mes-power transmission final [read-only].pdf

7/27/2019 MES-POWER TRANSMISSION final [Read-Only].pdf

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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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6

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


Belt Drive


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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.



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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


Types of Flat belt drives:

1. Open belt drive

2. Crossed-belt drive

Flat Belt


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Open belt drive

It is employed when the two parallel shafts have to rotate

in the same direction.


Open belt drive


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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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Crossed belt drive


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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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.


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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Stepped cone pulley



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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


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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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


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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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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 [{/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


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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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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2

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


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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2

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.


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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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



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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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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.


This results in:

Loss of power

Decrease in the velocity ratio

Creep in Flat belt drive


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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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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


V-Belt Drive


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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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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


Problems on belt drive


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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


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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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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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)


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


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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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


V-Belt Drive


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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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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


Chain Drives


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Chain Drives


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


Silent Chain


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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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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)


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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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


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


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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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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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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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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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.


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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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.


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


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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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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z

z

z Simple Gear Train


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


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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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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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



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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.



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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.



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


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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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


Circular pitch, pc= dT

pc= 25050

pc=15.7 mm


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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


Velocity ratio of a gear drive,

=N1 T2

T1N2

= N2T1

T2N1 X

= 48080

30X

= 180 rpm

Solution:


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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 = ?


=T2

T1

N1

N2

= N2T1

T2N1 X

= 360 rpm

Velocity Ratio =N1

N2

= 1:1.8

Solution:


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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=?


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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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.


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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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.


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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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:


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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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


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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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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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.


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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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.


(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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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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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.


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



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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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mes-power transmission final [read-only].pdf

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