ME306-Fall 2013- Chapter (3)- Velocities Analysis of Mechanisms.pdf

Chapter (3): Velocity Analysis of Mechanisms ME306 - Fall 2013

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Chapter (3)

Velocities Analysis of Mechanisms

(Relative Velocity Method)


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Chapter (3)

Velocities Analysis of Mechanisms ( )

(Relative Velocity Method)

3.1 Introduction

The study of motion of various parts of the mechanisms and machines is important for

determining their positions, velocities and accelerations at different moments. In this

chapter, the velocity analysis for the mechanisms can be carried out by graphical method

(Velocity polygon) based on the relative velocity method.

3.2 Velocity Analysis

Velocity is the rate of change of displacement. Velocity can be linear velocity or angular

velocity.

Linear velocity dx

vdt

, and angular velocity d

dt

.

The relation between the linear velocity and angular velocity can be obtained as:

x r leading to dx d

v r rdt dt

(where

d

dt

).

In the kinematic analysis of mechanisms and machines, the velocities of different links or

parts can be obtained by using one the following methods:

(i) Relative velocity method or velocity polygon method

(ii) Instantaneous centre method

(iii) Analytical method

In this course, the velocity polygon method is discussed and used for determining the

velocities and accelerations of various mechanisms.

3.3 Relative Velocity Method

Consider a rigid link OA of length rOA (or OA) rotates about a fixed point O with a uniform

angular velocity OA (rad/sec) in clockwise direction as illustrated in Figure (1). For a small

rotation through an angle d in a time dt , the point A will move along the arc AA as shown.


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A

O

B

OA

A

O

d

OA d

(a) (b)

A'

va

(c)

o ab

rOA

The velocity of point A relative to the fixed point O can be expressed as:

ao OA OAr d

v rdt

(Absolute velocity)

Absolute Velocity is defined as the velocity of a point with respect to a fixed point.

The velocity of A is OA OAr and is perpendicular to OA. This is can be represented by a vector

oa as shown in Figure (1c).

Consider a point B located on the link OA. The velocity of point B is b OA OBv r

perpendicular to OB and is represented by a vector ob. From the Figure (1):

OA

OA

OBob OB

oa OA OA

bo b

ao a

v v OB

v v OA

It is noted from the diagram that, the point b divided the vector oa in the same ratio the

point B divides the link OA. Then, the proportionality law is very useful in the drawing of the

velocity and acceleration polygons in order to find the relative velocities and accelerations of

the points on the links or parts.

Note that: Usually small letters (oa, ob,) are used to represent the vectors in velocity polygon

while the capital letters (OA, OB,.) are used to represent the length of the links in space

diagrams.

The application of the relative velocity method is illustrated through the following examples.


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3.3.1 Example 1- Slider-crank Mechanism

Consider a slider-crank mechanism used in the reciprocating engine in which the crankshaft

OA rotates with uniform angular velocity (rad/sec) in clockwise direction as illustrated in

Figure (2). The slider P moves on a fixed guide G. The crank OA and slider P are joined by a

connecting rod AP. The velocities of these links of the mechanism can be obtained by

drawing the velocity polygon.

Figure (2): Velocity analysis of slider-crank mechanism

(1) Drawing the given mechanism by using a suitable scale.

(2) For a given angular velocity (rad/sec), determine the velocity of point A relative to

the fixed point O, i.e., a o ao aov v v v ( 0ov is the velocity of the fixed point O).

(3) To start drawing the velocity polygon, choose a point o and draw the vector oa

perpendicular to the link OA to some suitable scale as illustrated in Figure (2b).

(4) The velocity of the slider P relative to the point A ( pav ) is perpendicular to the link

AP. Then from point a (velocity polygon), draw a vector ap of unknown magnitude

but in the perpendicular direction to the link AP.

(5) The velocity of the slider P relative to guide G ( pgv ) is along the line of the cylinder.

Since there is no relative motion between G and O, the point g on the velocity polygon

can be assumed to be at the point o (Fig. 2b). Then from point g, draw a line parallel

to the cylinder (or OG) to intersect vector ap at the point p. Therefore, the vector gp

represents the velocity of the slider P.

(6) The velocity of point B located on the connecting rod AP can be obtained by using the

following relation:

O

A

P

G

B

OA

a (a) Slider-crank mechanism (b) Velocity polygon

b


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ab AB ABab ap

ap AP AP

The point b on the velocity vector ap can be located as shown in the velocity polygon (Fig.

2b). The vector ob is then represented the absolute velocity of point B.

3.3.2 Rubbing Velocity at a Pin Joint

The links or parts in the mechanisms are usually connected together by a pin joints as shown

in Figure (3).

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(a) Links move in opposite direction

2

Link 1 Link 2

pin1

(b) Links move in the same direction

2

Link 1 Link 2

pin

Figure (3): Pin joint of two links

Let 1 is the angular velocity of link 1 (rad/sec).

2 is the angular velocity of link 2 (rad/sec).

r is the radius of the pin at the joint.

The rubbing velocity at the pin is defined as the algebraic difference between the angular

velocities of the two links that connected at the pin joint. The rubbing velocity rubv at the pin

joint can be obtained as:

(a) 1 2rub pinv r when the two links are moved in the same direction.

(b) 1 2rub pinv r when the two links are moved in the opposite direction.

Example (1): The crank of a slider-crank mechanism is rotating at 300rpm in the clockwise

direction. The length of the crank is 150mm and connecting rod is 600mm long. When the

crank has turned through an angle of 45o with the horizontal as shown in Figure (4), find:

(1) The velocity of the slider p,

(2) The angular velocity of the connecting rod AP,


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(3) The velocity of the mid-point C of the connecting rod AP, and

(4) The rubbing velocity at pin joints O and A if the radius of the pin joint is 15mm.

A

OP

OAC

45o

Space diagram G

Figure (4): Slider-crank mechanism

Solution:

Given: OA = 150mm, AP= 600mm, and AC=CP = 300mm.

Crank speed N=300rpm 2

300 31.42 rad / sec (clockwise)60

OA

(1) The velocity of point A with respect to fixed point O is:

31.42 0.15 4.713 m / secA OAv OA

(2) Scale to draw the velocity polygon: 1 m/sec= 1 cm.

(3) The relative velocity of point P with respect to A is drawn to AP.

(4) The velocity of slider P with respect to O is drawn to OP. Vectors ap and op intersect at

point p.

Table: Velocity analysis for different links

Velocity Magnitude

(m/sec) Direction Vector

Scaled length (cm)

vc 4.713 to OC oc 4.713

vpc unknown to PC cp unknown

vp unknown along PO op unknown

Scale for velocity polygon: 1m/sec=1cm


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A

OP

C

45o

(a) Space diagram

o

a

vPA

p

vA=vAO OA

vP

vA=vAO

vPA AP

vP PO

(b) Velocity polygon

cvC

Figure (5): Velocity polygon

(5) The absolute velocity of point C on the connecting rod is obtained by the relation:

ac AC ACac ap

ap AP AP

From the velocity polygon:

(i) The velocity of P with respect to A is obtained from the polygon as:

vPA = vector ap = 3.40 cm = 3.40 cm1.0 (scale) = 3.40 m/sec.

(ii) The velocity of P vpo= vp = vector op = 4.00cm = 4.00cm1.0 (scale) = 4.00 m/sec.

(iii) The angular velocity of the connecting rod AP is obtained by:

3.405.670 rad / sec

0.60

APAP

v

AP

(iv) The velocity of point c is represented by the vector oc, i.e.,

vc = oc = 4.100 cm = 4.100cm 1.0 (scale)= 4.10 m/sec.

(v) The rubbing velocity at point A (same directions) is determined as:

31.42 5.670 0.015 0.3863 m / secrub OA AP pinv r .

Example (2): In a four-bar mechanism AABCD, AD is fixed link and is 15.0cm long. The crank

AB is 4.0cm long and rotates at 120rpm clockwise, while the link CD=8.0cm oscillates about


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D. BC and AD are of equal lengths. Find the angular velocity of the link CD when the angle

BAD is 60o.

B

DA

AB

C

60o

Figure (6): Four-bar Mechanism

Given: AD=15.0cm , AB= 4.0cm, CD=8.0cm, BC=AD=15.0cm

Speed of the crank NAB=120rpm (clockwise)

Solution:

(1) Draw the space diagram to a suitable scale (10 mm=1 cm).

(2) Calculate the given velocity:

2120 4 50.24 cm / sec

60B BA ABv v AB

(3) Scale to draw the velocity polygon: 10 cm/sec=1cm

50.24= 5.024cm

10Bv Bv AB


Velocity Magnitude (cm/sec)

Direction Vector Scaled length*

(cm)

VB 50.24 to AB ab 5.024

VBC unknown to BC bc unknown

VC unknown CD cd unknown

*Scale for velocity diagram: 10cm/sec = 1 cm

From the velocity diagram:

The velocity of point C : 3.80 cmCv cd


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3.80 cm / sec 10 38.0 cm / secC CDv v , C Dv v CD CDv v

Then the angular velocity of the link CD is obtained as:

38.0

4.75 rad / sec (clockwise)8.0

CDCD CD CD

v cdv CD

CD CD .

(b) Velocity diagram

a,d

B

DA

AB

C

60o

vC AB

vB AB vCB BC

vBA=vB=5.024cm AB

b

c

vCB BC

vCD=vC AB

(a) Space mechanism

Example (3): In the mechanism shown in Figure (6), the crank OA rotates at 20rpm

anticlockwise and gives motion to the sliders B and D. The dimensions of the links are given

as: OA=300mm, AB=1200mm, BC= 450mm and CD=450mm. For the given configuration,

find: (a) velocity of the sliders at B and D, and (b) the angular velocity of CD.

DD

A

O

OAC

30oBB

1050mm

Figure (6)


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Solution:

Given: 2

20 2.10 rad / sec60

OA

.

(1) Scale to draw the space diagram for the mechanism: 100mm=1cm.

(2) The velocity of the crank OA: 300

2.10 0.630 m / sec1000

A AO OAv v OA .

(3) Scale for velocity diagram: 0.1 m/sec=1cm.

6.30 cmAv OA


Velocity Magnitude

(m/sec) Direction Vector

Scaled length* (cm)

VA 0.630 to OA oa 6.30

VBA unknown to AB ab unknown

VB unknown OB ob unknown

VD unknown the path of the slider D

od unknown

VDC unknown to CD cd unknown

*Scale for velocity diagram: 0.1m/sec = 1 cm

(4) From the fixed point o, we draw a vector oa perpendicular to the crank OA (i.e.,

oa=vc=6.30cm).

(5) From point a, we draw a vector ab AB (i.e., ab=vBA), and from point o, we draw a

vector ob OB (i.e., ob=vBO). The point b is the intersection of the vectors ab and ob.

(6) We locate the point c on the link AB by using the following ratio relation:

450(5.60) 2.10 cm

1200

ab AB BCbc ab

bc BC AB

(7) From point c, we draw a vector cd to the link CD, from point o, we draw a vector od

to the path of the slider D. The point d is the intersection of the two vectors cd and

od.


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(b) Velocity diagram a

vA

vA=6.30cm AB

vB OB

(a) Space mechanism D

A

O

OAC

30o

BvBA

ob

vBA BA

cd

vC vD path of the slider D

vDC DC

vB

vD

vDC


The velocity of the slider B: 3.90 cm 3.90 cm 0.1 0.390 m / secB Bv ob v

The velocity of the slider D: 2.45 cm 2.45 cm 0.1 0.245 m / secD Dv od v

The angular velocity of the link CD: 3.60 0.1

0.80 rad / sec0.45

CDCD

v cd

CD CD

.

Example (4): In the mechanism shown in Figure (8), the crank AB rotates about fixed point

A at a uniform speed of 120 rev/min (clockwise direction). The link DC oscillates about a

fixed point D, which is connected to AB by the coupler BC. The slider P moves in the

horizontal guides being driven by the link EP.

For the given configuration, Find:

(1) Velocity of the sliding block P,

(2) Angular velocity of the link DC,

(3) Rubbing velocity of the pin joint C which is 50mm in diameter.

Given: AB=150mm, BC=CD=450mm, DE=150mm, EP=375mm.

Figure (7): Space mechanism and velocity diagram


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45o

P

375mm

D

A

375mm

100mm

B

C

E

AB

Figure (8)

Solution:

Given: 2

120 12.57 rad / sec60

AB

(1) Space mechanism: 50mm=1cm

AB=3.0 cm, BC=CD=9.0cm, DE=3.0cm and EP=7.5cm.

(2) Velocity B Av v 12.57 m / secBA ABv AB .

(3) Scale for Velocity diagram: 0.4m/sec=1cm

m / sec

cm0.4

Bv

link AB.


(i) Velocity of the sliding block P: 1.85 cm =1.85 0.4 0.74 m / secBv ab

(ii) Angular velocity of link DC:

5.6 0.44.98 rad / sec

0.45

DCDC

v dc

DC DC

(Anticlock wise)

(iii) Rubbing velocity of the pin joint C:

5.5 0.44.89 rad / sec

0.45

BCBC

v bc

BC BC

(Anticlock wise)

34.98 4.89 0.025 0.225 10 m / secrub BC DC pinv r .


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45o

P

D

A

B

C

E

AB

(b) Velocity diagram (scale 0.4m/sec=1cm)

a,d

vP path of the slider P

(a) Space mechanism (scale 50mm=1cm)

VB=4.72cm AB

VB AB

VCB BC VDC CD

VPE PE

b

VDC

VCB

vP path of the slider P

vP

vPE

c

e

PECD

BC

p

Example (5):

ME306-Fall 2013- Chapter (3)- Velocities Analysis of Mechanisms.pdf

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