Ppt mech 5sem_dom

DYNAMICS OF MACHINERY
U5MEA19
Prepared by
Mr.Shaik Shabbeer
Mr.Vennishmuthu.V
Assistant Professor, Mechanical Department
VelTech Dr.RR & Dr.SR Technical
University

UNIT I : FORCE ANALYSIS
Rigid Body dynamics in general plane motion – Equations of
motion - Dynamic force analysis - Inertia force and Inertia
torque – D’Alemberts principle - The principle of
superposition - Dynamic Analysis in Reciprocating Engines –
Gas Forces - Equivalent masses - Bearing loads - Crank shaft
Torque - Turning moment diagrams - Fly wheels –Engine
shaking Forces - Cam dynamics - Unbalance, Spring, Surge
and Windup.

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.

D’Alembert’s principle.
 D’Alembert’s 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 (FT)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 masses-
Balancing 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 wasn’t completely round and if it didn’t rotate evenly
about it’s central axis, then he had a problem!
What the problem he had?
The wheel would vibrate causing damage to itself and it’s
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 rotor’s 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 = m r c 1 1
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
19 - 36
• 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.

FREE VIBRATIONS OF PARTICLES. SIMPLE HARMONIC
MOTION
19 - 37
• 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,
( )
0=+
−=+−==
kxxm
kxxkWFma st

δ
• General solution is the sum of two particular solutions,
( ) ( )tCtC
t
m
k
Ct
m
k
Cx
nn ωω cossin
cossin
21
21
+=






+





=
• x is a periodic function and ωn is the natural circular
frequency of the motion.
• C1 and C2 are determined by the initial conditions:
( ) ( )tCtCx nn ωω cossin 21 += 02 xC =
nvC ω01 =( ) ( )tCtCxv nnnn ωωωω sincos 21 −== 

FREE VIBRATIONS OF PARTICLES. SIMPLE
HARMONIC MOTION
19 - 38
( )φω += txx nm sin
==
n
n
ω
π
τ
2
period
===
π
ω
τ 2
1 n
n
nf natural frequency
( ) =+= 2
0
2
0 xvx nm ω amplitude
( ) == −
nxv ωφ 00
1
tan phase angle
• Displacement is equivalent to the x component of the sum of two vectors
which rotate with constant angular velocity
21 CC

+
.nω
02
0
1
xC
v
C
n
=
=
ω

FREE VIBRATIONS OF PARTICLES. SIMPLE HARMONIC
MOTION
19 - 39
( )φω += txx nm sin
• 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.
( )
( )2sin
cos
πφωω
φωω
++=
+=
=
tx
tx
xv
nnm
nnm

( )
( )πφωω
φωω
++=
+−=
=
tx
tx
xa
nnm
nnm
sin
sin
2
2


SIMPLE PENDULUM (APPROXIMATE
SOLUTION)
19 - 40
• 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.
for small angles,
( )
g
l
t
l
g
n
n
nm
π
ω
π
τ
φωθθ
θθ
2
2
sin
0
==
+=
=+
:tt maF =∑
• Consider tangential components of acceleration and
force for a simple pendulum,
0sin
sin
=+
=−
θθ
θθ
l
g
mlW



SIMPLE PENDULUM (EXACT
SOLUTION)
19 - 41
0sin =+ θθ
l
gAn exact solution for
leads to
( )
∫
−
=
2
0
22
sin2sin1
4
π
φθ
φ
τ
m
n
d
g
l
which requires numerical solution.






=
g
lK
n π
π
τ 2
2

SAMPLE PROBLEM
19 - 42
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.
• 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.

SAMPLE PROBLEM
19 - 43
mkN6mkN4 21 == kk
• Springs in parallel:
- determine the spring constant for equivalent
spring
mN10mkN10 4
21
21
==
+==
+=
kk
P
k
kkP
δ
δδ
- apply the approximate relations for the
harmonic motion of a spring-mass system
n
n
n
m
k
ω
π
τ
ω
2
srad14.14
kg20
N/m104
=
===
s444.0=nτ
( )( )srad4.141m040.0=
= nmm xv ω
sm566.0=mv
2
sm00.8=ma( )( )2
2
srad4.141m040.0=
= nmm axa

SAMPLE PROBLEM
19 - 44
mkN6mkN4 21 == kk • 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
n
m
k
ω
π
τ
ω
2
srad93.6
kg20
400N/m2
=
===
s907.0=nτ
( )( )srad.936m040.0=
= nmm xv ω
sm277.0=mv
2
sm920.1=ma( )( )2
2
srad.936m040.0=
= nmm axa
mN10mkN10 4
21
21
==
+==
+=
kk
P
k
kkP
δ
δδ

FREE VIBRATIONS OF RIGID BODIES
19 - 45
• If an equation of motion takes the form
0or0 22
=+=+ θωθω nn xx 
the corresponding motion may be considered
as simple harmonic motion.
• Analysis objective is to determine ωn.
( ) ( )[ ] mgWmbbbmI ==+= ,22but 2
3
222
12
1
0
5
3
sin
5
3
=+≅+ θθθθ
b
g
b
g 
g
b
b
g
n
nn
3
5
2
2
,
5
3
then π
ω
π
τω ===
• For an equivalent simple pendulum,
35bl =
• Consider the oscillations of a square plate
( ) ( ) θθθ  ImbbW +=− sin

SAMPLE PROBLEM
19 - 46
k
A cylinder of weight W is suspended as
shown.
Determine the period and natural
frequency of vibrations of the cylinder.
• 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.
• Substitute the kinematic relations to arrive
at an equation involving only the angular
displacement and acceleration.

SAMPLE PROBLEM
19 - 47
• From the kinematics of the system, relate the linear
displacement and acceleration to the rotation of the cylinder.
θrx = θδ rx 22 ==
θα rra ==θα 
= θ
ra =
• Based on a free-body-diagram equation for the equivalence of
the external and effective forces, write the equation of motion.
( ) :∑∑ = effAA MM ( ) αIramrTWr +=− 22
( )θδ rkWkTT 2but 2
1
02 +=+=
• Substitute the kinematic relations to arrive at an equation
involving only the angular displacement and acceleration.
( )( ) ( )
0
3
8
22 2
2
1
2
1
=+
+=+−
θθ
θθθ
m
k
mrrrmrkrWWr


m
k
n
3
8
=ω k
m
n
n
8
3
2
2
π
ω
π
τ ==
m
k
f n
n
3
8
2
1
2 ππ
ω
==

SAMPLE PROBLEM
19 - 48
s13.1
lb20
n =
=
τ
W
s93.1=nτ
The disk and gear undergo torsional
vibration with the periods shown.
Assume that the moment exerted by the
wire is proportional to the twist angle.
Determine a) the wire torsional spring
constant, b) the centroidal moment of
inertia of the gear, and c) the maximum
angular velocity of the gear if rotated
through 90o
and released.
• 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.
• With the natural frequency and moment of
inertia for the disk known, calculate the
torsional spring constant.
• With natural frequency and spring
constant known, calculate the moment of
inertia for the gear.
• Apply the relations for simple harmonic
motion to calculate the maximum gear
velocity.

SAMPLE PROBLEM
19 - 49
s13.1
lb20
n =
=
τ
W
s93.1=nτ
• 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.
( ) :∑∑ = effOO MM
0=+
−=+
θθ
θθ
I
K
IK


K
I
I
K
n
nn π
ω
π
τω 2
2
===
• With the natural frequency and moment of inertia for
the disk known, calculate the torsional spring
constant.
2
2
2
2
1 sftlb138.0
12
8
2.32
20
2
1
⋅⋅=











== mrI
K
138.0
213.1 π= radftlb27.4 ⋅=K

SAMPLE PROBLEM
19 - 50
s13.1
lb20
n =
=
τ
W
s93.1=nτ
radftlb27.4 ⋅=K
K
I
I
K
n
nn π
ω
π
τω 2
2
===
• With natural frequency and spring constant
known, calculate the moment of inertia for the
gear.
27.4
293.1
I
π= 2
sftlb403.0 ⋅⋅=I
• Apply the relations for simple harmonic motion
to calculate the maximum gear velocity.
nmmnnmnm tt ωθωωωθωωθθ === sinsin
rad571.190 =°=mθ
( ) 





=





=
s93.1
2
rad571.1
2 π
τ
π
θω
n
mm
srad11.5=mω

PRINCIPLE OF CONSERVATION OF
ENERGY
19 - 51
• Resultant force on a mass in simple harmonic motion
is conservative - total energy is conserved.
constant=+VT
=+
=+
222
2
2
12
2
1 constant
xx
kxxm
nω

• Consider simple harmonic motion of the square plate,
01 =T ( ) ( )[ ]
2
2
1
2
1 2sin2cos1
m
m
Wb
WbWbV
θ
θθ
≅
=−=
( ) ( )
( ) 22
3
5
2
1
22
3
2
2
12
2
1
2
2
12
2
1
2
m
mm
mm
mb
mbbm
IvmT
θ
ωθ
ω


=
+=
+= 02 =V
( ) 00 222
3
5
2
12
2
1
2211
+=+
+=+
nmm mbWb
VTVT
ωθθ bgn 53=ω

SAMPLE PROBLEM
19 - 52
Determine the period of small
oscillations of a cylinder which rolls
without slipping inside a curved
surface.
• 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.

SAMPLE PROBLEM
19 - 53
01 =T ( )( )
( )( )2
cos1
2
1
mrRW
rRWWhV
θ
θ
−≅
−−==
( ) ( )
( ) 22
4
3
2
2
2
2
1
2
12
2
1
2
2
12
2
1
2
m
mm
mm
rRm
r
rR
mrrRm
IvmT
θ
θθ
ω


−=





 −
+−=
+= 02 =V
• Apply the principle of conservation of energy between
the positions of maximum and minimum potential
energy.
2211 VTVT +=+

SAMPLE PROBLEM
19 - 54
2211 VTVT +=+
( ) ( ) 0
2
0 22
4
3
2
+−=−+ m
m rRmrRW θ
θ 
( )( ) ( ) ( )22
4
3
2
2 mnm
m rRmrRmg ωθ
θ
−=−
rR
g
n
−
=
3
22
ω
g
rR
n
n
−
==
2
3
2
2
π
ω
π
τ
01 =T ( )( )22
1 mrRWV θ−≅
( ) 22
4
3
2 mrRmT θ−= 02 =V
• Solve the energy equation for the natural frequency
of the oscillations.

FORCED VIBRATIONS
19 - 55
:maF =∑
( ) xmxkWtP stfm =+−+ δωsin
tPkxxm fm ωsin=+
( ) xmtxkW fmst =−+− ωδδ sin
tkkxxm fm ωδ sin=+
Forced vibrations - Occur
when a system is subjected
to a periodic force or a
periodic displacement of a
support.
=fω forced frequency

FORCED VIBRATIONS
19 - 56
[ ] txtCtC
xxx
fmnn
particulararycomplement
ωωω sincossin 21 ++=
+=
( ) ( )222
11 nf
m
nf
m
f
m
m
kP
mk
P
x
ωω
δ
ωωω −
=
−
=
−
=
tkkxxm fm ωδ sin=+
tPkxxm fm ωsin=+
At ωf = ωn, forcing input is in
resonance with the system.
tPtkxtxm fmfmfmf ωωωω sinsinsin2
=+−
Substituting particular solution into governing equation,

SAMPLE PROBLEM
19 - 57
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.
• 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.

SAMPLE PROBLEM
19 - 58
• The resonant frequency is equal to the natural
frequency of the system.
ftslb87.10
2.32
350 2
⋅==m
( )
ftlb000,36
inlb30007504
=
==k
W = 350 lb
k = 4(350
lb/in)
rpm549rad/s5.57
87.10
000,36
==
==
m
k
nω
Resonance speed = 549
rpm

SAMPLE PROBLEM
19 - 59
W = 350 lb
k = 4(350
lb/in) rad/s5.57=nω
• Evaluate the magnitude of the periodic force due to
the motor unbalance. Determine the vibration
amplitude from the frequency ratio at 1200 rpm.
( ) ftslb001941.0
sft2.32
1
oz16
lb1
oz1
rad/s125.7rpm1200
2
2
⋅=













=
===
m
f ωω
( )( )( ) lb33.157.125001941.0 2
12
6
2
==
== ωmrmaP nm
( ) ( )
in001352.0
5.577.1251
300033.15
1 22
−=
−
=
−
=
nf
m
m
kP
x
ωω
xm = 0.001352 in. (out of
phase)

DAMPED FREE VIBRATIONS
19 - 60
• With viscous damping due to fluid friction,
:maF =∑ ( )
0=++
=−+−
kxxcxm
xmxcxkW st

δ
• Substituting x = eλt
and dividing through by eλt
yields the characteristic equation,
m
k
m
c
m
c
kcm −





±−==++
2
2
22
0 λλλ
• Define the critical damping coefficient such that
nc
c m
m
k
mc
m
k
m
c
ω220
2
2
===−





• All vibrations are damped to some degree by
forces due to dry friction, fluid friction, or internal
friction.

DAMPED FREE VIBRATIONS
19 - 61
• Characteristic equation,
m
k
m
c
m
c
kcm −





±−==++
2
2
22
0 λλλ
== nc mc ω2 critical damping coefficient
• Heavy damping: c > cc
tt
eCeCx 21
21
λλ
+= - negative roots
- nonvibratory motion
• Critical damping: c = cc
( ) tnetCCx ω−
+= 21 - double roots
- nonvibratory motion
• Light damping: c < cc
( ) ( )tCtCex dd
tmc
ωω cossin 21
2
+= −
=





−=
2
1
c
nd
c
c
ωω damped frequency

DAMPED FORCED VIBRATIONS
19 - 62
( )[ ] ( )( )[ ]
( )( )
( )
=
−
=
=
+−
==
2
222
1
2
tan
21
1
nf
nfc
nfcnf
m
m
m
cc
cc
x
kP
x
ωω
ωω
φ
ωωωω
δ
magnification
factor
phase difference between forcing and steady
state response
tPkxxcxm fm ωsin=++  particulararycomplement xxx +=

ELECTRICAL ANALOGUES
19 - 63
• Consider an electrical circuit consisting of an inductor,
resistor and capacitor with a source of alternating
voltage
0sin =−−−
C
q
Ri
dt
di
LtE fm ω
• Oscillations of the electrical system are analogous to
damped forced vibrations of a mechanical system.
tEq
C
qRqL fm ωsin
1
=++ 

19 - 64
• The analogy between electrical and mechanical
systems also applies to transient as well as steady-
state 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 - 65
• The electrical system analogy provides a means of
experimentally determining the characteristics of a given
mechanical system.
• For the mechanical system,
( ) ( ) 0212112121111 =−++−++ xxkxkxxcxcxm 
( ) ( ) tPxxkxxcxm fm ωsin12212222 =−+−+ 
• For the electrical system,
( ) 0
2
21
1
1
21111 =
−
++−+
C
qq
C
q
qqRqL 
( ) tE
C
qq
qqRqL fm ωsin
2
12
12222 =
−
+−+ 
• The governing equations are equivalent. The characteristics
of the vibrations of the mechanical system may be inferred
from the oscillations of the electrical system.

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
 -
constanta.......
4
3
3
2
2
1
====
a
a
a
a
a
a

(2)
 Critical damping
 No real oscillation
 Time taken for the displacement to become effective
zero is a minimum

(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 ωt, then
 x = A cos (ωt - φ)
 where φ is the phase lag of x behind F

PHASE (2)
 Figure
 1. As f → 0, φ → 0
 2. As f → ∞, φ → π
 3. As f → f0, φ → π/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 child’s swing
 Destruction of the Tacoma Bridge
 Sound:
 An opera singer shatters a wine glass
 Resonance tube
 Kundt’s 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

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 air-
vane 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 ω 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, Y-
axis, precession axis and Z-axis, the couple or torque axis .

Ppt mech 5sem_dom

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