Prof' B'Dinesh Prabhu PESCEMandya

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Prof' B'Dinesh Prabhu PESCEMandya

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Title: Prof' B'Dinesh Prabhu PESCEMandya

1
CHAPTER-7Flywheel Camshaft
2
Flywheel
3
Flywheel

Necessity
In a combustion engine, especially
in one with one or two cylinders, energy is
imparted to the crankshaft intermittently, in
order to keep it rotating at a fairly uniform
speed under a substantially constant load, it is
necessary to provide it with a flywheel.

4
Flywheel

In a single cylinder engine(4 Stroke), in which
there is only one power stroke in two revolutions
of the crankshaft, a considerable fraction of
energy generated per cycle is stored in the
flywheel, the proportion thus stored decreases
with an increase in the No. of cylinders
In a 4 cylinder engine about 40 of the energy of
the cycle is temporarily stored.

5
Flywheel

However,
not all of this energy goes into flywheel
During the 1st half of the power stroke, when
energy is being supplied in excess by the burning
gases, all of the reciprocating parts of the
engine are being accelerated absorb energy
besides, the rotating parts other than the
flywheel also have some flywheel capacity, this
reduces the proportion of the energy of the cycle
which must be stored in the flywheel.

6
Flywheel

In a 6 cylinder engine the proportion of the
energy which must be absorbed returned by the
moving parts amounts to about 20.
The greater the No. of cylinders the smaller the
flywheel capacity required per unit of piston
displacement, because the overlap of power
strokes is greater besides other rotating parts
of the engine have greater inertia.

7
Flywheel

However, the flywheel has by far the greatest
inertia even in a multi cylinder engine.
Aside from its principle function, the fly wheel
serves as a member of the friction clutch, it
usually carries also the ring gear of the
electric starter.

8
Flywheel
9
Flywheel
10
Flywheel

Energy accumulator
Energy re-distributor
A flywheel serves as a reservoir which stores
energy during the period when the supply of
energy is more than the requirement releases,
it during the period when the requirement of
energy is more than supply.
A flywheel helps to keep the crankshaft rotating
at a uniform speed

11
Flywheel

In internal combustion engines,
the energy is developed during one stroke and the
engine is to run for the whole cycle on the
energy produced during this one stroke.
the energy is developed, only during power stroke
which is much more than the engine load and no
energy is being developed during suction,
compression and exhaust strokes in case of 4
stroke engines during compression in case of 2
stroke engines.

12
Flywheel

The excess energy developed during
power stroke is absorbed by the flywheel and
releases it to the crankshaft during other
strokes in which no energy is developed, thus
rotating the crankshaft at a uniform speed.
When the flywheel absorbs energy, its speed
increases and when it releases, the speed
decreases. Hence a flywheel does not maintain a
constant speed, it simply reduces the fluctuation
of speed.

13
Flywheel

The function of a governor in engine is
entirely different from that of a flywheel
Governor regulates the mean speed of an engine
when there are variations in the load,
e.g., when the load on the engine increases it
becomes necessary to increase the supply of
Working fluid. On the other hand, when the load
decreases, less working fluid is required.
The governor automatically controls the
supply, of working fluid to the engine with the
varying load condition and keeps the mean speed
within certain limits.

14
Flywheel

The flywheel does not maintain constant speed
It simply reduces the fluctuation of speed.
A flywheel controls the speed variations caused
by the fluctuation of the engine turning moment
during each cycle of operation.
It does not control the speed variations caused
by the varying load.

15
Flywheel

Capacity Diameter
The flywheel capacity of a given mass increases
with its distance from the axis of rotation
consequently, if the flywheel is made large in
diameter it need not be so heavy.
On the other hand there are two reasons for
limiting the diameter.
At high rpm, flywheel is subjected to disruptive
or bursting force, by keeping down the
diameter, the F O S can be kept high.
As it is cast or cast pressed steel housing,
this need not weigh so much if the diameter is
smaller

16
Flywheel
A Flywheel is given a high rotational inertia
i.e., most of its weight is well out from the axis
17
Construction of Flywheels
18
Construction of Flywheels

The flywheels of smaller size (up to 600 mm
diameter) are casted in one piece.
The rim hub are joined together by means of
web.
The holes in the web may
be made for handling purposes.

19
Construction of Flywheels

In case the flywheel is of larger size (up to
2.5m diameter), the arms are made instead of web.
The number of arms depends upon the size of
flywheel its speed of rotation.

20
Construction of Flywheels

The split flywheels are above 2.5m diameter
are usually casted in two piece.
It has advantage of relieving shrinkage stresses
in arms due to unequal rate of cooling of casting.

Maximum Fluctuation of Speed
Coefficient of Fluctuation of Speed

22
Flywheel

The Maximum Fluctuation of Speed
the difference between the maximum
minimum speeds during a cycle. i.e., (N1-N2)
Where, N1Maximum speed in r.p.m. during the
cycle,
N2Minimum speed in r.p.m. during the cycle
The Coefficient of Fluctuation of Speed is the
ratio of the maximum fluctuation of speed to the
mean speed.

23
Flywheel
24

Fluctuation of Energy

The Fluctuation of Energy may be determined by
the turning moment diagram for one complete cycle
of operation
25
Flywheel

turning moment is zero when the crank angle is
zero
It rises to a max. value when crank angle reaches
90 and it is again zero when crank angle is 180.

26
Flywheel

work doneturning moment x angle turned
The area of the turning moment diagram represents
the work done per revolution
Engine is assumed to work against the mean
resisting torque,
Since it is assumed that
work done by the turning moment / revolution
work done against the mean resisting torque
Therefore,
area of rectangle (aAFe) is proportional to work
done against the mean resisting torque.

27
Flywheel

When crank moves from 'a' to 'p'
work done by the engine is area aBp,
whereas the energy required area aABp.
In other words, the engine has done less work
than the requirement. This amount of energy is
taken from the flywheel and hence the speed of
the flywheel decreases.

28
Flywheel

Now the
crank moves from p to q,
work done by the engine area pBbCq
requirement of energy area pBCq
Therefore the engine has done more work than the
requirement.
This excess work is stored in the flywheel and
hence the speed of the flywheel increases while
the crank moves from p to q.

29
Flywheel

Similarly when the crank moves from q to r,
more work is taken from engine than is
developed area CcD.
To supply this loss, the flywheel gives up some
of its energy and thus the speed decreases while
the crank moves from q to r.
As the crank moves from r to s,
excess energy is again developed area DdE
the speed again increases.
As the piston moves from s to e,
again there is a loss of work the speed
decreases.

30
Flywheel

The variations of energy above below the mean
resisting torque line are called Fluctuation of
energy.
The areas BbC, CcD, DdE etc. represent
fluctuations of energy.

31
Flywheel

Engine has a max. speed either at q or at s.
This is due to the fact that flywheel absorbs
energy while the crank moves from p to q and from
r to s.
Engine has a minimum speed either at p or at r.
The reason is that flywheel gives out some of
its energy when the crank moves from a to p and q
to r.
The difference between the maximum and the
minimum energies is known as maximum fluctuation
of energy.

32
Flywheel
33
Flywheel

The Fluctuation of Energy the variation of
Energy above below mean resisting torque line.

34
Flywheel

For a 4 Stroke IC Engine,
During suction stroke, since the pr. inside the
engine cylinder is less than the atmospheric pr.,
a negative loop is formed
During compression stroke, the work is done on
the gases, there a higher negative loop is
obtained
In the working stroke, the fuel burns the gases
expand, therefore a large positive loop is formed
During exhaust stroke, the work is done on the
gases, therefore a negative loop is obtained

35
Maximum Fluctuation of Energy
36

Maximum
Fluctuation of Energy

37
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38

Coefficient of
Fluctuation of Energy

39
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40

Energy stored in a flywheel

41
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42
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43
Example-1
44
Example-1

The turning moment diagram for a
Petrol engine is drawn to following scales
Turning moment, 1 mm 5 Nm Crank angle, I mm
1
The turning moment diagram repeats itself at
every half revolution of the engine and the areas
above and below mean turning moment line, taken
in order are 295, 685, 40, 340, 960, 270 mm2
Determine the mass of 300 mm diameter flywheel
rim when the coefficient of fluctuation of speed
is 0.3 and the engine runs at 1800 r.p.m..
Also determine the cross-section of the rim when
the width of the rim is twice of thickness.
Assume density of rim material as 7250 kg/m3.

45
Example-1

Solution
Given
? 7250 kg/m3
Cs 0.3 0.003
D 300 mm or
R 150 mm
0.15 m
N 1800r.p.m. or
? 2 p x 1800 /60 188.5rad/s
m, b, t ?

? E I?2Cs m.R2.?2Cs m ?V ?.(pD.A)
?.(2pR.A) A b x t, b2t
46
Example-1

Mass of the flywheel (m kg)
As per given scale,
on the turning moment diagram,
1 mm2 5 Nm x 10 5 x(10 xp /180) 0.087 Nm
Max. fluctuation of energy ? E m.R2.?2Cs
Let the total energy at A E,
from fig., Energy at B E 295
Energy at C E 295 - 685 E - 390
Energy at D E - 390 40 E - 350
Energy at E E - 350 - 340 E - 690
Energy at F E - 690 960 E 270
Energy at G E 270 - 270 EEnergy at A

47
Example-1

. Maximum energy E295 at B
Minimum energy E-690 at E
Maximum fluctuation of energy,
i.e. ? E Max. energy Min. energy
(E 295) - (E - 690) 985 mm2
985 x 0.08786 Nm
Also, Maximum fluctuation of energy,
i.e. ? E 86 m.R2.?2Cs
m (0.15)2 (188.5)2 (0.003) 2.4xm
. m 86/2.4 35.8 kgAns.

48
Example-1

Cross-section of the flywheel rim
Mass of the flywheel rim, m Vol. x density
(A x 2pR) x?
Let, Width of rim, b2 x t, thickness of rim
. Cross-sectional area of rim, Ab x t 2t x t
2 t2
mass of the flywheel rim (m) 35.8 A x 2pRx ?
2 t2 X2 p x 0.15 x 7250
13668 t2
. t2 35.8/13668 0.0026
or t 0.051 m 51mm.Ans.
b 2 t 2 x 51 102 mm.Ans.

49
Flywheel

Example-2

50
Example-2

A single cylinder, single acting,4
stroke oil engine develops 20 kW at
300 r.p.m.
The work done by the gases during expan. stroke
is 2.3 times the work done on the gases during
the compression and work done during the suction
and exhaust strokes is negligible.
The speed is to be maintained within 1.
Determine the mass moment of inertia of the
flywheel.

51
Example-2

Solution Given
P 20kW
20x 103 W
N 300r.p.m. or
? 2px300/60
31.42rad/s
Cs1 or
?1- ?2 1?

? E I?2Cs
52
Example-2
53
Example-2
54
Example-2

The work done during expan. stroke
is shown by triangle ABC in Fig., in which
base AC p radians height BF Tmax
. Work done during expansion stroke,
WE 14160 (1/2) X p X Tmax 1.571 Tmax
Or Tmax 14160/1.571 9013 Nm
Height above the mean torque line,
BG BF - FG Tmax- Tmean
9013 - 636.5 8376.5 Nm

55
Example-2
56
Example-2
57
Stresses in a Flywheel Rim
58
Stresses in a Flywheel Rim

A flywheel, consists of a rim at which major
portion of mass or weight is concentrated.
Following types of stresses are induced in the
rim
1. Tensile stress due to centrifugal force,
2. Tensile bending stress caused by the
restraint of the arms, and
3. The shrinkage stresses due to unequal rate of
cooling of casting. This stress is taken care
of by a factor of safety.

59
Stresses in Rim

1. Tensile stress due to centrifugal force
The tensile stress in the rim due
to the centrifugal force, assuming
that rim is unstrained by arms, is
determined in a similar way as a
thin cylinder subjected to internal pr..
Tensile stress,
st ?.R2.?2 ?.v2 ...(v ?.R)
(when ? is in kg/m3 and v is in m/s, then st will
be in N/m2 or Pa)
Note From the above expression the mean diameter
(D) of the flywheel may be obtained by using the
relation vpDN / 60

60
Stresses in Rim
61
Stresses in Rim
62

Example-3

63
Example-3

A multi-cylinder engine is to run at a
constant load at a speed of 600 r.p.m.
On drawing the crank effort diagram
to scale of 1 mm 250 Nm and 1 mm 3,
The areas in mm2 above below mean torque line
are 160,- 172, 168,- 191, 197,- 162 mm2
The speed is to be kept within 1 of the mean
speed of the engine. Calculate the necessary
moment of inertia of the flywheel.
Determine suitable dimensions for cast iron
flywheel with a rim whose breadth is twice its
radial thickness. The density of cast iron is
7250 kg/m3, and its working stress in tension is
6MPa.
Assume that rim contributes 92 of flywheel
effect.

64
Example-3

Solution. Given
N 600 r.p.m. or
? 2p x 600/60 62.84 rad/s
? 7250 kg/m3 st 6 MPa 6 x 106 N/m2

? E I?2Cs st?.v2 v (pDN)/60 ? E
m.R2.?2Cs m ?V ?.(pD.A) ?.(2pR.A) A b x
t, b2t
65
Example-3

Moment of inertia of the flywheel
We know that,
Maximum fluctuation of energy, ?E Ix?2xCs
Scale for the turning moment is
1 mm 250 Nm scale for the crank angle is
1 mm 3 3xp/180 p/60 rad, therefore
1 mm2 on the turning moment diagram
i.e., 1 mm2 250 x p/60 13.1 Nm

66
Example-3

Let total energy at A E. . from Fig.,
Energy at B E 160
Energy at C E 160 - 172 E - 12 Energy at D
E - 12 168 E 156
Energy at E E 156 - 191 E 35 (min.
energy)
Energy at F E - 35 197 E 162 (max.
energy)
Energy at G E 162 - 162 E Energy at A
Max. fluctuation of energy,
? E Max. energy Min. energy
(E 162) - (E - 35) 197 mm2
? E 197 x 13.12581 N-m

67
Example-3

Since the fluctuation of speed is1
of the mean speed (?),
therefore total fluctuation of speed,
?1-?2 2 ? 0.02?
coefficient of fluctuation of speed,
Cs (?1-?2)/? 0.02
maximum fluctuation of energy, ?E,
?E, 2581 I.?2.Cs I (62.84)2 0.02 79xI
. I 2581/79 32.7 kgm2..Ans

68
Example-3

Dimensions of a flywheel rim
We know that, Mass of the flywheel rim,
m Vol. x density 2pR x A x ? pD x (b x t )x
? b2t (given)
Peripheral velocity (v) mean diameter (D)
We know that, tensile stress, st?.v2
i.e., st6 x 106?.v2 7250 X v2
So, v2 (6 x 106)/7250 827.6 or v 28.76 m/s
We also know that, peripheral velocity,v
(pDN)/60
i.e., v28.76 (pDN)/60 (pDx600)/60 31.42 D
So D 28.76/31.42 0.915 m 915 mm Ans.

69
Example-3

Mass of flywheel rim
Since rim contributes 92 of flywheel effect,
. Energy of flywheel rim, Erim 0.92 x total
Energy of the flywheel, E
Max. fluctuation of energy, ?E E x 2 Cs
i.e., ?E 2581 E x 2 Cs E x 2 x 0.02 0.04 E
. E 2581/ 0.04 64525 N-m
Energy of flywheel rim, Erim 0.92 E
(v?R) 0.92 x 64525 59363 Nm
Also, Erim (1/2)I?2 (1/2)mk2?2 (1/2)mR2?2
(1/2)xmxv2
i.e., 59363 1/2 x m x v2 1/2 x m (28.76)2
413.6 m
. m 59363/413.6 143.5 kg

70
Example-3

The mass of the flywheel rim may also
be obtained by using following relations.
Since the rim contributes 92 of the flywheel
effect,
Irim 0.92xIflywheel or m.k2 0.92 x 32.7 30
kgm2
Since radius of gyration, k R D/2 0.915/2
0.4575m,
i.e., m(30/k2) (30/0.45752) (30/0.209)143.5kg
(? E) rim 0.92 (? E )flywheel
m. V2.CS 0.92 (? E )flywheel
m (28.76)2x0.02 0.92x 2581
16,55xm 2374.5
or m 2374.5/16.55 143.5kg

71
Example-3

m 59363/413.6 143.5 kg
Also, mass of the flywheel rim,
m (b x t )x pD x ? b2t (given)
143.5 b x t x pD x ?
2 t X t X p x 0.915 x 7250 41686 t2
t2 143.5/41686 0.00344
t 0.0587 say 0.06 m 60 mm Ans.
b 2 t 2 x 60 120 mm Ans.

Example-4

73
Example-4

An otto cycle engine develops 50 kW at
150 r.p.m. with 75 explosions per minute.
The change of speed from the commencement to the
end of power stroke must not exceed 0.5 of mean
on either side.
Design a suitable rim section having width four
times the depth so that the hoop stress does not
exceed 4 MPa.
Assume that the flywheel stores 16/15 times the
energy stored by the rim and that the workdone
during power stroke is 1.40 times the workdone
during the cycle. Density of rim material is 7200
kg/m3

74
Example-4

Solution.
Given
P 50 kW
50 X 103W
N 150r.p.m.
n 75
st 4 MPa
4 x 106 N/m2
? 7200 kg/m3

75
Example-4

We know that, Mass of the flywheel rim,
m Vol. x density2pR x A x ?
i.e., m(b x t )x pD x ? b4t (given)
Further, Energy of the flywheel rim,
Erim (1/2)I?2 (1/2)mk2?2 (1/2)mR2?2
(v?R)
(1/2)xmxv2 Erim(15/16)E given
And Max. fluctuation of energy, ?E E x 2 Cs
hoop Stressst?v2 vpDN/60

76
Example-4

Tmean transmitted by the engine or flywheel.
Power transmitted, P (2xpxNxTmean)/60
i.e., 50x103 (2xpx150xTmean)/60 15.71 Tmean
Tmean 50 x 103/15.71 3182.7 N-m
Workdone/cycle Tmeanx ? 3182.7 x 4 p 40000
Nm
Or The workdone per cycle for a 4 stroke
engine,
Workdone l cycle (Px60)/(No. of
explosions/min)
(Px60)/n (50000x60)/7540000 Nm
. Workdone during power stroke
1.4xWorkdone/cycle 1.4x40000 56000 N-m

77
Example-4

The workdone during power stroke is shown
by ?le ABC in Fig
in which base AC p radians and height BF Tmax
. Workdone during working stroke 1/2xpXTmax
1.571 Tmax
. Also Workdone during working stroke56000 Nm
. Tmax(56000/1.571) 35646Nm
Height above the mean torque line,
BG BF-FG Tmax-Tmean
35646-3182.7 32463.3Nm

78
Example-4
79
Example-4

Mean diameter of the flywheel (D)
Hoop stress, st?v2
i.e., 4 x 106 ?.v2 7200 X v2
. v2 4 X 106/7200 556
or v 23.58 m/s
Peripheral velocity, v pDN/60
i.e., 23.58 pDN/60 pDx150/60 7.855 D
D 23.58/7.855
D3 m Ans.

80
Example-4

Cross-sectional dimensions of the rim
Cross-sectional area of the rim,
Ab x t 4t x t 4t2 (b4t..given)
Since N1 - N2 0.5 N either side, (..given)
. total fluctuation of speed,
N1 - N2 1 of mean speed 0.01 N
coefficient of fluctuation of speed,
Cs. (N1 - N2)/N 0.01
E Total energy of the flywheel.
Maximum fluctuation of energy (? E),
46480 Ex 2 Cs E x 2 x 0.01 0.02 E
. E 46480/0.02 2324 x 103 Nm

81
Example-4

Since the energy stored by the flywheel is
16/15 times the energy stored by rim,
Therefore the energy of the rim,
Erim (15/16)E (15/16)x 2324 x 103 2178.8 x
103Nm
Also Erim (1/2)xmxv2
i.e., 2178.8 X 103 (1/2)xmx(23.58)2 278xm
. m 2178.8 x103 / 278 7837 kg
Also, mass of the flywheel rim, m A x pD x ?
i.e., 7837 A x pD x ? 4 t 2 X p X 3 x 7200
271469 t2
or t2 7837 / 271469 0.0288
or t 0.17 m 170 mm Ans.
b 4 t 4 x 170 680 mmAns.

82
Example-5
83
Example-5
84
Example-5
? E I?2Cs st?.v2 v (pDN)/60 ? E
m.R2.?2Cs m ?V ?.(pD.A) ?.(2pR.A) A b x
t, b2t
85
Example-5
86
Example-5
87
Example-5
88
Example-5
89
Example-5
90
Example-5
91
Example-5
92

Stresses in Flywheel Arms

93
Stresses in Flywheel Arms

Following stresses are induced
in the arms of a flywheel
Tensile stress due to centrifugal force acting on
the rim.
Bending stress due to torque transmitted from the
rim to the shaft or from the shaft to the rim.
Shrinkage stresses due to unequal rate of cooling
of casting. These stresses are difficult to
determine.

94
Stresses in Flywheel Arms

Tensile stress due to centrifugal force
Due to the centrifugal force acting on the rim,
the arms will be subjected to direct tensile
stress whose magnitude is,
. Tensile stress in the arms,
st1 (3/4)st (3/4)?xv2

95
Stresses in Flywheel Arms

Bending stress due to torque transmitted
T Maximum torque transmitted by the shaft,
Z Section modulus for the c/s of arms
R Mean radius of rim
r Radius of the hub
n Number of arms

96
Stresses in Arms
97

Design of Flywheel Arms

98
D/n of Arms
99

Design of Shaft, Hub and Key

100
Design of Shaft, Hub and Key
101
Shaft, Hub and Key
102
Design of Shaft, Hub and Key
103
Example-6
104
Example-6

Design and draw a cast iron flywheel
used for a four stroke I.C engine
developing180 kW at 240 r.p.m.
The hoop or centrifugal stress developed in the
flywheel is5.2 MPa, the total fluctuation of
speed is to be limited to 3 of the mean speed.
The work done during the power stroke is 1/3
more than the average work done during the whole
cycle.
The max. torque on shaft is twice the mean
torque. The density of cast iron is 7220 kg/m3.

105
Example-6
106
Example-6
107
Example-6
108
Example-6
109
Example-6
110
Example-6
111
Example-6
112
Example-6
113
Example-6
114
Example-6
115
Example-6
116
Example-6
117
Camshaft
118
Camshaft

The camshaft provides a means of actuating the
opening controlling the period before closing,
both for inlet as well as exhaust valves
It also provides a drive for the ignition
distributor and the mechanical fuel pump
The camshaft receives its motion from the
crankshaft, from which all of the accessories
also must be driven.

119
4-Stroke Single Cylinder SI Engine
120

4-Stroke Single CylinderSI Engine With2
different locations of camshaft

121
Multi Cylinder Engine
122
Camshaft

An ignition-unit
oil-pump drive for an engine in which the pump
is outside the crankcase

123
Camshaft

The camshaft receives its motion from the
crankshaft, from which all of the accessories
also must be driven
In the case of passenger car engines these
accessories usually include a fuel pump, an oil
pump, a water pump, a generator, an ignition
unit, and a fan
In cars equipped with hydraulic power steering a
drive must be provided for an additional pump

124
Camshaft

The camshaft consists of a number of cams at
suitable angular positions for operating the
valves at approximate timings relative to the
piston movement and in a sequence according to
the selected firing order.
There are two lobes on the camshaft for each
cylinder of the engine one to operate the intake
valve and the other to operate the exhaust valve.

125
Camshaft
126
Camshaft
127
Camshaft

Usually there is an integral spiral toothed gear
on the camshaft to drive the distributor and the
oil pump.
The fuel pump is operated from an integral
eccentric or a bolt-on eccentric.
In some cases, however, distributors are driven
directly from the camshaft end

128
Camshaft

Material
The camshaft is forged from alloy steel or cast
from hardenable cast iron and is case hardened
A typical cast iron alloy for a camshaft would
consist of
3.3 carbon, 2 silicon, 0.65 manganese, 0.65
chromium, 0.25 molybdenum the remainder iron

129
Camshaft

Camshaft drive
The camshaft rotates at half the crankshaft speed
so as to open and close the valves once in every
two revolutions of crankshaft
Drive from crankshaft to cam shaft may be
Gear drive
Chain drive or
Toothed belt drive

130
Camshaft

Gear drives, need no tensioner, but are noisy and
are suitable only for camshafts mounted close to
the crankshafts. Obviously gear is not suitable
for overhead camshafts.

131
Camshaft
Timing gears are made of cast iron, steel,
aluminum or laminated fiber.
132
Camshaft

Majority of passenger car engines now have a
chain drive, those that have gear drive have
one of the gears made of non-metallic material.
In a chain drive a separate idler gear has to be
provided. Moreover, a long chain drive tends to
whip to avoid which a tensioning device is used,
which may be a roller or spring-loaded steel
strip or a rubber pad attached to a spring-loaded
piston.

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Camshaft

Chains are made with links of alloy steel with
ground case hardened pins.
Sprockets are made of nylon, aluminum, steel or
iron.

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Camshaft
The latest type of drive is, by means of a
toothed rubber belt, which is made of rubber
molded onto a non-stretching cord. Such belts
operating relatively silently, do not need any
lubrication and are becoming increasingly popular
for both petrol as well as diesel engines with
overhead camshaft. Periodic tensioning checks
are all that is required for these.
135

Toothed Drive Belt for Double overhead Camshaft

136
Cam

A cam is a mechanical member for transmitting a
desired motion to a follower by direct contact.
The driver is called the cam, and the driven
member is called the follower.
The cam, may remain stationary, or translate
oscillate or rotate whereas the follower may
translate or oscillate

137
Cam

Types of Cams
Cams may be classified in the following three
ways
(i) Follower motion,
e.g. dwell-rise-dwell (D-R-D) dwell-rise
return-dwell (D-R-R-D) or rise-return-rise
(R-R-R).
(ii) Cam shape
e.g. wedge, radial, globoidal, cylindrical,
conical, spherical.
(iii) Manner of constraint of the follower.
Constraint may be obtained either by spring
loading to keep the follower in contact with the
cam surface, or by position drive.

138
Cam