kom Unit-II

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Unit II: Kinematics of Linkage Mechanisms KOM

Unit II Kinematics of Linkage Mechanisms

Displacement, velocity and acceleration analysis of simple mechanisms - Graphicalmethod velocity and acceleration polygons - velocity analysis using instantaneouscentres - kinematic analysis of simple mechanisms - Coincident points - Corioliscomponent of acceleration - Introduction to linkage synthesis problem.

Introduction:

Displacement: The displacement of a body is its change of position with reference

to a certain fied point. Velocity: It is the rate of change of displacement of a body with respect to time. It isvector !uantity.

Linear velocity: It is the rate of change of displacement of a body along a straightline with respect to time. Its units are m"s

Angular velocity: It is the rate of change of angular position of a body with respectto time. Its units are rad"s

Acceleration: It is the rate of change of velocity of a body with respect to time. It isvector !uantity.

Angular Acceleration: It is the rate of change of angular velocity of a body withrespect to time.

Velocity Analysis:

#elocity analysis involves the linear velocities of various points ondifferent links of a mechanism as well as the angular velocities of the links.

The velocity analysis is the prere!uisite for acceleration analysis whichfurther leads to force analysis of various links of a mechanism.

Methods of Velocity Analysis:

1 !raphical Method $a% &elative velocity method $b% Instantaneous centre method" Analytical Method

Note: Graphical methods given accurate solution. 'nalytical methods givenapproimate solution.

#onfiguration diagram $or% &pace diagram:

' machine or a mechanism is represented by a skeleton or a line diagram,commonly known as a configuration diagram

Mechanism #onfiguration diagram

'elative velocity of ()o *odies Moving in &traight Lines:

Consider two bodies ' and ( moving parallel lines in the same direction with

absolute velocities v' and v(. $v' ) v(%

&elative velocity of ' with respect to ( v'( * #ector difference of v' and v( + +

B A AB vvv −=

&elative velocity of ( with respect to ' v(' * #ector difference of v( and v' + +

A B BA vvv −=

AB B A BA vvvv −=

−−=

+ +

&elative velocity of point ' with respect to ( $v'(% and the relative velocity of point ( with respect to ' are e!ual in magnitude but opposite in direction.

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Motion of a link:

#elocity of any point on a link with respect to another point on the same link isalways perpendicular to the line oining these points on the configuration or spacediagram.

'ngular velocity of the link '(

#elocity of ( with respect to ' v(' ,

'( +++

×== ω abv BA ---------------- $i%

#elocity of $any point% C with respect to ' vC',

'C +++

×== ω ocvCA ------------- $ii%/rom e!uation $i% and $ii%

AB

AC

AB

AC

ab

ac

v

v

BA

CA ===.

. +++

+++

ω

ω

'u++ing velocity at a pin ,oint:

&ubbing velocity is defined as the algebric sum between the angular velocities of

the two links which are connected by pin oints, multiplied by the radius of the pin.

Consider two links 0' and 0( connected by a pin oint at 0 as shown in diagram. 1- 'ngular velocity of the link 0' or 'ngular velocity of the point ' withrespect to 0. 2- 'ngular velocity of the link 0( or 'ngular velocity of the point ( withrespect to 0. r &adius of the pin

'ccording to the direction &ubbing velocity in the pin oint 0* $1 - 2%.r, if the links move in the same direction

* $1 3 2%.r , if the links move in the opposite direction.

-ote: 4hen the pin connects one sliding member and the other turning member, theangular velocity of the sliding member is 5ero. In such cases, &ubbing velocity at the pin oint * r

1 In a four +ar chain A*#D. AD is fi/ed and is 10 mm long (he crank A* is

2mm long and rotates at 1" rpm clock)ise. )hile the link #D 3 4 mm

oscillates a+out D *# and AD are of e5ual length 6ind the angular velocity of

link #D )hen angle *AD 3 78

&olution:

6peed N (' * 127 r.p.m.'ngular velocity (' * 2 89 127":7 * 12.;:< rad"s6ince the length of crank '( * =7 mm * 7.7= m.Therefore velocity of B with respect to A or #elocity of B, $because A is a fied point%,v(' * v( * (' 9 AB * 12.;:< 9 7.7= * 7.;7> m"s

#onfiguration $or space% diagram:

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Velocity diagram:

vector ab * v(' * v( * 7.;7> m"s(y measurement,

vCD * vC * vector dc * 7.>%

&olution:

#onfiguration diagram:

#onfiguration diagram: Velocity diagram:

V a * ωa OA * 1; 9 7.2 * > rn"s

$i% 'ngular velocity of link (D

= bd/BD * 2.?;"7.2: * 11.> rad"s $counter-clockwise%

'ngular velocity of link BC * bc/BC * 3 1=% 97.71; * 7.>< rn"s

$ii% The velocity of slider D * hd * m/s

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= 6igure sho)s the link mechanism of a 5uick return mechanism of the slotted

lever type. the various dimensions of )hich are OA 3 2 mm. OP 3" mm. AR

3 > mm. RS 3 =mm 6or the configuration sho)n determine the velocity of

the cutting tool at & and the angular velocity of the link RS. (he crank OP

rotates at "1 rpm

&olution:

Velocity diagram:

rad"s22:7

2172 =×= π ω OP

vop * 0@ *22 9 7.2 * =.= rn"s.

The velocity of the ram 6 * os $or gs% * =.; m"s.It is towards right for the given position of the crank.

'ngular velocity of link &6, clockwise rad"s:A.=7.>

1.====

R

vrsrsω

Acceleration:

The rate of change of velocity with respect to time is known as acceleration and actsin the direction of the change in velocity. Thus, acceleration is a vector !uantity.

Acceleration Diagram for a Link

Consider two points ' and ( on a rigid link as shown in /igure, Bet the point (moves with respect to ', with an angular velocity of rad"s and let rad"s2 be theangular acceleration of the link '(.

$a% Bink. $b% 'cceleration diagram.

'cceleration of a particle $whose velocity changes both in magnitude and direction%at any instant have the following two components

1. Centripetal or &adial component,2. Tangential component

Centripetal or radial component of the acceleration of B with respect to A,

AB

v ABa BA

r

BA

222 '(linkof Bength =×=×= ω ω .......

=

ABv ABω

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This radial component of acceleration acts perpendicular to the velocity v(',In other words, it acts parallel to the link AB!

Tangential component of the acceleration of B with respect to A, ABa t BA ×=×= α α '(linkof Bength

Tangential component of acceleration acts parallel to the velocity v('. In other words, it acts perpendicular to the link AB.69U'?LI-K M;#@A-I&M

#onfiguration diagram Velocity diagram:

Acceleration diagram:

&

-o

Vector Magnitude Direction &ense

1 r ba

a $or %a1 ba ( )

AB

ab2

E '( F '

" t baa $or% ba b1

9 '( '( or a1 ba or E ab Fb

= r cb

a $or% b1c b ( )

BC

bc2 E (C F(

2 t

cba $or% c bc1 ++ (C or b1c b ++

0 r cb

a $or% d1cd ( )

DC

dc2 E DC FD

7 t cb

a $or% cd c1 +++ DC or d1cd ++

&LID;'?#'A-K M;#@A-I&M

#elocity diagram 'cceleration diagram

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&

-o

Vector Magnitude Direction &ense

1 aao $or %o1a1 ( )OA

oa2

E 0' F 0

" r baa $or% a1 ba ( )

OA

ab2

E '( F '

= t baa $or% ba b1 + '( +

2 a bg $or% g1 b1 + E to line of motion of ( +

2 (he dimensions and configuration of the four +ar mechanism. sho)n in

6igure. are as follo)s: 1A 3 = mmB "* 3 =7 mmB A* 3 =7 mm. and 1" 3

7mm (he angle A1" 3 78 (he crank 1A has an angular velocity of 1

rads and an angular acceleration of = rads ". +oth clock)ise Determine the

angular velocities and angular accelerations of "*. and A* and the velocity

and acceleration of the ,oint *

!iven:

'@1 * 17 rad"s H'@1 * >7 rad"s2H@1' * >77 mm * 7.> mH@2( * '( *>:7 mm * 7.>: m

&olution:

Velocity diagram: Acceleration diagram

#elocity of ' with respect to @1 or velocity of ',v'@1 * v' * '@1 9 @1' * 17 9 7.> * > m"s

Velocity of * and angular velocities of "* and A*

p1a * v'@1 * v' * > m"s(y measurement, we find thatv(@2 * v( * vector p2 b * 2.2 m"sv(' * vector ab * 2.7; m"s

'ngular velocity of @2(,%$Clockwiserad"s1.:

>:.7

2.2

2

22 ===

B P

v BP B P ω

'ngular velocity of '(,

%$Clockwise;.Arad"s>:.7

7;.2===

AB

v BA ABω

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Acceleration of * and angular acceleration of "* and A*

Tangential component of the acceleration of ' with respect to @1,2

111 m"s ?>.7>7@ =×=×= Aa AP t

AP α

&adial component of the acceleration of ' with respect to @1,

221

2

1

2

11 m"s>7>.717 =×=×== A P

A P va AP AP r

AP ω

&adial component of the acceleration of ( with respect to '.

222

11.:Am"s>:.7

7;.2===

AB

va BAr BA

&adial component of the acceleration of ( with respect to @2,

22

2

2

2

21>.==m"s

>:.7

2.2===

B P

va BP r BP

&adial component of the acceleration of ' with respect to @1 ,2

1 m"s >7J == " #ar

AP

Tangential component of the acceleration of ' with respect to @1,2

1 m"s ?J== "aar

AP

(y measurement, a' * a'@1 * >1.: m"s2&adial component of the acceleration of ( with respect to ',

2r

(' 11.:Am"sayJ ==a&adial component of the acceleration ( with respect to @2,

2r

(@2

J

2 1>.==m"sa 5 p ==

2

2(@2 m"s2?.: bJJ pa === Ba

2m"s1>.:ybJ === B

t

BA aa

2

2 m"s2:.:5bJ == t

BP a

'ngular acceleration of @2(

ise%'nticlockw$rad"sA>:.7

:.2:

@

a 2

2

t

(@2@2( ===

Bα

'ngular acceleration of '(,

ise%'nticlockw$rad"sA>:.7

:.1>

'

a 2

t

(''( ===

Bα

0 (he crank of a slider crank mechanism rotates clock)ise at a constant speed

of =rpm (he crank is 10 mm and the connecting rod is 7 mm long

Determine: 1 Linear velocity and acceleration of the midpoint of the connecting

rod. and " Angular velocity and angular acceleration of the connecting rod. at a

crank angle of 208 from inner dead centre position

!iven:

6peed K(0 * >77 r.p.m. or (0 * 2 89 >77":7 * >1.=2 rad"sH0( * 1;7 mm * 7.1; mH(' * :77 mm * 7.: m&olution:

#onfiguration $or% space diagram:

Velocity diagram: Acceleration diagram:

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Binear velocity of ( with respect to 0 or velocity of (,v(0 * v( * 0( 9 0( * >1.=2 9 7.1; * =.A1> m"s

(y measurement from velocity diagram,#elocity of ' with respect to (, v'( * vector ba * >.= m" s and

#elocity of ', v'* vector oa * = m" s vD * vector od * =.1 m"s

Acceleration of the midpoint of the connecting rod

&adial component of the acceleration of ( with respect to 0 or the acceleration of (,

222

m"s1.1=.=

==== OB

v

aa

BO

B

r

BO

&adial component of the acceleration of ' with respect to (,

222

m"s>.1?:.7

=.>===

BA

va ABr AB

(y measurement from the acceleration diagramaD * vector oJ dJ * 11A m"s2

Angular velocity of the connecting rod

2rad"s:A.;:.7

=.>===

BA

v AB

ABω $'nticlockwise about ( %

Angular acceleration of the connecting rod

/rom the acceleration diagram, 2m"s17>=t ABa $(y measurement%4e know that angular acceleration of the connecting rod '(,

2rad"s:A.1A1:.7

17>===

BA

a t AB ABα $Clockwise about (%

7 6or the configuration sho)n +elo). calculate

$i% (he acceleration of the slider at B

$ii% (he acceleration of point E

$iii% (he angular acceleration of link AB.

OA rotates at " rads counter?clock)ise

Velocity diagram: Acceleration diagram:

&

-o

Vector Magnitude$ms"% Direction &ense

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1 r ao

a $or %o1a1 ( )1

=:== eo f e

$iii% clockwisecounterrad"s17=:.1

1:A 21 ===== AB

bb

AB

f a

t

ba

abα

Analysis of slider?crank mechanism +y analytical method:

After rotation of crank +y angle C

% - Bength of connecting rod between the centers,r - &adius of crank or crank pin circle,L - Inclination of connecting rod to the line of stroken - % "r -&atio of length of connecting rod to the radius of crank

Velocity of the piston

/rom the geometry * @M@ * 0@M N 0@ * $@MCM 3 CM0% N $@O 3 O0%

$@O *% cos PH O0* r cos Q%

* $% 3 r% N $% cos P 3 r cos Q%

%cos-$1%cos1$ φ θ % r

+−=

+−= %cos-$1

r %cos1$ φ θ

% r

[ ]%cos-n$1%cos1$ φ θ +−= r

/rom triangles C@OsinP * CO " @C

CO * @C sinPCO * l sinP

/rom triangles CO0sin Q * CO " 0C

CO * 0C sin QCO * r sin Q/rom e!u $2% and $>% CO * % sinP * r sin Q

% "r *sin Q" sinP4here n * % "r

n * sin Q"sinP

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()

(!)

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sin L * sin Q"n4e know that

( ) 21

2sin1cos φ φ −=

21

2

2sin1cos

−=

n

θ φ

Rpanding the above epression by binomial theorem, we get

.......sin

2

11cos

2

2

+×−=n

θ φ

... $Keglecting higher terms%

2

2

2

sincos1

n

θ φ =−

6ubstituting the value of $1 cos L% in e!uation $1%

( )

×+−=

2

2

2

sin cos1

n

nr " θ θ

( )

+−= nr 2sin

cos1

2 θ

θ

Differentiating e!uation $;% with respect to Q

+=

nr

d

d"

2

cossin2sin

θ θ θ

θ

+=

nr

d

d"

2

cossin2sin

θ θ θ

θ

+=

nr

2

2sinsin

θ θ

( )θ θ θ 2sincossin2 =

* #elocity of @ with respect to 0 $or% velocity of the piston @

dt

d

d

d"

dt

d"vv # PO

θ

θ ×=== ω

θ ×=

d

d"

$&atio of change of angular velocity * *dQ " dt%

6ubstituting the value of d"dQ from e!uation $:%, we have

+==

nr vv # PO

2

2sinsin

θ θ ω

Acceleration of the piston:

6ince the acceleration is the rate of change of velocity, therefore acceleration of the

piston @,ω

θ

θ

θ ×=×==

d

dv

dt

d

d

dv

dt

dva

# # # #

Differentiating e!uation $A% with respect to Q,

×+=

nr

d

dv #

2

22coscos

θ θ ω

θ

+=

nr

d

dv # θ θ ω

θ

2coscos

6ubstitute dv p"dQ value in e!uation $


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φ

ω θ φ

cos

cos ×=ndt

d

'ngular velocity of connecting rod

dt

d PC

φ ω =

φ

ω θ

cos

cos×=

n

4e know that

( ) 21

2

sin1cos φ φ −=

= nθ

φ

sin

sin

2

1

2

2sin1cos

−=

n

θ φ

2

1

2

2sin

1

cos

−

×=

n

n PC

θ

θ ω ω

( ) 2

122 sin

1

cos

θ

θ ω

−

×=

nn

n

( ) 21

22 sin

cos

θ

θ ω ω

−

=

n

PC

Angular acceleration of the connecting rod PC * @C * 'ngular acceleration of P +it, res#ect to C

dt

d PC PC

%$ω α =

4e know that

dt

d

d

d

dt

d PC PC θ

θ

ω ω ×=%$%$

ω θ

ω ω ×=

d

d

dt

d PC PC %$%$

=ω θ

dt

d

ω θ

ω

α ×= d d PC

PC

%$

4e know that

( ) 21

22sin

cos

θ

θ ω ω

−=

n PC

Differentiate above e!uation w.r.t

( )

−=

21

22 sin

cos%$

θ

θ ω

θ

ω

nd

d

dt

d PC

( ) ( ) ( ) ( )

( )

−

−×−×−−−

=

−

θ

θ θ θ θ θ θ

ω 22

12

1222

122

sin

%cossin2$sin2

1cossinsin

n

nn

( ) ( ) ( )( )

−

−+−−=

−

θ

θ θ θ θ θ ω

22

221

2221

22

sin

cossinsinsinsin

n

nn

( ) ( )( )

−−−−

−=−

θ

θ θ θ θ ω

22

221

2221

22

sin

cossinsinsin

n

nn

$Dividing and multiplying by $n2-sin2Q% 1"2%

( )

( )

−

−−

−= 2>22

222

sin

cossinsin

θ

θ θ θ ω

n

n

( )( )[ ]θ θ

θ

θ ω 222

2>

22

cossin

sin

sin+−

−

−= n

n

( )[ ]1

sin

sin 2

2>

22

−−

−= n

n θ

θ ω

$sin2Q3cos2Q*1%

( ) [ ]( ) 2

>22

22

sin

1sin

θ

θ ω ω

θ

ω α

−

−−=×=

n

n

d

d PC PC

The negative sign shows that the sense of the acceleration of the connecting rod issuch that it tends to reduce the angle L

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'ngular velocity of connecting rod( ) 2

122 sin

cos

θ

θ ω ω

−=

n PC

'ngular 'cceleration of connecting

( )

( ) 2>

22

22

sin

1sin

θ

θ ω ω

θ

ω α

−

−−=×=

n

n

d

d PC PC

6ince sin2Q is small as compared to n2, therefore it may be neglected in angular velocity e!uation and angular acceleration e!uations of the connecting rod

Therefore, 'ngular velocityn

PC

θ ω ω

cos=

'ngular 'cceleration( )

>

22 1sin

n

n #c

−−= θ ω α

'lso in above e!uation unity is small as compared to n"B hence the term unitymay +e neglected

'ngular 'ccelerationn

#c

θ ω α

sin2−=

#oriolis #omponent of Acceleration

Tangential component of acceleration of the slider B with respect to the coincident point C on the link is known as Coriolis component of acceleration and is always perpendicular to the link. Coriolis component of the acceleration of B with respect of C , vaa t

BC

c

BC ω 2==

4here * 'ngular velocity of the link OA, and v * #elocity of slider B with respect to coincident point C!

> A mechanism of a crank and slotted lever 5uick return motion is sho)n in

6igure If the crank rotates counter clock)ise at 1" rpm. determine for the

configuration sho)n. the velocity and acceleration of the ram D Also determine

the angular acceleration of the slotted lever #rank. A* 3 10 mm B &lotted arm.

9# 3 > mm and link #D 3 " mm

&olution:

K(' * 127 r.p.m(' * 2 8 9 127":7* 12.;A rad"s H'( * 1;7 mm * 7.1; mH 0C * A77 mm * 7.A mHCD * 277 mm * 7.2 m

#elocity of ( with respect to ', v(' * (' 9 '( * 12.;A 97.1; *1.? m"s

#onfiguration diagram: Velocity diagram:



13/21


(y measurement, vector obJ * 1.;; m"s

wkt, JJ AB

AC

v

v

ab

ac =

m"s2.1;'(

'Cvv

JabJac == m"s2.1;

'(

'Cv

JabJ ==

(y measurement from velocity polygonvD * vector od * 2.1; m"sv((J * vector bJb * 1.7; m"svDC * vector cd * 7.=; m"s

vC0

* vector oc * 2.1; m"s'ngular velocity of the link 0C or 0(J,

Acceleration of the ram D

I-&(A-(A-;9U& #;-('; M;(@9D

' link as a whole may be considered to be rotating about an imaginary centreor a given centre at a given instant. 6uch a centre has 5ero velocity or the link is atrest at this point. This is known as the instantaneous centre or centre of rotation. This

centre varies from instant to instant for different positions of the link. The locus of these centres is termed the centrode

Velocity of a oint on a Link

Consider two points A and B on a rigid link, having velocities va and v b respectively, as shown in /ig$a%. /rom A and B draw lines perpendicular to thedirections of motion and let them meet at -! Then - is the instantaneous centre ofrotation of the link AB for its given position. If ω is the instantaneous angular velocity of the link AB then

va * . -A and v b* %B! Thus

ab

v -A

-Bv .

= ++++++++++++ $i%

The velocity diagram for the link AB has been drawn in /ig. $b%. Triangles -AB andoab are similar. Sence

AB

ab

-B

ob

-A

oa ==

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ω ==== -C

v

AB

v

-B

v

-A

vcabba ++++++++++++ $ii%

4here C is any point on the link AB!

$a% Instantaneous centre $+% Velocity diagram

roperties of Instantaneous #entre

The properties of instantaneous centre are1. 't the instantaneous centre of rotation, one rigid link rotates instantaneously

relative to another for the configuration of the mechanism considered. .2. The two rigid links have no linear velocities relative to each other at theinstantaneous centre.>. The two rigid links have the same linear velocity relative to the third rigid link, or any other link.

-um+er of Instantaneous #entres

The number of instantaneous centres in a mechanism is e!ual to the number of possible combinations of two links. The number of instantaneous centres is

( )

2

1−= nn N

where n * number of links.

(ypes of Instantaneous #entres

The instantaneous centres for a mechanism are of the following types 1. /ied instantaneous centres 2. @ermanent instantaneous centres >. Keither fied nor permanent instantaneous centres.

Consider a four-bar mechanism shown in /ig $c%. /or this mechanism, n * =.Sence N * = $= - 1%"2 * :. The instantaneous centres are12, 1>, 1=, 2>, 2=, >=

6ig $c% Instantaneous centres in a four?+ar mechanism

The instantaneous centres 12 and 1= remain at the same place for the

configuration of the mechanism, and are, therefore, called fied instantaneouscentres.The instantaneous centres 2> and >= move when the mechanism moves, but the

oints are of permanent. Therefore, they are called permanent instantaneous centres. The instantaneous centres 1> and 2= vary with the configuration of the mechanismand are neither fied nor permanent instantaneous centres.

Location of Instantaneous #entres:

The following observations are !uite helpful in locating the instantaneous centres1. /or a pivoted or pin oint, the instantaneous centre for the two links lies on thecentre of the pin $/ig. d%.2. In a pure rolling contact of the two links, the instantaneous centre lies at their pointof contact $/ig. e%. This is because the relative velocity between the two links at the

point of contact is 5ero.>. In a sliding motion, the instantaneous centre lies at infinity in a direction perpendicular to the path of motion of the slider. This is because the sliding motion ise!uivalent to a rotary motion of the links with radius of curvature e!ual to infinity$/ig. f%. If the slider $link 2% moves on a curved surface $link 1%, then theinstantaneous centre lies at the centre of curvature of the curved surface $/ig. g U h%.

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6ig $d% 6ig $e% 6ig $f%

6ig $f% 6ig $g%

Location of instantaneous centreArnold?Kennedys (heorem

This theorem states that if three plane bodies have relative motion amongthemselves, their three instantaneous centres must lie on a straight line.

6ig$h% Arnold?Kennedy theorem

Method of Locating Instantaneous #entres:

The following procedure may be adopted to locate the instantaneous centres1. Determine the number of instantaneous centres from, N * n (n .)/0!

2. Vake a list of all the instantaneous centres by writing the link numbers in the firstrow and instantaneous centres in ascending order in columns. /or eample, for afour-bar chain shown below

>. Bocate the fied and permanent instantaneous centres by inspection. Tick markthese instantaneous centres.=. Bocate the remaining neither fied nor permanent instantaneous centres $circled% by using 'rnold-WennedyJs theorem. This is done by a circle diagram, as shown in/ig$i%. Vark points on a circle e!ual to the number of links in the mechanism. Xoin the points by solid lines for which instantaneous centres are known by inspection. Kow oin the points forming the other instantaneous centres by dotted lines. Theinstantaneous centre shall lie at the intersection of the lines oining the instantaneouscentres of the two adacent triangles of the dotted line.

6ig$i%

Determination of Angular Velocity of a Link

The angular velocities of two links vary inversely as the distances from their common instantaneous centre to their respective centres of rotation relative to theframe.

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4 In a pin ,ointed four +ar mechanism. as sho)n in 6igure. A* 3 = mm.

*#3#D3=7 mm. and AD 3 7 mm (he angle *AD 3 78 (he crank A*

rotates uniformly at 1 rpm in clock)ise Locate all the instantaneous centres

and find the angular velocity of the link *#.

&olution:

N '( *177 r.p.m or '( * 2 8 9 177":7 * 17.=A rad"s6ince the length of crank AB * >77 mm * 7.> m,Therefore velocity of point B on link AB, v( * '( 9 AB * 17.=A 9 7.> * >.1=1 m"s

( ) ( ):

2

1==

2

1=

−=

−=

nn N

Instantaneous centres 12, 1>, 1= 2>, 2= >=

Angular velocity of the link *#

Bet (C * 'ngular velocity of the link (C.6ince ( is also a point on link (C, therefore velocity of point ( on link (C, v( *(C 9 I1>(

(y measurement, I1> ( * ;77 mm * 7.; m

rad"s:.2.1=1

(I

v

1>

(

(C ===

E Locate all the instantaneous centres of the slider crank mechanism as sho)n

in 6igure (he lengths of crank 9* and connecting rod A* are 1 mm and 2

mm respectively If the crank rotates clock)ise )ith an angular velocity of 1

rads. find: $i% Velocity of the slider A. and $ii% Angular velocity of the connecting

rod A*

&olution:

0( * 17 rad" sH OB * 177 mm * 7.1 m Binear velocity of the crank OB, v0( * v( * 0( 9 OB * 17 9 7.1 * 1 m"s

Department of Mechanical, AAMEC Page 1$


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Kumber of instantaneous centre

( ) ( )

:2

1==

2

1 =−

=−

= nn N

Instantaneous centres 12, 1>, 1=, 2>, 2=, >=

(y measurement, we find thatI1> ' * =:7 mm * 7.=: mH andI1> ( * ;:7 mm * 7.;: m1 Velocity of the slider A

v' * #elocity of the slider '.

(I

v

'I

v

1>

(

1>

' =

m"s 7.

(

'( ===

art?A

1. 4hat is configuration diagramY 4hat is its useY $MayFune "1". '"4,-ovDec "1". '"4%

2. 4rite about rubbing velocity. $MayFune "1=. '"4"1%

Department of Mechanical, AAMEC Page 1%


18/21


>. Define rubbing velocity. 4hat will be the epression for rubbing velocity at a pin oint when two links are rotating in opposite directionY $-ovDec "1". '"4%

ZDefine rubbing velocity in a pin oint and write the e!uation for the same $MayFune "1". '"4%G

=. Define instantaneous centre $MayFune "12. '"4"1% ZDefine instantaneous centre of rotation. $MayFune "1=. '"2">%G;. Define number of instantaneous centre. $-ovDec "1=. '"4"1%G Z4rite down the epression for finding the number of instantaneous centres in a

mechanism. $MayFune "1=. '"4"1%G:.J6tate theJ direction of movement of slider and link for Coriolis component to be positive. $-ovDec "12. '"2">A. Identify the direction of Coriolis component for the given link $velocity vector !#p% that rotates at an angular velocity $% $/ig.$a% and /ig.$b% $-ovDec "1=.'"2">%

/ig.$a% /ig.$b%

%

>. The following data refer to the dimensions of the links of a four-bar mechanism'( * ;7 mmH (C * :: mmH CD * ;: mm and 'D $fied link% *177 mm. 't theinstant when 1D'( * :77, the link '( has an angular velocity of 17.; rad"s in thecounter clockwise direction. Determine the velocity of point C, velocity of point R onthe link (C while (R * =7 mm and the angular velocities of the links (C and CD.

'lso sketch the mechanism and indicate the data. $MayFune "1=. '"4"1%

=. The crank '( of four bar mechanism shown in figure rotates at :7 rpm clockwise.Determine the relative angular velocities of the coupler to the crank and the lever tothe coupler. /ind also the rubbing velocities al the surface of pins 2; mm radius andthe oints, ( and C. $


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$iv% velocity of slider D. $7mm long and rotates at 177 rpm clockwise while the link CD*:7mm oscillatesabout DH (C*127mm. [sing graphical method. /ind the angular velocity andangular acceleration of link (C when angle ('D*:77 $-ovDec "1". '"4%

11. @O&6 is a four bar chain with link @6 fied $figure%. The length of the links are

@O*:2.; mmH O&*1A; mmH &6*112.; mmH and @6*277 mm. The crank @O rotatesat 17 rad"s clockwise. Draw the velocity and acceleration diagram when angleO@6*:7\ and O and & lie on the same side of @6. /ind the angular velocity andangular acceleration of the links O& and &6. $1:% $MayFune "1=. '"2">%

12. ' four bar chain is represented by a !uadrilateral '(CD in which 'D is fiedand is 7.: m long. The crank '(* 7.>m long rotates in a clockwise direction at17rad"s and with an angular acceleration of >7 rad"s2, both clockwise.

Department of Mechanical, AAMEC Page 1'


20/21


The crank drives the link CD $*7.>: m% by means of the connecting link (C$*7.>:m%. The angle ('D * :77, [sing graphical method, determine the angular velocities and angular acceleration of CD and (C $MayFune "1". '"4%

1>. In the mechanism shown in /igure $below% the slider is moving to the right witha velocity of 1 m/s and an acceleration of 2.; m/s0! The dimensions of various linksare '( * >m inclined at =;\ with the vertical and (C* 1.; m inclined at =;\ with thehori5ontal. Determine

$i% 'cceleration of point ( and its magnitude of hori5ontal and vertical component$%

1=. Bocate the instantaneous centres of the slider crank mechanism shown in figure./ind the velocity of the slider. $


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1?. Differentiate low degree and high degree of compleity with suitable sketch. $:%$-ovDec "1=. '"4"1%


kom Unit-II

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