Ch 08 Theory Of Machine R.S.Khurmi

Since the point B moves with respect to A with an angular velocity of ω rad/s, therefore centripetal or radial component of the acceleration of B with respect to A , This radial componen

8.1 IntrIntrIntroductionoduction

We have discussed in the previous chapter thevelocities of various points in the mechanisms Now we shalldiscuss the acceleration of points in the mechanisms Theacceleration analysis plays a very important role in thedevelopment of machines and mechanisms

8.2

8.2 Acceleration Diagram for a LinkAcceleration Diagram for a Link

Consider two points A and B on a rigid link as shown

in Fig 8.1 (a) Let the point B moves with respect to A, with

an angular velocity of ω rad/s and let α rad/s2 be the angular

acceleration of the link AB.

(a) Link (b) Acceleration diagram.

Fig 8.1. Acceleration for a link.

Warping Machine

CONTENTS

We have already discussed that acceleration of a particle whose velocity changes both inmagnitude and direction at any instant has the following two components :

1 The centripetal or radial component, which is perpendicular to the velocity of theparticle at the given instant

2 The tangential component, which is parallel to the velocity of the particle at the giveninstant

Thus for a link A B, the velocity of point B with respect to A (i.e vBA) is perpendicular to the

link A B as shown in Fig 8.1 (a) Since the point B moves with respect to A with an angular velocity

of ω rad/s, therefore centripetal or radial component of the acceleration of B with respect to A ,

This radial component of acceleration acts perpendicular to the velocity vBA, In other words,

it acts parallel to the link A B

We know that tangential component of the acceleration of B with respect to A ,

BAt Length of the link

This tangential component of acceleration acts parallel to the velocity vBA In other words,

it acts perpendicular to the link A B.

In order to draw the acceleration diagram for a link A B, as shown in Fig 8.1 (b), from any point b', draw vector b'x parallel to BA to represent the radial component of acceleration of B with

respect to A i.e aBAr and from point x draw vector xa' perpendicular to B A to represent the tangential component of acceleration of B with respect to A i.e aBAt Join b' a' The vector b' a' (known as

acceleration image of the link A B) represents the total acceleration of B with respect to A (i.e aBA)and it is the vector sum of radial component (aBAr )and tangential component (aBAt )of acceleration

(a) Points on a Link (b) Acceleration diagram.

Fig 8.2. Acceleration of a point on a link.

Consider two points A and B on the rigid link, as shown in Fig 8.2 (a) Let the acceleration

of the point A i.e aA is known in magnitude and direction and the direction of path of B is given The acceleration of the point B is determined in magnitude and direction by drawing the acceleration

diagram as discussed below

1 From any point o', draw vector o'a' parallel to the direction of absolute acceleration at point A i.e a , to some suitable scale, as shown in Fig 8.2 (b).

2. We know that the acceleration of B with

respect to A i.e aBA has the following two

components:

(i) Radial component of the acceleration

of B with respect to A i.e aBAr , and

acceleration B with respect to A i.e aBAt These two

components are mutually perpendicular

3 Draw vector a'x parallel to the link A B

(because radial component of the acceleration of B

with respect to A will pass through AB), such that

vector a x′ =aBAr =vBA2 /AB

where vBA = Velocity of Bwith respect to A

Note: The value of vBA may be obtained by drawing the

velocity diagram as discussed in the previous chapter.

4 From point x , draw vector xb'

perpendicular to A B or vector a'x (because tangential

component of B with respect to A i.e aBAt ,is

perpendicular to radial component aBAr ) and

through o' draw a line parallel to the path of B to

represent the absolute acceleration of B i.e aB The

vectors xb' and o' b' intersect at b' Now the values

of aB and aBAt may be measured, to the scale

5 By joining the points a' and b' we may determine the total acceleration of B with respect

to A i.e aBA The vector a' b' is known as acceleration image of the link A B.

6 For any other point C on the link, draw triangle a' b' c' similar to triangle ABC Now vector b' c' represents the acceleration of C with respect to B i.e aCB, and vector a' c' represents the acceleration of C with respect to A i.e aCA As discussed above, aCB and aCA will each have twocomponents as follows :

(i) aCB has two components; aCBr andaCBt as shown by triangle b' zc' in Fig 8.2 (b), in which b' z is parallel to BC and zc' is perpendicular to b' z or BC.

(ii) aCA has two components ; aCAr andaCAt as shown by triangle a' yc' in Fig 8.2 (b), in which a' y is parallel to A C and yc' is perpendicular to a' y or A C.

7 The angular acceleration of the link AB is obtained by dividing the tangential components

of the acceleration of B with respect to A (aBAt )to the length of the link Mathematically, angular acceleration of the link A B,

A slider crank mechanism is shown in Fig 8.3 (a) Let the crank OB makes an angle θ with

the inner dead centre (I.D.C) and rotates in a clockwise direction about the fixed point O with

uniform angular velocity ωBO rad/s

∴ Velocity of B with respect to O or velocity of B (because O is a fixed point),

We know that centripetal or radial acceleration of B with respect to O or acceleration of B (because O is a fixed point),

(a) Slider crank mechanism (b) Acceleration diagram.

Fig 8.3. Acceleration in the slider crank mechanism.

The acceleration diagram, as shown in Fig 8.3 (b), may now be drawn as discussed below:

1 Draw vector o' b' parallel to BO and set off equal in magnitude of aBOr = aB, to somesuitable scale

2 From point b', draw vector b'x parallel to B A The vector b'x represents the radial component

of the acceleration of A with respect to B whose magnitude is given by :

2

a =v BA

Since the point B moves with constant angular velocity, therefore there will be no tangential

component of the acceleration

3 From point x, draw vector xa' perpendicular to b'x (or A B) The vector xa' represents the tangential component of the acceleration of A with respect to B i.e aABt

Note: When a point moves along a straight line, it has no centripetal or radial component of the acceleration.

4 Since the point A reciprocates along A O, therefore the acceleration must be parallel to velocity Therefore from o', draw o' a' parallel to A O, intersecting the vector xa' at a'.

Now the acceleration of the piston or the slider A (aA) and aABt may be measured to the scale

5 The vector b' a', which is the sum of the vectors b' x and x a', represents the total acceleration

of A with respect to B i.e aAB The vector b' a' represents the acceleration of the connecting rod A B.

6 The acceleration of any other point on A B such as E may be obtained by dividing the vector

b' a' at e' in the same ratio as E divides A B in Fig 8.3 (a) In other words

Example 8.1 The crank of a slider crank mechanism rotates clockwise at a constant speed

of 300 r.p.m The crank is 150 mm and the connecting rod is 600 mm long Determine : 1 linear velocity and acceleration of the midpoint of the connecting rod, and 2 angular velocity and angular acceleration of the connecting rod, at a crank angle of 45° from inner dead centre position.

Solution Given : NBO = 300 r.p.m or ωBO = 2 π × 300/60 = 31.42 rad/s; OB = 150 mm = 0.15 m ; B A = 600 mm = 0.6 m

We know that linear velocity of B with respect to O or velocity of B,

vBO = vB = ωBO × OB = 31.42 × 0.15 = 4.713 m/s

(Perpendicular to BO)

(a) Space diagram (b) Velocity diagram (c) Acceleration diagram.

Fig 8.4

1 Linear velocity of the midpoint of the connecting rod

First of all draw the space diagram, to some suitable scale; as shown in Fig 8.4 (a) Now the velocity diagram, as shown in Fig 8.4 (b), is drawn as discussed below:

1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of

B with respect to O or simply velocity of B i.e vBO or vB, such that

vector ob = vBO = vB = 4.713 m/s

2 From point b, draw vector ba perpendicular to B A to represent the velocity of A with respect to B i.e vAB , and from point o draw vector oa parallel to the motion of A (which is along A O)

to represent the velocity of A i.e v The vectors ba and oa intersect at a.

Note : This picture is given as additional information and is not a direct example of the current chapter.

Pushing with fluids

Oil pressure on upper side of piston

Ram moves

outwards

By measurement, we find that velocity of A with respect to B,

AB A

3 In order to find the velocity of the midpoint D of the connecting rod A B, divide the vector

ba at d in the same ratio as D divides A B, in the space diagram In other words,

bd / ba = BD/BA

Note: Since D is the midpoint of A B, therefore d is also midpoint of vector ba.

4 Join od Now the vector od represents the velocity of the midpoint D of the connecting rod i.e vD.

By measurement, we find that

vD = vector od = 4.1 m/s Ans.

Acceleration of the midpoint of the connecting rod

We know that the radial component of the acceleration of B with respect to O or the acceleration of B,

2 BO

(4.713)

148.1 m/s0.15

AB

(3.4)

19.3 m/s0.6

r v a BA

Now the acceleration diagram, as shown in Fig 8.4 (c) is drawn as discussed below:

1. Draw vector o' b' parallel to BO, to some suitable scale, to represent the radial component

of the acceleration of B with respect to O or simply acceleration of B i.e aBOr oraB,such that

vectoro b′ ′ =aBOr =aB =148.1 m/s2

Note: Since the crank OB rotates at a constant speed, therefore there will be no tangential component of the acceleration of B with respect to O.

2 The acceleration of A with respect to B has the following two components:

(a) The radial component of the acceleration of A with respect to B i.e aABr ,and

(b) The tangential component of the acceleration of A with respect to B i.e aABt These twocomponents are mutually perpendicular

Therefore from point b', draw vector b' x parallel to A B to represent aABr =19.3 m/s2and

from point x draw vector xa' perpendicular to vector b' x whose magnitude is yet unknown.

3 Now from o', draw vector o' a' parallel to the path of motion of A (which is along A O) to represent the acceleration of A i.e aA The vectors xa' and o' a' intersect at a' Join a' b'.

4 In order to find the acceleration of the midpoint D of the connecting rod A B, divide the vector a' b' at d' in the same ratio as D divides A B In other words

b d b a′ ′ ′ ′ =/ BD BA/

Note: Since D is the midpoint of A B, therefore d' is also midpoint of vector b' a'.

5. Join o' d' The vector o' d' represents the acceleration of midpoint D of the connecting rod

i.e aD

By measurement, we find that

a = vector o' d' = 117 m/s2Ans.

2 Angular velocity of the connecting rod

We know that angular velocity of the connecting rod A B,

3.45.67 rad/s (Anticlockwise about )0.6

v

B BA

Angular acceleration of the connecting rod

From the acceleration diagram, we find that

AB

103171.67 rad/s (Clockwise about )0.6

t

a

B BA

Example 8.2 An engine mechanism is shown in Fig 8.5 The crank CB = 100 mm and the

connecting rod BA = 300 mm with centre of gravity G, 100 mm from B In the position shown, the crankshaft has a speed of 75 rad/s and an angular acceleration of 1200 rad/s2 Find:1 velocity of

G and angular velocity of AB, and 2 acceleration of G and angular acceleration of AB.

Fig 8.5 Solution Given : ωBC = 75 rad/s ; αBC = 1200 rad/s2, CB = 100 mm = 0.1 m; B A = 300 mm

= 0.3 m

We know that velocity of B with respect to C or velocity of B,

vBC =vB = ω ×BC CB =75×0.1=7.5 m/s (Perpendicular to BC)

Since the angular acceleration of the crankshaft, αBC = 1200 rad/s2, therefore tangential

component of the acceleration of B with respect to C,

aBCt = αBC ×CB =1200×0.1=120 m/s2

Note: When the angular acceleration is not given, then there will be no tangential component of the acceleration.

1 Velocity of G and angular velocity of AB

First of all, draw the space diagram, to some suitable scale, as shown in Fig 8.6 (a) Now the velocity diagram, as shown in Fig 8.6 (b), is drawn as discussed below:

1 Draw vector cb perpendicular to CB, to some suitable scale, to represent the velocity of

B with respect to C or velocity of B (i.e vBC or vB), such that

vector cb= v =v =7.5 m/s

2 From point b, draw vector ba perpendicular to B A to represent the velocity of A with respect to B i.e vAB , and from point c, draw vector ca parallel to the path of motion of A (which is

along A C) to represent the velocity of A i.e vA.The vectors ba and ca intersect at a.

3. Since the point G lies on A B, therefore divide vector ab at g in the same ratio as G divides

A B in the space diagram In other words,

ag ab/ = AG AB/

The vector cg represents the velocity of G.

By measurement, we find that velocity of G,

v = vector cg = 6.8 m/s Ans.

From velocity diagram, we find that velocity of A with respect to B,

vAB = vector ba = 4 m/s

We know that angular velocity of A B,

AB AB

413.3 rad/s (Clockwise)0.3

v BA

(a) Space diagram (b) Velocity diagram.

Fig 8.6

2 Acceleration of G and angular acceleration of AB

We know that radial component of the acceleration of B with

respect to C,

*

2 BC

BC

(7.5)

562.5 m/s0.1

r v a CB

AB

453.3 m/s0.3

r v a BA

Now the acceleration diagram, as shown in Fig 8.6 (c), is drawn

as discussed below:

1 Draw vector c' b'' parallel to CB, to some suitable scale, to

represent the radial component of the acceleration of B with respect to C,

i.e aBCr ,such that

vector c b′ ′′ =aBCr =562.5 m/s2

2 From point b'', draw vector b'' b' perpendicular to vector c' b'' or CB to represent the tangential component of the acceleration of B with respect to C i.e aBCt , such that

2 BC

3 Join c' b' The vector c' b' represents the total acceleration of B with respect to C i.e aBC.

4 From point b', draw vector b' x parallel to B A to represent radial component of the acceleration of A with respect to B i.e ABr

a such that vectorb x′ =aABr =53.3 m/s2

5 From point x, draw vector xa' perpendicular to vector b'x or B A to represent tangential component of the acceleration of A with respect to B i.e aABt ,whose magnitude is not yet known

6 Now draw vector c' a' parallel to the path of motion of A (which is along A C) to represent the acceleration of A i.e aA.The vectors xa' and c'a' intersect at a' Join b' a' The vector b' a'

represents the acceleration of A with respect to B i.e aAB

(c) Acceleration diagram.

Fig 8.6

* When angular acceleration of the crank is not given, then there is no aBCt In that case, aBCr =aBC=aB, as discussed in the previous example.

7 In order to find the acceleratio of G, divide vector a' b' in g' in the same ratio as G divides

B A in Fig 8.6 (a) Join c' g' The vector c' g' represents the acceleration of G.

By measurement, we find that acceleration of G,

t

a BA

Example 8.3 In the mechanism shown in Fig 8.7, the slider C is

moving to the right with a velocity of 1 m/s and an acceleration of 2.5 m/s2.

The dimensions of various links are AB = 3 m inclined at 45° with the

vertical and BC = 1.5 m inclined at 45° with the horizontal Determine: 1 the

magnitude of vertical and horizontal component of the acceleration of the

point B, and 2 the angular acceleration of the links AB and BC.

Solution Given : vC = 1 m/s ; aC = 2.5 m/s2; A B = 3 m ; BC = 1.5 m

First of all, draw the space diagram, as shown in Fig 8.8 (a), to some

suitable scale Now the velocity diagram, as shown in Fig 8.8 (b), is drawn as

discussed below:

1. Since the points A and D are fixed points, therefore they lie at one place in the velocity diagram Draw vector dc parallel to DC, to some suitable scale, which represents the velocity of slider C with respect to D or simply velocity of C, such that

vector dc = vCD = vC = 1 m/s

2 Since point B has two motions, one with respect to A and the other with respect to C, therefore from point a, draw vector ab perpendicular to A B to represent the velocity of B with respect to A , i.e vBA and from point c draw vector cb perpendicular to CB to represent the velocity

of B with respect to C i.e vBC .The vectors ab and cb intersect at b.

(a) Space diagram (b) Velocity diagram (c) Acceleration diagram.

We know that radial component of acceleration of B with respect to C,

2 BC

BC

(0.72)

0.346 m/s1.5

r v a CB

BA

(0.72)

0.173 m/s3

r v a AB

Now the acceleration diagram, as shown in Fig 8.8 (c), is drawn as discussed below:

1 *Since the points A and D are fixed points, therefore they lie at one place in the acceleration diagram Draw vector d' c' parallel to DC, to some suitable scale, to represent the acceleration of C with respect to D or simply acceleration of C i.e aCD or aC such that

4. The acceleration of B with respect to A will also have two components, i.e one radial component of B with respect to A (a r

BA) and other tangential component of B with respect to A (a t

5 From point y, draw vector yb' perpendicular to vector a'y or AB to represent aBAt The

vector yb' intersect the vector xb' at b' Join a' b' and c' b' The vector a' b' represents the acceleration

of point B (aB) and the vector c' b' represents the acceleration of B with respect to C.

1 Magnitude of vertical and horizontal component of the acceleration of the point B

Draw b' b'' perpendicular to a' c' The vector b' b'' is the vertical component of the acceleration

of the point B and a' b'' is the horizontal component of the acceleration of the point B By measurement, vector b' b'' = 1.13 m/s2 and vector a' b'' = 0.9 m/s2 Ans.

2 Angular acceleration of AB and BC

By measurement from acceleration diagram, we find that tangential component of acceleration

of the point B with respect to A ,

We know that angular acceleration of A B,

AB BA 2

1.410.47 rad/s3

t

a AB

Example 8.4 PQRS is a four bar chain with link PS fixed The lengths of the links are PQ

= 62.5 mm ; QR = 175 mm ; RS = 112.5 mm ; and PS = 200 mm The crank PQ rotates at 10 rad/s clockwise Draw the velocity and acceleration diagram when angle QPS = 60° and Q and R lie on the same side of PS Find the angular velocity and angular acceleration of links QR and RS.

Solution Given : ωQP = 10 rad/s; PQ = 62.5 mm = 0.0625 m ; QR = 175 mm = 0.175 m ;

R S = 112.5 mm = 0.1125 m ; PS = 200 mm = 0.2 m

We know that velocity of Q with respect to P or velocity of Q,

vQP = vQ = ωQP × PQ = 10 × 0.0625 = 0.625 m/s

(Perpendicular to PQ)

Angular velocity of links QR and RS

First of all, draw the space diagram of a four bar chain, to some suitable scale, as shown in

Fig 8.9 (a) Now the velocity diagram as shown in Fig 8.9 (b), is drawn as discussed below:

(a) Space diagram (b) Velocity diagram (c) Acceleration diagram.

Fig 8.9

1. Since P and S are fixed points, therefore these points lie at one place in velocity diagram Draw vector pq perpendicular to PQ, to some suitable scale, to represent the velocity of Q with respect to P or velocity of Q i.e vQP or vQ such that

vector pq = vQP = vQ = 0.625 m/s

2 From point q, draw vector qr perpendicular to QR to represent the velocity of R with respect to Q (i.e vRQ) and from point s, draw vector sr perpendicular to S R to represent the velocity

of R with respect to S or velocity of R (i.e vRS or vR) The vectors qr and sr intersect at r By

measurement, we find that

vRQ = vector qr = 0.333 m/s, and vRS = vR = vector sr = 0.426 m/s

We know that angular velocity of link QR,

QR RQ

0.3331.9 rad/s (Anticlockwise)0.175

v RQ

and angular velocity of link R S,

RS RS 0.426 3.78 rad/s (Clockwise)

0.1125

v SR

Angular acceleration of links QR and RS

Since the angular acceleration of the crank PQ is not given, therefore there will be no tangential component of the acceleration of Q with respect to P.

We know that radial component of the acceleration of Q with respect to P (or the acceleration

r v a QR

(0.426)

1.613 m/s0.1125

SR

The acceleration diagram, as shown in Fig 8.9 (c) is drawn as follows :

1 Since P and S are fixed points, therefore these points lie at one place in the acceleration diagram Draw vector p'q' parallel to PQ, to some suitable scale, to represent the radial component

of acceleration of Q with respect to P or acceleration of Q i.e aQPr ora such thatQ

vector p q′ ′ =aQPr =aQ =6.25 m/s2

2 From point q', draw vector q' x parallel to QR to represent the radial component of acceleration of R with respect to Q i.e aRQr such that

vector q x′ =aRQr =0.634 m/s2

3 From point x, draw vector xr' perpendicular to QR to represent the tangential component

of acceleration of R with respect to Q i.e aRQt whose magnitude is not yet known

4 Now from point s', draw vector s'y parallel to S R to represent the radial component of the acceleration of R with respect to S i.e aRSr such that

vector s y′ =aRSr =1.613 m/s2

5 From point y, draw vector yr' perpendicular to S R to represent the tangential component

of acceleration of R with respect to S i.e aRSt

6 The vectors xr' and yr' intersect at r' Join p'r and q' r' By measurement, we find that

aRQt =vectorxr′=4.1 m/s and2 aRSt =vectoryr′=5.3 m/s2

We know that angular acceleration of link QR,

4.123.43 rad/s (Anticlockwise)

t

a SR

Example 8.5 The dimensions and

configuration of the four bar mechanism, shown in

Fig 8.10, are as follows :

P1A = 300 mm; P2B = 360 mm; AB = 360

mm, and P1P2 = 600 mm.

The angle AP1P2 = 60° The crank P1A has

an angular velocity of 10 rad/s and an angular

acceleration of 30 rad/s2, both clockwise.

Determine the angular velocities and angular

accelerations of P2B, and AB and the velocity and

acceleration of the joint B.

Solution Given : ωAP1 = 10 rad/s ; αAP1 = 30 rad/s2; P1A = 300 mm = 0.3 m ; P2B = A B =

360 mm = 0.36 m

We know that the velocity of A with respect to P1 or velocity of A ,

vAP1 = vA = ωAP1 × P1A = 10 × 0.3 = 3 m/s

Velocity of B and angular velocitites of P2B and AB

First of all, draw the space diagram, to some suitable scale, as shown in Fig 8.11 (a) Now the velocity diagram, as shown in Fig 8.11 (b), is drawn as discussed below:

1 Since P1 and P2 are fixed points, therefore these points lie at one place in velocity diagram

Draw vector p1 a perpendicular to P1A , to some suitable scale, to represent the velocity of A with

respect to P1 or velocity of A i.e vAP1 or vA, such that

vector p1a = vA P1 = vA = 3 m/s

2. From point a, draw vector ab perpendicular to AB to represent velocity of B with respect

to A (i.e vBA) and from point p2 draw vector p2b perpendicular to P2B to represent the velocity of B

with respect to P2 or velocity of B i.e vBP2 or vB The vectors ab and p2b intersect at b.

By measurement, we find that

v AB

Acceleration of B and angular acceleration of P2B and AB

We know that tangential component of the acceleration of A with respect to P1,

Radial component of the acceleration of B with respect to A

2 BA

BA

(2.05)

11.67 m/s0.36

r v a AB

(2.2)

13.44 m/s0.36

1 Since P1 and P2 are fixed points, therefore these points

will lie at one place, in the acceleration diagram Draw vector

p1' x parallel to P1A , to some suitable scale, to represent the

radial component of the acceleration of A with respect to P1,

such that

vector p x′ =a r =30 m/s

2 From point x, draw vector xa' perpendicular to P1A to

represent the tangential component of the acceleration of A with

respect to P1, such that

vector xa′ =aA P1t =9 m/s2

3 Join p1' a' The vector p1' a' represents the acceleration

of A By measurement, we find that the acceleration of A ,

aA = aAP1 = 31.6 m/s2

4 From point a', draw vector a' y parallel to A B to represent the radial component of the acceleration of B with respect to A , such that

vectora y′ =aBAr =11.67 m/s2

5. From point y, draw vector yb' perpendicular to A B to represent the tangential component

of the acceleration of B with respect to A (i.e aBAt ) whose magnitude is yet unknown

6 Now from point p2′,draw vector p z2′ parallel to P

2B to represent the radial component

of the acceleration B with respect to P2, such that

7 From point z, draw vector zb' perpendicular to P2B to represent the tangential component

of the acceleration of B with respect to P2 i.e aBP2t

8 The vectors yb' and zb' intersect at b' Now the vector p2' b' represents the acceleration of

B with respect to P2 or the acceleration of B i.e aBP2 or aB By measurement, we find that

aBP2 = aB = vector p2' b' = 29.6 m/s2Ans.

Also vector yb′=aBAt =13.6 m/s , and vector2 zb′= aBP 2t =26.6 m/s2

We know that angular acceleration of P2B,

P2B BP2 2

2

26.673.8 rad/s (Anticlockwise)0.36

t

a AB

Example 8.6 In the mechanism, as shown in Fig 8.12, the crank OA rotates at 20 r.p.m.

anticlockwise and gives motion to the sliding blocks B and D The dimensions of the various links are OA = 300 mm; AB = 1200 mm; BC = 450 mm and CD = 450 mm.

Fig 8.12

For the given configuration, determine : 1 velocities of sliding at B and D, 2 angular

velocity of CD, 3 linear acceleration of D, and 4 angular acceleration of CD.

Solution Given : NAO = 20 r.p.m or ωAO = 2 π × 20/60 = 2.1 rad/s ; OA = 300 mm = 0.3 m ;

A B = 1200 mm = 1.2 m ; BC = CD = 450 mm = 0.45 m

Bicycle is a common example where simple mechanisms are used.

Note : This picture is given as additional information and is not a direct example of the current chapter.

We know that linear velocity of A with respect to O or velocity of A ,

vAO = vA = ωAO × O A = 2.1 × 0.3 = 0.63 m/s (Perpendicular to O A)

1 Velocities of sliding at B and D

First of all, draw the space diagram, to some suitable scale, as shown in Fig 8.13 (a) Now the velocity diagram, as shown in Fig 8.13 (b), is drawn as discussed below:

(a) Space diagram.

(b) Velocity diagram (c) Acceleration diagram.

Fig 8.13

1. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of

A with respect to O (or simply velocity of A ), such that

By measurement, we find that velocity of sliding at B,

∴ Angular velocity of CD,

CD DC

0.370.82 rad/s (Anticlockwise)

0.45

v CD

BA

(0.54)

0.243 m/s1.2

r v a AB

DC

(0.37)

0.304 m/s0.45

r v a CD

Now the acceleration diagram, as shown in Fig 8.13 (c), is drawn as discussed below:

1 Draw vector o' a' parallel to O A, to some suitable scale, to represent the radial component

of the acceleration of A with respect to O or simply the acceleration of A , such that

vectoro a′ ′ =aAOr = aA =1.323 m/s2

2 From point a', draw vector a' x parallel to A B to represent the radial component of the acceleration of B with respect to A , such that

vectora x′ =aBAr =0.243 m/s2

3 From point x, draw vector xb' perpendicular to A B to represent the tangential component

of the acceleration of B with respect to A (i.e aBAt ) whose magnitude is not yet known

4. From point o', draw vector o' b' parallel to the path of motion of B (which is along BO) to represent the acceleration of B (aB) The vectors xb' and o' b' intersect at b' Join a' b' The vector

a' b' represents the acceleration of B with respect to A

5 Divide vector a' b' at c' in the same ratio as C divides A B in the space diagram In other

∴ Angular acceleration of CD,

CD

1.282.84 rad/s (Clockwise)0.45

t

a CD

Example 8.7 Find out the acceleration of the slider D and

the angular acceleration of link CD for the engine mechanism shown

in Fig 8.14.

The crank OA rotates uniformly at 180 r.p.m in clockwise

direction The various lengths are: OA = 150 mm ; AB = 450 mm;

First of all draw the space diagram, to some suitable scale,

as shown in Fig 8.15 (a) Now the velocity diagram, as shown in Fig 8.15 (b), is drawn as discussed

vector oa = vAO = vA = 2.83 m/s

2 Since the point B moves with respect to A and also with respect to P, therefore draw vector ab perpendicular to A B to represent the velocity of B with respect to A i.e vBA ,and from point

p draw vector pb perpendicular to PB to represent the velocity of B with respect to P or velocity of

B (i.e vBP or vB) The vectors ab and pb intersect at b.

3 Since the point C lies on PB produced, therefore divide vector pb at c in the same ratio as

C divides PB in the space diagram In other words, pb/pc = PB/PC.

All dimensions in mm.

Fig 8.14

4 From point c, draw vector cd perpendicular to CD to represent the velocity of D with respect to C and from point o draw vector od parallel to the path of motion of the slider D (which is vertical), to represent the velocity of D, i.e vD.

By measurement, we find that velocity of the slider D,

vD = vector od = 2.36 m/s Velocity of D with respect to C,

vDC = vector cd = 1.2 m/s Velocity of B with respect to A ,

vBA = vector ab = 1.8 m/s and velocity of B with respect to P, vBP = vector pb = 1.5 m/s

Acceleration of the slider D

We know that radial component of the acceleration of A with respect to O or acceleration

BA

(1.8)

7.2 m/s0.45

r v a AB

BP

(1.5)

9.4 m/s0.24

r v a PB

DC

(1.2)

2.2 m/s0.66

r v a CD

Now the acceleration diagram, as shown in Fig 8.15 (c), is drawn as discussed below:

1 Since O and P are fixed points, therefore these points lie at one place in the acceleration diagram Draw vector o' a' parallel to O A, to some suitable scale, to represent the radial component

of the acceleration of A with respect to O or the acceleration of A (i.e aAOr or aA), such that

4 Now from point p', draw vector p' y parallel to PB to represent the radial component of the acceleration of B with respect to P (i.e aBPr ), such that

2 BP

vector p y′ =a r =9.4 m/s

5. From point y, draw vector yb' perpendicular to vector b'y or PB to represent the tangential component of the acceleration of B, i.e aBPt The vectors xb' and yb' intersect at b' Join p' b' The vector p' b' represents the acceleration of B, i.e a

6 Since the point C lies on PB produced, therefore divide vector p'b' at c' in the same ratio

as C divides PB in the space diagram In other words, p'b'/p'c' = PB/PC

7 From point c', draw vector c'z parallel to CD to represent the radial component of the acceleration of D with respect to C i.e aDCr ,such that

By measurement, we find that acceleration of D,

aD = vector o'd' = 69.6 m/s2 Ans.

t

a CD

Example 8.8 In the toggle mechanism shown in Fig 8.16, the slider D is constrained to

move on a horizontal path The crank OA is rotating in the counter-clockwise direction at a speed

Fig 8.16

of 180 r.p.m increasing at the rate of 50 rad/s2 The dimensions of the various links are as follows:

OA = 180 mm ; CB = 240 mm ; AB = 360 mm ; and BD = 540 mm.

For the given configuration, find 1 Velocity of slider D and angular velocity of BD, and

2 Acceleration of slider D and angular acceleration of BD.

Solution Given : NAO = 180 r.p.m or ωAO = 2 π × 180/60 = 18.85 rad/s ; O A = 180 mm

= 0.18 m ; CB = 240 mm = 0.24 m ; A B = 360 mm = 0.36 m ; BD = 540 mm = 0.54 m

We know that velocity of A with respect to O or velocity of A ,

vAO = vA = ωAO × O A = 18.85 × 0.18 = 3.4 m/s

(Perpendicular to O A)

1 Velocity of slider D and angular velocity of BD

First of all, draw the space diagram to some suitable scale, as shown in Fig 8.17 (a) Now the velocity diagram, as shown in Fig 8.17 (b), is drawn as discussed below:

1 Since O and C are fixed points, therefore these points lie at one place in the velocity diagram Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of A with respect to O or velocity of A i.e vAO or vA, such that

3 From point b, draw vector bd perpendicular to BD to represent the velocity of D with respect to B i.e vDB, and from point c draw vector cd parallel to CD (i.e., in the direction of motion

of the slider D) to represent the velocity of D i.e vD

By measurement, we find that velocity of B with respect to A ,

vBA = vector ab = 0.9 m/s Velocity of B with respect to C,

vBC = vector cb = 2.8 m/s Velocity of D with respect to B,

vDB = vector bd = 2.4 m/s and velocity of slider D, vD = vector cd = 2.05 m/s Ans.

2 Acceleration of slider D and angular acceleration of BD

Since the angular acceleration of OA increases at the rate of 50 rad/s2, i.e αAO = 50 rad/s2,

therefore

Tangential component of the acceleration of A with respect to O,

a t = α ×OA=50×0.18=9 m/s2

Radial component of the acceleration of A with respect to O,

2 AO

AO

(3.4)

63.9 m/s0.18

r v a OA

BA

(0.9)

2.25 m/s0.36

r v a AB

BC

(2.8)

32.5 m/s0.24

r v a CB

DB

(2.4)

10.8 m/s0.54

r v a BD

Now the acceleration diagram, as shown in Fig 8.17 (c), is drawn as discussed below:

1 Since O and C are fixed points, therefore

these points lie at one place in the acceleration

diagram Draw vector o'x parallel to O A, to some

suitable scale, to represent the radial component

of the acceleration of A with respect to O i.e.

AOr ,

a such that

vectoro x′ =aAOr =63.9 m/s2

2. From point x , draw vector xa'

perpendicular to vector o'x or O A to represent the

tangential component of the acceleration of A with

respect to O i.e aAOt ,such that

vector x a′ =aAOt =9 m/s2

3 Join o'a' The vector o'a' represents the

total acceleration of A with respect to O or

5 From point y, draw vector yb' perpendicular to vector a'y or A B to represent the tangential component of the acceleration of B with respect to A i.e aBAt whose magnitude is yet unknown

6. Now from point c', draw vector c'z parallel to CB to represent the radial component of the acceleration of B with respect to C i.e aBCr ,such that

vectorc z′ =aBCr =32.5 m/s2

7 From point z, draw vector zb' perpendicular to vector c'z or CB to represent the tangential component of the acceleration of B with respect to C i.e aBCt The vectors yb' and zb' intersect at b'.

Join c' b' The vector c' b' represents the acceleration of B with respect to C i.e aBC

8. Now from point b', draw vector b's parallel to BD to represent the radial component of the acceleration of D with respect to B i.e aDBr ,such that

9 From point s, draw vector sd' perpendicular to vector b's or BD to represent the tangential component of the acceleration of D with respect to B i.e aDBt whose magnitude is yet unknown.

10. From point c', draw vector c'd' parallel to the path of motion of D (which is along CD)

to represent the acceleration of D i.e aD The vectors sd' and c'd' intersect at d'.

By measurement, we find that acceleration of slider D,

Solution Given : ωAO1 = 100 rad/s ; O1A = 100 mm = 0.1 m

We know that linear velocity of A with respect to O1, or velocity of A ,

vAO1 = vA = ωAO1 × O1A = 100 × 0.1 = 10 m/s (Perpendicular to O1A )

1 Linear velocity of the point E on bell crank lever

First of all draw the space diagram, as shown in Fig 8.19 (a), to some suitable scale Now the velocity diagram, as shown in Fig 8.19 (b), is drawn as discussed below:

1. Since O1, O2 and O3 are fixed points, therefore these points are marked as one point in the

velocity diagram From point o1, draw vector o1a perpendicular to O1A to some suitable scale, to

represent the velocity of A with respect to O or velocity of A , such that

vector o a = v = v = 10 m/s

2 From point a, draw vector ac perpendicular to A C to represent the velocity of C with respect to A (i.e vCA) and from point o3 draw vector o3c perpendicular to O3C to represent the

velocity of C with respect to O3 or simply velocity of C (i.e vC) The vectors ac and o3c intersect at

point c.

(b) Velocity diagram (c) Acceleration diagram.

Fig 8.19

3 Since B lies on A C, therefore divide

vector ac at b in the same ratio as B divides A C in

the space diagram In other words, ab/ac = AB/AC

4 From point b, draw vector bd

perpendicular to BD to represent the velocity of D

with respect to B (i.e vDB), and from point o2 draw

vector o2d perpendicular to O2D to represent the

velocity of D with respect to O2 or simply velocity

of D (i.e vD) The vectors bd and o2d intersect at d.

5. From point o2, draw vector o2e

perpendicular to vector o2d in such a way that

o2e/o2d = O2E/O2D

By measurement, we find that velocity of

point C with respect to A ,

vCA = vector ac = 7 m/s Velocity of point C with respect to O3,

vCO3 = vC = vector o3c = 10 m/s

Velocity of point D with respect to B,

v = vector bd = 10.2 m/s

(a) Space diagram.

Warping machine uses many mechanisms.

Velocity of point D with respect to O2,

vDO2 = vD = vector o2d = 2.8 m/s

and velocity of the point E on the bell crank lever,

vE = vEO2 = vector o2e = 5.8 m/s Ans.

2 Acceleration of the points E and B

Radial component of the acceleration of A with respect to O1 (or acceleration of A ),

2 AO1

1

10

1000 m/s0.1

CA

7

70 m/s0.7

r v a AC

DB

(10.2)

693.6 m/s0.15

r v a BD

a

O E

Now the acceleration diagram, as shown in Fig 8.19 (c), is drawn as discussed below:

1 Since O1, O2 and O3 are fixed points, therefore these points are marked as one point in the

acceleration diagram Draw vector o1' a' parallel to O1A , to some suitable scale, to represent the

radial component of the acceleration of A with respect to O1 (or simply acceleration of A ), such that

2

vectoro a′ ′ =a r =a =1000 m/s

2. From point a', draw a'x parallel to A C to represent the radial component of the acceleration

of C with respect to A (i.e aCAr ), such that

2

vector a x′ =a r =70 /s

3. From point x, draw vector xc' perpendicular to A C to represent the tangential component

of the acceleration of C with respect to A (i.e aCAt ), the magnitude of which is yet unknown

4 From point o3', draw vector o3' y parallel to O3C to represent the radial component of the

acceleration of C with respect to O3 (i.e aCO3r ), such that

2 CO3

vector o y′ =a r =500 /s

5. From point y, draw vector yc' perpendicular to O3C to represent the tangential component

of the acceleration of C with respect to O (i.e a t ) The vectors xc' and yc' intersect at c'.

6. Join a' c' The vector a' c' represents the acceleration of C with respect to A (i.e aCA).

7. Since B lies on A C, therefore divide vector a'c' at b' in the same ratio as B divides AC in the space diagram In other words, a'b'/a'c' = AB/AC Join b' o2' which represents the acceleration of

point B with respect to O2 or simply acceleration of B By measurement, we find that

Acceleration of point B = vector o2' b' = 440 m/s2Ans.

8 Now from point b', draw vector b' z parallel to BD to represent the radial component of the acceleration of D with respect to B (i.e aDBr ), such that

2 DB

vectorb z′ =a r =693.6 m/s

9. From point z, draw vector zd' perpendicular to BD to represent the tangential component

of the acceleration of D with respect to B (i.e aDBt ), whose magnitude is yet unknown

10 From point o2' , draw vector o2' z1 parallel to O2D to represent the radial component of

the acceleration of D with respect to O2 (i.e aDO 2r ), such that

2

vector o z′ =a r =39.2 m/s

11 From point z1, draw vector z1d' perpendicular to O2D to represent the tangential component

of the acceleration of D with respect to O2 (i.e aDO 2t ) The vectors zd' and z1d' intersect at d'.

12. Join o2' d' The vector o2'd' represents the acceleration of D with respect to O2 or simply

acceleration of D (i.e aDO2 or aD)

13 From point o2', draw vector o2' e' perpendicular to o2' d' in such a way that

o e o d′ ′ ′ ′ =O E O D

Note: The point e' may also be obtained drawing aEOr 2and aEOt 2as shown in Fig 8.19 (c).

By measurement, we find that acceleration of point E,

aE = aEO2 = vector o' 2 e' = 1200 m/s2Ans.

3 Angular acceleration of the bell crank lever

By measurement, we find that the tangential component of the acceleration of D with respect

Example 8.10 A pump is driven from an engine

crank-shaft by the mechanism as shown in Fig 8.20 The

pump piston shown at F is 250 mm in diameter and the

crank speed is 100 r.p.m The dimensions of various links

are as follows:

OA = 150 mm ; AB = 600 mm ; BC = 350 mm ;

CD = 150 mm; and DE = 500 mm.

Determine for the position shown : 1 The velocity of

the cross-head E, 2 The rubbing velocity of the pins A

and B which are 50 mm diameter 3 The torque required

at the crank shaft to overcome a presure of 0.35 N/mm 2 ,

and 4 The acceleration of the cross-head E.

All dimensions in mm.

Fig 8.20

Solution Given : NAO = 100 r.p.m or ωAO = 2 π × 100/60 = 10.47 rad/s; OA = 150 mm = 0.15 m ;

A B = 600 mm = 0.6 m ; BC = 350 mm = 0.35 m ; CD = 150 mm = 0.15 m ; DE = 500 mm = 0.5 m

We know that velocity of A with respect to O or velocity of A ,

vAO = vA = ωAO × O A = 10.47 × 0.15 = 1.57 m/s (Perpendicular to O A)

1 Velocity of the cross-head E

First of all, draw the space diagram, to some suitable scale, as shown in Fig 8.21 (a) Now the velocity diagram, as shown in Fig 8.21 (b), is drawn as discussed below:

(a) Space diagram (b) Velocity diagram (c) Acceleration diagram.

Fig 8.21

1 Since O and C are fixed points, therefore these points are marked as one point in the velocity diagram Now draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of A with respect ot O or the velocity of A , such that

vector oa = vAO = vA = 1.57 m/s

2 From point a, draw vector ab perpendicular to A B to represent the velocity of B with respect to A (i.e vBA), and from point c draw vector cb perpendicular to CB to represent the velocity

of B with respect to C (i.e vBC) The vectors ab and cb intersect at b.

By measurement, we find that

vBA = vector ab = 1.65 m/s

and vBC = vB = vector cb = 0.93 m/s

3 From point c, draw vector cd perpendicular to CD or vector cb to represent the velocity of

D with respect to C or velocity of D, such that

By measurement, we find that velocity of E with respect to D,

v = vector de = 0.18 m/s

and velocity of the cross-head E,

vEO = vE = vector oe = 0.36 m/s Ans.

2 Rubbing velocity of the pins at A and B

We know that angular velocity of A with respect to O,

ωAO = 10.47 rad/s (Anticlockwise)

Angular velocity of B with respect to A ,

BA BA

1.652.75 rad/s0.6

v AB

and angular velocity of B with respect to C,

BC BC

0.932.66 rad/s0.35

v CB

3 Torque required at the crankshaft

Given: Pressure to overcome by the crankshaft,

pF = 0.35 N/mm2Diameter of the pump piston

Let FA = Force required at the crankshaft at A

Assuming transmission efficiency as 100 per cent,

Work done at A = Work done at F

Acceleration of the crosshead E

We know that the radial component of the acceleration of A with respect to O or the acceleration of A ,

2 AO

(1.57)

16.43 m/s0.15

OA

Radial component of the acceleration of B with respect to A ,

2 BA

BA

(1.65)

4.54 m / s0.6

r v a AB

BC

(0.93)

2.47 m/s0.35

r v a CB

ED

(0.18)

0.065 m/s0.5

r v a DE

Now the acceleration diagram, as shown in Fig 8.21 (c), is drawn as discussed below:

1. Since O and C are fixed points, therefore these points are marked as one point in the acceleration diagram Draw vector o'a' parallel to O A, to some suitable scale, to represent the radial component of the acceleration of A with respect to O or the acceleration of A , such that

vectoro a′ ′ =aAOr =aA =16.43 m/s2

2 From point a', draw vector a'x parallel to A B to represent the radial component of the acceleration of B with respect to A (i.e aBAr ), such that

vector a x′ =aBAr =4.54 m/s2

3. From point x, draw vector xb' perpendicular to A B to represent the tangential component

of the acceleration of B with respect to A (i.e aBAt ) whose magnitude is yet unknown

4 Now from point c', draw vector c' y parallel to CB to represent the radial component of the acceleration of B with respect to C (i.e aBCr ), such that

2 BC

vectorc y′ = a r = 2.47 m/s

5 From point y, draw vector yb' perpendicular to CB to represent the tangential component

of the acceleration of B with respect to C (i.e aBCt ) The vectors yb' and xb' intersect at b' Join c'b' and a'b' The vector c'b' represents the acceleration of B with respect to C (i.e aBC) or the acceleration

of B (i.e aB) and vector a'b' represents the acceleration of B with respect to A (i.e aBA)

By measurement, we find that

aBC = aB = vector c'b' = 9.2 m/s2

and aBA = vector a'b' = 9 m/s2

6 From point c', draw vector c'd' perpendicular to CD or vector c'b' to represent the acceleration of D with respect to C or the acceleration of D (i.e aDC or aD), such that

Note: Since the magnitude of a r

ED is very small, therefore the points d' and z coincide.

8 From point z, draw vector ze' perpendicular to DE to represent the tangential component

of the acceleration of E with respect to D (i.e aEDt ) whose magnitude is yet unknown

9. From point o', draw vector o'e' parallel to the path of motion of E (which is vertical) to represent the acceleration of E The vectors ze' and o'e' intersect at e'.