contents · 2020-02-04 · four bar chain or quadric cycle chain ... 1. beam engine (crank and...
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Contents
UNIT: 1............................................................................................................... Error! Bookmark not defined.
INTRODUCTION: .............................................................................................................................................. 7
Machine: .............................................................................................................................................................. 7
Structure: ............................................................................................................................................................. 7
Kinematic link: .................................................................................................................................................... 8
Types of Links: .................................................................................................................................................... 9
kinematic Pair: ..................................................................................................................................................... 9
Classification of Kinematic Pairs: ....................................................................................................................... 9
1. According to the type of relative motion between theelements. .............................................................. 9
Kinematic Chain: ............................................................................................................................................... 10
Mechanism......................................................................................................................................................... 12
Types of Kinematic Chains ............................................................................................................................... 13
Four Bar Chain or Quadric Cycle Chain ........................................................................................................... 13
Single Slider Crank Chain ................................................................................................................................. 14
Double Slider Crank Chain................................................................................................................................ 14
Grubler’s Criterion for Plane Mechanisms: ....................................................................................................... 14
Types of Constrained Motions: ......................................................................................................................... 14
Degrees of Freedom for Plane Mechanisms: ..................................................................................................... 15
b. ........................................................................................................................................................................ 16
INVERSION:..................................................................................................................................................... 16
APPLICATIONS: .............................................................................................................................................. 17
1. Beam engine (crank and lever mechanism) ............................................................................................ 17
2. Coupling rod of a locomotive (Double crankmechanism). .................................................................... 17
INVERSIONS OF SINGLE SLIDER CRANK CHAIN: ................................................................................. 17
Pendulum pump or Bull engine. .................................................................................................................... 17
Crank and slotted lever quick return motion mechanism: ............................................................................. 17
INVERSIONS OF DOUBLE SLIDER CRANK CHAIN ................................................................................ 18
Elliptical trammels. ........................................................................................................................................ 18
METHODS FOR DETERMINING THE VELOCITY OF A POINT ON A LINK: ....................................... 18
RELATIVE VELOCITY OF TWO BODIES MOVING IN STRAIGHT LINES ........................................... 18
Motion of a Link ................................................................................................................................................ 20
Hence velocity of any point on a link with respect to another point on the same link is always perpendicular
to the line joining these points on the configuration (or space) diagram. ...................................................... 20
Velocities in Slider Crank Mechanism .............................................................................................................. 22
PROBLEMS: ..................................................................................................................................................... 23
GIVEN : ............................................................................................................................................................. 23
RESULT: ........................................................................................................................................................... 24
Example 2. ......................................................................................................................................................... 24
Solution:............................................................................................................................................................. 25
Linear velocity of the slider D ....................................................................................................................... 25
Angular velocity of the link BD..................................................................................................................... 26
RESULTS: ......................................................................................................................................................... 26
Example 3 .......................................................................................................................................................... 26
Given: ................................................................................................................................................................ 28
SOLUTION: ...................................................................................................................................................... 28
1. Velocity of F .............................................................................................................................................. 28
2. Velocity of sliding of CE in thetrunnion ................................................................................................ 30
3. Angular velocity ofCE ............................................................................................................................ 30
RESULTS: ......................................................................................................................................................... 31
SOLUTION : ..................................................................................................................................................... 31
RESULT : .......................................................................................................................................................... 32
GIVEN : ............................................................................................................................................................. 33
Velocity of slider S (cutting tool) .................................................................................................................. 33
Angular velocity of link RS ........................................................................................................................... 34
RESULTS : ........................................................................................................................................................ 34
GIVEN: .............................................................................................................................................................. 35
SOLUTION: ...................................................................................................................................................... 35
1. Velocity of slider D .................................................................................................................................... 35
2. Angular velocities of links AB, CB andBD ........................................................................................... 36
3. Velocities of rubbing on the pins A andD .............................................................................................. 37
4. Torque applied to the crankOA .............................................................................................................. 37
Velocity in Mechanisms (Instantaneous Centre Method) Introduction: ........................................................... 38
VELOCITY OF A POINT ON A LINK BY INSTANTANEOUS CENTRE METHOD:............................... 39
Properties of the Instantaneous Centre: ............................................................................................................. 42
LOCATION OF INSTANTANEOUS CENTRES: .......................................................................................... 42
PROBLEMS: ..................................................................................................................................................... 43
Given .............................................................................................................................................................. 43
Solution: ......................................................................................................................................................... 43
Location of instantaneous centers .................................................................................................................. 43
RESULT: ........................................................................................................................................................... 45
Given: ................................................................................................................................................................ 46
Location of instantaneous centers .................................................................................................................. 46
1. Velocity of the slider .............................................................................................................................. 47
Velocity of point E on the bell crank lever .................................................................................................... 50
Acceleration Diagram for a Link ....................................................................................................................... 51
1. Linear velocity of the midpoint of the connecting rod .............................................................................. 45
Acceleration of the midpoint of the connecting rod ...................................................................................... 46
2. Angular velocity of the connecting rod ..................................................................................................... 47
Angular acceleration of the connecting rod ................................................................................................... 47
Angular acceleration of BD ........................................................................................................................... 51
Velocity of the ram E ..................................................................................................................................... 52
Acceleration of the ram E .............................................................................................................................. 53
LIMITING ANGLE OF FRICTION ................................................................................................................. 59
ANGLE OF REPOSE ........................................................................................................................................ 59
PROBLEMS: ..................................................................................................................................................... 60
SCREW FRICTION .......................................................................................................................................... 62
SCREW JACK .................................................................................................................................................. 63
minor diameter. .................................................................................................................................................. 68
Example 2. A 150 mm diameter valve, against which a steam pressure of 2 MN/m2 is acting, is closed by
means of a square threaded screw 50 mm in external diameter with 6 mm pitch. If .................................... 69
TORQUE REQUIRED LOWERING THE LOAD BY A SCREW JACK : .................................................... 71
Torque required on the screw ........................................................................................................................ 74
Ratio of the torques required to raise and lower the load .............................................................................. 74
Efficiency of the machine .............................................................................................................................. 75
FRICTION OF PIVOT AND COLLAR BEARING ........................................................................................ 76
FLAT PIVOT BEARING:................................................................................................................................. 77
1. Considering uniform pressure ................................................................................................................ 78
6 ......................................................................................................................................................................... 80
2. CONSIDERING UNIFORMWEAR: ..................................................................................................... 82
POWER ABSORBED IN FRICTION: ......................................................................................................... 84
FRICTION CLUTCHES: .................................................................................................................................. 84
SINGLE DISC OR PLATE CLUTCH: ............................................................................................................. 85
MULTIPLE DISC CLUTCH ............................................................................................................................ 86
SOLUTION: .................................................................................................................................................. 88
1. FINAL SPEED OF THE MOTOR ANDROTOR .................................................................................. 90
TYPES OF BELTS ............................................................................................................................................ 93
V-BELT DRIVE ................................................................................................................................................ 94
ADVANTAGES AND DISADVANTAGES OF V-BELT DRIVE OVER FLAT BELT DRIVE .................. 94
Advantages ..................................................................................................................................................... 94
Disadvantages ................................................................................................................................................ 95
ROPE DRIVE: ................................................................................................................................................... 95
TYPES OF FLAT BELT DRIVES ................................................................................................................... 96
RATIO OF DRIVING TENSIONS FOR FLAT BELT DRIVE ....................................................................... 97
ANGLE OF CONTACT .................................................................................................................................... 98
CENTRIFUGAL TENSION ........................................................................................................................... 100
INITIAL TENSION IN THE BELT ............................................................................................................... 101
MAXIMUM TENSION IN THE BELT ......................................................................................................... 102
CONDITON FOR MAXIMUM POWER TRANSMISSION ........................................................................ 102
LENGTH OF AN OPEN BELT: ..................................................................................................................... 103
POWER TRANSMITTED BY A BELT: ....................................................................................................... 109
SOLUTION: .................................................................................................................................................... 110
SOLUTION: .................................................................................................................................................... 111
Stress in the for an open belt drive: ............................................................................................................. 111
Torque on the shaft larger pulley: ................................................................................................................ 116
Torque on the shaft of smaller pulley .......................................................................................................... 116
Power transmitted ........................................................................................................................................ 116
Power lost in friction .................................................................................................................................... 116
1. Minimum width of belt ............................................................................................................................ 117
GIVEN : ........................................................................................................................................................... 119
SOLUTION: .................................................................................................................................................... 119
Power transmitted ........................................................................................................................................ 119
GIVEN: ............................................................................................................................................................ 121
SOLUTION; .................................................................................................................................................... 121
Number of V-belts ....................................................................................................................................... 123
INTRODUCTION ........................................................................................................................................... 124
ADVANTAGES AND DISADVANTAGES OF GEAR DRIVE .................................................................. 125
Advantages ................................................................................................................................................... 125
Disadvantages .............................................................................................................................................. 125
CLASSIFICATION OF TOOTHED WHEELS.............................................................................................. 126
4. According to position of teeth on the gear surface. .............................................................................. 128
GEAR TRAINS: .............................................................................................................................................. 130
TYPES OF GEAR TRAINS............................................................................................................................ 130
SIMPLE GEAR TRAIN .................................................................................................................................. 130
COMPOUND GEAR TRAIN ......................................................................................................................... 132
EPICYCLIC GEAR TRAIN ........................................................................................................................... 134
1. Tabular method. .................................................................................................................................... 135
2. Algebraic method. ................................................................................................................................ 136
1. Tabular method ........................................................................................................................................ 137
Speed of gear B when gear A makes 300 r.p.m. clockwise ......................................................................... 138
GIVEN : ........................................................................................................................................................... 140
SOLUTION: .................................................................................................................................................... 140
1. Sketch the arrangement ........................................................................................................................ 140
2. Number of teeth on wheels A and ........................................................................................................ 140
3. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A infixed .............................. 140
4. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m.
counterclockwise.......................................................................................................................................... 141
CAM INTRODUCTION: ................................................................................................................................ 142
1. According to the surface in contact. ..................................................................................................... 142
(a) Knife edge follower. ......................................................................................................................... 142
(b) Roller follower. ................................................................................................................................. 142
(c) Flat faced or mushroom follower. ..................................................................................................... 142
(d) Spherical faced follower ................................................................................................................... 143
2. According to the motion of the follower. ............................................................................................. 143
CLASSIFICATION OF CAMS ...................................................................................................................... 144
1. Radial or disc cam. ............................................................................................................................... 144
2. Cylindrical cam..................................................................................................................................... 144
NOMENCALTURE: ....................................................................................................................................... 144
MOTION OF THE FOLLOWER .................................................................................................................... 146
DISPLACEMENT, VELOCITY AND ACCELERATION DIAGRAMS WHEN THE FOLLOWER MOVES
WITH UNIFORM VELOCITY ...................................................................................................................... 146
DISPLACEMENT, VELOCITY AND ACCELERATION DIAGRAMS WHEN THE FOLLOWER MOVES
WITH SIMPLE HARMONIC MOTION ........................................................................................................ 147
DISPLACEMENT, VELOCITY AND ACCELERATION DIAGRAMS WHEN THE FOLLOWER MOVES
WITH UNIFORM ACCELERATION AND RETARDATION..................................................................... 148
CONSTRUCTION OF CAM PROFILE FOR A RADIAL CAM .................................................................. 150
(a) Profile of the cam when the axis of follower passes through the axis of camshaft .......................... 151
(b) Profile of the cam when the axis of the follower is offset by 20 mm from the axis of the cam ....... 152
(a) Profile of the cam when the line of stroke of the valve rod passes through the axis of the cam shaft
154
Example :3. ...................................................................................................................................................... 157
SOLUTION; .................................................................................................................................................... 157
Maximum velocity of the follower during out stroke and return stroke: ..................................................... 159
Maximum acceleration of the follower during out stroke and return stroke: .............................................. 159
Construction..................................................................................................................................................... 159
Example 5. ....................................................................................................................................................... 162
BALANCING .................................................................................................................................................. 164
INTRODUCTION: .......................................................................................................................................... 164
BALANCING OF A SINGLE ROTATING MASS BY A SINGLE MASS ROTATING IN THE SAME
PLANE ............................................................................................................................................................ 165
BALANCING OF A SINGLE ROTATING MASS BY TWO MASSES ROTATING IN DIFFERENT
PLANES: ......................................................................................................................................................... 166
1. When the plane of the disturbing mass lies in between the planes of the two balancing Masses ........ 167
2. When the plane of the disturbing mass lies on one end of the planes of the balancing Masses ........... 168
BALANCING OF SEVERAL MASSES ROTATING IN THE SAME PLANE: ......................................... 169
Analytical method ........................................................................................................................................ 170
2. Graphical method ..................................................................................................................................... 171
BALANCING OF SEVERAL MASSES ROTATING IN DIFFERENT PLANES: ..................................... 171
couple vectors are drawn radically outwards for the masses on one side of the reference plane and radially
inward for the masses on the other side of the reference plane. .................................................................. 173
BALANCING OF RECIPROCATING MASSES: ......................................................................................... 173
PRIMARY AND SECONDARY UNBALANCED FORCES OF RECIPROCATING MASSES................ 174
PARTIAL BALANCING OF UNBALANCED PRIMARY FORCE IN A RECIPROCATINGENGINE ... 175
BALANCING OF PRIMARY FORCES OF MULTI-CYLINDER IN-LINE ENGINES ............................. 177
‘For a multi-cylinder engine, the primary forces may be completely balanced by suitably arranging the crank
angles, provided that the number of cranks are not less than four’. ................................................................ 178
BALANCING OF V-ENGINES ..................................................................................................................... 179
Considering primary forces .......................................................................................................................... 180
Considering secondary forces ...................................................................................................................... 182
BALANCING OF RADIAL ENGINES (DIRECT AND REVERSE CRANKS METHOD ) ...................... 184
Considering the primary forces .................................................................................................................... 184
Hence, for primary effects the mass m of the reciprocating parts at P may be replaced by two masses at C
and C’ each of magnitude m/2. .................................................................................................................... 185
Example 1 ........................................................................................................................................................ 186
SOLUTION: ................................................................................................................................................ 186
1. Magnitude of the masses at A and D ....................................................................................................... 194
2. Distance between planes A and D ........................................................................................................... 196
Reciprocating mass and the relative angular position for each of the inner cranks ..................................... 198
Maximum secondary unbalanced force ....................................................................................................... 199
1. Unbalanced primary and secondary forces ........................................................................................... 200
2. Unbalanced primary and secondary couples ........................................................................................ 202
Maximum unbalanced primary couple ........................................................................................................ 204
Maximum unbalanced secondary couple ..................................................................................................... 205
UNIT-5 VIBRATION ..................................................................................................................................... 210
TYPES OF VIBRATORY MOTION.............................................................................................................. 210
NATURAL FREQUENCY OF FREE LONGITUDINAL VIBRATIONS .................................................... 211
1. Equilibrium Method ............................................................................................................................. 211
2. Energy method ...................................................................................................................................... 213
FREQUENCY OF FREE DAMPED VIBRATIONS (VISCOUS DAMPING) ............................................. 216
3. WHEN THE ROOTS ARE EQUAL (CRITICALDAMPING) ............................................................... 223
DAMPING FACTOR OR DAMPING RATIO .............................................................................................. 224
FREE TORSIONAL VIBRATIONS OF A TWO ROTOR SYSTEM ........................................................... 238
TORSIONALLY EQUIVALENT SHAFT ..................................................................................................... 243
INTRODUCTION:
DEFINITION
The subject Theory of Machines may be defined as that branch of Engineering-
science, which deals with the study of relative motion between the various parts of a machine, and
forces which act on them. The knowledge of this subject is very essential for an engineer in designing
the various parts ofmachine.
The Theory of Machines may be sub-divided into the following four branches :
1. Kinematics. It is that branch of Theory of Machines which deals with the relative motion between
the
various parts of the machines’
2. Dynamics. It is that branch of Theory of Machines which deals with the forces and their effects,
whileacting
upon the machine parts in motion.
3. Kinetics. It is that branch of Theory of Machines which deals with the inertia forces which arise
from the combined effect of the mass and motion of the machineparts.
4. Statics. It is that branch of Theory of Machines which deals with the forces and their effects while
the ma chine parts are at rest. The mass of the parts is assumed to benegligible.
Machine:
A machine consists of a number of parts or bodies we shall study the mechanisms of the
various parts or bodies from which the machine is assembled. This is done by making one of the parts
as fixed, and the relative motion of other parts is determined with respect to the fixed part.
Structure:
It is an assemblage of a number of resistant bodies (known as members) having no relative
motion between them and meant for carrying loads having straining action. A railway bridge, a roof
truss, machine frames etc., are the examples of a structure.
Kinematic link:
Each part of a machine, which moves relative to some other part, is known as a kinematic link
(or simply link) or element. A link may consist of several parts, which are rigidly fastened together,
so that they do not move relative to one another. For example, in a reciprocating steam engine piston,
piston rod and crosshead constitute one link ; connecting rod with big and small endbearings
constitute a second link ; crank, crank shaft and flywheel a third link and the cylinder, engine frame
and main bearings a fourth link.
A link or element need not to be a rigid body, but it must be a resistant body. A body is said to be a
resistant body if it is capable of transmitting the required forces with negligible deformation. Thus a
link should have the following two characteristics:
1. It should have relative motion,and
2. It must be a resistantbody.
Types of Links:
In order to transmit motion, the driver and the follower may be connected by the following
three types of links:
1. Rigid link. A rigid link is one which does not undergo any deformation while transmitting motion.
Strictly speaking, rigid links do not exist. However, as the deformation of a connecting rod, crank etc.
of a reciprocating steam engine is not appreciable, they can be considered as rigidlinks.
2. Flexible link. A flexible link is one which is partly deformed in a manner not to affect
the transmission of motion. For example, belts, ropes, chains and wires are flexible links and transmit
tensile forcesonly.
3. Fluid link. A fluid link is one which is formed by having a fluid in a receptacle andthe
motion is transmitted through the fluid by pressure or compression only, as in the case of hydraulic
presses, jacks andbrakes.
kinematic Pair:
The two links or elements of a machine, when in contact with each other, are said to form
a pair. If the relative motion between them is completely or successfully constrained (i.e. in a definite
direction), the pair is known as kinematic pair. Let us discuss the various types of constrained
motions.
Classification of Kinematic Pairs:
The kinematic pairs may be classified according to the following considerations:
1. According to the type of relative motion between theelements.
The kinematic pairs according to type of relative motion between the elements may be
classified as discussed below:
(a) Sliding pair. When the two elements of a pair are connected in such a way that one can only slide
relative to the other, the pair is known as a sliding pair. The piston and cylinder, cross-head and
guides of a reciprocating steam engine, ram and its guides in shaper, tail stock on the lathe bed etc.
are the examples of a sliding pair. A little consideration will show that a sliding pair has a completely
constrainedmotion.
(b) Turning pair. When the two elements of a pair are connected in such a way that one can only
turn or revolve about a fixed axis of another link, the pair is known as turning pair. A shaft with
collars at both ends fitted into a circular hole, the crankshaft in a journal bearing in an engine, lathe
spindle supported in head stock, cycle wheels turning over their axles etc. are the examples of a
turning pair. A turning pair also has a completely constrainedmotion.
(c) Rolling pair. When the two elements of a pair are connected in such a way that one rolls over
another fixed link, the pair is known as rolling pair. Ball and roller bearings are examples of rolling
pair.
(d) Screw pair. When the two elements of a pair are connected in such a way that one element can
turn about the other by screw threads, the pair is known as screw pair. The lead screw of a lathe with
nut, and bolt with a nut are examples of a screwpair.
(e) Spherical pair. When the two elements of a pair are connected in such a way that one element
(with spherical shape) turns or swivels about the other fixed element, the pair formed is called a
spherical pair. The ball and socket joint, attachment of a car mirror, pen stand etc., are the examples
of a sphericalpair.
2. According to the type of contact between the elements. The kinematic pairs according to the type
of contact between the elements may be classified as discussedbelow:
(a) Lower pair. When the two elements of a pair have a surface contact when relative motion takes
place and the surface of one element slides over the surface of the other, the pair formed is known as
lower pair. It will be seen that sliding pairs, turning pairs and screw pairs form lowerpairs.
(b) Higher pair. When the two elements of a pair have a line or point contact when relative motion
takes place and the motion between the two elements is partly turning and partly sliding, then the pair
is known as higher pair. Pair of friction discs, toothed gearing, belt and rope drives, ball and roller
bearings and cam and follower are the examples of higherpairs.
3. According to the type of closure. The kinematic pairs according to the type of closure between the
elements may be classified as discussedbelow:
(a) Self closed pair. When the two elements of a pair are connected together mechanically in such a
way that only required kind of relative motion occurs, it is then known as self closed pair. The lower
pairs are self closedpair.
(b) Force - closed pair. When the two elements of a pair are not connected mechanically but are kept
in contact by the action of external forces, the pair is said to be a force-closed pair. The cam and
follower is an example of force closed pair, as it is kept in contact by the forces exerted by spring and
gravity.
Kinematic Chain:
When the kinematic pairs are coupled in such a way that the last link is joined to the first
link to transmit definite motion (i.e. completely or successfully constrained motion), it is called a
kinematic chain. In other words, a kinematic chain may be defined as a combination of kinematic
pairs, joined in such a way that each link forms a part of two pairs and the relative motion between
the links or elements is completely or successfully constrained. For example, the crank- shaft of an
engine forms a kinematic pair with the bearings which are fixed in a pair, the connecting rod with the
crank forms a second kinematic pair, the piston with the connecting rod forms a third pair and the
piston with the cylinder forms a fourth pair. The total combination of these links is a kinematicchain.
If each link is assumed to form two pairs with two adjacent links, then the relation between
the number of pairs ( p ) forming a kinematic chain and the number of links ( l ) may be expressed in
the form of an equation :
l = 2 p – 4
. . . (i)
Since in a kinematic chain each link forms a part of two pairs, therefore there will be as many links
as the number of pairs.
Anotherrelationbetweenthenumberoflinks(l)andthenumberof joints ( j ) which
constitute a kinematic chain is given by the expression:
The equations (i) and (ii) are applicable only to kinematic chains, in which lower pairs are used.
These equations may also be applied to kinematic chains, in which higher pairs are used. In that
case each higher pair may be taken as equivalent to two lower pairs with an additional element or
link.
Let us apply the above equations to the following cases to determine whether each of
them is a kinematic chain or not.
1. Consider the arrangement of three links A B, BC and C A with pin joints aA,B andC.
Number of links, l =3
Number of pairs, p=3
and number of joints, j =3
From equation (i),l=2p-3
3=2*3-4=2
L.H.S. > R.H.S
Now from equation (ii),`
i.e. L.H.S. >R.H.S
Mechanism
When one of the links of a kinematic chain is fixed, the chain is known as mechanism. It
may be used for transmitting or transforming motion e.g. engine indicators, typewriter etc
A mechanism with four links is known as simple mechanism, and the mechanism with
more than four links is known as compound mechanism. When a mechanism is required to
transmit power or to do some particular type of work, it then becomes a machine. In such cases, the
various links or elements have to be designed to withstand the forces (both static and kinetic)safely.
A little consideration will show that a mechanism may be regarded as a machine in which each part
is reduced to the simplest form to transmit the required motion.
Forces Acting in a Mechanism
Consider a mechanism of a four bar chain, as shown in Fig. Let force FA newton is acting at the
joint A in the
direction of the velocity of A (vAm/s) which is perpendicular to the link D A. Suppose a force FB
newton is transmitted to the jointB in the direction of the velocity of B (i.e. vB m/s) whichis
perpendicular to the link CB. If weneglect
the effect of friction and the change of kinetic energy of the link (i.e.), assuming the efficiency of
transmission as 100%), then by the principle of conservation of energy,
Types of Kinematic Chains
The most important kinematic chains are those which consist of four lower pairs, each pair
being a sliding pair or a turning pair. The following three types of kinematic chains with four lower
pairs are important from the subject point of view
1. Four bar chain or quadric cyclicchain,
2. Single slider crank chain,and
3. Double slider crankchain.
These kinematic chains are discussed, in detail, in the following articles
Four Bar Chain or Quadric Cycle Chain
We have already discussed that the kinematic chain is a combination of four or more
kinematic pairs, such that the relative motion between the links or elements is completely
constrained. The simplest and the basic kinematic chain is a four bar chain or quadric cycle chain.It
consists of four links, each of them forms a turning pair at A, B, C and D. The four links may be of
different lengths. According to Grashof ’s law for a four bar mechanism, the sum of the shortest
and longest link lengths should not be greater than the sum of the remaining two link lengths if
there is to be continuous relative motion between the twolinks.
A very important consideration in designing a mechanism is to ensure that the input
crank makes a complete revolution relative to the other links. The mechanism in which no link
makes a complete revolution will not be useful. In a four bar chain, one of the links, in particular the
shortest link, will make a complete revolution relative to the other three links, if it satisfies the
Grashof ’s law. Such a link is known as crank or driver. A D (link 4) is a crank. The link BC (link
2) which makes a partial rotation or oscillates is known as lever or rocker or follower and the link
CD (link 3) which connects the crank and lever is called connecting rod or coupler. The fixed link
A B (link 1) is known as frame of the mechanism. When the crank (link 4) is the driver, the
mechanism is transforming rotary motion into oscillatingmotion.
Single Slider Crank Chain
A single slider crank chain is a modification of the basic four bar chain. It consist of one sliding
pair and three turning pairs. It is, usually, found in reciprocating steam engine mechanism. This type
of mechanism converts rotary motion into reciprocating motion and vice versa. In a single slider
crank chain, as shown the links 1 and 2, links 2 and 3, and links 3 and 4 form three turning pairs
while the links 4 and 1 form a sliding pair.
The link 1 corresponds to the frame of the engine, which is fixed. The link 2 corresponds to
the crank ; link 3 corresponds to the connecting rod and link 4 corresponds to cross-head. As the
crank rotates, the cross-head reciprocates in the guides and thus the piston reciprocates in the
cylinder.
Double Slider Crank Chain
A kinematic chain which consists of two turning pairs and two sliding pairs is known as
double slider crank chain. We see that the link 2 and link 1 form one turning pair and link 2 and
link 3 form the second turning pair. The link 3 and link 4 form one sliding pair and link 1 and link 4
form the second sliding pair.
Grubler’s Criterion for Plane Mechanisms:
The Grubler’s criterion applies to mechanisms with only single degree of freedom joints
where the overall movability of the mechanism is unity. Substituting n = 1 and h = 0 in Kutzbach
equation, we have
1 = 3 (l – 1) – 2 j or 3l – 2j – 4 = 0
This equation is known as the Grubler's criterion for plane mechanisms with constrained motion.
A little consideration will show that a plane mechanism with a movability of 1 and only single
degree of freedom joints cannot have odd number of links. The simplest possible mechanisms of
this type are a four bar mechanism and a slider-crank mechanism in which l = 4 and j = 4
Types of Constrained Motions:
Following are the three types of constrained motions:
1. Completely constrained motion. When the motion between a pair is limited to a
definite direction irrespective of the direction of force applied, then the motion is said to be a
completely constrained motion. For example, the piston and cylinder (in a steam engine) form a pair
and the motion of the piston is limited to a definite direction (i.e. it will only reciprocate) relative to
the cylinder irrespective of the direction of motion of the crank. The motion of a square bar in a
square hole, and the motion of a shaft with collars at each end in a circular hole, are also examples
of completely constrainedmotion.
2. Incompletely constrained motion. When the motion between a pair can take place in more
than one direction, then the motion is called an incompletely constrained motion. The change in the
direction of impressed force may alter the direction of relative motion between the pair. A circular
bar or shaft in a circular hole, is an example of an incompletely constrained motion as it may either
rotate or slide in a hole. These both motions have no relationship with theother.
3. Successfully constrained motion. When the motion between the elements, forming a
pair,issuchthattheconstrainedmotionisnotcompletedbyitself,butbysomeothermeans,then
the motion is said to be successfully constrained motion. Consider a shaft in a foot-step bearing.
The shaft may rotate in a bearing or it may move upwards. This is a case of incompletely con
strained motion. But if the load is placed on the shaft to prevent axial upward movement of the
shaft,thenthemotionofthepairissaidtobesuccessfullyconstrainedmotion.Themotionofan
I.C. engine valve (these are kept on their spring) and the piston reciprocating inside an engine
cylinder seat by a are also the examples of successfully constrained motion.
Degrees of Freedom for Plane Mechanisms:
In the design or analysis of a mechanism, one of the most important concern is the
number of degrees of freedom (also called movability) of the mechanism. It is defined as the
number of input parameters (usually pair variables) which must be independentlycontrolled in order
to bring the mechanism into a useful engineering purpose. It is possible to determine the number of
degreesof
freedom of a mechanism directly from the number of links and the number and types of joints
which it includes.
Consider a four bar chain, as shown in Fig., A little consideration will show that only one variable
such as ϴis needed to define the relative positions of all the links. In other words, we say that the
number of degrees of freedom of a four bar chain is one. Now, let us consider a five bar chain, as
shown in Fig., In this case two variables such as ϴ1 and ϴ2 are needed to define completely the
relative positions of all the links. Thus, we say that the number of degrees of freedom is * two.
In order to develop the relationship in general, consider two links A B and CD in a plane motion as
shown in fig I.
The link AB with co-ordinate system O X Y is taken as the reference link (or fixed link). The
position of point P on the moving link CD can be completely specified by the three variables, i.e.
the co-ordinates of the point P denoted by x and y and the inclination θ of the link CD with X-axis
or link A B. In other words, we can say that each link of a mechanism has three degrees of freedom
before it is connected to any other link. But when the link CD is connected to the link A B by a
turning pair at A , the position of link CD is now determined by a single variable θ and thus has one
degree of freedom.
Slider crank–crank rocker mechanism:
A slider –mechanism in which OA is the crank moving with uniform angular velocity in the
clockwise direction. At point B, a slider moves on the fixed guide G. AB is the coupler joining A at
B. It is required to find the velocity of the slider atB.
Writing the velocity vectorequation,
Vel. of B rel. to O = vel. Of B to A + vel. Of A rel. to O
Vbo= vba+vao; vbg=vao+vba
Vbo is replaced by vbg as O and G are two points on fixed link with zero relative between them.
Take the vector vao which is completely known.
Vao= ὼ.OA; ┴ to OA
Vba is ┴ AB, draw a line ┴ AB through a;
Through g draw a line parallel to the motion of B. the intersection of the two lines locates the point
b.
For the given configuration, the coupler AB has angular velocity in the counter- clockwise
direction, the magnitude being vba/ BA.
INVERSION:
We have already discussed that when one of links is fixed in a kinematic chain, it is called a
mechanism. So we can obtain as many mechanisms as the number of links in a kinematic chain by
fixing, in turn, different links in a kinematic chain. This method of obtaining different mechanisms
by fixing different links in a kinematic chain, is known as inversion of the mechanism.
It may be noted that the relative motions between the various links is not changed in any manner
through the process of inversion, but their absolute motions (those measured with respect to the
fixed link) may be changed drastically.
APPLICATIONS:
INVERSIONS OF FOUR BAR CHAIN
Though there are many inversions of the four bar chain, yet the following are important from the
subject point of view :
1. Beam engine (crank and lever mechanism)
A part of the mechanism of a beam engine (also known as crank and lever mechanism) which
consists of four links . In this mechanism, when the crank rotates about the fixed centre A, the lever
oscillates about a fixed centre D. The end E of the lever CDE is connected to a piston rod which
reciprocates due to the rotation of the crank. In other words, the purpose of this mechanism is to
convert rotary motion into reciprocating motion.
2. Coupling rod of a locomotive (Double crankmechanism).
The mechanism of a coupling rod of a locomotive (also known as double crank
mechanism) which consists of four links. In this mechanism, the links AD and BC (having equal
length) act as cranks and are connected to the respective wheels. The link CD acts as a coupling rod
and the link A B is fixed in order to maintain a constant centre to centre distance between them. This
mechanism is meant for transmitting rotary motion from one wheel to the other wheel.
INVERSIONS OF SINGLE SLIDER CRANK CHAIN:
We have seen in the previous article that a single slider crank chain is a four-link
mechanism. We know that by fixing, in turn, different links in a kinematic chain, an inversion is
obtained and we can obtain as many mechanisms as the links in a kinematic chain. It is thus
obvious, that four inversions of a single slider crank chain are possible. These inversions are found
in the following
Mechanisms.
Pendulum pump or Bull engine.
In this mechanism, the inversion is obtained by fixing the cylinder or link 4 (i.e. sliding pair). In this
case, when the crank (link 2) rotates, the connecting rod (link 3) oscillates about a pin pivoted to the
fixed link 4 at A and the piston attached to the piston rod (link 1) reciprocates. The duplex pump
which is used to supply feed water to boilers have two pistons attached to link 1.
Crank and slotted lever quick return motion mechanism:
This mechanism is mostly used in shaping machines, slotting machines and in rotary internal
combustion engines. this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, The
link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB
revolves with uniform angular speed about the fixed centre C. A sliding block attached to the
crankpin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point
A . A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates
along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC
produced.
In the extreme positions, AP1 and AP2 are tangential to the circle and the cutting tool is at the end
of the stroke. The forward or cutting stroke occurs when the crank rotates from the position CB1 to
CB2 (or through an angle â) in the clockwise direction. The return stroke occurs when the crank
rotates from the position CB2 to CB1 (or through angle á)
in the clockwise direction. Since the crank has uniform angular speed.
INVERSIONS OF DOUBLE SLIDER CRANK CHAIN
The following three inversions of a double slider crank chain are important from the subject point of
view :
Elliptical trammels.
It is an instrument used for drawing ellipses. This inversion is obtained by fixing the
slotted plate (link 4),The fixed plate or link 4 has two straight grooves cut in it, at right angles to
each other. The link 1 and link 3, are known as sliders and form sliding pairs with link 4. The link A
B (link 2) is a bar which forms turning pair with links 1 and 3. When the links 1 and 3 slide along
their respective grooves, any point on the link 2 such as P traces out an ellipse on the surface of link
4, A little consideration will show that AP and BP are the semi-major axis and semi-minor axis of
the ellipse respectively.
METHODS FOR DETERMINING THE VELOCITY OF A POINT ON A LINK:
Though there are many methods for determining the velocity of any point on a link in a
mechanism whose direction of motion (i.e. path) and velocity of some other point on the same link
is known in magnitude and direction, yet the following two methods are important from the subject
point of view.
1. Relative velocity method,and
2. Instantaneous centremethod.
Velocity in mechanisms (relative velocity method)
RELATIVE VELOCITY OF TWO BODIES MOVING IN STRAIGHT LINES
Here we shall discuss the application of vectors for the relative velocity of two bodies moving along
parallel lines
and inclined lines, as shown in Fig. 1 (a) and 2 (a)
respectively. Consider two bodies A and B moving along parallel lines in the same direction with
absolute velocities vAand
vB such that vA> vB, as shown in Fig. 1 (a). The relative velocity of A with respect to B ,
From Fig.1 (b), the relative velocity of A with respect to B (i.e. vAB) may be written in the vector
form as follows:
Now consider the body B moving in an inclined direction as shown in Fig. 2 (a). The relative
velocity of A with respect to B may be obtained by the law of parallelogram of velocities or triangle
law of velocities. Take any fixed point o and draw vector oa to represent vA in magnitude and
direction to some suitable scale. Similarly, draw vector ob to represent vB in magnitude and
direction to the same scale. Then vector ba represents the relative velocity of A with respect to B as
shown in Fig. 2 (b). In the similar way as discussed above, the relative velocity of A with respect to
B,
From above, we conclude that the relative velocity of point A with respect to B (vAB) and the
relative velocity of point B with respect A (vBA) are equal in magnitude but opposite in direction,
i.e.
Motion of a Link
Consider two points A and B on a rigid link A B, asshown in Fig 3 (a). Let one of the extremities (B) of
the linkmove relative to A , in a clockwise direction. Since the distance from A to B remains the same,
therefore there can be norelative motion between A and B, along the line A B. It is thus obvious, that the
relative motion of B
with respect to A must be perpendicular to A B.
Hence velocity of any point on a link with respect to another point on the same link is always
perpendicular to the line joining these points on the configuration (or space) diagram.
Thus, we see from equation (iii), that the point c on the vector ab divides it in the same ratio as C
divides the link A B.
Note: The relative velocity of A with respect to B is represented by ba, although A may be a fixed
point. The motion between A and B is only relative. Moreover, it is immaterial whether the link
moves about A in a clockwise direction or about B in a clockwise direction. Velocity of a Point on
aLinkbyRelativeVelocityMethodTherelativevelocitymethodisbasedupontherelativevelocity of the
various points of the link as discussed in Art..3. Consider two points A and B on a link as shown in
Fig..4 (a). Let the absolute velocity of the point A i.e. vA is known in magnitude and direction and
the absolute velocity of the point B i.e. vB is known in direction only. Then the velocity of B may
be determined by drawing the velocity diagram as shown in Fig. 4 (b). The velocity diagram is
drawn as follows:
1. Take some convenient point o, known as thepole.
2. Through o, draw oa parallel and equal to vA, to some suitablescale.
3. Through a, draw a line perpendicular to A B of Fig. 4 (a). This line will representthe
velocity of B with respect to A , i.e.vBA.
4. Through o, draw a line parallel to vBintersecting the line of vBA atb.
5. Measure ob, which gives the required velocity of point B ( vB), to thescale
Velocities in Slider Crank Mechanism
In the previous article, we have discused the relative velocity method for the velocity of any point
on a link, whose direction of motion and velocity of some other point on the same link is known.
The same method may also be applied for the velocities in a slider crank mechanism. A slider crank
mechanism is shown in Fig. 5 (a). The slider A is attached to the connecting rod A B. Let theradius
of crank OB be r and let it rotates in a clockwise direction, about the point O with uniform angular
velocity ὼ rad/s. Therefore, the velocity of B i.e. vB is known in magnitude and direction. The
slider reciprocates along the line of stroke A O. The velocity of the slider A (i.e. vA) may be
determined by relative velocity method as discussed below:
1.From any pointo, draw vector ob parallel to the direction of vB (or perpendicular to OB) such
that ob = vB= ὼ.r, to some suitable scale, as shown in Fig. 5(b).
2. Since A B is a rigid link, therefore the velocity of A relative to B is perpendicular to A B.Now
draw vector ba perpendicular to A B to represent the velocity of A with respect to B i.e.vAB.
3. From point o, draw vector oa parallel to the path of motion of the slider A (which is along AO
only). The vectors ba and oa intersect at a. Now oa represents the velocityof the slider A i.e. vA, to
the scale. The angular velocity of the connecting rod A B (ὼAB) maybe determined as follows:
The direction of vector ab (or ba) determines the sense of ὼAB which shows that it is
anticlockwise.
PROBLEMS:
Example 1.
In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm long and
rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and AD are of
equal length. Find the angular velocity of link CD when angle BAD =60°.
GIVEN :
NBA=120 r.p.m
ὼ=2π×120/60
=12.568 rad/s
BAD =60°
CD = 80 mm
SOLUTION:
Since the length of crank A B = 40 mm = 0.04 m, therefore velocity of B with respect to A or
velocity of B, (because A is a fixed point),
vBA= vB=ὼBA× A B = 12.568 × 0.04 = 0.503 m/s
First of all, draw the space diagram to some suitable scale, as shown in Fig. 6 (a). Now the velocity
diagram, as shown in Fig. 6(b), is drawn as discussed below :
1. Since the link A D is fixed, therefore points a and d are taken as one point in the velocity
diagram. Draw vector ab perpendicular to B A, to some suitable scale, to represent the velocity ofB
with respect to A or simply velocity of B (i.e. vBA or vB) suchthat
Vector ab = vBA= vB= 0.503 m/s
2. Now from point b, draw vector bc perpendicular to CB to represent the velocity of C with
respect to B (i.e. vCB) and from point d, draw vector dc perpendicular to CD to represent the
velocity of C with respect to D or simply velocity of C (i.e. vCD or vC). The vectors bc and dc
intersect atc.
By measurement, we find that
vCD= vC = vector dc = 0.385 m/s
we know that velocity of link CD,
RESULT:
ὼCD=4.8 rad/s
Example 2.
In Fig.7, the angular velocity of the crank OA is 600 r.p.m. Determine the linear velocity of the
slider D and the angular velocity of the link BD, when the crank is inclined at an angle of 75° to the
vertical. The dimensions of various links are : OA = 28 mm ; AB = 44 mm ; BC 49 mm ; and BD
= 46 mm. The centre distance between the centres of rotation O and C is 65 mm. The path of travel
of the slider is 11 mm below the fixed point C. The slider moves along a horizontal path and OC is
vertical.
Given
NAO=600 r.p.m
AB = 44 mm
OA = 28 mm
BC = 49 mm
BD = 46mm
ὼAO=2π×600/60=62.84 rad/s
Solution:
since OA=28mm=0.028, therefore velocity of A with respect of O or velocity of O (because O is a
fixed point),
VAO =VA =ὼAO× OA=62.84 ×0.028=1.76m/s
Linear velocity of the slider D
First of all draw the space diagram, to some suitable scale, as shown in Fig. 8 (a). Now the velocity
diagram, as shown in Fig. 8 (b), is drawn as discussed below :
1. Since the points O and C are fixed, therefore these points are marked as one point, in thevelocity
diagram. Now from point o, draw vector oa perpendicular to O A, to some suitable scale, to
represent the velocity of A with respect to O or simply velocity of A suchthat
vector oa= vAO= vA = 1.76 m/s
2. From point a, draw vector ab perpendicular to A B to represent the velocity of B with respect A
(i.e. vBA) and from point c, draw vector cb perpendicular to CB to represent the velocityof B with
respect to C or simply velocity of B (i.e. vBC or vB). The vectors ab and cb intersect atb.
3. From point b, draw vector bd perpendicular to BD to represent the velocity of D with respect toB
(i.e. vDB) and from point o, draw vector od parallel to the path of motion of the slider D which is
horizontal, to represent the velocity of D (i.e. vD). The vectors bd and od intersect atd.
By measurement, we find that velocity of the slider D,
vD= vector od = 1.6 m/s
Angular velocity of the link BD
By measurement from velocity diagram, we find that velocity of D with respect to B,
vDB= vector bd = 1.7 m/s
Since the length of link BD = 46 mm = 0.046 m, therefore angular velocity of the link BD,
RESULTS:
VD=1.6 m/s
ὼBD=36.96 rad/s
Example 3
In a mechanism shown in Fig. 9, the crank OA is 100 mm long and rotates clockwise about O at
120 r.p.m. The connecting rod AB is 400 mm long. At a point C on AB, 150 mm from A, the rod CE
350 mm long is attached. This rod CE slides
in a slot in a trunnion at D. The end E is connected by a link EF, 300 mm long to the horizontally
moving slider F.
For the mechanism in the position shown, find 1. velocity of F, 2. velocity of sliding of CE .
Given:
VAO=120r.p.m
ὼAO=2π×120/60=4π rad/s
SOLUTION:
Since the length of crank O A = 100 mm= 0.1 m, therefore velocity of A with respect to O or
velocity of A (because O is a fixed point),
vAO= vA= ὼAO× O A = 4 π× 0.1 = 1.26 m/s
1. Velocity of F
First of all draw the space diagram, to some suitable scale, as shown in Fig. 10 (a). Now the velocity
diagram, as shown in Fig. 10 (b), is drawn as discussed below
1. Draw vector oa perpendicular to A O, to some suitable scale, to represent the velocity of Awith
respect to O or simply velocity of A (i.e. vAOor vA), suchthat
vector oa= vAO= vA= 1.26 m/s
2. From point a, draw vector ab perpendicular to A B to represent the velocity of B with respect toA
i.e. vBA, and from point o draw vector ob parallel to the motion of B (which moves along BO only)
to represent the velocity of B i.e. vB. The vectors ab and ob intersect at b.
3. Since the point C lies on A B, therefore divide vector ab at c in the same ratio as C divides A Bin
the space diagram. In otherwords,
ac/ab = AC/A B
4. From point c, draw vector cd perpendicular to CD to represent the velocity of D with respect toC
i.e. vDC, and from point o draw vector od parallel to the motion of CD, which moves along CD
only, to represent the velocity of D, i.e. vD.
5. Since the point E lies on CD produced, therefore divide vector cd at e in the same ratio asE
divides CD in the space diagram. In other words,
cd/ce= CD/CE
6. From point e, draw vector ef perpendicular to EF to represent the velocity of F with respect toE
i.e. vFE, and from point o draw vector of parallel to the motion of F, which is along FD to represent
the velocity of F i.e. vF.
By measurement, we find that velocity of F,
vF= vector of = 0.53 m/s
2. Velocity of sliding of CE in thetrunnion
Since velocity of sliding of CE in the trunnion is the velocity of D, therefore velocity of sliding of
CE in the trunnion
= vector od = 1.08 m/s
3. Angular velocity ofCE
By measurement, we find that linear velocity of C with respect to E,
vCE= vector ec = 0.44 m/s
Since the length CE = 350 mm = 0.35 m, therefore angular velocity of CE,
RESULTS:
VF=0.53m/s
VCE=0.44m/s
ὼCE=1.26 rad/s
Example 4. In a mechanism as shown in Fig. 7.15, the various dimensions are : OC = 125mm ; CP
= 500 mm ; PA = 125 mm ; AQ = 250 mm and QE = 125 mm.
The slider P translates along an axis which is 25 mm vertically below point O. The crank OC
rotates uniformly at 120 r.p.m. in the anti-clockwise direction. The bell crank lever AQE rocks
about fixed centre Q.
Given
NCO=120 r.p.m
ὼCO=12.57 rad/s
SOLUTION :
We know that linear velocity of C with respect to O or velocity of C, (because O is as fixed point)
vCO= vC= ὼCO× OC = 12.57 × 0.125 = 1.57 m/s
First of all, draw the space diagram, as shown in Fig. 7.12(a) to some suitable scale. Nowthe
velocity diagram, as shown in Fig. 12 (b) is drawn as discussed below:
1. Since the points O and Q are fixed, therefore these points are taken as one point in thevelocity
diagram. From point o, draw vector oc perpendicular to OC, to some suitable scale, to represent
the velocity of C with respect to O or velocity of C, suchthat
vector oc = vCO= vC= 1.57 m/s
2. From point c, draw vector cp perpendicular to CP to represent the velocity of P with respect toC
(i.e. vPC) and from point o, draw vector op parallel to the path of motion of slider P (which is
horizontal) to represent the velocity of P (i.e. vP). The vectors cp and op intersect atp.
3. From point p, draw vector pa perpendicular to PA to represent the velocity of A with respect to
P (i.e. vAP) and from point q, draw vector qa perpendicular to Q A to represent the velocity of A
(i.e. vA). The vectors pa and qa intersect ata.
4. Now draw vector qe perpendicular to vector qa in such a waythat
QE/QA = qe/qa
By measurement, we find that the velocity of point E,
vE = vector oe = 0.7 m/s
RESULT :
VE= 0.7 m/s
Example 5. Fig. 13 shows the structure of Whitworth quick return mechanism used in
reciprocating machine tools. The
various dimensions of the tool are as follows : OQ = 100 mm ; OP = 200 mm, RQ =150 mm and
RS = 500 mm.
The crank OP makes an angle of 60° with the vertical. Determine the velocity of the slider S
(cutting tool) when the crank rotates
at 120 r.p.m. clockwise. Find also the angular velocity of the link RS and the velocity of the sliding
block T on the slotted lever QT.
GIVEN :
NPO= 120 r.p.m. or ὼPO= 2 π × 120/60 = 12.57 rad/s
SOLUTION :
Since the crank OP = 200 mm = 0.2 m, therefore velocity of P with respect to O or velocity of P
(because O is a fixed point),
vPO= vP= ὼPO× OP = 12.57 × 0.2 = 2.514 m/s
Velocity of slider S (cutting tool)
First of all draw the space diagram, to some suitable scale, as shown in Fig. 14 (a). Now the velocity
diagram, as shown in Fig. 14 (b) is drawn as discussed below :
1. Since O and Q are fixed points, therefore they are taken as one point in the velocity diagram.
From point o, draw vector op perpendicular to OP, to some suitable scale, to represent thevelocity
of P with respect to O or simply velocity of P, suchthat
vector op = vPO= vP= 2.514 m/s
2. From point q, draw vector qt perpendicular to QT to represent the velocity of T with respect toQ
or simply velocity of T (i.e. vTQ or vT) and from point p draw vector pt parallel to the path of
motion of T (which is parallel to TQ) to represent the velocity of T with respect to P (i.e. vTP). The
vectors qt and pt intersect att
3. Since the point R lies on the link TQ produced, therefore divide the vector tq at r in the same
ratio as R divides TQ, in the space diagram. In otherwords,
qr/qt = QR/QT
The vector qr represents the velocity of R with respect to Q or velocity of R (i.e.vRQ or vR).
4. From point r, draw vector rs perpendicular to R S to represent the velocity of S with respect toR
and from point o draw vector or parallel to the path of motion of S (which is parallel to QS) to
represent the velocity of S (i.e vS). The vectors rs and os intersect ats.
By measurement, we find that velocity of the slider S (cutting tool),
vS= vector os = 0.8 m/s
Angular velocity of link RS
From the velocity diagram, we find that the linear velocity of the link R S,
vSR= vector rs = 0.96 m/s
Since the length of link R S = 500 mm = 0.5 m, therefore angular velocity of link R S,
RESULTS :
VS=0.8m/s,
ὼRS=0.92 rad/s,
VTP=0.85 m/s.
Example 6 In the toggle mechanism, as shown in Fig. 15. the slider D is constrained to move on a
horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed of 180
r.p.m. The dimensions of various links are as follows :
OA = 180 mm ; CB = 240 mm ; AB = 360 mm ; and BD = 540 mm.
For the given configuration, find : 1. Velocity of slider D, 2. Angular velocity of links AB, CB and
BD; 3. Velocities of rubbing on the pins of diameter 30 mm at A and D, and 4. Torque applied to
thecrank
GIVEN:
NAO=180 r.p.m
ὼAO=18.85 rad/s
SOLUTION:
Since the crank length O A = 180 mm = 0.18 m, therefore velocity of A with respect to O or
velocity of A (because O is a fixed point),
vAO= vA= ὼAO× O A = 18.85 × 0.18 = 3.4 m/s
1. Velocity of slider D
First of all draw the space diagram, to some suitable scale, as shown in Fig. 16(a)Now the
velocity diagram, as shown in Fig. 16 (b), is drawn as discussed below :
1. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of Awith
respect to O or velocity of A (i.e. vAOor vA,) suchthat
vector oa= vAO= vA= 3.4 m/s
2. Since point B moves with respect to A and also with respect to C, therefore draw vector ab
perpendicular to A B to represent the velocity of B with respect to Ai.e. vBA, and draw vector cb
perpendicular to CB to represent the velocity of B with respect to C, i.e. vBC. The vectors ab andcb
intersect atb.
3. From point b, draw vector bd perpendicular to BD to represent the velocity of D with respect toB
i.e. vDB, and from point c draw vector cd parallel to the path of motion of the slider D (which is
along CD) to represent the velocity of D, i.e. vD. The vectors bd and cd intersect at d.
By measurement, we find that velocity of the slider D,
vD= vector cd = 2.05 m/s
2. Angular velocities of links AB, CB andBD
By measurement from velocity diagram, we find that
Velocity of B with respect to A ,
vBA= vector ab = 0.9 m/s
Velocity of B with respect to C,
vBC= vB = vector cb = 2.8 m/s
and velocity of D with respect to B,
vDB= vector bd = 2.4 m/s
We know that A B = 360 mm = 0.36 m ; CB = 240 mm = 0.24 m and BD = 540 mm = 0.54 m.
3. Velocities of rubbing on the pins A andD
Given : Diameter of pins at A and D,
DA= DD = 30 mm = 0.03 m
Radius, rA= rD = 0.015 m
We know that relative angular velocity at A
= ὼBC – ὼBA+ ὼDB= 11.67 – 2.5 + 4.44 = 13.61 rad/s
and relative angular velocity at D
=ὼDB = 4.44 rad/s
Velocity of rubbing on the pin A
= 13.61 × 0.015 = 0.204 m/s = 204 mm/s
and velocity of rubbing on the pin D
= 4.44 × 0.015 = 0.067 m/s = 67 mm/s
4. Torque applied to the crankOA
Let TA= Torque applied to the crank O A, in N-m
Power input or work supplied at A
= TA× ὼAO= TA × 18.85 = 18.85 TAN-m
We know that force at D,
FD= 2 kN = 2000 N
Power output or work done by D,
= FD × vD = 2000 × 2.05 = 4100 N-m
Assuming 100 per cent efficiency, power input is equal to poweroutput.
18.85TA=4100 or TA = 217.5N-m
RESULTS:
TA=217.56 N-m
ὼAB=2.5 rad/s, ὼCB=11.67 rad/s
Velocity in Mechanisms (Instantaneous Centre Method)
Introduction:
Sometimes, a body has simultaneously a motion of rotation as well as translation, Such as wheel
of a car, a sphere rolling (but not slipping) on the ground. Such a Motion will have the combined
effect of Rotation and translation. Consider a rigid link AB, which moves from its initial position
AB to A1B1 as shown in Fig
The instantaneous centre method is convenient and easy to apply in simple mechanisms, whereas
the relative velocity method may be used to any configuration diagram.
In actual practice, the motion of link A B is so gradual that it is difficult to see the two separate
motions. But we see the two separate motions, though the point B moves faster thanthepoint A .
Thus, this combined motion of rotation and translation of the link A B may be assumed to be a
motion of pure rotation about some centre I, known as the instantaneous centre of rotation (also
called centre or virtualcentre).
The position of instantaneous centre may be located as discussed below:
Since the points A and B of the link has moved to A1 and B1 respectively under the motion of
rotation (as assumed above), therefore the position of the centre of rotation must lie on the
intersection of the right bisectors of chords A A1 and B B1.
Let these bisectors intersect at I as shown in Fig.,
which is the instantaneous centre of rotation or virtual centre of the link A B. From above, we see
that the position of the link AB goes on changing, therefore the centre about which the motion is
assumed to take place (i.e. the instantaneous centre of rotation also goes on changing. Thus the
instantaneous centre of a moving body may be defined as that centre which goes on changing from
one instant to another. The locus of all such instantaneous centres is known as centrode. A line
drawn through an instantaneous centre and perpendicular to the plane of motion is called
instantaneous axis. The locus of this axis is known as axode.
VELOCITY OF A POINT ON A LINK BY INSTANTANEOUS CENTRE METHOD:
The instantaneous centre method of analysing the motion in a mechanism is based upon the concept
that any displacement of a body (or a rigid link) having motion in one plane, can be considered as a
pure rotational motion of a rigid link as a whole about some centre, known as instantaneous centre
or virtual centre of rotation. Consider two points A and B on a rigid link. Let vA and vB be the
velocities of points A and B, whose directions are given by angles α and β as shown in Fig If vA is
known in magnitude and direction and vB in direction only, then the magnitude of vB may be
determined by the instantaneous centre method as discussed below :
Draw A I and BI perpendiculars to the directions vA and vB respectively. Let these lines intersect at
I, which is known as instantaneous centre or virtual centre of the link. The complete rigid link is to
rotate or turn about the centre I. Since A and B are the points on a rigid link, therefore there cannot
be any relative motion between them along the lineA B.
From the above equation, we see that
1. If vA is known in magnitude and direction and vB in direction only, then velocity of point Bor
any other point C lying on the same link may be determined in magnitude anddirection.
2. The magnitude of velocities of the points on a rigid link is inversely proportional to thedistances
from the points to the instantaneous centre and is perpendicular to the line joining the point tothe
instantaneous centre.
Properties of the Instantaneous Centre:
The following properties of the instantaneous centre are important from the subject point of view:
1. A rigid link rotates instantaneously relative to another link at the instantaneous centre forthe
configuration of the mechanismconsidered.
2. The two rigid links have no linear velocity relative to each other at the instantaneous centre. At
this point (i.e. instantaneous centre), the two rigid links have the same linear velocity relative to the
third rigid link. In other words, the velocity of the instantaneous centre relative to any third rigid
link will be same whether the instantaneous centre is regarded as a point on the first rigid link oron
the second rigidlink.
Types of Instantaneous Centres:
The instantaneous centres for a mechanism are of the following three types:
1. Fixed instantaneouscentres,
2. Permanent instantaneous centres,and
3. Neither fixed nor permanent instantaneouscentres.
LOCATION OF INSTANTANEOUS CENTRES:
The following rules may be used in locating the instantaneous centres in a mechanism:
1. When the two links are connected by a pin joint (or pivot joint), the instantaneous centre lieson
the centre of the pin. Such a instantaneous centre is of permanent nature, but if one of the links is
fixed, the instantaneous centre will be of fixedtype.
2. When the two links have a pure rolling contact (i.e. link 2 rolls without slipping upon the fixed
link 1 which may be straight or curved), the instantaneous centre lies on their point of contact.The
velocity of any point A on the link 2 relative to fixed link 1 will be perpendicular to I12A and is
proportional to I12A.
3. When the two links have a sliding contact, the instantaneous centre lies on the common normal at
the point of contact. We shall consider the following threecases:
(a) When the link 2 (slider) moves on fixed link 1 having straight surface the instantaneouscentre
lies at infinity and each point on the slider have the samevelocity.
(b) When the link 2 (slider) moves on fixed link 1 having curved surface the instantaneous
centre lies on the centre of curvature of the curvilinear path in the configuration at that instant.
(c)When the link 2 (slider) moves on fixed link 1 having constant radius of curvature the
instantaneous centre lays at the centre of curvature i.e. the centre of the circle, for all configuration
of the links.
PROBLEMS:
Example 1. In a pin jointed four bar mechanism, as shown in Fig, AB = 300 mm, BC = CD = 360
mm, and AD = 600 mm. The angle BAD = 60°. The crank AB rotates uniformly at 100 r.p.m.
Locate all the instantaneous centers and find the angular velocity of the link BC.
Given
NAB= 100 r.p.m or
ὼAB= 2 π 100/60= 10.47 rad/s
BAD = 60°
BC = CD = 360 mm
AD = 600mm
AB = 300mm
Solution:
Since the length of crank A B = 300 mm = 0.3 m, therefore velocity of point B on link A B,
VAB=ὼ x AB= 10.47 x 0.3= 3.141 m/s
Location of instantaneous centers
The instantaneous centers are located as discussed below:
1. Since the mechanism consists of four links (i.e. n = 4 ), therefore number of
instantaneous centers,
2. For a four bar mechanism, the book keeping table may be drawn as discussed in fig1
3. Locate the fixed and permanent instantaneous centers by inspection. These centers area12,
I23, I34 and I14, as shown in Fig1
4. Locate the remaining neither fixed nor permanent instantaneous centers by Arnold
Kennedy’s theorem. This is done by circle diagram as shown in Fig. 2. Mark four
points (equal to the number of links in a mechanism) 1, 2, 3, and 4 on the circle.
5. Join points 1 to 2, 2 to 3, 3 to 4 and 4 to 1 to indicate the instantaneous centre already
located i.e. I12, I23, I34 and I14.
6. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1. The side 13, common to both triangles, is
responsible for completing the two triangles. Therefore the instantaneous centre I13 lies on the
intersection of the lines joining the points I12 I23 and I34I14 as shown in Fig. 1 Thus centre I13is
located. Mark number 5 (because four instantaneous centers have already been located)
on the dotted line 1 3.
7. Now join 2 to 4 to complete two triangles 2 3 4 and 1 2 4. The side 2 4, common to both
triangles, is responsible for completing the two triangles. Therefore centre I24 lies on the
intersection of the lines joining the points I23I34 and I12I14 as shown in Fig.1. Thus centre I24is
located. Mark number 6 on the dotted line 2 4. Thus all the six instantaneous centers are located.
Angular velocity of the link
Let ὼBC= angular velocity of the link
Since B is also a point on link BC, therefore velocity of point B on link BC
VB=ὼBC x I13B
by measurements , we find that I13B = 500 mm = 0.5
RESULT:
ὼBC=6.282 m/s
Example 2. Locate all the instantaneous centers of the slider crank mechanism as shown in Fig..
The lengths of crank OB and connecting rod AB are 100 mm and 400 mm respectively. If the crank
rotates clockwise with an angular velocity of 10 rad/s, find: 1. Velocity of the slider A, and 2.
Angular velocity of the connecting rod AB.
Given:
ὼOB=10 rad/sec
OB=100mm=0.1mm
AB=400mm=0.4mm
SOLUTION:
We know that linear velocity of the crank OB,
VOB= vB = ὼOB × OB= 10×0.1=1 m/s
Location of instantaneous centers
The instantaneous centers in a slider crank mechanism are located as discussed below:
1. since there are four links(i.e. n=4), therefore the number of instantaneouscentres,
N= n (n-1) /2
= 4 (4-1) / 2 = 6
2. For a four link mechanism, the book keeping table may be drawn as discussed.
3. Locate the fixed and permanent instantaneous centers by inspection. These centers area12,
I23 and I34 as shown in fig 2. Since the slider (link 4) moves on a straight surface (link 1),
therefore the instantaneous centre I14 will be at infinity.
4. 4. Locate the other two remaining neither fixed nor permanent instantaneous centers, by
Arnold Kennedy’s theorem. This is done by circle diagram as shown in fig 3. Mark four
points 1, 2, 3 and 4 (equal to the number of links in a mechanism) on the circle to indicate
I12, I23, I34 and14.
5. 5. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1 in the circle diagram. The side 1 3,
common to both triangles, is responsible for completing the two triangles. Therefore
the centre I13 will lay on the intersection of I12I23 and I14I34, produced if necessary.
Thus centre I13 is located. Join 1 to 3 by a dotted line and mark number 5 omit.
6. Join 2 to 4 by a dotted line to form two triangles 2 3 4 and 1 2 4. The side 2 4, common tooth
triangles, is responsible for completing the two triangles. Therefore the centre I24 lies on the
intersection of I23I34 and I12I14. Join 2 to 4 by a dotted line on the circle diagram and mark
number 6 on it. Thus all the six instantaneous centers are located.
By measurement, we find that
I13A = 460 mm = 0.46 m ; and I13B = 560 mm = 0.56 m
1. Velocity of the slider
Let vA = Velocity of the slider A.
We know that vA /I13 A =vB /I13B
=0.46/0.56 = 0.82 m/s
2. Angular velocity of the connecting road
Let ὼAB= angular velocity of the connecting road
We know that vA /I13 A =vB /I13 B = ὼAB
RESULT:
ὼAB= 1/0.56= 1.78 rad/s
ὼAB=1.78 rad/s
Example 3. The mechanism of a wrapping machine, as shown in Fig. has the following
dimensions:
O1A = 100 mm; AC = 700 mm; BC = 200 mm; O3C = 200 mm; O2E = 400 mm; O2D = 200 mm
and BD = 150 mm.
The crank O1A rotates at a uniform speed of 100 rad/s. Find the velocity of the point E of the bell
crank lever by instantaneous centre method.
Given:
O1A = 100 mm; AC = 700 mm; BC = 200 mm; O3C = 200 mm; O2E = 400 mm; O2D = 200 mm
and BD = 150 mm
ὼ O1A=100 rad/s
solution:
We know that the linear velocity of crank O1A ,
VO1 A=VA=ὼO1A ×O1A=100×0.1=10m/s
Now let us locate the required instantaneous centers as discussed below :
1. Since the mechanism consists of six links (i.e. n = 6), therefore number of instantaneouscentres,
2. Since the mechanism has 15 instantaneous centers, therefore these centers may be listed in
the book keeping table, as discussed
3. Locate the fixed and the permanent instantaneous centers by inspection. These centers area12,
I23, I34, I35, I14, I56 and I16 as shown inFig1
4. LocatetheremainingneitherfixednorpermanentinstantaneouscentresbyAronholdKennedy’s
theorem. This is done by circle diagram as shown in Fig. 2 Mark six points on the circle (i.e. equal
to the number of links in a mechanism), and join 1 to 2, 2 to 3, 3 to 4, 3 to 5, 4 to 1, 5 to 6, and 6 to
1, to indicate the fixed and permanent instantaneous centers i.e. I12, I23, I34, I35, I14, I56, and I16
respectively.
5. Join 1 to 3 by a dotted line to form two triangles 1 2 3 and 1 3 4. The side 1 3, common to both
triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I13lies
on the intersection of the lines joining the points I12I23 and I14I34 produced if necessary.
Thus centre I13 is located. Mark number 8 (because seven centers have already been located) on the
dotted line 1 3.
6. Join 1 to 5 by a dotted line to form two triangles 1 5 6 and 1 3 5. The side 1 5, common to both
triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I15lies
on the intersection of the lines joining the points I16I56 and I13I35 produced if necessary. Thus
centre I15 is located. Mark number 9 on the dotted line 15.
Note: For the given example, we do not require other instantaneous centers.
By measurement, we find that
I13A = 910 mm = 0.91 m ; I13B = 820 mm = 0.82 m ; I15B = 130 mm = 0.13 m ;
I15D = 50 mm = 0.05 m ; I16D = 200 mm = 0.2 m ; I16E = 400 mm = 0.4 m
Velocity of point E on the bell crank lever
Let vE= Velocity of point E on the bell crank lever,
vB= Velocity of point B, and
vD= Velocity of point D.
We know thatA/I13A=VB/I13B
RESULTS:
VE = 6.92m/s
VD = 3.46m/s
VE = 6.92m/s
ACCELERATION IN MECHANISMS:
we have discussed in the velocities of various points in the mechanisms. Now we shall discuss the
acceleration of points in the mechanisms. The acceleration analysis plays a very important role in
the development of mechanisms and mechanisms.
Acceleration Diagram for a Link
Consider two points A and B on a rigid link as shown in Fig. 1 (a). Let the point B moves with
respect to A, with
an angular velocity of ὼ rad/s and let α rad/s2 be the angular acceleration of the link AB.
We have already discussed that acceleration of a particle whose velocity changes both in magnitude
and direction at any instant has the following two components :
1. The centripetal or radial component, which is perpendicular to the velocity of the particle at the
given instant.
2. The tangential component, which is parallel to the velocity of the particle at the given instant.
Thus for a link A B, the velocity of point B with respect to A (i.e. vBA) is perpendicular to the link
A B as shown in Fig. 1 (a). Since the point B moves with respect to A with an angular velocity
of ὼ rad/s, therefore centripetal or radial component of the acceleration of B with respect to A,
This radial component of acceleration acts perpendicular to the velocity vBA, In other words, it acts
parallel to the link A B .
We know that tangential component of the acceleration of B with respect to A ,
This tangential component of acceleration acts parallel to the velocity vBA. In other words, it acts
perpendicular to the link A B.
In order to draw the acceleration diagram for a link A B, as shown in Fig. 1 (b), from any point b',
draw vector box parallel to BA to represent the radial component of acceleration of B with
ACCELERATION OF A POINT ON A LINK:
Consider two points A and B on the rigid link, as shown in Fig.2 (a). Let the acceleration
of the point A i.e. atA is known in magnitude and direction and the direction of path of B is
given. The acceleration of the point B is determined in magnitude and direction by drawing the
acceleration diagram as discussed below.
1. From any point o', draw vector ova' parallel to the direction of absolute acceleration
at point A i.e. atA, to some suitable scale, as shown in Fig. 2(b).
2. We know that the acceleration of B with respect to A i.e. aBA has the following two components:
(I)Radial component of the acceleration of B with respect toile andBA.(ii)Tangential
component of the acceleration B with respect to i.e. . These two aBA. Components are
mutually perpendicular.
3. Draw vector ax parallel to the link A B (because radial component of the acceleration of With
respect to A will pass through AB), suchthat
where vBA=Velocity of B with respect toe
4. From point x, draw vector be' perpendicular to A B or vector ax (because tangential component
of B with respect to A i.e. aBA, perpendicular to radial component aBAr) and through o' draw a
line parallel to the path of B to represent the absolute acceleration of B i.e. aB. The vectors band
o' b' intersect at b'. Now the values of aB and aBAt may be measured, to thescale.
5. By joining the points a' and b' we may determine the total acceleration of B with respect to A
i.e. aBA. The vector a' b' is known as acceleration image of the link AB.
6. For any other point C on the link, draw triangle a' b' c' similar to triangle ABC. Now vector b' c'
represents the acceleration of C with respect to B i.e. aCB, and vector a' c' represents the
acceleration of C with respect to A i.e. aCA. As discussed above, aCB and aCA will each have two
components as follows:
7. The angular acceleration of the link AB is obtained by dividing the tangential components of
the acceleration of B with respect to A t the length of the link. Mathematically, angular
(alba) acceleration of the linkable,
AE 2201 mechanics of
Problems:
Example 1. The crank of a slider crank mechanism rotates clockwise at a constant speed of 300
r.p.m. The crank is 150 mm and the connecting rod is 600 mm long. Determine : 1. Linear
velocity and acceleration of the midpoint of the connecting rod, and 2. angular velocity and
angular acceleration of the connecting rod, at a crank angle of 45° from inner dead centre
position.
GIVEN:
NBO=300 r.p.m
OB=150mm=0.15mm
BA=600mm=0.6mm
ὼBO=31.42 rad/s
SOLUTION:
We know that linear velocity of B with respect to O or velocity of B,
vBO = vB = ὼBO × OB = 31.42 × 0.15 = 4.713 m/s
1. Linear velocity of the midpoint of the connecting rod
First of all draw the space diagram, to some suitable scale; as shown in Fig. 3 (a). Now the
velocity diagram, as shown in Fig. 3 (b), is drawn as discussed below:
1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of With
respect to O or simply velocity of B i.e. vBO or vB, suchthat
vector ob = vBO = vB = 4.713 m/s
2. From point b, draw vector ba perpendicular to B A to represent the velocity of A with respect
to B i.e. vAB, and from point o draw vector oa parallel to the motion of A (which is along A O)
to represent the velocity of A i.e. vA. The vectors ba and oa intersect ata.
By measurement, we find that velocity of A with respect to B,
vAB=vector ba =3.4 m / s
A v vector oa 4 m / s 3. In order to find the velocity of the midpoint D of the connecting rod A B, divide the vector at d in the same ratio as D divides A B, in the space diagram. In other words,
bad / ba = BD/BA
AE 2201 mechanics of
4. Join do. Now the vector do represents the velocity of the midpoint D of the connecting
rowdier. vD.
By measurement, we find that
vD = vector do = 4.1 m/s
Acceleration of the midpoint of the connecting rod
We know that the radial component of the acceleration of B with respect to O or the acceleration
of B,
2. The acceleration of A with respect to B has the following two components:
(a) The radial component of the acceleration of A with respect to B i.e. aAB,rand
(b) The tangential component of the acceleration of A with respect to B i.e. aABt.These two
components are mutually perpendicular
AE 2201 mechanics of
Therefore from point b', draw vector b' x parallel to A B to represent aAB 19.3 m/s and from
point x draw vector ax' perpendicular to vector b' x whose magnitude is yet unknown.
3. Now from o', draw vector o' a' parallel to the path of motion of A (which is along A O)to
represent the acceleration of A i.e. atA. The vectors ax' and o' a' intersect at a'. Join alb'.
4. In order to find the acceleration of the midpoint D of the connecting rod A B, divide the vector
a' b' at d' in the same ratio as D divides A B. In other words
bad/boa=BD/BA
5. Join o' d'. The vector o' d' represents the acceleration of midpoint D of the connecting rod i.e.
aD.
By measurement, we find that
aD = vector o' d' = 117 m/s2
2. Angular velocity of the connecting rod
We know that angular velocity of the connecting rod A B
Angular acceleration of the connecting rod From the acceleration diagram, we find that
We know that angular acceleration of the connecting rod A B,
RESULTS:
VD=4.1m/s,
ὼAB=5.67rad/s
αAB=171.67 rad/s
Example 2. In the toggle mechanism shown in Fig. 4, the slider D is constrained to move on a
horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed of 180
r.p.m. increasing at the rate of 50 rad/s2. The dimensions of the various links are as follows:
OA = 180 mm ; CB = 240 mm ; AB = 360 mm ; and BD = 540 mm.
For the given configuration, find 1. Velocity of slider D and angular velocity of BD, and 2.
Acceleration of slider D and angular acceleration of BD.
AE 2201 mechanics of
GIVEN:
NAO=180 r.p.m
OA=180mm
CB=240mm
AB=360mm
SOLUTION:
We know that velocity of A with respect to O or velocity of A ,
vAO = vA = ὼAO × O A = 18.85 × 0.18 = 3.4 m/s
1. Velocity of slider D and angular velocity of BD
First of all, draw the space diagram to some suitable scale, as shown in Fig. 5 (a). Now
the velocity diagram, as shown in Fig. 5 (b), is drawn as discussed below:
1. Since O and C are fixed points, therefore these points lie at one place in the velocity
diagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity
of A with respect to O or velocity of A i.e. vAO or vA, suchthat
vector oa = vAO = vA = 3.4 m/s
AE 2201 mechanics of
2. Since B moves with respect to A and also with respect to C, therefore draw vector ab
perpendicular to A B to represent the velocityof B with respect to A i.e. vBA, and draw vector
cob perpendicular to CB to represent the velocity of B with respect to C i.e. vBC. The vectors ab
and cob intersect atb.
3. From point b, draw vector bad perpendicular to BD to represent the velocity of D with
respect to B i.e. vDB, and from point c draw vector cod parallel to CD (i.e., in the direction of
motion of the slider D) to represent the velocity of D i.e.vD.
By measurement, we find that velocity of B with respect to A ,
vBA = vector ab = 0.9 m/s
Velocity of With respect to C,
vBC = vector cob = 2.8 m/s
Velocity of D with respect to B,
and velocity of slider D,
Angular velocity of BD
vDB = vector bad = 2.4m/s
vD = vector cod = 2.05m/s
We know that the angular velocity of BD,
2. Acceleration of slider D and angular acceleration of BD
Since the angular acceleration of OA increases at the rate of 50 rad/s2, i.e. α AO= 50 rad/s2,
AE 2201 mechanics of
therefore
Tangential component of the acceleration of A with respect to O,
Now the acceleration diagram, as shown in Fig. 5(c), is drawn as discussed below:
1. Since O and C are fixed points, therefore these points lie at one place in the
acceleration diagram. Draw vector ox parallel to O A, to some suitable scale, to represent
the radial component of the acceleration of A with respect too
2. From point x, draw vector ax' perpendicular to vector ox or O A to represent the tangential
component of the acceleration of A with respect too.
3. Join ova'. The vector ova' represents the totalaccelerationof A withrespect to O or
acceleration of A i.e. aAO oraA
4. Now from point a', draw vector a'y parallel to A B to represent the radial component of
the acceleration of B with respect to A
5. From point y, draw vector by' perpendicular to vector a'y or A B to represent the tangential
component of the acceleration of B with respect to A whose magnitude is yet unknown
6. Now from point c', draw vector co’s parallel to CB to represent the radial component of
the acceleration of B with respect tic
AE 2201 mechanics of
7. From point z, draw vector zebu' perpendicular to vector co’s or CB to represent the tangential
component of the acceleration of B with respect to C. The vectors by' and zebu' intersect at b'
aBC Join c' b'. The vector c' b' represents the acceleration of B with respect to C i.e.BC.
8. Now from point b', draw vector by’s parallel to BD to represent the radial component of the
acceleration of D with respect tub
9. From point s, draw vector sad' perpendicular to vector by’s or BD to represent the tangential
component of the acceleration of D with respect tub.
10. From point c', draw vector cod' parallel to the path of motion of D (which is along
CD)to represent the acceleration of D i.e. aD. The vectors sad' and cod' intersect tad'.
By measurement, we find that acceleration of slider D,
aD = vector cod' = 13.3 m/s2
Angular acceleration of BD
By measurement, we find that tangential component of the acceleration of D with respect to B,
RESULTS:
aD=13.3 m/s2
ὼBD=4.5 rad/s
Example 3.
Fig. 6 shows the mechanism of a radial valve gear. The crank OA turns uniformly at 150 r.p.m
and is pinned at A to rod AB. The point C in the rod is guided in the circular path with D as
centre and DC as radius. The dimensions of various links are:
OA = 150 mm ; AB = 550 mm ; AC = 450 mm ; DC = 500 mm ; BE = 350 mm.
Determine velocity and acceleration of the ram E for the given position of the mechanism.
AE 2201 mechanics of
GIVEN :
NAO = 150 r.p.m.
ὼAO =0.15mm
A B = 550 mm = 0.55 m
A C = 450 mm = 0.45 m
DC = 500 mm = 0.5 m
BE = 350 mm = 0.35 m
SOLUTION:
We know that linear velocity of A with respect to O or velocity of A ,
vAO = vA = ὼAO × O A = 15.71 × 0.15 = 2.36 m/s
Velocity of the ram E
First of all draw the space diagram, as shown in Fig. 7 (a), to some suitable scale. Now the
velocity diagram, as shown in Fig. 7(b), is drawn as discussed below:
1. Since O and D are fixed points, therefore these points are marked as one point in the velocity
diagram. Draw vector oaperpendicularto O A, to some suitable scale, to represent the velocity
of A with respect to O or simply velocity of A , suchthat
vector oa vAOvA 2.36 m/s
2. From point a, draw vector ac perpendicular to A C to represent the velocity of C with respect
AE 2201 mechanics of
to A (i.e. vCA), and from point d draw vector dc perpendicular to DC to represent the velocityof
With respect toDorsimplyvelocityofC(i.e.vCDorC).The vectors ac and dc intersect at c.
3. Since the point B lies on A C produced, therefore divide vector ac at b in the same ratio abs
divides A C in the space diagram. In other words ac:cb = AC:CB. Join ob. The vector ob
represents the velocity of B (i.e.vB)
4. From point b, draw vector be perpendicular to be to represent the velocity of E with respect
to B (i.e. vEB), and from point o draw vector one parallel to the path of motion of the ram E
(which is horizontal) to represent the velocity of the ram E. The vectors be and one intersect ate.
By measurement, we find that velocity of C with respect to A,
vCA = vector ac = 0.53 m/s
Velocity of C with respect to D,
Velocity of E with respect to B,
vCD = vC= vector dc = 1.7 m/s
vEB = vector be = 1.93 m/s
and velocity of thermal, vE = vector one = 1.05m/s
Acceleration of the ram E
We know that the radial component of the acceleration of A with respect to O or the acceleration
of A ,
AE 2201 mechanics of
The acceleration diagram, as shown in Fig. 7 (c), is drawn as discussed below
1. Since O and D are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector ova' parallel to O A, to some suitable scale, to represent the
radial component of the acceleration of A with respect to O or simply the acceleration of A,
such that
2. From point d', draw vector do’s parallel to DC to represent the radial component of
the acceleration of C with respect to D, suchthat
3. From point x, draw vector act' perpendicular to DC to represent the tangential component of
the acceleration of C with respect told
4. Now from point a', draw vector a'y parallel to A C to represent the radial component of
the acceleration of C with respect to A , suchthat
5. From point y, draw vector yuck' perpendicular to AC to represent the tangential component of
acceleration of C with respect toe
6. Join arc'. The vector arc' represents the acceleration of C with respect toe
7. Since the point Bison A C produced, therefore divide vector arc' at b' in the same ratios
AE 2201 mechanics of
B divides A C in the space diagram. In other words, a'
c' : c' b' = A C : CB
8. From point b', draw vector b' z parallel to BE to represent the radial component of the
acceleration of E with respect to B, suchthat
9. From point z, draw vector zed' perpendicular to BE to represent the tangential component of
the acceleration of E with respect tub.
10. From point o', draw vector one' parallel to the path of motion of E (which is
horizontal)to represent the acceleration of the ram E. The vectors zed' and one' intersect ate'.
By measurement, we find that the acceleration of the ram E,
RESULTS:
ae=3.1 m/s2
VE=1.05m/s
UNIT-2
FRICTION
AE 2201 mechanics of
When a solid body slides over a stationary solid body, a force is exerted at the surface of contact
by the stationary body on the moving body. This force is called the force of friction and is
always acting in the direction opposite to the motion. The property of the bodies by virtue of
which a force is exerted by a stationary body on the moving body to resist the motion of the
moving body is called friction. Friction acts parallel to the surface of contact and depends of
contact and depends upon the nature of surface of contact.
1. Static friction. It is the friction, experienced by a body, when arrest.
2. Dynamic friction. It is the friction, experienced by a body, when in motion. The dynamic
friction is also called kinetic friction and is less than the static friction. It is of the following three
types:
(a)Sliding friction. It is the friction, experienced by a body, when it slides over another body.
(b)Rolling friction. It is the friction, experienced between the surfaces which has balls or rollers
interposed between them.
(c)Pivot friction. It is the friction, experienced by a body, due to the motion of rotation as in case
of foot step bearings.
The friction may further be classified as :
1. Friction between unlubricated surfaces, and
2. Friction between lubricated surfaces.
These are discussed in the
followingarticles.Friction Between
UnlubricatedSurfaces
The friction experienced between two dry and unlubricated surfaces in contact is known as dry or
solid friction. It is due to the surface roughness. The dry or solid friction includes the sliding
friction and rolling friction as discussed above.
Friction Between Lubricated Surfaces
When lubricant (i.e. oil or grease) is applied between two surfaces in contact, then the friction
may be classified into the following two types depending upon the thickness of layer of a
lubricant.
1. Boundary friction (or greasy friction or non-viscous friction). It is the friction, experienced
between the rubbing surfaces, when the surfaces have a very thin layer of lubricant. The
thicknessofthisverythinlayerisofthemoleculardimension.In thistypeoffriction,athinlayer of
lubricant forms a bond between the two rubbing surfaces. The lubricant is absorbed on the
surfaces and forms a thin film. This thin film of the lubricant results in less friction between
them. The boundary friction follows the laws of solidfriction.
2. Fluid friction (or film friction or viscous friction). It is the friction, experienced between the
rubbing surfaces, when the surfaces have a thick layer of the lubricant. In this case, the actual
surfaces do not come in contact and thus do not rub against each other. It is thus obvious that
fluid friction is not due to the surfaces in contact but it is due to the viscosity and oiliness of the
lubricant.
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Note : The viscosity is a measure of the resistance offered to the sliding one layer of the lubricant
over an adjacent layer. The absolute viscosity of a lubricant may be defined as the force required
to cause a plate of unit area to slide with unit velocity relative to a parallel plate, when the two
plates are separated by a layer of lubricant of unit thickness.
The oiliness property of a lubricant may be clearly understood by considering two lubricants of
equal viscosities and at equal temperatures. When these lubricants are smeared on two different
surfaces, it is found that the force of friction with one lubricant is different than that of the other.
This difference is due to the property of the lubricant known as oiliness. The lubricant which
gives lower force of friction is said to have greater oiliness.
Limiting Friction
Consider that a body A of weight W is lying on a rough horizontal body B as shown in Fig. 1
(a). In this position, the body A is in equilibrium under the action of its own weight W , and the
normal reaction RN (equal to W ) of B on A . Now if a small horizontal force P1 is applied to the
body A acting through its centre of gravity as shown in Fig. 1 (b), it does not move because of
the frictional force which prevents the motion. This shows that the applied force P1 is exactly
balanced by the force of friction F1 acting in the opposite direction.
If we now increase the applied force to P2 as shown in Fig. 1 (c), it is still found to be in
equilibrium. This means that the force of friction has also increased to a value F2 = P2. Thus
every time the effort is increased the force of friction also increases, so as to become exactly
equal to the applied force. There is, however, a limit beyond which the force of friction cannot
increase as shown in Fig..1 (d). After this, any increase in the applied effort will not lead to any
further increase in the force of friction, as shown in Fig. .1 (e), thus the body A begins to move in
the direction of the applied force. This maximum value of frictional force, which comes into
play, when a body just begins to slide over the surface of the other body, is known as limiting
force of friction or simply limiting friction. It may be noted that when the applied force is less
than the limiting friction, the body remains at rest, and the friction into play is called static
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friction which may have any value between zero and limiting friction.
Laws of Static Friction
Following are the laws of static friction :
1. The force of friction always acts in a direction, opposite to that in which the body tendsto
move.
2. The magnitude of the force of friction is exactly equal to the force, which tends the bodyto
move.
3. The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction(RN)
between the two surfaces. Mathematically
F/RN= constant
4. The force of friction is independent of the area of contact, between the two surfaces.
5. The force of friction depends upon the roughness of the
surfaces. Laws of Kinetic or Dynamic Friction
Following are the laws of kinetic or dynamic friction :
1.The force of friction always acts in a direction, opposite to that in which the body is moving.
2.The magnitude of the kinetic friction bears a constant ratio to the normal reaction between the
two surfaces. But this ratio is slightly less than that in case of limiting friction.
3.For moderate speeds, the force of friction remains constant. But it decreases slightly with the
increase of speed.
Laws of Solid Friction
Following are the laws of solid friction :
1. The force of friction is directly proportional to the normal load between the surfaces.
2. The force of friction is independent of the area of the contact surface for a given normal load.
3. The force of friction depends upon the material of which the contact surfaces are made.
4.The force of friction is independent of the velocity of sliding of one body relative to the
other body.
Laws of Fluid Friction
Following are the laws of fluid friction :
1. The force of friction is almost independent of the load.
2. The force of friction reduces with the increase of the temperature of the lubricant.
3. The force of friction is independent of the substances of the bearing surfaces.
4. The force of friction is different for different
lubricants. Coefficient of Friction
It is defined as the ratio of the limiting friction (F) to the normal reaction (RN) between the two
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bodies. It is generally denoted by µ. Mathematically, coefficient of friction,
µ= F/RN
LIMITING ANGLE OF FRICTION
Consider that a body A of weight (W ) is resting on a horizontal plane B, as shown in Fig. 2.If a
horizontal force P is applied to the body, no relative motion will take place until the applied force
P is equal to the force of friction F, acting opposite to the direction of motion.
The magnitude of this force of friction is F = µ.W =µ.RN, where RN is the normal reaction.
In the limiting case, when the motion just begins, the body will be in equilibrium under the
action of the following three forces :
1. Weight of the body (W),
2. Applied horizontal force (P),and
3. Reaction (R) between the body A and the plane.
The reaction R must, therefore, be equal and opposite to the resultant of W and P and will be
inclined at an angle Φ to the normal reaction RN. This angle Φ is known as the limiting angle of
friction.
It may be defined as the angle which the resultant reaction R makes with the normal reaction RN.
From Fig. 2, tan Φ = F/RN =µ. RN/ RN =µ
ANGLE OF REPOSE
Consider that a body A of weight (W ) is resting on an inclined plane B, as shown in Fig. 3. If the
angle of inclination α of the plane to the horizontal is such that the body begins to move down
the plane, then the angle α is called the angle of repose.
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A little consideration will show that the body will begin to move down the plane when the angle
of inclination of the plane is equal to the angle of friction (i.e. = Φ. This may be proved as
follows :
The weight of the body (W ) can be resolved into the following two components :
1. W sin α, parallel to the plane B. This component tends to slide the body down the plane.
2. W coos α, perpendicular to the plane B. This component is balanced by the normal
reaction (RN) of the body And the plane B. The body will only begin to move down the
plane, when
PROBLEMS:
Example 10.1. A body, resting on a rough horizontal plane required a pull of 180 N inclined at
30º to the plane just to move it. It was found that a push of 220 N inclined at 30º to the plane just
moved the body. Determine the weight of the body and the coefficient of friction.
Given:
ϴ=30°
Ppull= 180 N
Ppush=220 N
SOLUTION:
W = Weight of the body in nektons,
RN= Normal reaction,
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µ = Coefficient of friction, and
F= Force of friction
First of all, let us consider a pull of 180 N. The force of friction (F) acts towards left as shown in
Fig
Resolving the forces horizontally
F=P coos ϴ
F= 180 coos 30º = 180 × 0.866 = 156 N
Now resolving the forces vertically,
RN= W – 180 sin 30º = W – 180 × 0.5 = (W – 90) N
We know that F =µ.RN or 156 =µ (W–90) ........................ (I)
Now let us consider a push of 220 N. The force of friction (F) acts towards right as shown in
Fig.
Resolving the forces horizontally,
F= 220 coos 30º = 220 × 0.866 = 190.5 N
Now resolving the forces vertically,
RN= W + 220 sin 30º = W + 220 × 0.5 = (W + 110) N
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We know that F =µ.RN or 190.5 = µ(W+110) ........................ (ii)
From equation (I) and (ii)
W=1000 N and µ=0.1714
RESULTS:
W=1000 N
µ=0.1714
SCREW FRICTION
The screws, bolts, studs, nuts etc. are widely used in various machines and structures for
temporary fastenings. These fastenings have screw threads, which are made by cutting a
continuous helical groove on a cylindrical surface. If the threads are cut on the outer surface of a
solid rod, these are known as external threads. But if the threads are cut on the internal surface
of a hollow rod, these are known as internal threads. The screw threads are mainly of two types
i.e. V-threads and square threads. The V-threads are stronger and offer more frictional
resistance to motion than square threads. Moreover, the V-threads have an advantage of
preventing the nut from slackening. In general, the-
threads are used for the purpose of tightening pieces together e.g. bolts and nuts etc. But the
square threads are used in screw jacks, vice screws etc. The following terms are important for the
study of screw :
1. Helix. It is the curve traced by a particle, while describing a circular path at uniform
speed and advancing in the axial direction at a uniform rate. In other words, it is the curve traced
by a particle while moving along a screw thread
2. Pitch. It is the distance from a point of a screw to a corresponding point on the next
thread, measured parallel to the axis of the screw.
3. Lead. It is the distance, a screw thread advances axially in one turn.
4. Depth of thread. It is the distance between the top and bottom surfaces of a thread(also
known as crest and root of thread).
5. Single-threaded screw. If the lead of a screw is equal to its pitch, it is known as single
threaded screw.
6. Multi-threaded screw. If more than one thread is cut in one lead distance of a screw, its
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known as multi-threaded screw e.g. in a double threaded screw, two threads are cut in one lead
length. In such cases, all the threads run independently along the length of the rod.
Mathematically,
Lead = Pitch × Number of threads
7. Helix angle. It is the slope or inclination of the thread with the horizontal. Mathematically,
SCREW JACK
The screw jack is a device, for lifting heavy loads, by applying a comparatively smaller effort at
its handle. The principle, on which a screw jack works is similar to that of an inclined plane
shows a common form of a screw jack, which consists of a square threaded rod (also called
screw rod or simply screw) which fits into the inner threads of the nut. The load, to be raised or
lowered, is placed on the head of the square threaded rod which is rotated by the application of
an effort at the end of the lever for lifting or lowering the load.
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Torque Required Lifting the Load by a Screw Jack
If one complete turn of a screw thread by imagined to be unwound, from the body of the
screw and developed, it will form an inclined plane as shown in Fig
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Since the principle on which a screw jack works is similar to that of an inclined plane, there-
for the force applied on the lever of a screw jack may be considered to be horizontal as shown in
Fig
Since the load is being lifted, therefore the force of friction (F =µRN) will act downwards.
All the forces acting on the screw are shown in Fig
Resolving the forces along the plane,
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Example 1 An electric motor driven power screw moves a nut in a horizontal plane against a
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force of 75 ken at a speed of 300 mm/min. The screw has a single square thread of 6 mm pitch
on a major diameter of 40 mm. The coefficient of friction at the screw threads is 0.1. Estimate
power of the motor.
*The nominal diameter of a screw thread is also known as outside diameter or major
diameter.
**The core diameter of a screw thread is also known as inner diameter or root diameter or
minor diameter.
Given:
W = 75 ken = 75 × 103 N
v = 300 mm/min
p = 6 mm
d0 = 40 mm
solution
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Result:
Power of the motor = 1.108 kW
Example 2. A 150 mm diameter valve, against which a steam pressure of 2 MN/m2 is acting,
is closed by means of a square threaded screw 50 mm in external diameter with 6 mm pitch. If
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the coefficient of friction is 0.12 ; find the torque required to turn the handle.
Given :
D = 150 mm = 0.15 mm = 0.15 m ;
Ps = 2 MN/m2 = 2 × 106 N/m2;
D0 = 50 mm
; p = 6 mm
solution:
Result:
T=134.2 N-m
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TORQUE REQUIRED LOWERING THE LOAD BY A SCREW JACK :
We have discussed in Art. 10.18, that the principle on which the screw jack works is similar to
that of an inclined plane. If one complete turn of a screw thread be imagined to be unwound from
the body of the screw and developed, it will form an inclined plane as shown in Fig
p = Pitch of the screw,
d = Mean diameter of the screw,
α = Helix angle,
P = Effort applied at the circumference of the screw to lower the
load,
W = Weight to be lowered, and
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MAXIMUM EFFICIENCY OF A SCREW JACK:
PROBLEMS:
Example 1. The pitch of 50 mm mean diameter threaded screw of a screw jack is 12.5 mm. The
coefficient of friction between the screw and the nut is 0.13. Determine the torque required on
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the screw to raise a load of 25 ken, assuming the load to rotate with the screw. Determine the
ratio of the torque required to raise the load to the torque required to lower the load and also
the efficiency of the machine.
Given :
d = 50 mm ;
p = 12.5mm
W = 25 ken = 25 × 103 N
SOLUTION:
Torque required on the screw
We know that the torque required on the screw to raise the load,
T1 = P × d/2 = 5305 × 50/2 = 132 625 N-mm
Ratio of the torques required to raise and lower the load
We know that the force required on the screw to lower the load,
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Efficiency of the machine
We know that the efficiency
RESULTS:
T1=132 625 N-mm
Efficiency =0.3377=37.7%
Example 2. The mean diameter of a Whitworth bolt having V-threads is 25 mm. The pitch of the
thread is 5 mm and the angle of V is 55º. The bolt is tightened by screwing a nut whose mean
radius of the bearing surface is 25 mm. If the coefficient of friction for nut and bolt is 0.1 and for
nut and bearing surfaces 0.16 ; find the force required at the end of a spanner 0.5 m long when
the load on the bolt is 10 kind
Given:
d = 25 mm
p = 5 mm
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W = 10 ken = 10 × 103 N
SOLUTION:
RESULT:
P1=124.6 N
FRICTION OF PIVOT AND COLLAR BEARING
The rotating shafts are frequently subjected to axial thrust. The bearing surfaces such as pivot and
collar bearings are used to take this axial thrust of the rotating shaft. The propeller shafts of
ships, the shafts of steam turbines, and vertical machine shafts are examples of shafts which carry
an axial thrust. The bearing surfaces placed at the end of a shaft to take the axial thrust are known
as pivots. The pivot may have a flat surface or conical surface as shown in Fig. 4 (a) and (b)
respectively. When the cone is truncated, it is then known as truncated or trapezoidal pivot as
shown in Fig. 4 (c). The collar may have flat bearing surface or conical bearing surface, but the
flat surface is most commonly used. There may be a single collar, as shown in Fig. 4 (d) or
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several collars along the length of a shaft, as shown in Fig. 4 (e) in order to reduce the intensity
of pressure.
In modern practice, ball and roller thrust bearings are used when power is being transmitted and
when thrusts are large as in case of propeller shafts of ships.
A little consideration will show that in a new bearing, the contact between the shaft and bearing
may be good over the whole surface. In other words, we can say that the pressure over the
rubbing surfaces is uniformly distributed.
But when the bearing becomes old, all parts of the rubbing surface will not move with the same
velocity, because the velocity of rubbing surface increases with the distance from the axis of the
bearing. This means that wear may be different at different radii and this causes to alter the
distribution of pressure. Hence, in the study of friction of bearings, it is assumed that
1. The pressure is uniformly distributed throughout the bearing surface, and
2. The wear is uniform throughout the bearing surface
FLAT PIVOT BEARING:
When a vertical shaft rotates in a flat pivot bearing (known as foot step bearing), as shown in
Fig. 5 the sliding friction will be along the surface of contact between the shaft and the bearing.
W = Load transmitted over the bearing surface,
R = Radius of bearing surface,
p = Intensity of pressure per unit area of bearing surface between rubbing surfaces, and
µ= Coefficient of friction.
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We will consider the following two cases :
1. When there is a uniform pressure ;and
2. When there is a uniform wear.
1. Considering uniform pressure
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2. Considering uniform wear
We have already discussed that the rate of wear depends upon the intensity of pressure (p) and
the velocity of rubbing surfaces (v). It is assumed that the rate of wear is proportional to the
product of intensity of pressure and the velocity of rubbing surfaces (i.e.p.v..). Since the velocity
of rubbing surfaces increases with the distance (i.e. radius r) from the axis of the bearing,
therefore for uniform wear
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CONICAL PIVOT BEARING:
6
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Consider a small ring of radius r and thickness dr. Let dl is the length of ring along the cone,
such that
1. CONSIDERING UNIFORM PRESSURE:
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2. CONSIDERING UNIFORMWEAR:
let prbe the normal intensity of pressure at a distance r from the central axis. We know that, in
case of uniform wear, the intensity of pressure varies inversely with the distance.
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Example 1. A conical pivot supports a load of 20 ken, the cone angle is 120º and the intensity
of normal pressure is not to exceed 0.3 N/mm2. The external diameter is twice the internal
diameter. Find the outer and inner radii of the bearing surface. If the shaft rotates at 200 r.p.m.
and the coefficient of friction is 0.1, find the power absorbed in friction. Assume uniform
pressure.
GIVEN :
W = 20 ken = 20 ×
103 N = 200 r.p.m
SOLUTION
OUTER AND INNER RADII OF THE BEARING SURFACE:
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POWER ABSORBED IN FRICTION:
RESULTS
T=301.76 N-m
P = 6.22kW
FRICTION CLUTCHES:
A friction clutch has its principal application in the transmission of power of shafts
and machines which must be started and stopped frequently. Its application is also found in
cases in which power is to be delivered to machines partially or fully loaded. The force of
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friction is used to start the driven shaft from rest and gradually brings it up to the proper speed
without excessive slipping of the friction surfaces. In automobiles, friction clutch is used to
connect the engine to the driven shaft. In operating such a clutch, care should be taken so that
the friction surfaces engage easily and gradually brings the driven shaft up to proper speed. The
proper alignment of the bearing must be maintained
and it should be located as close to the clutch as possible. It may be noted that
1. The contact surfaces should develop a frictional force that may pick up and hold the
load with reasonably low pressure between the contact surfaces.
2. The heat of friction should be rapidly dissipated and tendency to grab should be ata
minimum.
3. The surfaces should be backed by a material stiff enough to ensure a reasonably
uniform distribution of pressure.
The friction clutches of the following types are important from the subject point of view :
1. Disc or plate clutches (single disc or multiple disc clutch),
2. Cone clutches, and
3. Centrifugal clutches.
We shall now discuss, these clutches, in detail, in the following pages. It may be noted that
the disc and cone clutches are based on the same theory as the pivot and collar bearings.
SINGLE DISC OR PLATE CLUTCH:
A single disc or plate clutch, as shown in Fig.1, consists of a clutch plate whose both sides are
faced with a friction material (usually of Ferro do). It is mounted on the hub which is free to
move axially along the spines of the driven shaft. The pressure plate is mounted inside the clutch
body which is bolted to the flywheel. Both the pressure plate and the flywheel rotate with the
engine crankshaft or the driving shaft. The pressure plate pushes the clutch plate towards the
flywheel by a set of strong springs which are arranged radically inside the body. The three levers
(also known as release levers or fingers) are carried on pivots suspended from the case of the
body. These are arranged in such a manner so that the pressure plate moves away from the
flywheel by the inward movement of a thrust bearing. The bearing is mounted upon a forked
shaft and moves forward when the clutch pedal is pressed. When the clutch pedal is pressed
down, its linkage forces the thrust release bearing to move in towards the flywheel and pressing
the longer ends of the levers inward. The levers are forced to turn on their suspended pivot and
the pressure plate moves away from the flywheel by the knife edges, thereby compressing the
clutch springs.
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This action removes the pressure from the clutch plate and thus moves back from the flywheel
and the driven shaft becomes stationary. On the other hand, when the foot is taken off from the
clutch pedal, the thrust bearing moves back by the levers. This allows the springs to extend and
thus the pressure plate pushes the clutch plate back towards the flywheel. The axial pressure
exerted by the spring provides a frictional force in the circumferential
direction when the relative motion between the driving and driven members tends to take place.
If the torque due to this frictional force exceeds the torque to be transmitted, then no slipping
takes place and the power is transmitted from the driving shaft to the driven shaft.
MULTIPLE DISC CLUTCH
A multiple disc clutch, as shown in Fig. 2, may be used when a large torque is to be
transmitted. The inside discs (usually of steel) are fastened to the driven shaft to permit axial
motion (except for the last disc).
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The outside discs (usually of bronze) are held by bolts and are fastened to the housing which is
keyed to the driving shaft. The multiple disc clutches are extensively used in motor cars,
machine tools etc. n1= Number of discs on the driving shaft, and
n2= Number of discs on the drive shaft.
Example 1Determine the maximum, minimum and average pressure in plate clutch
when the axial force is 4 kind The inside radius of the contact surface is 50 mm and the
outside radius is 100 mm. Assume uniform wear
GIVEN :
W = 4 ken = 4 × 103 N ;
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r2 = 50 mm;
r1 = 100mm
SOLUTION: Maximum pressure
Minimum pressure
Aérage pressure
RESULTS:
Pmin=0.1273 N/mm2
Pad=0.17 N/mm2
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Example 2. A rotor is driven by a co-axial motor through a single plate clutch, both sides of the
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plate being effective. The external and internal diameters of the plate are respectively 220 mm
and 160 mm and the total spring load pressing the plates together is 570 N. The motor armature
and shaft has a mass of 800 kg with an effective radius of gyration of 200 mm. The rotor has a
mass of 1300 kg with an effective radius of gyration of 180 mm. The coefficient of friction for the
clutch is 0.35. The driving motor is brought up to a speed of 1250 r.p.m. when the current is
switched off and the clutch suddenly engaged. Determine
1. The final speed of motor and rotor, 2. The time to reach this speed, and 3. The kinetic energy
lost during the period of slipping.
How long would slipping continue if it is assumed that a constant resisting torque of 60 N-m
were present? If instead of a resisting torque, it is assumed that a constant driving torque of 60
N-m is maintained on the armature shaft, what would then be slipping time?
GIVEN :
d1 = 220 mm or r1 = 110 mm
d2 = 160 mm or r2 = 80 mm
W = 570 N
m1 = 800 kg
k1 = 200 mm = 0.2 m
m2 = 1300 kg
k2 = 180 mm = 0.18 m
SOLUTION:
1. FINAL SPEED OF THE MOTOR ANDROTOR
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2. TIME TO REACH THISSPEED
3. KINETIC ENERGY LOST DURING THE PERIOD OFSLIPPING
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TIME OF SLIPPING ASSUMING CONSTANT RESISTING TORQUE
TIME OF SLIPPING ASSUMING CONSTANT DRIVING TORQUE OF 60 N-M
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BELT, ROPE AND CHAIN DRIVES:
BELT:
INTRODUCTION
The belts or ropes are used to transmit power from one shaft to another by means of pulleys
which rotate at the same speed or at different speeds. The amount of power transmitted depends
upon the following factors :
1. The velocity of the belt.
2. The tension under which the belt is placed on the pulleys.
3. The arc of contact between the belt and the smaller pulley.
4. The conditions under which the belt is used. It may be noted that
(a)The shafts should be properly in line to insure uniform tension across the belt section.
(b)The pulleys should not be too close together, in order that the arc of contact on the smaller
pulley may be as large as possible.
(c) The pulleys should not be so far apart as to cause the belt to weigh heavily on the shafts,
thus increasing the friction load on the bearings
(d) A long belt tends to swing from side to side, causing the belt to run out of the pulleys,
which in turn develops crooked spots in the belt.
(e) The tight side of the belt should be at the bottom, so that whatever sag is present on the
loose side will increase the arc of contact at the pulleys.
( f ) In order to obtain good results with flat belts, the maximum distance between the shafts
should not exceed 10 meters and the minimum should not be less than 3.5 times the diameter of
the larger pulley.
. Selection of a Belt Drive
Following are the various important factors upon which the selection of a belt drive depends:
1. Speed of the driving anddrivenshafts, 2.Speed reduction ratio,
3. Power tobetransmitted, 4.Centre distance between the shafts,
5. Positivedriverequirements, 6. Shafts layout,
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7. Spaceavailable,and 8. Service conditions.
Types of Belt Drives
The belt drives are usually classified into the following three groups :
1. Light drives. These are used to transmit small powers at belt speeds up to about 10 m/s, as
in agricultural machines and small machine tools.
2. Medium drives. These are used to transmit medium power at belt speeds over 10 m/s but up
to 22 m/s, as in machine tools.
3. Heavy drives. These are used to transmit large powers at belt speeds above 22 m/s,
assign compressors and generators.
TYPES OF BELTS
Though there are many types of belts used these days, yet the following are important from the
subject point of view :
1. Flat belt. The flat belt, as shown in Fig. 1(a), is mostly used in the factories and workshops,
where a moderate amount of power is to be transmitted, from one pulley to another when the
two pulleys are not more than 8 metresapart.
2. V-belt. The V-belt, as shown in Fig. 1(b), is mostly used in the factories and workshops,
where a moderate amount of power is to be transmitted, from one pulley to another, when
the two pulleys are very near to each other.
3. Circular belt or rope. The circular belt or rope, as shown in Fig. 1 (c), is mostly used in the
factories and workshops, where a great amount of power is to be transmitted, from one pulley
to another, when the two pulleys are more than 8 meters apart. If a huge amount of power is to
be transmitted, then a single belt may not be sufficient. In
such a case, wide pulleys (for V-belts or circular belts) with a number of grooves are used. Then
a belt in each groove is provided to transmit the required amount of power from one pulley to
another.
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V-BELT DRIVE
We have already discussed that a V-belt is mostly used in factories and workshops where a great
amount of power is to be transmitted from one pulley to another when the two pulleys are very
near to each other.
The V-belts are made of fabric and cords molded in rubber and covered with fabric and rubber,
as shown in Fig. (a). These belts are molded to a trapezoidal shape and are made endless. These
are particularly suitable for short drives i.e. when the shafts are at a short distance apart. The
included angle for the V-belt is usually from 30° – 40°. In case of flat belt drive, the belt runs
over the pulleys whereas in case of V-belt drive, the rim of the pulley is grooved in which the
V-belt runs. The effect of the groove is to increase the frictional grip of the V-belt on the pulley
and thus to reduce the tendency of slipping. In order to have a good grip on the pulley, the V-belt
is in contact with the side faces of the groove and not at the bottom. The power is transmitted by
the *wedging action between the belt and the V-groove in the pulley
A clearance must be provided at the bottom of the groove, as shown in Fig. (b), in order
to prevent touching to the bottom as it becomes narrower from wear. The V-belt drive, may be
inclined at any angle with tight side either at top or bottom. In order to increase the power output,
several V- belts may be operated side by side. It may be noted that in multiple V-belt drive, all
the belts should stretch at the same rate so that the load is equally divided between them. When
one of the set of belts break, the entire set should be replaced at the same time. If only one belt is
replaced, the new unworn and unstressed belt will be more tightly stretched and will move with
different velocity
ADVANTAGES AND DISADVANTAGES OF V-BELT DRIVE OVER FLAT BELT
DRIVE
Following are the advantages and disadvantages of the V-belt drive over flat belt drive.
Advantages
1. The V-belt drive gives compactness due to the small distance between the centers of
pulleys. 2.The drive is positive, because the slip between the belt and the pulley groove is
negligible. 3.Since the V-belts are made endless and there is no joint trouble, therefore the
drive is smooth.
4. It provides longer life, 3 to 5 years.
5. It can be easily installed and removed.
6. The operation of the belt and pulley disquiet.
7. The belts have the ability to cushion the shock when machines are started.
8. The high velocity ratio (maximum 10) may be obtained.
9. The wedging action of the belt in the groove gives high value of limiting ratio of tensions.
Therefore the power transmitted by V-belts is more than flat belts for the same coefficient
of friction, arc of contact and allowable tension in the belts.
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10. The V-belt may be operated in either direction with tight side of the belt at the toper
bottom. The centre line may be horizontal, vertical or inclined.
Disadvantages
1.The V-belt drive cannot be used with large centre distances.
2.The V-belts are not as durable as flat belts.
3.The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
4. Since the V-belts are subjected to certain amount of creep, therefore these are not
suitable for constant speed application such as synchronous machines, and timing devices.
5.The belt life is greatly influenced with temperature changes, improper belt tension
and mismatching of belt lengths.
6.The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above50m/s
ROPE DRIVE:
The rope drives are widely used where a large amount of power is to be transmitted, from one
pulley to another, over a considerable distance. It may be noted that the use of flat belts is limited
for the transmission of moderate power from one pulley to another when the two pulleys are not
more than 8 meters apart. If large amounts of power are to be transmitted by the flat belt, then it
would result in excessive belt cross-section. It may be noted that frictional grip in case of rope
drives is more than that in V-drive. One of the main advantage of rope drives is that a number of
separate drives may be taken from the one driving pulley. For example, in many spinning mills,
the line shaft on each floor is driven by ropes passing directly from the main engine pulley on the
ground floor.
The rope drives use the following two types of ropes :
1. Fiber ropes, and 2. Wire ropes. The fiber ropes operate successfully when the pulleys are
about 60 meters apart, while the wire ropes are used when the pulleys are up to 150 meters
apart. FIBRE ROPES
The ropes for transmitting power are usually made from fibrous materials such as hemp,
manila and cotton. Since the hemp and manila fibers are rough, therefore the ropes made from
these fibers are not very flexible and possesses poor mechanical properties. The hemp ropes have
less strength as compared to manila ropes. When the hemp and manila ropes are bent over the
sheave (or pulley), there is some sliding of fibers, causing the rope to wear and chafe internally.
In order to minimize this defect, the rope fibers are lubricated with a tar, tallow or graphite. The
lubrication also makes the rope moisture proof. The hemp ropes are suitable only for hand
operated hoisting machinery and as the ropes for lifting tackle, hooks etc.
The cotton ropes are very soft and smooth. The lubrication of cotton ropes is not necessary.
But if it is done, it reduces the external wear between the rope and the grooves of its sheaves.
It may be noted that manila ropes are more durable and stronger than cotton ropes. The cotton
ropes are costlier than manila ropes.
Note :The diameter of manila and cotton ropes usually ranges from 38 mm to 50 mm. The size of
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the rope is usually designated by its circumference or ‘girth’.
ADVANTAGES OF FIBRE ROPE DRIVES
The fiber rope drives have the following advantages :
1. They give smooth, steady and quiet service.
2. They are little affected by outdoor conditions.
3. The shafts may be out of strict alignment.
4. The power may be taken off in any direction and in fractional parts of the whole amount.
5. They give high mechanical efficiency.
TYPES OF FLAT BELT DRIVES
The power from one pulley to another may be transmitted by any of the following types of belt
drives:
1. Open belt drive. The open belt drive, as shown in Fig. 2, is used with shafts arranged
parallel and rotating in the same direction. In this case, the driver A pulls the belt from one side
(i.e. lower side RQ) and delivers it to the other side (i.e. upper side L M). Thus the tension in
the
lower side belt will be more than that in the upper side belt. The lower side belt (because of more
tension) is known as tight side whereas the upper side belt (because of less tension) is known as
slack side.
2. Crossed or twist belt drive. The crossed or twist belt drive, as shown in Fig. 3, is used with
shafts arranged parallel and rotating in the opposite directions. In this case, the driver pulls the
belt from one side (i.e. RQ) and delivers it to the other side (i.e. L M). Thus the tension in the
belt RQ will be more than that in the belt L M. The belt RQ (because of more tension) is known
as tight side, whereas the belt L M (because of less tension) is known as slack side, as shown
infix. 3
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RATIO OF DRIVING TENSIONS FOR FLAT BELT DRIVE
Consider a driven pulley rotating in the clockwise direction
T1= Tension in the belt on the tight side,
T2= Tension in the belt on the slack side, and
= Angle of contact in radians (i.e. angle subtended by the arc A B, along which the belt touches
the pulley at the centre).
Now consider a small portion of the belt PQ, subtending an angle ϴat the centre of the pulley
as shown in Fig.. The belt PQ is in equilibrium under the following forces :
1. Tension T in the belt apt,
2. Tension (T + T) in the belt at,
3. Normal reaction RN,and
4. Frictional force, F = µ × RN , where µ is the coefficient of
friction between the belt and pulley.
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ANGLE OF CONTACT
When the two pulleys of different diameters are connected by means of an open belt then the
angle of contact or lap (ϴ) at the smaller pulley must be taken into consideration.
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A little consideration will show that when the two pulleys are connected by means of a crossed
Belt then the angle of contact or lap (ϴ) on both the pulleys is same
CENTRIFUGAL TENSION
Since the belt continuously runs over the pulleys, therefore, some centrifugal force is caused,
whose effect is to increase the tension on both, tight as well as the slack sides. The tension
caused by centrifugal force is called centrifugal tension. At lower belt speeds (less than 10
m/s), the centrifugal tension is very small, but at higher belt speeds (more than 10 m/s), its
effect is considerable and thus should be taken into account.
Consider a small portion PQ of the belt subtending an angle d the centre of the pulley
m = Mass of the belt per unit length in kg,
v = Linear velocity of the belt in m/s,
r = Radius of the pulley over which the belt runs in meters, and
TC= Centrifugal tension acting tangentially at P and Q in nektons
We know that length of the belt PQ
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INITIAL TENSION IN THE BELT
When a belt is wound round the two pulleys (i.e. driver and follower), its two ends are joined
together ; so that the belt may continuously move over the pulleys, since the motion of the belt
from the driver and the follower is governed by a firm grip, due to friction between the belt and
the pulleys. In order to increase this grip, the belt is tightened up. At this stage, even when the
pulleys are stationary, the belt is subjected to some tension, called initial tension. When the
driver starts rotating, it pulls the belt from one side (increasing tension in the belt on this side)
and delivers it to the other side (decreasing the tension in the belt on that side). The increased
tension in one side of the belt is called tension in tight side and the decreased tension in the other
side of the belt is called tension in the slack side.
T0= Initial tension in the belt,
T1= Tension in the tight side of the belt,
T2= Tension in the slack side of the belt, and
α=Coefficient of increase of the belt length per unit force
A little consideration will show that the increase of tension in the tight side
= T1 – T0
and increase in the length of the belt on the tight side
= (T1 – T0)
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Similarly, decrease in tension in the slack side
= T0 – T2
and decrease in the length of the belt on the slack side
= α T0 –T2)
Assuming that the belt material is perfectly elastic such that the length of the belt remains
constant, when it is at rest or in motion, therefore increase in length on the tight side is equal to
decrease in the length on the slack side. Thugs, équation équations (i) and (ii),
α (T1 – T0) = α (T0– T2) or T1 – T0 = T0 – T2
MAXIMUM TENSION IN THE BELT
A little consideration will show that the maximum tension in the belt (T) is equal to the total
tension in the tight side of the belt (Tt1
σ = Maximum safe stress in N/mm2,
b = Width of the belt in mm,and
t = Thickness of the belt imam.
We know that maximum tension in the belt,
T= Maximum stress × cross-sectional area of belt = σ. b. t
When centrifugal tension is neglected, then
T (or Tt1) =T1, i.e. Tension in the tight side of the belt and when centrifugal tension is
considered, then
T (or Tt1) =T1 + TC
CONDITON FOR MAXIMUM POWER TRANSMISSION
We know that power transmitted by a belt,
P = (T1 – T2) v
T1= Tension in the tight side of the belt in nektons,
T2= Tension in the slack side of the belt in nektons, and
v = Velocity of the belt in m/s.
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LENGTH OF AN OPEN BELT:
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We have already discussed in ART that in an open belt drive, both the pulleys rotate in the
same direction as shown in Fig.
r1 and r2= Radii of the larger and smaller pulleys,
x= Distance between the centers of two pulleys (i.e. O1O2), and
L= Total length of the belt.
Let the belt leaves the larger pulley at E and G and the smaller pulley at F and H as shown in
Fig.. Through O2, draw O2M parallel to FE.
From the geometry of the figure, we find that O2M will be perpendicular to1E.
Let the angle MO2O1= radians.
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LENGTH OF AN CROSS BELT:
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It may be noted that the above expression is a function of (r1 + r2). It is thus obvious that if sum
of the radii of the two pulleys be constant, then length of the belt required will also remain
constant, provided the distance between centers of the pulleys remain unchanged.
POWER TRANSMITTED BY A BELT:
The driving pulley (or driver) A and the driven pulley (or follower) B. We have already
discussed that the driving pulley pulls the belt from one side and delivers the same to the other
side. It is thus obvious that the tension on the former side (i.e. tight side) will be greater than the
latter side (i.e. slack side)
T1 and T2= Tensions in the tight and slack side of the belt respectively in
nektons,
r1 and r2= Radii of the driver and follower respectively, and
v = Velocity of the belt in m/s.
The effective turning (driving) force at the circumference of the follower is the difference
between the two tensions (i.e. T1 – T2).
Work done per second = (T1 – T2) v N-m/s and power transmitted,
P = (T1 – T2) v W
A little consideration will show that the torque exerted on the driving pulley is (T1 – T2) r1.
Similarly, the torque exerted on the driven pulley i.e. follower is (T1 – T2) r2
PROBLEMS:
Example 1. Two pulleys, one 450 mm diameter and the other 200 mm diameter are on parallel
shafts 1.95 m apart and are connected by a crossed belt. Find the length of the belt required
and the angle of contact between the belt and each pulley.
What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if
the maximum permissible tension in the belt is 1 ken, and the coefficient of friction between
the
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belt and pulley is 0.25 ?
GIVEN :
d1 = 450 mm = 0.45 m or r1 = 0.225 m ;
d2 = 200 mm = 0.2 m
x = 1.95 m
N1 = 200 r.n.
T1 = 1 kan = 1000 N
µ = 0.25
SOLUTION:
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RESULTS:
P=2.74 kW
L=4.975 m
ϴ=3.477 rad
Example. 2. A shaft rotating at 200 r.p.m. drives another shaft at 300 r.p.m. and transmits
6 kW through a belt. The belt is 100 mm wide and 10 mm thick. The distance between the
shafts is 4m. The smaller pulley is 0.5 m in diameter. Calculate the stress in the belt, if it is 1.
an open belt drive, and 2. across belt drive. Take µ =0.3.
Given : N1 = 200 r.p.m.
N2 = 300 r.p.m.
P = 6 kW = 6 × 103 W
B = 100 mm ;
T = 10 mm
x = 4 m
D2 = 0.5 m
µ = 0.3
SOLUTION:
Stress in the for an open belt drive:
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Stress in the for a cross belt:
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RESULT:
σ = 1.184 Map
Example 3. A leather belt is required to transmit 7.5 kW from a pulley 1.2 m in diameter,
running at 250 r.p.m. The angle embraced is 165° and the coefficient of friction between the belt
and the pulley is 0.3. If the safe working stress for the leather belt is 1.5 Map, density of leather
1 Mg/m3 and thickness of belt 10 mm, determine the width of the belt taking centrifugal tension
into account
GIVEN:
P = 7.5 kW = 7500 W
d = 1.2 m
N = 250 r.p.m
SOLUTION:
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RESULT:
b = 65.8mm
Example 4.An open belt drive connects two pulleys 1.2 m and 0.5 m diameter, on parallel shafts
4 meters apart. The mass of the belt is 0.9 kg per meter length and the maximum tension is not to
exceed 2000 N. The coefficient of friction is 0.3. The 1.2 m pulley, which is the driver, runs at
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200 r.p.m. Due to belt slip on one of the pulleys, the velocity of the driven shaft is only 450 r.p.m.
Calculate the torque on each of the two shafts, the power transmitted, and power lost in
friction. What is the efficiency of the drive ?
GIVEN:
d1 = 1.2 m or r1 = 0.6 m
d2 = 0.5 m or r2 = 0.25 m
x = 4 m
m = 0.9 kg/m;
T = 2000 N
µ = 0.3
N1 = 200 r.p.m.
N2 = 450 r.p.m.
SOLUTION:
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Torque on the shaft larger pulley:
We know that torque on the shaft of larger pulley,
TL= (T1 – T2) r1 = (1858 – 762) 0.6 = 657.6 N-m
Torque on the shaft of smaller pulley
We know that torque on the shaft of smaller pulley,
TS= (T1 – T2) r2 = (1858 – 762) 0.25 = 274 N-m
Power transmitted
We know that the power transmitted,
P = (T1 – T2) v = (1858 – 762) 12.57 = 13780 W
= 13.78 kW
Power lost in friction
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Efficiency of the drive
RESULTS:
TL= 657.6 N-m
TS= 274 N-m
P = 13.78 kW
P1= 13.78 kW
P2=12.91kW
Example5.. An open belt running over two pulleys 240 mm and 600 mm diameter connects
two parallel shafts 3 meters apart and transmits 4 kW from the smaller pulley that rotates at 300
r.p.m. Coefficient of friction between the belt and the pulley is 0.3 and the safe working tension is
10N per mm width. Determine : 1. minimum width of the belt, 2. initial belt tension, and 3. length
of the belt required.
GIVEN
d2 = 240 mm = 0.24 m ; d1 = 600 mm = 0.6 m ; x = 3 m ; P = 4 kW = 4000 W;
N2 = 300 r.p.m. ; µ = 0.3 ; T1 = 10 N/mm width
SOLUTION:
1.Minimum width of belt
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2. Initial belt tension
3. Length of the belt
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Example 5.. A belt drive consists of two V-belts in parallel, on grooved pulleys of the
same size. The angle of the groove is 30°. The cross-sectional area of each belt is 750 mm2 and
µ. = 0.12. The density of the belt material is 1.2 Mg/m3 and the maximum safe stress in the
material is 7 Map. Calculate the power that can be transmitted between pulleys 300 mm
diameter rotating at 1500 r.p.m. Find also the shaft speed in r.p.m. at which the power
transmitted would be maximum.
GIVEN :
d = 300 mm = 0.3 m N = 1500 r.p.m. SOLUTION:
Power transmitted
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shaft speed:
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RESULTS:
N1=2809 r.p.m
P=171.752
Kwan
Example 6. A compressor, requiring 90 kW is to run at about 250 r.p.m. The drive is by
V-belts from an electric motor running at 750 r.p.m. The diameter of the pulley on the
compressor shaft must not be greater than 1 meter while the centre distance between the pulleys
is limited to 1.75 meter. The belt speed should not exceed 1600 m/min.
Determine the number of V-belts required transmitting the power if each belt has a cross
-sectional area of 375 mm2, density 1000 kg/m3 and an allowable tensile stress of 2.5 Map. The
groove angle of the pulley is 35°. The coefficient of friction between the belt and the pulley is
0.25. Calculate also the length required of each belt.
GIVEN:
P = 90 kW
N2 = 250 r.p.m.
N1 = 750 r.p.m.
d2 = 1 m
x = 1.75 m
v = 1600 m/min = 26.67 m/s
a = 375 mm2 = 375 × 10–6m2
SOLUTION;
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Number of V-belts
Length of each belt
RESULTS:
L=5.664m
T2=67.36 N
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UNIT-3
GEARING AND CAMS
Gear profile and geometry – Nomenclature of spur and helical gears – Gear trains:
Simple, compound gear trains and epicyclical gear trains - Determination of speed and
torque - Cams – Types of cams – Design of profiles – Knife edged, flat faced and roller ended
followers with and without offsets for various types of follower motions
INTRODUCTION
We have discussed in the previous chapter, that the slipping of a belt or rope is a common
phenomenon, in the transmission of motion or power between two shafts. The effect of slipping
is to reduce the velocity ratio of the system. In precision machines, in which a definite velocity
ratio is of importance (as in watch mechanism), the only positive drive is by means of gears or
toothed wheels. A gear drive is also provided, when the distance between the driver and the
follower is very small. Friction Wheels The motion and power transmitted by gears is
cinematically equivalent to that transmitted by friction wheels or discs. In order to understand
how the motion can be Transmitted by two toothed wheels, consider two plain circular wheels
A and B mounted on shafts, having sufficient rough surfaces and pressing against each other
as shown in Fig Let the wheel A be keyed to the rotating shaft and the wheel B to the shaft, to be
rotated. A little consideration will show, that when the wheel A is rotated by a rotating shaft, it
will rotate the wheel B in the opposite direction as shown infix.(a)
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The wheel B will be rotated (by the wheel A) so long as the tangential force exerted by the wheel
A does not exceed the maximum frictional resistance between the two wheels. But when the
tangential force (P) exceeds the *frictional resistance (F), slipping will take place between the
two wheels. Thus the friction drive is not a positive drive. In order to avoid the slipping, a
number of projections (called teeth) as shown in Fig.(b), are provided on the periphery of the
wheel A , which will fit into the corresponding recesses on the periphery of the wheel B. A
friction wheel with the teeth cut on it is known as toothed
wheel or gear. The usual connection to show the toothed wheels is by their **pitch circles.
Note : Cinematically, the friction wheels running without slip and toothed gearing are identical.
But due to the
possibility of slipping of wheels, the friction wheels can only be used for transmission of small
powers
ADVANTAGES AND DISADVANTAGES OF GEAR DRIVE
The following are the advantages and disadvantages of the gear drive as compared to belt,
rope and chain drives :
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.
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2. The error in cutting teeth may cause vibrations and noise during operation.
CLASSIFICATION OF TOOTHED WHEELS
The gears or toothed wheels may be classified as follows :
1. According to the position of axes of the shafts. The axes of the two shafts between which
the motion is to be transmitted, maybe
(a) Parallel, (b) Intersecting, and (c) Non-intersecting adnoun-parallel.
The two parallel and co-planar shafts connected by the gears is shown in Fig. 1. These
gears are called spur gears and the arrangement is known as spur gearing. These gears have
teeth parallel to the axis of the wheel as shown in Fig. 1. Another name given to the spur gearing
is helical gearing, in which the teeth are inclined to the axis. The single and double helical gears
connecting parallel shafts are shown in Fig. 1 (a) and (b) respectively. The double helical gears
are known as herringbone gears. A pair of spur gears are cinematically equivalent to a pair of
cylindrical discs, keyed to parallel shafts and having a line contact.
The two non-parallel or intersecting, but coplanar shafts connected by gears is shown in Fig.
1(c). These 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.
The two non-intersecting and non-parallel i.e. non-coplanar shaft connected by gears is shown
in Fig. 1(d). These gears are called skew bevel gears or spiral gears and the arrangement is
known as skew bevel gearing or spiral gearing. This type of gearing also have a line contact, the
rotation of which about the axes generates the two pitch surfaces known as hyperboloids.
Notes : (a)When equal bevel gears (having equal teeth) connect two shafts whose axes are
mutually perpendicular, then the bevel gears are known as miters.
(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.
(c) The worm gearing is essentially a form of spiral gearing in which the shafts are usually at
right angles.
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2. According to the peripheral velocity of the gears. The gears, according to the
peripheral velocity of the gears may be classified as:
(a) Low velocity, (b) Medium velocity, and (c) High velocity.
The gears having velocity less than 3 m/s are termed as low velocity gears and gears having
velocity between 3 and 15 m/s are known as medium velocity gears. If the velocity of gears is
more than 15 m/s, then these are called high speed gears.
3. Accordingtothetypeofgearing.Thegears,accordingtothetypeofgearingmaybe
classified as:
(a) External gearing, (b) Internal gearing, and (c) Rack and pinion.
In external gearing, the gears of the two shafts mesh externally with each other as shown in Fig.
2(a). The larger of these two wheels is called spur wheel and the smaller wheel is called pinion.
In an external gearing, the motion of the two wheels is always unlike, as shown in Fig. 2(a).
In internal gearing, the gears of the two shafts mesh internally with each other as shown in Fig.
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2 (b). The larger of these two wheels is called annular wheel and the smaller wheel is called
pinion. In an internal gearing, the motion of the two wheels is always like, as shown in Fig. 2(b).
Sometimes, the gear of a shaft meshes externally and internally with the gears in a *straight line,
as shown in Fig. 2 Such type of gear is called rack and pinion. The straight line gear is called
rack and the circular wheel is called pinion. A little consideration will show that with the help of
a rack and pinion, we can convert linear motion into rotary motion and vice-versa as shown in
Fig. 2
4. According to position of teeth on the gear surface.
The teeth on the gear surface may be
(a) straight, (b) inclined, and (c) curved.
We have discussed earlier that the spur gears have straight teeth where as helical gears have their
teeth inclined to the wheel rim. In case of spiral gears, the teeth are curved over the rim surface
NOMENCLATURE
The following terms, which will be mostly used in this chapter, should be clearly understood
at this stage. These terms are illustrated in Fig. 3
1. Pitch circle. It is an imaginary circle which by pure rolling action, would give the same
motion as the actual gear.
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2. Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is usually
specified by the pitch circle diameter. It is also known as pitch diameter.
3. Pitch point. It is a common point of contact between two pitch circles.
4. Pitch surface. It is the surface of the rolling discs which the meshing gears have replaced
at the pitch circle.
5. Pressure angle or angle of obliquity. It is the angle between the common normal to two gear
teeth at the point of contact and the common tangent at the pitch point. It is usually denoted by Φ
6. Addendum. It is the radial distance of a tooth from the pitch circle to the top of the tooth .
7. Dedendum. Itis the radial distance of a tooth from the pitch circle to the bottom of the tooth.
8. Addendum circle. It is the circle drawn through the top of the teeth and is concentric with
the pitchcircle.
9. Dedendum circle. It is the circle drawn through the bottom of the teeth. It is also called
rootcircle.
10. Circular pitch. It is the distance measured on the circumference of the pitch circle from a
point of one tooth to the corresponding point on the next tooth. It is usually denoted by Pc.
Circularpitch, Pc=ad/T
A little consideration will show that the two gears will mesh together correctly, if the two wheels
have the same circular pitch.
11. Diametric pitch. It is the ratio of number of teeth to the pitch circle diameter in millimeters.
It is denoted by Pd. Pd=ad1/T1.
12. Module. It is the ratio of the pitch circle diameter in millimeters to the number of teeth.
It is usually denoted by m.Mathematically,
Module = D/T
13. Clearance. It is the radial distance from the top of the tooth to the bottom of the tooth, in a
meshing gear. A circle passing through the top of the meshing gear is known as clearance circle.
14. Total depth. It is the radial distance between the addendum and the addendum circles of a
gear. It is equal to the sum of the addendum anddedendum.
15. Working depth. It is the radial distance from the addendum circle to the clearance circle. It is
equal to the sum of the addendum of the two meshing gears.
16. Tooth thickness. It is the width of the tooth measured along the pitch circle.
17. Tooth space . It is the width of space between the two adjacent teeth measured along the
pitch circle.
18. Backlash. It is the difference between the tooth space and the tooth thickness, as measured
along the pitch circle. Theoretically, the backlash should be zero, but in actual practice some
backlash must be allowed to prevent jamming of the teeth due to tooth errors and thermal
expansion.
19. Face of tooth. It is the surface of the gear tooth above the pitch surface.
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20. Flank of tooth. It is the surface of the gear tooth below the pitch surface.
21. Top land. It is the surface of the top of the tooth.
22. Face width. It is the width of the gear tooth measured parallel to its axis.
23. Profile. It is the curve formed by the face and flank of the tooth.
24. Fillet radius. It is the radius that connects the root circle to the profile of the tooth.
25. Path of contact. It is the path traced by the point of contact of two teeth from the beginning
to the end of engagement.
26. *Length of the path of contact. It is the length of the common normal cut-off by the
addendum circles of the wheel and pinion.
27. ** Arc of contact. It is the path traced by a point on the pitch circle from the beginning to
the end of engagement of a given pair of teeth. The arc of contact consists of two parts,i.e.
(a) Arc of approach. It is the portion of the path of contact from the beginning of the
engagement to the pitch point.
(b) Arc of recess. It is the portion of the path of contact from the pitch point to the end of the
engagement of a pair of teeth.
Note : The ratio of the length of arc of contact to the circular pitch is known as contact ratio i.e.
number of pairs of teeth in contact.
GEAR TRAINS:
Introduction
Sometimes, two or more gears are made to mesh with each other to transmit power from one
shaft to another. Such a combination is called gear train or train of toothed wheels. The nature
of the train used depends upon the velocity ratio required and the relative position of the axes of
shafts. A gear train may consist of spur, bevel or spiral gears.
TYPES OF GEAR TRAINS
Following are the different types of gear trains, depending upon the arrangement of wheels :
1. Simple gear train, 2. Compound gear train, 3. Reverted gear train, and 4. Epicyclical gear train.
In the first three types of gear trains, the axes of the shafts over which the gears are mounted are
fixed relative to each other. But in case of epicyclical gear trains, the axes of the shafts on which
the gears are mounted may move relative to a fixed axis.
SIMPLE GEAR TRAIN
When there is only one gear on each shaft, as shown in Fig.4, it is known as simple gear train.
The gears are represented by their pitch circles. When the distance between the two shafts is
small, the two gears 1 and 2 are made to mesh with each other to transmit motion from one shaft
to the other, as shown in Fig. 4 (a). Since the gear 1 drives the gear 2, therefore gear 1 is called
the driver and the gear 2 is called the driven or follower.
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It may be noted that the motion of the driven gear is opposite to the motion of driving gear. Since
the speed ratio (or velocity ratio) of gear train is the ratio of the speed of the driver to the speed
of the driven or follower and ratio of speeds of any pair of gears in mesh is the inverse of their
number of teeth, therefore It may be noted that ratio of the speed of the driven or follower to the
speed of the driver is known as train value of the gear train. Mathematically
From above, we see that the train value is the reciprocal of speed ratio.
Sometimes, the distance between the two gears is large. The motion from one gear to
another, in such a case, may be transmitted by either of the following two methods:
1. By providing the large sized gear, or 2. By providing one or more intermediate gears.
A little consideration will show that the former method (i.e. providing large sized gears) is very
inconvenient and uneconomical method ; whereas the latter method (i.e. providing one or more
intermediate gear) is very convenient and economical. It may be noted that when the number of
intermediate gears are odd, the motion of both the gears (i.e. driver and driven or follower) is like
as shown in Fig. 4 (b). But if the number of intermediate gears are even, the motion of the driven
or follower will be in the opposite direction of the driver as shown in Fig.4 (c).
Now consider a simple train of gears with one intermediate gear as shown in Fig. 4 (b).
Let N1= Speed of driver in r.p.m.,
N2= Speed of intermediate gear in r.p.m.,
N3= Speed of driven or follower in r.p.m.,
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T1= Number of teeth on driver,
T2= Number of teeth on intermediate gear, and
T3= Number of teeth on driven or follower.
Since the driving gear 1 is in mesh with the intermediate gear 2, therefore speed ratio for these
two gears is
Similarly, as the intermediate gear 2 is in mesh with the driven gear 3, therefore speed ratio for
these two gears is
The speed ratio of the gear train as shown in Fig. 13.1 (b) is obtained by multiplying the
equations (I) and (ii).
Similarly, it can be proved that the above equation holds good even if there are any number of
intermediate gears. From above, we see that the speed ratio and the train value, in a simple train
of gears, is independent of the size and number of intermediate gears. These intermediate gears
are called idle gears, as they do not affect the speed ratio or train value of the system. The idle
gears are used for the following two purposes :
1. To connect gears where a large centre distance is required, and
2. To obtain the desired direction of motion of the driven gear (i.e. clockwise or anticlockwise).
COMPOUND GEAR TRAIN
When there are more than one gear on a shaft, as shown in Fig. 5, it is called a compound train
of gear. We have seen in Art. 5 that the idle gears, in a simple train of gears do not affect the
speed ratio of the system. But these gears are useful in bridging over the space between the
driver and the driven. But whenever the distance between the driver and the driven or follower
has to be bridged over by intermediate gears and at the same time a great ( or much less) speed
ratio is required, then the advantage of intermediate gears is intensified by providing compound
gears on intermediateshafts.Inthiscase,eachintermediateshafthastwogearsrigidlyfixedtoitsothat
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they may have the same speed. One of these two gears meshes with the driver and the other with
the driven or follower attached to the next shaft as shown in Fig.5
In a compound train of gears, as shown in Fig. 5 the gear 1 is the driving gear mounted on
shaft A, gears 2 and 3 are compound gears which are mounted on shaft B. The gears 4 and 5 are
also compound gears which are mounted on shaft C and the gear 6 is the driven gear mounted on
shaft D.
Let N1= Speed of driving gear 1,
T1= Number of teeth on driving gear 1,
N2,N3..., N6= Speed of respective gears in r.p.m., and
T2,T3..., T6= Number of teeth on respective gears.
Since gear 1 is in mesh with gear 2, therefore its speed ratio is
N1 /N2=T2/ T1
Similarly, for gears 3 and 4, speed ratio is
N3 /N4=T4/ T3
and for gears 5 and 6, speed ratio is
N5 /N6=T6/ T5
The speed ratio of compound gear train is obtained by multiplying the equations (I), (ii) and (iii),
The advantage of a compound train over a simple gear train is that a much larger speed reduction
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from the first shaft to the last shaft can be obtained with small gears.
If a simple gear train is used to give a large speed reduction, the last gear has to be very large.
Usually for a speed reduction
in excess of 7 to 1, a simple train is not used and a compound train or worm gearing is employed
EPICYCLIC GEAR TRAIN
We have already discussed that in an epicyclical gear train, the axes of the shafts, over which the
gears are mounted, may move relative to a fixed axis. A simple epicyclical gear train is shown in
Fig. 6, where a gear A and the arm C have a common axis at O1about which they can rotate.
The gear B meshes with gear A and has its axis on the arm at O2, about which the gear B can
rotate. If the arm is fixed, the gear train is simple and gear A can drive gear B or vice- versa, but
if gear A is fixed and the arm is rotated about the axis of gear A (i.e. O1), then the gear B is
forced to rotate upon and around gear
A . Such a motion is called epicyclical and the gear trains arranged in such a manner that one or
more of their members move upon and around another member are known as epicyclical gear
trains (epic. means upon and cyclic means around). The epicyclical gear trains may be simple or
compound. The epicyclical gear trains are useful for transmitting high velocity ratios with gears
of
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moderate size in a comparatively lesser space. The epicyclical gear trains are used in the back
gear of lathe, differential gears of the automobiles, hoists, pulley blocks, wrist watches
etc.Velocity Ratios of Epicyclical Gear Train
The following two methods may be used for finding out the velocity ratio of an epicyclical gear
train.
1. Tabular method, and 2. Algebraic method.
These methods are discussed, in detail, as follows:
1.Tabular method.
Consider an epicyclical gear train as shown in
Fig. Let TA = Number of teeth on gear A , and
TB = Number of teeth on gear B.
First of all, let us suppose that the arm is fixed. Therefore the axes of both the gears are also
fixed relative to each other. When the gear A makes one revolution anticlockwise, the gear B
will make *TA / TB revolutions, clockwise. Assuming the anticlockwise rotation as positive and
clockwise as negative, we may say that when gear A makes + 1revolution, then the gear B will
make (– TA/ TB) revolutions. This statement of relative motion is entered in the first row of the
table(seeTable13.1).Secondly,ifthegearAmakes+xrevolutions,thenthegearBwillmake
– x × TA/ TB revolutions. This statement is entered in the second row of the table. In other
words, multiply the each motion (entered in the first row) by x. Thirdly, each element of an
epicyclical train is given + y revolutions and entered in the third row. Finally, the motion of each
element of the gear train is added up and entered in the fourth row. A little consideration will
show that when two conditions about the motion of rotation of any two elements are known, then
the unknown speed of the third element may be obtained by substituting the given data in the
third column of the four throw.
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2. Algebraic method.
In this method, the motion of each element of the epicyclical train relative to the arm is set down
in the form of equations. The number of equations depends upon the number of elements in the
gear train. But the two conditions are, usually, supplied in any epicyclical train viz. some element
is fixed and the other has specified motion. These two conditions are sufficient to solve all the
equations ; and hence to determine the motion of any element in the epicyclical gear train. Let
the arm C be fixed in an epicyclical gear train as shown in Fig. 13.6. Therefore speed of the gear
A relative to the arm C.
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PROBLEMS;
Example 1In an epicyclical gear train, an arm carries two gears A and B having 36 and 45
teeth respectively. If the arm rotates at 150 r.p.m. in the anticlockwise direction about the
centre
of the gear A which is fixed, determine the speed of gear B. If the gear A instead of being fixed,
makes 300 r.p.m. in the clockwise direction, what will be the speed of gear B ?
GIVEN :
TA= 36 ; TB= 45 ; NC= 150 r.p.m.
SOLUTION:
We shall solve this example, by tabular method .
1. Tabular method
First of all prepare the table of motions as given below :
Speed of gear B when gear A is fixed
Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,
y = + 150 r.p.m.
Also the gear A is fixed, therefore
x + y=0 or x = – y = – 150r.p.m.
Bac i. Electromechanical engineering.theory of machines of mechanisms
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Speed of gear B when gear A makes 300 r.p.m. clockwise
Since the gear A makes 300 r.p.m. Clockwise, therefore from the fourth row of the table,
x + y = – 300 or x = – 300 – y = – 300 – 150 = – 450 r.p.m.
Speed of gear B,
EXAMPLE:2. train, the arm A carries two gears B and C and a compound gear D - E. The gear
B meshes with gear E and the gear C meshes with gear D. The number of teeth on gears B, C
and D are 75, 30 and 90 respectively. Find the speed and direction of gear C when gear B is
fixed and the arm A makes 100 r.p.m. Clockwise.
GIVEN :
TB= 75 ; TC= 30 ; TD = 90 ;
NA= 100 r.p.m. (clockwise)
SOLTUION;
The reverted epicyclical gear train . First of all, let us find the number of teeth on gear E (TE). Let
dB, dC, dandy dEbe the pitch circle diameters of gears B, C, D and E respectively. From the
geometry of the figure,
Since the number of teeth on each gear,
dB+ dE= dC+ d
for the same module, are proportional to their
pitch circle diameters, therefore
TB+ TE= TC+ TD
TE= TC+ TD– TB= 30 + 90 – 75 = 45
The table of motions is drawn as follow
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Example:3.In an epicyclical gear train, the internal wheels A and B and compound wheels C
and D rotate independently about axis O. The wheels E and F rotate on pins fixed to the arm G.
E gears with A and C and F gears with B and D. All the wheels have the same module and the
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number of teeth are : TC= 28; TD= 26; TE= TF= 18. 1. Sketch the arrangement ; 2. Find the
number of teeth on A and B ; 3. If the arm G makes 100 r.p.m. clockwise and A is fixed, find the
speed of B ; and 4. If the arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m. counter
clockwise ; find the speed of wheel B.
GIVEN :
TC= 28 ; TD= 26 ; TE= TF= 18
SOLUTION:
1. Sketch the arrangement
The arrangement is shown in Fig.
2. Number of teeth on wheels A and
Let TA= Number of teeth on wheel A ,and
TB= Number of teeth on wheel B.
If dA, dB, dC, d, dEand dFare the pitch circle diameters of wheels A , B, C, D, E and F
respectively, then from the geometry of Fig.
dA= dC+ 2 dEand
dB = d+ 2dF
Since the number of teeth are proportional to their pitch circle diameters, for the same
module, therefore
TA= TC+ 2 TE= 28 + 2 × 18 = 64
and TB= TD+ 2 TF= 26 + 2 × 18 =62
3. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A infixed
First of all, the table of motions is drawn as given below
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Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the table,
y = – 100
Also, the wheel A is fixed, therefore from the fourth row of the table,
x + y = 0 or x = – y = 100
speed of wheel B=4.2 r.p.m
4. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m.
counterclockwise
Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the
table
y = – 100
Also the wheel A makes 10 r.p.m. counter clockwise, therefore from the fourth row of the table,
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CAM INTRODUCTION:
A cam is a rotating machine element which gives reciprocating or oscillating motion to
another element known as follower. The cam and the follower have a line contact and constitute
ahigherpair.Thecamsareusuallyrotatedatuniformspeedbyashaft,but thefollowermotionis pre-
determined and will be according to the shape of the cam. The cam and follower is one of the
simplest as well as one of the most important mechanisms found in modern machinery today.
The cams are widely used for operating the inlet and exhaust valves of internal combustion
engines, automatic attachment of machineries, paper cutting machines, spinning and weaving
textile machineries, feed mechanism of automatic lithest.
Classification of Followers
The followers may be classified as discussed below:
1. According to the surface in contact.
The followers, according to the surface in contact, are as follows:
(a) Knife edge follower.
When the contacting end of the follower has a sharp knife edge, it is called a knife edge follower,
as shown in Fig.1 (a). The sliding motion takes place between the contacting surfaces (i.e. the
knife edge and the cam surface). It is seldom used in practice because the small area of
contacting surface results in excessive wear. In knife edge followers, a considerable side thrust
exists between the follower and the guide.
(b) Roller follower.
When the contacting end of the follower is a roller, it is called a roller follower, as shown in Fig.
1 (b). Since the rolling motion takes place between the contacting surfaces (i.e. the roller and the
cam), therefore the rate of wear is greatly reduced. In roller followers also the side thrust exists
between the follower and the guide. The roller followers are extensively used where more space
is available such as in stationary gas and oil engines and aircraft engines.
(c) Flat faced or mushroom follower.
When the contacting end of the follower is a perfectly flat face, it is called a flat-faced follower,
as shown in Fig. 1 (c). It may be noted that the side thrust between the follower and the guide is
much reduced in case of flat faced followers. The only side thrust is due to friction between the
contact surfaces of the follower and the cam. The relative motion between these surfaces is
largely of sliding nature but wear may be reduced by off-setting the axis of the follower, as
shown in Fig. 1 (f ) so that when the cam rotates, the follower also rotates about its own axis. The
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flat faced followers are generally used where space is limited such as in cams which operate the
valves of automobile engines.
(d) Spherical faced follower
When the contacting end of the follower is of spherical shape, it is called a spherical faced
follower, as shown in Fig.1 (d). It may be noted that when a flat-faced follower is used in
automobile engines, high surface stresses are produced. In order to minimize these stresses, the
flat end of the follower is machined to a spherical shape.
2. According to the motion of the follower.
The followers, according to its motion, are of the following two types:
(a) Reciprocating or translating follower. When the follower reciprocates in guides as the cam
rotates uniformly, it is known as reciprocating or translating follower. The followers as shown in
Fig. 1 (a) to (d) are all reciprocating or translating followers.
(b) Oscillatingorrotatingfollower.Whentheuniformrotarymotionofthecamisconverted
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into predetermined oscillatory motion of the follower, it is called oscillating or rotating follower.
The follower, as shown in Fig .1 (e), is an oscillating or rotating follower.
3. According to the path of motion of the follower. The followers, according to its path of
motion, are of the following two types:
(a) Radial follower. When the motion of the follower is along an axis passing through the centre
of the cam, it is known as radial follower. The followers, as shown in Fig.1 (a) to (e), are all
radial followers.
(b) Off-set follower. When the motion of the follower is along an axis away from the axis of the
cam centre, it is called off-set follower. The follower, as shown in Fig.1 (f), is an off-set
follower.
CLASSIFICATION OF CAMS
Though the cams may be classified in many ways, yet the following two types are important
from the subject point of view:
1. Radial or disc cam.
In radial cams, the follower reciprocates or oscillates in a direction perpendicular to the cam axis.
The cams as shown in Fig.2 are all radial cams.
2. Cylindrical cam.
In cylindrical cams, the follower reciprocates or oscillates in a direction parallel to the cam axis.
The follower rides in a groove at its cylindrical surface. A cylindrical grooved cam with a
reciprocating and an oscillating follower is shown in Fig.2 (a) and (b) respectively.
NOMENCALTURE:
Fig. shows a radial cam with reciprocating roller follower. The following terms are important in
order to draw the cam profile.
1. Base circle. It is the smallest circle that can be drawn to the cam profile.
2. Tracepoint.Itisareferencepointonthefollowerandisusedtogeneratethepitchcurve.In
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case of knife edge follower, the knife edge represents the trace point and the pitch curve
corresponds to the cam profile. In a roller follower, the centre of the roller represents the trace
point.
3. Pressure angle. It is the angle between the direction of the follower motion and a normal to
the pitch curve. This angle is very important in designing a cam profile. If the pressure angle is
too large, a reciprocating follower will jam in its bearings.
4. Pitch point. It is a point on the pitch curve having the maximum pressure angle.
5. Pitch circle. It is a circle drawn from the centre of the cam through the pitch points.
6. Pitch curve. It is the curve generated by the trace point as the follower moves relative to the
cam. For a knife edge follower, the pitch curve and the cam profile are same whereas for a roller
follower, they are separated by the radius of the roller.
7. Prime circle. It is the smallest circle that can be drawn from the centre of the cam and tangent
to the pitch curve. For a knife edge and a flat face follower, the prime circle and the base circle
are identical. For a roller follower, the prime circle is larger than the base circle by the radius of
the roller.
8. Liftorstroke.Itisthemaximumtravelofthefollowerfromitslowestpositiontothetopmost
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position.
MOTION OF THE FOLLOWER
The follower, during its travel, may have one of the following motions.
1. Uniform velocity, 2. Simple harmonic motion, 3. Uniform acceleration and retardation,
and 4. Cyclical motion
DISPLACEMENT, VELOCITY AND ACCELERATION DIAGRAMS WHEN THE
FOLLOWER MOVES WITH UNIFORM VELOCITY
The displacement, velocity and acceleration diagrams when a knife-edged follower moves with
uniform velocity are shown in Fig. 4 (a), (b) and (c) respectively. The abscissa (base) represents
the time (i.e. the number of seconds required for the cam to complete one revolution) or it may
represent the angular displacement of the cam in degrees. The ordinate represents the
displacement, or velocity or acceleration of the follower. Since the follower moves with uniform
velocity during its rise and return stroke, therefore the slope of the displacement curves must be
constant. In other words, AB1 and C1D must be straight lines. A little consideration will show
that the follower remains at rest during part of the cam rotation. The periods during which the
follower remains at rest are known as dwell periods, as shown by lines B1C1 and DE in Fig. 4
(a). From Fig. 4 (c), we see that the acceleration or retardation of the follower at the beginning
and at the end of each stroke is infinite.
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This is due to the fact that the follower is required to start from rest and has to gain a velocity
within no time. This is only possible if the acceleration or retardation at the beginning and at the
end of each stroke is infinite. These conditions are however, impracticable. In order to have the
acceleration and retardation within
the finite limits, it is necessary to modify the conditions which govern the motion of the follower.
This may be done by rounding off the sharp corners of the displacement diagram at the
beginning and at the end of each stroke, as shown in Fig. 5 (a). By doing so, the velocity of the
follower increases gradually to its maximum value at the beginning of each stroke and decreases
gradually to zero at the end of each stroke as shown in Fig. 5 (b). The modified displacement,
velocity and acceleration diagrams are shown in Fig. 5. The round corners of the displacement
diagram are usually parabolic curves because the parabolic motion results in a very low
acceleration of the follower for a given stroke and cam speed.
DISPLACEMENT, VELOCITY AND ACCELERATION DIAGRAMS WHEN THE
FOLLOWER MOVES WITH SIMPLE HARMONIC MOTION
The displacement, velocity and acceleration diagrams when the follower moves with simple
harmonic motion are shown in Fig. 6 (a), (b) and (c) respectively. The displacement diagram is
drawn as follows:
1. Draw a semi-circle on the follower stroke as diameter.
2. Divide the semi-circle into any number of even equal parts (say eight).
3. Divide the angular displacements of the cam during out stroke and return stroke into the same
number of equal parts.
4. The displacement diagram is obtained by projecting the points as shown in Fig..6 (a). The
velocity and acceleration diagrams are shown in Fig.6 (b) and (c) respectively. Since the follower
moves with a simple harmonic motion, therefore velocity diagram consists of a sine curve and
the acceleration diagram is a cosine curve.
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We see from Fig. 6 (b) that the velocity of the follower is zero at the beginning and at the end of
its stroke and increases gradually to a maximum at mid-stroke. On the other hand, the
acceleration of the follower is maximum at the beginning and at the ends of the stroke and
diminishes to zero at mid-stroke.
DISPLACEMENT, VELOCITY AND ACCELERATION DIAGRAMS WHEN THE
FOLLOWER MOVES WITH UNIFORM ACCELERATION AND RETARDATION
The displacement, velocity and acceleration diagrams when the follower moves with uniform
acceleration and retardation are shown in Fig. 8 (a), (b) and (c) respectively. We see that the
displacement diagram consists of a parabolic curve and may be drawn as discussed below :
1. Divide the angular displacement of the cam during outstroke ( èO) into any even number of
equal parts (say eight) and draw vertical lines through these points as shown in Fig. 8(a).
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2. Divide the stroke of the follower (S) into the same number of equal even parts.
3. Join Aar to intersect the vertical line through point 1 at B. Similarly, obtain the other points C,
D etc. as shown in Fig. 8 (a). Now join these points to obtain the parabolic curve for the out
stroke of the follower.
4. In the similar way as discussed above, the displacement diagram for the follower during
return stroke may be drawn. Since the acceleration and retardation are uniform, therefore the
velocity varies directly with the time. The velocity diagram is shown in Fig. 8 (b).
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CONSTRUCTION OF CAM PROFILE FOR A RADIAL CAM
In order to draw the cam profile for a radial cam, first of all the displacement diagram for the
given motion of the follower is drawn. Then by constructing the follower in its proper position at
each angular position, the profile of the working surface of the cam is drawn. In constructing the
cam profile, the principle of kinematic inversion is used, i.e. the cam is imagined to be stationary
and the follower is allowed to rotate in the opposite direction to the cam rotation.
Theconstructionofcamprofilesfordifferenttypesoffollowerwithdifferenttypesof
motions are discussed in the following examples.
Example 1. A cam is to give the following motion to a knife-edged follower :
1. Outstroke during 60° of cam rotation ;2. Dwell for the next 30° of cam rotation ;
3. Return stroke during next 60° of cam rotation, and 4. Dwell for the remaining 210° of cam
rotation. The stroke of the follower is 40 mm and the minimum radius of the cam is 50 mm. The
follower moves with uniform velocity during both the outstroke and return strokes. Draw the
profile of the cam when (a) the axis of the follower passes through the axis of the cam shaft, and
(b) the axis of the follower is offset by 20 mm from the axis of the cam shaft.
First of all, the displacement diagram, as shown in Fig.10, is drawn as discussed in
the following steps:
1. Draw a horizontal line AX = 360° to some suitable scale. On this line, mark AS = 60° to
represent outstroke of the follower, ST = 30° to represent dwell, TP = 60° to represent return
stroke and PX = 210° to represent dwell.
2. Draw vertical line AY equal to the stroke of the follower (i.e. 40 mm) and complete the
rectangle as shown in Fig.10.
3. Divide the angular displacement during outstroke and return stroke into any equal number of
even parts (say six) and draw vertical lines through each point.
4. Since the follower moves with uniform velocity during outstroke and return stroke, there- for
the displacement diagram consists of straight lines. Join AG and.
5. The complete displacement diagram is shown by AGHPX in Fig.10.
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(a) Profile of the cam when the axis of follower passes through the axis of camshaft
The profile of the cam when the axis of the follower passes through the axis of the cam shaft,
as shown in Fig.11, is drawn as discussed in the following steps :
1. Draw a base circle with radius equal to the minimum radius of the cam (i.e. 50 mm) with O as
centre.
2. Since the axis of the follower passes through the axis of the cam shaft, therefore mark trace
point A, as shown in Fig. 11.
3. From OA, mark angle AOS = 60° to represent outstroke, angle SOT = 30° to
represent dwell and angle TOP = 60° to represent return stroke.
4. Dividetheangulardisplacementsduringoutstrokeandreturnstroke(i.e.angleAOSandangle
TOP) into the same number of equal even parts as in displacement diagram.
5. Join the points 1, 2, 3 ...etc. and 0 ,1 , 2 , 3 , ... etc. with centre O and produce beyond the base
circle as shown inFig.11.
6. Now set off 1B, 2C, 3D ... etc. and 0H,1J ... etc. from the displacement diagram.
7. Join the points A, B, C,... M, N, P with a smooth curve. The curve AGHPA is the complete
profile of theca.
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Notes : The points B,C,D ....... L, M, and N may also be obtained as follows:
1. Mark AY = 40 mm on the axis of the follower, and set of Ab, Ac, Ad... etc. equal to the
distances 1B,2C,3D ..... etc. as in displacement diagram.
2. From the centre of the cam O, draw arcs with radii Ob, Oc, Do etc. The arcs intersect the
produced lines O1, O2... etc. at B,C,D ...... L, M, and N.
(b) Profile of the cam when the axis of the follower is offset by 20 mm from the axis of the
cam
shaft
The profile of the cam when the axis of the follower is offset from the axis of the cam shaft, as
shown in Fig.12, is drawn as discussed in the following steps :
1. Draw a base circle with radius equal to the minimum radius of the cam (i.e. 50 mm)with
O as centre.
2. Draw the axis of the follower at a distance of 20 mm from the axis of the cam,
which intersects the base circle ata.
3. Join AO and draw an offset circle of radius 20 mm with centre.
4. From OA, mark angle AOS = 60° to representoutstroke,angle SOT = 30° to represent
dwell and angle TOP = 60° to represent return stroke.
5. Divide the angular displacement during outstroke and return stroke (i.e. angle AS and
angle TOP) into the same number of equal even parts as in displacement diagram.
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6. Now from the points 1, 2, 3 ...etc. and 0 ,1 , 2 ,3 ... etc. on the base circle, draw
tangents to the offset circle and produce these tangents beyond the base circle as shown inFig.12.
7. Now set off 1B, 2C, 3D ... etc. and 0H,1J ... etc. from the displacement diagram.
8. Join the points A, B, C ...M, N, and P with a smooth curve. The curve AGHPA is the complete
profile of theca.
Example :2.A cam, with a minimum radius of 25 mm, rotating clockwise at a uniform speed
is to be designed to give a roller follower, at the end of a valve rod, motion described below :
1. To raise the valve through 50 mm during 120° rotation of the cam;
2. To keep the valve fully raised through next30°;
3. To lower the valve during next 60°;and
4. To keep the valve closed during rest of the revolution i.e. 150°;
The diameter of the roller is 20 mm and the diameter of the cam shaft is 25 mm. Draw the profile
of the cam when (a) the line of stroke of the valve rod passes through the axis of the cam shaft,
and (b) the line of the stroke is offset 15 mm from the axis of the cam shaft. The displacement of
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the valve, while being raised and lowered, is to take place with simple harmonic motion.
Determine the maximum acceleration of the valve rod when the cam shaft rotates at 100 r.p.m.
Draw the displacement, the velocity and the acceleration diagrams for one complete revolution
of theca.
Given :
S = 50 mm = 0.05 m ;
N = 100 r.p.m.
Since the valve is being raised and lowered with simple harmonic motion, therefore the
displacement diagram, as shown in Fig. 13 (a), is drawn in the similar manner as discussed in the
previous example.
(a) Profile of the cam when the line of stroke of the valve rod passes through the axis of
the cam shaft
The profile of the cam, as shown in Fig. 14 is drawn as discussed in the following steps :
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1. Draw a base circle with centre O and radius equal to the minimum radius of theca (
i.e. 25 mm).
2. Draw a prime circle with centre O and radius,
3. Draw angle AOS = 120° to represent rising or out stroke of the valve, angle SOT = 30° to
represent dwell and angle TOP = 60° to represent lowering or return stroke of the valve.
4. Divide the angular displacements of the cam during rising and lowering of the valve (i.e.
angle AOS and TOP ) into the same number of equal even parts as in displacement diagram.
5.Join the points 1, 2, 3, etc. with the centre O and produce the lines beyond prime circle as
shown in Fig.14
6. Set off 1B, 2C, 3D etc. equal to the displacements from displacement diagram.
7. Join the points A, B, C ... N, P, A. The curve drawn through these points is known as pitch
curve.
8. From the points A, B, C ... N, P, draw circles of radius equal to the radius of the roller.
9. Join the bottoms of the circles with a smooth curve as shown in Fig. 14 This is the
required profile of theca.
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(b) Profile of the cam when the line of stroke is offset 15 mm from the axis of the cam shaft
The profile of the cam when the line of stroke is offset from the axis of the cam shaft, as
shown in Fig. 15may be drawn as discussed in the following steps:
1. Draw a base circle with centre O and radius equal to 25mm.
2. Draw a prime circle with centre O and radius OA = 35mm.
3. Draw an off-set circle with centre O and radius equal to 15mm.
4. JoinOA.FromOAdrawtheangulardisplacementsofcami.e.drawangleAOS=120°,angle
SOT = 30° and angle TOP = 60°.
5. Divide the angular displacements of the cam during rising and lowering of the valve into
the same number of equal even parts (i.e. six parts ) as in displacement diagram.
6. From points 1, 2, 3 .... etc. and 0 ,1 , 3 , ...etc. on the prime circle, draw tangents to the
offset circle.
7. Set off 1B, 2C, 3D... etc. equal to displacements as measured from displacement diagram .
8. By joining the points A, B, C ... M, N, P, with a smooth curve, we get a pitch curve.
9. Now A, B, C...etc. as centre, draw circles with radius equal to the radius of roller.
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10. Join the bottoms of the circles with a smooth curve as shown in Fig.15. This is the
required profile of theca.
Example :3.
A cam rotating clockwise at a uniform speed of 1000 r.p.m. is required to give a roller follower
the motion defined below :
1. Follower to move outwards through 50 mm during 120° of cam rotation,
2. Follower to dwell for next 60° of cam rotation,
3. Follower to return to its starting position during next 90° of cam rotation,
4. Follower to dwell for the rest of the cam rotation.
The minimum radius of the cam is 50 mm and the diameter of roller is 10 mm. The line of stroke
of the follower is off-set by 20 mm from the axis of the cam shaft. If the displacement of the
follower takes place with uniform and equal acceleration and retardation on both the outward
and return strokes, draw profile of the cam and find the maximum velocity and acceleration
during out stroke and return stroke.
SOLUTION;
Since the displacement of the follower takes place with uniform and equal acceleration and
retardation on both outward and return strokes, therefore the displacement diagram, as shown in
Fig. 16 is drawn in the similar manner as discussed in the previous example.
But in this case, the angular displacement and stroke of the follower is divided into eight equal
parts Now, the profile of the cam, as shown in Fig. 17, is drawn as discussed in the following
steps :
1. Draw a base circle with centre O and radius equal to the minimum radius of the cam (i.e. 50
mm).
2. DrawaprimecirclewithcentreOandradiusOA=Minimumradiusofthecam+radiusof
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roller = 50 + 5 = 55 mm
3. Draw an off-set circle with centre O and radius equal to 20mm.
4. Divide the angular displacements of the cam during out stroke and return stroke into eight
equal parts as shown by points 0, 1, 2 . . . and 0 ,1 , 2 . . . etc. on the prime circle in Fig.17
5. From these points draw tangents to the off-set circle.
6. Set off 1B, 2C, 3D . . . etc. equal to the displacements as measured from the displacement
diagram.
7. By joining the points A, B, C . . . T, U, A with a smooth curve, we get a pitch curve.
8. Now from points A, B, C . . . T, U, draw circles with radius equal to the radius of the roller.
9. Join the bottoms of these circles with a smooth curve to obtain the profile of the cam as shown
in Fig.17
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Maximum velocity of the follower during out stroke and return stroke:
We know that angular velocity of the cam,
Maximum acceleration of the follower during out stroke and return stroke:
We know that the maximum acceleration of the follower during out stroke,
Example 4. Construct the profile of a cam to suit the following specifications :
Cam shaft diameter = 40 mm ; Least radius of cam = 25 mm ; Diameter of roller = 25 mm;
Angle of lift = 120° ; Angle of fall = 150° ; Lift of the follower = 40 mm ; Number of pauses are
two of equal interval between motions. During the lift, the motion is S.H.M. During the fall the
motion is uniform acceleration and deceleration. The speed of the cam shaft is uniform. The line
of stroke of the follower is off-set 12.5 mm from the centre of the cam.
Construction
First of all the displacement diagram, as shown in Fig18, is drawn as discussed in the
following steps :
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1. Since the follower moves with simple harmonic motion during lift (i.e. for 120° of cam
rotation), therefore draw the displacement curve ADG in the similar manner as discussed.
2. Since the follower moves with uniform acceleration and deceleration during fall (i.e. for 150°
of cam rotation), therefore draw the displacement curve HLP consisting of double parabola as
discussed .Now the profile of the cam, when the line of stroke of the follower is off-set 12.5 mm
from the centre of the cam, as shown in Fig. 19, is drawn as discussed in the following steps:
1. Draw a base circle with centre O and radius equal to the least radius of cam (i.e. 25mm).
2. Draw a prime circle with centre O and radius, OA = Least radius of cam + radius of roller =
25 + 25/2 = 37.5mm
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3. Draw a circle with centre O and radius equal to 20 mm to represent the camshaft.
4. Draw an offset circle with centre O and radius equal to 12.5mm.
5. Join OA. From OA draw angular displacements of the cam, i.e. draw angle AOS = 120° to
represent lift of the follower, angle SOT = 45° to represent pause, angle TOP = 150° to represent
fall of the follower and angle POA = 45° to represent pause
6. Divide the angular displacements during lift and fall (i.e. angle AOS and TOP) into
the same number of equal even parts (i.e. six parts) as in the displacement diagram.
7. From points 1, 2, 3 . . . etc. and 0 , 1 , 2 , 3. . . etc. on the prime circle, draw tangents to
the off-set circle.
8. Set off 1B, 2C, 3D . . . etc. equal to the displacements as measured from the displacement
diagram.
9. By joining the points A, B, C . . . M, N, P with a smooth curve, we get a pitch curve.
10. Now with A, B, C . . . etc. as centre, draw circles with radius equal to the radius of roller.
11. Join the bottoms of the circles with a smooth curve as shown in Fig. 19This is the required
profile of theca.
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Example 5.
It is required to set out the profile of a cam with oscillating follower for the
following motion :
(a) Follower to move outward through an angular displacement of 20° during 90° of cam
rotation ; (b) Follower to dwell for 45° of cam rotation ;(c) Follower to return to its original
position of zero displacement in 75° of cam rotation ; and (d) Follower to dwell for the
remaining period of the revolution of theca.
The distance between the pivot centre and the follower roller centre is 70 mm and the roller
diameter is 20 mm. The minimum radius of the cam corresponds to the starting position of the
follower as given in (a). The location of the pivot point is 70 mm to the left and 60 mm above the
axis of rotation of the cam. The motion of the follower is to take place with S.H.M. during out
stroke and with uniform acceleration and retardation during return stroke.
Construction
We know that the angular displacement of the roller follower, Since the distance between the
pivot centre and the roller centre (i.e. radius A1A) is 70 mm, therefore length of arc AA2, as
shown in Fig. 20, along which the displacement of the roller actually takes place.
Since the angle is very small, therefore length of chord AA2 is taken equal to the length of arc
AA2. Thus in order to draw the displacement diagram, we shall take lift of the follower equal to
the length of chord AA2i.e. 24.5 mm.
The follower moves with simple harmonic motion during out stroke and with uniform
acceleration and retardation during return stroke. Therefore, the displacement diagram, as shown
in Fig. 21, is drawn in the similar way as discussed .
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The profile of the cam, as shown in Fig. 22, is drawn as discussed in the following steps :
1. First of all, locate the pivot point A1 which is 70 mm to the left and 60 mm above the axis of
theca.
2. Since the distance between the pivots centre A1 and the follower roller centre A is 70 mm and
the roller diameter is 20 mm, therefore draw a circle with centre A and radius equal to the radius
of roller i.e. 10mm.
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3. We find that the minimum radius of the cam 60 10 50 mm , Radius of the prime circle, A
= Min. radius of cam + Radius of roller = 50 + 10 = 60 mm
4. Now complete the profile of the cam in the similar way as discussed
UNIT-4
BALANCING
Static and dynamic balancing – Single and several masses in different planes – Balancing
of reciprocating masses- primary balancing and concepts of secondary balancing – Single
and multi cylinder engines (Inline) – Balancing of radial V engine – direct and reverse
crank method
INTRODUCTION:
The high speed of engines and other machines is a common phenomenon now-a-days. It is,
therefore, very essential that all the rotating and reciprocating parts should be completely
balancedasfaraspossible.Ifthesepartsarenotproperlybalanced,thedynamicforcesaresetup.
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These forces not only increase the loads on bearings and stresses in the various members, but
also produce unpleasant and even dangerous vibrations. In this chapter we shall discuss the
balancing of unbalanced forces caused by rotating masses, in order to minimize pressure on the
main bearings when an engine is running. Balancing of Rotating Masses We have already
discussed, that whenever a certain mass is attached to a rotating shaft, it exerts some centrifugal
force, whose effect is to bend the shaft and to produce vibrations in it. In order to prevent the
effect of centrifugal force, another mass is attached to the opposite side of the shaft, at such a
position so as to balance the effect of the centrifugal force of the first mass. This is done in such
way that the centrifugal forces of both the masses are made to be equal and opposite. The process
of providing the second mass in order to counteract the effect of the centrifugal force of the first
mass is called balancing of rotating masses.
The following cases are important from the subject point of view:
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 different masses rotating in the same plane.
4. Balancing of different masses rotating in different planes.
We shall now discuss these cases, in detail, in the following pages.
BALANCING OF A SINGLE ROTATING MASS BY A SINGLE MASS ROTATING IN
THE SAME PLANE
Consider a disturbing mass m1 attached to a shaft rotating at ὼ rad/s as shown in Fig. Let r1 be
the radius of rotation of the mass m1 (i.e. distance between the axis of rotation of the shaft and
the centre of gravity of the mass m1). We know that the centrifugal force exerted by the mass m1
on the shaft This centrifugal force acts radically outwards and thus produces bending moment on
the shaft. In order to counteract the effect of this force, a balancing mass (m2) may be attached in
the same plane of rotation as that of disturbing mass (m1) such that the centrifugal forces due to
the two masses are equal and opposite.
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r2 = Radius of rotation of the balancing mass m2 (i.e. distance between the
axis of rotation of the shaft and the centre of gravity of mass m2 ).
Centrifugal force due to mass m2,
FC2= m2 ὼ 2.r 2
BALANCING OF A SINGLE ROTATING MASS BY TWO MASSES ROTATING IN
DIFFERENT PLANES:
We have discussed in the previous article that by introducing a single balancing mass in the same
plane of rotation as that of disturbing mass, the centrifugal forces are balanced. In other words,
the two forces are equal in magnitude and opposite in direction. But this type of arrangement for
balancing gives rise to a couple which tends to rock the shaft in its bearings. Therefore in order
to put the system in complete balance, two balancing masses are placed in two different planes,
parallel to the plane of rotation of the disturbing mass, in such a way that they satisfy the
following two conditions of equilibrium.
1. The net dynamic force acting on the shaft is equal to zero. This requires that the line of action
of three centrifugal forces must be the same. In other words, the centre of the masses of the
system must lie on the axis of rotation. This is the condition for static balancing.
2. The net couple due to the dynamic forces acting on the shaft is equal to zero. In other words,
the algebraic sum of the moments about any point in the plane must be zero. The conditions (1)
and (2) together give dynamic balancing. The following two possibilities may arise while
attaching the two balancing masses:
1. The plane of the disturbing mass may be in between the planes of the two balancing masses,
and
2. Theplaneofthedisturbingmassmaylieontheleftorrightofthetwoplanescontainingthe
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balancing masses. We shall now discuss both the above cases one by one.
1. When the plane of the disturbing mass lies in between the planes of the two balancing
Masses
Consider a disturbing mass m lying in a plane A to be balanced by two rotating masses m1 and
m2 lying in two different planes L and M as shown in Fig. 2. Let r, r1 and r2 be the radii of
rotation of the masses in planes A, L and M respectively.
Let.
l1 = Distance between the planes A and L,
l2 = Distance between the planes A and M, and
l = Distance between the planes L and M.
We know that the centrifugal force exerted by the mass m in the plane A,
FC= Mr.2.r
Similarly, the centrifugal force exerted by the mass m1 in the plane L,
FC1= m1.ὼ2.r1
and, the centrifugal force exerted by the mass m2 in the plane M,
FC2= m2.ὼ2.r2
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Since the net force acting on the shaft must be equal to zero, therefore the centrifugal force on
the disturbing mass must be equal to the sum of the centrifugal forces on the balancing masses,
therefore
Now in order to find the magnitude of balancing force in the plane L (or the dynamic force at the
bearing Q of a shaft), take moments about P which is the point of intersection of the plane M and
the axis of rotation. Therefore
Similarly, in order to find the balancing force in plane M (or the dynamic force at the bearing P
of a shaft), take moments about Q which is the point of intersection of the plane L and the axis of
rotation. Therefore
It may be noted that equation (I) represents the condition for static balance, but in order to
achieve dynamic balance, equations (ii) or (iii) must also be satisfied.
2. When the plane of the disturbing mass lies on one end of the planes of the balancing
Masses
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In this case, the mass m lies in the plane A and the balancing masses lay in the planes L and M, as
shown in Fig.3. As discussed above, the following conditions must be satisfied in order to
balance the system, i.e.
FC +FC2=F C1
Now, to find the balancing force in the plane L (or the dynamic force at the bearing Q of a shaft),
take moments about P which is the point of intersection of the plane M and the axis of rotation.
Therefore
FC1×l=FC×l2
Similarly, to find the balancing force in the plane M (or the dynamic force at the bearing P of a
shaft), take moments about Q which is the point of intersection of the plane L and the axis of
rotation. Therefore
FC2×l=FC×l1
BALANCING OF SEVERAL MASSES ROTATING IN THE SAME PLANE:
Consider any number of masses (say four) of magnitude m1, m2, m3 and m4 at distances of r1,
r2, r3 and r4 from the axis of the rotating shaft. Let masses with the horizontal line OX, as shown
in Fig..4(a) Let these masses rotate about an axis through O and perpendicular to the plane of
paper, with a constant angular velocity of ὼ rad/s. The magnitude and position of the balancing
mass may be found out analytically or graphically as discussed below
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Analytical method
The magnitude and direction of the balancing mass may be obtained, analytically, as discussed
below :
1. First of all, find out the centrifugal force* (or the product of the mass and its radius of
rotation) exerted by each mass on the rotating shaft.
2. Resolve the centrifugal forces horizontally and vertically and find their sums, i.e. ΣH
and ΣV. We know that Sum of horizontal components of the centrifugal forces
ΣH=m1.r1.coosϴ1+ m2.r2.coosϴ2+…..
And sum of vertical components of the centrifugal forces.
ΣV=m1.r1.sinϴ1+ m2.r2.sinϴ2+…..
3. Magnitude of the resultant centrifugal force
FC =√ (ΣH) 2+ (ΣV) 2
4. If ϴis the angle, which the resultant force makes with the horizontal, then
tanϴ=ΣV/ΣH
5. The balancing force is then equal to the resultant force, but in opposite direction.
6. Now find out the magnitude of the balancing mass, suchthat
FC= mar
M=balancing mass, and
r = its radius of rotation.
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2. Graphical method
The magnitude and position of the balancing mass may also be obtained graphically as discussed
below :
1. First of all, draw the space diagram with the positions of the several masses, as shown
in Fig..4(a).
2. Find out the centrifugal force (or product of the mass and radius of rotation) exerted by each
mass on the rotating shaft.
3. Now draw the vector diagram with the obtained centrifugal forces (or the product of the
masses and their radii of rotation), such that ab represents the centrifugal force exerted by the
mass m1 (or m1.r1) in magnitude and direction to some suitable scale. Similarly, draw bc, candy
de to represent centrifugal forces of other masses m2, m3 and m4 (or m2.r2, m3.r3 and4.r4).
4. Now, as per polygon law of forces, the closing side age represents the resultant force
in magnitude and direction, as shown in Fig. 4(b).
5. The balancing force is, then, equal to the resultant force, but in opposite direction.
6. Now find out the magnitude of the balancing mass (m) at a given radius of rotation (r), such
that
BALANCING OF SEVERAL MASSES ROTATING IN DIFFERENT PLANES:
When several masses revolve in different planes, they may be transferred to a reference plane
(briefly written as R.P.), which may be defined as the plane passing through a point on the axis
of rotation and perpendicular to it. The effect of transferring a revolving mass (in one plane) to a
reference plane is to cause a force of magnitude equal to the centrifugal force of the revolving
mass to act in the reference plane, together with a couple of magnitude equal to the product of
the force and the distance between the plane of rotation and the reference plane. In order to have
a complete balance of the several revolving masses in different planes, the following two
conditions must be satisfied:
1. The forces in the reference plane must balance, i.e. the resultant force must be zero.
2. The couples about the reference plane must balance, i.e. the resultant couple must be zero.
Let us now consider four masses m1, m2, m3 and m4 revolving in planes 1, 2, 3 and 4
respectively as shown in Fig. 5(a). The relative angular positions of these masses are shown in
the end view [Fig. 5 (b)]. The magnitude of the balancing masses mLand m in planes L and M
may be obtained as discussed below:
1. Take one of the planes, say L as the reference plane (R.P.). The distances of all the other
planes to the left of the reference plane may be regarded as negative, and those to the right as
positive.
2. Tabulate the data as shown in Table 21.1. The planes are tabulated in the same order in which
they occur, reading from left to right.
3. Acouplemayberepresentedbyavectordrawnperpendiculartotheplaneofthecouple.The
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couple C1 introduced by transferring m1 to the reference plane through O is proportional to
m1.r1.l1 and acts in a plane through Om1 and perpendicular to the paper.
The vector representing this couple is drawn in the plane of the paper and perpendicular to Om1
as shown by OC1 in Fig.5 (c). Similarly, the vectors OC2, OC3 and OC4 are drawn
perpendicular to Om2, Om3 and Om4 respectively and in the plane of the paper.
4. The couple vectors as discussed above, are turned counter clockwise through a right angle for
convenience of drawing as shown in Fig.5(d). We see that their relative positions remains
unaffected. Now the vectors OC2, OC3 and OC4 are parallel and in the same direction as Om2,
Om3random4,while the vector OC1is parallel to Om1burin*oppositedirection.Hencethe
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couple vectors are drawn radically outwards for the masses on one side of the reference plane
and radially inward for the masses on the other side of the reference plane.
5. NowdrawthecouplepolygonasshowninFig.5(e).Thevectordo represents the
balanced couple. Since the balanced couple CM is proportional to mM.rM.lM,therefore
From this expression, the value of the balancing mass mM in the plane M may be obtained, and
the angle of inclination Φ of this mass may be measured from Fig. 5 (b).
6. Now draw the force polygon as shown in Fig. 5 ( f ). The vector eon (in the direction from e too
) represents the balanced force. Since the balanced force is proportional to mL.rL, therefore,
From this expression, the value of the balancing mass mL in the plane L may be obtained and
the angle of inclination α of this mass with the horizontal may be measured from fig5(b).
BALANCING OF RECIPROCATING MASSES:
The various forces acting on the reciprocating parts of an engine. The resultant of all the forces
acting on the body of the engine due to inertia forces only is known as unbalanced force or
shaking force. Thus if the resultant of all the forces due to inertia effects is zero, then there will
be no unbalanced force, but even then an unbalanced couple or shaking couple will be present.
Consider a horizontal reciprocating engine mechanism as shown in Fig.1.
FI = Inertia force due to reciprocating parts,
FN = Force on the sides of the cylinder walls or normal force acting on
the cross-head guides, and
FB = Force acting on the crankshaft bearing or main bearing.
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Since FR and FI are equal in magnitude but opposite in direction, therefore they balance each
other. The horizontal component of FB (i.e. FBH) acting along the line of reciprocation is also
equal and opposite to FI. This force FBH = FU is an unbalanced force or shaking force and
required to be properly balanced. The force on the sides of the cylinder walls (FN) and the
vertical component of FB (i.e. FBV) are equal and opposite and thus form a shaking couple of
magnitude FN × x or FBV × x. From above we see that the effect of the reciprocating parts is to
produce a shaking force and a shaking couple. Since the shaking force and a shaking couple vary
in magnitude and direction during the engine cycle, therefore they cause very objectionable
vibrations. Thus the purpose of balancing the reciprocating masses is to eliminate the shaking
force and a shaking couple. In most of the mechanisms, we can reduce the shaking force and a
shaking couple by adding appropriate balancing mass, but it is usually not practical to eliminate
them completely. In other words, the reciprocating masses are only partially balanced.
PRIMARY AND SECONDARY UNBALANCED FORCES OF RECIPROCATING
MASSES
Consider a reciprocating engine mechanism as shown in Fig
m = Mass of the reciprocating parts,
l = Length of the connecting rod PC,
r = Radius of the crank OC,
ϴ= Angle of inclination of the crank with the line of stroke PO,
ὼ = Angular speed of the crank,
n = Ratio of length of the connecting rod to the crank radius = l / r.
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We have already discussed in Art. that the acceleration of the reciprocating parts is
approximately given by the expression
PARTIAL BALANCING OF UNBALANCED PRIMARY FORCE IN A
RECIPROCATINGENGINE
The primary unbalanced force (m ὼ2r coosϴ) may be considered as the component of the
centrifugal force produced by a rotating mass m placed at the crank radius r, as shown in Fig.2
The primary force acts from O to P along the line of stroke. Hence, balancing of primary force is
considered as equivalent to the balancing of mass m rotating at the crank radius r.
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This is balanced by having a mass B at a radius b, placed diametrically opposite to the crank pin
C. We know that centrifugal force due to mass,
=B.ὼ2.b
and horizontal component of this force acting in opposite direction of primary force
= B.ὼ2.bcos ϴ
The primary force is balanced, if
B.ὼ2.bcos ϴ= m.ὼ2.rcos ϴ
A little consideration will show, that the primary force is completely balanced if Bib = m.r, but
the centrifugal force produced due to the revolving mass
B, has also a vertical component (perpendicular to the line of stroke) of magnitude B ὼ2bsinϴ.
This force remains unbalanced. The maximum value of this force is equal to B2b when ϴis 90°
and 270°, which is same as the maximum value of the primary force m2r . From the above
discussion, we see that in the first case, the primary unbalanced force acts along the line of stroke
whereas in the second case, the unbalanced force acts along the perpendicular to the line of
stroke. The maximum value of the force remains same in both the cases. It is thus obvious, that
the effect of the above method of balancing is to change the direction of the maximum
unbalanced force from the line of stroke to the perpendicular of line of stroke. As a compromise
let a fraction ‘c’ of the reciprocating masses is balanced, such that
comer = Bib
Unbalanced force along the line of stroke
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BALANCING OF PRIMARY FORCES OF MULTI-CYLINDER IN-LINE ENGINES
The multi-cylinder engines with the cylinder centre lines in the same plane and on the same side
of the centre line of the crankshaft, are known as In-line engines. The following two conditions
must be satisfied in order to give the primary balance of the reciprocating parts of a multi-
cylinder engine :
1. The algebraic sum of the primary forces must be equal to zero. In other words, the primary
force polygon must *close ;and
2. The algebraic sum of the couples about any point in the plane of the primary forces must be
equal to zero. In other words, the primary couple polygon must close.
We have already discussed, that the primary unbalanced force due to the reciprocating masses is
equal to the component, parallel to the line of stroke, of the centrifugal force produced by the
equal mass placed at the crankpin and revolving with it. Therefore, in order to give the primary
balance of the reciprocating parts of a multi-cylinder engine, it is convenient to imagine the
reciprocating masses to be transferred to their respective crankpins and to treat the problem as
one of revolving masses.
Notes : 1. For a two cylinder engine with cranks at 180°, condition (1) may be satisfied, but this
will result in an unbalanced couple. Thus the above method of primary balancing cannot be
applied in this case.
2. For a three cylinder engine with cranks at 120° and if the reciprocating masses per cylinder are
same, then condition (1) will be satisfied because the forces may be represented by the sides of
an equilateral triangle. However, by taking a reference plane through one of the cylinder centre
lines, two couples with non-parallel axes will remain and these cannot vanish vector ally. Hence
the above method of balancing fails in this case also.
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* The closing side of the primary force polygon gives the maximum unbalanced primary force
and the closing side of the primary couple polygon gives the maximum unbalanced primary
couple
3. For a four cylinder engine, similar reasoning will show that complete primary balance is
possible and it follows that
‘For a multi-cylinder engine, the primary forces may be completely balanced by suitably
arranging the crank angles, provided that the number of cranks are not less than four’.
BALANCING OF SECONDARY FORCES OF MULTI-CYLINDER IN-LINE ENGINES
When the connecting rod is not too long (i.e. when the obliquity of the connecting rod is
considered), then the secondary disturbing force due to the reciprocating mass arises.
We have discussed in Art. that the secondary force,
This expression may be written as
As in case of primary forces, the secondary forces may be considered to be equivalent to the
component, parallel to the line of stroke, of the centrifugal force produced by an equal mass
placed at the imaginary crank of length r / 4n and revolving at twice the speed of the actual crank
(i.e. 2ὼ) as shown in Fig. 3
Thus, in multi-cylinder in-line engines, each imaginary secondary crank with a mass attached to
the crankpin is inclined to the line of stroke at twice the angle of the actual crank. The values of
the secondary forces and couples may be obtained by considering the revolving mass. This is
done in the similar way as discussed for primary forces. The following two conditions must be
satisfied in order to give a complete secondary balance of an engine :
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1. The algebraic sum of the secondary forces must be equal to zero. In other words, the secondary
force polygon must close, and
2. The algebraic sum of the couples about any point in the plane of the secondary forces must be
equal to zero. In other words, the secondary couple polygon must close.
Note : The closing side of the secondary force polygon gives the maximum unbalanced
secondary force and the closing side of the secondary couple polygon gives the maximum
unbalanced secondary couple.
BALANCING OF V-ENGINES
Consider a symmetrical two cylinder V-engine as shown in Fig. 22.33, The common crank OC is
driven by two connecting rods PC and QC. The lines of stroke OP and OQ are inclined to the
vertical OY, at an angle α as shown in Fig 5
m = Mass of reciprocating parts per cylinder,
l = Length of connecting rod,
r = Radius of crank,
n = Ratio of length of connecting rod to crank radius = l / r
ϴ= Inclination of crank to the vertical at any instant,
ὼ= Angular velocity of crank.
We know that inertia force due to reciprocating parts of cylinder 1, along the line of stroke
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and the inertia force due to reciprocating parts of cylinder 2, along the line of stroke
The balancing of V-engines is only considered for primary and secondary forces* as discussed
below :
Considering primary forces
We know that primary force acting along the line of stroke of cylinder 1
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Considering secondary forces
We know that secondary force acting along the line of stroke of cylinder 1
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BALANCING OF RADIAL ENGINES (DIRECT AND REVERSE CRANKS METHOD )
The method of direct and reverse cranks is used in balancing of radial or V-engines, in which the
connecting rods are connected to a common crank. Since the plane of rotation of the various
cranks (in radial or V-engines) is same, therefore there is no unbalanced primary or secondary
couple.
Consider a reciprocating engine mechanism as shown in Fig. 5. Let the crank OC (known as the
direct crank) rotates uniformly a ὼ radians per second in a clockwise direction. Let at any instant
the crank makes an angle ϴ with the line of stroke OP. The indirect or reverse crank OC is the
image of the direct crank OC, when seen through the mirror placed at the line of stroke. A little
consideration will show that when the direct crank revolves in a clockwise direction, the reverse
crank will revolve in the anticlockwise direction. We shall now discuss the primary and
secondary forces due to the mass (m) of the reciprocating parts at P.
Considering the primary forces
We have already discussed that primary force is m.ὼ2.r,cos ϴThis force is equal to the
component of the centrifugal force along the line of stroke, produced by a mass (m) placed at the
crank pin C. Now let us suppose that the mass (m) of the reciprocating parts is divided into two
parts, each equal to m / 2.
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It is assumed that m / 2 is fixed at the direct crank (termed as primary direct crank) pin C and
m/ 2 at the reverse crank (termed as primary reverse crank) pin C’ , as shown in Fig. 5 We
know that the centrifugal force acting on the primary direct and reverse crank
Hence, for primary effects the mass m of the reciprocating parts at P may be replaced by two
masses at C and C’ each of magnitude m/2.
Considering secondary force
We know that secondary force
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In the similar way as discussed above, it will be seen that for the secondary effects, the mass (m)
of the reciprocating parts may be replaced by two masses (each m/2) placed at D and D’ such
that OD = OD = r/4n. The crank OD is the secondary direct crank and rotates at 2ὼ rad/s in the
clockwise direction, while the crank OD’ is the secondary reverse crank and rotates at 2 ὼ rad/s
in the anticlockwise direction as shown in Fig.6
Example 1
Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg respectively. The
corresponding radii of rotation are 0.2 m, 0.15 m, 0.25 m and 0.3 m respectively and the angles
between successive masses are 45°, 75° and 135°. Find the position and magnitude of the
balance mass required, if its radius of rotation is 0.2m.
Given : m1 = 200 kg ; m2 = 300 kg ; m3 = 240 kg ; m4 = 260 kg ; r1 = 0.2 m ;
r2 = 0.15 m ; r3= 0.25 m ; r4 = 0.3 m ; ϴ1 = 0° ;ϴ2 = 45° ;ϴ3 = 45° + 75° = 120° ;
ϴ4 = 45° + 75°+ 135° = 255° ; r = 0.2 m
SOLUTION:
M=balancing mass and
ϴ= The angle which the balancing mass makes with m
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Since the magnitude of centrifugal forces are proportional to the product of each mass and its
radius , therefore
m1.r1=200 *0.2=40kg-m
m2.r2=300 *0.15=45kg-m
m3.r3=240 *0.25=60kg-m
m4.r4=260 *0.3=78kg-m
The problem may, now, be solved either analytically or graphically. But we shall solve the
problem by both the methods one by one.
Analytical method :
The space diagram is shown in fig
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Graphical method:
The magnitude and the position of the balancing mass may also be found graphically as
discussed below :
1. First of all, draw the space diagram showing the positions of all the given masses
as shown infix
2. Since the centrifugal force of each mass is proportional to the product of the mass
and radius, therefore
m1.r1 = 200 × 0.2 = 40 kg-m
m2.r2 = 300 × 0.15 = 45 kg-m
m3.r3 = 240 × 0.25 = 60 kg-m
m4.r4 = 260 × 0.3 = 78 kg-m
3. Now draw the vector diagram with the above values, to some suitable scale, as shown in
Fig. The closing side of the polygon age represents the resultant force. By measurement, we find
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that age = 23 kg-m.
4. The balancing force is equal to the resultant force, but opposite in direction as shown
in Fig. Since the balancing force is proportional to m.r,therefore
m × 0.2 = vector ea = 23 kg-m or m = 23/0.2 = 115 kg
By measurement we also find that the angle of inclination of the balancing mass (m) from
the horizontal mass of 200 kg,
ϴ= 201°
Example2. A shaft carries four masses A, B, C and D of magnitude 200 kg, 300 kg, 400 kg and
200 kg respectively and revolving at radii 80 mm, 70 mm, 60 mm and 80 mm in planes measured
from A at 300 mm, 400 mm and 700 mm. The angles between the cranks measured anticlockwise
are A to B 45°, B to C 70° and C to D 120°. The balancing masses are to be placed in planes X
and Y. The distance between the planes A and X is 100 mm, between X and Y is 400 mm and
between Y and D is 200 mm. If the balancing masses revolve at a radius of 100 mm, find their
magnitudes and angular positions.
Given : mA = 200 kg ; mB = 300 kg ; mC = 400 kg ; mD = 200 kg ; rA = 80 mm = 0.08m ; rB =
70 mm = 0.07 m ; rC = 60 mm = 0.06 m ; rD = 80 mm = 0.08 m ; rX = rY = 100 mm = 0.1 m
mX = Balancing mass placed in plane X, and
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mY = Balancing mass placed in plane.
The position of planes and angular position of the masses (assuming the mass A as horizontal)
are shown in Fig. a)and (b)respectively.
Assume the plane X as the reference plane (R.P.). The distances of the planes to the right of plane
X are taken as + vet while the distances of the planes to the left of plane X are taken as – vet. The
data may be tabulated as shown instable
The balancing masses mX and mY and their angular positions may be determined graphically as
discussed below:
1. First of all, draw the couple polygon from the data given in Table (column 6) as Shown in
Fig.(c)tosomesuitablescale.Thevector d o represents the balance Couple. Since the
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balanced couple is proportional to 0.04 mY, therefore by measurement,
2. Now draw the force polygon from the data given in Table (column 4) as shown infix. (d). The
vector eon represents the balanced force. Since the balanced force is proportional to 0.1 mX,
therefore by measurement,
mX = 355 kg
Example 3. A, B, C and D are four masses carried by a rotating shaft at radii 100,
125, 200 and 150 mm respectively. The planes in which the masses revolve are spaced 600 mm
apart and the mass of B, C and D are 10 kg, 5 kg, and 4 kg respectively.
Find the required mass A and the relative angular settings of the four masses so that the
shaft shall be in complete balance.
Given : rA = 100 mm = 0.1 m ; rB = 125 mm = 0.125 m ; rC = 200 mm = 0.2 m ; rD = 150 mm
= 0.15 m ; mB = 10 kg ; mC = 5 kg ; mD = 4 kg
SOLUTION:
The position of planes is shown in Fig. (a). Assuming the plane of mass A as the
Reference plane (R.P.), the data may be tabulated as below:
First of all, the angular setting of masses C and D is obtained by drawing the couple
polygon from the data given in Table 21.4 (column 6). Assume the positionofmass B in the
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Horizontal direction OB as shown in Fig. (b). Now the couple polygon as shown in Fig. (c) is
drawn as discussed below :
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1. Draw vector ob in the horizontal direction (i.e. parallel to OB) and equal to 0.75kg-m2, to
some suitablescale.
2. From point’s o and b, draw vectors oce and bck equal to 1.2 kg-m2 and 1.08 kg-
m2respectively. These vectors intersect act.
3. Now in Fig.(b), draw OC parallel to vector oce and OD parallel to vector bck.
By measurement, we find that the angular setting of mass C from mass B in the anticlockwise
direction, i.e.
BOC =240°
and angular setting of mass D from mass B in the anticlockwise direction, i.e.
BOD =100°
In order to find the required mass A (mA) and its angular setting, draw the force polygon to
some suitable scale, as shown in Fig. (d), from the data given in Table (column 4).
Since the closing side of the force polygon (vector do) is proportional to 0.1 mA, therefore
by measurement,
0.1 mA = 0.7 kg-m2 or mA = 7 kg
Example 4.A shaft carries four masses in parallel planes A, B, C and D in this order along its
length. The masses at B and C are 18 kg and 12.5 kg respectively, and each has an eccentricity
of 60 mm. The masses at A and D have an eccentricity of 80 mm. The angle between the masses
atBandCis100°andthatbetweenthemassesatBandAis190°,bothbeingmeasuredinthe
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same direction. The axial distance between the planes A and B is 100 mm and that between B
and C is 200 mm. If the shaft is in complete dynamic balance, determine : 1. The magnitude of
the masses at A and D ; 2. the distance between planes A and D ; and 3. the angular position of
the mass tad.
Solution. Given : mB = 18 kg ; mC = 12.5 kg ; rB = rC= 60 mm = 0.06 m ;
rA = rD = 80 mm = 0.08 m ; BOC = 100° ; BOA = 190° SOLUTION:
1. Magnitude of the masses at A and D
The position of the planes and angular position of the masses is shown in Fig. (a) and (b)
respectively. The position of mass B is assumed in the horizontal direction, i.e. along OB. Taking
the plane of mass A as the reference plane, the data may be tabulated as below :
First of all, the direction of mass D is fixed by drawing the couple polygon to some suitable
scale, as shown in Fig. (c), from the data given in Table (column 6). The closing side of the
couple polygon (vector c o ) is proportional to 0.08 mD.x. By measurement, we find that0.08
mD.x = vector c o = 0.235kg-m2
In Fig. (b), draw OD parallel to vector o to fix the direction ofmassD .......... (I)
Now draw the force polygon, to some suitable scale, as shown in Fig. (d), from the
data given in Table (column 4), as discussed below :
1. Draw vector ob parallel to OB and equal to 1.08kg-m.
2. From point b, draw vector bc parallel to OC and equal to 0.75kg-m.
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3. For the shaft to be in complete dynamic balance, the force polygon must be closed
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figure. Therefore from point c, draw vector cod parallel to OA and from point o draw vector do
parallel to OD. The vectors cod and do intersect at d. Since the vector cod is proportional to 0.08
mA, therefore by measurement
0.08 mA = vector cod = 0.77 kg-m or mA = 9.625 kg
and vector do is proportional to 0.08 mD, therefore by measurement,
0.08 mD = vector do = 0.65 kg-m or mD = 8.125 kg
2. Distance between planes A and D
From equation (I),
0.08 mD.x = 0.235 kg-m2
0.08 × 8.125 × x = 0.235 kg-m2 or 0.65 x = 0.235
=361.5 mm
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Example 6.A four crank engine has the two outer cranks set at 120° to each other, and their
reciprocating masses are each 400 kg. The distance between the planes of rotation of
adjacent cranks are 450 mm, 750 mm and 600 mm. If the engine is to be in complete primary
balance, find the reciprocating mass and the relative angular position for each of the inner
cranks. If the length of each crank is 300 mm, the length of each connecting rod is 1.2 m and the
speed of rotation is 240 r.p.m., what is the maximum secondary unbalanced force ?
Solution. Given : m1 = m4 = 400 kg ; r = 300 mm = 0.3 m ; l = 1.2 m ; N = 240 r.p.m. or = 25.14
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rad/s
Reciprocating mass and the relative angular position for each of the inner cranks
Let m2 and m3 = Reciprocating mass for the inner cranks 2 and 3 respectively, and
ϴ2 andϴ3 = Angular positions of the cranks 2 and 3 with respect to crank 1 respectively.
The position of the planes of rotation of the cranks and their angular setting are shown in Fig. (a)
and(b)respectively.Takingtheplaneofcrank2asthereferenceplane,thedatamaybetabulated as below:
Since the engine is to be in complete primary balance, therefore the primary couple polygon and
the primary force polygon must close. First of all, the primary couple polygon, as shown in Fig.
(c), is drawn to some suitable scale from the data given in Table (column 6), in order to find the
reciprocating mass for crank 3. Now by measurement, we find that
0.225 m3196kg-m or m3= 871kg
and its angular position with respect to crank 1 in the anticlockwise direction,
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ϴ3= 326°
Now in order to find the reciprocating mass for crank 2, draw the primary force polygon, as
shown in Fig. (d), to some suitable scale from the data given in Table (column 4).
Now by measurement, we find that
0.3 m2 =284 kg-m or m2 = 947kg
and its angular position with respect to crank 1 in the anticlockwise direction,
ϴ2 = 168°
Maximum secondary unbalanced force
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The secondary crank positions obtained by rotating the primary cranks at twice the angle, is
shown in Fig. (e). Now draw the secondary force polygon, as shown in Fig. ( f ), to
some suitable scale, from the data given in Table (column 4). The closing side of the polygon
shown dotted in Fig. ( f ) represents the maximum secondary unbalanced force.
By measurement, we find that the maximum secondary unbalanced force is proportional to
582 kg-m.
Maximum secondary unbalanced force 91 960N = 91.96 ken
Example 7. The cranks and connecting rods of a 4-cylinder in-line engine running at 1800
r.p.m. are 60 mm and 240 mm each respectively and the cylinders are spaced 150 mm apart. If
the cylinders are numbered 1 to 4 in sequence from one end, the cranks appear at intervals of
90° in an end view in the order 1-4-2-3. The reciprocating mass corresponding to each cylinder
is 1.5 kg. Determine : 1. Unbalanced primary and secondary forces, if any, and 2.Unbalanced
primary and secondary couples with reference to central plane of the engine.
Given :
N = 1800 r.p.m.or2 × 1800/60 = 188.52 rad/s ; r = 60 mm = 0.6 m ; l = 240 mm = 0.24 m;
m = 1.5 kg
1. Unbalanced primary and secondary forces
The position of the cylinder planes and cranks is shown in Fig. (a) and (b) respectively. With
reference to central plane of the engine, the data may be tabulated as below :
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TheprimaryforcepolygonfromthedatagiveninTable(column4)isdrawnasshowninFig.(c). Since the
primary force polygon is a closed figure, therefore there are no unbalanced primary forces.
The secondary crank positions’, taking crank 3 as the reference crank, is shown in Fig. (e). From
the secondary force polygon as shown in Fig. ( f ), we see that it is a closed Figure. Therefore
there are no unbalanced secondary forces.
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2. Unbalanced primary and secondary couples
The primary couple polygon from the data given in Table (column 6) is drawn as shown in Fig.
(d). The closing side of the polygon, shown dotted in the figure, represents unbalanced primary
couple. By measurement, we find the unbalanced primary couple is proportional to 0.19 kg-m2.
Unbalanced primary couple,
U.P.C = 0.19 × 2 = 0.19 (188.52)2 = 6752 N-m
The secondary couple polygon is shown in Fig. 22.1 (g). The unbalanced secondary couple
is shown by dotted line. By measurement, we find that unbalanced secondary couple is
proportional to 0.54 kg-m2.
Unbalanced secondary couple, = 4798 N-m
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Example 8 A five cylinder in-line engine running at 750 r.p.m. has successive cranks 144° apart,
the distance between the cylinder centre lines being 375 mm. The piston stroke is 225 mm and
the ratio of the connecting rod to the crank is 4. Examine the engine for balance of primary and
secondary forces and couples. Find the maximum values of these and the position of the central
crank at which these maximum values occur. The reciprocating mass for each cylinder
is 15 kg.
SOLUTION:
Assuming the engine to be a vertical engine, the positions of the cylinders and the cranks are
shown in Fig. (a), (b) and (c). The plane 3 may be taken as the reference plane and the crank 3 as
the reference crank. The data may be tabulated as given in the following table
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Now, draw the force and couple polygons for primary and secondary cranks as shown in Fig. (d),
(e), ( f ), and (g). Since the primary and secondary force polygons are close, therefore the engine
is balanced for primary and secondary forces
Maximum unbalanced primary couple
We know that the closing side of the primary couple polygon [shown dotted in Fig. (e)] gives the
maximum unbalanced primary couple. By measurement, we find that maximum unbalanced
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primary couple is proportional to 1.62 kg-m2.
Maximum unbalanced primary couple,
U.P.C. = 1.62 × 2 = 1.62 (78.55)2 = 9996 N-m
We see from Fig. (e) [shown by dotted line] that the maximum unbalanced primary couple
occurs when crank 3 is at 90° from the line of stroke.
Maximum unbalanced secondary couple
We know that the closing side of the secondary couple polygon [shown dotted in Fig. (g)] gives
the maximum unbalanced secondary couple. By measurement, we find that maximum
unbalanced secondary couple is proportional to 2.7kg-m2.
Maximum unbalanced secondary couple. = 4165 N-m
We see from Fig. (g) that if the vector representing the unbalanced secondary couple (shown by
dotted line) is rotated through 90°, it will coincide with the line of stroke. Hence the original
crank will be rotated through 45°. Therefore, the maximum unbalanced secondary couple occurs
when crank 3 is at 45° and at successive intervals of 90° (i.e. 135°, 225° and 315°) from the line
of stroke.
Example 9The firing order in a 6 cylinder vertical four stroke in-line engine is 1-4-2-6-3-5. The
piston stroke is 100 mm and the length of each connecting rod is 200 mm. The pitch distances
between the cylinder centre lines are 100 mm, 100 mm, 150 mm, 100 mm, and 100 mm
respectively. The reciprocating mass per cylinder is 1 kg and the engine runs at 3000 r.p.m.
Determine the out-of-balance primary and secondary forces and couples on this engine, taking a
plane midway between the cylinder 3 and 4 as the reference plane.
Given : L = 100 mm or r = L / 2 = 50 mm = 0.05 m ; l = 200 mm ; m = 1 kg ;
N = 3000 r.p.m.
SOLUITON:
The position of the cylinders and the cranks are shown in Fig. (a), (b) and (c). With the reference
plane midway between the cylinders 3 and 4, the data may be tabulated as given in the following
table :
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Now, draw the force and couple polygons for the primary and secondary cranks as shown in Fig.
(d), (e), ( f ) and (g). From Fig. (d) and (e), we see that the primary and secondary force polygons
are closed figures, therefore there are no out-of-balance primary and secondary forces. Thus the
engine is balanced for primary and secondary forces. Also, the primary and secondary couple
polygons, as shown in Fig. ( f ) and (g) are closed figures, therefore there are no out-of-balance
primary and secondary couples. Thus the engine is balanced for primary and secondary couples.
Example10.A vie-twin engine has the cylinder axes at right angles and the connecting rods
operate a common crank. The reciprocating mass per cylinder is 11.5 kg and the crank radius is
75 mm. The length of the connecting rod is 0.3 m. Show that the engine may be balanced for
primary forces by means of a revolving balance mass. If the engine speed is 500 r.p.m. What is
the value of maximum resultant secondary force ?
SOLUTION:
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Example 11 The reciprocating mass per cylinder in a 60° V-twin engine is 1.5 kg. The stroke
and connecting rod length are 100 mm and 250 mm respectively. If the engine runs at 2500
r.p.m., determine the maximum and minimum values of the primary and secondary forces. Also
find out the crank position corresponding these values.
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UNIT-5
VIBRATION
Free, forced and damped vibrations of single degree of freedom systems – Force
transmitted to supports – Vibration isolation – Vibration absorption – Tensional vibration of
shaft – Single and multi rotor systems – Geared shafts – Critical speed of shaft
INTRODUCTION
When elastic bodies such as a spring, a beam and a shaft are displaced from the equilibrium
position by the application of external forces, and then released, they execute a vibratory
motion. This is due to the reason that, when a body is displaced, the internal forces in the form
of elastic or strain energy are present in the body. At release, these forces bring the body to its
original position. When the body reaches the equilibrium position, the whole of the elastic or
strain energy is converted into kinetic energy due to which the body continues to move in the
opposite direction. The whole of the kinetic energy is again converted into strain energy due to
which the body again returns to the equilibrium position. In this way, the vibratory motion is
repeated indefinitely.
Terms Used in Vibratory Motion
The following terms are commonly used in connection with the vibratory motions
1. Period of vibration or time period. It is the time interval after which the motion is repeated
itself. The period of vibration is usually expressed in seconds.
2. Cycle. It is the motion completed during one time period.
3. Frequency. It is the number of cycles described in one second. In S.I. units, the frequency is
expressed in hertz (briefly written as Hz) which is equal to one cycle per second.
TYPES OF VIBRATORY MOTION
The following types of vibratory motion are important from the subject point of view :
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1. Free or natural vibrations. When no external force acts on the body, after giving it an initial
displacement, then the body is said to be under free or natural vibrations. The frequency of the
free vibrations is called free or natural frequency.
2. Forced vibrations. When the body vibrates under the influence of external force, then the
body is said to be under forced vibrations. The external force applied to the body is a periodic
disturbing force created by unbalance. The vibrations have the same frequency as the applied
force.
Note : When the frequency of the external force is same as that of the natural vibrations,
resonance takes place.
3. Damped vibrations. When there is a reduction in amplitude over every cycle of vibration, the
motion is said to be damped vibration. This is due to the fact that a certain amount of energy
possessedbythevibratingsystemisalwaysdissipatedinovercomingfrictionalresistancestothe
motion.
Types of Free Vibrations
The following three types of free vibrations are important from the subject point of view :
1. Longitudinal vibrations, 2. Transverse vibrations, and 3. Tensional vibrations.
Consider a weightless constraint (spring or shaft) whose one end is fixed and the other end
carrying a heavy disc, as shown in fig. This system may execute one of the three above
mentioned types of vibrations.
1. Longitudinal vibrations. When the particles of the shaft or disc moves parallel to the axis of
the shaft, as shown in then the vibrations are known as longitudinal vibrations.
In this case, the shaft is elongated and shortened alternately and thus the tensile and compressive
stresses are induced alternately in the shaft.
2. Transverse vibrations. When the particles of the shaft or disc move approximately
perpendicular to the axis of the shaft, as shown in Fig. 23.1 (b), then the vibrations are known
as transverse vibrations. In this case, the shaft is straight and bent alternately and bending
stresses are induced in the shaft.
3. Tensional vibrations*. When the particles of the shaft or disc move in a circle about the axis
of the shaft, as shown in Fig. 23.1 (c), then the vibrations are known as Tensional vibrations.
In this case, the shaft is twisted and untwisted alternately and the Tensional shear stresses are
induced in the shaft.
NATURAL FREQUENCY OF FREE LONGITUDINAL VIBRATIONS
The natural frequency of the free longitudinal vibrations may be determined by the following
three methods :
1. Equilibrium Method
Consider a constraint (i.e. spring) of negligible mass in an unstrained position, as shown in
Let s = Stiffness of the constraint. It is the force required to produce unit displacement in
the direction of vibration. It is usually expressed inn/m.
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m = Mass of the body suspended from the constraint in kg,
W = Weight of the body in nektons = mug,
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2. Energy method
We know that the kinetic energy is due to the motion of the body and the potential energy is
with respect to a certain datum position which is equal to the amount of work required to move
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the body from the datum position. In the case of vibrations, the datum position is the mean or
equilibrium position at which the potential energy of the body or the system is zero. In the free
vibrations, no energy is transferred to the system or from the system. Therefore the summation of
kinetic energy and potential energy must be a constant quantity which is same at all the times. In
other words,
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FREQUENCY OF FREE DAMPED VIBRATIONS (VISCOUS DAMPING)
We have already discussed that the motion of a body is resisted by frictional forces. In vibrating
systems,theeffectoffrictionisreferredtoasdamping.Thedampingprovidedbyfluidresistance is
known as viscous damping. We have also discussed that in damped vibrations, the amplitude of
the resulting vibration gradually diminishes. This is due to the reason that a certain amount of
energy is always dissipated to overcome the frictional resistance. The resistance
to the motion of the body is provided partly by the medium in which the vibration takes place and
partly by the internal friction, and in some cases partly by a dash pot or other external damping
device.
Consider a vibrating system, as shown in which a mass is suspended from one end of the spiral
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spring and the other end of which is fixed. A damper is provided between the mass and the rigid
support.
Let m = Mass suspended from the spring,
s = Stiffness of the spring,
x = Displacement of the mass from
the mean position at time t,
δ= Static deflection of the spring
= mug/s, and
c = Damping coefficient or the damping
force per unit velocity.
Since in viscous damping, it is assumed that the frictional resistance to the motion of the body is
directly proportional to the speed of the movement, therefore
Damping force or frictional force on the mass acting in opposite direction to the motion of the
mass
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3. WHEN THE ROOTS ARE EQUAL (CRITICALDAMPING)
If (c\2m)2=s\m then the radical becomes zero and the two roots k1 and k2 are equal. This is a
case of critical damping. In other words, the critical damping is said to occur when frequency of
damped vibration (fd) is zero (i.e. motion is a periodic). This type of damping is also avoided
because the mass moves back rapidly to its equilibrium position, in the shortest possible time.
For critical damping, equation (ii) may be written as
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DAMPING FACTOR OR DAMPING RATIO
The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc) is
known as damping factor or damping ratio. Mathematically,
The damping factor is the measure of the relative amount of damping in the existing system
with that necessary for the critical damped system.
Logarithmic Decrement
It is defined as the natural logarithm of the amplitude reduction factor. The amplitude
reduction factor is the ratio of any two successive amplitudes on the same side of the mean
position. If x1 and x2 are successive values of the amplitude on the same side of the mean
position,
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NATURAL FREQUENCY OF FREE TORSIONAL VIBRATIONS:
Consider a shaft of negligible mass whose one end is fixed and the other end carrying a disc as
shown in fig
Let
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If a mass whose mass moment of inertia is equal to IC/ 3 is placed at the free end and the
constraint is assumed to be of negligible mass, then Total kinetic energy of the constraint
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Hence the two systems are dynamically same. Therefore the inertia of the constraint may be
allowed for by adding IC/ 3 to the mass moment of inertia I of the disc at the free end. From the
above discussion, we find that when the mass moment of inertia of the constraint IC and the mass
moment of inertia of the disc I are known, then natural frequency of vibration,
k = Radius of gyration of rotor, and
I = Mass moment of inertia of rotor = make2
A little consideration will show that the amplitude of vibration is zero at A and maximum at B,
as shown in . It may be noted that the point or the section of the shaft whose amplitude of
tensional vibration is zero, is known as node. In other words, at the node, the shaft remains
unaffected by the vibration.
FREE TORSIONAL VIBRATIONS OF A TWO ROTOR SYSTEM
Consider a two rotor system as shown in fig. It consists of a shaft with two rotors at its ends. In
this system, the tensional vibrations occur only when the two rotors A and B move in opposite
directions i.e. if A moves in anticlockwise direction then B moves in clockwise direction at
the same instant and vice versa. It may be noted that the two rotors must have the same
frequency. We see from Fig. that the node lies at point N. This point can be safely assumed as a
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fixed end and the shaft may be considered as two separate shafts N P and N Q each fixed to one
of its ends and carrying rotors at the free ends.
l = Length of shaft,
lA = Length of part NP
i.e. distance of node from rotor A,
lB = Length of part NQ, i.e. distance
of node from rotor B,
IA = Mass moment of inertia of rotor,
IB = Mass moment of inertia of rotor B,
d = Diameter of shaft,
J = Polar moment of inertia of shaft, and
C = Modulus of rigidity for shaft material.
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Natural frequency of tensional vibration for rotor A,
From equations (iii) and (iv), we may find the value of lA and lB and hence the position of node.
Substituting the values of lA or lB in equation (I) or (ii), the natural frequency of tensional
vibration for a two rotor system may be evaluated.
FREE TORSIONAL VIBRATIONS OF A THREE ROTOR SYSTEM
Consider a three rotor system as shown is Fig. 24.6 (a). It consists of a shaft and three rotors A, B
and C. The rotors A and C are attached to the ends of a shaft, whereas the rotor B is attached in
between A and C. The torsional vibrations may occur in two ways that is with either one node or
two nodes. In each case, the two rotors rotate in one direction and the third rotor rotates in
opposite direction with the same frequency. Let the rotors A and C of the system, as shown in
fig rotate in the same direction and the rotor B in opposite direction. Let the nodal points or
nodes of such a system lies at N1 and N2 as shown in Fig. b). As discussed in Art., the shaft may
be assumed as a fixed end at the nodes.
l1 = Distance between rotors A and B,
l2 = Distance between rotors B and C,
lA = Distance of node N1 from rotor, lC
= Distance of node N2 from rotor C, IA
= Mass moment of inertia of rotor, IB =
Mass moment of inertia of rotor,
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IC = Mass moment of inertia of rotor C,
d = Diameter of shaft,
J = Polar moment of inertia of shaft, and
C = Modulus of rigidity for shaft material.
Natural frequency of tensional vibrations for rotor A,
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On substituting the value of lA from equation (iv) in the above expression, a quadratic equation
in lC is obtained. Therefore, there are two values of lC and correspondingly two values of lA.
One value of lA and the corresponding value of lC gives the position of two nodes. The
frequency obtained by substituting the value of lA or lC in equation (i) or (iii) is known as
two node frequency. But in the other pair of values, one gives the position of single node and the
other is beyond the physical limits of the equation. In this case, the frequency obtained is known
an fundamental frequency or single node frequency
It may be noted that
1. When the rotors A and B rotate in the same direction and the rotor C in the opposite
direction, then the tensional vibrations occur with a single node, as shown in Fig. (b). In this case
lA > l1i.e. the node lies between the rotors B and C, but it does not give the actual value of the
node.
2. When the rotors B and C rotate in the same direction and the rotor A in opposite direction,
then the tensional vibrations also occur with a single node as shown in Fig. (c). In this case lC >
l2i.e. the node lies between the rotors A and B, but it does not give the actual value of the node.
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3. When the amplitude of vibration for the rotor A (a1) is known, then the amplitude of rotor
B,
TORSIONALLY EQUIVALENT SHAFT
we have assumed that the shaft is of uniform diameter. But in actual practice, the shaft may
have variable diameter for different lengths. Such a shaft may, theoretically, be replaced by an
equivalent shaft of uniform diameter.
Consider a shaft of varying diameters as shown in Fig. (a). Let this shaft is replaced by an
equivalent shaft of uniform diameter d and length l as shown in Fig. (b).These two shafts must
have the same total angle of twist when equal opposing torques T are applied at their opposite
ends.
d1, d2 and d3 = Diameters for the lengths l1, l2 and l3 respectively,
1, 2 and 3 = Angle of twist for the lengths l1, l2 and l3 respectively,
= Total angle of twist, and
J1 J1 and J3 = Polar moment of inertia for the shafts of diameters d1, d2and
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d3 respectively.
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Let the node of the equivalent system lies at N as shown in fig (c) then the natural frequency of
Tensional vibration of rotor A,
and, natural frequency of the Tensional vibration of rotor B,
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Example 1. An electric motor is to drive a centrifuge, running at four times the motor speed
through a spur gear and pinion. The steel shaft from the motor to the gear wheel is 54 mm
diameter and L meter long ; the shaft from the pinion to the centrifuge is 45 mm diameter and
400 mm long. The masses and radii of gyration of motor and centrifuge are respectively 37.5 kg,
100 mm ; 30 kg and 140mm.
Neglecting the inertia effect of the gears, find the value of L if the gears are to be at the node for
tensional oscillation of the system and hence determine the frequency of Tensional oscillation.
Assume modulus of rigidity for material of shaft as 84 GN/m2.
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