PURPOSE

This booklet contains the basic lecture notes that will be delivered between November 18th and December 16th. These notes are not comprehensive and will need to be supplemented by your own notes taken during the lectures. You are additionally advised to read around the topics to further enhance your learning opportunity. On completion of this section you will be assessed on your learning by means of written assignment which will give you an opportunity to provide evidence for the following learning outcomes and assessment criteria: LO3 Be able to determine the dynamic parameters of power transmission system elements

3.1 determine the dynamic parameters of a belt drive. 3.2 determine the dynamic parameters of a friction clutch. 3.3 determine the holding torque and power Transmitted through compound and epicyclic gear trains.

Table of Contents

 

1. Belt Drives 02

Belt Drive Basics 03

Centripetal belt tension 06

Power transmitted by a belt drive 09

Relationships between tensions, friction and angle of lap 13

‘V’ belt sections 18

Maximum power transmission 21

Solutions to exercises 24

2. Clutches 26

Dog clutches 27

Plate clutch 28

Conical clutch 35

Solutions 39

3. Gear Trains 40

Simple Gear Trains 41

Power transmission 43

Compound gear trains 45

Epicyclic gear trains 50

Solutions 54

1 

POWER TRANSMISSION    This outcome requires that you become familiar with the theory behind belt

drives, clutches and gear trains. You will need to be familiar with the

mathematical concepts of transposition of formulae, solution of equations,

trigonometry, exponential functions and differential calculus. The latter

topic is needed for a complete understanding of the derivation of some of

the relationships. It is not necessary for the use of the relationships.   You will be made familiar with differential calculus within Unit 2: Analytical

Methods for Engineers.   Similarly there are some science based topics that you will need to be

familiar with. If you have completed a National Certificate/Diploma in

Manufacturing or Mechanical Engineering then you will be aware of them. If

not you may need to refer to a text book that has been written to cover the

contents of that syllabus.   This background will also help you with the Unit 3: Engineering Science.

     This first outcome will be presented to you in three sections.

You will need approximately 5 hours to study each section and to complete

the exercises.

Completion of the assignment will be additional to this time.   The sections are:  

 1. BELT DRIVES

2. CLUTCHES

3. GEAR TRAINS

2 

 

1. BELT DRIVES  1.1 INTRODUCTION.  

Belt drives have been around for a very long time. Their use is to

enable movement of one shaft to be transmitted to another shaft for

some reason or another. The ends of the shafts are fitted with

pulleys that the belt passes over, and the groove in the pulleys

reflects the shape of the belt cross section.   

The whole principle behind the arrangement is that friction exists

between the belt material and the pulley material. The driving

pulley effectively drags the driven pulley along with it as it rotates.

Clearly if there is insufficient friction between the belt and the

pulley then the belt will start to slip.   

Belt drives are not as positive a way of transmitting movement and

therefore power, as the other methods such as gears or linkages.   

However, belt drives are really useful when the distance between the

shafts is large, or when space considerations are important. They are

also cheap and easy to install and maintain, and provide a certain

amount of overload protection.   

Two conditions are really important for belt drives to be used:

Firstly the pulleys MUST be in line.

Secondly the shafts housing the pulleys MUST be parallel.   

You can get away with a little bit of misalignment, but too much will

result in poor performance, excess wear and noise.

3 

1.2 BELT DRIVE BASICS   

The following diagram shows a simple belt drive arrangement with

two pulleys of different sizes. The smaller pulley is the driving pulley

and the larger is the driven pulley. The reason for this will come

later, so be patient!   

As the pulleys are of different sizes the belt will be in contact with

different amounts of the pulley surface for each one. We need to

work out the contact angle with the belt for each pulley; this is

called the angle of lap.   

The angle of lap will be different for each of the pulleys, as you will

see from the diagram.

        

D    

C A

L2

B  

L1 O

         

FIG 1.2.1   

L1 is the angle of lap on the smaller pulley and L2 is that for the

bigger pulley.

4 

The distance between the centres of the pulleys is denoted by the

letter Z. If you examine the triangle OAB you will see that the angle

AOB is a right angle, as OA has been drawn parallel to CD, and CD is

the tangent to both circles.   

The small angle AOB can be found from the trigonometric

relationships for a right angled triangle.   

Sine AOB is determined from opposite divided by hypotenuse, so in

this case it is AB divided by Z. However AB is the difference in the

radii of the two pulleys, given by R-r.

We can say that sine AOB = (R-r)/z ------------------1.2.1

The angle of lap for the small (driving) pulley, L1, will be 1800 less

twice the angle AOB.   

The angle of lap for the larger (driven) pulley, L2, will be 1800 plus

twice the angle AOB.   

We will express these angles of lap in radians as this will be

necessary in later parts of our study.   

Example 1.2.a   

Two parallel shafts have pulleys attached to their ends. These

pulleys are in line and are 200mm and 300mm in diameter. The

centres of the shaft are 800mm apart.

Determine the angles of lap of a belt around the two pulleys

assuming that there is no slack.

Express these angles in both degrees and radians.

5 

Solution   

As the two pulleys have diameters of 200mm and 300mm they will

have radii of 100mm and 150mm.

The difference in their radii is therefore 150-100mm, which is 50mm.

The distance between centres of the shaft is 800mm

The sine of the small angle AOB is therefore 50/800, or 0.0625

The small angle AOB is 3.58330

The angle of lap on the small pulley is 1800-2x3.58330=172.83440

The angle of lap on the large pulley is 1800+2x3.58330=187.16660

Conversion to radians.   

As 3600 is equal to 2π radians, 10 is equal to 2π radians.

360   

172.83440 will be 172.8344x2π radians = 3.0165 radians.

360   

187.16660 will be 187.1666x2π radians =3.2667 radians.

360   

ANY ANGLES OF LAP USED IN FURTHER WORK MUST BE IN RADIANS   

Exercise 1.2.b   

In each case below calculate the angles of lap on both pulleys in

degrees and radians.   

1.2.b.1 r=250 R=300 Z=600

1.2.b.2 r=300 R=500 Z=1000

1.2.b.3 r=400 R=600 Z=2500

1.2.b.4 r=125 R=400 Z=750

1.2.b.5 r=50 R=90 Z=300

6 

δθ

1.3 CENTRIPETAL BELT TENSION   

The diagram below shows part of a belt that is stretched over two

pulleys of the same diameter. Remember that the belt has to be

stretched to produce the frictional force needed to create the drive.

The bits of the belt in contact with the pulleys must move in a

circular path as they go around the outside of the pulleys.   

As the belt is moving in a circular path it will be subject to

centripetal force provided by the tension in the belt.   

Examine this diagram:      

Tc   

ω        

Tc

FIG 1.3.1

  

The belt has a mass of m kg/m length, and it is flat.

r is the mean radius of the belt in contact with the pulley.

Examining the small element subtending the angle δθ at the centre of

the pulley gives us the following;

Length of the element =r.δθ

Mass of the element =m.r.δθ

velocity of the element =ω

centripetal force on element =ω2r. m.r.δθ

however as v=ωr

centripetal force on element =mv2.δθ

7 

The tension in the belt is TC and this can be resolved into two radial

components each of which is TC.sine δθ/2, and this is equal to the

centripetal force acting.

As the angle δθ is small, then sine δθ/2 is equal to δθ/2 in radians.

We have as a summary

Centripetal force on the element=m.v2.δθ which equals 2.TC.δθ/2

or TC=m.v

2 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐1.3.1

  

Example 1.3.a   

A flat pulley belt is transmitting power between two pulleys that are

500mm in diameter and are running at 1800 rev/min. The pulley belt

has a cross section of 40mmx4mm and is made of material with a

density of 1200kg/m3.

 Determine the centripetal tension in this belt under those conditions.

  

Solution   

Firstly we need to determine the linear speed, v, of the belt.

We know that the relationship between linear and angular velocity is

given by v=ωr, so we have that

ω=1800x2π/60=188.5rad/s

r=0.25m

so v=ωr=188.5x0.25=47.125m/s

The mass per metre is determined from the cross section and the

density.

m=40x103 mx4x103 mx1200kg/m3=0.192kg/m   

The centripetal belt tension is m.v2

In this case 0.192kg/m. (47.125m/s)2 =426.39N

(REMEMBER 1KG.M/S2 =1N)

8 

Exercise 1.3.b   

Complete the gaps in the following table:    

radius 

mm 

section 

mm x mm 

density 

kg/m3 

speed 

rev/min 

centrifugal   tension 

N 

400  5x20  1000  2400 

500  4x15  1200  500 

450  5x15    1800  600 

300  6x?  1500  1500  400 

  5x30  1200  2100  600     

This exercise needs you to move the equations around to make a

different item the subject.   

The equations that you need to use are all those that I used in the

example, most of which should be familiar to you from previous work.   

Just to help I have listed them below:      

v=ω.r

m=density x cross sectional area

TC=m.v2

     

Please be careful with the use of your units. That is the most likely

source of error.

9 

1.4. POWER TRANSMITTED BY A BELT DRIVE   

The entire purpose of the belt drive is to transmit motion, and with it

power, from one shaft to another. Power can only be transmitted if

there is a difference in tension in the belt as it is passing around the

pulleys.   

We normally refer to the tight side tension and the slack side tension.

Each of these has to have a value otherwise the implication is that

the belt is broken! It is usual to pre-tension the belt is order to

maintain a value of tension once the belt is running. This is the initial

tension. After all, if you examine the diagram below you will see

that, when running, the tight side tension will increase over the

initial tension, and the slack side will reduce.

  

TIGHT SIDE               

SLACK  SIDE   

FIG 1.4.1   

It is quite easy to determine that the initial tension is half way

between the tight and slack side tensions.

Using TT for tight side tension, and TS for slack side tension, then TI,

the initial tension is given by

TI =(TT+TS) /2. ------- 1.4.1

10 

We only need three items of information to calculate the power that

a belt drive can transmit. The first is the tight side tension, TT, the

second is the slack side tension, TS, and the third is the belt speed, v.

The relationship is that

Power = (TT – TS).v --------- 1.4.2   

Example 1.4.1   

A pulley arrangement is set up with a tight side tension of 1500N and

a slack side tension of 700N. The belt is to run at 10m/s.

What power will this arrangement transmit?   

Solution   

TT is 1500N and TS is 700N. The difference between them is 800N.

The belt speed is 10m/s, so the power transmitted is 800N x10 m/s.   

The numbers in the calculation give us a value of 800x10=8000.

The units used give us N.m/s, which is the same as a Watt.

The power transmitted is therefore 8000W, or 8kW.   

NOTE

If you are unfamiliar with the units being used you should refer to a

text book which will explain them to you.   

Any text which has been written for level 3 engineering science

should help.

11 

Example 1.4.2   

It is more usual to want to know how to set something up to give the

required performance. Let us consider the situation where we need

to determine the initial tension in the belt to give us the power

output we need at a given speed.   

A belt drive is to transmit 20.25kW when running at 15m/s belt

speed.

The maximum tension that the belt material can withstand is 3000N.

What should be the initial tension in the belt?

Solution   

We have been given the value of the tight side tension, or at least

the maximum tension that the belt material can withstand. The

relationship between power transmitted and belt tension has been

given to you in equation 1.4.2

P=(TT-TS).v

Substituting the values that we have gives   

20250=(3000-TS).15

divide both sides by 15

hence 1350 = 3000-TS

rearrange

so TS = 3000-1350 =1650N

the initial tension is the average of the tight and slack side tensions

TI = (3000+1650)/2

TI = 2325N

12 

Exercise 1.4.a   

Fill in the gaps in the following table.    

TT TS TI v P

2000   1500 20  

  800 1200 25

2400 2000   25  

2800   2100   35   

NOTE: All tensions in N, all belt speeds in m/s and all power in kW.     

All the above problems solved by using        

P=(TT‐TS).v 

13 

1.5. RELATIONSHIP BETWEEN TENSIONS, FRICTION AND ANGLE OF LAP.  

The derivation of the relationship between these quantities requires

a good working knowledge of differential calculus.

I will refer you to the following texts that explain the derivation

thoroughly;   

RIX, M.A Mechanical Science IV Hodder and Stoughton. 1985.   

ISBN 0-340-36785-7

BOLTON, W Mechanical Science Blackwell Scientific . 1993.   

ISBN 0-632-03579-X    

Although the texts are old, they still give an appropriate treatment of

the work.   

We have already come across the facts that there needs to be a

difference in tensions between the two parts of the belt for power to

be transmitted.

We also know that when pulleys are of different sizes the angle of lap

will be different for each pulley.

The difference in tensions between the two parts of the belt is due to

the friction between the pulleys and the belt material.   

We are using TT as the tight side tension and TS as the slack side

tension.

If the angle of lap of the belt on the pulley is θ radians and the

coefficient of friction between the belt material and the pulley

material is µ, then the following can be shown;

(TT –mv

2) = (T

S‐mv

2).e

μθ‐‐‐‐‐‐‐‐1.5.1

14 

The term mv2 is the centripetal tension that you came across earlier

in section 1.3. Its effect is to reduce the amount of tension available

to transmit power.

We will look at an example now to see the effect of this centripetal

tension.   

Example 1.5.1   

Two pulleys are mounted on parallel shafts that are 500mm apart.

The smaller driving pulley is 200mm in diameter, and the driven

pulley is 250mm in diameter. The shaft housing the driving pulley is

rotating at 2400 rev/min.

The pulley belt is flat with a cross section measuring 25mmx4mm.

The belt material has a density of 1250kg/m3, and the coefficient of

friction between the belt and pulley is 0.28.

If the maximum tension that the belt can withstand is 4000N,

determine the power that can be transmitted under these conditions.   

Solution   

Firstly we will calculate the angle of lap of the belt on each pulley.   

Equation 1.2.1 tells us that the sine of the small angle we need is

given by (R-r)/Z, which in this case is (125-100)/500.

The sine of the small angle is 25/500 = 0.05, making the angle 2.8660.   

The angle of lap on the small pulley is therefore

180-2x 2.866=174.2860

and on the large pulley

180+2x2.866=185.7320

  

When converted to radians, these become 3.042 and 3.242 respectively.

15 

When using equation 1.5.1 above we need to select an angle of lap.

The smaller one, usually on the driving pulley, is chosen because slip

between the belt and pulley is most likely where there is the shortest

length of contact between the two, and this is on the smallest pulley.

A slipping belt will cause loss of power.   

Secondly we will calculate the belt speed, in m/s.

If the small pulley has a radius of 100mm, and is rotating at 2400

rev/min, then

2400 rev/min = 2400/60 rev/s = 40 rev/s

and   

so

  

40 rev/s = 40x2xπ rad/s = 251.327 rad/s

v=ω.r, giving v= 251.327x0.1 = 25.1327 m/s

 

 

The centripetal tension in this belt will be mv2, or in this case if the

values are substituted, TC = 1250 x25x10-3x4x10-3x(25.1327)2 N.

 TC = 78.96N ( say 80 for our purposes)

Using equation 1.5.1 (previously),

We will obtain that   

(TT – 80) = (TS-80). e0.28 x 3.042

or (TT – 80) = (TS-80). e0.85176

  

hence (TT – 80) = (TS-80).2.3438.   

Now we substitute the value for TT, which is 4000N.

We get 4000 – 80 = (TS-80). 2.3438

16 

or 3920/2.3438 = ( TS-80).

so TS = 1672.5 + 80 = 1752.5N   

Thirdly we can now calculate the power that could be transmitted

under these conditions.   

Using equation 1.4.2 we find that   

P = (TT – TS).v   

Using our values we obtain that  

 P = (4000-1752.5)x25.1327 W

 

 P = 56.486kW

    

If the centripetal tension is ignored, the following changes occur.

The relationship becomes TT = T

S. e

µθ -----1.5.2

Substituting reveals that TS = 4000/2.3438  

 i.e. TS = 1706.6N

and from 1.4.2 we get

p = (4000-1706.6)x25.1327W  

 P = 57.639kW

    

There is little difference in the answers at low speeds, and it is

usual to use the second approach as it is easier.

17 

Exercise 1.5.a  

 Two pulleys with diameters of 400mm and 600mm are connected by a

flat belt with a cross-section measuring 40mm x 5mm. The belt

material has a density of 1500kg/m3, and the pulleys are 1.2m apart.

The coefficient of friction between the materials is 0.32

The maximum tension that the belt can withstand is 5000N.

Task

Select a set of values, (four or five), of driving (smallest) pulley

speed between 2400rev/min and 18000 rev/min.

 Calculate the power that can be transmitted for each value of the

driving pulley speed.

 Do this twice, once taking centripetal tension into account, and the

other time ignoring it.

 Plot your answers on a graph and draw your conclusions.

18 

1.6 ‘V’ BELT SECTIONS  

 The flat belt provides the most common solution to drive

requirements. The belts are relatively cheap as they are easy to

produce and the pulleys used are of a simple section. The biggest

problem is that they will have a limited power transmission capability

as slip will occur at fairly low loads.

 For a more positive location the use of ‘V’ or round belts is

recommended, or at greater expense the toothed or notched belts.

 Round or ‘V’ belts require grooved pulleys, which are more expensive

than those used for flat belts.

 A theoretical analysis of the behaviour of ‘v’ and round belts will

show that the relationships that were developed for flat belt drives

still apply. Assuming that the ‘v’ and round section belts rest in their

grooves and do not touch its bottom, the only change required is to

use the ‘virtual coefficient of friction’ in the relationship

TT = TS.e(µv)θ

 where (µv) is the virtual coefficient and is given by µ/sine β where β

is the half angle of the pulley groove.  

 The diagram below shows the arrangement.

                    

FIG 1.6.1.

19 

Example 1.6.1  

 Two pulleys are 100mm and 150mm in diameter and their centres are

400mm apart. If the coefficient of friction between the materials is

0.3, and the maximum tensile force in the belt is to be 2000N,

determine the power that can be transmitted for a belt speed of

24m/s if

 a). the belt is flat

 b). the belt is ‘v’ with an included angle of 400

 Solution

 

 Smallest angle of lap is calculated from

 

 sine of small angle = (75-50)/400

Small angle = 3.5830

Smallest angle of lap = 180-2x3.583 = 172.8340

 172.8340 = 3.0165 rads.

 Flat belt

 TT = TS. eµθ 2000 = TS.e0.3x3.0165

 TS = 809.12N

 

 Power = (2000-809.12).24 = 28.58kW

 

 ‘V’ belt

 TT = TS.e(µv).θ (µv=0.3/sine200) = 0.877

 µv.θ = 0.877x3.0165 = 2.646

hence TS = 2000/14.097 = 141.9N

Power = 44.59Kw

20 

Exercise 1.6.a  

 Two pulleys of 300mm and 500mm diameters are 800mm apart. The

belt material has µ = 0.35, and the maximum tension it can withstand

is 1500N. Determine the power transmitted at a small pulley speed of

1200rev/min if

 a) the belt is flat

 b) the belt is round and the pulley has an included angle of 360

21 

  

1.7. MAXIMUM POWER TRANSMISSION  

 It should now be clear that there are a lot of factors that affect the

amount of power that can be transmitted by a belt drive.

 We have the following so far:

 

 Belt tensions-tight, slack and initial.

Belt and pulley material-friction.

Physical arrangement-size and spacing.

Speed of rotation.

There is a further analysis that we can carry out to find the condition

that allows for maximum power transmission when the belt section,

material, pulley sizes and spacing have been determined.

Let us refer back to equation 1.5.1. To remind you;

(TT –mv2) = (TS-mv2).eµθ ----------------- 1.5.1

Making TS the subject  

TS = (TT –mv2) + mv2

eµØ

   

so TT –TS = TT - [(TT –mv2) + mv2] eµØ

Power = (TT-TS)v  

P = TTv - [(TTv –mv3) + mv3] eµØ

22 

dP = TT - [(TT –3mv2) + 3mv2] dv eµØ

   

For maximum power dP/dv is zero, so  

 

TT = [(TT –3mv2) + 3mv2] eµØ

TT eµØ = (TT –3mv2) + 3mv2 eµØ

  

This condition is satisfied when TT = 3mv2

This means that maximum power is transmitted when the tight

side tension is three times the centripetal tension.     

Example 1.7.1.  

 Two pulleys are 200mm and 260mm in diameter. They are mounted

on two parallel shafts with their centres 600mm apart. The flat belt

running over these pulleys has a section measuring 25mm by 3mm,

and is made from a material with µ = 0.3, and a density of 1200kg/m3

If the maximum tension that the belt can withstand is 2000N,

determine the speed of the smaller pulley when maximum power is

being transmitted.

 Solution

 

 Once again our first task is to determine the angle of lap. As we

know we need the value for the smaller pulley as this is where the

slip will occur.

Angle of lap = 180-2sine-1 (130-100)/600

23 

Angle of lap = 3.0416 rad  

 so µθ = 3.0416x0.3 = 0.91428, and

 eµθ = 2.49

 The condition for maximum power transmission is when the tight side

tension is three times the centripetal tension i.e. TT =3mv2

so 2000= 3mv2

therefore mv2 = 667N

As the belt has a cross section of 25x3, and a density of 1200kg/m3,

the value of m = 75x10-6 x 1200 =0.09

v2 = 667/0.09 = 7407.4

 v = 86.07m/s

 

 If the rotational speed of the small pulley is w rev/min then we have

that (w.2π.0.1)/60 = 86.07 or w = 8218.5 rev/min

   

Clearly this was a fairly lengthy calculation as you had to work

through the derivation of equation 1.7.3. You can just assume that

relationship from now on.

 Exercise 1.7.a

  

Select two rotational speeds within 200 rev/min below, and two

rotational speeds within 200 rev/min above, the speed calculated in

the example above for the small pulley developing maximum power.   

Calculate the power transmitted by the belt for those four conditions

and show that the above example does give the speed for maximum

power.

24 

 L1, degrees

 L1, radians L2, degrees L2, radians

 170.44

 2.975 189.56 3.308

 156.92

 2.739 203.074 3.544

 170.82

 2.981 189.18 3.302

 136.98

 2.391 223.02 3.892

 164.68

 2.874 195.32 3.409

1.8 SOLUTIONS TO ALL THE EXERCISES

Exercise 1.2.a

                    

Exercise 1.3.a  

radius 

mm 

section 

mm x mm 

density 

kg/m3 

speed 

rev/min 

centrifugal   tension 

N 

400  5x20  1000  2400  1010.65 

500  4x15  1200  1591.55  500 

450  5x15  1111.9  1800  600 

300  6x20  1500  1500  400 

262.5  5x30  1200  2100  600  

Exercise 1.4.a  

TT TS TI v P

2000 1000 1500 20 20

1600 800 1200 31.25 25

2400 2000 2200 25 10

2800 1400 2100 25 35

25 

  Series1 Series2 Poly. (Series1) Linear (Series2)PO

WE

R K

W

Exercise 1.5.a  

 This is best solved by using a spreadsheet.

 

  

POWER VERSES SPEED  

450

 400

 350

 300

 250

 200

 150

 100

 50

 0

0 1000 2000 3000 4000 5000 6000 7000

SPEED REV/MIN  

Exercise 1.6.a  

 Flat belt Power = 18kW

 

 Round belt Power = 19.642kW

    Exercise 1.7.a

 

 Again this is probably best solved using a spreadsheet and producing a graph

that should show maximum power at 7250 (ish) rev/min.

26  

2. CLUTCHES  2.1. INTRODUCTION

 

 In the last section we looked at the transmission of power between

two parallel shafts using a belt. Transmitting power between two

items of plant is often not as simple as connecting them together

using a belt. Often they need to be connected rigidly to avoid slip.

This rigid connection may involve shafts that are parallel, or those

that are in line.

 There are plenty of instances where in-line direct coupling is suitable

e.g. between the compressor and the turbine in a gas turbine unit, or

between a turbine and an alternator in an electrical generation unit.

 There are also instances when the power transmission between the

two items that are connected needs to be suspended for a while,

probably to accommodate a variation in load requirements.

 We are going to examine the situation where there are two coaxial

shafts that need to be disconnected and then quickly reconnected for

some reason. There are lots of different ways of doing that, some of

which are cruder than others. A device that carries out that function

is called a clutch.

 Somehow the two shafts will need to be separated for a short while,

and then reconnected. A suitable method would be to install a

device that consists of two surfaces, one of which is attached to the

ends of each of the shafts and these surfaces can be easily separated

and easily reconnected. When they are in contact they form a solid

coupling.

 It is not so much the fact that the two parts can be disconnected and

reconnected that is the difficulty, but the manner in which this takes

place.

27  

2.2 DOG CLUTCHES  

 These are the crudest form of clutch in that they consist of two parts

of a hollow cylinder which is cut to have a series of teeth.

 The diagram below shows the arrangement.

                       

As you can see this form of clutch will involve a shock load as the two

parts are brought into contact. There will be a considerable amount

of noise and excessive wear of the teeth in use. Practically there will

need to be clearance between the teeth on each part of the clutch so

that there is room for them to engage.

 Replacing the square section teeth with V shaped teeth will reduce

the problems when connecting, but increase the tendency for the two

halves to be thrown apart in use.

 Clearly an arrangement that allows the gradual connection of the two

shafts is much better, but this still requires that there is a high level

of friction between the two surfaces.

 We therefore need

 

 a) a high coefficient of friction between the two surfaces

 

 b) A high normal force holding them together.

28  

The most common arrangement is the plate clutch, which in its

simplest form, uses two flat circular plates in contact.

 2.3. THE PLATE CLUTCH

 

 The diagram below shows two plates held together by an axial force

of W Newtons.  

                          

Examining the right hand plate, we will consider a small ring segment

of the surface at a radius of r, thickness δr.

 Area of the ring segment = 2πr. δr

29  

If a pressure p exists between the two surfaces, then

Force on the ring segment = p. 2πr. δr ----------------------- 2.3.1

For a coefficient of friction of µ,

Frictional force F = µ. p. 2πr. δr --------------------------------- 2.3.2

This force is at right angles to the ring segment and so the moment of

this force about the centre of the plate is given by

Moment = F.r, which gives µ. p. 2πr. δr.r --------------------- 2.3.4

Rearranging equation 2.3.2 slightly, and equation 2.3.4 to gather the

terms in r together we get

Frictional force F = 2π.µ. p. r. δr -------------------------------- 2.3.5

Moment M = 2π.µ. p. r2. δr. --------------------------------------- 2.3.6

 Both of these last two equations show what is happening on the ring

segment. To find how this can be adapted to cover the whole plate

we must integrate between the two limits.

 If you do not follow the mathematics of the next bit, do not worry.

You will cover integration in Analytical Methods. Just accept the

final equations for now, and ensure that you can use those

equations.

 From 2.3.5

    

Total Force on the plate = W = 2π. ∫ p.r.δr ---------------------- 2.3.7

 between the limits of r2 and r1.

And from 2.3.6

30  

  

Total Moment = T = 2π.µ. ∫ p.r2.δr --------------------------------------- 2.3.8

 between the limits of r2 and r1.

 

 NOTE THE INCLUSION OF P WITHIN THE INTEGRAL!!!

 

 What is needed now is the relationship between p and the radius r.

There are two fundamental approaches that can be used.

a) The assumption that the pressure exerted by the force W is

constant over the surface of the plate. This is perfectly reasonable

when the clutch is new, and the friction material is not worn.

 This analysis is known as the uniform pressure condition, and

assumes that p does not vary over the surface of the plate.

 Uniform pressure assumes that p is constant with r, and so the

integral in equation 2.3.7 works out to be 2.π.p.r2./2

when the limits are included gives W = π.p.(r22 – r1

2) ------- 2.3.9

 The integral in equation 2.3.8 works out to the expression

 T = 2.πµ.p.(r2

3 – r13)/3 ------------------------------------------ 2.3.10

 By eliminating p between the equations we arrive at

 T = 2.µ.W.(r2

3 – r13)/3.(r2

2 – r12). ------------------------------ 2.3.11

 b) The assumption that wear on the plate surface is proportional to

the product of force times speed is appropriate for surfaces that are

not new. As the force on the plate is directly related to the pressure

acting, and the speed of the surface depends upon the radius from

the centre, we get the analysis known as the uniform wear condition

31  

which tells us that the product of pressure times radius is a constant

i.e. p.r = k. From this we can see that p = k/r and make this

substitution in the equations 2.3.7 and 2.3.8.

 As p = k/r, the substitution in equation 2.3.7 will give us

    

W = 2.π.∫r.k/r δr between the limits of r2 and r1

W = 2.π.k.(r2 – r1) ----------------------------------------------- 2.3.12

And substituting in 2.3.8 will give

T = 2.π.µ .∫r2.k/r δr between the limits of r2 and r1

T = π.µ.k(r22 – r1

2). ---------------------------------------------- 2.3.13

Eliminating k between the two equations will simplify the expression

to

T= µ.W.( r2 + r1)/2 ---------------------------------------------- 2.3.14

 If you have found the last few pages a bit daunting do not worry. You

are not required to be able to derive the equations, but you are

expected to know where they do come from and how to use them.

The mathematics will become clearer to you later in the course when

you have done a bit more of the Analytical Methods unit.

 The analysis above assumes only two surfaces in contact. The usual

single pate clutch uses both sides of the plate and as a result the

torque figures will be doubled for a given spring force and speed.

32  

centre plate friction liners  

pressure plate

diaphragm spring    

clutch engaged clutch disengaged

 

 flywheel  

 FIG 2.3.3

 

    

Example 2.3.1  

 A single plate clutch has a pair of frictional surfaces with an inside

diameter of 120mm and an outside diameter of 200mm. The

coefficient of friction between the friction material and the plates is

0.6.

 Assuming the uniform pressure approach what spring force is

necessary to enable transmission of 15W at 1500rev/min?

 Solution

 

 The uniform pressure analysis gives us the equation 2.3.11

T = 2.µ.W.(r23 – r1

3)/3.(r22 – r1

2).

Firstly we need to determine the value of T that is needed.  

 Using P = T.ω we get 15000 = T. 1500.2.π/60

From which T = 95.5Nm

Substituting the values in 2.3.11 gives us

95.5 = 2x2x0.6xW(0.13 -0.063)/3x(0.12 – 0.062)

33  

The inclusion of the 2 at the front of the equation is to allow for the

fact that there are two friction surfaces.

 95.5 = 2x4xW(0.000784)/3.(0.0064)

 

 95.5 = 0.098.W

 

 From which it is easy to determine that W = 95.5/0.098 = 974.5N

 

 This figure is the spring force needed to hold the clutch plates

together to allow the transmission of the power without the clutch

plates slipping. This spring force is usually provided by a series of

helical compression springs whose characteristics are such that when

the clutch is assembled they are compressed enough to provide more

than the required figure to allow for lower speed operation.

 Example 2.3.2

 

 We could repeat the example above using the constant wear

approach. In this case we will need to use equation 2.3.14

T= µ.W.( r2 + r1)/2

We already know that T = 95.5 so the calculation becomes much

shorter.  

 95.5 = 2x0.6.W.0.16/2 as r2 = 0.1 and r1 = 0.06

 

 W = 95.5x2/0.6x0.16x2 = 994.8N

 

 As you can see there is not a lot of difference between the results

from the two approaches. As long as you use an operational factor of

safety of three, or more, you are unlikely to risk damage to the

clutch in normal operation.

34  

Exercise 2.3.a  

 A multi-plate clutch has four plates with friction surfaces on each

side. Each spring used provides a spring force of 300N, and the

assembly has six springs in it. The internal radius of the friction

material is 40mm and the external radius is 100mm.

 The coefficient of friction between the materials is estimated to be

0.45

 Using both the constant pressure approach and the constant wear

approach determine the power that this clutch could transmit when

running at 2400rev/min.

35  

2.4. CONICAL CLUTCHES  

 The conical clutch is an alternative to the plate clutch as the

construction is somewhat simpler.

 The basis of the conical clutch is that there is a single pair of friction

surfaces in contact as shown in the diagrams below.                           

The force W pushes the cone into the matching taper forcing the two

friction surfaces together. The lower diagram is a magnification of a

part of the touching surfaces.

 The lower diagram shows an incremental ring segment at a radius of

r, and with a radial thickness of δr. The normal pressure caused by

the force W is p.

 For the incremental ring the actual length of friction material in

contact is given by δr/sine α.

 The area of friction material in contact is therefore 2.π.r. δr/sine α.

And the normal force, N, = p. 2.π.r. δr/sine α.

36  

The axial component of this force is N.sine α, which means that this

incremental axial force is given by

 N.sine α, or sine α. p. 2.π.r. δr/sine α. = p. 2.π.r. δr

   

The total axial force, W = 2π∫p.r. δr between the limits of r2 and r1.

 Continuing the analysis further will enable us to derive the

relationships that we will use. The method is exactly the same as

demonstrated for the plate clutch, and therefore it is not necessary

to repeat it here.

 Using the uniform pressure approach we will find that

T = 2.µ.W. (r23 – r1

3)/(r22 –r1

2).3 sine α ------------------------------------- 2.4.1

And for the uniform wear condition

T = W.µ.(r2 + r1)/2. sine α ------------------------------------------------------ 2.4.2     

YOU SHOULD NOTICE THE SIMILARITY BETWEEN THE EQUATIONS

2.3.11 AND 2.4.1, AND BETWEEN EQUATIONS 2.3.14 AND 2.4.2  

 The only difference is the inclusion of the sine of the cone angle.

 

 This is the same effect as we saw with the belt drives when we

changed from flat to vee.

 We need to work through an example to see how to deal with a cone

clutch, and compare it to a similarly dimensioned plate clutch.

37  

Example 2.4.a  

 A cone clutch has a friction surface whose smallest radius is 120mm

and largest is 165mm. The cone angle is 150.

 The maximum axial load is 2kN, and the clutch is running at 20rev/s.

The coefficient of friction is 0.4.

What is the power transmitted under these conditions?

Try both uniform pressure and uniform wear approaches.

Solution

Uniform pressure  

T = 2.µ.W. (r23 – r1

3) / (r22 –r1

2).3 sine α  

Substituting values will give us that  

T = 2x 0.4x 2000.(0.1653 – 0.123)/(0.1652 – 0.122).3 sine 150

 Giving us a value of T = 444Nm

 

 P = T.ω which gives us P = 444. 20.2.π watts = 55.79kW

 

 Uniform wear

 

 T = W.µ.(r2 + r1)/2. sine α

 Substituting values will give us that

 T = 2000x0.4x(0.165 + 0.12)/2.sine 150

 Giving a value of T = 440.5N, from which P = 55.35kW

 

 Again you can see that similar values are obtained from each

approach.

38  

The difference in performance between the plate and cone clutch

can be demonstrated by the removal of the sine function.

 So, for a plate clutch with only one friction face, and the same

details as the cone clutch above we will get that

 T = 2.µ.W. (r2

3 – r13) / (r2

2 –r12).3 for uniform pressure

giving T = 2.0.4.2000.(0.1653 – 0.123) / (0.1652 – 0.122).3

T = 115N and P= 14.45kW

And also that  

 T= µ.W.( r2 + r1)/2 for uniform wear

Giving T = 0.4.2000.(0.165 + 0.12) / 2

T = 114N and P = 14.32kW

You can see that the cone clutch transmits far more power. This is

because the slope of the cone gives a greater surface area in contact

for the same radial dimensions as a plate clutch. The cone clutch is

bound to be longer than its plate equivalent, but in the case above

the plate clutch would need to have two pairs of friction surfaces to

give a similar performance. (444 /115 = 3.86, = 4 to nearest number).

 Exercise 2.4.a

 A cone clutch has an angle of 200 and the friction surface is 80mm

long with a mean radius of 100mm. The coefficient of friction is 0.3,

and the axial load is 2500N.

 Determine the power that this clutch can transmit before slipping if it

is running at 2400rev/min. Consider both approaches.

39  

2.5 SOLUTIONS TO ALL THE EXERCISES

Exercise 2.3.a

Uniform pressure  

 P = 120.825kW

 

 Uniform wear

 

 P = 113.85kW

    

Exercise 2.4.a

Uniform pressure

P = 55.39kW

Uniform wear  

 P = 55.04kW

40  

3. GEAR TRAINS  

 3.1 INTRODUCTION

 

 In a previous section you have looked at belt drives which enable the

transmission of power from one shaft to another parallel shaft. Belt

drives are used frequently because they are simple to install and

cheap to create. They are used when the distance between the two

shafts makes other methods inappropriate. Shock loadings can be

accommodated as the belt will slip to avoid damage.

 If the shafts are close together, and the speed relationship between

the two is important, together with the efficiency of the

transmission, then a gear train may be more suitable.

 Gear trains are more expensive and more cumbersome than belts, but

they are more precise.

 It is important to realise that there are numerous types of gear train.

Gears used to transmit power between parallel shafts which have

their teeth cut parallel to the axis of the shafts and are known as

spur gears. Sometimes the teeth are cut on a helix and these give us

the helical gears used for high power transmission with less noise and

wear than spur gears. Gears used to transmit motion between

inclined shafts are bevel gears.

 Other common arrangements are the rack and pinion and the worm

and wheel. The former transmits motion from circular to linear, and

the latter enables transmission between to shafts at right angles.

 Our studies will only encompass the spur gear. The teeth on a spur

gear are cut so that the sides are curved to allow for smoother

meshing and less slip than if they were straight. For gear teeth to

mesh properly they need to have the same diametrical pitch, that is

the number of teeth divided by the pitch circle diameter has to be

the same.

41  

The figure below shows some of the terminology used with spur

gears.

   

pitch point

circular pitch pitch circle            

FIG 3.1.1       

3.2. SIMPLE GEAR TRAINS  

 A simple gear train is the term used to describe the arrangement

when each of the shafts involved only has one spur gear mounted on

it. The total number of gears used makes no difference. If each

shaft has only one gear it does not matter how many shafts are in the

train, it is still simple.

 Two examples are shown below

42  

As we have stated the teeth on each gear will only mesh if they have

the same diametrical pitch. Let us consider two meshing spur gears,

one with 20 teeth and one with 50 teeth. It follows from what we

said previously that if the smaller gear is 20mm in diameter, then the

larger one must be 50mm in diameter (one tooth per mm !).

 Suppose that the smaller gear is running at 10 rev/min. Then in one

minute it will have pushed 10x20 teeth on the bigger gear, i.e. 200.

As the bigger gear has 50 teeth on it, it will have completed 200/50

revolutions in the same time. It will be running at 4 rev/min.

 We can conclude that the number of teeth times the speed must be

the same for all the meshing gears in the train.

 Example 3.2.a

 

 Two parallel shafts are 400mm apart and need to be connected by

two spur gears. One shaft needs to run seven times faster than the

other.

 Determine the number of teeth on each gear if the diametrical pitch

is to be 0.4 teeth/mm.

 Solution

 

 As the speed ratio is seven, then it follows that the slower gear must

be seven times the size of the faster gear. The gap between the

centres of the gears must be eight times the radius of the smaller

gear.

 The radius of the smaller gear must be 400mm divided by eight, i.e.

50mm and that of the bigger gear 350mm. As each gear has 0.4 teeth

per mm of diameter the small gear must have 50x0.4 = 20 teeth, and

the bigger one 350x0.4 = 140 teeth.

43  

3.2.1. POWER TRANSMISSION  

 Let us now examine what happens to the power transmitted by two

spur gears in contact.

 The diagram below represents two meshing spur gears with radii of r

and R respectively. As the are meshing the gears must have the same

diametrical pitch so the number of teeth on each gear is n and N, and

their speeds are S and s.

 At the point on the teeth in contact there must exist an equal and

opposite force. This force will be tangential to the pitch circle. Note

that that is only absolutely true at one instant, but we will assume

that it is right otherwise the analysis becomes too complicated.                          

The torque provided by the smaller wheel is Fr, and this is increased

to FR on the bigger wheel. However because the speeds are also

dependent upon the number of teeth, but the number of teeth is

directly related to the diameter(and hence radius) we get that the

power provided by the smaller gear is FrS, and that received by the

bigger gear is FRs.

 Since sN = nS, and the number of teeth is proportional to the radius,

we deduce that skR = krS where k is the diametrical pitch. Making r

44  

the subject gives us r = sR/S, and substituting in the expression for

the power developed in the smaller gear we get Power = F(sR/S)S

which simplifies to give the same expression as that for the power in

the larger gear.

 Theoretically there is 100% power transfer, but in practice the design

of the gears means that there will be some loss. Generally the ratio

of the input power to the output power is known as the efficiency.

The loss of efficiency is due to noise, heat generation, slip and wear.

Let us now examine what happens when we have more than two spur

gears in a train. We will consider four gears A, B, C and E in contact

in that order. They will only mesh, as we know, if their diametrical

pitch is the same.

 The speed of gear A is SA and the number of teeth is TA. Similar

terminology applies to the other gears.

 SB = SA.TB/TA

 

 also SC = SB.TC/TB

and SE = SC.TE/TC

putting these together we get SE = SA.TE/TA  

 We can conclude that the input and output speeds are only affected

by the size of the first and last gears in the train. All that the

intermediate ones do is change the direction of rotation. If there are

only two shafts that need to be connected there is seldom any point

in having any more than three gears in the train. When three are used

the middle one is to reverse the direction of rotation of the output

shaft and is known as an idler gear.

45  

3.3 COMPOUND GEAR TRAINS  

 A compound gear train is the term to describe the situation when one

of the shafts in the train holds more than one gear.

 See the diagram below.

                          

If we label the shafts A, B and C from the left then the speed of shaft

B is only dependent on the number of teeth on the two meshing gears

on shafts A and B. The speed of shaft C is dependent on the number

of teeth on each of the two meshing gears on shafts B and C and also

the speed of shaft B.

 Example 3.3.a

  

A compound train such as shown in FIG 3.3.1 has an input shaft speed

of 2400rev/min. The number of teeth on the gear on shaft A is 36.

This gear meshes with one on shaft B that has 54 teeth. Also on shaft

B is another gear with 18 teeth. This gear meshes with a gear on

shaft C that has 72 teeth.

Determine the output speed of shaft C.

46  

Solution   

As the first meshing gear on shaft B has 54 teeth, and it meshes with

the 36 teeth on shaft A, then it will rotate at 36/54 times the speed

of shaft A.

2400x36/54 = 1600rev/min.

The second gear on shaft B must also be rotating at 1600 rev/min and

so shaft C will be rotating at 1600x18/72 = 400rev/min.   

Whilst the above example is easy enough to work through we must

not forget the condition that the meshing gears must have the same

diametrical pitch. If the diametrical pitch of the teeth is 0.5, then

the first gear has a diameter of 72mm and therefore a radius of

36mm, the second gear has a radius of 54mm. The small gear on the

same shaft has a radius of 18mm and the final gear has a radius of

72mm. The distances between shafts A and B will be 90mm, and

between B and C will also be 90mm.   

Example 3.3.b   

The exercise above gave us the information that the distance

between shafts A and B and between shafts B and C were the same.

This is particularly important if the input and output shafts need to

be in line.

What does need our attention is the further demands made when the

overall speed ratio between input and output is also specified.   

A compound gear train is shown in the following diagram. The output

shaft, O, is in line with the input shaft I. The intermediate shaft

carrying the two meshing gears is 120mm away. The output shaft is

required to run at eight times the input speed.

Determine a suitable selection of gears to fulfil these requirements.

47  

   

I O          

B C  

     

Let us assume that the input shaft gear has a number of teeth

indicated by TI and that the output shaft has a number of teeth

indicated by TO. The speed of the intermediate shaft is given by SB

which is also the same as SC.

So we know that SB = SI.TI /TB.

We know that SC = SB and that SO = SC.TO /TC

From the few relationships above we can determine that

SO = SI.TI.TO/TB.TC and since SO = 8SI it is logical to conclude that

TI.TO/TB.TC = 8.

Our next step is to create the condition for the input and output

shafts to be in line. For that we know TI + TB = TO + TC.   

We now have four unknowns but only two equations. So there is not

a unique solution to this problem, but a range of possible solutions.

We have not got the actual diametrical pitch but that does not

matter as the ratios that we calculate will stand for any value of

diametrical pitch. We will assume a diametrical pitch of one so that   

TI + TB = TO + TC = 120

48  

Let us assume that TI = 30. Then TB must be 90.

Substituting the values in the two equations above we get

TO + TC = 120 and that 30.TO /90.TC = 8

From which TO = 24.TC and we can deduce that TC = 120/25 = 4.8   

We immediately run into a problem as it is not possible to have a

fractional number of teeth on a gear.

Solving this problem could develop into a guessing game unless we

find a method of eliminating some of the guesswork.   

Selecting a value of TI will immediately tell us the value for TB and

we therefore have the ratio of TI /TB. As we know that the overall

speed ratio is 8 we can determine the ratio TB/TC and hence the

number of teeth on each of those gears.

Total number of teeth = 120 Ti   Tb   Ti/Tb 8/(Ti/Tb) 120/To To Tc   5   115 0.043478 184 185 0.648649 119.3514   10   110 0.090909 88 89 1.348315 118.6517   15   105 0.142857 56 57 2.105263 117.8947   20   100 0.2 40 41 2.926829 117.0732   25   95 0.263158 30.4 31.4 3.821656 116.1783   30   90 0.333333 24 25 4.8 115.2   35   85 0.411765 19.42857 20.42857 5.874126 114.1259   40   80 0.5 16 17 7.058824 112.9412   45   75 0.6 13.33333 14.33333 8.372093 111.6279   50   70 0.714286 11.2 12.2 9.836066 110.1639   55   65 0.846154 9.454545 10.45455 11.47826 108.5217   60   60 1 8 9 13.33333 106.6667   65   55 1.181818 6.769231 7.769231 15.44554 104.5545   70   50 1.4 5.714286 6.714286 17.87234 102.1277   75   45 1.666667 4.8 5.8 20.68966 99.31034   80   40 2 4 5 24 96   85   35 2.428571 3.294118 4.294118 27.94521 92.05479   90   30 3 2.666667 3.666667 32.72727 87.27273   95   25 3.8 2.105263 3.105263 38.64407 81.35593   100   20 5 1.6 2.6 46.15385 73.84615   105   15 7 1.142857 2.142857 56 64   110   10 11 0.727273 1.727273 69.47368 50.52632   115   5 23 0.347826 1.347826 89.03226 30.96774

49  

Looking at this chart reveals that there are only two occasions when

the number of teeth on each gear is a whole number and therefore a

possibility.

We will just check that all is well.

If TI is 80 then TB is 40 and SB is twice SI

SC is twice SI and SO is four times SC giving us our required speed ratio

of 8.    

Exercise 3.3.a  

 Using the compound gear arrangement as shown in FIG 3.3.2

determine a suitable selection of gears for an output speed of five

times the input speed with a distance between input/output and

intermediate shafts of 90 teeth.

 (I suggest that you use a spreadsheet approach as I have shown you)

50  

3.4 EPICYCLIC GEAR TRAINS  

 An epicyclic gear train is probably the most difficult to visualise.

They are employed where relatively small gear ratios are required,

but a large amount of power is to be transmitted. Mostly the gears

are helical for reduction of noise and wear etc, but this does mean

that they are expensive to make.

 An epicyclic gear train comprises a single gear attached to the end of

the input shaft. This is called the sun gear. A spider consisting of

one or more arms is free to rotate about the main axis of the input

shaft. On the end of each arm is mounted a planet gear each of

which is in contact with the sun gear. The sun and planet gears are

inside of an annulus and the planet gears are in contact with the

annulus.

 The diagram below shows the arrangement.

  

annulus  planet

  

planet    

arm  sun

 

  

sun

 

  

arm          

FIG 3.4.1 In order to calculate the overall gear ratio of an epicyclic gear train

we must firstly examine what happens in a simple case.

 If all the gears have the same diametrical pitch then it is easy to see

from FIG 3.4.1 that TS + 2.TP = TA where T stands for the number of

teeth, and the subscripts refer to sun, planet and annulus.

51  

In the diagram below we have an arm supporting a sun gear and a

planet gear.                   

FIG 3.4.2    

Suppose that the sun wheel is fixed so that it cannot rotate. Then

rotating the arm around the sun causes the planet to rotate. If we

adopt a sign convention that states clockwise rotation to be positive,

then 1 revolution of the arm clockwise will cause TS /TP revolutions of

the planet. T refers to the number of teeth and the subscripts are as

previously.

 Similarly if the arm is held rigidly then one revolution of the sun will

cause the planet to rotate by - TS /TP revolutions.

 Locking the arm and gears together and rotating the entire assembly

anticlockwise will mean that all the sun, planets and arm will rotate

by -1 revolution.

 By using two or more of these conclusions we can see what happens

in various circumstances.

 Take the epicyclic gear train shown in FIG 3.4.1.

52  

Example 3.4.a  

 The epicyclic gear train shown in the figure has a sunwheel with 100

teeth and the planet wheels have 25 teeth. Determine the gear ratio

when the annulus is fixed.

 Solution

 

 Firstly we need to lock the arm and then give the annulus one

revolution clockwise.

 Operation

 Arm Annulus Sun

 Planet

 Lock arm and rotate

annulus by +1

revolution

 0 +1 -TA/TS

 + TA/TP

    

Then we give the whole lot 1 revolution anticlockwise  

 Operation

 Arm Annulus Sun

 Planet

 rotate whole assembly

by -1 revolution

 -1 -1 -1

 -1

    

Add the two together  

 Operation

 Arm Annulus Sun

 Planet

 Adding both situations

 -1 0 -(1 + TA/TS)

 TA/TP -1

    

The number of teeth on the annulus must be 100 + 2x25 = 150  

 Therefore when the arm completes one revolution anticlockwise, the

sun gear will have completed 1 + 150/100 revolutions anticlockwise.

53  

This arrangement is therefore a reduction gear train with a ratio of

2.5.

 Exercise 3.4.a

 

 Calculate the gear ratio for an epicyclic gear train similar to the one

in the example above if the annulus has 180 teeth and each planet

wheel has 30 teeth. Assume the annulus to be fixed.

54  

 Operation

 Arm Annulus Sun

 Planet

 Adding both situations

 -1 0 -(1 + TA/TS)

 TA/TP -1

3.5 SOLUTIONS TO ALL THE EXERCISES

3.4.a

      

Substituting numbers gives  

 TA = 180, TP = 30 hence TS = 120

 

 The gear ratio is therefore 2.5, with the arm and sun rotating in the

same direction.

         

If you have read and understood all the previous work, and you have

completed the examples correctly then you should be ready to

attempt the assignment.

 Well done!