ICR Velocity Analysis Graphical Method, Theory of Machine

Velocity Analysis
Graphical Method
ICR of mechanism
Instantaneous Center of Rotation
Theory of Machine I

1. VELOCITY ANALYSIS OF SIMPLE
MECHANISMS: GRAPHICAL
METHODS
INSTANTANEOUS CENTER OF
ROTATION(ICR) METHOD
By
Mr. KeshavR. Pagar
Assistant Professor
Mechanical Engineering Department
GCOERC, Nashik -09
Email: [email protected]

2. OVERVIEW OFPRESENTATION
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 onICR Method

3. METHODSFORDETERMINING THEVELOCITY
OFAPOINTONALINK
1.Instantaneous center of rotation
(ICR) method
2. Relative velocity method

4. INSTANTANEOUS
CENTER OF
ROTATION(ICR)
Aanybodyhassimultaneouslyamotionof
rotationaswellastranslation,suchaswheelof
acar,asphererolling(notsleeping)ona
ground,suchmotionhavecombinedeffectof
rotationandtranslation.
Thecombinedmotionofrotationand
translationofthelinkmaybeassumedtobea
motionofpurerotationaboutsomecentreI,
knownastheinstantaneouscentreof
rotation(alsocalledcentrodorvirtual
centre).

5. VELOCITYANALYSISBYINSTANT-
CENTERMETHOD
Theinstantaneouscentermethodof
analyzingthemotioninamechanismis
basedupontheconceptthatany
displacementofabody(orarigidlink)
havingmotioninoneplane,canbe
consideredasapurerotationalmotionofa
rigidlinkasawholeaboutsomecentre,
knownasinstantaneouscentreorvirtual
centreofrotation.

6. SPACEANDBODYCENTRODES
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. NUMBEROFINSTANTANEOUS CENTRESINA
MECHANISM
Thenumberofinstantaneouscentersina
constrainedkinematicchainisequaltothe
numberofpossiblecombinationsoftwolinks.
Mathematically,numberofinstantaneous
centers,

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

9. TYPESOFINSTANTANEOUS 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 centresI12 and I14 are called
the fixed instantaneous centresas they remain in
the same place for all configurations of the
mechanism.
The instantaneous centresI23 and I34 are the
permanent instantaneous centresas they move
when the mechanism moves, but the joints are of
permanent nature.
The instantaneous centresI13 and I24 are neither
fixed nor permanent instantaneous centers as they
vary with the configuration of the mechanism.

11. LOCATION OFINSTANTANEOUS CENTRES
1.Whenthetwolinksare
connectedbyapinjoint(or
pivot joint), the
instantaneouscentrelies
onthecentreofthepinas
showninFig(a).
Suchainstantaneouscentre
isofpermanentnature,but
ifoneofthelinksisfixed,
theinstantaneouscentre
willbeoffixedtype.

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 centrelies
on their point of contact,
as shown in Fig(b).

13. 3.When the two links have a
sliding contact, the
instantaneous centrelies 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
centrelies 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
centrelies on the centreof
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 centreof
curvature i.e. the centre
of the circle, for all
configuration of the
links.

16. KENNEDY(ORTHREECENTRESIN
LINE) THEOREM
The Kennedy’s theorem states that if three bodies
move relatively to each other, they have three
instantaneous centresand lie on a straight line.
Ibcmust lie on the line joining Iaband Iac

17. Consider Ibclying outside the line joining Iaband Iac.
Now Ibcbelongs to both the links B and C.
Consider Ibcϵlink B: VBC must be perpendicular to the
line joining Iaband Ibc.
Consider Ibcϵlink C:VBC must be perpendicular to the
line joining Iacand Ibc.
But Ibcis a unique point; and hence, regardless of
whether it ϵlink B or Link C, it should have a unique
velocity (magnitude and direction). This is possible
only when the three instantaneous centres, namely,
Iab,Iacand Ibclie on the same straight line.
The exact location of Ibconthe line IabIacdepends on
the directions and magnitudes of the angular
Velocities of B and C relative to A.

18. METHODOFLOCATINGINSTANTANEOUS
CENTRESINAMECHANISM
1. First of all, determine the number of instantaneous
centres(N) by using the relation.

19. 2. Make a list of all the instantaneous centresin a
mechanism.
3.Locate the fixed and permanent instantaneous
centresby 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
centresare 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
centreI13 will lie on the intersection of I12 I23 and I14
I34, produced if necessary, on the mechanism. Thus the
instantaneous centreI13 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 centreI24 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 centresare located.
If theremorenumberofICRstobelocatedrawaline
inacirclediagraminawaythatitshouldformtwo
adjacent triangles and by extending the lines from IC
the required IC will get at intersection point.

23. FORMULAE TOFINDVELOCITYOFLINK.
Relation between the angular velocity of links.
??????
�
??????
�
=
??????
1�??????
��
??????
1�??????
��
If value of??????
2is known and we have to find the value of ??????
4
then, the equation can be written as,
??????
2
??????
4
=
??????
14??????
24
??????
12??????
24
Where ??????
14??????
24is the distance between the ICR ??????
14????????????????????????
24
Similarly ??????
12??????
24is the distance between the ICR
??????
12????????????????????????
24
To find velocity of link
V=??????x r
Where??????-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 thecircleinto“n” parts.(n-No. of links)
Join theICRsby solid line which arealreadylocated.
Now theremaining ICRsaretobefound 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 passingtheICRs
of two sidesof one triangle.
Draw another line from ICRs
sides of other triangle.
The intersection point will
represents theICtobelocate.
Repeat theprocedure to find n
number of ICRs.

25. EX1 :INAPINJOINTEDFOURBARMECHANISM, ASSHOWNINFIG, AB = 300
MM, BC = CD = 360 MM, ANDAD = 600 MM. THEANGLEBAD = 60°. THE
CRANKAB ROTATESUNIFORMLY AT100 R.P.M. LOCATEALLTHE
INSTANTANEOUS CENTRESANDFINDTHEANGULAR VELOCITYOFTHELINKBC.
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 centresin a
mechanism.

27. 3. Locate the fixed and permanent instantaneous
centresby inspection. These centresare I12, I23, I34
and I14.
4.Locate the remaining neither fixed nor permanent
instantaneous centresby 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
Line2From: I14 & I34
The Intersectionof
this line1&2will be
ICRI13.

30. For I24 join 13 in a way that this divides the diagram
into two adjacent triangle.
Triangle 1 : 2-1-4Sides: I12 & I14
Triangle 2 : 2-3-4 Sides: I23 & I34
Line 1 From: I12 & I14
Line2From: I23 & I34
The Intersectionof
this line1&2will be
ICRI24.
All ICRs are located.

31. Angular velocity of the link BC
We know the formula
??????
�
??????
�
=
??????
1�??????
��
??????
1�??????
��
??????
2= 10.47 rad/sec (known) ??????
3is to be found
So
??????
3
??????
2
=
??????
12??????
23
??????
13??????
23
Measurethedistancebetween
I12 I23=AB=0.3m
I13 I23= 0.5m (by Measurement from diagram)
Then by putting values
??????
3
10.47
=
0.3
0.5
??????
3
=6.282 rad/sec

32. Ex 2 : Locate all the instantaneous centresof 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 ??????
��= 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 centremethod.

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

36. Ex.4:Locatealltheinstantaneouscentresofthe
mechanismasshowninFig.Thelengthsofvariouslinks
are:AB=150mm;BC=300mm;CD=225mm;and
CE=500mm.WhenthecrankABrotatesinthe
anticlockwisedirectionatauniformspeedof240r.p.m.;
find1.VelocityofthesliderE,and2.Angularvelocityof
thelinksBCandCE.
[Ans.1.6m/s;2.4rad/s;6.6rad/s]

37. Ex. 5: Locate all the instantaneous centresfor 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:Amechanism,asshownin
Fig,hasthefollowingdimensions:
O1A=60mm;AB=180mm;O2B
=100mm;O2C=180mmandCD
=270mm.ThecrankO1Arotates
clockwiseatauniformspeedof120
r.p.m.TheblockDmovesinvertical
guides.Find,byinstantaneous
centremethod,thevelocityofDand
theangularvelocityofCD.
[Ans. 0.08 m/s ; 1.43 rad/s]

39. Ex.7:Thelengthsofvariouslinksofamechanism,as
showninFig.are:OA=0.3m;AB=1m;CD=0.8m;
andAC=CB.Determine,forthegivenconfiguration,
thevelocityofthesliderDifthecrankOArotatesat60
r.p.m.intheclockwisedirection.Alsofindtheangular
velocityofthelinkCD.Useinstantaneouscentre
method.
[Ans. 480 mm/s ; 2.5 rad/s]