Método de Klein

21

Analysis by

Analytical MethodsUNIT 5 ANALYSIS BY ANALYTICAL METHODS

Structure 5.1 Introduction

Objectives

5.2 Analytical Method for Velocity and Acceleration 5.2.1 Approximate Method for Slider Crank Mechanism

5.2.2 General Method for Velocity and Acceleration

5.3 Motion Referred to Motion Frames of Reference 5.3.1 Relative Motion of Two Points

5.3.2 Plane Motion of a Link

5.4 Method of Relative Velocity and Acceleration

5.5 Alternative Method of Determining Coriolis’ Component of Acceleration

5.6 Klein’s Construction for Determining Velocity and Acceleration of Slider Crank Mechanism

5.7 Summary

5.8 Key Words

5.9 Answers to SAQs

5.1 INTRODUCTION

In unit 4, you have studied instantaneous centre method for determining velocity of any point in a mechanism. This method is good for determining velocity and cannot be used determining acceleration. Also, if lines drawn for determining instantaneous centre are parallel, they cannot locate instantaneous centre. Even if these lines are nearly parallel, they will meet at a large distance to locate instantaneous centre within the limited size of the paper. In this unit, you will be explained relative velocity method and analytical method and how they can be used to determine velocity and then acceleration of a point in a mechanism. Acceleration is required for determining inertia forces which are required in dynamic analysis of a mechanism.

Objectives After going through this unit you should be able to

• analyse plane motion,

• analyse plane motion with moving frames of references,

• determine magnitude of Corioli’s acceleration and its direction,

• determine velocity and acceleration of any point in a mechanism, and

• apply Klein’s construction for determination of acceleration of slider in slider crank mechanism.

22

Motion Analysis of Planar Mechanism and Synthesis

5.2 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION

The analytical method can be used for determining velocity and acceleration of any point in a mechanism and angular acceleration of any link.

5.2.1 Approximate Method for Slider Crank Mechanism The slider crank mechanism OAB is shown in Figure 5.1. Let l and r be lengths of the connecting rod AB and crank OA respectively. Let x be the distance of the piston pin from its farthest position B, (for this point OA and AB are aligned).

Figure 5.1

( ) ( ) ( ) ( cos cosx r l OC CB r l r l= + − + = + − θ + φ)

or (1 cos ) (1 cos ) (1 cos ) (1 cos )lx r l rr

⎧ ⎫= − θ + − φ = − θ + − φ⎨ ⎬⎩ ⎭

Also AC sin sinr l= θ = φ

∴ sin sinrl

φ = θ

Also 2

2 22cos 1 sin 1 sinr

lφ = − φ = − θ2

∴

12 222cos 1 sinr

l

⎡ ⎤φ = − θ⎢ ⎥

⎢ ⎥⎣ ⎦

By Binomial theorem

2 4 6

2 41 1 1cos 1 sin sin sin2 8 16

r r rl l l

⎛ ⎞ ⎛ ⎞ ⎛ ⎞φ = − θ − θ − θ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

6

or 2 4 6

2 4 61 1 1(1 cos ) sin sin sin . . .2 8 16

r r rl l l

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− φ = θ + θ + θ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Substituting (1 – cos φ) in expression of x you will get

3

2 41 1(1 cos ) sin sin . . .2 8

r rx rl l

⎡ ⎤⎛ ⎞ ⎛ ⎞= − θ + θ + θ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

In this mechanism, rl

is approximately 0.25 and higher powers of rl

⎛⎜ shall have

negligible value.

⎞⎟

⎝ ⎠

Therefore, 21(1 cos ) sin2

rx rl

⎡ ⎤⎛ ⎞≈ − θ + θ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

O

A

3φ

B

4

B1

2 r

θ

1

l

C

x

23

Analysis by

Analytical MethodsVelocity of slider, dx dx dVdt d dt

θ= = ×

θ

ddtθ= ω (angular speed of crank)

Therefore, dxVd

= ωθ

or 1sin 2 sin cos2

rV rl

⎡ ⎤⎛ ⎞= ω θ + θ θ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

or 1sin sin 22

rV rl

⎛ ⎞⎛ ⎞= ω θ + θ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Let crank rotate at the constant angular velocity.

Acceleration of slider, dV dV dadt d dt

θ= = ×

θ

or 2 cos cos 2ra rl

⎡ ⎤= ω θ + θ⎢ ⎥⎣ ⎦

5.2.2 General Method for Velocity and Acceleration We start with plane curvilinear motion using polar coordinates to denote the position vector ‘r’ and its angular coordinate ‘θ’ measured from a fixed reference axis as shown in Figure 5.2. Let er and eθ unit vectors in radial and transverse directions respectively. These unit vectors have positive directions in increasing r and θ. The vectors er and eθ are parallel to the positive senses of Vr and Vθ. Both these unit vectors will swing through the angle in the time dt. As shown in figure, their tips move through the distances | d er | = 1 × dθ = dθ in the direction of eθ, and through 1d e d dθ = × θ = θ in the direction opposite to that of er. Therefore,

, wherer rd d d ddt d dt dtθ

θ θ= × = θ →

θe e e θ

and ( )r rd d ddt d dtθ θ θ= × = θ − = − θ

θe e

e e

Figure 5.2

The velocity of a particle P which has position vector ‘r’ is given by

( ) rr r

dd d drV r rdt dt dt dt

= = = +er e e

θθ = +e e e e = + r t rr r V Vθ θ

VrVθ

er

eθ

V

P

Yr

X

deθ

eθ

der

dθ dθ

er

θZ θ

24


where Vr and Vθ are the components of velocity in radial direction and transverse direction. The acceleration of particle P is given by

( )rdV da r rdt dt θ= = + θe e

( ) (rd dr rdt dt

)θ= + θe e

( ) ( )rr

dd r dr dr rdt dt dt dtθ θ= + + θ + θ

ee e e

( )rdr r r rdtθ

θ θ θ⎛ ⎞= + θ + θ + θ + θ⎜ ⎟⎝ ⎠

ee e e e

22r rr r r rθ θ= + θ + θ − θe e e e

2( ) ( 2 )rr r r r θ= − θ + θ + θe e

r ra aθ θ= +e e

where ar represents radial component and aθ represents transverse component of acceleration.

2 and 2ra r r a r rα= − θ = θ + θ

Here, r → Acceleration of slider along slotted link,

r θ → Centripetal acceleration, 2

r θ → Tangential component of acceleration, and

2 θ → Coriolis’ component of acceleration. r

SAQ 1 Explain Corioli’s component of acceleration and what is its magnitude?

Example 5.1

The length of the crank of quick return crank and slotted lever mechanism is 15 cm and it rotates at 10 rod/s in counter clockwise sense. For the configuration shown in Figure 5.3 determine angular acceleration of link BD.

(a) (b)

A D

B 600

ω = 10 rad/sec

O

ω

A

(c)

Figure 5.3

B 600

O

C

A

ω

480eθ

er

θ′θ′′

180

D 420

BE

600

15 13

θ32.5 cm

25

Analysis by

Analytical MethodsSolution

In Figure 5.3(b), o13 13tan or 1832.5 7.5 40

AEBE

θ = = = θ =+

Also, o13 or 42.06 cm

sin sin 90AB AB AB= = =

θ

The motion of point A is common to both OA and BD. The velocity and acceleration of A are given by

15 10 1.5 m/s100AV OA= × ω = × =

2 215 10 15 m/s100Aa OA= × ω = × =

The components of VA along BD are perpendicular to it are given by

o1.5 cos 48 1.0 m/srV r= = − = −

o

o 1.5 sin 481.5 sin 48V rrθ = θ = ∴ θ =

or o1.5 sin 48 2.65 rad/s42.06

100

θ = =

The transverse component of acceleration of A is give by

o osin 42 15 sin 42 10.037 m/sAa aθ = − = − = − 2

)Also ( ) 2 (a r rθ = θ + θ

or 42.0610.037 2 ( 1.0) (2.65)100

− = θ + −

or 25.30 10.037 11.26 rad/s0.4206−

θ = = −

Therefore, angular acceleration of BD is 11.26 rad/s2 in clockwise sense.

5.3 MOTION REFERRED TO MOVING FRAMES OF REFERENCE

The moving frame may, in general, translate and rotate as well as accelerate linearly or angularly. The purpose of the following treatment is to arrive at a systematic procedure for refering motion of a point with respect to a frame of reference if its motion is known in relation to another moving with respect to the former.

5.3.1 Relative Motion of Two Points Consier two points P1 and P2 moving with velocities V1 and V2 and accelerations a1 and a2. The velocity of P1 with respect to P2 is given and acceleration of P1 with respect to P2 is given by

12 1 2= −a a a

V V 12 1 2V= −

Z a B1B

26


Figure 5.4

Also 21 2 2 12V V V V= − = −

and 21 2 1 12a a a a= − = −

5.3.2 Plane Motion of a Link A motion is said to be a plane motion if all the points in the body stay in the same and parallel planes. The concept of plane motion enables us to consider only one of the parallel planes and analyse the motion of the points lying in that plane.

The parallel planes may not appear to be identical in shape and size but there is no difficulty because any plane of the body can be hypothetically extended by a massless extension of the rigid body for the purpose of kinematic analysis.

The mechanisms use links. Consider a link PQ which is shown displaced in Figure 5.5 to position P′ Q′ in a general plane motion. The motion may be considered to comprise translation of an arbitrary point on the link plus a rotation about an axis perpendicular to the plane and passing through that point. In an example in Figure 5.5 eight combinations are equivalent.

Q

P P′

Q′

Q1

Figure 5.5(a) : Translation from PQ to P′Q1 and Rotation about P′

θ

P′

Q

P1

Q′

P

θ

Figure 5.5(b) : Translation from PQ to P1 Q′ and Rotation about Q′

P1

P′

C

P

C′

θ

27

Analysis by

Analytical Methods

Figure 5.5(c) : Translation from PQ to P1 Q1 and Rotation about C′

Figure 5.5(d) : Translation from PQ to P1 Q1 and Rotation about R′

Q1 Q′

P′ P

Q

θ

Figure 5.5(e) : Rotation from PQ to PQ1 and Translation to P′ Q′

Figure 5.5(f) : Rotation from PQ to P1Q and Translation to P′ Q′

P1 P′

Q′Q

C′

Q1

C

P

θ

Figure 5.5(g) : Rotation from PQ to Pi Q1 and Translation to about P′ Q′

P′

O O′

P P1

Q′

θ

Q Q1

P′

Q′

O

P

O′

θ

θ

P1 P′

Q′ Q

P

θ

28


Figure 5.5(h) : Rotation from PQ to P1 Q1 and Translation to P′ Q′

The translation and rotation are commutative. In Figures 5.5(e) and (f) show equal amount of translation and rotation but in Figures 5.5(a) and (b) translation is done first and rotation later. Similarly, Figures 5.5(a) and (e), (c) and (g), (d) and (h) are commutative.

It may also be noted that, whatever be the mode of combination, the amount of rotation is same. The angular velocity of every point on the link is the same. It is, therefore, the customary to use the term angular velocity of the link rather than of any particular point on it.

The fact that a general plane motion can be thought of as a superpositon of translation and rotation is a special case of Chasle’s theorem. The theorem, in general, states that any general motion of a rigid body can be considered as an appropriate superposition of a translational motion and a rotational motion.

SAQ 2 Determine velocity of top point of a rolling wheel if centre of the wheel is moving with velocity ‘v’.

5.4 METHOD OF RELATIVE VELOCITY AND ACCELERATION

For this purpose, a link of a general shape may be considered to start with. It is shown in Figure 5.6. Let there by any arbitrary two points A and B moving with velocities VA and VB. The velocity of B relative to A may be represented by VB BA. Let relative velocity VBA makes an angle θ with line joining A and B. VBA can be resolved into two components VBA cos θ along AB and VBA sin θ perpendicular to AB. Since the link is rigid, distance AB remains constant. Therefore, component of velocity of B relative to A along AB cannot exist.

Therefore, cos 0 or cos 0BAV θ = θ =

or, 2π

θ =

This means a rigid link has rotation relative to point A, as well as translation along velocity of A.

VBA

aB

VB

B

b′ a BBB

o′

a′

ct

29

Analysis by

Analytical Methods

Figure 5.6

Therefore, the direction of VBA is perpendicular to the line joining A and B. If VA is known in magnitude and direction and for VB only direction is known, the magnitude can be determined by drawing a polygon as stated below :

B

(a) First the velocity of A, i.e. VA is plotted by assuming a suitable scale from an arbitrary selected point o which represents fixed reference. Let it be represented by ‘oa’.

(b) Now, draw a line through o parallel to the direction of velocity of B, i.e. VB. B

(c) Next, draw a line through point ‘a’ which is parallel to the perpendicular to the line AB which meets the direction of VB at point ‘b’. B

In this velocity polygon oab, ‘ob’ represents velocity VB in magnitude and direction. B

After drawing velocity polygon, we can proceed for determination of magnitude of acceleration of point B. If direction of acceleration of B is known and acceleration of point A is known in magnitude and direction. The vector equation can be written as follows :

B A B= +a a a A

t

Since t c cBA BA BA B A BA B= + ∴ = + +a a a a a a a A

Where aBA is acceleration of B relative to A; tBAa is tangential component of aBA and c

BAa is centripetal component of aBA.

2

c BABA

VaAB

=

It is directed from B to A along AB whereas tBAa has direction perpendicular to AB.

Acceleration polygon can be drawn as follows :

(a) Plot acceleration of A in magnitude and direction by assuming suitable scale. aA is represented by o′a′.

(b) Plot centripetal component cBAa by drawing line parallel to AB and represent

its magnitude according to the vector equation. cBAa is represented by a′a′1.

(c) From draw a line perpendicular to 1a 1aa′ or parallel to the perpendicular to

AB to represent direction of tBAa .

(d) Draw a line parallel to the direction of acceleration of B, i.e. aB from fixed reference o′ to meet the line representing direction of

B

tBAa . They meet at

b′ . o′b′ represents acceleration of B in direction and magnitude.

Example 5.2

30

In a slider crank mechanism shown in Figure 5.7, the crank OA rotates at 600 rpm. Determine acceleration of slider B when crank is at 45o. The lengths of crank OA and connecting rod AB are 7.5 cm and 30 cm respectively.


Figure 5.7

b′ o′

a′ a′1

aA

b

a

o

A

B O 45

aB ω r

0

Solution

The angular velocity of crank OA, 2 600 6.28 rad/s60

π ×ω = = . The crank OA is a

rotating body about fixed centre O. Therefore, the velocity of point A is given by

6.28 7.5 47.1 cm/sAV OA= ω × = × = or 0.471 m/s

Select suitable position of pole o which represents fixed reference. Draw a line oa perpendicular to OA to represented velocity of A, i.e. VA in magnitude and direction. From point a, draw a line perpendicular to AB to represent direction of relative velocity VBA. Now draw another line parallel to the motion of the slider B from O to represent direction of the velocity of slider VB to meet another line through a at b. Thus ob represents velocity of B in magnitude and direction.

B

Since crank OA rotates at uniform angular speed, therefore, acceleration of A will be centripetal acceleration.

2 2

2(0.471) 2.96 m/s7.5100

c AA A

VOA

= = = =a a

It is directed from A towards O.

c tB A BA A BA BA= + = + +a a a a a a

2

c BABC

VAB

=a along AB directed from B to A

2(3.425) 39.17 m/s

0.3cBC = =a

Select a suitable position of pole O which represents fixed reference. Plot acceleration of A, i.e. aA by drawing parallel to OA and representing its magnitude by a suitable scale. This is represented by o′a′. Now, draw a line from ‘a′’ parallel to AB for c

BAa and represent its magnitude. This is represented by a′a′1. Now from

a′1 draw a line perpendicular to a′a′1 or AB to represent direction of tBAa . Draw a

line from o′ parallel to the motion of slider B to represent direction of motion of B. This line from o′ will meet another line from a′1 at b′. The acceleration of slider B is represented by o′b′ in magnitude and direction and it gives

2215 m/sBa =

Example 5.3

31

Analysis by

Analytical MethodsA four bar chain O2 AB O4 is shown in Figure 5.8. The point C is on link AB. The crank O2A rotates in clockwise sense with 100 rad/s and angular acceleration 4400 rad/s2. The dimensions are shown in Figure 5.8(a). Determine acceleration of point C and angular acceleration of link O4B.

(a)

(b) (c)

O′2, O′4

i′,O′

b′ a′

c′

b′

b′

a′

75 mm

A B3

C 28 mm 80 mm

2

4 37 mm

α = 4400 rad/sec2

ω = 100 rad/sec 530

125 mm

O4

1

1 O2

c O2, O4

b

a

Figure 5.8

Solution Draw configuration diagram to the scale.

The velocity of point A, 275 100 7.5 m/s

1000AV O A= × ω = × =

Velocity Diagram For drawing velocity polygon the following steps may be followed :

(a) Assume suitable scale say 1 cm → 5 m/s. (b) Plot velocity of A, VA perpendicular to O2A. It is represented by

o2a in velocity polygon Figure 5.8(b). (c) Draw a line from point a perpendicular to AB to represent

direction of VBA. (d) Draw a line from o2 perpendicular to O4B to represent direction

of VB to meet line representing direction of VB BA at point b. o2b represents VBB in magnitude and direction.

(e) For determining velocity of C, plot point c on ab such that

‘ac’ AC abAB

= × .

From velocity polygon 6.5 m/sBAV =

and 9 m/sBV =

Acceleration Diagram

For drawing acceleration polygon, the following steps may be followed :

(a) Assume suitable scale depending on value of centripetal acceleration of A which is 2

2cAa O A= ω . Or

32


2 275 (100) 750 m/s1000

cAa = × = . A scale 1 cm → 250 m/s2 may

serve the purpose.

(b) Draw a line parallel to O2A and plot cAa which will be

represented by 3 cm. It is represented by o2′ a1′ in polygon.

(c) Draw a line perpendicular to O2A from a′1 in the sense of angular acceleration to represent tangential acceleration which

is 22

75 4400 300 m/s1000

tAa O A= × α = × = . It will be denoted

by a′1 a′ in polygon. o′2 a′ can be joined to get acceleration of A which is represented by o′2 a′ in magnitude and direction.

(d) Draw a line parallel to AB from a′ and plot centripetal component of acceleration c

BAa which is given by 2 2

2(6.5) 528.15 m/s0.08

c BABA

VaAB

= = = . Plot magnitude of cBAa . It

is represented by a′ b′1.

(e) Draw a line perpendicular to AB from b′1 to represent direction of tangential component of acceleration t

BAa .

(f) Draw a line parallel to O4B from o′2 to represent centripetal component of acceleration of B. The magnitude is given by

2 22

4

9 2189.189 m/s0.037

c BB

VaO B

= = = . It will be denoted by

8.76 cm and it is represented by o′2 b′2.

(g) Draw a line perpendicular to O4B or o′2 b′2 to represent direction of tangential component of acceleration of B to meet the line representing direction t

BAa at b′. Join a′b′ which represents aBA.

Join o′2 to b′ to get acceleration of B, i.e. aB. B

(h) To determine acceleration of C, plot a point c′ on line a′b′ such

that 28 72 2.5280

ACa c a bAB

′ ′ ′ ′= × = × = . Joint o′2 with c′ and

o′2 c′ represent acceleration of C in magnitude and direction. From acceleration polygon Figure 5.8(c), ac = 1400 m/s2.

(i) Angular acceleration of link O4B is given by

42

43479.3 rad/s

tB

O Ba

O Bα = =

The acceleration polygon is shown in Figure 5.8(c).

Example 5.4

Determine acceleration of slider D in a combined four bar chain and slider crank mechanism shown in Figure 5.9(a). The dimensions are shown in the figure. The crank OA rotates at 240 rpm in counterclockwise sense.

c

aDB

taDB

taBC

taBA

d′1

d VD

VAVB

VDB

O,C

b′1

b′

aBBB

aBAB

a′

caBA

aBDBB

a V

33

Analysis by

Analytical Methods

(a) (b)

A

B

D

3

2

49 mm

44 mm 65 mm

28 mm O

11

C

750

(c)

Figure 5.9

Solution Plot configuration diagram to the suitable scale. Velocity Diagram

Velocity of A, 22 240 0.7037 m/s

60AV O A π ×= × =

It is perpendicular to OA in sense of rotation. (a) Plot velocity of A, VA by assuming a suitable scale say

1 cm → 0.2 m/s. It is perpendicular to OA and represented by ‘oa’ on velocity polygon.

(b) From a, draw a line perpendicular to AB to represent direction of VBA.

(c) From o, draw a line perpendicular to BC to represent direction of velocity of B. Extend it if necessary to meet the line representing direction of relative velocity VBA. These two lines join at b.

(d) From b, draw a line perpendicular to BD to represent direction of VDB.

(e) From o, draw a line parallel to line of motion of slider D to meet line representing direction of VDB at d.

(f) In velocity polygon od represents velocity of slider D From velocity polygon VB = 0.5 m/s, VB BA = 0.4 m/s, VDB = 0.602 m/s Velocity polygon is shown in Figure 5.9(b).


34


Acceleration of A, 22 240 17.686 m/s60

cA Aa a OA π ×⎛ ⎞= = × =⎜ ⎟

⎝ ⎠ Select a

suitable scale, say 1 cm → 5 m/s2.

(a) Draw a line parallel to OA to represent acceleration of A, aA. It is represented by o′a′.

(b) Plot centripetal component of acceleration aBA, i.e. cBAa by

drawing line parallel to AB. 2

23.636 m/sc BBA

VaAB

= =

It is represented by a′ b′1.

(c) From b′1, draw a line perpendicular to AB to represent direction of tangential component of acceleration aBA.

(d) From o′ draw a line o′ b′2 parallel to BC to represent centripetal component of acceleration of B which has magnitude given by

225.102 m/sc B

BVaBC

= =

(e) From b′2, draw a line perpendicular to BC or o′ b′2 to represent direction of tangential component of acceleration of B, i.e. t

Ba .

This line meets direction of tBAa at b′.

(f) From b′, draw a line b′ d′1 parallel to BD to represent centripetal component of acceleration of aDB, i.e. . c

DBa

227.878 m/sc DB

DBVaBD

= =

(g) From d′1, draw a line perpendicular to b′ d′1 or BD to represent direction of tangential component of acceleration of aDB, i.e.

. tDBa

(h) From o′ draw a line parallel to the line of motion of slider D to meet direction of at d′. t

DBa

o′ d′ represents acceleration of slider in magnitude and direction. Magnitude of acceleration of slider

231.5 m/sDa =

Example 5.5

Figure 5.10(a) shows Andreau Variable Stroke engine Mechanism in which links 2 and 7 have pure rolling motion. The dimensions of various links are indicated in the Figure 5.10. Determine acceleration of slider D if link 2 rotates at 1800 rpm.

Solution

Draw configuration diagram to the scale.

Velocity of point A, 2 1800 3.581 m/s60AV OA π ×

= × = .

The velocity of point B will also be equal to VA.

Velocity Diagram

35

Analysis by

Analytical MethodsA

The vector equations are as follows :

C A C= +V V V

and C B CB= +V V V

D C DC= +V V V

1

Q

7 B

C

D

5

3

A2 Q300

300

15076 mm

38 mm 62 mm

38 mm dia

63 mm dia

Wheel Wheel

6

4 D

(a)

d

a

C

b

o,q

d′1

c

b′

c′1

d′

a′

o′,q′ c′

(b) (c)

Figure 5.10

Velocity polygon is shown in Figure 5.10(a).

(a) Assume suitable scale say 1 cm → 1 m/s.

(b) Assume suitable position of pole o, and plot VA and VB perpendicular to OA and QB respectively. They are represented by oa and ob.

B

(c) Draw lines to represent directions of VCA and VCB perpendicular to AC and BC from points a and b respectively to meet at point c.

(d) Draw line to represent direction of VDC perpendicular to DC at point c.

(e) Draw line parallel to motion of slider from o to represent direction of motion of slider to meet direction of VDC at d.

1.28 m/sCAV =

1.3 m/sCBV =

5.2 m/sDCV =


36

Vector equations are as follows : Motion Analysis of Planar Mechanism and Synthesis c t

C A CA CA= + +a a a a

Also c tC B CB CB= + +a a a a

c tD C DC DC= + +a a a a

Both the wheels are rotating at uniform angular speed

∴ 2

2674.9 m/sc AA A

Va aOA

= = =

∴ 2 2(3.581) 407.1 m/s

0.0315c B

B BVa aOB

= = = =

2 2

21.28 26.425 m/s0.062

c CACA

VaAC

= = =

2 2

21.3 22.236 m/s0.076

c CBCB

VaBC

= = =

Assuming scale for acceleration as 1 cm → 200 m/s2.

(a) Assuming suitable position for o′, plot aA and aB parallel to OA and BQ respectively. They are represented in acceleration polygon by o′ a′ and o′ b′ respectively.

B

(b) From a′ and b′ plot and parallel to AC and BC. They are represented by a c″

cCAa c

CBa2 and b′ c′1 respectively.

(c) Draw lines perpendicular to AC from c″2 and perpendicular to BC from c′1 to represent directions of and . These lines meet at c′.

tCAa t

CBa

(d) Plot by drawing line parallel to DC from c′. This is represented by c′ d′

cDCa

1.

(e) Now draw a line from d′1 perpendicular to DC to represent direction of . t

DCa

(f) From o′, draw a line parallel to motion of slider D to meet direction of at d′. t

DCa

o′ d′ represents acceleration of slider D in magnitude and direction.

Acceleration of slider = 1320 m/s2.

5.5 ALTERNATIVE METHOD OF DETERMINING CORIOLIS’ COMPONENT OF ACCELERATION

In rigid body rotation, distance of a point from the axis of rotation remains fixed. If a link rotates about a fixed centre and ,at the same time, a point moves over it along the link, the absolute acceleration of the point is given by the vector sum of

(a) the absolute acceleration of the coincident point relative to which the point, under consideration, is moving;

(b) acceleration of the point relative to the coincident point; and

37

Analysis by

Analytical Methods(c) Coriolis’ component of acceleration.

Figure 5.11 shows a link 2 rotating at constant angular speed say ω2, it moves from position OC to OC1. During this interval of time, the slider link 3 moves outwards from position B to B B2 with constant velocity VBA (A is point on link 2 shich coincides with B which is on link 3). The motion of the slider 3 from B to B2 may be considered in the following three stages :

(a) B to A1 due to rotation of link 2, (b) A1 to BB1 due to outward velocity VBA, (c) BB1 to B2B due to acceleration perpendicular to the link which is the Coriolis’

component of acceleration.

1 2 2 1Arc Arc ArcB B DB= − DB

2 1Arc ArcDB A= − A

Also 1 2 1 1 2Arc = BAB B A B V t= δ θ δ ω tδ

)

r= ω

22= (BAV tω δ

The tangential component of the velocity perpendicular to the link is say Vt and it is given by

V t

In this case ω has been assumed constant and slider moves with constant velocity. Therefore, tangential velocity of point B on the slider 3 will result in uniform increase in tangential velocity because of uniform increase in value of r in the above equation.

(b) (c)

(d) (e)

3

2

C

D

ω2

A1

δθ

B1

C1

B2

ω2

2ω2,VBA

VBA

ω2

VBA2ω2,VBA

B on link 3

ink 2 δθ

VBA

(a) Figure 5.11

To result in uniform increase in value of Vt, there has to exist constant acceleration perpendicular to link 2.

Therefore, 21 2

1 ( )2

B B a t= δ (From Unit 1)

where a is the acceleration.

VBA

2ω2,VBA

ω2

VBA

VBA

2ω2,VBA

ω2

VBA

O

A on l

1

38


∴ 2 22

1 ( ) ( )2 BAa t V tδ = ω δ

22 BAa V= ω

This is Coriolis’ component of acceleration and will be denoted by corBAa .

Therefore, cor22BA Ba V= ω A

The directional relationship of VBA and 2 ω2 VBA is shown in Figures 5.11(b), (c), (d) and (e). If slider moves towards centre of rotation o, its velocity can be transmitted to other side so that it is directed outwards.

The direction of Coriolis’ component of acceleration is given by the direction of the relative velocity vector for the two coincident points rotated by 90o in the direction of the angular velocity of the rotation of the link.

If the angular velocity ω2 and the velocity VBA are varying, they will not affect the expression of the Coriolis’ component of acceleration but their instant values will be used in determining the magnitude and this value will be applicable at that instant. SAQ 3

What are the necessary and sufficient conditions for the Coriolis’ component of acceleration to exist?

Example 5.6

A cam and follower mechanism is shown in Figure 5.12(a), the dotted line shows the path of point B (on the follower). The cam rotates at 100 rad/s. Draw the velocity and acceleration diagram for the mechanism and determine the linear acceleration of the follower. Minimum radius of cam = 30 mm and maximum lift = 35 mm.

(b)

(a) (c)

Figure 5.12

Solution

taBA

caAOVBA

2ω,VBA

b′1

aB

o′

a′

b′

a

4

3

O

1

2

B on folower

A on cam. path

45 mm

30o

o

b

VBA VB

VBA

2ω2

VA

39

Analysis by

Analytical MethodsLinear velocity of point A which is on the cam, VA = ω × OA.

or, 45100 4.5 m/s1000AV = × =

B A BA= +V V V

Velocity Diagram


(b) Plot velocity of A by drawing a line perpendicular to OA from pole o. It is represented by oa.

(c) From o, draw a line parallel to motion of follower to represent direction of velocity of follower.

(d) Draw a line from ‘a’, parallel to the motion of B which is parallel to the tangent at cam profile to meet line representing direction of velocity of follower at b. ob represents velocity of follower in magnitude and direction

1.75 m/s and 4.85 m/sB BAV V= =

Velocity polygon is shown in Figure 5.12(b).


cor corc tB A BA A BA BAa a a a a a a a= + + = + + +

22

2

(4.5)45 450 m/s

1000A

AVaOA

= = =

20c BA

BAVa = =∞

cor 22 2 4.85 100 970 m/sBAa V= × ω = × × =

(a) Select suitable scale say 1 cm → 200 m/s.

(b) Plot aA by drawing a line parallel to OA from o′ and length equal to 2.25 cm.

(c) Now plot acor perpendicular to profile of cam and length equal to 4.85 cm. It is represented by a′ b′1.

(d) Now draw a line perpendicular to a′ b′1 to represent direction of tBAa .

(e) Draw another line from o′ to represent direction of motion of B to meet direction of t

BAa at b′.

o′ b′ represents acceleration of B in magnitude and direction.

23 200 600 m/sBa = × =

The acceleration polygon is shown in Figure 5.12(c).

Example 5.7

A quick return motion mechanism is shown in Figure 5.13(a). The crank rotates at 20 rad/s. Determine angular acceleration of the slotted link 3.

Solution

The configuration diagram is drawn to the scale according to the given dimensions.

40

Let point B (on the link 3) coincides point A (on the crank 2). Velocity of point A is perpendicular to OA and


15 20 3 m/s100AV OA= × ω = × =

A B AB= +V V V

(c)

(a) Figure 5.13

Velocity Diagram


(b) Plot velocity VA by drawing line from o perpendicular to OA. It is represented by oa.

(c) Draw a line from o perpendicular to slotted link 3 to represent direction of velocity VB. B

(d) From a, draw a line parallel to slotted link 3 to represent VAB which represents movement of slider. This line meets direction of VB at b. B

Here ob represents velocity of B in magnitude and direction and ba represents velocity of slider VAB in magnitude and direction.

1.5 m/s ; 2.6 m/sB ABV V= =

The velocity polygon is shown in Figure 5.13(b).

Acceleration Diagram cor corc t t c

A B AB B B AB AB= + + = + + + +a a a a a a a a a

aBA

aB

craAB

caBC

taBC

a′

b′1

b′′ C

3

2

O1

4

15 cm

B On Link 2 A On Link 2 and 4

35 cm

25 cm

a

b

O,C

(b)

VAB

aA

b′

o′c′

41

Analysis by

Analytical Methods2 2

23 60 m/s0.15

c AA A

Va aOA

= = = =

2 2221.5 9 m/s ; 0

0.25c cB AB A

V Va aBC

BB= = = = =

∞

cor 21.52 2 2.6 31.20.25

BAB

Va VBC

= × = × × = m/s

(a) Assume suitable scale say 1 cm → 10 m/s2.

(b) Plot aA by drawing line parallel to OA from a suitable point o′. It is represented by o′ a′.

(c) Plot acor (Coriolis, component of acceleration) as per direction determined in Figure 5.13(c) and according to the vector equation. It is represented by b″ a′. It is perpendicular to link 3.

(d) From b″, draw a line perpendicular to b″ a′, i.e. parallel to link 3 to represent direction of motion of slider ( )t

ABa .

(e) Starting from o′, plot cBa by drawing line parallel to link 3. It is

represented by o′ b′1.

(f) From b′1, draw a line perpendicular to o′ b′1, i.e. perpendicular to link 3 to represent direction of t

Ba and meet line representing

direction of acceleration tABa at b′.

Here o′ b′ represents acceleration of B in magnitude and direction and b′ b″ represents acceleration of slider in magnitude and direction. Vector represented by b′1 b′ represents tangential component of acceleration of B

283.5 m/stBa =

∴ 83.5BCBC × α =

or Angular acceleration of link 3 83.5( )0.25BCα =

or 2334 rad/sBCα =

Example 5.8

Figure 5.14(a) shows Whitworth Quick Return Mechanism. The dimensions are written in the Figure 5.14. Determine velocity and acceleration of slider D when crank rotates at 120 rpm uniformly in the sense indicated in Figure 5.14.

Solution

The velocity of A, 2 2 1200.2 2.51 m/s60 60A

NV OA π π ×= × = × = .

Draw configuration diagram to the scale. The point B is on the slotted link which coincides with point A on link OA.

Velocity Diagram

A B AB= +V V V

BC

VV QQB

= × C

C

and D C D= +V V V

42


C

Q

Q

D

B

A

10 cm

15 cm

26 cm

20 cm

50 cm

VQB= ba

o,q

c

d

a

VO

VB

b VAB

(a) (b)

a

b′′

b′

b′1

o′,q′ d′

c′d′1

= 2ωaB VABr

AB ωQB

(c) (d)

Figure 5.14

(a) Plot VA by selecting proper scale say 1 cm → 0.5 m/s by a line perpendicular to OA. It is represented by oa.

(b) From o, draw a line perpendicular to OB for direction of VB. B

(c) From a, draw a line parallel to QB to represent direction of VAB, i.e. sliding velocity of slider to meet direction of velocity of B, i.e. VB at b. The vector ob represents V

B

BB in magnitude and direction.

2.42.4 m/s 0.15 1.38 m/s0.26B CV V= ∴ = × =

(d) Plot VC perpendicular to QC from o. It is represented by oc. (e) Draw a line parallel to motion of slider from o. (f) Draw a line perpendicular to CD from c to meet direction of velocity

of slider at d. The velocity of slider in magnitude and direction is given by od.

2

22.510.75 m/s, 31.5 m/s0.20D ABV V= = =


Acceleration of A, 2 2

22.51 31.5 m/s0.20

AA

VaOA

= = =

43

Analysis by

Analytical Methods cor corc t c tA B AB B B AB AB= + + = + + + +a a a a a a a a a

2 2

22.4 22.15 m/s ; 00.26

c cB AB A

V Va aQB

= = = = =∞

BB

cor 22.42 2 0.9 16.61 m/s0.26

BAB

Va VQB

= × = × × =

Acceleration of B can be determined as explained in Example 5.7. Assume suitable scale say 1 cm → 5 m/s2. Extend o′ b′ to the other side of o′ (C is on other side of B) upto c′ such that

QCo c o bQB

′ ′ ′ ′= ×

c tD C DC C DC D= + = + +a a a a a a C

2 2

20.8 1.28 m/s0.5

c DCDC

Va

CD= = =

(a) Draw a line parallel to CD and plot . It is represented by c′ d′

cDCa

1.

(b) Draw a line perpendicular to c′ d′1, i.e. perpendicular to CD to represent direction of . t

DCa

(c) Draw a line from o′ parallel to motion of slider D to meet direction of at d′. t

DCa

The acceleration of slider is represented by o′ d′ in magnitude and direction

26.5 m/sDa =

Example 5.9

A swivelling point mechanism is shown in Figure 5.15. If the crank OA rotates at 200 rpm, determine the acceleration of sliding of link DE in the trunion. The following dimensions are known AB = 18 cm, DE = 10 cm, EF = 10 cm, AD = DB, BC = 6 cm, OC = 15 cm, OA = 2.5 cm.

Solution

The configuration diagram is plotted to the scale say 1 cm → 4 cm.

Velocity Diagram

; andB A BA S D SD F E F= + = + = +V V V V V V V V V E

The point S is on the link DE which slides in the swivel block.

2 2.5 2 200 0.52 m/s60 100 60A

NV OAπ π ×= × = × =

Velocity of S related to D, VSD is perpendicular to SD. Velocity of S related to fixed block is parallel to link DE. Select a suitable scale say 1 cm → 0.2 m/s.

(a) Plot VA. It is represented by oa in velocity polygon (Figure 5.15(b)).

(b) Draw a line perpendicular to BC to represent direction of VB from o.

B

(c) Draw a line perpendicular to AB from a to represent direction of VBA to meet direction of VB at b. B

44


(d) Plot point d on ab such that ADad abAB

= .

(e) Plot de such that DEde dsDS

= .

(f) From e, draw ef perpendicular EF to represent direction of VFE. (g) Draw a line parallel to motion of slider to meet direction of VFE

at f. VSD = VS = 0.32 m/s, VDE = 0.84

(a)

4508.5 cm

7.5 cm A

C

E

B

O

D

SQ

12 cm

F

(b) (c)

Figure 5.15


Acceleration of A, 2 2

2(0.52) 10.816 m/s2.5100

AA

VaOA

= = =

c tB A BA B= + + Aa aa a

Also, c tB B B= +a a a

2 22 2

2(0.32) (0.44); 1.075 m/s180.04100

C CSD BASD BA

V Va aSD AB

= = = = =

2

cor 2(0.48)2 ; 3.84 m/s0.1

CDE S Ba V a= ω × = =

2(0.84)2 2

0.1DE

SV V 0.32DE

= × = × ×

a a cor C t corS SD SD SD= + = + +a a a a

Select suitable scale say 1 cm → 2 m/s2.

f

ea

d

b

s

O, C

s′′

s′

b′′

O′,C′

b′

d′

a′b′

S′1

45

Analysis by

Analytical Methods(a) Plot acceleration polygon for four bar chain OABC as explained

in earlier examples.

(b) Plot point d′ on a′ b′ such that ADa d a bAB

′ ′ ′ ′= .

(c) Plot parallel to DS. It is represented by d′ s′CSDa 1.

(d) Draw a line perpendicular to d′ s′1 for direction . tSDa

(e) Plot acor by drawing a line perpendicular to DE. It is represented by o′ s″.

(f) Draw a line perpendicular to o′ s″ for direction of sliding acceleration as to meet direction at s′. Join o′ s′ which represents acceleration of point S.

tSDa

28 m/ssa = .

5.6 KLEIN'S CONSTRUCTION FOR DETERMINING VELOCITY AND ACCELERATION OF SLIDER CRANK MECHANISM

There are some special methods for determining velocity and acceleration of slider in slider crank chain. Case I

In this case, crank rotation is assumed uniform. The configuration diagram of slider crank chain is shown in Figure 5.16(a). The method is illustrated in following steps :

(a) Draw configuration diagram at a suitable scale. (b) Extend PC if necessary to meet perpendicular to stroke line at M. (c) Draw a circle with C as centre and CM as radius. (d) Draw another circle with CP as diameter. (e) Draw common chord KL of these two circles and extend if necessary

to meet line of stroke at N. OCQN is acceleration polygon. (f) NO gives acceleration of piston at a scale equal to ω2 × configuration

scale.

(b)

(a) (c)

450P

X

K

C M Q x

N N1

L O O1

O′ P′

C′ P′′

X

O′

O P′

X

C′ P′′

Figure 5.16

Proof

In the acceleration polygon OCQN

46


OC → Centripetal acceleration of C,

CQ → Centripetal acceleration of P relative to C,

QN → Tangential component of acceleration, and

NO → Acceleration of piston.

Join K with P and C. The triangle CKP and CQK are similar

∴ CQ CKCK CP

=

∴ 2CKCQ

CP=

But CK CM=

Therefore, 2CMCQ

CP=

The acceleration according to the methods discussed earlier has been drawn in Figure 5.16(b). The acceleration polygon o′ c′ p″ p′ appears to be similar to OCQN. If ω is angular velocity of crank, VC = OC ω and similarly velocity of piston VP = ω OM.

Also, PCV CM= ω

Therefore, it indicates that CQ represents centripetal acceleration 2

PCVPC

⎛ ⎞⎜⎜⎝ ⎠

⎟⎟ to the

scale ω2 × configuration scale. Thus these two polygons are similar and ON represents acceleration of the piston to the above mentioned scale. For any point X on the connecting rod, draw a line through X parallel to the line of stroke to intersect CN at x and acceleration of x is given by ‘ox’.

Case II

In this case, crank rotates and has angular acceleration. The acceleration polygon is reproduced in Figure 5.16(c). The point Q is determined in the same way as in Case I. From O, draw a line perpendicular to CO and plot tangential acceleration OO1. Join C with O1. Through O1, draw a line parallel to line of stroke to meet KL and N1. The acceleration polygon is O1 C Q N1 and O1 N1 is acceleration of the piston. For a point X on CP join x (determined in Case I) with O1. This acceleration polygon is similar to Figure 5.16(c).

SAQ 4 Determine acceleration of piston by using Klein’s construction for crank angle Oo,

2π and π.

47

Analysis by

Analytical Methods5.7 SUMMARY

The motion of a point can always be related to any other point through relative velocity and relative acceleration. A general plane motion which has neither pure translation nor pure rotation can be systematically analysed by moving frame of references. The resulting motion is the vector sum of rotating motion and translation. The motion of any point in a mechanism can be related to a point whose motion is known. Therefore, starting from a point of known motion, the motion of any other point can be determined with the help of relative velocity and relative acceleration. The direction of relative velocity is always normal to the line joining the two points (because of rigid body concept). If distance between these two points does not change, the link is subjected to centripetal component and tangential component of acceleration. If the distance tends to change, because points are not connected with each other, there may be Coriolis’ component and sliding component of acceleration in addition to the earlier component of acceleration. The Coriolis’ component of acceleration is equal to twice the product of angular speed and sliding velocity is magnitude and perpendicular to rotating line. The motion can be determined analytically as well as graphically. You will know more about analysis in the next unit.

5.8 KEY WORDS

Relative Velocity : Velocity of a point in relation to any other point.

Relative Acceleration : Acceleration of a point in relation to any other point.

Coriolis’ Component of : It is equal to twice the product of angular Acceleration velocity and sliding velocity over which slider is sliding and acts perpendicular to link.

Polygon : It is a closed figure made by several lines.

Klein’s Construction : It is an alternative method which was devised by Prof. Klein for determining acceleration of piston in slider crank mechanism.

5.9 ANSWERS TO SAQs

SAQ 1

Refer the text.

SAQ 2

It will be vector sum of velocities of translation which is V and velocity due to

rotation which is V rr

⎛ ×⎜⎝ ⎠

⎞⎟ equal to V. Therefore, velocity of required point is 2 V

since both of them have same sense and same line of action.

SAQ 3

Available in text.

SAQ 4

Figure 5.17 shows Klein’s construction for various position of crank.

For θ = 0 Acceleration is given by N1 0.

48

Motion Analysis of Planar Mechanism and Synthesis 2

πθ = Acceleration is given by ON3.

θ = π Acceleration is given by ON2.

Figure 5.17

Outer Dead-Centre

N3 N2

C2,Q2

M3,C3 K2

P1 P3 P2 N1

K1

L1

C1,Q1

Innerdead Centre

O

Education

Método de Klein