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MECHANICAL ENGG. SCEINCE
PARTA
( PROPERTIES OF STEAM, STEAM TURBINES,REFRIGERATION, IC ENGINES)
PARTB
1) POWER TRANSMISSION2) CASTING3) FORGING4) MACHINE TOOLS5) WELDING
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Dept. of Mech & Mfg. Engg.
3
TRANSMISSION OF POWER
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What is a transmission system?The rotational motion can be transmitted fromone mechanical element to the other with thehelp of certain systems known astransmission system (Drives).
These systems is employed to drive a device
directly or transmit the rotational motion tovarious parts of the machine within itself.
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The one that drives is called the driving
system and the other which is driven is
called the driven system.
Round rods called shafts are used to
transmit the rotational motion.7
Dept. of Mech & Mfg. Engg. 8
Methods of Drive
Machines may be driven by any one of the
following two methods:
1. Individual Drive
2. Group Drive
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Individual Drive (Self-contained Drive)
Each machine has its own electric motor and starter.
The motor may drive the machine shaft through belt,
chain, gears etc..
It is used on machines that require considerable
power, operating at a full load.
10
Group Drive (Common Drive)
If several machine tools receive power from a common powerful motor
which runs a main shaft or overhead shaft is called group or common
drive.
The main shaft runs from one end to the other end of the shop.
The Main shaft drives another shaft called countershaft, which in turn
drives the machine drive shaft.
It is most suitable where power consumption of individual
machines is extremely variable.
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Classification of transmission systems is done depending on,
distance between them, speed & power as,
Types of drives
i. Belt drive.
ii. Chain drive.
iii. Gear drive.
iv. Rope drive.
Flat belt.---- 1.open 2.crossed
V - belt
Dept. of Mech & Mfg. Engg. 12
Belt Drive
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Dept. of Mech & Mfg. Engg. 13
Belt Drive (Friction Drive)
It is one of the most common and effective devices of
transmitting power or rotary motion between two parallel
shafts. It consists of two pulleys over which a thin
inextensible (endless) band is passed encircling both of
them.
Uses :Mills & Factories,specially when the distance between
them is not very great.
In a belt drive arrangement, there is a driverpulley mounted on the driving shaft,
and the driven pulley(follower) to which thepower has to be transmitted is mounted onthe driven shaft.
The pull or the tension on one of the sides of
the belt should be more than the other side
for the belt to move.
Dept. of Mech & Mfg. Engg. 14
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The portion of the belt which is having less tensionis calledthe slack sideand the one which is having more(higher)tension is called the tight side.
This depends on the direction of rotation of the driving pulley.i.e.
clockwise rotation- lowerside tight & upperside slack.
This arrangement increases the angle of contact of the belt onthe driven side and therefore the capacity of the drive.
Due to slip the pulley rotate at a lesser speed which reduces thepower transmission, hence belt drives are said to benot apositive typeof power transmission system.
The effective pulling power of the belt that causes the rotationof the driven pulley is the difference in tensions on the slackand tight sides. 15
Dept. of Mech & Mfg. Engg. 16
Types of Flat belt drives:
1. Open belt drive
2. Crossed-belt drive
Flat Belt
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Dept. of Mech & Mfg. Engg. 17
Open belt drive
It is employed when the two parallel shafts have to rotate
in the same direction.
Dept. of Mech & Mfg. Engg. 18
Open belt drive
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Dept. of Mech & Mfg. Engg. 19
Open belt drives
When the shafts are placed far apart, thelower side of the belt should be the tight sideand the upper side must be the slack side.
When the upper side becomes the slack side,it will sag due to its own weight and thusincreases the arc of contact.
20
Flat belt drives of the open system
can have:
Their shaft axes either horizontal or inclined.
They should never be vertical
(:.the centrifugal force developed in the belt
combined with the force of gravity causes the
belt to stretch and tend to leave the rim).
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Dept. of Mech & Mfg. Engg. 21
Crossed belt drive
Dept. of Mech & Mfg. Engg. 22
CROSSED BELT DRIVEIt is employed when:
Two parallel shafts have to rotate in the opposite
direction.
At the junction where the belt crosses, it rubs against
itself and wears off.
To avoid excessive wear, the shafts must be placed at a
maximum distance from each other
Operated at very low speeds.
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Pulley
Pulleys are used to transmit power from one shaft to the other at a
moderate distance away by means of a belt or strap running over
them.
They may be made of cast iron, wrought iron, pressed steel, wood.
What is crowning in a pulley?
It is the process of keeping the diameter of the rim greater at the
center than at the edges.
The effect of crowning is to keep the belt in a central position.
24
What is crowning in a pulley?
When the flat belt on cylindrical pulley is off-center and the pulley rotating, the belt quicklymoves up to the largest radius at the top ofthe crown and stays there.
The crown is important to keep the belt"tracking" stable, preventing the belt from"walking off" the edge of the pulley.
A crowned pulley eliminates the need for
pulley flanges and belt guide rollers.
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Dept. of Mech & Mfg. Engg. 25
About Crowning
When a flat belt runs over two pulleys, only one of them
needs to be crowned to achieve lateral stability.
The amount of curvature required in actual machinery is
small.
The method works for belts of leather or rubberized fabric
that have some elasticity.
26
Benefits of Crowning the pulley
Crowning of pulleys provides an automatic correction to
mis-tracking caused by transient forces that are applied to
the belt.
Without crowning these transient forces cause the belt to
be displaced without consistent means of returning to its
normal path.
This can cause belt edge cupping and wear.
For this reason it is wise to select a conveyor with pulley
crowned.
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Pulley crowning *
Critical dimensions:
Crowning of pulleys should not exceed 25mm on the
dia. / mtr of width
Width of the pulley should be 1/4th greater than width of
the belt.
Dept. of Mech & Mfg. Engg. 28
Types of pulleys
Stepped cone pulley (Speed cone)
Fast and loose pulleys
Guide pulley (Right angled drive)
Jockey pulley
Grooved pulley
Wrought-iron pulley
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Dept. of Mech & Mfg. Engg. 29
Stepped cone pulley
Dept. of Mech & Mfg. Engg. 30
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Dept. of Mech & Mfg. Engg. 31
Dept. of Mech & Mfg. Engg. 32
Stepped cone pulley
f
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Speed cone pulleys or stepped cone pulleys
are cast iron pulleys having several steps of different
diameters on which a belt may run.They are used for varying the velocity ratio between a
pair of parallel shafts by simply shifting the belt fromone step of the pulley to the other
i.e, When speed of the driven shaft is to be changedvery frequently
Used in lathe, drilling m/c etc..
Integral casting
One set of stepped cone pulley is mounted in reverse on the driven shaft
33
34
Fast and Loose pulley
When many machines obtain the drive from a
main driving shaft,
Run some machines intermittently without having
to Start and stop the main driving shaft
Fast pulley
Securely keyed or fixed firmly to the machine shaft
Loose pulley (with brass bush)
Mounted freely on the machine shaft Rotates/revolve freely
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Fast and Loose pulley
Dept. of Mech & Mfg. Engg. 36
WorkingWhen the belt is on the fast pulley, Power transmitted to the machine shaft
When machine shaft is to be brought to rest, Belt is shifted from fast pulley to loose pulley
Note:
1. Axial movement of the loose pulley towardsfast pulley is prevented .
2. Axial movement of the loose pulley awayfastpulley is also prevented.
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Jockey Pulley (rider pulley):
* Are used to Increase the arc of contact, the
tension & the power transmission.If
Center distance is small
One pulley is very small
Arc of contact small/less
The idler pulleys are mounted near the smaller pulleyand always ride on the slack side of the belt
38
Jockey Pulley
fk
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Dept. of Mech & Mfg. Engg. 39
Use:
To connect non-parallel shafts those which intersect and
those which do not intersect to guide the belt in to the
proper plane
When two shafts to be connected are close together
Guide pulley (Right angled drive)
Dept. of Mech & Mfg. Engg. 40
Guide pulley (Right angled drive)
Guide Pulley
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Dept. of Mech & Mfg. Engg. 41
The effect of groove is to increase the frictional grip
of the rope on the pulley & thus reduce the tendency to slip.
The groves are V-shaped.
Angle between 2 faces: 400 600
Uses:
Used in V-belts, rope.
Transmission of large powers over great distances
Grooved Pulley
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Dept. of Mech & Mfg. Engg. 43
Wrought-iron pulley
Light, strong and durable
Entirely free from initial strains
To facilitate the errection of pulleys on the main shaft,
they are usually made in halves and parts are securely
bolted together.
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46
Length of a belt (*)
Open belt drive:
h
(r1+ r2)L = + 2
+(r1 - r2)
2
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Let the two pulleys P and Q are connected by an open
belt.
From the centre C2 of the smaller pulley draw a line C2Gparallel to CD.
Let r1 ,r2 be the radii of larger and smaller pulleys P and Q.
Let X=Distance between the centers of the two pulleys.
From the geometry of the belt drive shown in fig., thelength of the belt is given by,
L=Arc Length ABC + Length CD + Arc Length DEF + LengthFA
=2[Arc Length BC+ Length CD + Arc Length DE]
=2[{/2 + } r1 + Length CD + {/2 - } r2]= 2 [{/2 + } r1 + Length GC2+ {/2 - } r2] ---(:CD=GC2)
=2[{/2 + }r1+X Cos + {/2- } r2 ]--- GC2/X=Cos47
=2[ /2 (r1+ r2 )+ (r1- r2) + X Cos](1).
= (r1+ r2 )+2 (r1- r2 )+2 X CosFrom the triangle GC1C2
Sin = (r1- r2 ) / X
=Sin-1 r1- r2/ X = r1- r2/ X (: is small).2.
Cos= [1- Sin2
]1/2
= [1-1/2 Sin2
] (By Binomial theorem andneglecting higher powers)
= [1-1/2{(r1- r2) / X }2
] ..3.
Substituting(2) and (3) in (1)
L= (r1+ r2 )+2 (r1- r2) / X (r1- r2) +2X [1- (r1- r2) 2/2X2 ]
= (r1+ r2 )+2 (r1- r2)2/X + 2X- (r1- r2)2/ X
L= (r1+ r2 )+ (r1- r2)2 / X + 2X *48
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49
Length of a belt (*)
L =e
+(r1+ r2)+ 2
(r1 + r2)2
Crossed belt drive:
50
Let the two pulleys P and Q are connected by an open belt.
From the centre C2 of the smaller pulley draw a line C2Gparallel to CD.
Let r1 ,r2 be the radii of larger and smaller pulleys P and Q.
Let X=Distance between the centers of the two pulleys.
From the geometry of the belt drive shown in fig., the lengthof the belt is given by,
L=Arc Length ABC + Length CD + Arc Length DEF + LengthFA
=2[Arc Length BC+ Length CD + Arc Length DE]
=2[{/2 + } r1 + Length CD + {/2 + } r2]= 2 [{/2 + }( r1 + r2) + Length CD]
= 2 [{/2 + }( r1 + r2) + Length GC2 ---- ---(:CD=GC2)
=
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= 2 [{/2 + } ( r1 + r2) + X Cos ] ..(:.GC2/2= Cos )
= [( +2 ) ( r1 + r2) + 2X Cos ] ..1
From the triangle GC1C2
Sin = (r1 + r2 ) / X
=Sin-1
r1 + r2/ X = r1 + r2 / X (: is small).2.
Cos= [1- Sin2
]1/2
= [1-1/2 Sin2
] (By Binomial theorem and neglecting higher powers)
= [1-1/2{(r1 + r2) / X }2
] ..3.
Substituting(2) and (3) in (1)
L = [ + 2{ (r1+ r2 ) / X } ] (r1+ r2 ) +2X [1- (r1+ r2)2 / 2X2
]
= (r1+ r2 )+2 (r1+ r2)2/X + 2X- (r1+ r2)2/ X
L = (r1+ r2 )+ (r1+ r2)2 / X + 2X.*
51
Dept. of Mech & Mfg. Engg. 52
Define Velocity Ratio of Belt Drive.(Speed Ratio)
The velocity ratio of a belt drive is defined
as the ratio of the speed of the driven
pulley to the speed of the driving pulley.
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Dept. of Mech & Mfg. Engg. 53
Obtain the expression for velocity ratio of belt
drive.
Letd1= Diameter of the driving pulley (mm)d2= Diameter of the driven pulley (mm)N1= Speed of the driving pulley (Revolutions/min
OR RPM)N2= Speed of the driven pulley (Revolutions/min
or RPM)
If there is no relative slip between the pulleys and theportions of the belt which are in contact with them,
the speed at every point on the belt will besame
54
The circumferential speeds of the driving and driven
pulleys and the linear speed of the belt are equal.
belttheof
dLinearspee
pulleydrivingtheof
speedntialCircumfere
pulleydrivenof
speedntialCircumfere= =
= d1N1 = d2N2
= d1N1 = d2N2
Velocity Ratio = N2 / N1 = d1 / d2
VelocityRatio pulleydrivingtheSpeedof
pulleydriventheofSpeed
pulleydriventheDiameterof
pulleydrivingtheofDiameter==
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Effect of Thickness of belt on the velocity ratio:
If THICKNESS OF THE BELT IS CONSIDERED,then
The circumferential speed should be the mean speedat the centre of the belt thickness.
[Linear speed
of the belt]= [Mean Circumferential speed of the driving pulley] =[Meancircumferential speed of the Driven pulley] .
= [(d1+t)N1] = [(d2+t)N2]
(d1+t)N1 = (d2+t)N2
Velocity ratio =N2/N1= d1+t /d2+t55
Dept. of Mech & Mfg. Engg. 56
Initial tension in belt drive
Definition
It is a uniform tension that exists initially when the drive
is not in motion. It is designated as To.
Formula:
To =T1 + T2
2
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57
Derive the expression for the ratio of tensions in
belt drive.
The driving pulley drives the driven pulley only if one sideof the belt has higher tension than the other side.
The figure shows a driven pulley rotating in clockwise
direction.
The polygon of forces acting on the element is represented by
the closed quadrilateral as shown in figure.
Consider a small element AB of belt,
T1= Higher tension,
T2= Lower tension,
= angle subtended by the element of AB
T =tension on the slack side of the belt.
= co-efficient of friction between the belt surface and pulley rim
58
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59
Let the tension in the tight side of the belt element AB
be greater than the slack side by T.
Therefore the tension in the tight side of the belt
element is T +T.
If R is the normal reaction exerted by the pulley on the
element of the belt. Then,
The force of friction R acts perpendicular to the
normal reaction R in the direction opposite to thedirection of motion as shown in figure.
Dept. of Mech & Mfg. Engg. 60
Element AB will be in equilibrium only whenfollowing forces act on it
1. Tension T on the slack side at A
2. Tension T +T on the tight side at B
3. Normal reaction R
4. Frictional force R acting perpendicular to R
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Dept. of Mech & Mfg. Engg. 61
Resolving all the forces in the direction of R.
R =
2
SinT + ( )
+ 2
SinTT
=
22
SinT +
2
SinT
For small angles the following assumptions can be made.
Sin /2 = /2 & Tx /2 is neglected.
R= 2T2
R =T ------------------------------ (1)
62
Resolving all the forces perpendicular to R
R = ( )
+
2
CosTT -
2
CosT
=
2
CosT +
2
CosT -
2
CosT
=
2
CosT
For small angles Cos /2 = 1
R = T ---------------------- (2)
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63
Substituting equation (1) in (2)
T = T
T
T=
Integrating between 0 and and tension T between T2 and T1
=
0
1
2
T
T T
T
log e2
1
T
T=
2
1
T
T= e
Where = Angle of contact in radians=Coefficient of friction.
Taking log of the previous eqn.,
logT1/T2 = log eWhere, e=2.718 Base of Napierian Logarithms
logT1/T2 = log 2.718
=0.4343
Dept. of Mech & Mfg. Engg. 64
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65
Slip
What is slip?
The sliding motion of the belt which causes a relative motion
between the pulley and the belt.
This occurs when the force to be transmitted by the driver is
greater than the force of friction.
(The difference between the actual speed of the driven pulley
and that calculated by the velocity ratio equation).
The driver pulley rotates along with the beltdue to a firm frictional grip between itssurface and the belt.
Sometimes this frictional grip is not sufficientso it causes some forward motion of thedriver pulley without carrying the belt with it.
Sometimes the belt moves faster in theforward direction, without carrying the driven
pulley with it. The difference between the linear speeds of
the pulley rim and the belt is a measure ofslip, this slip is expressed as a percentage.
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Effect of Slip on Velocity Ratio
The slip is expressed as percentage of the speed
Let S1,S2 be the % of slip between driving & drivenpulley and the belt
So total % of slip S = S1+S2
Circumferential speed of the driving pulley=d1N1
Considering the % Slip S1 between the drivingpulley and the belt passing over it,
Reduced Linear Speed of the belt because of the slip
S1 = d1N1 X [100-S1/100 ]
67
The circumferential speed of the belt on thedriven pulley when slip S2 occurs between thebelt and its rim is given by,
d2N2 =[speed of the belt on the Driven Pulley ] x [100-S2/100 ]
N2/N1=d1/d2 x [100-S /100 ]
If the thickness of the belt is considered, then
N2/N1=(d1+t)/(d2+t)/x [100-S /100 ]
68
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Dept. of Mech & Mfg. Engg. 69
Creep in Flat belt drive
The two tensions T2 &T1are not equal inmagnitude(T1>T2).Hence the stretch will be
different in different sides.
The phenomena of alternate stretchingand
contractionof the belt results in a relative
motion between the belt and the pulley
surface. This relative motionis called creep.
Dept. of Mech & Mfg. Engg. 70
This results in:
Loss of power
Decrease in the velocity ratio
Creep in Flat belt drive
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Dept. of Mech & Mfg. Engg. 71
Power transmitted in a belt drive
P = (T1-T2) * v
1,000kW
v= d N in m/sec
T1, T2 in Newtons
60*1000
Horse-power transmitted by a belt *
The effective pull of the belt is equal to
T1-T2=P
where P is called the driving force/driving tension
In kg.
If v is the linear velocity in metre per minute,then
Work transmitted per minute=Pxv kgfm.
Therefore, hp=Pxv/4500
If the belt passes over a pulley having the diameter d inmetre & which makes n revolutions per minute thenSpeed of belt,v=circumference x rev.per minute,
v=dn
Therefore hp=P x dn/4500 72
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Dept. of Mech & Mfg. Engg. 73
Disadvantages of flat belts
Power transmitted is less
Exact velocity ratio cannot be maintained
Slip & creep causes loss of power
Large power cannot be transmitted effectively
Dept. of Mech & Mfg. Engg. 74
V-Belt Drive
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Dept. of Mech & Mfg. Engg. 75
V-Belt Drive
Widely used form of belt drives in power transmission.
(0.5kW up to 150 kW)
They are made out of rubber & fibrous material.
They run in the V-grooves made in the pulleys & Power transmission can
be increased by using several belts placed side by side
The wedging action of the belts in the V-grooves enable
them to transmit high torques.
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Dept. of Mech & Mfg. Engg. 77
Transmit greater power
Permit large speed ratios
No slipping of the belt from the pulley
Maintenance is low
V-Belt Drive Advantages over flat beltdrive
Dept. of Mech & Mfg. Engg. 78
Problems on belt drive
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Dept. of Mech & Mfg. Engg. 79
L = (r1+ r2 )+ (r1- r2)2 / X + 2X---- (Open drive)
L = (r1+ r2 )+ (r1- r2)2
/ X + 2X---- (Cross drive)
logT1/T2 =0.4343 ( Ratio of Tension in belt
=Radians=X *( /180 )
, = Coefficient of friction)
Linear speed of the belt = d1N1 = d2N2
Velocity Ratio = N2 / N1 = d1 / d2Velocity ratio =N2/N1= d1+t /d2+t
Sin = (r1- r2 ) / X =Sin-1 r1- r2/ X
=180-2 = * ( /180 )
INITIAL TENSION To = (T1 + T2 ) / 2
Dept. of Mech & Mfg. Engg. 80
1) Simple problems like r1, r2 and center distance X will be
given, find out Length of open drive and cross rive melts)
2) The sum of diameters of two pulley is 1000mm and the
pulleys are connected by belt. If the pulleys rotate at 600 rpm
and 1800 rpm, det. The diameters of each pulley
3) In cross drive belt diff in tension between tight side and slack
side is 1200N, if the angle of contact () is 160 degree, and =
0.28, find tension in tight and slack side
4) The driven pulley of 400mm dia of a belt drive runs at
200rpm, the anle of lap is 165 deg, co-efficient of friction is 0.25,
find the power transmitted , if the initial tension is not to exceed
10 kN.
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Dept. of Mech & Mfg. Engg. 81
1) An electric motor provides 6 KW power to an open belt
drive. The diameter of the motor pulley is 200mm and it
rotates at 900 rpm. Calculate tight and slack side tension
in the belt if the ratio of tension is 2.
Solution:
P = 6kW
d1 = 200 mm
n1 = 900 rpm
2
1
T
T= 2.
82
P = 6kW, d1 = 200 mm, n1 = 900 rpm , 2
1
T
T
= 2.
Linear velocity of belt v =
1000.60
11Nd
=
1000.60
900.200.
= 9.425 m/sec
Power P =
T1 T2 =
v
P.1000=
425.9
6.1000= 636.6 N
1000
)(21 vTT
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Dept. of Mech & Mfg. Engg. 83
By data,
2
1
T
T= 2.
We get, 2T2 T2 = 636.6N
Slack side tension T2 =
Tight side tension T1= 2T2 = 1273.2 N
636.6 N
84
2) A leather belt transmits 20kW power from a pulley of
750mm diameter which runs at 500 rpm. The belt is
in contact with the pulley over an arc of 1600 and
the coefficient of friction between the belt and the
pulley is 0.3. Find the tension on each side of the
open belt drive.
Solution: P = 20 k W
d = 750 mm
n = 500 rpm
= 1600
= 0.3
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85
Linear velocity of belt v =1000.60
nd
=
1000.60
500.750.
= 19.635 m / sec
Power P =1000
)( 21 vTT
T1 T2 =v
p.1000
=635.19
20.1000= 1018.6 N .(1)
Dept. of Mech & Mfg. Engg. 86
By data,
2
1
T
T= e
= e ((0.3 ) (160) * ( /180 ))= 2.311 .(2)
From equations (1) and (2)
2.311 T2 T2 = 1018.6
Slack side tension T2 = 776.96 N
Tight side tension T1 = 776.96 (2.311) = 1795.6 N
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87
3) Power is transmitted by an open belt drive from a
pulley 300 mm. diameter running at 600rpm. to a pulley
500 mm. in diameter. The distance between the centre
lines of the shaft is 1m. and the coefficient of friction in
the belt drive is 0.25. If the safe pull in the belt is not to
exceed 500 N, determine the power transmitted by the
belt drive.
Solution: d1 = 300 mm.
n1 = 600 rpm
d2 = 500mm
c= 1m. = 1000 mm
= 0.25
T1 = 500 N
Dept. of Mech & Mfg. Engg. 88
Problems on slip in belt:-
The engine shaft running at 200 rpm, isrequired to drive a generator at 300 rpm, bymeans of a flat belt drive. Pulley on thedriving shaft has 500mm diameter, det. Thediameter of the pulley on the generator shaft ifthe belt thickness is 8mm and slip is 4%.
Back
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Dept. of Mech & Mfg. Engg. 89
Disadvantages of flat belts
Power transmitted is less
Exact velocity ratio cannot be maintained
Slip & creep causes loss of power
Large power cannot be transmitted effectively
Dept. of Mech & Mfg. Engg. 90
V-Belt Drive
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Dept. of Mech & Mfg. Engg. 92
V-Belt Drive Widely used form of belt drives in power transmission.
(0.5kW up to 150 kW)
They are made out of rubber & fibrous material.
They run in the V-grooves made in the pulleys & Power transmission can
be increased by using several belts placed side by side
The wedging action of the belts in the V-grooves enable
them to transmit high torques.
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Dept. of Mech & Mfg. Engg. 93
Transmit greater power
Permit large speed ratios
No slipping of the belt from the pulley
Maintenance is low
V-Belt Drive Advantages over flat beltdrive
Dept. of Mech & Mfg. Engg. 94
Chain Drives
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Dept. of Mech & Mfg. Engg. 95
Chain Drives
Dept. of Mech & Mfg. Engg. 96
Over comes the disadvantages of the beltdrive
Can be used from 3m to upto 8m centredistances (transmits upto 100kW)
Used in agricultural machinery, bicycles,motor cycles etc..
Two types of chains used in power transmission:1. Roller Chain2. Silent Chain
Chain Drives
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97
Roller Chain
Dept. of Mech & Mfg. Engg. 98
Silent Chain
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Dept. of Mech & Mfg. Engg. 99
Dept. of Mech & Mfg. Engg. 100
Roller Chain
A chain drive consisting of a chain and two sprockets
Widely used in low or medium speed power
transmission systems
This type of chain is employed in bicycles,
motorcycles, machine tools etc..
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Dept. of Mech & Mfg. Engg. 101
Consists of a series of toothed plates pinned
together in rows across the width of the chain
Advantage:
- Smooth and noiseless operation at high velocities
Silent Chain (Inverted Tooth Chain)
Dept. of Mech & Mfg. Engg. 102
Positive non-slip drives
Efficiency is high
Employed for small as well as large centre distances up to.
8m
Permit high velocity ratio
Transmit more power than belt drives
They produce less load on shafts compared to belt drives Maintenance is low
Chain Drive Advantages
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Dept. of Mech & Mfg. Engg. 103
Driving and driven shafts should be in perfect
alignment
Requires good lubrication
High initial cost
Chain Drive Disadvantages
Rope Drive
When centre distances are greater than 10 m
Power to be transmitted is more than 200 HP
Used in lifts, hoists etc
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Gear Drives
Dept. of Mech & Mfg. Engg. 106
Gear
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When considerable power has to be transmitted over ashort centre positively with a constant velocity ratio
gear drives are preferred.Gear drives possess a very prominent role in
mechanical power transmission.
A gear is a toothed wheel with the teeth cut on theperiphery of a cylinder or a cone, or on ellipticaldiscs.
They are mounted on the axles or shafts and keyed tothem.
Two gears are mounted on the individual shafts
107
What are the different types of gears used in
gear drives?1. Spur Gears - For Parallel Axes shafts.
2. Helical Gears - For both Parallel and Non-parallel
and non-intersecting axes shafts.
3. Spiral Gears - For Non-parallel and Non-intersectingaxes shafts.
4. Bevel Gears - For Intersecting Axes shafts.
5. Worm Gears - For Non-Parallel and Non-co-planaraxes shafts.
6. Rack and Pinion - Rotary motion For converting intolinear motion.
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109
Dept. of Mech & Mfg. Engg. 110
Spur Gears When the axes of the driving and driven shafts are
parallel and co-planar.
The teeth of the gear wheels are parallel to the axes
The contact between the mating gears will be
along a line ,Can transmit higher power.
Noise will be very high.
Applications:
o Machine tools,
o Automobile gear boxes and in
o All general cases of power transmission where geardrives are preferred.
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Dept. of Mech & Mfg. Engg. 111
Dept. of Mech & Mfg. Engg. 112
Spiral Gears
Used to connect onlytwo non-parallel, non-
intersecting shafts
There is a pointcontact in spiral gears
Because of the point contact the spiral gears are
more suitable for transmitting less power.
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Dept. of Mech & Mfg. Engg. 113
114
Helical Gears
Similar to the spur gears But teeth are cut in the form of the helix around the
gear
Used for transmitting power between two parallel
shafts and also between non -parallel, non-intersecting
shafts.
Contact between the mating gears will be along a
curvilinear path.
Helical gears are preferred to spur gears when smooth
and quiet running at higher speeds are necessary.
Generally they are used in automobile power
transmission.
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Dept. of Mech & Mfg. Engg. 115
Dept. of Mech & Mfg. Engg. 116
Bevel gears
Used when the axes of the two shafts are inclined
to one another, and intersect when produced.
Teeth are cut on the conical surfaces.
The most common examples of power
transmission are those in which the axes of the
two shafts are at right angles to each other.
When two bevel gears have their axes at rightangles and are of equal sizes, they are called
Miter gears.
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Dept. of Mech & Mfg. Engg. 117
118
Rack and Pinion
Used when a rotary motion is to be converted intoa linear motion.
Rack is a rectangular bar with a series of straightteeth cut on it.
Theoretically rack is considered to be a spur gearof infinite diameter.
Application:
Machine tools, such as, lathe, drilling, planingmachines,
Some steep rail tracks, where the teeth of thelocomotive wheel mesh with a rack embedded inthe ground, offering the locomotive improvedtraction.
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119
Worm Gear
Worm gearing is a Special form of screw gearing,i.e Its action is like screw and nut.
This type of gearing is used for transmitting power between non-intersecting and
non- paraIlel shafts. This is a gearing in which teeth have a line contact.
Axes of driving and driven shafts are at right angles.
A simple worm gear combination consists of a screw meshing with a helical gear.
Velocity ratio does not depend upon the diameters of the worm and gear but upon
the ratio of no. of teeth on the worm gear to the no. of threads on the worm.
Wormgearing is used to provide high angular velocity reduction.
Used in machine tools like Lathe, Drill, Milling to get large velocity ratio.
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What are the Advantages and Disadvantages of
Gear Drives?
Advantages;1. They are positive non-slip drives.
2. Most convenient for very small centre distances.
3. By using different types of gears, it will be possible to
transmit the power when the axes of the shafts are not
only parallel, but even when non-parallel, intersecting,
non-intersecting and co-planar or non-coplanar.
4. The velocity ratio will remain constant throughout. '5. They can be employed conveniently for low, medium and
high power transmission.
Dept. of Mech & Mfg. Engg. 122
6. Any velocity ratio as high as, even upto 60 : 1 canbe obtained.
7. They have very high transmission efficiency.
8. Gears can be cast in a wide range of both metallic
and non-metallic materials.
9. If required gears may be cast integral with the
shafts.
10. Gears are employed for wide range of applications
like in watches, precision measuring instruments,
machine tools, gear boxes fitted in automobiles,
aero engines, etc.
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123
Disadvantages
1. They are not suitable for shafts of very largecentre distances.
2. They always require some kind of lubrication.
3. At very high speeds noise and vibrations will be
more.
4. They are not economical because of the
increased cost of production of precision gears.
5. Use of large number of gear wheels in gear
trains increases the weight of the machine.
Figure:
Gear Tooth Profile
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Spur Gear elements
Figure:
Circular pitch (p): It is the distance from a point on one teethto the corresponding point on the next tooth measured alongthe pitch circle.
If T is the number of teeth on a gear and d is the diameter ofthe pitch circle,
Circular pitch, pc= d / T, where T is the no.of teeth on gear.
Module(m): It is the ratio of the pitch circle diameter of a gearto the number of teeth on a gear
m=d/TDiametral pitch (Pd)= Ratio of module ie., 1/m is the Pd.Pd= T/d
Spur gear elements
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Dept. of Mech & Mfg. Engg. 127
Define velocity ratio of Gear drive
The velocity ratio of a gear drive is defined as
the ratio of the speed of the driven gear to the
speed of the driving gear.
Dept. of Mech & Mfg. Engg. 128
Obtain an expression for gear drive.Let
d1 = pitch circle diameter of the driving gear
d2 = pitch circle diameter of the driven gear
T1 = Number of teeth on the driving gear
T2 = Number of teeth on the driven gear.
N1 =speed of the driving gear in revolutions
per minute.
N2 = speed of the driven gear in revolutions
per minute.
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129
Since there is no slip between the pitch cylinders of the two
gear wheels,
The linear speed of the two pitch cylinders must be equal.
gearDrivingthengrepresenticylinder
pitchtheofspeedLinear
=
geardrivenngrepresenticylinder
pitchtheofspeedLinear
d1N1 = d2N2
1
2
N
N=
2
1
d
d
(1)
The circular pitch for both the meshing gears remains same.
i.e. pc =1
1
T
d=
2
2
T
d
Dept. of Mech & Mfg. Engg. 130
i.e.,
2
1
d
d
2
1
T
T
= ..(2)
From equations (1) and (2)
Velocity Ratio of a Gear Drive =1
2
N
N=
2
1
d
d=
2
1
T
T
Velocity ratio of the worm and worm wheel is expressed as:
Speed of the WormSpeed of the WormWheel
Number of Teeth on Worm
Wheel
Number of Threads on theWorm
==Velocityratio
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Dept. of Mech & Mfg. Engg. 131
A gear train is an arrangement of number of
successively meshing gear wheels through
which the power can be transmitted between
the driving and driven shafts.
What do understand by a gear train?
132
The gear wheels used in gear train may be spur ,
bevel or helical etc.
The different types of gear trains are:
1. Simple gear train.
2. Compound gear train.
3. Reverted gear train.
4. Epicyclic Gear train.
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133
Simple Gear Train
The idler gears act as the intermediategears to establish the drive between thedriver and the driven gears , but theychange the direction of rotation of thedriven gear.
Even number of idler gears will rotate thedriven gear in the opposite direction to thatof the driving gear.
Odd number of idler gears will rotate thedriven gear in the same direction as that ofthe driving gear.
134
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Dept. of Mech & Mfg. Engg. 135
z
z
z Simple Gear Train
Dept. of Mech & Mfg. Engg. 136
Gear meshing
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Draw a neat sketch of a simple gear train and derive anexpression for the velocity ratio of the same.
In a simple gear train a series
of gear wheels are mounted
on different shafts between
the driving and driven shafts
each shaft carrying only one
gear.
A Driving gear
B - Intermediate gear (idler gear)
C- Intermediate gear(idler gear)
D Driven gear
D
C
Dept. of Mech & Mfg. Engg. 138
LetNA = speed in RPM of gear A
NB = speed in RPM of gear B
NC = speed in RPM of gear C
ND = speed in RPM of gear D
TA = Number of teeth of gear A
TB = Number of teeth of gear B
TC = Number of teeth of gear C
TD = Number of teeth of gear D
D
C
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Dept. of Mech & Mfg. Engg. 139
i. A drives B
ii. B drives C
iii. C drives D
A
B
N
N
=B
A
T
T
B
C
N
N=
C
B
T
T
C
D
N
N=
D
C
T
T
D
C
A
Dept. of Mech & Mfg. Engg. 140A
D
N
N
D
A
T
T=Velocity Ratio
.
.
B
A
TT
C
B
TT
D
C
TT= . .
Velocity Ratio =
A
D
N
N
Substituting from (i), (ii) and (iii).
.
A
D
N
NVelocity Ratio =
C
D
N
N
B
C
N
N
A
B
N
N= . .
Velocity ratio between the driving and driven gears is given by,
D
C
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Dept. of Mech & Mfg. Engg. 141
Draw a neat sketch of a compound gear train and derive
an expression for the velocity ratio of the same.
A compound gear train isone in which each shaft
carries two or more gears
and keyed to it.
When Velocity Ratio isvery high simple geartrain found difficultbecause of small centerdistance.
Gear B,C Compound gear
Dept. of Mech & Mfg. Engg. 142
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143
Gear A drives B,
B
A
A
B
T
T
N
N= .(1) A
D
Since gears B and C are keyedto the same shaft,
NB = Nc but TB Tc
Gear C drives D,
D
C
C
D
T
T
N
N=
Both of them rotate at thesame speed
(2)
c
B
144
Velocity ratio between driving and driven gear
C
D
A
D
N
N
N
N= .
A
C
N
N=
Substituting from (1) and (2)
D
C
A
D
TT
NN =
B
A
TTVelocity ratio = .
A
D
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Advantages of Gear Drives
1. It is a positive drive and is used to connectclosely spaced shafts.
2. High efficiency, compactness, reliable service,more life, simple operation and lowmaintenance.
3. It can transmit heavier loads than other drivesand can be used where precise timing is
desired.
Dept. of Mech & Mfg. Engg.
145
Disadvantages: 1. They are not suitable for large centre
distance because the drive becomesbulky.
2. High production cost.
3. Due to errors and inaccuracies in theirmanufacture, the drive may become noisyaccompanied by vibrations at high speeds.
Dept. of Mech & Mfg. Engg. 146
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A counter shaft has a pulley 1200 mm. diameter keyed to it and it is to
have a speed of200 rpm. It is to be driven by an electric motor whichhas a speed of 1000 rpm. What diameter pulley should be fitted to the
electric motor? Find the velocity ratio and the speed of the belt. Solution:
d2= 1200 mm, n2= 200 rpm, n1= 1000 rpm.
Dept. of Mech & Mfg. Engg. 147
Dept. of Mech & Mfg. Engg. 148
Gear Drive Problems
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FORMULAS TO BE USED
SIMPLE GEAR TRAIN:-
COMPOUND GEAR TRAIN
Circular pitch, pc= d / T, where T is the no.of teeth ongear, d= Pitch circle diameter
Train value = 1/ velocity ratioDept. of Mech & Mfg. Engg. 149
A
D
N
N
D
A
T
T
=Velocity Ratio
D
C
A
D
T
T
N
N=
B
A
T
TVelocity ratio = .
Module m = d mmT
Dept. of Mech & Mfg. Engg. 150
CONDITIONS TO SOLVE PROBLEM
1. To get maximum velocity ratio Select the
driving gear with maximum number of teeth
& diven gear with minimum number of teeth
2.To get minimum velocity ratio Select the
driving gear with minimum number of teeth
& diven gear with maximum number of teeth
3.Mesh two gears with same diametral pitch
4. spur gear has to be meshed with spur gear and
helical gear has to be meshed with helical gears
But spur gear can be compounded with helical gear with
Same shaft.
Maximum velocity ratio will be obtained only in compound
Gear ratio.
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Dept. of Mech & Mfg. Engg. 151
1) A gear wheel has 50 teeth of module 5mm. Find the pitch
circle diameter and the circular pitch.
Given:
module, m= 5mm
T or z= 50
Pc=?
d =?
Solution:
Module m = dT
5 = d50
d=250 mm
Dept. of Mech & Mfg. Engg. 152
Circular pitch, pc= dT
pc= 25050
pc=15.7 mm
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Dept. of Mech & Mfg. Engg. 153
2) Two gear wheels having 80 teeth and 30 teeth mesh
with each other. If the smaller gear wheel runs at 480
rpm, find the speed of the larger wheel.
Given:
Larger Gear wheel Smaller Gear Wheel
T1= 80 T2= 30
N1=? N2=480 rpm
Dept. of Mech & Mfg. Engg. 154
Velocity ratio of a gear drive,
=N1 T2
T1N2
= N2T1
T2N1 X
= 48080
30X
= 180 rpm
Solution:
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Dept. of Mech & Mfg. Engg. 155
3) A gear wheel of 20 teeth drives another gear wheel
having 36 teeth running at 200 rpm. Find the speed
of the driving wheel and the velocity ratio.
Given:
Driving wheel Driven Wheel
T1= 20 T2= 36
N1=? N2=200 rpm
velocity Ratio = ?
Dept. of Mech & Mfg. Engg. 156
=T2
T1
N1
N2
= N2T1
T2N1 X
= 360 rpm
Velocity Ratio =N1
N2
= 1:1.8
Solution:
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Dept. of Mech & Mfg. Engg. 157
4) In a simple train of gears, A has 30 teeth, B has 40
teeth, C has 60 teeth and D has 40 teeth. If A makes
36 rpm, find the rpm of the gear C and D.
Given:
TA= 30, TB= 40, TC= 60, TD= 40, NA=36 rpm
NC=? ND=?
Dept. of Mech & Mfg. Engg. 158
Given:
TA= 30, TB= 40, TC= 60, TD= 40, NA=36 rpm
NC=? ND=?
=NA TC
TANC
Solution:
=NA TD
TAND
= NATC
TANC X = 18 rpm
= NATD
TAND X = 27 rpm
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Dept. of Mech & Mfg. Engg. 159
5) A compound gear consists of 4 gears A,B,C and D
and they have 20, 30, 40 and 60 teeth respectively. A
is fitted on the driver shaft, and D is fitted on the driven
shaft , B and C are compound gears, B meshes with
A, and C meshes with D. If A rotates at 180 rpm, find
the rpm of D.
Dept. of Mech & Mfg. Engg. 160
Given:
TA= 20, TB= 30, TC= 40, TD= 60, NA=180 rpm
ND=?
=NA
ND
TD
TC XTB
TA
=NDTD
TC NAXTB
TA X
= 80 rpm
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Dept. of Mech & Mfg. Engg. 161
6) A compound gear train consists of 6 gears A,B,C,D,E
and F, they have 20, 30, 40, 50, 60, 70 teeth
respectively. A is fitted to the first shaft and meshed with
B. B and C are fitted to the second shaft and C is
meshed with D. D and E are fitted to the third shaft and
E is meshed with F which is fixed to another shaft. A
rotatates at 210 rpm, find speed of F.
Dept. of Mech & Mfg. Engg. 162
Given:
TA= 20, TB= 30, TC= 40, TD= 50, TE=60, TF=70,
NA=210 rpm
NF=?
cv
Compound train of wheels
Gear AGear F Gear B,C
Gear D,E
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Dept. of Mech & Mfg. Engg. 163
F A
A B
N T
N T= . C
D
T
T
E
F
T
T.
Given:
TA= 20, TB= 30, TC= 40, TD= 50, TE=60, TF=70,
NA=210 rpm
NF=?
20
30
F
A
N
N= . 40
50
60
70.
= 96 rpmNF
cv
Solution:
Dept. of Mech & Mfg. Engg. 164
7) A compound gear train is formed by 4 gears P,Q,R and S.
Gear P meshes with gear Q and gear R meshes with
gear S. Gears Q and R are compounded. P is connected
to driving shaft and S connected to the driven shaft and
power is transmitted. The details of the gear are,
Gears P Q R S
No. of Teeth 30 60 40 80
If the gear S were to rotate at 60 rpm. Calculate the
speed of P. Represent the gear arrangement
schematically.
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Dept. of Mech & Mfg. Engg. 165
SOLUTION:
Velocity ratio =
S
R
P
S
T
T
N
N
=.
Q
P
T
T
PN
60=
80
40 .60
30
Speed of P, NP = 240 rpmGear R,TR=40
Gear P,TP=30
Gear Q,TQ=60
Gear S,TS=80
Gear arrangement
Dept. of Mech & Mfg. Engg. 166
8) A gear train consists of six gears A,B,C,D,E,F and
they contain 20,30,40,50,80,100 teeth respectively.
Show the arrangement of the gears to obtain
i. Maximum velocity ratio
ii. Speed reduction of 6.
N
cv
AD C
B
FE
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Dept. of Mech & Mfg. Engg. 167
A D E F
D B C A
N T T T
N T T T=
50 80 100
40 30 20
A
D
N
N= = 16.66
Solution:i. Maximum velocity ratio
cv
A(driven)D(driver) F
E
CB
Given:
A= 20
B= 30
C= 40
D= 50
E= 80
F= 100
168
ii. Speed reduction of 6.
cv
A
FC
E
D
CF A D
A B E F
TN T T
N T T T=
20 40 50
30 80 100
A
D
N
N= = 1/6
Given:
A= 20
B= 30
C= 40
D= 50
E= 80
F= 100
Driver
Driven
B
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Dept. of Mech & Mfg. Engg. 169
9)Five spur gears of 20,30,50,80 and 120 teeth are
available along with 3 helical gears of 30, 60 and 80
teeth of all having same diametral pitch. Show the
arrangement of gears to get maximum possible
velocity ratio using maximum numbers of wheels
from the above set of gears.
Dept. of Mech & Mfg. Engg. 170
Given: S1= 20 H1= 30S2= 30 H2= 60
S3= 50 H3= 80
S4= 80
S5= 120
Maximum possible velocity ratio = ?
(Max)120
V.R. =20
80
30
80
30
= 42.6
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Dept. of Mech & Mfg. Engg. 171
Dept. of Mech & Mfg. Engg. 172
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Dept. of Mech & Mfg. Engg. 173
10) There are 5 wheels having 20, 40, 60, 80 and 100
teeth with a diametral pitch of 3 and another set of 4
wheels of diametral pitch of 2 having 20, 40, 60 and 70
teeth. Sketch an arrangement to get maximum velocity
ratio using maximum number of wheels from the
above lot. Also mention the conditions used.
Dept. of Mech & Mfg. Engg. 174
(Max) 100V.R. =20
= 52.5
6040
Maximum number of wheels used = 8
Conditions:
1. Select the driving gear with maximum number of teeth &
driven gear with minimum number of teeth
2. Mesh two gears with same diametral pitch
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Dept. of Mech & Mfg. Engg. 175
Dept. of Mech & Mfg. Engg. 176
11) Two parallel shafts are to be connected by a gear
drive. They are very nearly 1m apart and their velocity
ratio is to be exactly 9:2. If the pitch of the gears is 57
mm, find the number of teeth in each of the two
wheels and distance between the shafts.
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Dept. of Mech & Mfg. Engg. 177
Let d1 and d2 be the pitch diameter of the 2 gears
Distance between the gears = = 1m
Given=
d2
d1
N1
N2
2
9=
.1
.22
9=d1 d2
d1=1636 mm
d2=363.6 mm
d1+d22
178
d1=1636 mm d2=363.6 mm
Circular pitch, Pc= d1T1 =d2
T2 =57 mm
=57
T1d1
=90.18
=57
T2d2
=20.04
Take T1 and T2 as 90 and 20 respectively.
PcT1
d1= =1633 mm
PcT2
d
2= =363 mm
Exact centre distance =d1+d2
2
=998 mm = 0.998m
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Gears with following parameters are available
Gear A has a module of 2 and 50 teeth, Gear B has a
diameter of 201mm and 67 teeth, Gear C has a moduleof 4 and 152mm diameter, Gear D has a diameter of50mm and 25 teeth, Gear E has a module of 3 and 100teeth, Gear F has a (Gear B has a diameter of 201mmand 67 teeth) module of 4 and is 350mm in diameter,Gear G has a module of 3 and diameter of 126mm, GearH has a diameter of 60mm and 30 teeth, Gear J has55teeth and a diameter of 110mm.Determine anarrangement to obtain the lowest speed possible forthe driven shaft , if the power is transmitted with fourshafts? If the driver gear rotates at 225rpm, determine
the speed of the driven shaft.
Dept. of Mech & Mfg. Engg. 179
GEARS A&B, C & D are integral & concentric
wheels. The number of teeth are Ta=80,Tb=66, Tc=55, Td=108, the wheel B & C meshwith each other. Wheel D drives wheel Ehaving Te=60 teeths. If wheel A runs at 400rpm, det. Speed of E.
Ans:-
Ne=864 rpm
Dept. of Mech & Mfg. Engg. 180
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Dept. of Mech & Mfg. Engg. 181
Dept. of Mech & Mfg. Engg. 182
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Dept. of Mech & Mfg. Engg. 183
Dept. of Mech & Mfg. Engg. 184
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Dept. of Mech & Mfg. Engg. 185
Dept. of Mech & Mfg. Engg. 186
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Dept. of Mech & Mfg. Engg. 187
Dept. of Mech & Mfg. Engg. 188
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Dept. of Mech & Mfg. Engg. 189
Dept. of Mech & Mfg. Engg. 190
Wrought-iron pulley
Light, strong and durable
Entirely free from initial strains
To facilitate the errection of pulleys on the main shaft,
they are usually made in halves and parts are securely
bolted together.
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