Dynamics of Machines (ALL UNITS)

ALL UNITS

1. Dynamics of Machinery
ME6505
Preparedby
Dr.K.Periasamy
Professor,
MechanicalDepartment
KongunaduCollegeofEngg.andTechnology.

2. UNIT I : FORCE ANALYSIS
Rigid Body dynamics in general plane motion –Equations of motion -Dynamic
force analysis -Inertia force and Inertia torque –D’Alembertsprinciple -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.

3. 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 analysingthe 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.

4. 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 (F
p): 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 F
Qperpendicular 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).

5. Forces acting on the connecting rod
•Inertia force of the reciprocating parts (F
1) 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 (F
C)
•The radial force (F
R) parallel to crank and
•The tangential force (F
T) acting perpendicular to crank

7. •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).

8. 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.

9. 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.

10. UNBALANCE:
Theconditionwhichexistsinarotorwhenvibratoryforce
ormotionisimpartedtoitsbearingsasaresultof
centrifugalforcesiscalledunbalanceortheuneven
distributionofmassaboutarotor’srotatingcentreline.
BALANCING:
Balancingisthetechniqueofcorrectingoreliminatingunwantedinertiaforces
ormomentsinrotatingorreciprocatingmassesandisachievedbychanging
thelocationofthemasscentres.
Theobjectivesofbalancinganenginearetoensure:
1.Thatthecentreofgravityofthesystemremainsstationeryduringa
completerevolutionofthecrankshaftand
2.Thatthecouplesinvolvedinaccelerationofthedifferentmovingparts
balanceeachother.

11. Types of balancing:
a)StaticBalancing:
i)Staticbalancingisabalanceofforcesduetoactionofgravity.
ii)Abodyissaidtobeinstaticbalancewhenitscentreofgravity
isintheaxisofrotation.
b)Dynamicbalancing:
i)Dynamicbalanceisabalanceduetotheactionofinertiaforces.
ii)Abodyissaidtobeindynamicbalancewhentheresultant
momentsorcouples,whichinvolvedintheaccelerationof
differentmovingpartsisequaltozero.
iii)Theconditionsofdynamicbalancearemet,theconditionsof
staticbalancearealsomet.

12. BALANCING OF ROTATING MASSES
Whenamassmovesalongacircularpath,it
experiencesacentripetalaccelerationandaforceis
requiredtoproduceit.Anequalandoppositeforce
calledcentrifugalforceactsradiallyoutwardsandis
adisturbingforceontheaxisofrotation.The
magnitudeofthisremainsconstantbutthedirection
changeswiththerotationofthemass.

13. Inarevolvingrotor,thecentrifugalforceremainsbalancedaslongas
thecentreofthemassofrotorliesontheaxisofrotationoftheshaft.
Whenthisdoesnothappen,thereisaneccentricityandanunbalance
forceisproduced.Thistypeofunbalanceiscommoninsteamturbine
rotors,enginecrankshafts,rotorsofcompressors,centrifugalpumps
etc.

14. Theunbalanceforcesexertedonmachinemembersaretimevarying,impart
vibratorymotionandnoise,therearehumandiscomfort,performanceofthe
machinedeteriorateanddetrimentaleffectonthestructuralintegrityofthe
machinefoundation.
Balancinginvolvesredistributingthemasswhichmaybecarriedoutby
additionorremovalofmassfromvariousmachinemembers.Balancingof
rotatingmassescanbeof
1.Balancingofasinglerotatingmassbyasinglemassrotatinginthesame
plane.
2.Balancingofasinglerotatingmassbytwomassesrotatingindifferent
planes.
3.Balancingofseveralmassesrotatinginthesameplane
4.Balancingofseveralmassesrotatingindifferentplanes

15. 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.

16. r=radiusofrotationofthemassm
Thecentrifugalforceexertedbymassm1ontheshaftisgivenby,
F=mrc11
Thisforceactsradiallyoutwardsandproducesbendingmomentontheshaft.In
ordertocounteracttheeffectofthisforceFc1,abalancingmassm2maybe
attachedinthesameplaneofrotationofthedisturbingmassm1suchthatthe
centrifugalforcesduetothetwomassesareequalandopposite.
BALANCINGOFASINGLEROTATINGMASSBYTWOMASSESROTATING
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.

18. Balancing Multi-cylinder Engines, Balancing V-engines

22. 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

23. Introduction
19 -23
•Mechanical vibrationis 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 periodof the vibration.
•Number of cycles per unit time defines the frequencyof the vibrations.
•Maximum displacement of the system from the equilibrium position is the
amplitudeof 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 dampedto
some degree.

24. Free Vibrations of Particles. Simple Harmonic Motion
19 -24
•If a particle is displaced through a distance x
mfrom its
equilibrium position and released with no velocity, the
particle will undergo simple harmonic motion, 
0

kxxm
kxxkWFma
st


•General solution is the sum of two particular solutions, tCtC
t
m
k
Ct
m
k
Cx
nn
 cossin
cossin
21
21



















•xis a periodic functionand 
nis the natural circular
frequencyof the motion.
•C
1and C
2are determined by the initial conditions: tCtCx
nn  cossin
21  02xC nvC 
01  tCtCxv
nnnn  sincos
21 

25. Free Vibrations of Particles. Simple Harmonic Motion
19 -25  txx
nmsin 
n
n



2
period


2
1
n
n
n
f
natural frequency  
2
0
2
0
xvx
nm

amplitude 

n
xv
00
1
tan
phase angle
•Displacement is equivalent to the xcomponent of the sum of two vectors
which rotate with constant angular velocity 21
CC

 .
n 02
0
1
xC
v
C
n




26. Free Vibrations of Particles. Simple Harmonic Motion
19 -26  txx
nmsin
•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





tx
tx
xv
nnm
nnm
  
 




tx
tx
xa
nnm
nnm
sin
sin
2
2


27. Types of Vibratory motion:

28. Transverse vibration

30. Sample Problem
19 -30mkN6mkN4
21
 kk
•Springs in parallel:
-determine the spring constant for equivalent
springmN10mkN10
4
21
21



kk
P
k
kkP


-apply the approximate relations for the
harmonic motion of a spring-mass systemn
n
n
m
k




2
srad14.14
kg20
N/m10
4

 s 444.0
n
   srad 4.141m 040.0

nmmxv sm566.0
m
v 2
sm00.8
ma   
2
2
srad 4.141m 040.0

nmm axa

31. Free Vibrations of Rigid Bodies
19 -31
•If an equation of motion takes the form0or0
22
 
nn
xx 
the corresponding motion may be considered
as simple harmonic motion.
•Analysis objective is to determine 
n.  mgWmbbbmI  ,22but
2
3
222
12
1 0
5
3
sin
5
3
 
b
g
b
g
 g
b
b
g
n
nn
3
5
2
2
,
5
3
then 


 
•For an equivalent simple pendulum,35bl
•Consider the oscillations of a square plate  ImbbW  sin

32. Forced Vibrations
19 -32:maF  xmxkWtP
stfm  sin tPkxxm
fm
sin  xmtxkW
fmst   sin tkkxxm
fm
sin
Forced vibrations-Occur
when a system is subjected
to a periodic force or a
periodic displacement of a
support.
f

forced frequency

33. Forced Vibrations
19 -33  txtCtC
xxx
fmnn
particulararycomplement
 sincossin
21 
    
222
11
nf
m
nf
m
f
m
m
kP
mk
P
x


 




 tkkxxm
fm
sin tPkxxm
fm
sin
At 
f= 
n, forcing input is in
resonancewith the system.tPtkxtxm
fmfmfmf  sinsinsin
2

Substituting particular solution into governing equation,

34. Sample Problem
19 -34
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.

35. Sample Problem
19 -35
•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)rpm 549rad/s 5.57
87.10
000,36


m
k
n

Resonance speed = 549
rpm

36. Sample Problem
19 -36
W= 350 lb
k = 4(350
lb/in)rad/s 5.57
n

•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
oz 16
lb 1
oz 1
rad/s 125.7 rpm 1200
2
2

















m
f
    lb 33.157.125001941.0
2
12
6
2

 mrmaP
nm    
in 001352.0
5.577.1251
300033.15
1
22





nf
m
m
kP
x

x
m= 0.001352 in. (out of
phase)

37. Damped Free Vibrations
19 -37
•With viscous dampingdue to fluid friction,:maF  
0

kxxcxm
xmxcxkW
st


•Substituting x = e
lt
and dividing through by e
lt
yields the characteristic equation,m
k
m
c
m
c
kcm 






2
2
22
0lll
•Define the critical damping coefficient such thatnc
c
m
m
k
mc
m
k
m
c
220
2
2






•All vibrations are damped to some degree by
forces due to dry friction, fluid friction, or internal
friction.

38. Damped Free Vibrations
19 -38
•Characteristic equation,m
k
m
c
m
c
kcm 






2
2
22
0lll 
ncmc 2
critical damping coefficient
•Heavy damping: c > c
ctt
eCeCx
21
21
ll

-negative roots
-nonvibratory motion
•Critical damping: c = c
c 
t
n
etCCx


21
-double roots
-nonvibratory motion
•Light damping: c < c
c
 tCtCex
dd
tmc
 cossin
21
2

 









2
1
c
nd
c
c

damped frequency

39. Damped Forced Vibrations
19 -39    
 
 






2
2
2
2
1
2
tan
21
1
nf
nfc
nfcnf
m
m
m
cc
cc
x
kP
x





magnification
factor
phase difference between forcing and steady
state responsetPkxxcxm
fm
sin particulararycomplement
xxx 

41. )38.2(
32
4
0
d
I

 )39.2(
32
4
0
l
Gd
l
GIM
k
t
t


 Polar Moment of Inertia:
Torsional Spring Constant:

42. 1)Ifthecrosssectionoftheshaftsupportingthediscisnotcircular,an
appropriatetorsionalspringconstantistobeused.
2)Thepolarmassmomentofinertiaofadiscisgivenby:
3)Animportantapplication:inamechanicalclockg
WDDh
J
832
44
0


where ρis the mass density
h is the thickness
D is the diameter
W is the weight of the disc

43. 43
where ω
nis given by Eq. (2.41) and A
1and A
2can be
determined from the initial conditions. If
•General solution of Eq. (2.40) can be obtained:)44.2(sincos)(
21 tAtAt
nn   )45.2()0()0( and )0(
00 



 t
dt
d
tt
The constants A
1and A
2can be found:)46.2(/
02
01
nA
A





Eq. (2.44) can also represent a simple harmonic motion.

44. UNIT IV : FORCED VIBRATION
Response to periodic forcing -Harmonic Forcing -
Forcing caused by unbalance -Support motion –
Force transmissibility and amplitude transmissibility
-Vibration isolation.

45. 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.

46. 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

47. 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 = -bvwhere b = damping coefficient
Frequency of damped vibration < Frequency of undamped vibration

48. Types of Damped Oscillations (1)
Slight damping (underdamping)
Characteristics:
-oscillations with reducing amplitudes
-amplitude decays exponentially with time
-period is slightly longer
-Figure
-constant a.......
4
3
3
2
2
1

a
a
a
a
a
a

49. Types of Damped Oscillations (2)
Critical damping
No real oscillation
Time taken for the displacement to become effective zero is a minimum
Heavy damping (Overdamping)
Resistive forces exceed those of critical damping
The system returns very slowly to the equilibrium position

50. Magnification Factor or Dynamic Magnifier
•Itistheratioofmaximumdisplacementoftheforcedvibration(xmax
)tothedeflectionduetothestaticforceF(xo).Wehaveprovedin
thepreviousarticlethatthemaximumdisplacementorthe
amplitudeofforcedvibration,

51. •If there is no damping (i.e. if the vibration is undamped), then c = 0.
In that case, magnification factor
•At resonance, = n. Therefore magnification factor,

52. 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)

53. Characteristics of Forced Oscillation
In steady state, the system vibrates at the frequency of the driving force

54. 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
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
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

55. Resonance (1)
Resonance occurs when an oscillator is acted upon by a second driving
oscillator whose frequency equalsthe 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 leadsthat of the oscillator
by 90
Examples
Mechanics:
Oscillations of a child’s swing
Destruction of the Tacoma Bridge
Sound:
An opera singer shatters a wine glass
Resonance tube
Kundt’stubeElectricity
Radio tuning
Light
Maximum absorption of infrared waves by a NaClcrystal

56. 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

57. 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
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

58. Vibration Isolation and Transmissibility
•Theratiooftheforcetransmitted(FT)totheforceapplied(F)is
knownastheisolationfactorortransmissibilityratioofthespring
support.
1.Springforceorelasticforcewhichisequaltos.xmax,and
2.Dampingforcewhichisequaltoc..xmax.

59. 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

60. Governors
•EngineSpeedcontrol
ThispresentationisfromVirginiaTechandhasnotbeeneditedbyGeorgia
CurriculumOffice.
•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.
•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

61. 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.

62. Pneumatic Governors
•Sometimes called air-vane
governors, they are
operated by the stream of
air flow created by the
cooling fins of the
flywheel.

63. 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.

64. Mechanical Governor
•With this system, governor rpm is always directly
proportional to engine rpm.

65. 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.

66. Vacuum Governors
•Located between the carburetor and the intake manifold.
•It senses changes in intake manifold pressure (vacuum).
•As engine speed increases or decreases the governor closes
or opens the throttle respectively to control engine speed.

67. •HuntingHunting 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.
•Stabilityis 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.
•Sensitivityis 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.

68. Summary
•Governors are usually of the following types:
•Air-vane (pneumatic)
•Mechanical (centrifugal)
•Vacuum
•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.
•Small engine governors are used to:
•Maintain selected engine speed.
•Prevent over-speeding.
•Limit high and low speeds.

69. Gyroscope
Agyroscopeconsistsofarotormountedintheinnergimbal.Theinner
gimbalismountedintheoutergimbalwhichitselfismountedona
fixedframeasshowninFig.WhentherotorspinsaboutX-axiswith
angularvelocityωrad/sandtheinnergimbalprecesses(rotates)
aboutY-axis,thespatialmechanismisforcedtoturnaboutZ-axis
otherthanitsownaxisofrotation,andthegyroscopiceffectisthus
setup.Theresistancetothismotioniscalledgyroscopiceffect.

71. 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 .

75. GYROSCOPIC EFFECT ON SHIP

78. Stability of a Four Wheel Drive Moving in a Curved Path