TRANSMISSION OF POWER

1Dept. of Mech & Mfg. Engg.

TRANSMISSION OF POWER


What is a transmission system?

The rotational motion can be transmitted from one mechanical element to the other with the help of certain systems known as transmission system (Drive).


The methods of power transmission are (Types of drives)

i. Belt driveii. Chain drive.iii. Gear drive.iv. Rope drive.


Open belt drive It is employed when the two parallel shafts have to rotate in the same direction.


Open belt drive


Open belt drives

– When the shafts are placed far apart, the lower side of the belt should be the tight side and 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 thus increases the arc of contact.


Open belt drive


Flat belt drives of the open systemshould always have:

• Their shaft axes either horizontal or

inclined.

• They should never be vertical


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.


Crossed belt drive


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.


What is crowning in a pulley?

When the flat belt on cylindrical pulley is off-center

and the pulley rotating the belt quickly moves up to the

largest radius at the top of the 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.


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.


Benefits of Crowning of a 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

crowning.


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

Min. dia. of the belt should be at least 25 times thickness of

the belt used to run the pulley


Types of pulleys

• Stepped cone pulley (Speed cone)

• Fast and loose pulleys

• Guide pulley (Right angled drive)

• Jockey pulley

• Grooved pulley

• Wrought-iron pulley


Stepped cone pulley

• When speed of the driven shaft is to be changed

very frequently

• Used in lathe, drilling m/c etc..

• Integral casting

• One set of stepped cone pulley mounted in

reverse on the driven shaft

Link


Stepped cone pulley

f


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 to the machine shaft

Loose pulley (with brass brush)• Mounted freely on the machine shaft• Rotates freely


Fast and Loose pulley


WorkingWhen the belt is on the fast pulley,

– Power transmitted to the machine shaftWhen machine shaft is to be brought to rest,

– Belt is shifted from fast pulley to loose pulleyNote:1. Axial movement of the loose pulley towards

fast pulley is prevented

2. Axial movement of the loose pulley away fast pulley is prevented


Jockey PulleyIf • Center distance is small• One pulley is very small• Arc of contact smallThen■ Use idler pulley■ Placed on the slack side of the beltResult* Increase in arc of contact* Increase in tension* Increase in power transmission


Jockey Pulley

fk


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)


Guide pulley (Right angled drive)

Guide Pulley


The effect of groove is to increase the frictional grip

of the rope on the pulley.

This reduces 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


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.


Length of a beltOpen belt drive:

h(r1+ r2)L = + 2

+ (r1 - r2)2

29

Length of a belt

L =e

+(r1+ r2) (r1 + r2)2

Crossed belt drive:


Effect of sum of pulley diameter on the length of belt for open type

• Any variation in which (r1+ r2) is kept constant will vary

the length of the belt because of the term containing r1- r2

• If speed cones are connected by an open belt, the belt

will be slacker in some position than in others


Effect of sum of pulley diameter on the length of belt for crossed type

• Here r1 and r2 only occur in the form of sum,

• If sum is kept constant by varying r1 and r2, length of

the belt will be constant

In speed cones:

They are connected by crossed belt, hence

– Length of belt remains constant

– Sum of diameters of the corresponding steps should

be constant.


Equation for arc of contact (angle of lap) for open belt drive (theta)

= =Cos 2

r2-r1

Dd2-d1

2D

Dia. of the larger pulley - Dia. of the smaller pulley

Centre distance between the two pulleys=


Equation for arc of contact (angle of lap) for crossed belt drive (theta)

2

- = =r2 + r1

Dd2 + d1

2DCos

Dia. of the larger pulley + Dia. of the smaller pulley

2 Centre distance between the two pulleys=


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.


Obtain the expression for velocity ratio of belt drive. Let d1= 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 the portions of the belt which are in contact with them

The speed at every point on the belt will be same


The circumferential speeds of the driving and driven pulleys and the linear speed of the belt are equal.

belttheof

dLinearspee

pulleydrivingtheofspeedntialCircumfere

pulleydrivenof

speedntialCircumfere= =

= d1N1 = d2N2

= Πd1N1 = Πd2N2

Velocity Ratio = N2 / N1 = d1/d2

Velocity Ratio pulleydrivingtheSpeedof

pulleydriventheofSpeedpulleydriventheDiameterofpulleydrivingtheofDiameter==


Initial tension in belt drive

DefinitionIt is a uniform tension that exists initially when the drive

is not in motion. It is designated as To.

Formula: To =

T1 + T2

2


The polygon of forces acting on the element is represented by the closed quadrilateral as shown in figure.


Derive the expression for the ratio of tensions in belt drive. The driving pulley drives the driven pulley only if one side of the belt has higher tension than the other sideThe figure shows a driving pulley rotating in clockwise directionConsider a small element AB of belt,

T1= Higher tension, T2= Lower tension, δθ = angle subtended by the element of AB T =tension on the slack side of the belt. μ = co efficient of friction between the belt surface and

pulley rim


Let the tension in the tight side of the belt element AB

be greater than the slack side by δT.

Therefore the tension in the tight side of the belt

element is T +δT.

If R is the normal reaction exerted by the pulley on the

element of the belt.

Then,

The force of friction μR acts perpendicular to the

normal reaction R in the direction opposite to the

direction of motion as shown in figure.


Element AB will be in equilibrium only when following forces act on it

1. Tension T on the slack side at A

2. Tension T +δT on the slack side at A

3. Normal reaction R

4. Frictional force μR acting perpendicular to R


Resolving all the forces in the direction of R.

R =

2SinT +

2

SinTT

=

22 SinT +

2SinT

For small angles the following assumptions can be made.

Sin δθ/2 = δθ/2 & δT. δθ /2 is neglected.

R= 2T2

R =T δθ ------------------------------ (1)


Resolving all the forces perpendicular to R

μR =

2CosTT -

2

CosT

=

2CosT +

2CosT -

2CosT

=

2CosT

For small angles Cos δθ/2 = 1

μR = δT ---------------------- (2)

Substituting equation (1) in (2)

μT δθ = δT


Substituting equation (1) in (2)

μT δθ = δT

TT = μ δθ

Integrating δθ between 0 and θ and tension δT between T2 and T1

0

1

2

T

T TT

log e

2

1

TT

= μθ

2

1

T

T = e μ θ


Slip

What is slip?

–The sliding motion of the belt which

causes a relative motion between the

pulley and the belt.δT= μR The equation is,


Creep in Flat belt drive

• The phenomena of alternate stretching

and contraction of the belt results in a

relative motion between the belt and the

pulley surface.

• This relative motion is called creep.


This results in:

–Loss of power

–Decrease in the velocity ratio

Creep in Flat belt drive


Power transmitted in a belt drive

P = (T1-T2) * v

4500 HP

v= d N in m/min

T1, T2 in kgf


Power transmitted in a belt drive

P = (T1-T2) * v60,000

kW

v= d N in m/min

T1, T2 in Newton


What are the different types of gears used in gear drives? Explain. The different types of gears used are: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-intersecting

axes shafts.

4. Bevel Gears - For Intersecting Axes shafts.

5. Worm Gears - For Non-Parallel and Non-co-planar axes shafts.

6. Rack and Pinion - For converting Rotary motion into linear motion.



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 gear

drives are preferred.


Spiral Gears

• Used to connect only two non-parallel, non-

intersecting shafts

• There is a point contact in spiral gears

• Because of the point contact the spiral gears are

more suitable for transmitting less power.


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.Disadvantage:It produces end thrusts on the driving and driven shafts.


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 right angles and are of equal sizes, they are called Miter gears.


Rack and Pinion • Used when a rotary motion is to be converted into

a linear motion.• Rack is a rectangular bar with a series of straight

teeth cut on it.• Theoretically rack is considered to be a spur gear

of infinite diameter. Application:• Machine tools, such as, lathe, drilling, planing

machines,• Some steep rail tracks, where the teeth of the

locomotive wheel mesh with a rack embedded in the ground, offering the locomotive improved traction.


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 can be 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.


Disadvantages1. They are not suitable for shafts of very large

centre 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.


Define pitch and module of spur gear.

Circular pitch(p): It is the distance from a point on one teeth to the corresponding point on the next tooth measured along the pitch circle. Module(m): It is the ratio of the pitch circle diameter of a gear to the number of teeth on a gear.

i. e. m=d/Z


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 gearT2 = 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.


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

π d1N1 = π d2N2

1

2

NN

=

2

1

dd

……………… (1)

The circular pitch for both the meshing gears remains same.

i.e. pc =

1

1

Td

=2

2

Td


i.e.,

2

1

dd

2

1

TT

= ……………………..(2)

From equation (1) and (2)

Velocity Ratio of a Gear Drive =1

2

NN

=2

1

dd

=2

1

TT

Velocity ratio of the worm and worm wheel is expressed as:

Speed of the WormSpeed of the Worm Wheel

Number of Teeth on Worm WheelNumber of Threads on the Worm

==Velocity ratio


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?


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.


Draw a neat sketch of a simple gear train and derive an expression 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 gear carrying only one gear.

A → Driving gear B → Intermediate gear C → Intermediate gear

D → Driven gear Simple gear train

Gear AGear B

Gear C

Gear D


Let NA = speed in RPM of gear ANB = speed in RPM of gear BNC = speed in RPM of gear CND = speed in RPM of gear D

TA = Number of teeth of gear ATB = Number of teeth of gear BTC = Number of teeth of gear CTD = Number of teeth of gear D Simple gear

train

Gear AGear B

Gear C

Gear D


i. A drives B

ii. B drives C

iii. C drives D

A

B

NN

=B

A

TT

B

C

NN

=C

B

TT

C

D

NN

=D

C

TT

Simple gear train

Gear AGear B

Gear C

Gear D

70Dept. of Mech & Mfg. Engg.A

D

NN

D

A

TT

= Velocity Ratio

.

.

B

A

TT

C

B

TT

D

C

TT

= . .

Velocity Ratio = A

D

NN

Substituting from (i), (ii) and (iii) .

.

A

D

NN

Velocity Ratio =

C

D

NN

B

C

NN

A

B

NN

= . .

Velocity ratio between the driving and driven gears is given by,

Simple gear train

Gear AGear B

Gear C

Gear D


Draw a neat sketch of a compound gear train and derive

an expression for the velocity ratio of the same.

A compound gear train is one in which each shaft carries two or more gears and keyed to it.

Gear B →Compound gearGear C →Compound gear

Gear C

Gear A

Gear B

Gear D

Compound gear train


Gear A drives B,

B

A

A

B

TT

NN

……….(1)

Gear C

Gear A

Gear B

Gear D

Compound gear train

Since gears B and C are keyed to the same shaft,

NB = Nc but TB Tc

Gear C drives D,

D

C

C

D

TT

NN

Both of them rotate at the same speed

………(2)


Gear C

Gear A

Gear B

Gear D

Compound gear train

Velocity ratio between driving and driven gear

C

D

A

D

NN

NN

. A

C

NN

=

Substituting from (1) and (2)

D

C

A

D

TT

NN

B

A

TT

Velocity ratio = .


PROBLEMS1) 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 SNo. of

Teeth30 60 40 80

If the gear S were to rotate at 60 rpm. Calculate the speed of P. represent the gear arrangement schematically.


SOLUTION:

Velocity ratio =

S

R

P

S

TT

NN

.

Q

P

TT

PN60

=

8040 .

6030

Speed of P, NP = 240 rpmGear R,TR=40

Gear P,TP=30

Gear Q,TQ=60

Gear S,TS=80

Gear arrangement


2) 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

TT

= 2.


P = 6kW, d1 = 200 mm, n1 = 900 rpm , 2

1

TT

= 2.

Linear velocity of belt v =

1000.6011 Nd

=

1000.60900.200.

= 9.425 m/sec

Power P =

10000)( 21 vTT

T1 – T2 =

vP.1000

= 425.9

6.1000 = 636.6 N


By data,2

1

TT

= 2.

From equations, (1) & (2)

We get, 2T2 – T2 = 636.6N

Slack side tension T2 = 636.6 N

Tight side tension T1= 2T2 = 1273.2 N


3) 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



1000.60nd

=1000.60

500.750.

= 19.635 m / sec

Power P =1000

)( 21 vTT

T1 – T2 = vp.1000

=

635.1920.1000

= 1018.6 N …………….(1)


By data,

2

1

TT

= 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


4) 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



1000.60nd =

1000.60600.300.

Radius of driver pulley r1 =

21d

=

2300

= 150 mm.

Radius of driven pulley r2 =

22d

= 2

500= 250 mm.

Angle of lap on smaller pulley = - 2 sin -1

crr 12

(because r2 – r1)

= - 2 sin -1

1000150250

= 2.94 rad.


Power P =

100021 vTT

P =

1000425.98.239500

= 2.425 kW

Slack side tension T2 = eT1

085.2500

= 239.8 N

Ratio of tensions2

1

TT= e

= e (0.25 ) . (2.94) = 2.085


Higher tension is in the tight side of the belt and

Lower tension is in the slack side of the belt.

Back


The centrifugal force developed in the belt

combined with the force of gravity causes

the belt to stretch and tend to leave the rim

of the pulleys, thereby losing contact with

their rim surfaces.

Back


Integral casting– Single component having 3 or 4 pulleys of

different sizes one adjacent to another

How speed of driven shaft can be varied?Answer:By shifting the belt from one pair of pulley to other.

Back

88Dept. of Mech & Mfg. Engg. Back

How?

1.Abutting its (loose pulley) boss to that of fast pulley

2.Collar fixed to the machine shaft

TRANSMISSION OF POWER

Documents

belt crosses

open belt drives

flat belt drives

belt guide rollers

belt edge cupping

crossed belt driveit

crowned pulley

pulley flanges