VTU EDUSAT PROGRAMME -17

DYNAMICS OF MACHINES (10 ME 54)

ANALYSIS OF CAMS

1.0 INTRODUCTION

Based on the flank, cams are classified into two groups,

1. Circular arc cam

2. Straight edged cam or Tangent cam

Circular arc Cam:

When the flanks of the cam connecting the nose and base circles are of convex circular arc,

such cams are referred as circular arc cams.

Fig.1

Tangent cam:

When the flanks between the nose and base circles are of straight and tangential to both the

circles, then, the cams are called tangent cams.

Fig.2

These are usually symmetric about the centre line of the cam. Generally, the following

combinations of cam and follower are used.

(a) Circular arc cam with flat faced follower

(b) Tangent cam with reciprocating roller follower

1.1 CIRCULAR ARC CAM WITH FLAT FACED FOLLOWER

The Fig. 3 represents various main dimensions of circular arc cam.

Fig.3

(a) Expression for determining the displacement, velocity and acceleration of

the follower when flat face of the follower has contact on the circular flank

Fig.4

Let

r1=OB=Least base circle radius

r2=Nose circle radius

R=QD= Flank circle radius

d=Distance between the centres of cam and nose circles

α= Angle of ascent

φ=Angle of contact on circular flank

Displacement:

X = BC = OC – OB = DE - r1

= (QD-QE) – r1

= (R - OQ cos θ) – r1

= R-(R - r1) cosθ – r1

Velocity:

= x

= (R-r1) (sin θ) ω

From this equation, it is evident that, at the beginning of the ascent, the velocity is zero

(when θ=0 ) and it increases with θ. It will be maximum when the follower is just shift from circular

flank to circular nose.

Vmax = ω (R-r1) sinφ

Acceleration:

a =

a= ω2 ( R- r1) cos θ

It is obvious from the above equation that, at the beginning of the ascent when θ=0,

acceleration is maximum and it goes on decreasing and is maximum when θ=φ

(b)Expression for determining the displacement, velocity and acceleration of the

follower when flat face of the follower has contact on the nose

Fig.5

Let

r1=OB=Least base circle radius

r2=Nose circle radius

R=QD= Flank circle radius

d=Distance between centers of cam and nose circles

α= Angle of ascent

φ=Angle of contact on circular flank

Displacement:

x = BC = OC – OB = DE - r1

= (DP+PE) – r1

= r2+OP cos (α - θ) –r1

Velocity:

= x

The velocity is minimum when α = θ or (α - θ)=0. This happens when follower is at the apex

of circular nose and it is maximum when (α - θ) is maximum and it is so when the contact changes

from circular flank to circular nose ie., (α - θ) =φ

Acceleration:

a =

a= - ωd cos (α - θ) ω

a= - ω2d cos (α - θ)

Negative sign indicates retardation. It is maximum when (α - θ) = 0 i.e., when the follower is

at the apex of the nose and minimum (α - θ) is maximum i.e., when the follower changes contact

from circular flank to circular nose.

It may be noted that, as the contact between cam and follower passes through point D, the

acceleration of the follower suddenly changes from ω2 ( R- r1) cosφ to - ω

2d cos φ i.e., a sudden

change from positive acceleration to negative acceleration (retardation).

Note: Cosine rule

Fig. 6

Consider the triangle POQ in Fig. 7,

OQ = R-r1, QP = R-r2, OP = d

and apply cosine rule

Fig. 7

QP2= QO

2+PO

2-2QOxPOxCos POQ

(R-r2)2 = (R-r1)

2+d

2-2(R-r1)(d) cos (180-α)

= (R-r1)2+d

2+2(R-r1) d cosα

R2-2Rr2+r2

2 = R

2+r1

2-2Rr1+d

2+2Rdcosα -2r1dcosα

2Rr1-2Rr2-2Rdcosα = r12-r2

2+d

2-2r1dcosα

Acceleration of the follower at the beginning of lift/flank, θ =0

Acceleration of the follower at the end of contact with flank,θ = φ

Acceleration of the follower at the beginning of nose, (α - θ) =φ

Acceleration of the follower at the apex of nose, (α - θ) = 0

Example 1

A symmetrical circular arc cam operating a flat faced follower has the following particulars.

Least radius of cam = 30 mm; lift = 20 mm; Angle of lift = 75o; Nose radius = 5 mm; speed = 600

rpm

Find, (i) The principal dimensions of cam

(ii) The acceleration of the follower at the beginning of lift, at the end of contact with the

circular flank, at the beginning of contact with nose and at the apex of nose.

Solution,

r 1 = 15 mm, r 2 = 5 mm , 2α = 150o,

α = 75o,

N = 600 rpm

Fig. 8

We have,

Total lift + r1 = PO + r2

20 + 30 = d – 5

d = 45 mm

we have,

R = 82.42 mm

PQ = R-r2 = 82.41-5 = 77.41 mm

QO = R-r1 = 82.41 – 30 = 52.41 mm

To determine angle φ , consider triangle PQO,

xPO

= Sin 105x45/ (84.42 – 5) = 34.2

o

(i) At the beginning of the lift, when θ = 0

Acceleration, a = ω2 ( R- r1) cosθ

= x cos0

= 206.86 m/s2

(ii) At the end of the contract with flank, when θ = φ = 24o.33’

Acceleration, a = ω2 ( R- r1) cosφ

= 171.09 m/s

2

(iii) At the beginning of contact with nose,

Acceleration, a = - ω2d cos φ

= - 146.92 m/s

2 (iv) At the apex of nose,

Acceleration, a = - ω2d

= - 177.7 m/s

2

Example.2

The following particulars relate to a symmetrical circular cam operating a flat faced

follower.

Least radius = 16 mm; Nose radius = 3.2 mm; Distance between cam shaft centre and nose centre

= 25 mm; Angle of action of cam = 150o and cam shaft speed = 600 rpm.

Assuming that, there is no dwell between the ascent and descent, determine the lift of the valve, the

flank radius and the acceleration and retardation of the follower at a point where circular nose

merges into circular flank.

Solution :

r 1 = 16 mm, r 2 = 3.2 mm , OP = d = 25mm, 2α = 150o,

α = 75o,

N = 600 rpm

Fig. 9

We know, (i)Lift = BT = OT- OB

= OP+PT – OT

= d + r2 - r1

= 25 + 3.2 -16

Lift =x = 12.2 mm

(ii) Flank radius, R,

(iii) Flank angle, ф

We have, from triangle OQP,

= 29o.6’

( iv) Acceleration at the end of the contract with flank, when θ = = 29o.6’

a = ω2 ( R- r1) cosφ

= 129.39 m/s

2

(v)Retardation at the beginning of contact with nose,

Acceleration, a = - ω2d cos φ

= - 85.81 m/s

2

Example 3,

A symmetrical arc cam using flat faced follower has the following particulars,

Total lift = 25 mm

Least radius = 35mm

Angle of lift = 90o

Flank radius = 105 mm

Speed = 1200 rpm

Calculate (i) main dimensions of the cam

(ii) Acceleration of the follower at the beginning of the lift, at the end of contact

with flank, at the beginning of contact with nose and at the apex of nose.

Solution:

X= 25 mm, r 1 = 35 mm, α = 90

o , R= 105 mm, N = 1200 rpm

Fig. 10

Referring to Fig. 10, we have,

OQ = Flank radius – Least radius

OQ = R- r1 = 105 – 35 = 70 mm

Also,

OP = d = OR – RP = Least radius +lift –nose radius

OP = 35+25- r2 = 60 - r2 mm

PQ = Flank radius – nose radius

PQ = 105 - r2 mm

From triangle OPQ,

PQ2 = OP

2 + OQ

2 – 2 OP x OQ x cos (180 – a)

( 105 - r2)2

= (60 - r2)2 + 70

2 – 2 (60 - r2) x 70 cos (180 – 90)

1052

+ r22 – 210 r2 = 3600 + r2

2 -120 r2 + 4900

90 r2 = 2525

r2 = 28 mm

OP = d = ( 60 – r2) = ( 60 – 28) = 32 mm

PQ = 105 - r2 = 105 – 28 = 77 mm

From triangle OPQ,

OP/ sin ф = PQ/ sin (180 – a)

sin ф = OP/PQ sin (180 – a)

= 32.sin (180 – 90)/77

= 0.4156

ф = 24º 33’

(i) At the beginning of the lift, when θ = 0

Acceleration, a= ω2 ( R- r1) cosθ

= x cos0

= 1105.44 m/s2

(ii) At the end of the contract with flank, when θ = φ = 24o.33’

Acceleration, a= ω2 ( R- r1) cosф

= 1007.26 m/s

2

(iii) At the beginning of contact with nose,

Acceleration, a= - ω2d cos ф

= - 460.44 m/s

2 (iv) At the apex of nose,

Acceleration, a = - ω2d

= - 505.32 m/s

2

Example 4,

A suction valve of a 4 stroke petrol engine is operated by a symmetrical circular cam with flat faced

follower. The details are as follows.

Lift = 10 mm; Least radius = 20 mm; Nose radius = 2.5 mm; Crank angle when suction valve opens

after TDC = 4o; Crank angle when suction valves closes after BDC = 50

o; Cam shaft speed = 600

rpm.

Determine maximum velocity of the valve and its maximum acceleration and retardation.

Solution:

r 1 = 20 mm, r 2 = 2.5 mm , Lift =x= 10 mm, N = 600 rpm

Fig. 11 Valve timing diagram

Angular displacement of cam when suction valve is open = 180 – 4+50 = 226o

For four stroke engine, the speed of cam shaft is half the speed of crank shaft

Anglular displacement of cam shaft during opening of the valve = = 2α

Since the cam is a symmetrical, angle of ascent = angle of decent, α = = 56.5o

Fig. 12

From the above diagram,

OP + r2 = Lift + r1

OP = 20+10-2.5 = 27.5 mm = d

Flank radius,

R = 116.87 mm

Flank angle,

From triangle OQP,

= 11

o.45’

Velocity is maximum when α = ,

Vmax = ω ( R – r1) sin

=

= 1.22 m/s

Maximum acceleration of valve when θ = 0,

Acceleration, a= ω2 ( R- r1)

= 382.42 m/s

2

Retardation is maximum, when α-θ =0,

a = - ω2d

= - 108.57 m/s

2

Example 5

The following particulars refers to a symmetrical circular arc cam used to operate suction valve

mechanism of a four stroke petrol engine

Total lift = 10 mm

Least radius = 25mm

Nose radius = 5 mm

Suction valve opens 6o after TDC

Suction valve closes 40o after BDC

Engine Speed = 2000 rpm

Find (i) Maximum velocity of the valve

(ii) Maximum acceleration and retardation

(iii) Minimum force to be exerted by the spring to overcome inertia of the valve parts of mass 0.28

kg.

Solution:

r 1 = 25 mm, r 2 = 5 mm , Lift =x= 10 mm, N = 2000 rpm

For four stroke engine, the cam shaft speed is half of engine crank shaft speed

Cam shaft speed = ½ c Engine speed= ½ x2000= 1000 rpm= N

Fig. 13 Crank angle diagram

Angular displacement of cam when suction valve is open = 180 – 6+40 = 214o

For four stroke engine, the speed of cam shaft is half the speed of crank shaft

Anglular displacement of cam shaft during opening of the valve

= 2a

Since the cam is a symmetrical, angle of ascent = angle of decent,

Fig. 14

From the above diagram,

OP + r2 = Lift + r1

OP = 25+10- 5 = 30 mm = d

Flank radius,

R = 140. 95 mm

Flank angle,

From triangle OQP,

= 10

o.22’

(i) Maximum Velocity when follower leaves the flank α = ,

Vmax = ω ( R – r1) sin

= 2.15 m/s

(ii) Maximum acceleration of valve when θ = 0,

Acceleration, a= ω2 ( R- r1)

= 1271 m/s

2

Retardation is maximum, when α-θ =0,

a = - ω2d

= - 328.98 m/s2

(iii) Minimum force to be exerted by the spring to overcome inertia of the valve parts,

Minimum force = Mass x retardation

= 0.28 x 328.98

= 82.24 N

Example 6,

A flat faced valve is operated by a symmetrical circular arc cam. The straight line path of the tappet

passes through the cam axis. Total angle of action = 150o, lift = 6 mm, base circle diameter = 30

mm, Period of acceleration is half the period of retardation during the lift. The cam rotates at

1250 rpm. Determine the flank and nose radii and maximum acceleration and retardation during the

lift.

Solution:

r 1 = 15 mm, Lift =x= 6 mm, N = 1250 rpm, α = 75o

Fig.15

Let,

Acceleration period angle = and Retardation period angle = b

Acceleration = ½ Retardation

= ½ b

Consider triangle POQ,

β + (180-a) + = 180

1.5β = 75o

β = 500

= 25

0

We have, Lift + r1 = d + r2

d = Lift + r1 - r2

d = 6 + 15 - r2

d = 21- r2

Again, consider triangle POQ,

OQ/Sinb = OP/sinf = PQ/sin (180- a)

OQ/Sin50 = 21- r2/Sin25 = PQ/Sin (180- 75)

OQ = (21- r2) Sin50/Sin25 = 38 – 1.8 r2 -----(1)

Also,

OQ = (PQ) Sin50 / Sin(180-75)

But, from above Fig.,

PQ = (QO+OC’)-DP

= QO + r1 –r2

OQ = (QO + r1 – r2) Sin50 / Sin(180-75)

OQ = 0.793 QO + 11.9 – 0.793 r2

0.207 OQ = 11.9 – 0.793 r2

OQ = 57.5– 0.793 r2 ---------(ii)

From equations (i) and (ii)

38 – 1.8 r2 = 57.5– 0.793 r2

Threrefore, nose radius, r2 = 9.6 mm,

OQ = 38 – 1.8 x 9.6 = 20.7 mm

Distance between cetres of nose and base circles, d = 21- r2

d = 21- 9.6

d = 11.4 mm

Flank radius,

R = 35.7 mm

Maximum acceleration of valve ,

a= ω2 ( R- r1)

a = 447.55 m/s2

Maximum retardation ,

a = - ω2d

a = - 195.33 m/s2

1.2 TANGENT CAM WITH ROLLER RECIPROCATING FOLLOWER

Tangent cams are made with straight flanks. A tangent cam is shown in Fig.16. The flanks AB

and IH are straight lines and tangent to the base circle at A and I and tangent to nose at B and I. The

centre of the circular nose is P. The path of the centre of roller follower is shown by dotted line.

Fig.16

a. Expression for determination of displacement, velocity and acceleration of the

roller follower when in contact on the straight flank

Fig. 17

Let

r1= Least base circle radius

r2= Roller radius

r3=Nose circle radius

d=Distance between the cam and nose circles

L= (r1 + r3)

α= Angle of ascent

φ=Angle of contact of cam with straight flank

Displacement:

x = OG - OB

=

= OB ( - 1)

x = (r1 + r3) ( - 1)

Velocity:

= x

= (r1 + r3) ( - 0) x ω

v = ω (r1 + r3) ( )

From this equation, it is evident that, as increases, also increases where as

decreases. With that, the velocity increases. Velocity will be maximum when is maximum. It

happens when point of contact is just leaving the straight flank i.e., when .

vmax = ω (r1 + r3) ( )

Acceleration:

a =

a= ω (r1 + r3)

a= ω2 (r1 + r3)

a= ω2 (r1 + r3)

a= ω2 (r1 + r3)

Acceleration is minimum when is minimum. This is possible when is

minimum and is maximum. It is so, when θ=0 or the roller at the beginning of its lift along

the straight flank. Acceleration is maximum when the roller shifts from flank to nose circle i.e.,

when .

amin= ω2 (r1 + r3)

amax= ω2 (r1 + r3)

(b) Expression for determination of displacement, velocity and acceleration of

the roller follower when in contact with nose

Let

r1= Least base circle radius

r2= Roller radius

r3=Nose circle radius

d=Distance between the cam and nose circles

L= (r1 + r3)

α= Angle of ascent

φ=Angle of contact of cam with straight flank

Fig. 18

Displacement:

x = OJ – OG

x = (OP+PJ) – ( OE+EG)

= ( d+L) – (OP cosθ1+PG cosβ)

= ( d+L) – (d cosθ1 + L cosβ)

= L +d – d cosθ1 - L cosβ -----(i)

From right angled triangles, OEP and GEP

EP = GP sinβ = OP sin θ1

= L sinβ = d sin θ1

Squaring on both sides,

L2

sin2β = d

2 sin

2 θ1

L2(1- cos

2 β) = d

2 sin

2 θ1

L2- L

2 cos

2 β = d

2 sin

2 θ1

L2 cos

2 β = L

2- d

2 sin

2 θ1

L cos β = (L2- d

2 sin

2 θ1)

1/2

Substituting the above value in equation (i)

X = L +d – d cosθ1- (L2- d

2 sin

2 θ1)

1/2

Velocity:

v = x

= -d .-sin θ1. - (L2- d

2 sin

2 θ1)

-1/2 ( -d

2 2sin θ1.cos θ1)

= d sin θ1. - (L2- d

2 sin

2 θ1)

-1/2 ( d

2 sin 2θ1)

v= ω d

Acceleration:

a =

Multiply numerator and denominator by

Example 7

In a symmetrical tangent cam operating a roller follower, the least radius of cam is 30 mm and roller

radius is 17.5 mm. The angle of ascent is 75o , lift is 17.5 mm and the speed of cam is 600 rpm.

Calculate,

1.Principal dimensions of cam

2.The acceleration of the follower at the beginning of lift, where straight flank merges into the

circular nose and at the apex of the circular nose. Assume that, there is no dwell between ascent and

descent.

Solution:

r 1 = 30 mm, r 3 = 17.5 mm, Lift =x= 17.5 mm, N = 600 rpm, α = 75o

w = 2p600/60 = 62.83 rad/s

From Fig. 17, OP + PT = OC + CT

OP = OC + CT- PT

OP = r 1 + x - r 2

OP = 47.5 - r 2 -------------( i )

Fig. 19

From the above Fig. 19,

OQ + QA = OA

OQ = OA - QA

= r 1 - r 2

OQ = 30 - r 2 ---------(ii)

Consider a triangle QOP,

Cos a = OQ/OP

Substitute OQ & OP from equations (i) and (ii)

Fig.20 Fig. 21

Cos 75 = 30 - r 2/ 47.5 - r 2

r 2 = 23.8 mm

OP = d = 30 + 17.5 -23.8 = 23.7 mm

From the triangles, GOB and POQ,

tan = GB/OB = PQ/OB

= OP Sina/ OB

= d Sin 75/ r1 + r3

= 23.7 Sin 75/ 30 + 17.5

Φ = 25.6

Acceleration of the follower,

(i) At the beginning of the lift, i.e., when θ = 0

a = ω2 (r1 + r3)

= (62.83)2 (30 + 17.5) (2-1)

= 187.5 m/s2

(ii) At the end of flank, i.e., when θ = Φ = 25.6

a = ω2 (r1 + r3)

a = (62.83)2 (30 + 17.5) (2- cos

2 25.6/ cos

3 25.6)

a = 303.38 m/s2

(iii) Acceleration when in contact with nose, when θ1= a- Φ = 75 - 25.6 = 49.4

a = 57.6 m/s

2

Example 8

A tangent cam with 70 mm base circle diameter operates a roller follower of 30 mm diameter. The

angle between the tangential faces of the cam is 90o and these faces are joined by a nose circle of

8 mm radius The speed of the cam is 120 rpm . Calculate the acceleration of the roller centre, when

the roller leaves the straight flank and at the apex of cam.

Solution:

r 1 = 35 mm, r 3 = 15 mm, r2 = 8 mm N = 120 rpm, w = 2p120/60 = 12.56 rad/s

a = 180 -90-45 = 45o

Fig. 22

From the above Fig.

OQ + QA = OA

OQ = OA - QA

= r 1 - r 2

OQ = 35 - 8

OQ = 27 mm

Also , OP = d

Consider a triangle QOP,

Cos a = OQ/OP

Cos 45 = 27/OP

OP = 38.18 mm = d

From the triangles, GOB and POQ,

tan = GB/OB = PQ/OB

= OP Sina/ OB

= d Sin 75/ r1 + r3

= 38.18 Sin 75/ 35 + 15

Φ = 28.37o

Acceleration of the roller follower

(i) At the end of flank, i.e., when θ = Φ = 28.37

a = ω2 (r1 + r3)

a = (12.56)2 (35 + 15) (2- cos

2 28.37/ cos

3 28.37)

a = 14.19 m/s2

(ii) Roller at the apex,

a = - ω2

r (1+1/n) where, n= r2+r1/d =0.602

= -16 m/s2

1.3 Under Cutting of Cam

Generally, prime circle of a cam is proportioned to give a satisfactory pressure angle.

However, some times the follower may not be completing the desired motion. This happens if the

curvature of the pitch curve is too sharp. Fig 32 (a) represents the pitch curve of a cam, while (b)

shows generation the curve by roller follower.

(a) Fig. 23 (b)

Fig. 24

Fig.24 represents, a roller follower trying to generating pitch curve. It is seen that, the cam profile

loops over itself in order to realize the profile of the pitch curve. Since it is impossible to produce

such a cam profile, the result is that, the cam will be undercut and become a pointed cam. Now

when the roller follower will be made to move over this cam, it will not be producing the desired

motion.

It is seen that, the cam will be pointed if the radius of the roller is equal to the radius of curvature of

the pitch curve. Therefore, to have minimum radius of curvature of the cam profile , the radius of

curvature of the prime circle must always be greater than that of the radius of the roller.

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