Alem_B_mechanism_book.pdf

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Contents lll
5.2.2 Dependent Coordinates 119
i) Relative Coordinates 119
ii) Reference-Point Coordinates 122
iii) Natural Coordinates 125
iv) Mixed Coordinates 127
5.3 Constraint Equations 127
5.4 Kinematic Analysis 128
5.4.1 The Initial Position Problem 128
5.4.2 Velocity Analysis 129
5.5 Methods for Solving the Position, Velocity and Acceleration Equations 130
.lb •
5.5.1 Position Equation 130
5.5.2 Velocity and Acceleration Analysis 135
~ . - ,
Chapter € ~~S- .... . .......... . .·. . ................... . .. . . . . . .... ... . 146
6.1 Classification of Followers 146
6.1.1 Classification ofFollowers Based on Surface ofContact 147
6. 1.2 Classification of Followers Based on Type of Follower Motion 147
6. 1.3 Classification of Followers Based on Follower Line of Motion 148
6.2 Classification of Cams 148
6.3 Graphical Design ofCam Curves 149
6.3 .1 Disc Cam with a Flat-Faced Radial Follower 149
6.3.2 Disc Cam with a Radial Roller Follower 150
6.3.3 Disc Cam with Oscillating Follower 151
6.3.4 Positive-Return Cams 152
6.4 Nomenclature 153
6.5 Displacement Diagram 154
6.6 Types ofFollow~r Motion 154
6.6.1 Unifonn Motion 154
6.6.2 Modified Uniform Motion 156
6.6.3 Parabolic Motion 156
6.6.4 Simple Harmonic Motion 159
6.6.5 Cycloidal Motion 161
6.6.6 Advanced Cam Curves 163
6.6.7 Nonstandard Cam Curves 165
6.7 Analytical Cam Design 176
6.7.1 Disc Cam with Radial Flat-Faced Follower 176
6.7.2 Disc Cam with Radial Roller Follower 179
6.7.3 Maximum Pressure Angle 185
6.8 Tangent Cam with Reciprocating Roller Follower 189
6.8.1 Velocity and Acceleration Analysis ofTangent Cams 190

Contents vu
14.9 Balance ofV-Engines
14.10 Balance of Four-Bar Linkages
417
422
Chapter 15 GYROSCOPES ........ .... . .. ... . . . . . . ...... .-.............. 430
15.1 Precessional Motion
15.2 Gyroscopic Couple
15.3 Precessional Motion of a Thin Rod Rotating in the Vertical Plane about
430
431
a Horizontal Axis through Its Center 433
15.4 General Case: A Body Rotating mfd ACcelerating Simultaneously
about Each ofthe Principal Axes 435
'
15.5 Typical Examples ofthe Appl~cation ofPrecessional Motion 437
15.5.1 Effect ofGyrosco{t~ Motion on a Four-Wheel Drive Moving
along a Curved Path 437
14.5.2 Effect of Gyroscopic Motion on a Ship during Steering and
Pitching 443
15.5.3 Effect of Gyroscopic Motion on an Aircraft during Change of
Course 448
REFERENCES ......... .. ...... .. ... . . . ..... . . . .... . ................ . .. 454
INDEX ........ ... .. . ........ . .... .... .. .................... . . ......... 456
...

IV l'Y!eclzanisms ofMachinery
Chapter 7 UNIVERSAL JOINTS
.. . ....... . .. .... .. ... . ................. 200
7.1 Velocity Ratio ofShafts
7.2 Polar Angular Velocity Diagram
7.3 Coefficient ofSpeed Fluctuation
6.4 Angular Acceleration ofDriven Shaft
6.5 Double Hooke's Joint
200
203
204
205
206
Chapter ~ GOVERNORS ....
. . ........ . . .. .. . .... . . . .. .. . ... . ..... . 213
8.I Classification ofGovernors
8.1.1 Centrifugal Governors (Loaded Governors)
8.1.2 Governor Characteristics
8.1.2.1 Controlling force
8.1.2.2 Stability
8.1.2.3 Sensitivity
8.1.2.4 Isochronous Governors
8.1.2.5 Power ofa Governor
8.1 .2.6 Effort ofa Governor
8. 1.3 The Porter Governor
8.1.3.I Controlling Force of Porter Governor
8.1.3.2 Effort ofPorter Governor
8.1.3.3 Power ofthe Porter Governor
8.1.3.4 Effect ofFriction on the Porter Governor
8.1.4. Hartnel Governor
8.1.4.1 Controlling Force and Stability ofthe Hartne!
Governor
8.2 Shaft Governors
8.2.1 The Centrifugal Shaft Governor (Flywheel Type)
8.2.1.1 Condition ofisochronism
8.2.1.2 EffectofAngular Acceleration
8.2.2 Inertia Governors
.I
213
213
214
214
21.5
217
218
218
219
2!9
222
223
224
225
229
23 1
239
240
241
241
246
apter 9 GEAR TRAINS ............ . . ....... . . . . . . ....... . . . . ... . .. . . 256
9.1 Angular Velocity Ratio 256
9.2 Types ofGear Trains 259
9.3 Reverted Gear Train 259
9.4 Planetary Gear Trains 261
9.5 Methods ofAnaly~is ofPlanetary Gear Trains 262
9.5.1 Solution ofPlanetary Gear Trains by Formula Method 262
9.5.2 Solution ofPlanetary Gear Trains by Tabulation Method 266

I'
PREFACE
The textbook is an amplification and an upgrading ofan existing Teaching Material o~
the same subject by the same author which has been in use for over a decade in the
Department of Mechanical Engineering, Faculty of Technology, Addis Ababa
University. The textbook is based on the course Mechanics III - A1echanisms of
A1achinery offered to Mechanical Engineering students at the junior/senior level.
The coverage includes both analysis and synthesis of mechanisms. The primary
objective of writing the book is to equip mechanical engineering students with the
necessaryth~oretical backgroundto understand designtheory and analysis ofmachines.
Secondly, it aims to furnish practical applications ofmechanisms thereby introducing
the student to various mechanisms. With these two objectives in mind, it is envisaged
that the book vvill be ofsome help in assisting students and practicing engineers alike
in the design ofmechanisms and machines, beside being used as a standard text for the
course in mechanisms of machinery at the junior level.
The subject matter discussed in the textbook includes Introduction to Computer
Methods for Kinematic and Dynamic Analysis ofMulti-Body Systems, Gyroscopes,
Balancing ofRotating and Reciprocating Machines. The introduction of computer
methods for analysis ofmechanisms will help the studentto analyze mechanisms for the
whole cycle avoiding the phase-by-phase analysis. The treatment ofgyroscopic motion
will be useful to students in that they will be able to include gyroscopic effects in the
design of mechanisms which often arise in practice. The inclusion of balancing of
rotating and reciprocating machines will assist students in the analysis and design of
engine parts. For a one-semester undergraduate course, chapters 5, 11 and 15 can be
omitted and no discontinuity in the treatment ofthe subject matter will be felt.

x Mechanisms ofMachinery
The author has benefitted from discussions with colleagues and students who have used
the Teaching Material on the same subject previously and made valuable comments
regarding the presentation and formatting ofthe text and expresses his appreciation to
. -
all who have contributed in one way or the other. Special thanks are due to Dr.-Ing.
Leul Fisseh~ Associate Prof. ofMechanical Engineering, Faculty ofTechnology, Addis
Ababa University anct to Prof. Don Streit~ Professor of Mechanical Engineering,
Pennsylvania State University, for reading the manuscript and making valuable
comments and suggestions. Their input has been very helpful in the preparation ofthe
book. Special thanks are also in order to W/o Mestawet Worku who has typed the
manuscript. The author also thanks Ato Haftay Hailu who came up withthe formatting
of the sketches included in the text in AutoCAD. Needless to say, the author is
indebted to the very many authors in the same subject whom he has freely referenced.
The author is grateful to the Department of Mechanical Engineering, Faculty of
Technology, for making it possible to write the book. Thanks are also due to the
Research and Publications Office, Addis Ababa University, for granting financial
assistance for word processing ofthe manuscript.
.!
Alem Bazezew
July 2001

Chapter 1
INTRODUCTION
The subject Mechanisms ofMachinery deals with the study of motion and forces in
machinery devices that provide a wide vmecy offunctions. The subject matter treated
inhere includes synthesis and analysis ofmachinery.
1.1 BASIC DEFINITIONS
Mechanisms: Mechanisms is commonly defined as the division of machine design '
which is concerned with the kinematic design of links, cams, gears, gear trains etc.
Kinematic design is the design on the basis ofmotion requirements.
Kinematic chain: Kinematic chain the arrangement of links in a chain fashion which
permits relative motion between the links. Kinematic chains are classified into two: i)
open-loop kinematic chains in which the links are composed without closed branches,
and ii) closed-loop kinematic chains in which the links form closed loops.
A mechanism (Multibody system): A mechanism is a combination ofrigid bodies, or
links, so formed and connected that they may move upon each other with a definite
relative motion. The resistant bodies are connected by movable joints or pairing
elements. A mechanism is obtained by fixing one ofthe links of a kinematic chain to
the ground or frame. For example, in the slider-crank mechanism shown in Fig. 1.1,
link I is the ground or frame which is stationary, link 2 is the crank, link 3 the
connectingrodand link 4 the slider. The slider crank mechanism is commonly used in
internal combustion engines.
y
A
X
Fig. 1.1 Slider-crank mechanism
A machine: A machine is a mechanism or a collection ofmechanisms which transmits
force from the source of power to the resistance to be overcome. An example of a

2 Mechanisms ofMachinery
machine is an internal combustion engine which comprises a number ofmechanisms
like the slider-crank mechanism, the cam-shaft mechanism, and the flywheeL t'o
mention a few among others.
A link: A link is a rigid body having two or more pairing elements by means ofwhich
it may be connected to other bodies for the purpose oftransmitting force or motion. A
simple link has two pairing elements whereas a compound link has more than two
pairing elements. Figs. 1.2 (a) and (b) show a simple and a compound link, respectively.
(b)
Fig. 1.2 {a) Simple link, (b) Compoundlink
KinematicPair:A kinematicpair is a pair ofelements orrigid bodies, permanentlykept
in contact, so that there exists a relative movement between these elements. Kinematic
pairs arejoined bypairingelements which allowthe relative motion. In the slider-crank
mechanism shown in Fig. 1.1, links 1 and 2, 2 and 3, 3 and 4, are kinematic pairs
joined by pin joints; links 4 and 1 also form a kinematic pair joined by the sliding
element. A kinematic pair can have a maximum of five degrees of freedom and a
minimum of one degree of freedom. When the members are connected by surface of
contact the connection is known as a lower pair. Ifthe connection is a point or a line
contact, it is known as a higherpair.
A kinematic pair orsimply ajoint, permits certain degrees offreedom ofrelative motion
and prevents others. Class I joints allow one degree of freedom, Class II allows two
degrees offreedom, Class III allows three degrees of~eedom and so on.
A turningpair, revolutejoint, orpinjoint permits one link or member to rotate about
an axis in a single plane relative to the other, tqus 1Jle revolute joint has one degree of
freedom rotation about an axis. A prismaticjoint allows translatory motion along one
axis, thus has one degree offreedom. A cylindrical pair has two degrees offreedom
and allows rotational and translational motions about an axis. A sphericaljoint has
three degrees offreedom and allows rotation in space about three axes.
Inversion: If, in a mechanism, the link which was originally fixed is allowed to move
and another link becomes fixed, the mechanism is said to be inverted. The inversion
ofa mechanism does not change the motion ofits links relative to each other, but does

Introduction
change their absolute motion. The number ofpossible kinematic inversions is equal to
the number of links in a mechanism.
1.2 MOTION
Motioncanbe thought ofas a timeseries ofdisplacementsbetweensuccessive positions
ofa point or a particle. The position ofa point is defined as the vector from the origin
ofa specified reference coordinate system to the point.
1.2.1 Types of motion
a) Rectilinear motion: When a body_moves along a straight line·and does not rotate, it
is said to undergo rectilinear moti;Jn.
b) Plane motion: When the motion ofa body is confined to one plane only, the motion ·
is said to be plane motion. Plane motion can be either rectilinear or curvilinear.
When a body moves along a curved path, it is said to have curvilinear motion. Plane
motion can also be translatory, rotary or a combination ofboth to give general plane
motion.
c) Helical motion: when a body moves so that each point ofthe body has motion of
rotation
about a fixed axis and at the same time has translation parallel to the axis, the body is
said to undergo helical motion.
d) Spherical motion: When a body moves so that each point on the body has motion
about a fixed'point and remains at a constant distance from the fixed point, the motion
is said to
be"spherical motion.
e) Spacial motion: When the motion of a body is not confined to a plane, the body is
said to have spacial motion. Helical and spherical motion are special cases of space
motion.
1.2.2 Transmission of Motion
Motion is transmitted from one member to another in three ways:
a) by direct contact between two members;
b) through an intermediate link or a connecting rod;
c) by aflexible connector such as belt or chain.

In the transmission of motion one element of the mechanism must be a driver an<
another element must be a driven element or a follower.
Cycle: When parts ofa mechanism hf!.ve passedthrough ail possiblepositions accordin~
to prescribed laws and have assumed their original positions, they are said to have
completed a cycle ofmotion.
Period: A period is the time required to complete a cycle ofmotion.
Phase: A phase is the relative instantaneous position ofa mechanism at a given instant
during a cycle.
1.3 COORDINATE SYSTEMS
Coordinates are mathematical ways or models that describe the position and motion of;
a dynamic system or a mechanism. These are sets ofparameters selected so as to define
the position, velocity and acceleration ofa dynamical system at all times. There are two·
types of coordinates in use:
1.3.1 Independent Coordinates
These are the minimum number ofcoordinates that describe the position ofa system.
The number ofindependent coordinates is equal to the degree offreedom ofthe system.
B A
1 /
"
(a) (b) B
Fig. 1.3 Description ofa mechanism using independent coordinates
Independent coordinates are in general, not acceptable, because they do not describe the
position of a mechanism unequivocally. As can be noted in Fig. 1.3, for the same
position ofthe driver given by the angle e,the positions ofthe follower and connecting
rod are different; i.e. the linkage is not defined uniquely.

In the transmission of motion one element of the mechanism must be a driver and
another element must be a driven element or a follower.
Cycle: When parts ofa mechanismhfl.ve passed through ail possible positions according
to prescribed laws and have assumed their original positions, they are said to have
completed a cycle ofmotion.
Period: A period is the time required to complete a cycle ofmotion.
Phase: A phase is the relative instantaneous position ofa mechanism at a given instant
during a cycle.
1.3 COORDINATE SYSTEMS
Coordinates are mathematical ways or models that describe the position and motion of
a dynamic system or a mechanism. These are sets ofparameters selected so as to define
the position, velocity and acceleration ofa dynamical system at all times. There are two
types of coordinates in use:
1.3.1 Independent Coordinates
These are the minimum number ofcoordinates that describe the position ofa system.
The number ofindependent coordinates is equal to the degree offreedom ofthe system.
B A
(a) (b) B
Fig. 1.3 Description ofa mechanism using independent coordinaies
Independent coordinatesare in general, not acceptable, becausethey do not describe the
position of a mechanism unequivocally. As can be noted in Fig. 1.3, for the same
position of the driver given by the angle e' the positions ofthe follower and connecting
rod are different; i.e. the linkage is not defined uniquely.

Introduction 5
1.3.2 Dependent Coordinates
These are the number ofcoordinates (which are not independent) interrelated through
certain independent equations known as constraint equations. The number of
dependent coordir:ates is larger than the degree of freedom. For a system defined by
using n coordinates and paving m independent constraint equations, the number of
degrees of freedom f is given by
f =n - m (1.1)
1.4 DEGREE OF FREEDOM
The minimwn number of coordina~s required to fully describe the configuration of
mechanism is called the number ofdegrees offreedom. The degree of freedom of a
linkage in a plane is determined using the mobility criterion proposed by Gruebeler:
f =3(n - 1) - 2P1
- P2
where f = number ofdegrees offreedom ofthe linkage;
n = number of links;
P1 = number ofkinematic pairs having one degree offreedom;
P2 = number of kinematic pairs having two degrees of freedom.
For example, in the slider-crank mechanism shown in Fig. 1.1,
Therefore, the number ofdegree of freedom of the system is
.··
(1.2)

Chapter 2
LINKAGES
A linkage is a kinematic chain in which one of the links is fixed to the ground which
usually is the frame. A linkage permits relative motion between its links and may have
one or more degrees offreedom. A linkage with zero or negative degree offreedom is
a structure which does not allow any relative motion between the links.
There are very many linkages formed by different connections of members. The
members oflinkages are c01mected so as to produce the motion required by the design.
Inthis chapter we will consider basic linkages that have frequent applications in various
machines.
2.1 FOUR-BAR LINKAGE
The kinematic linkage shown in Fig. 2.1 is known as the four-bar linkage. The
conventional numbering system is to label the ground or frame as link I, and then to
number links clockwise around the mechanism "loop" as shown in Fig. 2.1.
B
./
_ r
_l
1
Fig. 2.1 Four-bar linkage
Link 1 is the frame or ground; generally, it is stationary.
Link 2 is the driver which may rotate or oscillate.
Link 3 is the coupler, sometimes known as the connecting rod. This link
undergoes general plane motion.
Link 4 is the follower or driven element, which may rotate or oscillate
depending on the rotary or oscillatory motion of link 2, and on link
dimensions.
These four links are joined by four revolutejoints.
Four-bar linkages are able to produce a variety ofnon-uniform motion and can transmit

Linkages 7
large forces. The links ofa four-bar mechanism should be proportioned in such a way
that locking is avoided.
B
Fig. 2.2 f:ocking ofjour-bar mechanism
r
For example, in Fig. 2.2, iflink 2 is the driving link, position A 'B '0 4 shows the case of
locking. For such a position link 4 can move in any ofthe two directions as indicated
in the figure.
2.1.1 Transmission angle
The angle y between the coupler 3 and the output link 4 (follower) is called the
transmission angle. The transmission angle y is shown in Fig. 2.3.
B
...........
2 ........... z
...........
...........
...........
___ _ _r_
t __ __ _
Fig. 2.3 Transmission angle
The equation for the transmission angle can be derived as follows. F!om Fig. 2.3
2 _ 2 2
2 8
_ 2 . 2
2
z - r1 + r2 - r1
r2
cos 2
- r3 -r-r4 - r3
r4
cosy
from which we obtain the transmission angle
(2.1)
In general, for good force transmission to the output link, the transmission angle should .

' I)
'/ 8 Mechanisms ofMachinery
be in the range 400 < y < 1400.
' ' /
-t--
I
/
B
Fig. 2.4 Range oftransmission angle
2.1.2 Motion of a four-bar mechanism
The type ofmotion that a four-bar linkage executes depends on the proportional size of
its links. There are three basic types ofmotion which a four-bar linkage can produce.
These basic types ofmotion are characterized by the terms crank-rocker to indicate that
link 2 rotates and link 4 oscillates; double crank to indicate that both the driver and
follower rotate; and double-rocker to indicate that both the driver and follower oscillate
through certain angles.
Grashoff's Law
The type ofmotion executed by a four-bar linkage can be determined by considering th~
arrangement of the links and their proportions. To determine whether a four-bar link
will operate as a crank-rocker, a double-crank Oli a double-rocker, Grashoff's law is
applied. Grashoffs law is stated as follows:
i) If the sum ofthe lengths ofthe longest and shortest links is less than or equal to the
sum ofthe lengths ofthe other two links, then
a) two different crank rockers will be formed when the shortest link is the crank
and either ofthe adjacent links is the fixed link;
b) a double crank will be formed when the shortest link is the fixed link;
c) a double rocker will be formed when the link opposite the shortest link is the
fixed link.
ii) Ifthe sum ofthe lengths ofthe longest and shortest links is greater than the sum

Linkages 9
of the lengths of the other two, only a double-rocker mechanism will be
formed.
2.1.3 Variations of the four-bar linkage
The four barlinkage may have various forms. Fig. 2.5 shows somevariations ofthe four
bar mechanism.
A
' / .
'- I · / 1 I
,_r..., //
---
(a)
(b)
B
Fig. 2.5 Variations ofthefour-bar mechanism
:/I . 2 SLIDER-CRANK MECHANISM:
A slider-crankmechanism is basically a four-bar mechanism with three revolutejoints,
or tum.in.g pairs and a prismatic joint or a sliding pair. Fig. 2.6 shows the basic slider
c;aflk mechanism.
In the slider-crank mechanism, commonly

ij
l
I
l
--
Linkages 9
of the lengths of the other two, only a double-rocker mechanism will be
formed.
2.1.3 Variations of the four-bar linkage
The four bar linkage may have various forms. Fig. 2.5 shows some variations ofthe four
bar mechanism.
A
......
"
y
04- 1··
't"- - I
'-, I ·, / / 1
- r:..--....._ ___ ..,
(a)
(b)
B
Fig. 2.5 Variations ofthejour-bar mechanism
/' ·. 2 SLIDER-CRANK MECHANISM:
A slider-crankmechanism is basically a four-bar mechanism with three revolute joints,
or tum.!Hg pairs and a prismatic joint or a sliding pair. Fig. 2.6 shows the basic slider
- cmn.k mechanism.
In the slider-crank mechanism, commonly

link I is theframe, considered t9 be fixed;
link 2 is the crank which is the driver;
link 3 is the connecting rod, the link between the driver and the follower;
link 4 is the slider which is the driven element.
The slider crank mechanism converts rotary motion into reciprocating motion and vice-
versa. Commonlythe slider-crankmechanism is applied ininternal combustionengines,
where in the application the slider, link 4, is the driver and the crank, link 2, is driven.
During a cycle there are two dead points A ' and A " in which the crank and the
connecting rod are in line. At the dead positions the crank can move in either direction
unless constrained by an external force. In the case ofan engine the external constraint
is provided by the remaining cylinders and a flywheel.
2.2.1 Inversion of the Slider-crank mechanism
It is known that a mechanism is formed by fixing one of the links of the kinematic
chain. Ifa different link is fixed in turn, an inversion ofthe mechanism in considerat*Pn
is formed. In any mechanism, as many inversions are obtained as the number of li~s
in the original mechanism. It should be noted that an inversion ofa mechanism does
not change the relative motion ofthe links, however, the absolute motion is altered.
In the slider crank mechanism, by fixing links other than link I, it is possible to obtain
three other inversions. Fig. 2.7 shows these various inversions of the slider crank
mechanism.
./
2~ ,
l /
·~
1 /, /
(a) (b)
~-
1 /
'
(c)
Fig. 2.7 Inversions ofthe slider-crank mechanism
I
I
I

Linkages 11
Fig. 2.7(a) shows an inversion ofthe slider-crank mechanism obtained by fixing link 2.
This mechanism is used in Whitworth quick-return mechanism. Figure 2.7(b) shows a
rotary engine or an oscillating engine which is obtained by fixing link 3. Figure 2.7(c)
shows a mechanism known as a bull-engine obtained by fixing the cylinder, link 4.
Inversions ofthe slider-crank mechanism are used in many applications like in quick-
return mechanisms, rotary internal combustion engines, etc.
2.3 THE SCOTCH YOKE
"" .
The Scotch-yoke mechanism, shown in Fig. 2.8, is widely used as a sine and cosine
generator, i.e. it is used to produce h~onic motion. It is also used to produce desired
vibrations.
----
/
/ ____,___
I
I
- - · .,...,....,...,~

:
........_ /
--
A'

lA
I
4 !??%!
1
Fig. 2.8 The Scotch yoke
The displacement ofthe slider x in moving form A to A ' is given by
X = r - rcose = r(l - cosO)
Substituting e= wt, the displacement is
x = r(1 - coswt)
The velocity of the follower is
v = dx = rwsinwt = rwsinB
dt
(2.2)
(2.3)
(2.4)

and the acceleration is:
2.4 QUICK-RET~ MECHANISMS
(2.5)
Quick return mechanisms are combinations of simple linkages which give a quick
return-stroke ofthe follower for a constant angular velocity ofthe driver. In the design
ofquick-return mechanisms, the ratio ofthe crank angle for the working stroke to that
of the return stroke is known as the time-ratio. The time-ratio for quick-return
mechanisms is always greater than unity to give a slower cutting stroke and a faster
return stroke.
2.4.1 Crank-shaper mechanism
Fig. 2.9 shows a schematic representation of the six-bar crank-shaper mechanism.
Links 1 - 4 ofthis mechanism form a variation ofthe slider-crank mechanism in which
the crank is held fixed as in Fig. 7.2(a).
0 B
B'
I
I
. I
A"
Fig. 2.9 Crank-shaper mechanism

Linkages 13
To produce a quick return, the time ratio should be larger than 1. For the direction of
motion indicated in the figure, the cutting stroke occurs when the crank rotates from
Oyl ' to Oyl " through an angle a, the idle stroke being when the crank moves from
Oyl" to Oyl" through the angle fJ.
For constant angular speed ofthe crank, the time ratio Q is given by
Q = time of cutting stroke = tw
time of return stroke tr
"' .
and for a constant angular velocity~2 of link 2,
a
Q--
fJ r
Length of stroke ofthe tool holder Cis given by
Length of stroke = B 1
B "
= 2B 1
D
= 20 B sin(a - 90)
4 2
From the above equation, the length ofthe stroke is found to be
O:zA
Length of stroke = 20 B - -
4 0204
2.4.2 Drag link
(2.6)
(2.7)
(2.8)
The drag link mechanism is developed by connecting two four-bar linkage in series.
Schematically, it is shown in Figure 2.10. Links 1, 2, 3 and comprise a four-bar drag
link mechanism in which both links 2 and 4 rotate 360 degrees. Links 1, 4, 5 and 6
comprise a four-bar crank-slider mechanism. The two four-bar mechanisms are
connected via the triangle-shaped link 4, known as ternary link, and they have a
common ground link 1. Assembling the two four-bar mechanisms in this way results
in a single six-bar mechanism.
For a constant angular velocity of link 2, link 4 will rotate at a non-uniform velocity.
For a clockwise rotation oflink 2 the cutting stroke makes an angle a (C '02C") and the
return stroke makes an angle fJ. Hence the time ratio is given by
Q = time of working stroke
time of return stroke
a
Q --
/3
(2.9)
(2.10)

B"
A
6
- - . -----+- . - -
D'
Fig. 2.10 Six-bar drag-link mechanism
2.4.3 Whitworth mechanism
The Whitworth mechanism is another variation of the slider crank mechanism in
which the crank is held fixed. Links 1, 2, 3 and 4 comprise an inverted slider-crank
mechanism. Links 1, 4, 5 and 6 comprise a crank-slider mechanism. These two four-
bar mechanisms are coupled through link 4. They also have a common ground resulting
in a six-bar mechanism. This mechanism is commonly used in shaping and slotting
machines.
-I
Fig. 2.11 Whitworth mechanism
When the driving link 0 2B moves from 0 2B" through an angle a. for the cutting stroke,
the cuttingtool D moves from D 'to D ". When the link 0 2B moves from 0 2B" to 0 2B '

Linkages 15
through an angle .B for the return stroke, D moves from D" to D '.
The crank angle ofthe cutting stroke and the return strokes are as shown in Fig. 2.11.
The time ratio is give by:
a.
Q =
p (2.11)
2.5 TOGGLE MECHANISMS
A simple toggle consists oftwo links which tend to line-up in a straight line at one point
·. ~ .
in their motion. Fig. 2.12 shows a_
.schematic representation of such links in toggle
mechanisms.
Fig. 2.12 A simple toggle
The mechanical advantage ofthe simple toggle shown in Fig. 2.12 is the velocity ratio
of the input point A to the output point B.
FB
Mechanical advantage =
FA
=
X VA
= tan(a) = (2.12)
y
As the angle a approaches 900, the links CA and AB come into toggle and the
mechanical advantage approaches infinity; which in practice is less than infinity due to
frictional effects. The simple toggle is used in punch presses, riveting machines, stone
crushers, etc.
2.5.1 Stone Crusher
The stone crusher shown in Fig. 2.13 uses two toggle linkages in series to obtain a high
mechanical advantage. When links 2 and 3 are in toggle, links 4 and 5are also in toggle
to produce the high crushing forces needed. When link 2 reaches the lowest point ofits
stroke, it comes into toggle with link 3 and at the same time links 4 and 5 come into
toggle with each other. This configuration results in a very large crushing force.

16 Mechanisms ofJ
l1aclzinery
p 6
:.....____,-G -
~ -
/)'
//;/////,'l/
//;
I
Fig. 2.13 Stone crusher
p
= 2 tana.
F
(2.13)
where P = resistance to be overcome
and F = required force to overcome the resistance
It can be observed that for a given force F, P increases as the angle a. diminishes.
2.5.2 Punch Press
The toggle mechanism for a punch press is shown schematically in Fig. 2.14. The large
punching force needed is obtained when the crank OjA. and connecting rod AB come
into toggle at the lower end of the punch stroke ex~ly at the time where it is most
needed.
2.5.3 Cold-heading rivet machine
The cold-heading riveting machine, shown in Fig. 2.15, is designed so as to give each
rivet two successive blows. The first blow is obtained at point 2. Following the first
blow, the hammer moves upwards a short distance to point 3; which then produces a
second blow by moving to point 4. After the second blow, the h8.IJ1..mer then moves a
long distance to point 1 during which a clearance is produced to move the work piece.
Both strokes are produced by one revolution ofthe crank and at tl1e lowest point ofeach
stroke links CB and BD are in toggle.

~ punching force
Fig. 2.14 Punch press
I
I
I
I
I
I
I
I
1 I
I
I
I
I
I
I
I
c
-- -- -- -- ~--
-- I

--- -
j

/
long stroke ----
.r...___--'*===!=== -=-3 2 4
! D '
short ~troke
Fig. 2.15 Cold~heading rivet machine
Linkages 17

2.6 STRAIGHT LINE MECHANISMS
These are mechanisms which can ge_
nerate straight lines from rotary motion. They are
designed so that a point on one ofthe links moves in a straight line without the need for
guides, thus converting rotary motion to straight line motion. Some examples of
straight line mechanisms are discussed below.
2.6.1 Watt mechanisms
The Watt mechanism, schematically shov.rn in Fig. 2.16, generates an approximate
straight line motion. If links 2 and 4 are ofequal length, the tracing point P traces an
approximate straight line with a symmetric figure 8 towards the ends of the stroke
length. The figure 8 is straighter if
AP
PB
1~
'~----
2
---1 A
2.6.2 Evans' Linkage
3
p
B~--
4
--~.,, 1
04
Fig. 2.16 Walt mechanism
.I
(2.14)
This lin..l.(age, shov.rn in Fig. 2.17, has an oscillating drive arm which has a maximum
operating angle a ofabout 400. For a relati-vely..short guided slide, a large reciprocating
output stroke DD' is obtained. The output motion ofpoint D is on a true straight line.
If the guide way in slide Fig. 2.17 is replaced by a link rotating about a..'1 axis~ the
motion obtained approaches a straight line ifthe link length is equal to the stroke with
a slight deviation.

I
I
I
~
I
I _,.---~
I
I . , .
I . 0
r-D.-,..- I 2 D
I / .
TI output stroke
'.
Fig. 2.17 Evan's linkage
2.6.3 Tchebichefrs mechanism
Linkages 19
Tchebicheff's straight line mechanism is shown in Fig. 2.18. Links 2 and 4 have equal
length, and the tracing point P is at the center oflink 3.
A'
-- - r
p·--'~--::---~__.
(
I
I
Fig. 2.18 Tchebichef's mechanism
The tracing point P occupies P ' when link 2 is inthe vertical position if

2.6.4 Peaucillier inversor
This is another mechanism that produces a straight line. The tracing point P moves in
a straight line. The geometry ofthe mechanism is given by:
AB = BC = CP = PA and
0,!1 = 0 2C and 0/)4 = 0 4B.
By symmetry, points 0 2, Band P always lie on a straight line. Under these conditions
(02B) · (02P) is a constant. See Figure 2.19.
A
p
c ·
Fig. 2.19 Peaucillier inversor
Curves described by Band Pare inverse ofeach other. For circular motion ofB that
passes through 0 2, P traces a straight line which is a circle of infinite radius,
perpendicular to 0 2 04• If 02 is located outside the circular path ofpoint B, then the
mechanism can be used to generate circular arcs of large radii.
2.6.5 The TchebicheffCombination of the Watt and Evans mechanisms
The Tchebicheffcombination ofthe Watt and Evans-linkages is shown in Fig. 2.20. As
indicated in the figure the tracing point D moves ip·it straight line.
E 6 0 6 '
~ 1 /·--~
04 4 I I 1
/
i !o
I
0 2 · - - ·- - ·
-·/
Fig. 2.20 Combination ofWatt andEvan's mechanisms

Linkages 21
2.6.6 D - drive mechanism
This mechanism, arranged as shown inFig. 2.21, is basically q. four bar linkage in which
the coupler is extended to include the tracing point D. The output-link point describes
a path closely resembling the letter D which contains an approximate straight portion
as part ofits cycle.
A
Fig. 2.21 D-drive
This motion is ideal for quickengagement and disengagement before and aftera straight
driving stroke. It is used as a film advancing mechanism in movie-film projectors.
2.7 PARALLEL MECHANISMS
These mechanisn:s are employed for producing parallel motions and reproducing
motions at different scales. Common examples of parallel mechanisms are the
pantograph and the drafting machine.
2.7.1 The Pantograph
The pantograph is used to enlarge or reduce trajectories to different scales. They are
commonly used in cutting tools to duplicate complicated shapes to desired scales. A
schematic representation ofthe pantograph is shown in Fig. 2.22. Links 2, 3, 4 and 5
form a parallelogram. Link 3 is extended to contain point C and point E lies on the
intersection oflines 0 2C and DB.
A pen attached at E reproduces the movement of C to a redu~.ed scale and vice versa;
i.e. the motion ofE is parallel to that of C.
To produce this parallel motion, the necessary condition to be satisfied is
0 2
C
=constant
0 2
E
(2.15)
for all positions ofC. Then, for any position ofC, triangle 0 2DE is similar to triaJ:l~l~--~;'~ ~.
CBE, thus
"-

A
2
E
D
Fig. 2.22 The pantograph
= constant
And the ratio ofthe sizes ofthe figures at C and E is
Size offigure at C = 0 2C
Size offigure at E 02
£
2.7.2 The Drafting Machine
(2.16)
(2.17)
Another application of parallel mechanisms is found in drafting machines. In the
drafting machine shown in Fig. 2.23, parallelograms 0 2AB04 and CDFE are coupled
by the ring ACBE. The straightedges attached to ring DE can be rotated and clamped
at any angle relative to the ring DE. Moreover, the straightedges can move to any
parallel position on the drawing board as shown in·the figure.
/
/
/ /
/ /
/ /
/ /
/ /
/ / '
/ / . --
/-;_......._,---
(/ /l--- -
-
' /
A
n·
1
.1
II
./
"I I
II
II
..d.l..
----( ):: - - - - - - - -:::~
--- --------
_ _ _ "'"'C./
--
E
Fig. 2.23 Drafting machine

Linkages 23
2.8 INTERMITTENT MOTION MECHANISMS
These mechanisms convert continuous motion into intermittent motion. Common
examples of intermittent motion mechanisms are the Geneva wheel and ratchet
mechanism.
2.8.1 Geneva Mechanism
This mechanism, shown in Fig. 2.24,,.Provides intermittent rotary motion. During one
'
revolution of the crank the Geneva wheel rotates through a fractional part of a
revolution, the amount of which pepertds upon the number ·of slots. The circular
segment attached to the crank locks the wheel against rotation when the roller is not
engaged. Angle f3 is half the angle subtended by adjacent slots:
(2.18)
where n = number ofslots
Letting r2 = the crank radius, the center-distance C between the center ofthe locking
device and the Geneva wheel is given by
c =
90dat.engagement
an disengagement

~~/
locking device
Fig. 2.24 Geneva mechanism
2.8.2 Locking-slide Geneva
(2.19)
In the locking-slide Geneva mechanism, see Fig. 2.25, pin P1 locks and unlocks the
Geneva wheel where as pin P2 rotates the Geneva wheel during the unlocked port_io~.'-
...-'

In the position shown in Fig. 2.25, the drive pin P2 is about to enter the slot to index the
Geneva, whereas the locking pin P1 is just clearing out the slot ofthe Geneva.
drive pin
p2
input crank
Fig. 2.25 Locking-slide mechanism
2.8.3 Ratchet Mechanism
This mechanism is used to produce intermittent circular motion from an oscillating or
reciprocating member and /or to allow rotational motion in one direction alone. Fig.
2.24 shows a ratchet mechanism. The pawl which engages the ratchet teeth is pivoted
at one end and the other end is shaped to fit the ratchet tooth flank. The pawl is kept in
contact with the wheel by means of a spring. At engagement, the pawl is in
compression. The driving pawl 3 induces intermittent circular motion of wheel 4.
Another pawl 5 prevents the wheel from turning backward in the undesired direction.
The line ofaction PNofthe driving pawl and tooth must pass between OA to guarantee
contactwith the tooth. The line ofaction ofthe locking pawl must pass between 0 and
B. This mechanism has a common application in counting devices.
./

Linkages 25
2.9 STEERING GEAR MECHANISM
The steering gear mechanism is used to change the direction of the wheel axle with
respect to the chassis which enables motion ofan automobile in any desired direction.
Commonly, steering in automobiles is done by means of the front wheels, the back
wheels having fixed direction with respect to the chassis.
The front wheels are mounted on the front axle and pivoted at points A and Bas shown
in Fig. 2.25. Points A and Bare fixed on the chassis. When the vehicle turns, left or
right, the front wheels and the respecti:Ve atdes tum about the respective pivot points.
The back a.,-xle remains fixed to the chassis and the wheels do not turn.
r
X a
. : outer front wheel I
Inner front wheel ---- 1 left tum ~
front axle
0~ - - - - - - - - - - - - - - - - - - - --.1".t----,,......---,.----,----__.,&;;___,..
I I ....-
J e
' ~)
' ---------
1 ~ //........ ...... ~
/ /
/ --------
/ /
/ /
/ /
/ ............
/ ......
/ ......
/ ......
/ ...... ----
/ _.,.-
/ /
/ /
/ /
/ ...........
/ /
/ /
_.-::/
p;....--
c~ - --- -- - -- - ---- - - -- -
back axle
c 7'
rear wheels
Fig. 2.27 Steering gear mechanism
b
I
i
I
_____l_
To avoid skidding or slipping of the wheels sideways, the front two wheels must turn
about the same instantaneous center C which lies on the axis of the back wheels.
Otherwise, skidding will result onthe front wheels whichcausesundue wearinthe tires.
The condition for correct steering is that all four wheels must turn about the same
instantaneous center. This condition is satisfied ifthe inner wheel axis makes a larger
turning angle ecompared to the outer wheel axis turning angle f3.

26 Mechanisms of.Maclzinery
Forthe leftturn indicated in Fig. 2.27, 8 > f3. The condition to be satisfied is obtained
as follows.
cote
EO X
= - --
oc b
cotfJ
AO X +a
- - =
oc b
and the condition for cotTect steering is obtained to be
a
cot{J - core = -
b
(2.20)
(2.21)
(2.22)
Equation (2.22) is the fundamental equation for correct steering which~ if satisfied,
eliminates skidding ofthe from wheels.
2.9.1 Ackerman Steering Gear
The Ackerman steering mechanism consists ofa four-barmecha..Tlismjoinedby revolme
joints as shown in Fig. 2.28. The shorter links QR ar1d PS are of equal length and are
connected to the front wheel axles by hinge joints. Links PQ and RS are of unequal
length.
left tum
'
0~ - - - - - - - - - - - - - - - - - - -.1"lf---:.,;,-- ...,.-----..:,.P.:::-tl.
8 /
/
'v/
/ /
/ __.,""
/ /
/ -
/ -
/ ...,..""
/ /
/ /
/ .-""
/ / __.,.-
/ /
/ /
/ /
/ __.,""
/ /
/ /
I -::;.-_... .
, P
c~--- - ----- -- ----- -- -
.I
Fig. 2.28 Ackerman steering mechanism
Vhen the vehicle moves along a straight path, links PQ andRSare parallel to each other
and links QR and PS are equally inclined to the longitudinal axis ofthe vehicle. Figure
2.26 shows the position ofthe vehicle when steering to the left. In this position, the
lines ofthe front axles intersect at Con the axis ofthe back wheels and links PO and
. -
RS are no longer parallel to each other.

Linkages 27
To satisfy the fundamental equation of steering, links PS and RS should be
proportioned suitably.
r

..
Chapter 3
VELOCITY ANALYSIS OF LINKAGES
Velocities and accelerations in mechanisms.are determined by different methods. The
basic methods of analysis discussed in this text are the following:
1- Velocity and acceleration analysis using vector mathematics in which velocity and
acceleration of a point are expressed relative to fixed or moving coordinates.
11 - Velocity and acceleration analysis using equations of relative motion which are
solved graphically by velocity and acceleration polygons or by using trigonometric
relations.
iii - Velocity and acceleration analysis by using complex numbers.
iv - Vectors velocity analysis using the instant-center method.
~ - · ...
/ '3.1. VELOCITY ANALYSIS BY VECTOR MATHEMATlCS
~ .."
Consider the motion ofpoint P moving with re~:tect to the x -y - z coordinate system,
which, in tum, moves relative to the X- Y - Z coo-:dinate system as shown in Fig. 3.1.
./
R P is the position vector ofP relative to the .X-Y-~~ystem,
R. is the position vector ofP relative to the X'-y-.t"systtm,
Rais the position vector of the origin of the moving CClordinate system x-y-z
relative to the fixed coordinate system X-Y-Z.
The position vector ofP relativ~ to the X-Y-Z system R P is expresse:.d as:
R = R + R
p 0
(3.1)
Introducing unit vectors i, j, and k along the x, y and z axes respectively,
R =xi + yj + zk (3.2)

Velocity Analysis of Link~ges 29
-..
y
p
X
z
Fig. 3.1 Position coordinates ofa moving point
Velocity ofP relative to the X-Y-Z coordinate system is
(3.3)
J_l0
== VJs the velocity ofthe origin ofx-y-z system relative to the fixed system.
Let
R. is given by:
R = !!_(xi + yj + zk)
dt
t - •
= (xi + Ji! + i k) + (x~ + yj + zk)
i i + yj + i k = v
and noting that
i = (!) X i
j = (!) x j
k = (!) X k
(3.4)
(3.5)
(3.6)
where w is the angular velocity vector of x - y - z system relative to X - Y- Z, the
second.term in the.expression for Ris
xi + yj -! zk =x(w xi) + y(m xj) + z(w xk)

30 Mechanisms qfMachinery
Upon simplification,
x( + y/ + zk = w x (xi + yj + zk)
= w xR
Thus, the velocity ofP relative to the moving coordinate system is
R = V + roxR
Therefore, the velocity Vp ofpoint P relative to the fixed system is:
VP = V 0
+ V + roxR
(3.7)
(3.8)
(3.9)
where V0 is velocity ofthe origin ofthe x-y-z system relative to the X-Y-Z system;
V is velocity ofpoint P relative to x-y-z system;
w is angular velocity ofthe x-y-z system relative to X-Y-Z system;
and R is position vector ofP with respect to the origin ofthe x-y-z system.
Example3.1
The mechanism shown in Fig. 3.2 has the following dimensions:
0:). =200mm , AB =1500mm , B04 =400mm , 0 2
0 4
=1350mm
~
Link 2 rotates at a constant angular velocity of 4 rad/s in the clockwise direction. For
the phase shown, determine the velocity ofpoint B and the angular velocity oflink 3.
y
Fig. 3.2
Solution:
.I
/
/
I
I
./
The coordinate systems X-Y and x-y are selected as shown in the Fig·. 3.2, where the
x-y system is moving relative to the fixed X-Y system.

········----. -·
Velocity Analysis ofLinkages 31
From the geometry of the mechanism
e = 134.4° and fJ = 4-1.-!0
Velocity ofpoint B can be determined from the equation
V8 = VA + V .+ roxR
where v A = velocity ofthe origin ofx-~ system
olb
w - angular velocity ofx-y syst~m
R = the position ofB rel?-tive to the origin of the x - y system, hence R = AB
J
VB = velocity ofpoint11;direction perpendicular to 0 ,(3, magnitude unknovvn.
Iv A I = 00~U? = 200·4n: = 800.1r mm/s . .,
v =0 because B is a fixed point in tlie x-y system
w xR magnitude unknown, and direction perpendicular to AB,
(1) =(1) '
.)
Introducing unit vectors i andj along the Xand Y axes respectively,
v B. = v 8 cos j 7 i + v 8
sin '57j
v A = v A cos -1-1.4 i.::. v A sin 4-1.4 j = 1795.li -J7j8.-lj mmls
w xR =(w xR) j
Substituting the above relations into the relative velocity equation,
v 8 cos 57 i :.. v 8 sin 57j = 1795.7i - 1758.4j + (w x R)j
Summing the i components:
I ) '
v 8 cos 57 = 1795. 7
V8 = 3297.0 mm/s
• l ! J ' _; !r1 Y .:
Summing thej components:
v 8 sin 57 =. -1758.4 + (w R)
' :
./

32 !Wee/zanisms ofMachinery
(w R) = 4523.5 mm/s
w3 = 3. OJ rad/s
3.2 Y elocity analysis by using equations of relative motion
3.2.1 Velocity of points on a common link
A and Bare two point on a common rigid linkAB as shown in Fig. 3.3 (a). The points
are moving with velocities V,~ and V8 respectively. Using the equation of rel~tive
motion, velocity of one point can be determined relative to the other.
A b
B (a) (b)
Fig. 3.3 (a) Velociry ofpoints on a common link, (b)velociry polygon
The velocity ofA is given by
V A = V B + VAIB
V a
....
3.10)
where VAlB is the relative velocity ofA relative ,t6 B. The relative velocity vector is
VAIB shown in Fig. J.3(b). In the relative velocity polygon ofFig. 3.3(b), all absolute
velocity vectors originate from the same point o2'
Note that the velocity ofA relative to B and the velocity ofB relative to A are equal in
magnitude, collinear and opposite in direction, i.e.
(3.1 1)
Example 3.2
Link 2 of the four bar linkage shown in Fi!Z. 3.4 is the drivincr link having a constant
- ::>
angular velocity w2 = 10 radls in the clockwise direction. For the phase shown,

determine the velocity of point B and angular velocity oflinks 3 and 4.
0 2 A = 100 mm, AB == 200 mm, O.B = 75 mm
b
2 ;
230
(a) (b)
Fig. 3.4 .
Solution:
From geometry ofthe mechanism, 8 =8.2° and 8 =22°
A and Bare points ori the common link AB. The velocity ofpoint B can be expressed
in terms ofthe velocity ofA and the velocity ofB relative to A as follows.
VA = VB + VA/8
Iv A I= .(00) ·w 2 and is perpendicular to 00
v 8 is perpendicular to 0 .;B with unknown magnitude
v BIA is perpendicular to AB and magnitude unknown.
The velocity polygon is shown in Fig. 3.4(b). The vectors are added tail to head as
shown in the figure.
Vector o~ represents v A = (00).w2 ·= 1000 mmls, and is perpendicular to 00.
v 8 is represented along the line through the point o2 perpendicular to

...,4.
.), Mechanisms of},fac/zinery
VB/A is along the line through a and perpendicular to AB. The intersection ofthese
lines is point b.
Vector o:;b gives V8 and vector ab gi·es v BIA .
L.ao2
b = 59 , !..o2
ab = 44.8 , f. o.,ba = 76.2
From the sine formula
v v
A 3
- - -
sin76.2 sin44.8
VB
sin4.:/.8
VA
=
sin76.2
VB = 725.6 mmls
=
sin76.2
VB/A
sin59
VB/A
sin.JY
xJOOO
=
sin76.2
VRU = 886.2
VB 725. 6
(!)~ = -- =
0 4B 75
(1)4 = 9.67 rad/s
VB!A
(l), = COAB = -- =
_,
AB
(l), = 4.441 radls
'
Example 3.3
882.6
200
.J
Fe-: an inteiV'al of its motion the piston rod ofthe hy~raulic cyl':<nder has a velccity of
1.5 n/s as shown in Fig. 3.5. At a certain instant, e =fJ =60°. For this in~t_ant
detem~ne the angular velocity of linko .A and link AB.
~ . ~
Solution
Coordinate sys~":m x - y and the corresponC.ing unit vectors i a..'ld j , respectively, are
introduced as shovn in Fig. 3.5.

y
/////////// /
Fig. 3.5
For the instant when 8 =j3 =60° the angle y =30°.
For the instant given,
VB = -1.5 i mls
V A =VA (cos30i + sin30j )
VAIB =VAlB (cos30i - sin30j)
The velocity of A relative to B is given by
V A =V B + VAIB
Or,
D
VA cos30 i + VA sin30 j = -1.5i + VAIB cos30 i - VAlB sin30 j
Collecting like terms
VA sin30 = -V
AlB sin30
VA cos30 = -1.5 + VAIB cos30
Solving the above equations simultaneously
VA = -0.866 mls
VAIB = -VA =0.866 mls

36 J
Yfechanisms ofMachinery
The angular velocity of link 0 ;i is
v_-1
w =- - =3.464 radls in the counterclockwise direction
0:!.-1 0.250
and the angular velocity oflink AB is
= V..t!B = I. 732 radls
W AB 0.500
in the clockwise direction
3.2.2 Velocity of a block sliding on a rotating link
Fig. 3.6(a) shows a block sliding on a rotating link. Block A slides on the rotating lipk
0 J3. The angular velocity w ofthe link and the velocity ofthe block are assumed to
be known.
y
~
y
~NA
X ~·
'///h
(a) (b)
Fig. 3.6 (a) Velocity ofa block sliding on a rotating link, (b) velocity polygon
To determine the velocity ofthe block, let A ' be a point on the link coincident with
the block A for the instant represented. Tqe-..rvelocity of A' relative to 0 is "
perpendicular to DB at A 1
• The velocity ofA relative to A I is along the link parallel
to OA '. Velocity ofA is determined from the relative velocity equation.
. "
(3.12)
The velocity polygon is sketched in Fig. 3.6(b).
Relative velocity of coincident particles on separate links is effected by physical
constraints such as guides.
Example 3.4
A cran.k-shaper mechanism is shown in Fig. 3.7. The dimensions ofthe various links

, are:
020~ = 350 mm, 0r4 = 100 mm, O~B = 550 mm and BC = 125 mm.
For the phase shown, the crank 0r4 makes an angle of 300 with the horizontal and
rotates at a constant angular velocity of 60 rpm in the counter clockwise direction.
Deterrnin~ the angular velocity oflink 0 .B ap.d velocity of C.
Solution:
I
I
- ~·

/
/

"
6
w2 = 60 rpm = 271: rad/s
3'
· A

-·;-
/
. I
(a)
Fig. 3.7
From geometry ofthe mechanism, for the phase shown
e= 12.2°, J3 = 26° and Ofi = 410 mm
a
(b)
Block A slides in the slot on link O,B. Considering a point A' on link O~B coincident .
with A for the instant represented, the relative velocity equation can be written as:

vA = (00) .w2 direction perpendicular to o r4.
= 100 x 2r.: = 200rr mmls
= 628 mmls
VA 1 = magnitude unknown, direction perpendicular to o ,;A 1
.
V,41
A 1 = magnitude unknown, direction along the slot on o .;B.
The velocity polygon, drawn to a suitable scale, is sho·wn in Fig. 3.7(b).
The fixed points 02
and o 2
are represented by the point o
2
, o-+.
Vector o p is drawn perpendicular to 0_r4 and represents VA .·
F!om o 2 , o"' vA1 is laid along the perpendicular to 0 ,r4 1
. ·
Through point a, oAlA 1 is laid along the slot parallel to o 4B.
The intersection ofthe last two lines is point a'.
I . d I . V
o2a gives DA 1 an o a g1ves rJ/.rJ 1 .
From the scale drawing:
o A I= 423 mm/s
From oAI, the angular velocity of link o 4
B, OJ.; , can be determined.
OJ =
.J
423
- - = 1.03 radls
410
.I
To determine the velocity of C, consider B and C , points on a common link. The
relative velocity equation is
vc = V B + VClB
-
Since B lies on o .r4- 1
produced, the velocity ofB is along o2
a 1
on the velocity polygon
proportional to its radial distance from D 4
•
v8 is represented by o
2
b.
Vc = magnitude unknown, direction along the horizontal slot.

Vc
1 8
= magnitude unknown, direction perpendicular to BC .
vc lies along a horizontal line through o 2 and vc 1 8 lies along a line
perpendicular to BC through point b. The intersection ofthese two lines
is point c where. vector o 2c represent Vc , and-vector cb represent v c / 8
.
From the scale drawing,
vc =608 mmls
Example 3.5 •. ,'
.
In the mechanism shown in Fig. f.g crank 0~ revolves at a constant angular velocity
of5 radls in the clockwise direction. Determine the angular velocity o'frod AB which
slides through the pivoted collar C for the instant when e =90°. O_;A =250 mm.
A
/
e/
_(_
·
600mm
Fig. 3.8
Solution
For e =900 the position of the mechanism is shown in Fig. 3.8(a). The rotating
coordinate system x- y is introduced with unit vectors i andj along the x- and y-~es,
respectively, as shown in the figure.
For the position when e = 90°, consider a pointC 1
fixed on link AB and coincident
with point C .
AC 1
=650 mm
AC1
= 650 i mm

40 ."'-fecltanisms ofMacltinery
B
.w2 = -5 k radls
fJ - -1 ( 250)
-tan --
600
·,
I
'-- ·
600mm
Fig. 3.8 (a)
--mo~
_ill_
'//Y/,
I
I
I
I
!
The relative velocity equation ofpointC 1
relative to A is "Written as
where
= -5k x(96.1i + 230.8})
·-
=1154i -480.5j mmls
Since the length rAe; remains constant,
V =r~c~ =0

={WAC I k) X ( -65Qi)
Substituting in the relative velocity equation
Ve1 =Jl54 i - 480. 5 j - 650 (J)Ael j
A> '
Collecting like terms and solving for.the unknowns
vel =1154 i mmls r
wAel = -0.74 k radls
3.2.3 Relative velocity of coincident particles at the point of contact of rolling
elements
Rolling contact exists when there is no sliding at the contact point between two links.
To satisfy this condition, the velocity component along the tangential direction must be
zero. For pure rolling contact oflinks 2 and 3 shown in Fig. 3.9, the points P2
on link
2 and p3 on link 3 have the same velocities i.e. vp =vp . ,....../.
2 3 ,,<'
n
t
Fig. 3.9 Rolling contact
Demonstration
Let a point P be common to the two links, 2 and 3, which have relative motion to each
other. The relative velocity equation may be written as:
The condition for pure rolling is that the relative velocity
VpIP = 0
3 2
(3.13)
(3. 14)
This condition is met when the point of contact lies on the line of the centers o 20j~ ·
r •

If v PIP were not zero, its direction would by along the tangent t- t, in which case
3 2
link 3 would slide relative to link 2 along the t- t direction.
For pure rolling of direct-contact mechanisms, the angular velocity ratio ofthe driver
and follower is inversely proportional to the ratio ofthe lengths of line segments into
which the point ofcontact divides the line ofcenters.
Rolling ofcircles or cylinders is a special case ofrolling motion. .As shown in Fig. 3.10,
links 2 and 3 roll with the point of contact P2 on link 2 and P3 on link 3.
For pure rolling ,
= (3.15)
Fig. 3.10 Rolling contact ofcircles
.·
Example 3.6
In Fig. 3.'1 l gear D (teeth not shown) rotates in the clock-wise direction about 02 with
a constant angular velocity of4 rad/s. The arm o ~ is mounted on an independent
shaft at 0 2 , and the srr:all gear at B meshes with gear D. Ifthe arm has a counter-
clock-wise angular velocity of 3 rad/s at the instant represented, determine the
corresponding angular velocity of gear B.

25 mm
3 I ~
Fig. 3.11
Solution
Consider the contact point C to be C2 as a point on link 2 and C.t on link 4. For pure
rolling contact
V C: =V C;
- <L oO
Vc, =w2 xR0~ =~32frj mmls
Hence, the velocity of Cas a point on gear A is
Using the relative velocity equation,
vc = VA + v
• C/A
I.e.
VA =w3 xR0~ =375 j m~ _
I
After substituting and simplifying we get
Vc lA = -775 j mmls
J

Force Analysis ofMachinery 331
11.10 GAS FORCES
For a four-stroke cycle the variation ofthe gas pressure in the combustion chamber or
cylinder for two revolutions ofthe crankshaft is shown in Fig. 11.14.
Themagnitude ofthe gas pressure is determinedfrom thermodynamic analysis. The gas
force on the piston is the product ofthe pressure and the piston head area.
Expansion Exhaust Intake Compression
,-..,
.§
z
'--'
~
;::l
Cl)
Cl)
~
~
Cl)
~
0
0 0
540° 720
Crank Angle (deg)
Fig. 11.14 Gas pressure infour-stroke cycle engine
To analyze the effect ofthe force P on the whole engine we start by assuming that all
the moving parts are massless which reduces the inertia forces and torques to zero. At
the same time, we assume that there is no friction as well.
y
A
t-----
p - --
/
X
Fig. 11.15 Gasforce acting on an engine piston
The gas force P is a function oftime varying with wt. The free-body-diagrams for a
particular phase of the mechanism, i.e. of the piston, connecting rod and crank are
shown in Fig. 11.16(a). Fig. 11.16(b) shows the forces acting on the frame.
As can be noted from the free-body-diagram ofthe crank, the crank is in equilibrium by
the couple T; formed by F~ and the frame force F~ . This torque is opposed to the
crank torque T2
.

44 J
lfecfzanisms ofMachinery
from which the angular velocity of gear B is obtained to be
VCJA
=31 radls cc-w
Exampie3.7
The circular cam shown in Fig. 3.12 is driven at an angular velocity of w2
= I 0 rad/s
clockwise. There is pure rolling contact between the cam and the roller link 3. Find the
angular velocity ofthe roller and the oscillating link 4 for the phase shown.
87.5
4
----~==~-----e
e±r~====~==~:~-
~-o~ l
'////, I
75
(a)
Fig. 3.12
Solution
From geometry _
ofthe linkage, for the phase shown
. . .
Q2C = 71.5 mm," 8 = 5~·~~·'!-Ild·l= 100.~
: ... ~ ........

Point Cis the point of rolling contact between the cam, link 2 and the roller, link 3.
Therefore,
v =vc
Cz J
To determine w3
and w4 , the velocity equations are written as follows:
and
V B = vA + V B/A
""
V Cz =vA + VCIA
VB is unknown·in ma&-nitude, direction perpendicular to o 4
B;
V A is known both in magnitude and direction,
IVA I = (0 0 ) .w2, direction perpendicular to 0 2A
= 30x 10 = 300 mmls
V BIA is unknown in magnitude with direction perpendicular to the line
joining A and B,·
Vc is unknown in magnitude with direction perpendicular to 0 2
C2
2
VC!A is unknown in magnitude, direction perpendicular to AC2 .
Using a convenient scale, the velocity polygon is drawn in Fig. 3.12(b).
VCIB =670 mm/s
= VC3/B = 670
(1)3
BC3
12. 5
co3 = 53. 6 rad/s
3.2.4 Relative velocity of crank and connecting rod
In the slider-crank mechanism shown in Fig. 3.1 3(a), let w2
be the angular velocity of
the crank o 2
A . The velocity of B can be determined using the velocity of point A as
the reference which can easily be determined.

A
(a) (b)
Fig. 3.13 (a) Motion ofthe slider-crank mechanism, (b) velocitypolygon
where v A is known both in magnitude and direction;
v 8 is known in direction, magnitude is unknown;
v BIA is known in direction, magnitude is unknown.
Example3.8
In the slider-crank mechanism of Fig. 3.14(a), the crank rotates at a constant angular
velocity of 1On rad/s clockwise. Determine the velocity ofthe slider B and the angular
velocity ofthe connecting rod (J) ] for the phase when e= 600.
O;rA = 150 mm andAB = 600 mm.
(a) (b)
Fig. 3.14
Solution:
From geometry ofthe ~lider crank mechanism, forB = 600, cp = 12.5°.
The relative velocity equation is written as:
V B = V A + VB/A
v
8
is unknown in magnitude, direction along the x-axis.

Velocity Analysis of Linkages 47
is known both in magnitude and direction
=(00) . (J)7
=(150)10rr = 4712.4 mm/s
vBIA is unknown in magnitude, direction perpendicular to AB.
velocity polygon is shown in Fig. 3.14(b) from which the following relations are
UCI•~u••-d•
sin72.5 sin77.5
V
4721.4 . 72 -
B = xszn .J I'
sin77.5
VB =4603..J mm/s
VB/A
sin30
VA
=---
sin77.5
V 4721.4 . "0
BIA = Xszn.J
sin77.5
VBIA =2413.4 mmls
2413.4
= - - -
600
ro. = 4. 02rad/s
.J
.5 Algebraic solution of the slider-crank mechanism
F'Referring to Fig. 3.13 and taking the origin of the coordinate system at the crank
• center, the position ofthe slider is defined by x. From the geometry ofthe mechanism,
and
~ ·
r sine =l 'simp
X =r cos8 + f COS(/)
where r is the crank radius,
I is length of connecting rod.
(3.17)
(3.18)

48 J
l-tecltanisms ofjrfadzinery
From the position ofthe mechanism it can be noted that
cosq;
= jt2
- (r sinB/
f
= 1 - (!.. sinB/
l (3.19)
and substituting for cos· q; in equation (3.18), the position ofthe slider is obtained to
be:
X = r COS8 - f
r ~
1 - (- sinB)-
f
(3.20)
Alternatively, equation (3.20) can be obtained from the law of cosines given by
l ) , ? e
- =r- + :c- - 2rx cos (19)
and by soiving quadratic equation for :~.
The velocity ofthe slide: is obtained by differentiating equations (~ .17) and (3.18) with
. ' ' . . c . e d dco
respect to t1me ana suostnutmQ 10r cos rn . szn an - ·
~ 'f' , dt
By differentiating equation (3.17) with respect to time the following relations are
obtained.
cosq; dq;
dt
t
= r COS<p d8
I dr
which yields which dq; to be
dt
dco
dt
e-
= r cosq; dB
l cosq; dt ./
From equation (3.18), differentiating with respect to time yields
~ -
... - . e de 1
dq;
-r szn - - simp
dt dt
(3.21)
(3.22)
N · h · · h 1 · f dro
otmg t atx 1s t e ve ocny o the slider and substituting for dt :the velocity of the
slider is obtained to be
r sinB r cosB
V = -rw sin8 - I w
1 I rn<:rn

: on simplification,
I
.e r sine case
V =-rm sm + I~; - ( 7siner (.., / '"')
.) __.)
~r smallvalues of ::.._, whic:1 usually is the case in slidercrank mechanisms, !..sin8 "" 0
I f
and the velocity is given by
V = - rw +
/ .....
r sin;e
2!
(3.24)
-tRework Example 3. 8 usmg the algebraic method ofsolution to determine the velocitv
ofthe slider B.
The velocity of the slider B is given by the equation
V
. e -r s~ne case
= - rw szn + ---;:====--===========
8
IFl~ sine)'
r =150 mm, l =600 mm, w2 =1On radls
For the phase when 8 = 60, the velocity of tr.e slider is
150·sin60 cas10
V8 = 150·1On sin60 + --;::::::========:::::====
(
150 )
2
600 · 1 - - sin{/)
600
from which the velocity of the slider is obtained to be
V8
=4616.4 mm/s

1V!eclzanisms oflvfacltinery
by using the approximate relation for whicn case the ratio ~ is small values of,
l
ulting in !...sinB <>< 0, the velocity of the slider B is given by
l
V8 =150·J07r (sin60 +
150
sinr2·60))
2(600) .
m which the velocity ofthe slider is obtained to be
V8
=4591.2 mmls
e variation between the exact solution and c.pproximate solution is smaller than one
·cent.
; Veiocity analysis by complex numbers
)St ofthe systems ofa..Tialysis using complex polar notation are based on the following
1damemallaw:
Jfche elements ofa mechanism are replaced byposition vecwrs such that their sum
is zero, rhen their time derivarives are also equal lO zero.
is law means that ifone takes any linkage or mechanism and replaces the members
the mechanism by vectors such that their sum is zero, then the sum ofthe velocity
ctors is zero, so also the sum of the acceleration vectors.
msider the slider-crank inversion shown in Fig. 3.15(a).
nk 2 is the driver (crank) and has a constant angular velo~ity w2 and for the instant
der consideration an angular position of e?. Dimensions ¢'linkages are asswned to
- '
known, so the angular position ofthe follower, link 4, can be obtained.
• ' qf
:placmg eacn link by a vector such that the position polygon closes as shown in Fig.
i5(b), a mathematicai expression for the summation law can be written as:
R + R -R =0
I 2 ~ (3.25)
1ere R = vector for the £rounded link
I ~ • '
R ., = vector for the crank
R .: = vector to determine the position ofiink 3. Note that the magnitude of R~
is variable.

(a'
'

(b)
Fig. 3.15 (a) Inverson ofthe slider-crank mechanism, (b) position polygon
, To solve the position equatim given by equa~ion (3.25), the vectors are represented in
the complex notation.
·;'The position of a particle on alink representedby a vector ~P as shown in Fig. 3.16
, may be expressed in any of the bllowing equival~nt forms:
R p :::: a + ib
R p =rp (cosB2
+ i sin82)
R P
;e,
:::: r p· e -
(3.26)
' where r P is the magnitude ofvector R P .
Imag.
Real
Fig. 3.16 Position vector ofa particle --
Using this complex representation, equation (3.25) is traL-,formed into
(3.27)
. Differentiating the above equation we obtain
(3.2~)

Note that r 1
and 81
are constants with time derivatives equal to zero and r4
is variable with a non-zero time derivative.
Let 8; = w;
=0 (3.29)
Separating equation (3.29) into real and imaginary terms, we get:
(3.3.0)
The unknown quantities in the above pair of equations are w4 and f4 . Solving for
w4
and f 4
we obtain
r4 =r2w2 sin(B4 f B)
(3.31)
Equation (3.31) represents the complete solution ofthe velocity ofthe linkage for any
angular position e2 of the driver in which the angular position of link r, e4 is
obtained as a function of B2
. .I
Notice that equation (3.31) can also be determi~d more directly by simply summing
horizontal and vertical vector components, and then taking the time derivative of the
resulting equations.
Example 3.10
Solve the problem of Example 3.8 by using complex numbers.
Solution
Replacing the members of the mechanism by vectors as shown in Fig. 3.17 the

Imag.
Real
Fig. 3.17
r
The crank 00 is replaced by vector R2 , the connecting rod by R
3
and the position of
the slider B is represented by R 1 . The angular positions ofthe vectors R
1
, R2
and R
3
are given by el, 82 and e3·, respectively.
For the vector diagram shown in Fig. 3.15, the summation law is:
R2 + RJ = R1
RI = r 1e
iSI
R2
iS
= r e 2
2
R3
. iS.
= r e )
3
Transforming equation (a) into the complex form:
Differentiating equation (b) with respect to time, we obtain:
Noting that
r; =0, r~ =0, e~ =0
equation (c) reduces to the following equation
f' i81 r · iBJl _ . f i81l
r 2cv21 ze 1 + r 3cv31 ze 1 - r 11e 1
Separating the real and imaginary terms ofequation (d)
r2w2 cose2 + r3cv3 cose4 =0
(a)
(b)
(c)
(d)
(e)

Solving equation (e) simultaneously for w3
and r~ ,
r2
cosB2
(JJ3 = -(JJ2
r3
cosB3
r1 ~ r1w1 ( ~sin81
:
From Example 3.5, B2 = 600 and 83 = 360-12.5 = 247.5° from which
= _1On;x 150 cos60
w
3
600 cos347. 5
(/)3 =4.02
r1
=150x107r(-sin60 +
r1 = -4603.4 mm/s
Example 3.11
cos60 sin347.5
cos347.5
In the mechanism shown in Fig. 3.18, the driver link 2 has a constant
velocity w2 rad/s rad/s in the clockwise direction. Determine the angular vel
of link 5 and the rate at which the length AB is changing for the phase shown.
Fig. 3.18
Solution
The links ofthe mechanism are represented by vectors as shown in Fig. 3.19 in which
the vectors representing the links are given by
R l =rl e
i81
R 2 =r2 e i82

R.
>
Fig. 3.19 .
Fig. 3.19, the summation law ofthe position vectors ofthe mechanism gives
the complex notation, the position equation is written as
= 0
geometry of the mechanism
scalar components ofequation (b) are
~;~Substituting for 81 and 83 , we get
-rl + r2 cose2 + r3 sin85 - r5 cos85 =0
r2 sin e2 - r3 cos85 - r5 sin85 = 0
.In solving for the unknowns r3
and 85
, from the second of equations (d) we get
r _ r1 sin82 - r5 sin85
3 -
cose5
Substituting for r3
in the first ofequations (d) yields
[
r2 sin82 - r5 sin85l .
-rl + r2 cose1 + szn85 - r5 cose5
cose5
=0
(a)
(b)
(c)
(d)
(e)

This equation is simplified to give
Having the position equation solved, the angular velocity of link 5, w5
can
determined by differentiating equation (b) which yields
From the equation of geometry 83
= 90° + 85 we get
8~ =8~
Hence, equation (g) can be written as
= 0
Separating the real and imaginary parts of equation (h), we get the scalar components
which can be solved simultaneously to yield the unknown velocity components; i.e.,
Substituting for 83
in terms of 85
and simplifying yields
.I
Solving equation (i) forw5 andr~ we get .
=r w ( (cos82 + sin8sJ
r~ , 2
• cos85
3.4 Analysis ofvelocity vectors by instant-center method
Coincident points on two links in motion that have the same absolute velocity relative
to a fixed link will have zero velocity relative to each other. At this instarit either link
will have pure rotation relative to the other link about the coincident point. Consider

2 and 3 which move relative to each other. Point A on a body 2 has a velocity
relative to pointA on body 3 and point Bon 3 has velocity v BJBz relative to point
body 2 as shown in Fig. 3.20. Perpendiculars to both velocities intersect at point
is the instantaneous center of rotation of body 2 relative to 3 or vice versa.
p may be considered as a point on body 3 about which boay 2 is instantaneously
or it may be considered as a point on body 2 about which body 3 is
.UU.JL......·
·-ously rotating. The velocity of P2 relative to P 3 or velocity of P 3 relative to
=0 (3.32)
y /
3
X
Fig. 3.20 Instantaneous center oftwo bodies with relative mOlion
An instantaneous center ofrotation is defmed as a point common to two links which
the same velocity in each link; i.e., it is a point at which the two bodies have no
velocity. It is also a point on one link about which another link is
rotating. For different combinations oflinks 1 and 2, Fig. 3.1-9'shows
locations ofthe instantaneous centers C.
2
'1-
3.4.1 Types of instantaneous centers
The instantaneous centers of a mechanism are ofthree types:
· 1. Fixed instantaneous centers:these instantaneous centers (I C.) remain fixed for
all configurations of the mechanism.
Permanent instantaneous centers: these instantaneous centers move with the
mechanism but the joints are ofpermanent nature.

J
2 and 3 which move relative to each other. Point A on a body 2 has a velocity
relative to pointA on body 3 and point Bon 3 has velocity v B/ Bz relative to point
body 2 as shown in Fig. 3.20. Perpendiculars to both velocities intersect at point
is the instantaneous center of rotation of body 2 relative to 3 or vice versa.
p may be considered as a point on body 3 about which boay 2 is instantaneously
may be considered as a point on body 2 about which body 3 is
rotating. The velocity of P2 relative to P 3 or velocity of P 3 relative to
=0 (3.32)
y r
..,
.)
X
Fig. 3.20 Instantaneous center oftwo bodies with relative motion
instantaneous center ofrotation is defined as a point common to two links which
the same velocity in each link; i.e., it is a point at which the two bodies have no
·ve velocity. It is also a point on one link about which another link is
ly rotating. For different combinations oflinks 1 and 2, Fig. 3.1-9'shows
locations ofthe instantaneous centers C. '-1
·
3.4.1 Types of instantaneous centers
The instantaneous centers of a mechanism are ofthree types:
. 1. Fixed instantaneous centers:these instantaneous centers (I C.) remain flxed for
all configurations of the mechanism.
Permanent instantaneous centers: these instantaneous centers move with the
mechanism but the joints are ofpermanent nature.

3. Neitherfixed norpermanent instantaneous centers: these instantaneous
vary with the configuration ofthe mechanism.
c
pin joint
over a
curved surface
roiling contact
center ofcurvature
ofpath C
, C at infinity
I
I
I
I
,,,,,,,,,,,,,,,,,,
I I'd'
s 1 mg contact
Ai 2 V
[iJ • A
,,<'
'' sliding contact
over a
curved surface
Fig. 3.21 Location ofvarious instantaneous centers
In the four bar linkage shown in Fig. 3.22, points 02, 0 4, A and B are the obvious
instantaneous centers for they satisfy both definitions for an instantaneous center. 02,
and 0 4 are fixed instantaneous centers; A and B are permanent ones.
'
Since for each pair of links there is one instantaneous center, for the four-bar linkage
shown, the number ofinstantaneous centers m
six, in accordance with the pairs oflinks .
given below:
Fig. 3.22 Permanent andfixed instantaneous centers ofafour-bar mechanism

Velocity A nalysis ofLinkages 59
the above pairing oflinks, it can be observed that there are six ways ofcombining
links for a four-bar mechanism and hence, there are six instantaneous centers.
In general for n links in a mechanism the number of instantaneous centers is:
n(n-1)
number of instantaneous centers =
2
(.., "")
.) ..J.)
3.4.2 The Arnhold - Kennedy Theorem of Three centers
Since an instantaneous center is a point ~omhlon to two links, it is usually denoted by
the number oflinks. For the four-bar li~age shown in Fig. 3.23, centers I2, 23, 34 and
14 are located by inspection.
r
'////,
I
I
I
I
I
f I"
" .J
I
I
I
I
I
I
'////,
l
Fig. 3.23 Location ofall instantaneous centers ofa four-bar mechanism
The intersection of the perpendiculars to the velocities of points A and B yield the
instantaneous center I3 which is the point about which link 3 appears to rotate relative
to link I or vice versa.
To locate the center 24, the Arnhold- Kennedy theorem is applied. The theorem states
that:
When three bodies move relative to one another they have three instantaneous
centers, all ofwhich lie on the same straight line.
Demonstration:
Consider the three bodies shown in Fig. 3.24. Link I is the stationary link and links 2
and 3 rotate about the centers 12 and 13.

2
/
/
/
/
/
/
· ~
12_/_ .
Q.,
'///,
' '
"""" 13
-+--lt_B;-
'////,
Fig. 3.24 Demonstration ofthe line-of-centers
Let us assume that the instantaneous center 23 lies at point C. Then the velocity of C ·
as a point on link 2 isVc and as a point on link 3 is 1· c both perpendicular to 0 2 C and
1 j
0 3C, respectively. The directions ofthe velocities obviously do not coincide, hence,
the velocity ofCas a point on link 2 is different from the velocity ofpoint C as a point
on link 3. Thus point C cannot be the instantaneous center. The only point that satisfies
the definition ofthe instantaneous center lies on the straight line 0 20 3• ·Tnus, the three
instantaneous centers lie on a straight line.
Going back to the discussion ofthe four-bar linkage shown in Fig. 3.21, to locate the
instantaneous center 24, we proceed as follows.
For links 1, 2 and 4 the instantaneous centers 12, 24 and 14 should lie on a straight line.
Similarly, for links 2, 3, and 4, centers 23, 34 and 24 should lie on another straight line,
where 24, common to both lines, is located at the intersection ofthese lines.
.I
~ .
In general, to determine the instantaneous centers for the four bar linkage we could
proceed as follows: ·
"
For each combination of Jlinks the known and unknown instantaneous centers
should lie on a straight line. Intersections of such lines that contain
instantaneous centers give the unknown instantaneous centers.
The results thus obtained are shown in a table.
1

Instantaneous centers
Known Unknown
Link
I, 2, 3 I2, 23 13
I, 2, 4 12, I4 24
..
], 3, 4 34, 14J> l 13
J
2, 3, 4 23, 3'4
... ..
24
orthe Whitworth mechanism shown in Fig. 3.25, locate all the instantaneous centers.
·umber of links in the mechanism n = 6, therefore, the number of instantaneous
lS
_number of instantaneous centers =n(n -I) =15
2
By presenting the known and unknown instantaneous centers in a tabular form for all
possible combinations o.f three links, one is able to determine all the instantaneous
-,.. ....
·•.centers.
•
.
~,The fixed and permanent instantaneous centers are located by inspection. Thus, the
instantaneous centers 12, 14, 16, 23, 34, 45, 56, are located by inspection, which are the
known instantaneous-centers.
The known and unknown instantaneous centers are tabulated below for all possible
combination ofthree links.

34 at infinity
16 at infinity
34 at infinity
.I
Fig. 3.25 ' .,
Considering intersections of lines containing known and unknown instantaneous
centers, all instantaneous centers whichare neither fixed norpermanentare located. ·For
example, the instantaneous center 13 can be determined from the intersection ofthe
lines containing 12, 23, and 13, and 14, and 34 and 13. For further location of other
instantaneous centers, 13 is known..
The above procedure is repeated to locate all the instantaneous centers shown in Fig.
3.25.

Instantaneous centers
Links Known unknown
1, 2, 3 12, 23 13
1. 2, 4 12. 14 24
1, 2, 5 12 25, 15
1, 2, 6 I2, 16 26
1, 3, 4 I4, 34 13
I, 3, 5 b - ~ I3, I5, 35
I. 3. 6 I6, 13, 36
1, 4, 5 I4, 45 ' '· ' 15
'
-
1, 4, 6 I4, I6 r 46
I, 5, 6 I6, 56 15
2, 3, 4 23, 34 24
2, 3, 5 23 25, 35
2, 3, 6 23 26, 36
2, 4, 5 45 24,25
2, 4, 6 24, 26, 46
2, 5, 6 56 25, 26
3, 4, 5 34, 45 35
3, 4, 6 34 36, 46
3, 5, 6 56 35, 46
4, 5, 6 45,56 46
Determination ofvelocity
~,,_,....,,.....,..........,. the four bar linkage shown in Fig. 3.26. Link 2 has a constant angular
It is required to determine the velocity of points like B, D and E for the
instant represented. The velocities are determined using the line-of-centers method as
Consider the straight line defined by the instantaneous eenters 14, 12 and 24. This is .
the line ofinstantaneous centers for links 1, 2 and 4 called the line-of-centers.

64 Mechanisms ofjl1aclzinery
Fig. 2.26 Line-of-centers ofafour-bar mechanism
Point 24 is a point common to links 2 and 4 and has the same velocity whether it is
considered as a point of link 2 or link 4. Since the velocity of point A on link 2 is
known, considering 24 as a point on link 2 extended, we can determine its velocity.
Rotate point A to A' on the line of centers. Since A and A' are ofequal distance from
0 2, their velocities are also equal in magnitude.
The velocity of24 which is a point on the line ofcenters will have the same direction
as the velocity of A' and its magnitude is determined from its distance from 02> by
constructing similar triangles.
.I
Now considering point 24 as a point on link 4 extended whose velocity is determined,
we can use it to find the velocity of any other p~int.9n the same link.
With 0"' as center point B is rotated to B ' on the line of centers. The velocity ofBand
B ' are equal in magnitude. Thevelocity ofB 'can be determined by constructing similar
triangles with apex at 0 -1· VE is determined in a similar manner.
To determine V0 , we consider link 3 which contains point D, link 2 which contains a
point whose velocity is known and a reference link, link 1.
These three links define the instantaneous centers 12, and 31 which are on the line of
centers.

ocity ofthe common instantaneous center 23 is that ofA and is known. Point D
rotated to the line of centers 23-12-13 to D ' about point 13 and its velocity is
!A Pt·enuu.LI.JU from similar triangles.
24 and 23 are called the transf
er points, they are centers of the moving linl<.
taneous center 13 is called the pivot point.
.line-of-centers method is summarized as follows:
Step 1-
Step 2-
Step 3-
Step -1-
Identify the link containing a point whose velocity is known, the
A> "
link containing the point whose velocity is to be determined and
:
a reference li~, usually the ground or the frame.
/'
Locate the three instantaneous centers defined by the three links
and draw the line of centers.
Consider the common instantaneous center as a point on the link
which contains the point whose velocity is known and by using
similar triangles find the velocity ofthe common instantaneous
center.
Next considerthe common instantaneous center as a point on the
link which contains the point whose velocity is unknown. From
similar triangles determine the velocity ofthe point.
Determine the angular velocity OJ ofthetam head AE ofthe rock-crasher shown in Fig.
3.17 in the position for.which 8 = 600. The crank 00 has angular speed of 60
rev/min. When A is at the bottom ofits cirde, B and D are on a horizontal line through
0 4 and lines AB and 0 6D are vertical. The dimensions of the various links are
00 =100 mm, AB =750 mm, 0 6D =BD = 0 J3 =375mm
Construct the given configuration graphically and use the method of instantaneous
center of zero velocity. ~
··

66 .Mechanisms ofMachinery
3
Fig. 3.27
Solution
The schematic representation of the toggle mechanism is shown in Fig 3.27. The
instantaneous centers are located by the Arnhold-Kennedy theorem.
To determine the velocity ofpointE proceed as follows:
First identify links I, 2, and 6 where link I is the reference (fixed link); link 2 is a link
which contains a point w}:10se velocity is known, i.e. VB is known, and link 6 contains
./
the.point whose velocity we want to determine. ,·
Now the instantaneous centers I6, I2 and 2q ar1- located defining the line of three
centers as shown in Fig. 3.28. From the velocityofB, the velocity ofB' on link 2 and
on the line of three centers is drawn, Through the tip and 0 2 a line is drawn.
Considering 26 as a point on link 2, a line is drawn parallel to V8
to intersect the line
through the tip ofV8 at a point which determines V(26
J.
Again, consider 26 as a point on link 6. The velocity ofE is determined from similar
triangles by determining the velocity ofE' where E' is the point E referred to the line
of centers.
VE =439 mm/s

Fig. 3.28
angular velocity of link 6 is determined to be
- v£
- -
AE
-
439
=1.17 radls
375
.The 'Vertical oscillatory motion ofthe plunger F shown in Fig. 3.29 is actuated by the
'pressure change in .
the hydraulic cylinder E. For the position when 8 = 600, the plunger
has a downward velocity of 2 mls. Determine the velocity of the roller A in its
.. horizontal guide and the angular velocity of link AD.

Fig. 3.29
Solution
~'
To apply the line-of-centers method to solve the problem, first identify a link which
contains a point whose velocity is kno~ For this problem, link 3 is identified which
contains point D whose vertical component ofveloc.itY is known.
(V l = V = 2 m/s
D'y F
~
- 17 In addition, link 2 is considered which contains point A whose velocity is to be
''
determined~ The third link identified is link 1 which is the ground link. Links 1, 2, and
3 define the instantaneous centers 12, 23 and 13 which must lie in a straight line.
Again, consider links 1, 3 and 5 which define the instantaneous centers 13, 35, and 13
which must lie in a straighnine.
Considering these two lines-of-centers, 13 must lie at the intersection as shown in Fig,
3.30.

; 12
A, /
'//////.f / /////// VA
15 at
(Yo)y
Fig. 3.30
velocity of roller and the angular velocity of link AD are determine as follows.
. V = (VD )y =_l__
D cos/3 ci:Js/3
angle/3 is obtained from Fig. 3.30 to be 30°. Hence, the velocity ofthe roller Dis
2
·. • VD =-- =2.31 mls
cos/3
angular velocity oflink 3 is obtained from
VD
(JJAD =(JJ3 =-
CD
-·
rom Fig. 3.30, CD =172.8 mm. Hence, the angular velocity oflink 3 is
=:
2.31
0.1728
= 13.36 rad/s

To determine the velocity of point A, we know th~t
v:4
-
AC
From the above relation we obtain
V = V AC = 2.i2 m/s
A D 'CD
./

BLEMS
In the linkage shown in Fig. P3.1, 0 2C has a constant clockwise angular
velocity w2
=2 rad/s during an interval ofmotion while the hydraulic cylinder
gives pin A a constant velocity of I. 2 mls to the right. For the position shown
where 0 2 C is vertical and BC is horizontal, calculate the angular velocity of
BC.
"" B ~ 400 mm C
-:;..,......------------=-.
I ·.·.
300 mm
A
Fig. P3.1
By using complex numbers, determine the general equation of the angular
velocity w8
c of link BC of problem 3-1.
In the crossed linkage shown in Fig. P3.2 , crank 2 is the driver and rotates at a
constant velocity w2
= I O
k rad/s . Determine the velocity of c;3
and C also
the angular velocity oflink 4 for the phase where e =53° . Use vector algebra..
Given data:
0~ = I50mm , AG3 = 300mm , AB = 450mm ,
AC =600mm , 0 4B =I50mm , 0 2
0 4
=450mm
y
A
Fig. P3.2
- - x
c

. 3.4 Solve problem 3.3 graphically.
3.5 Fig. P3.3 shows linkage Or4B04
C which consists ofthree moving links 2, 3,
4 and ground link 1. Assuming pure rolling of4 on 1, find the velocity ofpoint
B, and the angular velocity of links 3 and 4 when link 2 rotates at a constant
angular velocity (J)2 =1 rad/s' and e =65° .
0 _r4 = 900mm , AB = 1200mm , BC = 600mm ,
r 1
= 600mm , r4
= 450mm
Fig. P3.3
3.6 Fig. P3.4 shows a swivelling-joint mechanism. The dimensions ofthe various
links are:
0.r4 =25mm , AB = 180mm , AD = AB = 90mm
0 4B =50 mm , DF = EF = 100mm , 0 2
0 4
= 150mm
The crank 0 _r4 rotates at 191.rpm as shown in _the figure. Determine the
velocity ofsliding oflinkDE in the trunnion. Det~rfnine also the velocity ofthe
slider F. ·
3.7 In the Whitworth mechanism shown in Fig. P3~5, crank 0_r4- rotates at 120 rpm
in clockwise direction. Determine the angular velocity ofthe slotted link AB
and the velocity ofthe slider D for the phase when e = 60°. The dimensions
ofthe various links are:
0_r4 =200mm , 0.2
04
= JOOmm , 0 4
C =150mm , CD =500mm
Solve the problem
a) graphically;
b) by using vector algebra.
J

11
!
--'·
85 mm
I
' 75 mm
Fig. P3.4
Fig. P3.5
I
I
I
i
1120 mm

74 Mechanisms ofMaclzinery
3.8 For the mecaanism shown in Fig. P3.6, locate all the instantaneous centers of
zero velocirv.
Ifthe crank 0~ rotates at a co-
nstant clockwise angular speed of120 rpm, find
the velocity ofB, C and D and the angular velocity oflinks AB, BC and CD by
using the instantaneous cemermethod. The Dimensions ofthe various links are:
0~ =200mm , AB =1500mm , BC =600mm ,
O.:B =400mm , CD =500mm
!35o mm
Fig. P3.6
3.9 The mechanism shown in Fig. P3.7, is driven such that velocity of point C,
Vc =250 mm/S to the right:. Rolling contact is assumed between links 1 and
2, but slip is possible between links 2 and 3. For the position shown in the
figure, determine the angular velocity of link 3.
v .
.I
,
_I
I
X
I
I
65 mm I
Fig. 3.40 I
r-,
' - ' I
·.

Link 2 of the mechanism shown in Fig. P3.8, rotates at 2000 rpm. Using the
instantaneous centers and the line-of-centers method, determine the velocity of
points B, C and D for the phase shown where 0 ~ is horizontal. The
dimensions ofthe various links are:
0~ =50mm , AB =150mm , AC =50mm, CD = 125mm
Fig. P3.8
3.11 In the oscillating cylinder mechanism shown in Fig. P3.9 crank 02
A rotates at
a constant speed ofIOOn:rad/s clockwise. The cylinder oscillates about 0 4
. For
the phase when e =60° , using vector algebra, determine:
a) velocity ofpiston B relative to the cylinder walls;
b) angular velocity ofthe piston rod AB;
c) acceleration ofthe piston B relative to the cylinder walls;
d) angular acceleration ofthe piston rod AB.
Fii!. P3.9

3.12 In the Geneva wheel shown in Fig. P3.10 the pin P is an integral part of wheel
A and the locking plate B engages the radial slots in wheel C turning the wheel
Cone-fourth ofa revGl.ution for each revolution ofthe pin. At the engagement
position shown~ 8 =45°. For constantclockwise angular velocity w2
ofwheel
A determine the angular velocity ofwheel C in terms of8 and fJ .
If w2
=2 rad/s clockwise, determine the corresponding angular velocity w3
of wheel c for the position when e =20° .
/
a I_/
~------~·----~~~
Fig. P3.LO
3.13 For the mechanism shown in Fig. P3.11, link 0 2
B has an angular velocity
w2 = 4 radls in the .counter clockwise direction for the instant when
8 =f3 =60°. Pin A is fixed on link 0 2
B and slides in the circular slot on link
0 3C. Theradius ofcurvature ofthe slot is 150 mm and for the phase shown, the
tangent to the slot at the point ofcontact is parallel 0 2
A . Determine the angular
velocity oflink 0 3 C for the phase shown.
l

150mm
Fig. P3.11
Gear D rotates in the counterclockwise direction about 0 2
with a constant
angular velocity w2 =4 rad/s. The 900 sector 0 2A B is mounted on an
independent shaft at 0 2 , and each ofthe small gears at A and B meshes with
gear D. Ifthe sector has a counter clockwise angular velocity cv3
=3 rad/s at
the instant represented, determine the corresponding angularvelocity ofeach of
the small gears. Use the instantaneous center method.
25 mm
Fig. P3.12

---.i 3.15 For the linkage shown in Fig. P3.15, determine the angular velocity oflinks 3
and 6 when e =45°. The angular velocity of link 2 is 1 rad/s. Dimensions
ofthe links are:
Oy4 =50 mm, AB =·250 mm, AC = 75 mm,0 6
C = 150 mm
2
Fig. P3.13
.I

80 Mechanisms oflv!achinery
Substituting for i·, j. and k ,
v =(xi + jij + zk) + i(roxz) + y(ro><J} + i(roxk)
Letting
(xi + jij + ik) = a
and noting that
x· (wxi) + y (w xj) + z· (w xk) =mx(xi + yj + zk)
=mxV
the acceleration component V can be written as
V =a + w xV
The last term on the right hand side ofthe acceleration equation is
= rox V + rox(ro><R)
Substituting for V, wxR· and V0
, the acceleration equation becomes
aP = a 0 + a + ffixR + 2roxV + ro x(roxR)
The different acceleration components are:
(4.5)
(4.6)
(4.7)
(4.8)
(4.9)
(4.10)
is Coriolis ' component ofac9-dleration, sense normal to V;
wxwxR
where w
v
R
is acceleration ofthe origin·ofx-y-z relative to X-Y-Z system;
is acceleration ofP relative to x-y-z system;
is the tangential accele-ation ofa point fixed on the x-y-z system
coincident with P as the system rotates about 0 ;
is the normal component of acceleration ofthe point coincident
withP;
is angular velocity ofx-y-z system relative to X-Y-Z system;
is velocity ofP relative to x-y-z system; and
is position vector ofP.

Acceleration Analysis ofLinkages 81
For the mechanism shown in Fig. 4.2, link 2 rotates at a constant angular velocity
012
== 2 radls in the clockwise direction and slider D m-
oves to the right at a constant
linear velocity of150mmls.By using vector algebra determine the acceleration ofpoint
C. Given are:
0~ == 150mm, AB == 175mm , AC = JOOmm, BC =ED = 200mm
y
Fig. 4.2
Solution:
From the geometry ofthe mechanism
OJ =OJ k ==-2k radls
2 2
The velocity of point B can be determined as follows. Since B remains fixed with
respect to A and D
vB =.vA + VB/A
VB = V D + VBID
Velocity ofpoint C can be written as:
vc = V A + V C/A
vc = V B + VCIB
Combining equations (a) and (b) we obtain:
,
V A + V~= V D + V B
ID
(a)
(b)
(c)
(d)
(e)

VA is known in magnitude with direction perpendicular to 0~
= -300 i mm/s
.-
VB/A is unknown in magnitude with direction perpendicular to AB
VB/A = -VB/A j
VD is known both in magnitude and direction
V0 · = V0 sin53 i + V0 cos53j
=200 i + 150 j mm/s
VB
IDis unknown in magnitude, direction perpendicular to BD
VBID = -VBID cos7 i + V81D sin7 j
= -0.993 V810 i + 0.122 VBID j mmls
Substituting into equation (e)
-300 i -VB
/A j =200 i + 150 j -0.993 VBID i + 0.122 VB!D j
Collecting like terms:
i terms:
-300 =200 - 0.993 VBID
which yields
VBID =503.5 mm/s
Or,
VBID = -500 i + 61.4 j mm/s
j terms:
-VB/A =150 + 0.122• VBID
which gives
VB
/A = -211.4 mm/s
Or,
.1'

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