Chapter 3. velocity analysis (IC,GRAPHICAL AND RELATIVE VELOCITY METHOD)

1
Velocity Analysis of Linkages
Chapter 3
Mechanisms of Machinery
(MEng 3301)
Mechanical Engineering
Department
Prepared by: Addisu D. November, 2013

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Velocity Analysis of Linkages
 Velocities and accelerations in mechanisms are determined
by different methods.
i. Velocity and acceleration analysis using vector mathematics
 velocity and acceleration of a point are expressed
relative to fixed or moving coordinates.
i. Velocity and acceleration analysis using equations of
relative motion
 Can be solved graphically by velocity and acceleration
polygons or by using trigonometric relations.
i. Velocity and acceleration analysis by using complex
numbers.
ii. Vectors velocity analysis using the instant center method.

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3.1. Velocity Analysis by Vector Mathematics
 Consider the motion of point P moving with respect to the
x- y- z coordinate system, which in turn, moves relative to
the X-Y-Z coordinate system as shown.
 - is the position vector of P relative to the X-Y-Z system.
 - is the position vector of P relative to the x-y-z system
 - is the position vector of the origin of the moving
coordinate system x-y-z relative to the fixed coordinate
system X-Y-Z.
pR

R

oR


4
 The position vector of P relative to the X-Y-Z system Rp is
expressed as:
(3.1)
 Introducing unit vectors i, j, and k along the x, y, and z axes
respectively,
= xi + yj + zk (3.2)
 Velocity of P relative to the X-Y-Z coordinate system is
(3.3)
- is the velocity of the origin of x-y-z system
relative to the fixed system
(3.4)
RRRV opp

+==
oo VR

=
( )
)()( kzjyixkzjyix
zkyjxi
dt
d
R


+++++=
++=
RRR op

+=
R


5
Let
(3.5)
note that
(3.6)
Where ω is the angular velocity vector of x-y-z system relative
to X-Y-Z.
⇒
(3.7)
- Thus the velocity of P relative to the moving coordinate system
is
(3.8)
Vkzjyix =++ 
kk
jj
ii
×=
×=
×=
ω
ω
ω



R
zkyjxi
kzjyixkzjyix


×=
++×=
×+×+×=++
ω
ω
ωωω
)(
)()()(
RVR

×+= ω

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∴ the velocity of point P relative to the fixed system is:
= velocity of the origin of the x-y-z system relative to
the X-Y- Z system
= velocity of point p relative to x-y-z system
ω = angular velocity of the x-y-z system relative to X-Y-Z
system
= position vector of P with respect to the origin of the
x-y-z system
)9.3(RVVV op

×++= ω
oV

V

R


7
3.2. Velocity analysis by using equation of
relative motion
3.2.1. Velocity of points on a common link
 A and B are two points on a common rigid link AB as shown.
 The points are moving with velocities VA and VB
respectively.
 Using the equation of relative motion, velocity of one point
can be determined relative to the other.
VA = VB + VA/B (3.10)
Where VA/B = velocity of A relative to B
 All absolute velocity vectors originate from the same point
O2.
 Note that the velocity of A relative to B and the velocity of B
relative to A are equal in magnitude, collinear and opposite

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3.2.2. Velocity of a block sliding on a rotating
link
- As shown in the figure block A slides on the rotating link O2B.
- The angular velocity ω of the link and the velocity of the
block are assumed to be known.
-Let A’ be a point on the link coincident with the block A for
the instant represented.
-The velocity of A’ relative to O2 is perpendicular to O2B at A’.
-The velocity of A relative to A’ is along the link parallel to
O2A’
-VA = VA’ + VA/A’ (3.12)
-Relative velocity of coincident particles on separate links is

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3.2.3 . Relative velocity of crank and
connecting rod
 Let ω2 be the angular velocity of the crank O2A.
 Velocity of B can be determined using the velocity of point
A as the reference which can easily be determined.
VB = VA + VB/A
(3.16)
Where
VA is known both in magnitude and direction;
VB is known in direction, magnitude is unknown;

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3.3. Velocity Analysis by Complex
Numbers
 Most of the systems of analysis using complex polar
notation are based on the following fundamental law:
If the elements of a mechanism are replaced by position
vectors such that their sum is zero, then their time
derivatives are also equal to zero.
 This law means that if one takes any linkage or mechanism
and replaces the members of the mechanism by vectors
such that their sum is zero, then the sum of the velocity
vectors is zero, so also the sum of the acceleration vectors.
Considering the slider crank shown:
 Link 2 is the driver (crank) and has
a constant angular velocity ω2 & for
the instant under consideration an
angular position of θ2.
 Dimensions of linkages are assumed
to be known, so the angular position
of the follower, link 4, can be obtained.

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 Replacing each link by a vector such that the position
polygon closes as shown in figure (b), a mathematical
expression for the summation law can be written as:
Where R1 = vector for the ground link
R2 = vector for the crank
R4 = vector to determine the position of link 3. Note that
the magnitude of R4 is variable.
 The position of a particle on a link represented by a vector Rp
as shown below may be expressed in any of the following
equivalent forms:
 Using this complex representation, equation (3.25) is
transformed into
)25.3(0421 =−+ RRR
( )
2
)26.3(sincos 22
θ
θθ
i
pp
pp
p
erR
irR
ibaR
=
+=
+=
)27.3(0421
421 =−+ θθθ iii
ererer

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 Differentiating the above equation we obtain
 Note that r1 and θ1 are constants and r4 is variable.
 Let
 Separating equation (3.29) into real and imaginary terms:
 The unknown quantities in the above pair of equations are ω4
and
solving for these unknown variables:
)28.3(0442
44422 =−− θθθ
θθ iii
ereireir 
ii ωθ =
)29.3(0442
44422 =−− θθθ
ωω iii
ereireir 
)30.3(0sincossin
0cossincos
44444222
44444222
=+−−
=−−
θωθθω
θωθθω
rrr
rrr


4r
)31.3()cos(
)sin(
24
4
22
4
24224
θθ
ω
ω
θθω
−=
−=
r
r
rr

Definition
The definition of an instant center of a velocity is a point,
common to two bodies in plane motion, which the point has the
same instantaneous velocity in each body.
The instantaneous center of velocity is defined as the
instantaneous location of a pair of coincident points of two
different rigid bodies for which the absolute velocities of the
two points are equal.
It may also be defined as the location of a pair of coincident
points of two different rigid bodies for which the apparent
velocity of one of the points is zero as seen by an observer on
the other body.
3.4. Analysis of Velocity Vectors by
Instant-Center Method.
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Velocity Analysis by Instant-center Method
The instantaneous center method of analyzing the motion in
a mechanism is based upon the concept that any
displacement of a body (or a rigid link) having motion in
one plane, can be considered as a pure rotational motion of a
rigid link as a whole about some centre, known as
instantaneous centre or virtual centre of rotation.
Consider two points A and B on a rigid link. Let vA and vB be
the velocities of points A and B, whose directions are given
by angles α and β as shown in Fig. If vA is known in
magnitude and direction and vB in direction only, then the
magnitude of vB may be determined by the instantaneous
centre method as discussed below :
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Velocity Analysis by Instant-center Method
Draw AI and BI perpendiculars
to the directions vA and vB
respectively.
Let these lines intersect at I,
which is known as
instantaneous centre or virtual
centre of the link.
The complete rigid link is to
rotate or turn about the centre I.
Since A and B are the points on
a rigid link, therefore there
cannot be any relative motion
between them along the line
AB.
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Properties of the Instantaneous Centre
The following properties of the instantaneous centre are
important from the subject point of view :
 A rigid link rotates instantaneously relative to another link at
the instantaneous centre for the configuration of the
mechanism considered.
 The two rigid links have no linear velocity relative to each
other at the instantaneous centre. At this point (i.e.
instantaneous centre), the two rigid links have the same linear
velocity relative to the third rigid link.
 In other words, the velocity of the instantaneous centre relative
to any third rigid link will be same whether the instantaneous
centre is regarded as a point on the first rigid link or on the
second rigid link.
17

Number of Instantaneous Centres in a
Mechanism
The number of instantaneous centers in a constrained
kinematic chain is equal to the number of possible
combinations of two links. The number of pairs of links or
the number of instantaneous centers is the number of
combinations of n links taken two at a time.
Mathematically, number of instantaneous centers,
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Types of Instantaneous Centres
The instantaneous centers for a mechanism are of the
following three types :
 Fixed instantaneous centers,
 Permanent instantaneous centres, and
 Neither fixed nor permanent instantaneous centres.
The first two types i.e. fixed and permanent instantaneous
centres are together known as primary instantaneous
centres and the third type is known as secondary
instantaneous centres.
19

Consider a four
bar mechanism
ABCD.
The number of
instantaneous
centres (N) in a
four bar
mechanism is
given by
20

The instantaneous centres I12 and I14 are called the fixed
instantaneous centres as they remain in the same place for
all configurations of the mechanism.
The instantaneous centres I23 and I34 are the permanent
instantaneous centres as they move when the mechanism
moves, but the joints are of permanent nature.
The instantaneous centres I13 and I24 are neither fixed nor
permanent instantaneous centres as they vary with the
configuration of the mechanism.
21

Location of Instantaneous Centres
The following rules may be used in locating the
instantaneous centres in a mechanism :
1. When the two links are connected by a pin joint (or pivot
joint), the instantaneous centre lies on the centre of the pin.
Such a instantaneous centre is of permanent nature, but if one
of the links is fixed, the instantaneous centre will be of fixed
type.
22

1. When the two links have a pure rolling contact (i.e. link 2
rolls without slipping upon the fixed link 1 which may be
straight or curved), the instantaneous centre lies on their
point of contact. The velocity of any point A on the link 2
relative to fixed link 1 will be perpendicular to I12 A and is
proportional to I12 A . In other words
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1. When the two links have a sliding contact, the
instantaneous centre lies on the common normal at the
point of contact. We shall consider the following three
cases :
 When the link 2 (slider) moves on fixed link 1 having straight
surface, the instantaneous centre lies at infinity and each point
on the slider have the same velocity.
 When the link 2 (slider) moves on fixed link 1 having curved
surface, the instantaneous centre lies on the centre of curvature
of the curvilinear path in the configuration at that instant.
 When the link 2 (slider) moves on fixed link 1 having constant
radius of curvature, the instantaneous centre lies at the centre of
curvature i.e. the centre of the circle, for all configuration of
the links.
24

25

Aronhold Kennedy (or Three Centres in
Line) Theorem
The Arnhold- Kennedy
theorem states that:
When three bodies move
relative to one another
they have three
instantaneous centers, all
of which lie on the same
straight line.
For the four-bar linkage
shown in Fig., centers 12,
23, 34 and 14 are located
by inspection.
26

The intersection of the perpendiculars to the velocities of
points A and B yield the instantaneous center 13 which is
the point about which link 3 appears to rotate relative to
link 1 or vice versa.
Demonstration:
Consider the three bodies shown below. Link 1 is the stationary link
and links 2 and 3 rotate about the centers 12 and 13.
Let us assume that the instantaneous center 23 lies at point C. Then
the velocity of C as a point on link 2 is Vc2 and as a point on link 3
is Vc3 both perpendicular to 02C and 03C, respectively.
The directions of the velocities obviously do not coincide, hence,
the velocity of C as a point on link 2 is different from the velocity
of point C as a point on link 3.
Aronhold Kennedy (or Three Centres in Line)
Theorem
27

Thus point C cannot
be the instantaneous
center.
The only point that
satisfies the
definition of the
instantaneous center
lies on the straight
line 0203.
Thus, the three
instantaneous
centers lie on a
straight line.
Demonstration of Aronhold Kennedy Theorem
28

Method of Locating Instantaneous Centres
in a Mechanism
Consider a pin jointed four bar mechanism shown. The
following procedure is adopted for locating instantaneous
centres.
1. First of all, determine the number of instantaneous centres
(N) by using the relation
29

1. Make a list of all the instantaneous centres in a
mechanism. Since for a four bar mechanism, there are six
instantaneous centres, therefore these centres are listed as
shown in the following table (known as book-keeping
table).
1. Locate the fixed and permanent instantaneous centres by
inspection. I12 and I14 are fixed instantaneous centres and
I23 and I34 are permanent instantaneous centres.
Method of Locating Instantaneous Centres in a
Mechanism
30

1. Locate the remaining neither fixed nor permanent instantaneous
centres (or secondary centres) by Kennedy’s theorem. This is done
by circle diagram as shown in Fig. 6.8 (b). Mark points on a circle
equal to the number of links in a mechanism. In the present case,
mark 1, 2, 3, and 4 on the circle.
Mechanism
31

1. Join the points by solid lines to show that these centres are already
found. In the circle diagram [Fig. 6.8 (b)] these lines are 12, 23, 34
and 14 to indicate the centres I12, I23, I34 and I14.
2. In order to find the other two instantaneous centres, join two such
points that the line joining them forms two adjacent triangles in the
circle diagram. The line which is responsible for completing two
triangles, should be a common side to the two triangles. In Fig. 6.8
(b), join 1 and 3 to form the triangles 123 and 341 and the
instantaneous centre* I13 will lie on the intersection of I12 I23 and I14
I34, produced if necessary, on the mechanism. Thus the
instantaneous centre I 13 is located. Join 1 and 3 by a dotted line on
the circle diagram and mark number 5 on it. Similarly the
instantaneous centre I24 will lie on the intersection of I12 I14 and I23
I34, produced if necessary, on the mechanism. Thus I24 is located.
Join 2 and 4 by a dotted line on the circle diagram and mark 6 on it.
Hence all the six instantaneous centres are located.
Mechanism
32

Mechanism
33

Mechanism
35

VELOCITY ANALYSIS WITH INSTANT
CENTERS
Once the ICs have been found, they can be used to do a
very rapid graphical velocity analysis of the linkage.
From the definition of the instant center, both links sharing
the instant center will have identical velocity at that point.
36

Example 1. In a pin
jointed four bar
mechanism shown, AB =
300 mm, BC = CD = 360
mm, and AD = 600 mm.
The angle BAD = 60°.
The crank AB rotates
uniformly at 100 r.p.m.
Locate all the
instantaneous centres and
find the angular velocity
of the link BC.
VELOCITY ANALYSIS WITH INSTANT CENTERS
37

Example 1
Solution. Given : NAB = 100 r.p.m
or ωAB = 2 × 100/60 = 10.47 rad/s
Since the length of crank AB = 300 mm = 0.3 m, therefore
velocity of point B on link AB,
vB = ωAB × AB = 10.47 × 0.3 = 3.141 m/s
Location of instantaneous centres
The instantaneous centres are located as discussed below:
 Since the mechanism consists of four links (i.e. n = 4 ),
therefore number of instantaneous centres,
38

1. For a four bar mechanism, the book keeping table may be
drawn.
2. Locate the fixed and permanent instantaneous centres by
inspection. These centres are I12, I23, I34 and I14, as shown in
Fig. 6.10.
3. Locate the remaining neither fixed nor permanent
instantaneous centres by Aronhold Kennedy’s theorem.
This is done by circle diagram as shown in Fig. 6.11. Mark
four points (equal to the number of links in a mechanism)
1, 2, 3, and 4 on the circle.
39

40

1. Join points 1 to 2, 2 to 3, 3 to 4 and 4
to 1 to indicate the instantaneous
centres already located i.e. I12, I23, I34
and I14.
2. Join 1 to 3 to form two triangles 1 2 3
and 3 4 1. The side 13, common to
both triangles, is responsible for
completing the two triangles.
Therefore the instantaneous centre I13
lies on the intersection of the lines
joining the points I12 I23 and I34 I14 as
shown in Fig. 6.10. Thus centre I13 is
located. Mark number 5 (because four
instantaneous centres have already
been located) on the dotted line 1 3.41

1. Now join 2 to 4 to complete two
triangles 2 3 4 and 1 2 4. The side
2 4, common to both triangles, is
responsible for completing the two
triangles. Therefore centre I24 lies
on the intersection of the lines
joining the points I23 I34 and I12 I14
as shown in Fig. 6.10. Thus centre
I24 is located. Mark number 6 on
the dotted line 2 4. Thus all the six
instantaneous centres are located.
42

Angular velocity of the link BC
Let ωBC = Angular velocity of the link BC.
Since B is also a point on link BC, therefore velocity of
point B on link BC,
By measurement, we find that I13B = 500 mm = 0.5 m
43

Example 2.
Locate all the instantaneous centres of the slider crank
mechanism as shown in Fig. 6.12. The lengths of crank OB and
connecting rod AB are 100 mm and 400 mm respectively. If the
crank rotates clockwise with an angular velocity of 10 rad/s, find:
1. Velocity of the slider A, and 2. Angular velocity of the
connecting rod AB.
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45

46

47

48

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End of Chapter 3
Next Lecture
Chapter 4: Acceleration Analysis of
Linkages
57

Chapter 3. velocity analysis (IC,GRAPHICAL AND RELATIVE VELOCITY METHOD)

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