The document discusses the instantaneous center of rotation (ICR) method for analyzing the velocity of links in planar mechanisms. It defines ICR as the point about which the motion of a link can be approximated as a pure rotation. The document outlines how to locate ICRs for different joint types using Kennedy's theorem. It provides examples of using the ICR method to determine velocities and angular velocities of links in slider-crank and four-bar mechanisms.

1. VELOCITY ANALYSIS OF SIMPLE
MECHANISMS: GRAPHICAL
METHODS
INSTANTANEOUS CENTER OF
ROTATION (ICR) METHOD
By
Mr. Keshav R. Pagar
Assistant Professor
Mechanical Engineering Department
GCOERC, Nashik-09
Email: keshavpagar04@gmail.com

2. OVERVIEW OF PRESENTATION
 Instantaneous center of rotation (ICR)
method:
 Definition of ICR,
 Body and space centrode.
 Types of ICRs,
 Methods of locating ICRs (limit to only 6
link mechanisms),
 Kennedy’s Theorem,
 Numerical on ICR Method

3. METHODS FOR DETERMINING THE VELOCITY
OF A POINT ON A LINK
1. Instantaneous center of rotation
(ICR) method
2. Relative velocity method

4. INSTANTANEOUS
CENTER OF
ROTATION (ICR)
 A any body has simultaneously a motion of
rotation as well as translation, such as wheel of
a car, a sphere rolling (not sleeping) on a
ground, such motion have combined effect of
rotation and translation.
 The combined motion of rotation and
translation of the link may be assumed to be a
motion of pure rotation about some centre I,
known as the instantaneous centre of
rotation (also called centrod or virtual
centre).

5. 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.

6. SPACE AND BODY CENTRODES
 IC: a point in the body which may be considered
fixed at any particular moment
 Space centrode: locus of IC in space during a
definite motion of the body
 Body centrode: locus of IC relative to body itself
 Body centrode rolls without slipping over space
centrode.
 Click for more info:
 https://www.youtube.com/watch?v=-k3_k84yKow

7. 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.
 Mathematically, number of instantaneous
centers,

8. TYPES OF INSTANTANEOUS CENTRES
 Primary instant centers
1. Fixed instant centers
2. Permanent instant centers
 Secondary instant centers
1. Neither fixed nor permanent instant centers

9. TYPES OF INSTANTANEOUS CENTRES
 Consider a four bar mechanism ABCD.
 The number of instantaneous centres (N) in a four bar
mechanism is given by

10.  From above fig:
 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 centers as they
vary with the configuration of the mechanism.

11. LOCATION OF INSTANTANEOUS CENTRES
1. When the two links are
connected by a pin joint (or
pivot joint), the
instantaneous centre lies
on the centre of the pin as
shown in Fig(a).
 Such a instantaneous centre
is of permanent nature, but
if one of the links is fixed,
the instantaneous centre
will be of fixed type.

12.  2. When the two links
have a pure rolling contact
(i.e. link 2 rolls without
slipping upon the fixed
link 1 which may be
straight or curved), the
instantaneous centre lies
on their point of contact,
as shown in Fig(b).

13.  3. When the two links have a
sliding contact, the
instantaneous centre lies on
the common normal at the
point of contact. We shall
consider the following three
cases :
 (a) When the link 2 (slider)
moves on fixed link 1 having
straight surface as shown in
Fig.(c), the instantaneous
centre lies at infinity and
each point on the slider have
the same velocity.

14.  (b) When the link 2 (slider)
moves on fixed link 1 having
curved surface as shown in
Fig.(d),the instantaneous
centre lies on the centre of
curvature of the curvilinear
path in the configuration at
that instant.

15.  (c) When the link 2
(slider) moves on fixed
link 1 having constant
radius of curvature as
shown in Fig.(e), the
instantaneous centre
lies at the centre of
curvature i.e. the centre
of the circle, for all
configuration of the
links.

16. KENNEDY (OR THREE CENTRES IN
LINE) THEOREM
 The Kennedy’s theorem states that if three bodies
move relatively to each other, they have three
instantaneous centres and lie on a straight line.
 Ibc must lie on the line joining Iab and Iac

17.  Consider Ibc lying outside the line joining Iab and Iac.
Now Ibc belongs to both the links B and C.
 Consider Ibc ϵ link B: VBC must be perpendicular to the
line joining Iab and Ibc.
 Consider Ibc ϵ link C:VBC must be perpendicular to the
line joining Iac and Ibc.
 But Ibc is a unique point; and hence, regardless of
whether it ϵ link B or Link C, it should have a unique
velocity (magnitude and direction). This is possible
only when the three instantaneous centres, namely,
Iab,Iac and Ibc lie on the same straight line.
 The exact location of Ibcon the line Iab Iac depends on
the directions and magnitudes of the angular
Velocities of B and C relative to A.

18. METHOD OF LOCATING INSTANTANEOUS
CENTRES IN A MECHANISM
1. First of all, determine the number of instantaneous
centres (N) by using the relation.

19. 2. Make a list of all the instantaneous centres in a
mechanism.
 3. Locate the fixed and permanent instantaneous
centres by inspection.
I12 and I14 are fixed instantaneous centres &
I23 and I34 are permanent instantaneous
centres.

20. 4. 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.(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.

21. 5. Join the points by solid lines to show that these
centres are already found. In the circle diagram [Fig.(b)]
these lines are 12, 23, 34 and 14 to indicate the centres
I12, I23, I34 and I14.
6. 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. (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 I13 is located. Join 1 and 3 by a
dotted line on the circle diagram and mark number 5 on
it.
[ I13 will lie on the intersection of and I14-I34 ]

22. 7. 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.
[ I24 will lie on the intersection of I12-I14 and I23-I34 ]
Hence all the six instantaneous centres are located.
 If there more number of ICRs to be locate draw a line
in a circle diagram in a way that it should form two
adjacent triangles and by extending the lines from IC
the required IC will get at intersection point.

23. FORMULAE TO FIND VELOCITY OF LINK.
 Relation between the angular velocity of links.
𝜔 𝑥
𝜔 𝑦
=
𝐼1𝑦 𝐼𝑥𝑦
𝐼1𝑥 𝐼𝑥𝑦
If value of 𝜔2 is known and we have to find the value of 𝜔4
then, the equation can be written as,
𝜔2
𝜔4
=
𝐼14 𝐼24
𝐼12 𝐼24
Where 𝐼14 𝐼24 is the distance between the ICR 𝐼14 𝑎𝑛𝑑 𝐼24
Similarly 𝐼12 𝐼24 is the distance between the ICR
𝐼12 𝑎𝑛𝑑 𝐼24
 To find velocity of link
V=𝟂 x r
Where 𝟂- angular velocity of link (calculated from above
formula)
r- Length of link whose velocity is to be found.

24. Steps to Draw Circle Diagram:
 Draw a circle of any diameter.
 Divide the circle into “n” parts.(n- No. of links)
 Join the ICRs by solid line which are already located.
 Now the remaining ICRs are to be found by simple way,
 Those ICR you want to find join that 2 points in a way that
the line (Dotted) joining will divide the diagram into two
adjacent triangles.
 Draw one line passing the ICRs
of two sides of one triangle.
 Draw another line from ICRs
sides of other triangle.
 The intersection point will
represents the IC to be locate.
 Repeat the procedure to find n
number of ICRs.

25. EX 1 : IN A PIN JOINTED FOUR BAR MECHANISM, AS SHOWN IN FIG, AB = 300
MM, BC = CD = 360 MM, AND AD = 600 MM. THE ANGLE BAD = 60°. THE
CRANK AB ROTATES UNIFORMLY AT 100 R.P.M. LOCATE ALL THE
INSTANTANEOUS CENTRES AND FIND THE ANGULAR VELOCITY OF THE LINK BC.
Sol. Given : NAB = 100 r.p.m
ωAB = 2 π × 100/60 =10.47 rad/s
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
Draw the mechanism with given dimensions(or with scale)
(Draw the diagram by keeping distance to all sides)

26.  Location of instantaneous centres
1. The mechanism consists of four links (i.e. n = 4 ),
therefore number of instantaneous centres,
2. Make a list of all the instantaneous centres in a
mechanism.

27.  3. Locate the fixed and permanent instantaneous
centres by inspection. These centres are I12, I23, I34
and I14.
 4. Locate the remaining neither fixed nor permanent
instantaneous centres by Kennedy’s theorem. This is
done by circle diagram

28.  Mark four points (equal to the number of links in
a mechanism) 1, 2, 3, and 4 on the circle.

29.  For I13 join 13 in a way that this divides the diagram
into two adjacent triangle.
Triangle 1 : 1-2-3 Sides: I12 & I23
Triangle 2 : 1-4-3 Sides: I14 & I34
 Line 1 From: I12 & I23
 Line 2 From: I14 & I34
 The Intersection of
this line 1 & 2 will be
ICR I13.

30.  For I24 join 13 in a way that this divides the diagram
into two adjacent triangle.
Triangle 1 : 2-1-4 Sides: I12 & I14
Triangle 2 : 2-3-4 Sides: I23 & I34
 Line 1 From: I12 & I14
 Line 2 From: I23 & I34
 The Intersection of
this line 1 & 2 will be
ICR I24.
All ICRs are located.

31.  Angular velocity of the link BC
We know the formula
𝜔 𝑥
𝜔 𝑦
=
𝐼1𝑦 𝐼 𝑥𝑦
𝐼1𝑥 𝐼 𝑥𝑦
𝜔2= 10.47 rad/sec (known) 𝜔3 is to be found
So
𝜔3
𝜔2
=
𝐼12 𝐼23
𝐼13 𝐼23
Measure the distance between
I12 I23=AB=0.3m
I13 I23= 0.5m (by Measurement from diagram)
Then by putting values
𝜔3
10.47
=
0.3
0.5
𝜔3=6.282 rad/sec

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

33. ANSWER:
 VA= 0.82 m/s 𝜔 𝐴𝐵= 1.78rad/s

34.  Ex 3:The mechanism of a wrapping machine, as shown
in Fig, has the following dimensions : O1A = 100 mm;
AC = 700 mm; BC = 200 mm; O3C = 200 mm; O2E =
400 mm; O2D = 200 mm and BD = 150 mm. The crank
O1A rotates at a uniform speed of 100 rad/s. Find the
velocity of the point E of the bell crank lever by
instantaneous centre method.

35. ANSWER:
VB = 9.01m/s
VD = 3.46m/s
VE = 6.92m/s

36. Ex.4: Locate all the instantaneous centres of the
mechanism as shown in Fig. The lengths of various links
are : AB = 150 mm ; BC = 300 mm ; CD = 225 mm ; and
CE = 500 mm. When the crank AB rotates in the
anticlockwise direction at a uniform speed of 240 r.p.m. ;
find 1. Velocity of the slider E, and 2. Angular velocity of
the links BC and CE.
[Ans. 1.6 m/s ; 2.4 rad/s ; 6.6 rad/s]

37. Ex. 5: Locate all the instantaneous centres for the
crossed four bar mechanism as shown in Fig. The
dimensions of various links are : CD = 65 mm; CA = 60
mm ; DB = 80 mm ; and AB = 55 mm. Find the angular
velocities of the links AB and DB, if the crank CA
rotates at 100 r.p.m. in the anticlockwise direction
[Ans. 50 rad/s ; 27 rad/s]

38. Ex. 6: A mechanism, as shown in
Fig, has the following dimensions :
O1 A = 60 mm ; AB = 180 mm ; O2 B
= 100 mm ; O2 C = 180 mm and CD
= 270 mm. The crank O1 A rotates
clockwise at a uniform speed of 120
r.p.m. The block D moves in vertical
guides. Find, by instantaneous
centre method, the velocity of D and
the angular velocity of CD.
[Ans. 0.08 m/s ; 1.43 rad/s]

39. Ex. 7: The lengths of various links of a mechanism, as
shown in Fig. are : OA = 0.3 m ; AB = 1 m ; CD = 0.8 m ;
and AC = CB. Determine, for the given configuration,
the velocity of the slider D if the crank OA rotates at 60
r.p.m. in the clockwise direction. Also find the angular
velocity of the link CD. Use instantaneous centre
method.
[Ans. 480 mm/s ; 2.5 rad/s]