Kinematics of machines

10-06-2017Jahangirabad Institute of Technology 1
 By
Md Gulfaraz Alam
Assistant professor
Jahangirabad institute of technology

Kinematics of Mechanics
Unit I
Introduction of Mechanisms and Machines

 In the starting of Kinematics of Machines we need to
understand the concepts of ‘Kinematics’ and ‘Dynamics’.
 Kinematics of mechanisms is concerned with the motion
of the parts without considering how the influencing
factors (force and mass) affect the motion. Therefore,
kinematics deals with the fundamental concepts.
 Kinetics deals with action of forces on bodies.
 Dynamics is the combination of kinematics and kinetics.
 Dynamics of mechanisms concerns the forces that act on
the parts -- both balanced and unbalanced forces, taking
into account the masses and accelerations of the parts as
well as the external forces.

The basic of KOM is include “Mechanisms” and “Machines”.
The word Mechanism has many meanings. In kinematics, a
mechanism is a means of transmitting, controlling, or
constraining relative movement .
Movements which are electrically, magnetically,
pneumatically operated are excluded from the concept of
mechanism.
The central theme for mechanisms is rigid bodies connected
together by joints.

 A machine is a combination of rigid or resistant
bodies, formed and connected do that they move with
definite relative motions and transmit force from the
source of power to the resistance to be overcome.
 A machine has two functions: transmitting definite
relative motion and transmitting force. These
functions require strength and rigidity to transmit the
forces.

Mechanisms can be divided into planar mechanisms and
spatial mechanisms, according to the relative motion of the
rigid bodies.
In a planar mechanisms, all of the relative motions of the
rigid bodies are in one plane or in parallel(2D) planes.
If there is any relative motion that is not in the same plane
or in parallel planes(3D), the mechanism is called the
spatial mechanism.
In other words, planar mechanisms are essentially two
dimensional while spatial mechanisms are three
dimensional.

 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.

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 rigid links.
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 forces only.
Fluid link.
A fluid link is one which is formed by having a fluid in a receptacle
and the motion is transmitted through the fluid by pressure or
compression only, as in the case of hydraulic presses, jacks and
brakes.

 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.

 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 lower pairs.

 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.
 A pair of friction discs,
toothed gearing, belt and
rope drives, ball and roller
bearings and cam and
follower are the examples of
higher pairs.

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.
 The motion of a square bar
in a square hole, as shown
in Fig. 2, and the motion of
a shaft with collars at each
end in a circular hole, as
shown in Fig.

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.

3.Successfully constrained
motion
 When the motion between the
elements, forming a pair, is
such that the constrained
motion is not completed by
itself, but by some other
means, then the motion is
said to be successfully
constrained motion.
 The motion of an I.C. engine
valve and the piston
reciprocating inside an
engine cylinder are also the
examples of successfully
constrained motion.

 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.

 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 :
 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.
 Another relation between the number of links (l) and the number of joints ( j )
which constitute a kinematic chain is given by the expression :
 If in the example
 L.H.S>R.H.S That’ll be considered as “Locked Chain”.
 L.H.S=R.H.S That’ll be considered as “One D.O.F”.
 L.H.S<R.H.S That’ll be considered as “Unconstrained Chain”.
42  pl
2
2
3
 lj

 Degrees of freedom/mobility of a mechanism:
 It is the number of inputs (number of independent
coordinates) required to describe the configuration or
position of all the links of the mechanism, with respect
to the fixed link at any given instant.
 Grubler’s equation:
hjln  2)1(3

When one of links is fixed in a kinematic chain, its
called mechanism. So we can obtain many mechanism
by fixing, in turn, different links in kinematic chain.
This method is known as “inversion of mechanism”.

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, is shown in Fig.
 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.

 Coupling rod of a locomotive (Double crank
mechanism).
 Watt’s indicator mechanism (Double lever
mechanism).

10-06-2017Jahangirabad Institute of Technology

10-06-2017Jahangirabad Institute of Technology
 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.

1. Pendulum pump or Bull
engine.
2. Oscillating cylinder
engine
3. Rotary internal
combustion engine
4. Crank and slotted lever
quick return motion
mechanism.
5. Whitworth quick return
motion mechanism.

 Elliptical trammels
 Scotch yoke mechanism
 Oldham’s coupling

Velocity and Acceleration diagram

Velocity analysis of any mechanism can be carried out by
various methods.
1. By graphical method
2. By relative velocity method
3. By instantaneous method

 By Graphical Method
The following points are to be considered while solving problems
by this method.
1. Draw the configuration design to a suitable scale.
2. Locate all fixed point in a mechanism as a common point in
velocity diagram.
3. Choose a suitable scale for the vector diagram velocity.

4. The velocity vector of each rotating link is r
to the link.
5. Velocity of each link in mechanism has both magnitude and
direction. Start from a point whose magnitude and direction is
known.
6. The points of the velocity diagram are indicated by small letters.

To explain the method let us take a few
specific examples.
1. Four – Bar Mechanism:
In a four bar chain ABCD link AD is fixed
and in 15 cm long. The crank AB is 4 cm
long rotates at 180 rpm (cw) while link CD
rotates about D is 8 cm long BC = AD and
BAD| = 60o
. Find angular velocity of link
CD.
Configuration Diagram
60o
wBA
A D
B
C
15 cm
15 cm
8 cm

Velocity vector diagram
Vb = r = ba x AB = 4x
60
120x2π
= 50.24 cm/sec
Choose a suitable scale
1 cm = 20 m/s = ab
r
to CD
r
to BC
r
to AB
a, d
b
C Vcb

Vcb = bc
Vc = dc = 38 cm/s = Vcd
We know that V = ωR
Vcd = cD x CD
WcD = 75.4
8
38Vcd

CD
rad/s (cw)

Learning Outcomes:
• This session deals with velocity vector diagrams for
determining the velocity at different points in
different mechanisms like IC engine Mechanism and
Crank and slotted Lever Mechanism.

• Introduction.
• Definition of Displacement, Velocity and
Acceleration.
• Difference between absolute Velocity and
Relative Velocity.
• Steps to construct Velocity Vector diagram.
• Velocity Vector diagram for a Four Bar
Mechanism

1. Slider Crank Mechanism:
In a crank and slotted lover mechanism crank rotates
of 300 rpm in a counter clockwise direction. Find
(i) Angular velocity of connecting rod and
(ii) Velocity of slider.
Configuration diagram
60 mm
45o
A
B
150 mm

Step 1: Determine the magnitude and velocity of
point A with respect to 0,
VA = O1A x O2A = 60x
60
300x2
= 600  mm/sec
Step 2: Choose a suitable scale to draw velocity vector diagram.
O
Vaa
b
r
to AB r
to OA
Along sides B
Vab = ab
ba =
150BA
Vba
 r/s
Vb = ob velocity of slider
Note: Velocity of slider is along
the line of sliding.

3. Shaper Mechanism:
In a crank and slotted lever mechanisms crank O2A rotates
at  r/s in CCW direction. Determine the velocity of slider.
Configuration diagram
4
O1
O2
C
B
3
2
W
5
6
D Scale 1 cm = x m/s

Va = 2 x O2A
CO
cO
BO
bO
1
1
1
1

To locate point C 






BO
CO
bOcO
1
1
11
Scale 1 cm = x m/s
d O1O2
VDC
c
a
b
VBA
VAO2 = VA
VBO1

 To Determine Velocity of Rubbing
Two links of a mechanism having turning point will be connected
by pins. When the links are motion they rub against pin surface.
The velocity of rubbing of pins depends on the angular velocity of
links relative to each other as well as direction.

For example: In a four bar mechanism we have
pins at points A, B, C and D.
 Vra = ab x ratios of pin A (rpa)
+ sign is used  ab is CW and Wbc is CCW i.e. when angular
velocities are in opposite directions use + sign when angular
velocities are in some directions use - ve sign.
VrC = (bc + cd) radius r
VrD = cd rpd
Problems on velocity by velocity vector method (Graphical

Problems on velocity by velocity vector
method (Graphical solutions)
 Problem 1:
In a four bar mechanism, the dimensions of the links are as given
below:
AB = 50 mm, BC = 66 mm
CD = 56 mm and AD = 100 mm
At a given instant when o
60DAB|  the angular velocity of link
AB is 10.5 r/s in CCW direction.

Determine,
i) Velocity of point C
ii) Velocity of point E on link BC when BE = 40 mm
iii) The angular velocity of link BC and CD
iv) The velocity of an offset point F on link BC, if BF = 45 mm, CF
= 30 mm and BCF is read clockwise.
v) The velocity of an offset point G on link CD, if CG = 24 mm,
DG = 44 mm and DCG is read clockwise.
vi) The velocity of rubbing of pins A, B, C and D. The ratio of the
pins are 30 mm, 40 mm, 25 mm and 35 mm respectively.

 Solution:
Step -1: Construct the configuration diagram
selecting a suitable scale.
60o
A D
B
C
F
G
Scale: 1 cm = 20 mm

Step – 2: Given the angular velocity of link AB and its direction of
rotation determine velocity of point with respect to A (A is fixed
hence, it is zero velocity point).
Vba = BA x BA
= 10.5 x 0.05 = 0.525 m/s

Step – 3: To draw velocity vector diagram
choose a suitable scale, say 1 cm = 0.2 m/s.
 First locate zero velocity points.
 Draw a line r
to link AB in the direction of rotation of link AB
(CCW) equal to 0.525 m/s.
C
f
Ved
a, d
e, g
Vba = 0.525 m/s
b

 From b draw a line r
to BC and from d. Draw d line r
to CD
to interest at C.
 Vcb is given vector bc Vbc = 0.44 m/s
 Vcd is given vector dc Vcd = 0.39 m/s

Step – 4: To determine velocity of point E (Absolute
velocity) on link BC, first locate the position of point E
on velocity vector diagram. This can be done by taking
corresponding ratios of lengths of links to vector
distance i.e.
BC
BE
bc
be
  be =
BC
BE
x Vcb =
066.0
04.0
x 0.44 = 0.24 m/s
Join e on velocity vector diagram to zero velocity points a, d
vector de = Ve = 0.415 m/s.

Step 5: To determine angular velocity of links BC and CD, we
know Vbc and Vcd.
 Vbc = WBC x BC
 WBC = )(.sec/6.6
066.0
44.0
cwrad
BC
Vbc

Similarly, Vcd = WCD x CD
 WCD = s/r96.6
056.0
39.0
CD
Vcd
 (CCW)

Step – 6: To determine velocity of an offset point F
to CF from C on velocity vector diagram.
to BF from b on velocity vector diagram to
intersect the previously drawn line at ‘f’.
 From the point f to zero velocity point a, d and measure
vector fa/fd to get Vf = 0.495 m/s.

Step – 7: To determine velocity of an offset point.
to GC from C on velocity vector diagram.
to DG from d on velocity vector diagram to
intersect previously drawn line at g.
 Measure vector dg to get velocity of point G.
Vg = s/m305.0dg 

Step – 8: To determine rubbing velocity at pins
 Rubbing velocity at pin A will be
Vpa = ab x rad of pin A = 10.5 x 0.03 = 0.315 m/s
 Rubbing velocity at pin B will be
Vpb = (ab + cb) x rad of point at B.
[ab CCW and cbCW]
Vpb = (10.5 + 6.6) x 0.04 = 0.684 m/s.
 Rubbing velocity at point D will be
 cd x rpd of pin D
= 6.96 x 0.035 = 0.244 m/s

Cam follower

Cams convert rotary
oscillating or linear
motion into a linear or
reciprocating action to
carry out useful work.

• Plate cam
• Cylindrical cam
• Face cam

• Plate cam
• Cylindrical cam
• Face cam
• End cam

Plate cam
Sometimes called a disc,
radial or edge cam.
Made up of a flat plate or
disc with an edge profile to
transmit motion.

 Plate,Wedge or Disc cam

 Cylindrical cam
Sometimes called a barrel
cam. It’s curved surface has
a groove machined, within
which a follower is
contained. The movement
is parallel to cam axis.

 Cylindrical cam

Face cam
In its flat surface, this
rotary cam has a groove cut
within which a constrained
follower moves. The groove
ensures no need for a
return spring.

 Face cam

End cam
In this case the end of the
cylindrical cam has the
profile machined on the
end.

 End cam

Wedge or Knife type
 This has the advantage that it
can follow complicated
profiles. However, it is not
often used as it wears rapidly,
due to the high frictional
forces. It can also track
profiles in both directions.

Wedge or Knife type

Roller
 This has the advantage that wear is
minimised, due to the roller action.
However, the profile of the cam and
roller radius must not conflict. If the
profile is smaller than the radius of
the roller then the roller will follow
the incorrect path.

Roller

Flat ended
 This cannot be used for concave
profiles. It wears slower than a
knife-edge follower, since the
points of contact move across
the surface of the follower. These
are often offset and allowed to
spin.

Flat ended

• List three applications of different
types of cams.
• What could be used today instead of
cams? (Think of CNC machines.)
• What materials could be used for
cams?

 The motion of the cam follower
depends upon the profile or
shape of the cam. Therefore the
profile determines the resultant
action.
 Cams are expensive to make and
therefore the cam follower is
normally sacrificial - allowed to
wear.

 Three Types
• Uniform velocity
• Uniform acceleration and
retardation

 Three Types
• Uniform velocity
• Uniform acceleration and
retardation
• Simple Harmonic Motion

Draw a cam displacement diagram for the following cam:
This cam needs to rise for 50 mm and fall 50mm due to uniform velocity.

Draw a cam displacement diagram for the following cam:
This cam needs to rise for 50 mm and fall 50mm due to uniform velocity.
Repeat this cam but rise due to SHM and fall using uniform acceleration
and retardation

 Homework Questions?
 Name 3 types of cam
 Name 2 types of follower
 Describe uniform acceleration
and retardation.
 Name the other types of
profile

 Gear and Gear train

 Power transmission is the movement of energy from its
place of generation to a location where it is applied to
performing useful work
 A gear is a component within a transmission device that
transmits rotational force to another gear or device

1. According to the position of axes of the shafts.
a. Parallel
1.Spur Gear
2.Helical Gear
3.Rack and Pinion
b. Intersecting
Bevel Gear
c. Non-intersecting and Non-parallel
worm and worm gears

 Teeth is parallel to axis of
rotation
 Transmit power from one
shaft to another parallel shaft

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

 Simple gear train
 Compound gear train
 Planetary gear train
Simple Gear Train
 The most common of the gear train is the gear pair
connecting parallel shafts. The teeth of this type can
be spur, helical or herringbone.
 Only one gear may rotate about a single axis

 For large velocities,
compound
arrangement is
preferred
 Two or more gears
may rotate about a
single axis

 They have higher gear ratios.
 They are popular for automatic transmissions in
automobiles.
 They are also used in bicycles for controlling power of
pedaling automatically or manually.
 They are also used for power train between internal
combustion engine and an electric motor

Kinematics of machines

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