THEORY OF MACHINE

Prepared by
Dr. Jyoti Vimal
Assistant Professor, Mechanical
Engineering Department

Unit- I Gear:
Classification,
Terminology/ Nomenclature of spur gear
Law of gearing,
Forms of teeth,
Tooth profile, Cycloidal and Involute tooth forms,
path of contact, teeth in contact,
Interference. Spur, Helical, Spiral, Worm and Bevel gears.

GEAR..
Gears are toothed members which transmit power / motion between two shafts
by meshing without any slip. Hence, gear drives are also called positive drives.
In any pair of gears, the smaller one is called pinion and the larger one is called
gear immaterial of which is driving the other.
When pinion is the driver, it results in step down drive in which the output speed
decreases and the torque increases. On the other hand, when the gear is the
driver, it results in step up drive in which the output speed increases and the
torque decreases.

GEARS.
Advantages

1. It transmits exact velocity ratio.

2. It may be used to transmit large power.
3. It has high efficiency.
4. It has reliable service.
5. It has compact layout.

Disadvantages

1. The manufacture of gears require special

tools and equipment.
2. The error in cutting teeth may cause
vibrations and noise during operation.

CLASSIFICATION
According to the position of axes of the shafts.

Parallel

1.Spur Gear
2.Helical Gear
3.Rack and Pinion

Intersecting

Bevel Gear

Non-intersecting and Non-parallel

worm and worm gears

SPUR GEAR

Teeth is parallel to axis of rotation.

Transmit power from one shaft to another

parallel shaft.
Used in Electric screwdriver, oscillating
sprinkler, windup alarm clock, washing machine
and clothes dryer.

EXTERNAL AND INTERNAL

SPUR GEAR

HELICAL GEAR
The teeth on helical gears are cut at an angle to
the face of the gear .
This gradual engagement makes helical gears
operate much more smoothly and quietly than
spur gears.
One interesting thing about helical gears is that
if the angles of the gear teeth are correct, they
can be mounted on perpendicular shafts,
adjusting the rotation angle by 90 degrees.

HELICAL GEAR..

HERRINGBONE GEARS

To avoid axial thrust,

two helical gears of opposite
hand can be mounted side by side,
to cancel resulting thrust forces.
Herringbone gears
are mostly used on
heavy machinery.

RACK AND PINION

Rack and pinion gears are

used to convert rotation
(From the pinion) into
linear motion (of the rack)
A perfect example of this
is the steering system
on many cars.

BEVEL GEARS
o

The two non-parallel or intersecting, but

coplanar shafts connected by gears are called
bevel gears and the arrangement is known
as bevel gearing. The bevel gears, like spur
gears, may also have their teeth inclined to the
face of the bevel, in which case they are known as
helical bevel gears.
Bevel gears are useful when the direction of a shaft's
rotation needs to be changed

STRAIGHT AND SPIRAL BEVEL GEARS

NOTES:
(a) When equal bevel gears (having equal
teeth) connect two shafts whose axes are
mutually perpendicular,then the bevel gears
are known as mitres.
(b) A hyperboloid is the solid formed by
revolving a straight line about an axis (not
in the same plane), such that every point on the
line remains at a constant distance from the axis.

WORM AND WORM GEAR

Worm and worm gear pair consists of a worm, which is very

similar to a screw and a worm gear, which is a helical gear. They
are used in right-angle skew shafts. In these gears, the
engagement occurs without any shock. The sliding action
prevalent in the system while resulting in quieter operation
produces considerable frictional heat. High reduction ratios 8 to
400 are possible.
Efficiency of these gears is low anywhere from 90% to 40 %.
Higher speed ratio gears are non-reversible. Their precision
rating is fair to good. They need good lubrication for heat
dissipation and for improving the efficiency. The drives are very
compact.

NOMENCLATURE OF SPUR GEARS

NOMENCLATURE OF SPUR GEARS

Pitch surface: The surface of the imaginary rolling
cylinder (cone, etc.) that the toothed gear may be
considered to replace.
Pitch circle: A right section of the pitch surface.
Addendum circle: A circle bounding the ends of the
teeth, in a right section of the gear.
Root (or dedendum) circle: The circle bounding the
spaces between the teeth, in a right section of the gear.
Addendum: The radial distance between the pitch circle
and the addendum circle.
Dedendum: The radial distance between the pitch circle
and the root circle.
Clearance: The difference between the dedendum of one
gear and the addendum of the mating gear

NOMENCLATURE OF SPUR GEARS

Face of a tooth: That part of the tooth surface lying
outside the pitch surface.
Flank of a tooth: The part of the tooth surface lying
inside the pitch surface.
Circular thickness (also called the tooth thickness):
The thickness of the tooth measured on the pitch circle.
It is the length of an arc and not the length of a straight
line.
Tooth space: pitch diameter The distance between
adjacent teeth measured on the pitch circle.
Backlash: The difference between the circle thickness of
one gear and the tooth space of the mating gear.
Circular pitch (Pc) : The width of a tooth and a space,
measured on the pitch circle.
Pc = m

NOMENCLATURE OF SPUR GEARS

Diametral pitch (Pd): The number of teeth of a gear unit

pitch diameter. The diametral pitch is, by definition, the
number of teeth divided by the pitch diameter. That is,

Where
Pd = diametral pitch
T = number of teeth
D = pitch diameter

Module (m): Pitch diameter divided by number of teeth.

The pitch diameter is usually specified in inches or
millimeters; in the former case the module is the inverse
of diametral pitch.
m = D/T

LAW OF GEARING
The fundamental law of gearing states
that the angular velocity ratio between
the gears of a gear set must remain
constant throughout the mesh. This
amounts to the following relationship
1/ 2 = n1/n2 = d2/d1 = z2/z1
where the terminology for the above is as
follows

LAW OF GEARING
NOTE: In order to maintain constant angular
velocity ratio between two meshing gears, the
common normal of the tooth profiles, at all
contact points with in mesh, must always
pass through a fixed point on the line of
centers, called pitch point.
Although the two profiles have different
velocities V1 and V2 at point K, their velocities
along N1N2 are equal in both magnitude and
direction. Otherwise the two tooth profiles
would separate from each other. Therefore,
we have

O1 N1 1 O2 N 2 2

4.1

LAW OF GEARING.
1 O2 N 2

2 O1 N1

4.2

We notice that the intersection of the

tangency N1N2 and the line of center O1O2
is point P, and from the similar triangles

O1N1 P O2 N 2 P

4.3

LAW OF GEARING
Therefore, velocity
ratio

1 O2 P

2 O1 P

4.4

LAW OF GEARING.
From the equations 4.2 and
4.4, we can write,

1 O2 P
O N

2 2
2
O1 P
O1 N1

4.5

ratio of the radii of the two base

circles and also given by;

O1 N1 O1 P cos
O2 N 2 O2 P cos

and

4.6

LAW OF GEARING.
centre distance between the base
circles

O1O2 O1 P O2 P
O1 N1 O2 N 2

cos
cos
O1 N1 O2 N 2

cos

4 .7

= pressure angle or the angle of

obliquity. It is angle between the
common normal to the base circles
and the common tangent to the pitch
circles

FORMS OF TEETH,

(a)

(b)

(c)

Profiles satisfying the law of gearing, (a) involute (b) cycloidal and
(c) Circular arc

FORMS OF TEETH..

Involute Gear Tooth Profile

Involute is the path generated by the end of a thread as it

unwinds from a reel. In order to understand what is
involute, imagine a reel with thread wound in the clockwise
direction as in Fig.1.2. Tie a knot at the end of the thread.
In the initial position, the thread is at B0 with knot on the
reel at C0. Keeping the reel stationary, pull the thread and
unwind it to position B1. The knot now moves from C0 to
C1. If the thread is unwound to position B2 the knot moves
to C2 position. In repeated unwinding, the taut thread
occupies position B3, B4 while the knot moves to C3, C4
positions. Connect these points C0 to C4 by a smooth curve,
the profile obtained is nothing but an involute, the
illustration of which is given below.

INVOLUTE GEAR TOOTH PROFILE

This forms the left side part of the tooth profile. If similar process is
repeated with thread wound on the reel in anticlockwise direction in the
same position, it forms the right side part of the same tooth. The
completely formed involute tooth is shown in Fig.

Fig.The generation of
involute profile on right side

Fig. The generation of

involute profile on left side

INVOLUTE GEAR TOOTH PROFILE

The common normal to the profile at the contact point will be tangent to the base
circles. It passes through a fixed point lying at the intersection of the tangent to
the rolling/pitch circles and the line connecting the centres of the gear wheels.
This point is known as the pitch point. As the gears rotate the contact point travels
along the common tangent to the base circle. Hence this line is also known as the
line of action.

Involute gear tooth profile

appearance after generation

Gear meshing

CYCLOIDAL GEAR TOOTH

PROFILE
Cycloid is the locus of a point on
the circumference of a circle
when it rolls on a straight line
without slipping. If the circle rolls
on the outside of another circle or
inside of another circle gives rise
to epicycloid and hypocycloid
respectively. The profile of a
cycloidal tooth consists of two
separate curves or double
curvature. This tooth form also
satisfies the law of gearing or
conjugate action similar to an
involute gear.
Figure illustrating the
generation of cycloidal tooth

Static force analysis.

If components of a machine accelerate, inertia is
produced due to their masses. However, the
magnitudes of these forces are small compares to the
externally applied loads. Hence inertia effect due to
masses are neglected. Such an analysis is known as
static force analysis
What is inertia?
The property of matter offering resistance to any
change of its state of rest or of uniform motion in a
straight line is known as inertia.

conditions for a body to be in static and dynamic

equilibrium?

Necessary and sufficient conditions for static and

dynamic equilibrium are
Vector sum of all forces acting on a body is zero
The vector sum of the moments of all forces acting about any
arbitrary point or axis is zero.

Static force analysis and dynamic force analysis.

If components of a machine accelerate, inertia forces are

produced due to their masses. If the magnitude of these
forces are small compared to the externally applied loads,
they can be neglected while analysing the mechanism. Such
an analysis is known as static force analysis.
If the inertia effect due to the mass of the component is also
considered, it is called dynamic force analysis.

DAlemberts principle.
DAlemberts principle states that the inertia forces and
torques, and the external forces and torques acting on a body
together result in statical equilibrium.
In other words, the vector sum of all external forces and
inertia forces acting upon a system of rigid bodies is zero.
The vector sum of all external moments and inertia torques
acting upon a system of rigid bodies is also separately zero.

The principle of super position states that for linear

systems the individual responses to several disturbances or
driving functions can be superposed on each other to obtain
the total response of the system.
The velocity and acceleration of various parts of reciprocating
mechanism can be determined , both analytically and
graphically.

Dynamic Analysis in Reciprocating Engines-Gas Forces

Piston efforts (Fp): Net force applied on the piston , along
the line of stroke In horizontal reciprocating engines.It is also
known as effective driving force (or) net load on the gudgeon
pin.
crank-pin effort.
The component of FQ perpendicular to the crank is known as
crank-pin effort.
crank effort or turning movement on the crank shaft?
It is the product of the crank-pin effort (F T)and crank pin
radius(r).

Forces acting on the connecting rod

Inertia force of the reciprocating parts (F1) acting along the
line of stroke.
The side thrust between the cross head and the guide bars
acting at right angles to line of stroke.
Weight of the connecting rod.
Inertia force of the connecting rod (FC)

The radial force (FR) parallel to crank and

The tangential force (FT) acting perpendicular to crank

Determination of Equivalent Dynamical System of

Two Masses by Graphical Method

Consider a body of mass m, acting at G as

shown in fig 15.15. This mass m, may be replaced

by two masses m1 and m2 so that the system becomes

dynamical equivalent. The position of mass m1 may be fixed
arbitrarily at A. Now draw perpendicular CG at G, equal in
length of the radius of gyration of the body, kG .Then join AC
and draw CB perpendicular to AC intersecting AG produced
in

B. The point B now fixes the position of the second

mass m2. The triangles ACG and BCG are similar. Therefore,

Turning movement diagram or crank effort diagram?

It is the graphical representation of the turning movement or

crank effort for various position of the crank.
In turning moment diagram, the turning movement is taken
as the ordinate (Y-axis) and crank angle as abscissa (X axis).

UNIT II : BALANCING
Static and dynamic balancing - Balancing of rotating
masses Balancing reciprocating massesBalancing a single cylinder Engine - Balancing
Multi-cylinder Engines, Balancing V-engines, Partial balancing in locomotive Engines-Balancing
machines.

STATIC AND DYNAMIC BALANCING

When man invented the wheel, he very quickly learnt that if

it wasnt completely round and if it didnt rotate evenly
about its central axis, then he had a problem!
What the problem he had?
The wheel would vibrate causing damage to itself and its
support mechanism and in severe cases, is unusable.
A method had to be found to minimize the problem. The
mass had to be evenly distributed about the rotating
centerline so that the resultant vibration was at a minimum.

UNBALANCE:

The condition which exists in a rotor when vibratory

force or motion is imparted to its bearings as a result
of centrifugal forces is called unbalance or the
uneven distribution of mass about a rotors rotating
centreline.

BALANCING:

Balancing is the technique of correcting or eliminating

unwanted inertia forces or moments in rotating or
reciprocating masses and is achieved by changing the
location of the mass centres.
The objectives of balancing an engine are to ensure:
1. That the centre of gravity of the system remains stationery
during a complete revolution of the crank shaft and
2. That the couples involved in acceleration of the different
moving parts balance each other.

Types of balancing:

a) Static Balancing:
i) Static balancing is a balance of forces due to action of gravity.
ii) A body is said to be in static balance when its centre of gravity
is in the axis of rotation.
b) Dynamic balancing:
i) Dynamic balance is a balance due to the action of inertia forces.
ii) A body is said to be in dynamic balance when the resultant
moments or couples, which involved in the acceleration of
different moving parts is equal to zero.
iii) The conditions of dynamic balance are met, the conditions of
static balance are also met.

BALANCING OF ROTATING MASSES

When a mass moves along a circular path, it

experiences a centripetal acceleration and a force is
required to produce it. An equal and opposite force
called centrifugal force acts radially outwards and is
a disturbing force on the axis of rotation. The
magnitude of this remains constant but the direction
changes with the rotation of the mass.

In a revolving rotor, the centrifugal force remains balanced as long as

the centre of the mass of rotor lies on the axis of rotation of the shaft.
When this does not happen, there is an eccentricity and an unbalance
force is produced. This type of unbalance is common in steam turbine
rotors, engine crankshafts, rotors of compressors, centrifugal pumps
etc.

The unbalance forces exerted on machine members are time varying, impart
vibratory motion and noise, there are human discomfort, performance of the
machine deteriorate and detrimental effect on the structural integrity of the
machine foundation.
Balancing involves redistributing the mass which may be carried out by
addition or removal of mass from various machine members. Balancing of
rotating masses can be of
1. Balancing of a single rotating mass by a single mass rotating in the same
plane.
2. Balancing of a single rotating mass by two masses rotating in different
planes.
3. Balancing of several masses rotating in the same plane
4. Balancing of several masses rotating in different planes

BALANCING OF A SINGLE ROTATING MASS BY A SINGLE

MASS ROTATING IN THE SAME PLANE

Consider a disturbing mass m1 which is attached to a shaft rotating at rad/s.

r = radius of rotation of the mass m

The centrifugal force exerted by mass m1 on the shaft is given by,
F=mrc11
This force acts radially outwards and produces bending moment on the shaft. In
order to counteract the effect of this force Fc1 , a balancing mass m2 may be
attached in the same plane of rotation of the disturbing mass m1 such that the
centrifugal forces due to the two masses are equal and opposite.

BALANCING OF A SINGLE ROTATING MASS BY TWO MASSES ROTATING

There are two possibilities while attaching two balancing masses:

1. The plane of the disturbing mass may be in between the planes of
the two balancing masses.
2. The plane of the disturbing mass may be on the left or right side of
two planes containing the balancing masses.
In order to balance a single rotating mass by two masses rotating in different
planes which are parallel to the plane of rotation of the disturbing mass i) the
net dynamic force acting on the shaft must be equal to zero, i.e. the centre
of the masses of the system must lie on the axis of rotation and this is the
condition for static balancing ii) the net couple due to the dynamic forces
acting on the shaft must be equal to zero, i.e. the algebraic sum of the
moments about any point in the plane must be zero. The conditions i) and ii)
together give dynamic balancing.

Balancing Multi-cylinder Engines, Balancing V-engines

Problem 1.
Four masses A, B, C and D are attached to a shaft and revolve in the
same plane. The masses are 12 kg, 10 kg, 18 kg and 15 kg
respectively and their radii of rotations are 40 mm, 50 mm, 60 mm
and 30 mm. The angular position of the masses B, C and D are 60 ,
135 and 270 from mass A. Find the magnitude and position of the
balancing mass at a radius of 100 mm.
Problem 2:
The four masses A, B, C and D are 100 kg, 150 kg, 120 kg and 130 kg
attached to a shaft and revolve in the same plane. The corresponding
radii of rotations are 22.5 cm, 17.5 cm, 25 cm and 30 cm and the angles
measured from A are 45, 120 and 255. Find the position and
magnitude of the balancing mass, if the radius of rotation is 60 cm.

UNIT III : FREE VIBRATION

Basic features of vibratory systems - idealized
models - Basic elements and lumping of
parameters - Degrees of freedom - Single degree
of freedom - Free vibration - Equations of motion natural frequency - Types of Damping - Damped
vibration critical speeds of simple shaft - Torsional
systems; Natural frequency of two and three rotor
systems

INTRODUCTION
Mechanical vibration is the motion of a particle or body which
oscillates about a position of equilibrium. Most vibrations in
machines and structures are undesirable due to increased stresses
and energy losses.
Time interval required for a system to complete a full cycle of the
motion is the period of the vibration.
Number of cycles per unit time defines the frequency of the vibrations.
Maximum displacement of the system from the equilibrium position is the
amplitude of the vibration.
When the motion is maintained by the restoring forces only, the vibration
is described as free vibration. When a periodic force is applied to the
system, the motion is described as forced vibration.
When the frictional dissipation of energy is neglected, the motion
is said to be undamped. Actually, all vibrations are damped to
some degree.
19 - 65

FREE VIBRATIONS OF PARTICLES. SIMPLE HARMONIC

MOTION
If a particle is displaced through a distance xm from its
equilibrium position and released with no velocity, the
particle will undergo simple harmonic motion,

ma F W k st x kx
mx kx 0
General solution is the sum of two particular solutions,

x C1 sin

k
t C 2 cos
m

C1 sin n t C 2 cos n t

k
t
m

x is a periodic function and n is the natural circular

frequency of the motion.
C1 and C2 are determined by the initial conditions:

x C1 sin n t C 2 cos n t

v x C1 n cos n t C 2 n sin n t

19 - 66

C 2 x0
C1 v0 n

FREE VIBRATIONS OF PARTICLES. SIMPLE

HARMONIC MOTION

C1

v0
n

C 2 x0

Displacement is equivalent to the x component of the sum of two vectors C1 C 2
which rotate with constant angular velocity .
n
x xm sin n t

xm

v0 n 2 x02

amplitude

tan 1 v0 x0 n phase angle

2
period
n
1 n
fn

natural frequency
n 2

19 - 67

FREE VIBRATIONS OF PARTICLES. SIMPLE HARMONIC

MOTION
Velocity-time and acceleration-time curves can be
represented by sine curves of the same period as the
displacement-time curve but different phase angles.

x xm sin n t
v x

xm n cos n t

xm n sin n t 2
a x
xm n2 sin n t
xm n2 sin n t

19 - 68

SIMPLE PENDULUM (APPROXIMATE

SOLUTION)

Results obtained for the spring-mass system can be

applied whenever the resultant force on a particle is
proportional to the displacement and directed towards
the equilibrium position.

Consider tangential components of acceleration and

force for a simple pendulum,

Ft mat : W sin ml
g sin 0
l
for small angles,

g
0
l
m sin n t

n
19 - 69

2
l
2
n
g

SIMPLE PENDULUM (EXACT

SOLUTION)
g
An exact solution for

sin 0
l

leads to

l 2
d
n 4

g 0 1 sin 2 2 sin 2
m

which requires numerical solution.

2K
l
2

n

g

19 - 70

SAMPLE PROBLEM
For each spring arrangement, determine
the spring constant for a single
equivalent spring.
Apply the approximate relations for the
harmonic motion of a spring-mass
system.
A 50-kg block moves between vertical
guides as shown. The block is pulled
40mm down from its equilibrium position
and released.
For each spring arrangement,
determine a) the period of the vibration,
b) the maximum velocity of the block,
and c) the maximum acceleration of the
block.
19 - 71

SAMPLE PROBLEM
k1 4 kN m k2 6 kN m

Springs in parallel:
- determine the spring constant for equivalent
spring
- apply the approximate relations for the
harmonic motion of a spring-mass system
k
10 4 N/m
n

14.14 rad s
m
20 kg

n
P k1 k2
k

n 0.444 s

vm x m n

0.040 m 14.14 rad s

P
k1 k2

10 kN m 10 N m

19 - 72

2
n

vm 0.566 m s

am x m an2
0.040 m 14.14 rad s 2

am 8.00 m s 2

SAMPLE PROBLEM
k1 4 kN m k2 6 kN m

Springs in series:
- determine the spring constant for equivalent
spring
- apply the approximate relations for the harmonic
motion of a spring-mass system

n
n

k
2400N/m

6.93 rad s
m
20 kg
2
n

n 0.907 s

vm x m n
P k1 k2
k

P
k1 k2

10 kN m 104 N m

19 - 73

0.040 m 6.93 rad s

vm 0.277 m s

am x m an2
0.040 m 6.93 rad s 2

am 1.920 m s 2

FREE VIBRATIONS OF RIGID BODIES

If an equation of motion takes the form
2
x n
x 0

or

2
n
0

the corresponding motion may be considered

as simple harmonic motion.
Analysis objective is to determine n.
Consider the oscillations of a square plate

I
W b sin2 mb

2
2

1 m 2b 2b
but I 12

2 mb
3

, W mg

3g
3g

sin
0
5b
5b
3g
2
5b
then n
, n
2
5b
n
3g
For an equivalent simple pendulum,
19 - 74

l 5b 3

SAMPLE PROBLEM

From the kinematics of the system, relate

the linear displacement and acceleration
to the rotation of the cylinder.
Based on a free-body-diagram equation for
the equivalence of the external and
effective forces, write the equation of
motion.

A cylinder of weight W is suspended as Substitute the kinematic relations to arrive

shown.
at an equation involving only the angular
displacement and acceleration.
Determine the period and natural
frequency of vibrations of the cylinder.

19 - 75

SAMPLE PROBLEM
From the kinematics of the system, relate the linear
displacement and acceleration to the rotation of the cylinder.

x r

2 x 2r

a r

a r r

Based on a free-body-diagram equation for the equivalence of

the external and effective forces, write the equation of motion.

M A M A eff :

Wr T2 2r ma r I

but T2 T0 k 12 W k 2r
Substitute the kinematic relations to arrive at an equation
involving only the angular displacement and acceleration.

Wr 12 W 2kr 2r m r r 12 mr 2
8k

0
3m

19 - 76

8k
n
3m

2
3m
2
n
8k

fn

n 1 8k

2 2 3m

SAMPLE PROBLEM
Using the free-body-diagram equation for
the equivalence of the external and
effective moments, write the equation of
motion for the disk/gear and wire.
W 20 lb
n 1.13 s

n 1.93 s

With the natural frequency and moment of

inertia for the disk known, calculate the
torsional spring constant.

The disk and gear undergo torsional

vibration with the periods shown.
With natural frequency and spring
Assume that the moment exerted by the constant known, calculate the moment of
wire is proportional to the twist angle.
inertia for the gear.
Determine a) the wire torsional spring
Apply the relations for simple harmonic
constant, b) the centroidal moment of
motion to calculate the maximum gear
inertia of the gear, and c) the maximum
velocity.
angular velocity of the gear if rotated
through 90o and released.
19 - 77

SAMPLE PROBLEM
Using the free-body-diagram equation for the equivalence of
the external and effective moments, write the equation of
motion for the disk/gear and wire.

M O M O eff :
W 20 lb
n 1.13 s

n 1.93 s

K
I

K I
K
0
I

2
I
2
n
K

With the natural frequency and moment of inertia for

the disk known, calculate the torsional spring
constant.
2
1
20
8

2
I 12 mr 2

0.138 lb ft s
2 32.2 12
1.13 2
19 - 78

0.138
K

K 4.27 lb ft rad

SAMPLE PROBLEM
With natural frequency and spring constant
known, calculate the moment of inertia for the
gear.
I
1.93 2
I 0.403 lb ft s 2
4.27
W 20 lb
n 1.13 s

n 1.93 s

Apply the relations for simple harmonic motion

to calculate the maximum gear velocity.

m sin nt

m n sin nt

m m n

m 90 1.571 rad

K
I

2
I
2
n
K

K 4.27 lb ft rad
19 - 79

2
2
1.571 rad

1
.
93
s

m m

m 5.11 rad s

PRINCIPLE OF CONSERVATION OF
ENERGY

Resultant force on a mass in simple harmonic motion

is conservative - total energy is conserved.

T V constant

2
1 mx

2
2

12 kx 2 constant

x n2 x 2

Consider simple harmonic motion of the square plate,

T1 0

V1 Wb1 cos Wb 2 sin 2 m 2

12 Wb m2
2
T2 12 mvm2 12 I m
2
12 m bm 12

12

53 mb 2 m2

23 mb 2 m2

V2 0

T1 V1 T2 V2
19 - 80

0 12 Wb m2 12

53 mb 2 m2 n2 0

n 3 g 5b

SAMPLE PROBLEM
Apply the principle of conservation of
energy between the positions of maximum
and minimum potential energy.
Solve the energy equation for the natural
frequency of the oscillations.

Determine the period of small

oscillations of a cylinder which rolls
without slipping inside a curved
surface.

19 - 81

SAMPLE PROBLEM
Apply the principle of conservation of energy between
the positions of maximum and minimum potential
energy.

T1 V1 T2 V2
V1 Wh W R r 1 cos

T1 0

W R r m2 2

V2 0

2
T2 12 mvm2 12 I m

1 m R r 2
m
2

12

34 m R r 2 m2

19 - 82

1 mr R r 2

m
2
r
2

SAMPLE PROBLEM

Solve the energy equation for the natural frequency

of the oscillations.

T1 0

V1 W R r m2 2

T2 34 m R r 2m2

V2 0

T1 V1 T2 V2

m2 3
0 W R r
4 m R r 2 m2 0
2
m2 3
mg R r 4 m R r 2 m n 2m
2

n2
19 - 83

2
2 g

n
n
3 Rr

3 Rr
2 g

FORCED VIBRATIONS
Forced vibrations - Occur
when a system is subjected
to a periodic force or a
periodic displacement of a
support.

f forced frequency

F ma :

Pm sin f t W k st x mx

W k st x m sin f t mx

mx kx Pm sin f t

mx kx k m sin f t

19 - 84

FORCED VIBRATIONS
x xcomplementary x particular

C1 sin n t C 2 cos n t xm sin f t

Substituting particular solution into governing equation,

m 2f xm sin f t kxm sin f t Pm sin f t

xm

Pm
k m 2f

Pm k

1 f n

1 f n

mx kx Pm sin f t
mx kx k m sin f t
At f = n, forcing input is in
resonance with the system.
19 - 85

SAMPLE PROBLEM
The resonant frequency is equal to the
natural frequency of the system.
Evaluate the magnitude of the periodic
force due to the motor unbalance.
Determine the vibration amplitude from
the frequency ratio at 1200 rpm.
A motor weighing 350 lb is supported by
four springs, each having a constant 750
lb/in. The unbalance of the motor is
equivalent to a weight of 1 oz located 6
in. from the axis of rotation.
Determine a) speed in rpm at which
resonance will occur, and b) amplitude of
the vibration at 1200 rpm.
19 - 86

SAMPLE PROBLEM
The resonant frequency is equal to the natural
frequency of the system.
m

W = 350 lb
k = 4(350
lb/in)

350
10.87 lb s 2 ft
32.2

k 4 750 3000 lb in
36,000 lb ft
k
36,000

m
10.87
57.5 rad/s 549 rpm

Resonance speed = 549

rpm

19 - 87

SAMPLE PROBLEM
Evaluate the magnitude of the periodic force due to
the motor unbalance. Determine the vibration
amplitude from the frequency ratio at 1200 rpm.

f 1200 rpm 125.7 rad/s

1 lb
m 1 oz

16
oz

W = 350 lb
k = 4(350
lb/in)

n 57.5 rad/s

0.001941 lb s 2 ft
32.2 ft s 2

Pm man mr 2
0.001941
xm

126 125.7 2 15.33 lb

Pm k

1 f n

15.33 3000
1 125.7 57.5 2

0.001352 in
xm = 0.001352 in. (out of
phase)
19 - 88

DAMPED FREE VIBRATIONS

All vibrations are damped to some degree by
forces due to dry friction, fluid friction, or internal
friction.
With viscous damping due to fluid friction,

F ma :

W k st x cx mx
mx cx kx 0

Substituting x = et and dividing through by et

yields the characteristic equation,

m2 c k 0

c
c

2m
2m

k
m

Define the critical damping coefficient such that

19 - 89

cc

2
m

k
0
m

cc 2 m

k
2m n
m

DAMPED FREE VIBRATIONS

Characteristic equation,

c
c

2m
2m

m2 c k 0

k
m

cc 2m n critical damping coefficient

Heavy damping: c > cc

x C1e 1t C2 e 2t

- negative roots
- nonvibratory motion

Critical damping: c = cc

x C1 C 2t e nt

- double roots
- nonvibratory motion

Light damping: c < cc

x e c

d n
19 - 90

2m t

C1 sin d t C2 cos d t

1
cc

damped frequency

DAMPED FORCED VIBRATIONS

mx cx kx Pm sin f t

x xcomplementary x particular

xm
xm

Pm k

tan
19 - 91

2 c cc f n

1 f n

2 c c

2 2

n 2

magnification
factor

phase difference between forcing and steady

state response

ELECTRICAL ANALOGUES
Consider an electrical circuit consisting of an inductor,
resistor and capacitor with a source of alternating
voltage

E m sin f t L

di
q
Ri 0
dt
C

1
Lq Rq q E m sin f t
C
Oscillations of the electrical system are analogous to
damped forced vibrations of a mechanical system.

19 - 92

ELECTRICAL ANALOGUES
The analogy between electrical and mechanical
systems also applies to transient as well as steadystate oscillations.
With a charge q = q0 on the capacitor, closing the
switch is analogous to releasing the mass of the
mechanical system with no initial velocity at x = x0.
If the circuit includes a battery with constant voltage
E, closing the switch is analogous to suddenly
applying a force of constant magnitude P to the
mass of the mechanical system.

19 - 93

ELECTRICAL ANALOGUES
The electrical system analogy provides a means of
experimentally determining the characteristics of a given
mechanical system.
For the mechanical system,

m1x1 c1 x1 c2 x1 x 2 k1 x1 k 2 x1 x2 0
m2 x2 c2 x 2 x1 k 2 x2 x1 Pm sin f t
For the electrical system,

q1 q1 q2

0
C1
C2
q q
L2 q2 R2 q 2 q 1 2 1 Em sin f t
C2
L1q1 R1 q 1 q 2

The governing equations are equivalent. The characteristics

of the vibrations of the mechanical system may be inferred
from the oscillations of the electrical system.
19 - 94

UNIT IV : FORCED VIBRATION

Response to periodic forcing - Harmonic Forcing Forcing caused by unbalance - Support motion
Force transmissibility and amplitude transmissibility
- Vibration isolation.

DAMPING
a process whereby energy is taken from the
vibrating system and is being absorbed by the
surroundings.
Examples of damping forces:

internal forces of a spring,

viscous force in a fluid,
electromagnetic damping in galvanometers,
shock absorber in a car.

DAMPED VIBRATION (1)

The oscillating system is opposed by dissipative

forces.
The system does positive work on the
surroundings.
Examples:
a mass oscillates under water
oscillation of a metal plate in the magnetic field

DAMPED VIBRATION (2)

Total energy of the oscillator decreases with time

The rate of loss of energy depends on the
instantaneous velocity
Resistive force instantaneous velocity
i.e. F = -bv where b = damping coefficient
Frequency of damped vibration < Frequency of
undamped vibration

TYPES OF DAMPED OSCILLATIONS

(1)

Slight damping (underdamping)

Characteristics:
- oscillations with reducing amplitudes

- amplitude decays exponentially with time

- period is slightly longer
- Figure

a1 a2 a3

....... a constant
a2 a3 a4

TYPES OF DAMPED OSCILLATIONS

(2)

Critical damping

No real oscillation
Time taken for the displacement to become effective
zero is a minimum

TYPES OF DAMPED OSCILLATIONS

(3)

Heavy damping (Overdamping)

Resistive forces exceed those of

critical damping
The system returns very slowly to
the equilibrium position

EXAMPLE: MOVING COIL

GALVANOMETER

the deflection of the pointer is critically damped

EXAMPLE: MOVING COIL

GALVANOMETER

Damping is due to
induced currents flowing
in the metal frame
The opposing couple
setting up causes the coil
to come to rest quickly

FORCED OSCILLATION

The system is made to oscillate by periodic impulses

from an external driving agent
Experimental setup:

CHARACTERISTICS OF FORCED
OSCILLATION

Same frequency as the driver system

Constant amplitude
Transient oscillations at the beginning which
eventually settle down to vibrate with a constant
amplitude (steady state)

CHARACTERISTICS OF FORCED
OSCILLATION

In steady state, the system vibrates at the frequency

of the driving force

ENERGY

Amplitude of vibration is fixed for a specific

driving frequency
Driving force does work on the system at the
same rate as the system loses energy by doing
work against dissipative forces
Power of the driver is controlled by damping

AMPLITUDE

Amplitude of vibration depends on

the relative values of the natural

frequency of free oscillation
the frequency of the driving force
the extent to which the system is
damped

EFFECTS OF DAMPING

Driving frequency for maximum amplitude

becomes slightly less than the natural frequency
Reduces the response of the forced system

PHASE (1)

The forced vibration takes on the frequency of the

driving force with its phase lagging behind
If F = F0 cos t, then
x = A cos (t - )
where is the phase lag of x behind F

PHASE (2)

Figure
1. As f 0, 0
2. As f ,
3. As f f0, /2
Explanation

When x = 0, it has no tendency to move. maximum

force should be applied to the oscillator

PHASE (3)
When oscillator moves away from the centre, the
driving force should be reduced gradually so that the
oscillator can decelerate under its own restoring force
At the maximum displacement, the driving force
becomes zero so that the oscillator is not pushed any
further
Thereafter, F reverses in direction so that the
oscillator is pushed back to the centre

PHASE (4)

On reaching the centre, F is a

maximum in the opposite direction
Hence, if F is applied 1/4 cycle
earlier than x, energy is supplied
to the oscillator at the correct
moment. The oscillator then
responds with maximum
amplitude.

FORCED VIBRATION

Adjust the position of the load on the driving

pendulum so that it oscillates exactly at a
frequency of 1 Hz
Couple the oscillator to the driving pendulum by
the given elastic cord
Set the driving pendulum going and note the
response of the blade

FORCED VIBRATION

In steady state, measure the amplitude of forced

vibration
Measure the time taken for the blade to perform
10 free oscillations
Adjust the position of the tuning mass to change
the natural frequency of free vibration and repeat
the experiment

FORCED VIBRATION

Plot a graph of the amplitude of vibration at

different natural frequencies of the oscillator
Change the magnitude of damping by rotating
the card through different angles
Plot a series of resonance curves

RESONANCE (1)

Resonance occurs when an oscillator is acted upon by

a second driving oscillator whose frequency equals
the natural frequency of the system
The amplitude of reaches a maximum
The energy of the system becomes a maximum
The phase of the displacement of the driver leads that
of the oscillator by 90

RESONANCE (2)

Examples

Mechanics:

Oscillations of a childs swing

Destruction of the Tacoma Bridge

Sound:

An opera singer shatters a wine glass

Resonance tube
Kundts tube

RESONANCE

Electricity

Radio tuning

Light

Maximum absorption of infrared waves by a NaCl crystal

RESONANT SYSTEM

There is only one value of the driving frequency

for resonance, e.g. spring-mass system
There are several driving frequencies which give
resonance, e.g. resonance tube

RESONANCE: UNDESIRABLE

The body of an aircraft should not resonate with

the propeller
The springs supporting the body of a car should
not resonate with the engine

DEMONSTRATION OF RESONANCE

Resonance tube

Place a vibrating tuning fork above the mouth of the

measuring cylinder
Vary the length of the air column by pouring water
into the cylinder until a loud sound is heard
The resonant frequency of the air column is then
equal to the frequency of the tuning fork

DEMONSTRATION OF RESONANCE

Sonometer

Press the stem of a vibrating tuning fork against the

bridge of a sonometer wire
Adjust the length of the wire until a strong vibration
is set up in it
The vibration is great enough to throw off paper
riders mounted along its length

Oscillation of a metal plate in the

magnetic field

SLIGHT DAMPING

CRITICAL DAMPING

HEAVY DAMPING

AMPLITUDE

PHASE

BARTONS PENDULUM

DAMPED VIBRATION

RESONANCE CURVES

RESONANCE TUBE

A glass tube has a

variable water level
and a speaker at its
upper end

UNIT V :

GOVERNORS AND GYROSCOPES

Governors - Types - Centrifugal governors - Gravity

controlled and spring controlled centrifugal
governors Characteristics - Effect of friction Controlling Force .
Gyroscopes - Gyroscopic forces and Torques Gyroscopic stabilization - Gyroscopic effects in
Automobiles, ships and airplanes

GOVERNORS
Engine

Speed control

This presentation is from Virginia Tech and has not been edited by
Georgia Curriculum Office.

GOVERNORS
Governors serve three basic purposes:
Maintain a speed selected by the operator which
is within the range of the governor.
Prevent over-speed which may cause engine
damage.
Limit both high and low speeds.

GOVERNORS

Generally governors are used to maintain a fixed

speed not readily adjustable by the operator or to
maintain a speed selected by means of a throttle
control lever.
In either case, the governor protects against
overspeeding.

HOW DOES IT WORK?

If the load is removed on an operating engine, the
governor immediately closes the throttle.
If the engine load is increased, the throttle will
be opened to prevent engine speed form being
reduced.

EXAMPLE
The

governor on your
lawnmower maintains
the selected engine
speed even when you
mow through a clump
of high grass or when
you mow over no grass
at all.

PNEUMATIC GOVERNORS
Sometimes

called airvane governors, they

are operated by the
stream of air flow
created by the cooling
fins of the flywheel.

AIR-VANE GOVERNOR
When the engine experiences sudden increases in
load, the flywheel slows causing the governor to
open the throttle to maintain the desired speed.
The same is true when the engine experiences a
decrease in load. The governor compensates and
closes the throttle to prevent overspeeding.

CENTRIFUGAL GOVERNOR
Sometimes

referred to
as a mechanical
governor, it uses
pivoted flyweights
that are attached to a
revolving shaft or gear
driven by the engine.

MECHANICAL GOVERNOR
With

this system, governor rpm is always

directly proportional to engine rpm.

MECHANICAL GOVERNOR
If the engine is subjected to a sudden load that
reduces rpm, the reduction in speed lessens
centrifugal force on the flyweights.
The weights move inward and lower the spool and
governor lever, thus opening the throttle to
maintain engine speed.

VACUUM GOVERNORS
Located

between the carburetor and the intake

manifold.
It senses changes in intake manifold pressure
(vacuum).

VACUUM GOVERNORS
As

engine speed increases or decreases the

governor closes or opens the throttle respectively
to control engine speed.

HUNTING
Hunting is a condition whereby the engine speed
fluctuate or is erratic usually when first started.
The engine speeds up and slows down over and
over as the governor tries to regulate the engine
speed.
This is usually caused by an improperly adjusted
carburetor.

STABILITY
Stability is the ability to maintain a desired
engine speed without fluctuating.
Instability results in hunting or oscillating due to
over correction.
Excessive stability results in a dead-beat
governor or one that does not correct sufficiently
for load changes.

SENSITIVITY

Sensitivity is the percent of speed change

required to produce a corrective movement of the
fuel control mechanism.
High governor sensitivity will help keep the
engine operating at a constant speed.

SUMMARY

Small engine governors are used to:

Maintain

selected engine speed.

Prevent over-speeding.
Limit high and low speeds.

SUMMARY

Governors are usually of the following types:

Air-vane

(pneumatic)
Mechanical (centrifugal)
Vacuum

SUMMARY

The governor must have stability and sensitivity

in order to regulate speeds properly. This will
prevent hunting or erratic engine speed changes
depending upon load changes.

Gyroscope

A gyroscope consists of a rotor mounted in the inner gimbal. The inner

gimbal is mounted in the outer gimbal which itself is mounted on a
fixed frame as shown in Fig. When the rotor spins about X-axis with
angular velocity rad/s and the inner gimbal precesses (rotates)
about Y-axis, the spatial mechanism is forced to turn about Z-axis
other than its own axis of rotation, and the gyroscopic effect is thus
setup. The resistance to this motion is called gyroscopic effect.

GYROSCOPIC COUPLE

Consider a rotary body of mass m having radius of gyration k mounted on the

shaft supported at two bearings. Let the rotor spins (rotates) about X-axis with
constant angular velocity rad/s. The X-axis is, therefore, called spin axis, Yaxis, precession axis and Z-axis, the couple or torque axis .

GYROSCOPIC EFFECT ON SHIP

THANK YOU