft in its bearings i.e., considering the rotor a
span, this is the force required to cause a unit lateral displacement at mid-span of the simply supported
Thus
Where E is the Young's modulus, I is the second moment of area, and L is the length between the suppo
Thus, for equilibrium,
ie
where we have used to represent the natural frequency of the lateral vibration of the sp
shaft-rotor system. Thus when the rotational speed of the system coincides with the natural frequen
lateral vibrations, the shaft tends to bow out with a large amplitude. This speed is known as the critical
and it is necessary that such a resonance situation is avoided in actual practice. As discussed earlier
case of resonance, it takes some time for the amplitude to build up to a large value. Some of the tu
rotors whose operating speeds go beyond the critical speed are able to use this fact and rush-throug
critical speed. It is necessary to observe that, in synchronous whirl. the heavier side remains all the tim
the outer side. Thus when the shaft bends, an inner fibre is under compressive stress and outer fibre is
tensible stress but there is NO REVERSAL of stress.
Rayleighs Method
Fig 4.1 Multiple disks on a shaft

Rayleigh's method is based on the principle of conservation of energy. The energy in
an undamped system consists of the kinetic energy and the potential energy. The
kinetic energy T is stored in the mass and is proportional to the square of the
velocity. The potential energy U includes strain energy that is proportional to elastic
deformations and the potential of the applied forces. For a conservative system, the
total energy must remain constant. That is
Differentiating this expression, we get the equation of motion as follows.
Note that the amounts of kinetic and potential energy in the system may change with
time but their sum must remain constant. Thus if and are energies at time and
and are energies at time , then
For a shaft as shown in Fig.4.2 the potential energy is zero at the specific instant of
time when the mass is passing through its static equilibrium position and kinetic
energy is at its maximum . Similarly at the instant when the mass is at its
extreme position the kinetic energy is zero and the potential energy is at its
maximum . Thus we have the following relationship.
Therefore we have, considering all the disks on the shaft,
Where i=1, n represents summation over all the "n" disks.
So we get the frequency of natural vibration as,

Dunkerley's Empirical Method
When a shaft carries multiple disks it is always efficient to use this method.
Fig 4.2 Dunkerly's approximation for shaft
We consider only one force (wt of disk) acting on the shaft at a time. For each disk,
we find the corresponding natural frequency as . The Nutural
frequency of the shaft when all the loads(disks) act on the shaft simultaneously
can be found out by the using the formulae:

For each of the dub-systems ie shaft with only one disk, natural frequency is obtained
as where k is the lateral stiffness of the shaft in its bearings and m is the mass
of the disk.
To understand the basis of this method, we need to appreciate multi-d.o.f system
vibrations.
CRITICAL SPEEDS OF A LIGHT CANTILEVER SHAFT WITH A LARGE
HEAVY DISC AT ITS END
If a light shaft having two end supports has a central disc then the system
has been shown to have one critical speed. Even if the disc is not central, the system
will have one critical speed as long as we assume the mass of the disc to be
concentrated. If, however, the disc has mass as well as moment of inertia, and is not
central, then the system will have two critical speeds. The treatment given below is
for a light cantilever shaft having a disc which has mass as well as moment of inertia.
Since the critical speed is numerically equal to the natural frequency of lateral
vibrations, we will find the later for this system.
Consider the beam so as to be displaced from the equilibrium position as
shown in Fig. In this figure,
M = mass of the disc,
Mr2
= moment of inertia of the disc about an axis passing through
the CG of the disc and perpendicular to the plane of the paper.

y, y = displacement and acceleration of the CG of the disc,
,  = angular displacement and angular acceleration of the axis of
the disc due to bending
Further let
a11 = deflection of the CG of the disc per unit force acting on it in
the lateral direction = l3
/3EL
a22 = slope at the free end of the beam per unit moment acting on
the CG of the disc, in the plane of the paper = l/EI
a12 = deflection of the CG of the disc per unit moment acting on it
= slope at the free and of the beam per unit force acting on it
in a lateral direction = l2
/2EI
where l = length of the beam,
I = moment of inertia of the section of the beam about the neutral
axis.
E = modulus of elasticity of the material of the beam.
The inertia force and the inertia torque on the disc in the displaced position
are shown in Fig.8.6.1, along with their directions. These are as follows:
2
2 2 2
Inertiaforce My M y
Inertia torque Mr Mr
    

     
(1 )
where  is natural frequency for the principal mode of vibration of the
system.
Then the deflection at the CG of the disc and the rotation of the disc in the
plane of the paper are given by
2 2 2
11 12y a M y a Mr     (2)
2 2 2
21 22a M y a Mr      (3)
Eliminating y and  from the above two equations, and putting

3
2
2
Ml
g
3EI
3r
h
l

 



 

(4)
we have hg4
– 4(h + 1)g2
+ 4 = 0 (5)
giving the two natural frequencies as
2 2
1,2
2
g (h 1) (h 1) h
h
     
  
(6)
Figure (2) is a plot of the above equation and shows the variation of the two
natural frequencies of the system with the change in
2
2
3r
h
l
 
   
 
; it may be recalled
that r is radius of gyration of the disc about an axis passing through its CG and
perpendicular to the axis of the disc.
h = 0, corresponds to the concentrated mass
h  , corresponds to the disc having large radius of gyration with the
change in
2
2
3r
h
l
 
   
 
.
Transient Vibrations
INTRODUCTION
A system subjected to periodic excitation has two components of motion,
the transient and the steady state. In most of such cases the transient part is not
important as it dies out soon, and the steady state part is the one that persists.
However, where the excitation is of the aperiodic nature like a shock pulse or a
transient excitation, the response of the system is purely transient. After the duration
of the excitation, the system undergoes vibrations with its natural frequency with an
amplitude depending upon the type and duration of the excitation. It is in such cases
that the transient vibrations have importance. The practical examples of shock
excited transient vibrations are rock explosions, gunfires, loading or unloading of
packages by dropping them on hard floors, punching operations, automobiles at high

speeds passing over pits or curbs on the road, etc.
The use of Laplace transform is introduced in this chapter for the analysis of
systems subjected to shock pulses. The usual differential equations method or the so-
called classical method becomes very lengthy and cumbersome with transient
excitations of different shapes.
LAPLACE TRANSFORMATION
Laplace transform is a powerful mathematical tool that is extremely useful
in the solution of differential equations, and especially so, where transients are
involved. It is that branch of operational calculus wherein a function is transformed
from t (time) domain to a new s domain. The original differential equation in t
domain, by use of Laplace tranform, changes itself into an algebraic equation in s
domain. The solution of an algebraic equation is very easy as compared to that of a
differential equation. Once the solution is s domain is obtained, the process of
inverse transformation gives the solution back in t domain. Manipulation with
transformation and inverse transformation is facilitated by the use of table of
transform pairs which is given later in this section.
Laplace transform F(s) of a function f(t) is defines as
st
0
F(s) f(t)d dt


 
In shorthand it is generally written as
L [f(t)] = F(s)
The use of the basic definition of Laplace transform is illustrated below by
actually transforming a few common functions.
RESPONSE TO AN IMPULSIVE INPUT

Consider a damped spring mass system subjected to an impulse ˆF (t) , the
strength of the impulse being ˆF . Since the impulse acts for an extremely small
duration, its effect is to give an initial velocity to the mass given by
ˆF mdv
where dv is the change in velocity of the mass due to the impulse ˆF . If the system is
initially at rest, impulse gives it a starting velocity of
ˆF
dv
m

The initial displacement of the mass form the equilibrium position is zero
because of the extremely small duration of the impulse.
Thus the initial conditions for the mass are
x(0) 0
ˆF
x(0)
m
 


 

(1)
The differential equation for the system can now be written as
mx cx kx 0   (2)
The forcing function on the right has been taken to be zero since the impulse
effectively gives only the initial conditions obtained in Eq. (1).
Dividing Eq. (2) by m through out, it can be written as
2
n nx 2 x x 0     (3)
Taking the Laplace transform of the above equation, we have

2 2
n n[s X(s) sx(0) x(0)] 2 [sX(s) x(0)] X(s) 0       
Substituting the initial conditions of Eq. (1), and re-arranging, gives
2 2
n n
ˆF 1
X(s)
m s 2 s
 
  
     
(4)
In order to obtain the inverse transformation fro the above equation, the
expression on the right has to be re-arranged in one of the forms corresponding to the
transform pairs , for direct inversion. If 1  , the above equation is re-written in the
following form
 
2
n
22 2 2
n n n
1F
X(s)
m 1 (s ) 1
 
   
  
         
  
(5)
The inverse transform is,
t 2n
n
2
n
ˆF
x(t) e sin 1 t
m 1

   
  
(6)
which is the response of the system to an impulsive input.
RESPONSE TO A STEP INPUT
Figure shows a spring-mass-dashpot system subjected to a step force F0 u(t).
The magnitude of the force is constant at a value F0 for all time greater than or equal
to zero. The force is zero for 1 < 0. The differential equation of motion can be written
as
0mx cx kx F u(t)  

or 2 0
n n
F
x 2 x x u(t)
m
     (1)
Taking the Laplace transform of the above equation, we have
2 2 0
n n
F 1
[s X(s) sx(0) x(0)] 2 [sX(s) x(0)] X(s) .
m s
       
A second order system subjected to a finite step cannot have any initial
velocity or displacement. So, putting all initial conditions zero in the above equation,
and re-arranging, we have
0
2 2
n n
F 1
X(s)
m s(s 2 s )
 
  
     
(2)
The inverse transform of the above equation cannot be obtained straightway
from the tables. Hence splitting the right hand side into partial fractions, we have
0 n
2 2 2
n n n
F s 21 1
X(s)
m s s 2 s
  
  
      
The right had expression in the bracket above is still not invertable directly.
Assuming an underdamped system, i.e. 1  , the above equation is written as
follows:
   
2
n
2
0 n
2 2 2
2 2 2 2n
n n n n
. 1
1F (s )1 1
X(s)
m s
(s ) 1 (s ) 1
       
         
  
            
  
(3)
Inverse the transform of Eq. (3) can now be obtained directly from the table
and is given below
t t2 20 n n
n n2 2
n
F
x(t) 1 e cos 1 t e sin 1 t
m 1
 
 
         
    
Putting 2
nm = k, we finally have

t 2 20 n
n n
2
F
x(t) 1 e cos 1 t sin 1 t
k 1

            
     
(4)
For an undamped case, response equation can be written from the above
equation by putting 0  , or
0
n
F
x(t) [1 cos t]
k
   (5)
RESPONSE TO A PULSE INPUT
Pulse applications in engineering practice are very common. An explosion
occurring on a system with a comparatively larger natural period will be an impulse
while the same explosion occurring on a system with a smaller natural period will be
pulse. In this section two important types of pulses, rectangular and half sinusoidal,
are treated. The method lends itself for the analysis of any type of pulse for which a
mathematical equation can be written.
The vibratory systems considered in this section have been taken as
undamped systems to make the response equations simpler. Further, since most
physical systems are lightly damped and in most cases we are interested in maximum
displacements and accelerations, we will be slightly erring on the safer side in
neglecting small amount of damping.
Rectangular pulse
Consider a spring-mass system subjected to a rectangular pulse to height F0
and duration  as shown. The response equation can be written directly by

comparing the response of the system to a multi-step input by considering in this
case two equal and opposite steps, one at t = 0 and the other at t   . Therefore,
0 0
n n
F F
x(t) [1 cos t]u(t) [1 cos (t )]u(t )
k k
          (1)
The above equation can be written as the following two equations
0
n
0
n n
F
x(t) [1 cos t] for0 t
k
F
[cos (t ) cos t] for t
k

     

       

(2)
PHASE PLANE METHOD
A spring mass system with initial conditions X0 and V0, has its differential
equation written as
2
nx x 0   (1)
Its solution may be written as
nx Asin( t )  
where
2
2 0
0 2
n
V
A X 

and 1 n 0
0
X
tan
V
  
   
 
Differentiating equation (1) for velocity, we have

nx A ncos( t )   
or n
n
x
Acos( t )   

(2)
Squaring and adding Eq. (1) and (2), we have
2
2 2
n
x
x A
 
  
 
(3)
The above equation is a circle in a plane with coordinate axes x and n(x/ )
Its radius is A and centre at the origin. This is shown in Figure.. The starting point on
this displacement velocity plot is marked P1. At t1 seconds later the displacement and
velocity of the system are represented by point P2 where 1 2 n 1P OP t   radians.
From this diagram, the displacement and velocity phase of the motion are available
from the single point which corresponds to a particular time. This is the phase plane
plot. The horizontal projection of the phase trajectory on a time base gives the
displacement-time plot of the motion and is shown in Figure Similarly the vertical
projection on the time base will give the velocity-time plot of the motion.
It may be noted that the centre of the phase trajectory always lies on the x-
axis at a distance equal to the static equilibrium displacement of the system. In the
case discussed the static equilibrium displacement was zero and therefore the centre
of the circle was located at the origin. In case of a step force input F0, the static
equilibrium position suddenly changes through a distance F0/k. Thus the phase plane
plot for such a motion will be a circle whose centre lies F0/k above the centre. The
radius of this circle will be F0/k so that the trajectory starts from the origin
corresponding to zero initial conditions.
The use of the phase plane method is illustrated by the following examples
for systems subjected to multiple steps.
SHOCK SPECTRUM
The response of a spring-mass system to a particular pulse depends upon the
natural frequency of the system. The plot of the maximum response of the system
against the natural frequency of the system is called the shock spectrum of the

particular disturbance. The shock spectrum shows at a glance the natural frequencies
which cause large response amplitudes for the particular disturbance.
Non-Linear Vibrations
INTRODUCTION
Most physical systems can be represented by linear differential equations,
the types of which have been dealt with in the previous chapters. A general equation
of this type is
mx cx kx F(t)   (1)
In this equation which is for a linear system, the inertia force, the damping
force and the spring force are linear functions of x,x and x respectively. This is not
so in the case of non-linear systems. A general equation for a non-linear system is
mx ( )x f(x) F(t)    (2)
in which the damping force and the spring force are not linear functions of x and x.
There are quite some physical systems which have non-linear spring and damping
characteristics. Rubber springs and other similar isolators have spring stiffness which
increases with amplitude. Cast iron and concrete have spring stiffness which
decreases with amplitude. Examples of non-linear damping are dry friction damping
and material damping. Even so called linear systems tend to become non-linear with
larger amplitudes of vibration. The analysis of non-linear systems is comparatively
difficult. In certain cases there is no exact solution.
One major difference between the linear and non-linear systems is that the
law of superposition does not hold good for non-linear systems. Mathematically
speaking, if x1 is a solution of
1mx cx kx F (t)  
and x2 is a solution of
2mx cx kx F (t)  
then (x1 + x2) is a solution of

1 2mx cx kx F (t) F (t)   
This is not so in the case of non-linear systems. Even for the case of free
vibration any two known solutions of the non-linear system cannot be superimposed
to obtain a general solution.
PHASE PLANE
Phase plane was introduced in Sec. 9.6 for the case of linear systems. Here
we extend it for the case of non-linear systems.
Consider the differential equation
mx f(x) 0  (1)
The acceleration x can also be written as
dv
x v
dx

where v is the velocity of the particle. Substituting it in Eq. (1), we have
dv
mv f(x) 0
dx
 
or mv dv = –f(x) dx (2)
The above equation is intergrable directly. If v = V0 when x = X0, then the
integration of Eq. (2) gives
v x
V X0 0
mvdv f(x)dx   (2a)
or
22
0
0
mVmv
[F(x) F(X )]
2 2
  
The above equation is in accordance with the Law of Conservation of

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