Velocity Analysis using Pole Method

UNIT 5 ANALYSIS BY ANALYTICAL METHODS

Analysis by Analytical Methods

Structure 5.1

Introduction Objectives

5.2

5.3

Analytical Method for Velocity and Acceleration 5.2.1

Approximate Method for Slider Crank Mechanism

5.2.2

General Method for Velocity and Acceleration

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. 21

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). A

l r

2 3 θ

O

B φ

C

B1 4

1

x

Figure 5.1

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

l ⎧ ⎫ x = r (1 − cos θ) + l (1 − cos φ) = r ⎨(1 − cos θ) + (1 − cos φ) ⎬ r ⎩ ⎭

or Also

AC = r sin θ = l sin φ

r sin θ l

∴

sin φ =

Also

cos 2 φ = 1 − sin 2 φ = 1 −

∴

⎡ ⎤2 r2 cos φ = ⎢1 − 2 sin 2 θ ⎥ l ⎢⎣ ⎥⎦

r2 l2

sin 2 θ

1

By Binomial theorem

cos φ = 1 −

1 2

2

4

1 ⎛r⎞ 1 ⎛r⎞ 2 4 ⎜ ⎟ sin θ − ⎜ ⎟ sin θ − 8⎝l⎠ 16 ⎝l⎠ 2

or

4

6

⎛r⎞ 6 ⎜ ⎟ sin θ l ⎝ ⎠ 6

1 ⎛r⎞ 1 ⎛r⎞ 1 ⎛r⎞ 6 (1 − cos φ) = ⎜ ⎟ sin 2 θ + ⎜ ⎟ sin 4 θ + ⎜ ⎟ sin θ + . . . 2⎝l⎠ 8⎝l⎠ 16 ⎝ l ⎠

Substituting (1 – cos φ) in expression of x you will get 3 ⎡ ⎤ 1 ⎛r⎞ 1 ⎛r⎞ x = r ⎢(1 − cos θ) + ⎜ ⎟ sin 2 θ + ⎜ ⎟ sin 4 θ + . . .⎥ 2⎝l⎠ 8⎝l⎠ ⎣⎢ ⎦⎥

In this mechanism,

r ⎛r⎞ is approximately 0.25 and higher powers of ⎜ ⎟ shall have l ⎝l⎠

negligible value. Therefore, 22

⎡ ⎤ 1 ⎛r⎞ x ≈ r ⎢(1 − cos θ) + ⎜ ⎟ sin 2 θ ⎥ 2⎝l⎠ ⎣ ⎦

Velocity of slider, V =

dx dx d θ = × dt d θ dt

dθ =ω dt


(angular speed of crank)

dx dθ

Therefore,

V =ω

or

⎡ ⎤ 1 ⎛r⎞ V = r ω ⎢sin θ + ⎜ ⎟ 2 sin θ cos θ ⎥ 2⎝l⎠ ⎣ ⎦

or

⎛ ⎞ 1 ⎛r⎞ V = r ω ⎜ sin θ + ⎜ ⎟ sin 2 θ ⎟ 2⎝l⎠ ⎝ ⎠

Let crank rotate at the constant angular velocity.

dV dV d θ = × dt d θ dt

Acceleration of slider,

a=

or

r ⎡ ⎤ a = r ω2 ⎢cos θ + cos 2 θ⎥ l ⎣ ⎦

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 d eθ = 1 × d θ = d θ in the direction opposite to that of er. Therefore,

d er d er d θ dθ = × = θ eθ , where → θ dt dθ dt dt d eθ d eθ d θ = × = θ (− er ) = − θ er dt dθ dt

and

V Vθ deθ

Vr P

der

eθ

Y

er dθ

r

dθ er Z

eθ

θ

θ X

Figure 5.2

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

d er dr d dr = (r er ) = er + r dt dt dt dt = r er + r θ eθ = Vt er + Vθ eθ

V =

23


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

a=

dV d = (r er + r θ eθ ) dt dt

d d ( r er ) + (r θ eθ ) dt dt d er dr d r d er + r = + (θ eθ ) + r (θ eθ ) dt dt dt dt =

d eθ ⎞ ⎛ = θ eθ + θ r er + r (θ eθ ) + r θ eθ + r ⎜ dt ⎟⎠ ⎝ = r er + 2 r θ eθ + r θ eθ − r θ 2 er = ( r − r θ 2 ) er + ( r θ + 2 r θ ) eθ = ar er + aθ eθ

where ar represents radial component and aθ represents transverse component of acceleration. ar = r − r θ 2 and aα = r θ + 2 r θ

r → Acceleration of slider along slotted link,

Here,

r θ 2 → Centripetal acceleration,

r θ → Tangential component of acceleration, and 2r θ → Coriolis’ component of acceleration.

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. D

A

A

15 ω B

60

ω = 10 rad/sec

0

B

60

θ 32.5 cm

E

O

(a)

(b) 42 48

eθ

0

A

er θ′

θ′′ ω

B

18

C

0

60

0

O

(c) Figure 5.3

24

13 0

0

D


Solution

AE 13 13 = = or θ = 18o BE 32.5 + 7.5 40

In Figure 5.3(b),

tan θ =

Also,

13 AB = = AB or AB = 42.06 cm sin θ sin 90o

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

VA = OA × ω =

15 × 10 = 1.5 m/s 100

a A = OA × ω2 =

15 × 10 = 15 m/s 2 100

The components of VA along BD are perpendicular to it are given by Vr = r = − 1.5 cos 48o = − 1.0 m/s

1.5 sin 48o Vθ = r θ = 1.5 sin 48o ∴ θ = r o

or

1.5 sin 48 θ = = 2.65 rad/s 42.06 100

The transverse component of acceleration of A is give by aθ = − a A sin 42o = − 15 sin 42o = − 10.037 m/s 2

Also

aθ = (r θ) + 2 (r θ )

or

− 10.037 =

or

5.30 − 10.037 θ= = − 11.26 rad/s 2 0.4206

42.06 θ + 2 (− 1.0) (2.65) 100

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 a12 = a1 − a2 V12 = V1 − V2

25

Z a1 B

B


Figure 5.4

Also

V21 = V2 − V2 = − V12

and

a21 = a2 − a1 = − a12

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. P′

P

θ

Q′ Q1

Q

Figure 5.5(a) : Translation from PQ to P′Q1 and Rotation about P′ P1

P

P′

θ

Q′

Q

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

θ

26 C

C′


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

θ

O

O′ θ Q′ Q1

Q

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

θ

Q1

Q′

Q

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

P1

P′

θ

Q′

Q

Figure 5.5(f) : Rotation from PQ to P1Q and Translation to P′ Q′ P1

P

P′

θ C

C′

Q′

Q

Q1

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

P θ

27 O

O′


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 VBA. 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. B

Therefore,

VBA cos θ = 0 or

or,

θ=

cos θ = 0

π 2

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

28

aB aB B

B

o′

b′ a′


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.

(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

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 : a B = a A + a BA t c c a BA = a BA + a BA ∴ a B = a A + a BA + a Bt A

Since

t c Where aBA is acceleration of B relative to A; aBA is tangential component of aBA and aBA is centripetal component of aBA.

c aBA

2 VBA = AB

t It is directed from B to A along AB whereas aBA 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)

c Plot centripetal component a BA by drawing line parallel to AB and represent c its magnitude according to the vector equation. a BA is represented by a′a′1.

(c)

From a1 draw a line perpendicular to aa1′ or parallel to the perpendicular to t AB to represent direction of aBA .

(d)

Draw a line parallel to the direction of acceleration of B, i.e. aB from fixed t reference o′ to meet the line representing direction of a BA . They meet at b′ . o′b′ represents acceleration of B in direction and magnitude. B

Example 5.2

29


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. a A aB b′

o′

o ω

r

b O

45

B

0

aA

a′1

a′

Figure 5.7

Solution

2π × 600 = 6.28 rad/s . The crank OA is a 60 rotating body about fixed centre O. Therefore, the velocity of point A is given by The angular velocity of crank OA, ω =

V A = ω × OA = 6.28 × 7.5 = 47.1 cm/s 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. a A = a cA =

VA2 (0.471)2 = = 2.96 m/s 2 7.5 OA 100

It is directed from A towards O. c t a B = a A + a BA = a A + a BA + a BA

c = aBC

2 VBA along AB directed from B to A AB

c a BC =

(3.425) 2 = 39.17 m/s 0.3

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 c to AB for aBA and represent its magnitude. This is represented by a′a′1. Now from t a′1 draw a line perpendicular to a′a′1 or AB to represent direction of aBA . 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

aB = 215 m/s 2

Example 5.3

30

A 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.


O′2, O′4 3

A 28 mm

B

C 80 mm 4

75 mm 2

α = 4400 rad/sec

37 mm

i′,O′

O4

2

ω = 100 rad/sec 0 53

1 125 mm

O2

b′

1

a′

(a)

a′

c′

b b′

c

O2, O4

b′

a

(b)

(c) Figure 5.8

Solution

Draw configuration diagram to the scale. The velocity of point A, VA = O2 A × ω =

75 × 100 = 7.5 m/s 1000

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 VBA at point b. o2b represents VB in magnitude and direction. B

B

(e)

For determining velocity of C, plot point c on ab such that AC × ab . ‘ac’ = AB From velocity polygon and

VBA = 6.5 m/s VB = 9 m/s

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 a cA = O2 A ω2 . Or 31


75 × (100)2 = 750 m/s 2 . A scale 1 cm → 250 m/s2 may 1000 serve the purpose. a cA =

(b)

Draw a line parallel to O2A and plot a cA 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 75 × 4400 = 300 m/s 2 . It will be denoted is atA = O2 A × α = 1000 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 c component of acceleration aBA which is given by 2 VBA (6.5)2 c = = 528.15 m/s 2 . Plot magnitude of aBA . It 0.08 AB is represented by a′ b′1.

c aBA =

(e)

Draw a line perpendicular to AB from b′1 to represent direction t of tangential component of acceleration aBA .

(f)

Draw a line parallel to O4B from o′2 to represent centripetal component of acceleration of B. The magnitude is given by VB2 92 c aB = = = 2189.189 m/s 2 . It will be denoted by O4 B 0.037 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 t the line representing direction aBA 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 AC 28 × a ′ b′ = × 72 = 2.52 . Joint o′2 with c′ and that a′ c′ = AB 80 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

αO4 B =

aBt = 3479.3 rad/s 2 O4 B

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. d′1 c aDB

t aBA

b′

32 aB B

B

VD

t aBC aA B

B

O,C

d t aDB

b′1

VDB VB

c aBA a′ aDB B

B

VA


(a)

(b)

28 mm

44 mm

2

3

B

O 0

A

75

65 mm

49 mm C

11

D

(c) Figure 5.9

Solution

Plot configuration diagram to the suitable scale. Velocity Diagram

Velocity of A, VA = O2 A ×

2π × 240 = 0.7037 m/s 60

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, VBA = 0.4 m/s, VDB = 0.602 m/s B

Velocity polygon is shown in Figure 5.9(b). Acceleration Diagram

33

⎛ 2π × 240 ⎞ 2 Acceleration of A, a A = a cA = OA × ⎜ ⎟ = 17.686 m/s Select a 60 ⎝ ⎠ 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)

c Plot centripetal component of acceleration aBA, i.e. aBA by drawing line parallel to AB.

c aBA =

VB2 = 3.636 m/s 2 AB

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

aBc = (e)

VB2 = 5.102 m/s 2 BC

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. aBt . t This line meets direction of aBA at b′.

(f)

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

2 VDB = 7.878 m/s 2 BD

(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. t . aDB

(h)

From o′ draw a line parallel to the line of motion of slider D to t meet direction of aDB at d′. o′ d′ represents acceleration of slider in magnitude and direction. Magnitude of acceleration of slider aD = 31.5 m/s 2

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, VA = OA ×

2π × 1800 = 3.581 m/s . 60

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

34


The vector equations are as follows : VC = V A + VCA VC = VB + VCB

and

VD = VC + VDC C 0

30 76 mm 6

4 D 5

38 mm

0

15

62 mm D 3

B

7 0

30

Q 1 Wheel

2

Q

A 38 mm dia

Wheel

63 mm dia

(a) d′1 C c a b a′ o′,q′

c′

b′ c′1

o,q d

d′

(b)

(c) Figure 5.10

Velocity polygon is shown in Figure 5.10(a). (a)


(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.

(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.

B

VCA = 1.28 m/s VCB = 1.3 m/s VDC = 5.2 m/s


35


Vector equations are as follows : c t aC = a A + aCA + aCA c t aC = a B + aCB + aCB

Also

c t a D = aC + a DC + a DC

Both the wheels are rotating at uniform angular speed ∴

a A = a cA =

VA2 = 674.9 m/s 2 OA

∴

aB = aBc =

VB2 (3.581)2 = = 407.1 m/s 0.0315 OB

c aCA =

2 VCA 1.282 = = 26.425 m/s 2 AC 0.062

c aCB =

2 VCB 1.32 = = 22.236 m/s 2 BC 0.076

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)

c c From a′ and b′ plot aCA and aCB parallel to AC and BC. They are represented by a c″2 and b′ c′1 respectively.

(c)

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

(d)

c Plot aDC by drawing line parallel to DC from c′. This is represented by c′ d′1.

(e)

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

(f)

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

B

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

36

(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

(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 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 :


B

(a)

B to A1 due to rotation of link 2,

(b)

A1 to B1 due to outward velocity VBA,

(c)

B1 to B2 due to acceleration perpendicular to the link which is the Coriolis’ component of acceleration.

B

B

B

Arc B1 B2 = Arc DB2 − Arc DB1 = Arc DB2 − Arc AA1 Arc B1 B2 = A1 B1 δ θ = VBA δ t ω2 δ t

Also

= VBA ω2 (δ t ) 2

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

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. C

C1 2

B1

D

2ω2,VBA

B2 δθ

B on link 3

ω2

VBA 3 VBA

VBA

ω2

2ω2,VBA

A1

(b)

A on link 2

(c)

δθ

2ω2,VBA

VBA VBA

ω2

ω2

VBA

ω2

2ω2,VBA

O

(d)

VBA

(e)

1

(a) Figure 5.11

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

B1 B2 =

where a is the acceleration.

1 a (δ t ) 2 2

(From Unit 1) 37


1 a (δ t )2 = VBA ω2 (δ t )2 2

∴

a = 2 VBA ω2 cor This is Coriolis’ component of acceleration and will be denoted by aBA .

Therefore,

cor aBA = 2 ω2 VBA

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. t aBA

b′

b′1 4

aB

2ω,VBA A on cam. path

o′ c a AO

B on folower VBA

3 45 mm

a′ O

o

(b)

30

VBA

2

2ω2

b

1 VBA a

(a)

(c) Figure 5.12

Solution

38

VA

VB o

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

VA = 100 ×

or,


45 = 4.5 m/s 1000

VB = V A + VBA

Velocity Diagram

(a)


(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 VB = 1.75 m/s and VBA = 4.85 m/s

Velocity polygon is shown in Figure 5.12(b). Acceleration Diagram c t aB = a A + aBA + a cor = a A + aBA + aBA + a cor

(4.5) 2 aA = = 45 = 450 m/s 2 OA 1000 VA2

c = aBA

2 VBA =0 ∞

a cor = 2 VBA × ω = 2 × 4.85 × 100 = 970 m/s 2

(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)

t Now draw a line perpendicular to a′ b′1 to represent direction of aBA .

(e)

Draw another line from o′ to represent direction of motion of B to t meet direction of aBA at b′. o′ b′ represents acceleration of B in magnitude and direction. aB = 3 × 200 = 600 m/s 2

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.

39


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

VA = OA × ω =

15 × 20 = 3 m/s 100

V A = VB + V AB O 1

15 cm 2

4

A On Link 2 and 4

b B On Link 2

35 cm

O,C 3

(b)

25 cm

b′′

a

C a′

VAB

cr a AB aBA

(c)

aA

b′ aB o′c′

t aBC

c aBC

(a) Figure 5.13

b′1

Velocity Diagram

(a)


(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. VB = 1.5 m/s ; V AB = 2.6 m/s

The velocity polygon is shown in Figure 5.13(b). Acceleration Diagram a A = a B + a AB + a cor = a Bc + a Bt + a tAB + a cAB + a cor

40


V2 32 = A = = 60 m/s 2 OA 0.15

aA =

a cA

aBc =

VB2 1.52 V2 = = 9 m/s 2 ; a cAB = AB = 0 ∞ BC 0.25

a cor = 2 VAB ×

VB 1.5 = 2 × 2.6 × = 31.2 m/s 2 0.25 BC

(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 ( a tAB ) .

(e)

Starting from o′, plot aBc 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 aBt and meet line representing direction of acceleration a tAB 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 aBt = 83.5 m/s 2

∴

BC × α BC = 83.5

or Angular acceleration of link 3 (α BC ) =

83.5 0.25

α BC = 334 rad/s 2

or 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, VA = OA ×

2π N 2 π × 120 = 0.2 × = 2.51 m/s . 60 60

Draw configuration diagram to the scale. The point B is on the slotted link which coincides with point A on link OA. Velocity Diagram V A = VB + V AB

VC = and

VB × QC QB

VD = VC + VDC

41


c C 50 cm

15 cm

VO

D

Q

o,q 26 cm

10 cm

d

VB

Q 20 cm

B A b VAB

VQB= ba

a

(a)

(b)

b′′ b′ b′1

d′

o′,q′

r a AB = 2ωaB VAB

ωQB

c′

d′1

(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.

(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 VB in magnitude and direction.

B

B

B

VB = 2.4 m/s ∴ VC =

2.4 × 0.15 = 1.38 m/s 0.26

(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. VD = 0.75 m/s, VAB =

2.512 = 31.5 m/s 2 0.20


Acceleration of A, a A = 42

VA2 2.512 = = 31.5 m/s2 OA 0.20


a A = a B + a AB + a cor = a Bc + a Bt + a cAB + a tAB + a cor

aBc =

VB2 2.42 V = = 22.15 m/s 2 ; a cAB = AB = 0 QB 0.26 ∞

a cor = 2 VAB ×

VB 2.4 = 2 × 0.9 × = 16.61 m/s 2 0.26 QB

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

o′ c′ =

QC × o′ b′ QB c t a D = aC + a DC = aC + a DC + aD C

c aDC =

2 VDC 0.82 = = 1.28 m/s 2 0.5 CD

(a)

c Draw a line parallel to CD and plot aDC . It is represented by c′ d′1.

(b)

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

(c)

Draw a line from o′ parallel to motion of slider D to meet t direction of aDC at d′. The acceleration of slider is represented by o′ d′ in magnitude and direction aD = 6.5 m/s 2

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 VB = V A + VBA ; VS = VD + VSD and VF = VE + VFE

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

VA =

2π N 2.5 2 π × 200 × OA = × = 0.52 m/s 60 100 60

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.

(c)

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

B

B

43


AD . AB

(d)

Plot point d on ab such that ad = ab

(e)

Plot de such that de = ds

(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.

DE . DS

VSD = VS = 0.32 m/s,

VDE = 0.84

E

F

S B

Q

12 cm

D 7.5 cm A

8.5 cm

0

45

C

O

(a) s′′ O′,C′ a e s b′

b′′

d

d′ f

b

O, C

s′ b′

(b)

(c)

S′1

Figure 5.15


Acceleration of A, a A =

VA2 (0.52) 2 = = 10.816 m/s 2 2.5 OA 100

c a B = a A + a BA + a Bt A

Also,

a B = a Bc + a Bt C aSD =

2 VSD V2 (0.32)2 (0.44) 2 C ; aBA = = BA = = 1.075 m/s 2 18 0.04 SD AB 100

a cor = 2 ωDE × VS ; aBC = =2

(0.48) 2 = 3.84 m/s 2 0.1

VDE (0.84) 2 × VS = 2 × × 0.32 0.1 DE

C t aS = aSD + a cor = aSD + a SD + a cor

Select suitable scale say 1 cm → 2 m/s2. 44

a′

(a)

Plot acceleration polygon for four bar chain OABC as explained in earlier examples.

(b)

Plot point d′ on a′ b′ such that a′ d ′ = a′ b′

(c)

C Plot aSD parallel to DS. It is represented by d′ s′1.

(d)

t Draw a line perpendicular to d′ s′1 for direction aSD .

(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 t acceleration as to meet direction aSD at s′. Join o′ s′ which represents acceleration of point S.


AD . AB

as = 8 m/s 2 .

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. P′ O′ K X C

Q

M

C′

P′′

x

X N1 N

P

0

45 L

O′

O1

(b)

O

P′

O X

C′

(a)

P′′

(c) Figure 5.16

Proof

In the acceleration polygon OCQN

45

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 CK = CK CP

∴

CQ =

But

CK = CM

Therefore,

CQ =

CK 2 CP

CM 2 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,

VPC = ω CM

⎛V 2 ⎞ Therefore, it indicates that CQ represents centripetal acceleration ⎜ PC ⎟ to the ⎜ PC ⎟ ⎝ ⎠ 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, π and π. 2

46


5.7 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 Acceleration

: It is equal to twice the product of angular 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 ⎛V ⎞ rotation which is ⎜ × r ⎟ equal to V. Therefore, velocity of required point is 2 V r ⎝ ⎠ 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. 47


π 2

Acceleration is given by ON3.

θ=π

Acceleration is given by ON2.

θ=

M3,C3

K1

P1

P3

P2

N1

Innerdead Centre C1,Q1 O

C2,Q2 N3

N2 Outer Dead-Centre

L1

Figure 5.17

48

K2

Velocity Analysis using Pole Method

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