unit 4 analysis of cams.pdf

8/19/2019 unit 4 Analysis of CAMS.pdf

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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


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

the fol lower when f lat face of the fol lower has contact on the circular fl ank

Fig.4Let

r 1=OB=Least base circle radiusr 2=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 - r 1

= (QD-QE) – r 1

= (R - OQ cos θ) – r 1

= R-(R - r 1) cosθ – r 1

Velocity:

= x

= (R-r 1) (sin θ) ω


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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-r 1) sinφ

Acceleration:

a =

a= ω2 ( R- r 1) 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

foll ower when fl at face of the foll ower has contact on the nose

Fig.5


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Let

r 1=OB=Least base circle radius

r 2=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 flankDisplacement:

x = BC = OC – OB = DE - r 1

= (DP+PE) – r 1

= r 2+OP cos (α - θ) – r 1

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- r 1) cosφ to - ω2d cos φ i.e., a sudden

change from positive acceleration to negative acceleration (retardation).


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Note: Cosine rule

Fig. 6

Consider the triangle POQ in Fig. 7,

OQ = R-r 1, QP = R-r 2, OP = d

and apply cosine rule

Fig. 7

QP2= QO2+PO2-2QOxPOxCos POQ

(R-r 2)2 = (R-r 1)

2+d2-2(R-r 1)(d) cos (180-α)

= (R-r 1)2+d2+2(R-r 1) d cosα

R 2-2Rr 2+r 22 = R 2+r 1

2-2Rr 1+d2+2Rdcosα -2r 1dcosα

2Rr 1-2Rr 2-2Rdcosα = r 12-r 2

2+d2-2r 1dcosα

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


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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 = 75 o; 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 + r 1 = PO + r 220 + 30 = d – 5

d = 45 mm

we have,


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R = 82.42 mm

PQ = R-r 2 = 82.41-5 = 77.41 mm

QO = R-r 1 = 82.41 – 30 = 52.41 mm

To determine angle φ , consider triangle PQO,

xPO

= Sin 105x45/ (84.42 – 5)= 34.2o

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

Acceleration, a = ω2 ( R- r 1) 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- r 1) cosφ

= 171.09 m/s2

(iii) At the beginning of contact with nose,

Acceleration, a = - ω2d cos φ

= - 146.92 m/s2

(iv) At the apex of nose,

Acceleration, a = - ω2d

= - 177.7 m/s2

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


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Fig. 9

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

= OP+PT – OT

= d + r 2 - r 1= 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 f lank, when θ = = 29o.6’

a = ω2 ( R- r 1) cosφ

= 129.39 m/s2

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

Acceleration, a = - ω2d cos φ

= - 85.81 m/s2


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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 mmSpeed = 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 = 90o , R= 105 mm, N = 1200 rpm

Fig. 10

Referring to Fig. 10, we have,

OQ = Flank radius – Least radius

OQ = R- r 1 = 105 – 35 = 70 mm

Also,

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

OP = 35+25- r 2 = 60 - r 2 mm

PQ = Flank radius – nose radius

PQ = 105 - r 2 mm

From triangle OPQ,

PQ2 = OP2 + OQ2 – 2 OP x OQ x cos (180 – a)

( 105 - r 2)2 = (60 - r 2)

2 + 702 – 2 (60 - r 2) x 70 cos (180 – 90)

1052 + r 22 – 210 r 2 = 3600 + r 2

2 -120 r 2 + 4900

90 r 2 = 2525

r 2 = 28 mm

OP = d = ( 60 – r 2) = ( 60 – 28) = 32 mm PQ = 105 - r 2 = 105 – 28 = 77 mm

From triangle OPQ,


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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- r 1) 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- r 1) cosф

= 1007.26 m/s2

(iii) At the beginning of contact with nose,

Acceleration, a= - ω2d cos ф

= - 460.44 m/s2

(iv) At the apex of nose,

Acceleration, a = - ω2d

= - 505.32 m/s2

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 = 50o; 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


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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 + r 2 = Lift + r 1OP = 20+10-2.5 = 27.5 mm = d

Flank radius,

R = 116.87 mm

Flank angle,

From triangle OQP,

= 11o.45’

Velocity is maximum when α = ,

Vmax = ω ( R – r 1) sin

=

= 1.22 m/s


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Maximum acceleration of valve when θ = 0,

Acceleration, a= ω2 ( R- r 1)

= 382.42 m/s2

Retardation is maximum, when α-θ =0,

a = - ω2d

= - 108.57 m/s2

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 mmLeast 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,


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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


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We have, Lift + r 1 = d + r 2d = Lift + r 1 - r 2

d = 6 + 15 - r 2d = 21- r 2

Again, consider triangle POQ,

OQ/Sinb = OP/sinf = PQ/sin (180- a)OQ/Sin50 = 21- r 2/Sin25 = PQ/Sin (180- 75)

OQ = (21- r 2) Sin50/Sin25 = 38 – 1.8 r 2 -----(1)

Also,

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

But, from above Fig.,

PQ = (QO+OC’)-DP

= QO + r 1 – r 2

OQ = (QO + r 1 – r 2) Sin50 / Sin(180-75)

OQ = 0.793 QO + 11.9 – 0.793 r 2 0.207 OQ = 11.9 – 0.793 r 2

OQ = 57.5 – 0.793 r 2 ---------(ii)

From equations (i) and (ii)

38 – 1.8 r 2 = 57.5 – 0.793 r 2Threrefore, nose radius, r 2 = 9.6 mm,

OQ = 38 – 1.8 x 9.6 = 20.7 mm

Distance between cetres of nose and base circles, d = 21- r 2d = 21- 9.6

d = 11.4 mm

Flank radius,

R = 35.7 mm

Maximum acceleration of valve ,

a= ω2 ( R- r 1)

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


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Fig.16

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

roller f ollower when in contact on the straight fl ank

Fig. 17


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Let

r 1= Least base circle radius

r 2= Roller radius


d=Distance between the cam and nose circles

L= (r 1 + r 3)

α= Angle of ascent φ=Angle of contact of cam with straight flank

Displacement:

x = OG - OB

=

= OB ( - 1)

x = (r 1 + r 3) ( - 1)

Velocity:

= x

= (r 1 + r 3) ( - 0) x ω

v = ω (r 1 + r 3) ( )

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 (r 1 + r 3) ( )

Acceleration:

a =

a= ω (r 1 +

r 3)

a= ω2 (r 1 + r 3)

a= ω2 (r 1 + r 3)

a= ω2 (r 1 + r 3)

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


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

when .

amin= ω2 (r 1 + r 3)

amax= ω

2

(r 1 + r 3)

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

the roll er fol lower when i n contact with nose

Let

r 1= Least base circle radius

r 2= Roller radius


d=Distance between the cam and nose circlesL= (r 1 + r 3)

α= Angle of ascent

φ=Angle of contact of cam with straight flank

Fig. 18


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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 GEPEP = GP sinβ = OP sin θ1

= L sinβ = d sin θ1Squaring on both sides,

L2 sin2β = d2 sin2 θ1L2(1- cos2 β) = d2 sin2 θ1L2- L2 cos2 β = d2 sin2 θ1

L2 cos2 β = L2- d2 sin2 θ1L cos β = (L2- d2 sin2 θ1)

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Substituting the above value in equation (i)

X = L +d – d cosθ1- (L2- d2 sin2 θ1)1/2

Velocity:

v = x

= -d .-sin θ1. - (L2- d2 sin2 θ1)

-1/2 ( -d2 2sin θ1.cos θ1)

= d sin θ1. - (L2- d2 sin2 θ1)

-1/2 ( d2 sin 2θ1)

v= ω d

Acceleration:

a =

Multiply numerator and denominator by


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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


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From the above Fig. 19,

OQ + QA = OA

OQ = OA - QA

= r 1 - r 2OQ = 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 2r 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/ r 1 + r 3

= 23.7 Sin 75/ 30 + 17.5Φ = 25.6

Acceleration of the follower,

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

a = ω2 (r 1 + r 3)

= (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- cos2 25.6/ cos3 25.6)


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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/s2

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, r 2 = 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


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OQ = 27 mm

Also , OP = d

Consider a triangle QOP,

Cos a = OQ/OP

Cos 45 = 27/OPOP = 38.18 mm = d

From the triangles, GOB and POQ,

tan= GB/OB = PQ/OB

= OP Sina/ OB

= d Sin 75/ r 1 + r 3

= 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- cos2 28.37/ cos3 28.37)

a = 14.19 m/s2

(ii) Roller at the apex,

a = - ω2 r (1+1/n) where, n= r2+r 1/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)


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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.

REFERENCES

1.S.S.Rathan(2009),Theory of Machines,3rd edition, Tata MC Graw Hill Education Pvt.td, New

Delhi.

2.SadhuSingh(2012),Theory of Machines,3rdedition,Pearson, New Delhi.

3.Ballaneys(1988),Theory of Machines,16thedition,Khanna Publications, Delhi.

4.Sharma,C.S, Kamalesh Purohit(2006),Theory of Mechanisms and Machines, Prentice-Hall of

India Pvt. Ltd. New Delhi.

5.Malhotra and Guptha (2006),TheTheory of Machines,3rd edition Sathy Prakashan, New Delhi.


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6.Ashok A.G(2009), Mechanisms and Machine Theory, 2nd edition, PHI Learning Pvt. Ltd.

New Delhi.

unit 4 analysis of cams.pdf

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