PURPOSE These notes are not comprehensive and will need to ...
Transcript of PURPOSE These notes are not comprehensive and will need to ...
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!