Braking systems - Westfields Sports High School

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Engineering Studies
Preliminary Course
Stage 6
Braking systems
ES/S6 – Prelim 41082
P0021884
Acknowledgments
This publication is copyright Learning Materials Production, Open Training and Education Network –
Distance Education, NSW Department of Education and Training, however it may contain material from
other sources which is not owned by Learning Materials Production. Learning Materials Production
would like to acknowledge the following people and organisations whose material has been used.
Board of Studies, NSW
All reasonable efforts have been made to obtain copyright permissions. All claims will be settled in
good faith.
Materials development:
Peter Martin
Coordination:
Jeff Appleby
Edit:
Jeff Appleby, Stephen Russell
Illustrations:
Tom Brown, David Evans
DTP:
Matthew Britt, Carolina Barbieri
Copyright in this material is reserved to the Crown in the right of the State of New South Wales.
Reproduction or transmittal in whole, or in part, other than in accordance with provisions of the
Copyright Act, is prohibited without the written authority of Learning Materials Production.
© Learning Materials Production, Open Training and Education Network – Distance Education,
NSW Department of Education and Training, 1999. 51 Wentworth Rd. Strathfield NSW 2135.
Revised 2002
Module contents
Subject overview ................................................................................ iii
Module overview................................................................................ vii
Module components ................................................................ viii
Module outcomes ...................................................................... ix
Indicative time ............................................................................x
Resource requirements.............................................................. xi
Icons ................................................................................................... xiii
Glossary.............................................................................................. xv
Directive terms.................................................................................xxiii
Part 1: Development of braking systems and
engineering materials application – 1 .......................... 1–49
engineering materials application – 2 .......................... 1–33
Part 3: Engineering mechanics, hydraulics
and communications in braking systems – 1.............. 1–57
Part 4: Engineering mechanics, hydraulics
and communications in braking systems – 2.............. 1–47
Part 5: Engineering report for braking systems....................... 1–27
Bibliography........................................................................................29
Module evaluation .............................................................................31
i
ii
Subject overview
Stage 6 Engineering Studies Preliminary Course and HSC Course each
have five modules.
Engineering Studies Preliminary Course
Household appliances examines common appliances
found in the home. Simple appliances are analysed
to identify materials and their applications.
Electrical principles, researching methods and
techniques to communicate technical information are
introduced. The first student engineering report is
completed undertaking an investigation of materials
used in a household appliance.
Landscape products investigates engineering
principles by focusing on common products, such as
lawnmowers and clothes hoists. The historical
development of these types of products demonstrates
the effect materials development and technological
advancements have on the design of products.
Engineering techniques of force analysis are
described. Orthogonal drawing methods are
explained. An engineering report is completed that
analyses lawnmower components.
Braking systems uses braking components and
systems to describe engineering principles. The
historical changes in materials and design are
investigated. The relationship between internal
structure of iron and steel and the resulting
engineering properties of those materials is detailed.
Hydraulic principles are described and examples
provided in braking systems. Orthogonal drawing
techniques are further developed. An engineering
report is completed that requires an analysis of a
braking system component.
iii
Bio-engineering examines both engineering
principles and also the scope of the bio-engineering
profession. Careers and current issues in this field
are explored. Engineers as managers and ethical
issues confronted by the bio-engineer are
considered. An engineering report is completed that
investigates a current bio- engineered product and
describes the related issues that the bio-engineer
would need to consider before, during and after this
product development.
Irrigation systems is the elective topic for the
preliminary modules. The historical development of
irrigation systems is described and the impact of
these systems on society discussed. Hydraulic
analysis of irrigation systems is explained. The
effect on irrigation product range that has occurred
with the introduction of is detailed. An engineering
report on an irrigation system is completed.
iv
HSC Engineering Studies modules
Civil structures examines engineering principles as
they relate to civil structures, such as bridges and
buildings. The historical influences of engineering,
the impact of engineering innovation, and
environmental implications are discussed with
reference to bridges. Mechanical analysis of bridges
is used to introduce concepts of truss analysis and
stress/strain. Material properties and application are
explained with reference to a variety of civil
structures. Technical communication skills
described in this module include assembly drawing.
The engineering report requires a comparison of two
engineering solutions to solve the same engineering
situation.
Personal and public transport uses bicycles, motor
vehicles and trains as examples to explain
engineering concepts. The historical development of
cars is used to demonstrate the developing material
list available to the engineer. The impact on society
of these developments is discussed. The mechanical
analysis of mechanisms involves the effect of
friction. Energy and power relationships are
explained. Methods of testing materials, and
modifying material properties are examined. A
series of industrial manufacturing processes is
described. Electrical concepts, such as power
distribution, are detailed are introduced. The use of
freehand technical sketches.
Lifting devices investigates the social impact that
devices raging from complex cranes to simple car
jacks, have had on our society. The mechanical
concepts are explained, including the hydraulic
concepts often used in lifting apparatus. The
industrial processes used to form metals and the
methods used to control physical properties are
explained. Electrical requirements for many devices
are detailed. The technical rules for sectioned
orthogonal drawings are demonstrated. The
engineering report is based on a comparison of two
lifting devices.
v
Aeronautical engineering explores the scope of the
aeronautical engineering profession. Career
opportunities are considered, as well as ethical
issues related to the profession. Technologies unique
to this engineering field are described. Mechanical
analysis includes aeronautical flight principles and
fluid mechanics. Materials and material processes
concentrate on their application to aeronautics.
The corrosion process is explained and preventative
techniques listed. Communicating technical
information using both freehand and computer-aided
drawing is required. The engineering report is based
on the aeronautical profession, current projects and
issues.
Telecommunications engineering examines the
history and impact on society of this field. Ethical
issues and current technologies are described.
The materials section concentrates on specialised
testing, copper and its alloys, semiconductors and
fibre optics. Electronic systems, such as analogue
and digital, are explained and an overview of a
variety of other technologies in this field is
presented. Analysis, related to telecommunication
products, is used to reinforce mechanical concepts.
Communicating technical information using both
freehand and computer-aided drawing is required.
The engineering report is based on the
telecommunication profession, current projects and
issues.
Figure 0.1 Modules
vi
Module overview
This module will build upon the material covered in Household
appliances and introduce new concept relating to Braking systems.
Historical developments provides an overview of developments in
technology and society over time.
Engineering mechanics and hydraulics analyses friction in braking
systems, stress, strain, the modulus of elasticity, work, energy, power,
fluid mechanics, including Pascal’s Principle, pressure, hydraulics and
Archimedes’ Principle.
Engineering materials examines; wrought iron, steel and composites,
analyses the structure, properties, manufacturing methods, modification
of properties and uses as applied to braking systems, and investigates
materials testing of tensile strength, compressive strength and hardness.
Communications concentrates on visualisation and sketching of objects
using pictorial drawing. Isometric projection is extensively covered and
isometric circles explained. Orthogonal drawing is further extended,
including the application of AS1100 to braking systems and components.
CAD is also developed, with coordinate methods fully explained. Detail
drawings of components that give a full shape and size description are
provided.
The last part of this module involves an engineering report. This report
requires a detailed investigation of one braking component.
vii
Module components
Each module contains three components, the preliminary pages, the
teaching/learning section and additional resources.
•
The preliminary pages include:
–
module contents
–
subject overview
–
module overview
–
icons
–
glossary
–
directive terms.
Figure 0.2 Preliminary pages
•
The teaching/learning parts may
include:
–
part contents
–
introduction
–
teaching/learning text and tasks
–
exercises
–
check list.
Figure 0.3 Teaching/learning section
•
The additional information may
include:
–
module appendix
–
bibliography
–
module evaluation.
Additional
resources
Figure 0.4 Additional materials
Support materials such as audiotapes, video cassettes and computer disks
will sometimes accompany a module.
viii
Module outcomes
At the end of this module, you should be working towards being able to:
•
identify the scope of engineering and recognise current innovations
(P1.1)
•
explain the relationship between properties, uses and applications of
materials in engineering (P2.1)
•
use mathematical, scientific and graphical methods to solve
problems of engineering practice (P3.1)
•
develop written, oral and presentation skills and apply these to
engineering reports (P3.2)
•
apply graphics as a communication tool (P3.3)
•
describe developments in technology and their impact on engineering
products (P4.1)
describe the influence of technological change on engineering and its
effect on people (P4.2)
•
•
demonstrate the ability to work both individually and in teams (P5.1)
•
apply skills in analysis, synthesis and experimentation related to
engineering (P6.2).
Extract from Stage 6 Engineering Studies Syllabus, © Board of Studies, NSW, 1999.
Refer to <http://www.boardofstudies.nsw.edu.au> for original and current documents.
ix
Indicative time
The Preliminary course is 120 hours (indicative time) and the HSC
course is 120 hours (indicative time).
The following table shows the approximate amount of time you should
spend on this module.
Preliminary modules
Percentage of time
Number of hours
Household appliances
20%
24 hr
Landscape products
20%
24 hr
Braking systems
20%
24 hr
Bio-engineering
20%
24 hr
Elective: Irrigation systems
20%
24 hr
HSC modules
Percentage of time
Number of hours
Civil structures
20%
24 hr
Personal and public transport
20%
24 hr
Lifting devices
20%
24 hr
Aeronautical engineering
20%
24 hr
Telecommunications engineering
20%
24 hr
There are five parts in Braking systems. Each part will require about four to five
hours of work. You should aim to complete the module within 20 to 25 hours.
x
Resource requirements
You will need the following equipment for this module:
•
technical drawing equipment
–
rule, 0.5 mm pencil with B lead, protractor, set of compasses, drawing
board, tee-square, 60º-30º and 45º set squares, eraser, circle template
and radius curves.
•
Board of Studies approved calculator
•
access to resource materials including textbooks, newspapers and the Internet
•
access to a computer with a CAD program
•
brick or ream or paper
•
glue
•
fabric or sheet of garnet paper
•
2 large PET drink bottles
•
drill/skeawer/nail
•
cotton
•
bucket/wash tub
Note: The validity of some information provided on the Internet is
questionable. If you access information from sites that are reputable, the
information can be used confidently and quoted.
xi
xii
Icons
As you work through this module you will see symbols known as icons.
The purpose of these icons is to gain your attention and to indicate
particular types of tasks you need to complete in this module.
The list below shows the icons and outlines the types of tasks for Stage 6
Engineering studies.
Computer
This icon indicates tasks such as researching using an
electronic database or calculating using a spreadsheet.
Danger
This icon indicates tasks which may present a danger and
to proceed with care.
Discuss
This icon indicates tasks such as discussing a point or
debating an issue.
Examine
This icon indicates tasks such as reading an article or
watching a video.
Hands on
This icon indicates tasks such as collecting data or
conducting experiments.
Respond
This icon indicates the need to write a response or draw
an object.
Think
This icon indicates tasks such, as reflecting on your
experience or picturing yourself in a situation.
xiii
Research
This icon indicates you will need to do some
investigative work.
Return
This icon indicates exercises for you to return to your
teacher when you have completed the part. (OTEN OLP
students will need to refer to their Learner's Guide for
instructions on which exercises to return).
xiv
Glossary
As you work through the module you will encounter a range of terms that
have specific meanings. The first time a term occurs in the text it will
appear in bold.
The list below explains the terms you will encounter in this module.
anti-lock braking
system
abbreviated to ABS prevents wheels from locking
during emergency braking situations
absolute
coordinates
coordinates, used in CAD, that take all
measurements along the x and y axes from the
origin
alloy
the addition of another element or elements to a
metal used to change the properties of that metal
angle of friction
the angle that the resultant makes with the normal
when the friction force and the normal reaction are
replaced by a single force
Archimedes’
Principle
when a body is wholly or partially immersed in a
fluid, it is acted upon by an upthrust which is equal
to the weight of the fluid displaced
AS1100
the drawing standards used in Australia for all
technical drawings, such as mechanical and civil
engineering, survey and architectural drawings
back pedal brake
a common braking system for bicycles used in the
1950s requiring a freewheel system that enabled the
pedals to be pushed backwards to apply the braking
force to the rear wheel
batching
the combining or premixing of materials in
preparation for forming or manufacturing
components – used for composite or polymer based
components
xv
Brinell hardness
test
a hardness test that uses a hardened steel or
tungsten carbide ball indentor pressed into the
surface of a material for 10 to15 seconds – the loads
used are 500, 1 500 and 3 000 kg
buoyancy
for a body to float in a fluid, the upward thrust due
to the weight of the displaced fluid, must be equal
to the weight of the floating body, this upward
thrust is buoyancy
cable brakes
a braking system introduced by Daimler in 1899,
using a cable anchored to the chassis, and wound
around a drum
cast iron
cast iron is a ferrous metal generally containing
1.8% carbon to 4.0% carbon
cementite
a phase in the microstructure of steel consisting of
6.67% carbon dissolved in BCC iron – it is an
interstitial compound, Fe3C, that is extremely hard
and brittle
coefficient of
friction
the ratio of the limiting frictional resistance to the
normal reaction
composite
material
a composite material consists of two or more
materials combined to utilise the individual
properties of those materials to give distinctly
different service properties to the manufactured
composite
compression
moulding
used in the manufacture of components that are
made from thermosetting polymers or from
composites based upon thermosetting polymers; it
consists of compressing raw material into a mould
or cavity of the desired shape, and then applying
heat and pressure
compressive test
a test conducted on a prepared specimen, held in a
gripping device and a gradually increasing axial
load applied which shortens the specimen; the
applied load is plotted against the compression, to
produce a load-compression graph
compressive stress the internal resistance of a body to a deforming
force that is tending to shorten the body
continuous
precipitation
xvi
a precipitation of a new phase that completely
surrounds the existing equiaxed grains, forming a
continuous phase throughout the structure, an
example is the continuous precipitation of
cementite around the pearlite grains in a 1.2%
carbon steel
contracting band
brake
a braking system developed in the 1890’s in
response to the introduction of pneumatic tyres –
the main type operated on the principle of a steel
band acting externally on a hub or drum
deformed grains
the grains, visible in a microstructure, that have
been squashed and deformed as a result of cold
working
dendrites
the skeleton shaped grains formed during the
solidification of many metals; also a microstructural
feature, formed only when cooling an alloy from a
liquid, and consist of skeleton shaped grains which
are drawn using curved lines
detail drawing
an orthogonal drawing which gives a full size and
shape description of the component, it also includes
the material from which the component is to be
manufactured
drum brakes
a braking system introduced in 1902 by Louis
Renault that operated on the principle of two hinged
shoes being forced apart onto the inside of a
rotating drum
enlarging scale
a drawing scale that is used to enable small objects
to be drawn to a suitable scale on a piece of
drawing paper (a scale of 2:1 means that you draw
the object twice full size, whilst a scale of 10:1
means that you draw the object ten times full size)
equiaxed grains
a microstructural feature that shows grains that are
‘equiaxed’ or have equi-axes from the centre of the
grain
equilibrium
structure
the structure formed in a material as a result of a
slow rate of cooling which enables all reactions to
take place
eutectoid steel
a steel having a composition of 0.8% carbon
exploded
isometric
an exploded isometric drawing is a pictorial
drawing of an assembly in which the components
are drawn separated so that details of each
component can be seen
ferrite
a phase in the microstructure of steel, consisting of
carbon dissolved in Body Centred Cubic Structure
(BCC) a iron, up to a maximum of 0.025% at 723∞
C; it is an interstitial solid solution that is very soft,
ductile and malleable
full-section
a standard method of drawing used to show interior
details as visible outline
xvii
xviii
friction
the resistance to motion that occurs when two
surfaces slide or tend to slide over each other
grey cast iron
a cast iron produced when molten iron, containing
2.8% to 4.0% carbon, is slowly or moderately
cooled in a mould; the resultant structure has
graphite flakes in a pearlite or ferrite matrix – it is
very strong in compression, but weak in tension
half-section
a standard method of drawing used only with
symmetrical components, to show the interior
details on one side of the symmetry line as visible
outline, and the exterior details on the other side of
the symmetry line also as visible outline
hidden outline
lines that represent the edges of an object that
cannot be seen as visible outline when viewed from
the required direction; they are represented as thin
dark dashed lines, usually 0.25 mm thickness when
using A4 size paper
Hooke’s Law
extension is proportional to the applied load in a
tensile test
hydraulic system
a brake operating system using fluids to transfer
pressure throughout the system by the application
of Pascal’s Principle
inertia
the amount of matter in a body; it is also described
as the tendency of a body to remain at rest or, if
moving, remain in motion in a straight line
isometric
projection
a three dimensional pictorial drawing that uses
angles of 30∞-90∞-30∞
kinetic energy
the energy a body possesses due to its motion
limiting friction
the frictional resistance acting when a body is on
the point of moving
leading shoe
the shoe in a drum brake that tends to be pulled
against the drum surface due to the rotation of the
drum
malleable cast
iron
a cast iron produced when white cast iron is
reheated to 800∞C and soaked for 30 to 50 hours; a
moderate cooling rate produces graphite rosettes in
a pearlite matrix while a slow cooling rate produces
graphite rosettes in a ferrite matrix – it has higher
tensile properties than gray cast iron, and is also
tougher
matrix
the continuous phase in a material that holds the
other constituents together
mechanical
energy
a body’s capacity to do work
mechanical work
the work done when a force acts upon a body and
produces a displacement is mechanical work; it is
determined by the product of the force and the
displacement of the point of application of that
force
nodules
Carbon is deposited in nodular or spherical forms
orthogonal
drawing
a method of drawing utilising two dimensional
views and dimensions to give a shape and size
description of components – orthogonal drawing
must follow AS1100 Drawing Standards
part-section
a standard method of drawing used to show the
relevant interior details of part of the component as
visible outline
Pascal’s Principle
if the pressure at any point in a liquid that is
enclosed and at rest, is changed, then the pressure at
all points in the liquid is changes by the same
amount
pearlite
a microstructural constituent consisting of two
phases, ferrite and cementite, pearlite has a lamella
or plate like structure, alternating between plates of
ferrite and plates of cementite; it is drawn in a
microstructure to give the appearance of a finger
print
phase
a physically distinct, chemically homogeneous part
of a material
pictorial drawing
a three dimensional drawing used to show the
shape, and sometimes size description of a
component; isometric projection is one method of
drawing pictorials
pneumatic tyres
vulcanised rubber tubular tyres that use air to inflate
the tyre or inner tube
polar coordinates
coordinates used in CAD that take radial
measurements from the last point entered, using the
angle measured in a counterclockwise direction
from the positive x axis
xix
xx
potential energy
the energy a body possesses due to its position; it is
determined by the amount of work required to lift a
body through a vertical height
power
power is the time rate of doing work, and is
determined by the ratio of work done to the time
taken to do the work
pressure
pressure is force per unit area
reactive force
a force that acts as a response to an applied force or
applied forces; Newton, in his third law said that to
every action there is an equal and opposite reaction
reducing scale
a drawing scale that is used to enable large objects
to be drawn to scale on a piece of drawing paper (a
scale of 1:2 means that you draw the object half full
size, whilst a scale of 1:10 means that you draw the
object one tenth full size
relative
coordinates
coordinates used in CAD that take actual
measurements along the x and y directions from the
last point entered – negative values are frequently
used
Rockwell
hardness test
a hardness test that uses a variety of indentors,
including an industrial diamond cone, and a 1.5 mm
and 3 mm hardened steel ball, the indentor is
initially pressed into the surface of the material by a
minor load of 10 kg and the major load is then
applied
rosettes
Carbon is deposited around a central core with
radiating arms
service properties
the performance properties of a manufactured
component when being used for its designed
purpose
servo-assisted
the assistance in a drum brake of the rotating drum
that tends to pull the brake shoe against the rotating
surface of the drum
servo-assisted
brake
drum brakes that are designed so the leading shoe
or shoes are pulled in towards the braking surface
and thus increase the braking force
shape description
a full definition of the shape of a component in
technical drawing, using a drawing or a number of
views of that component
shear stress
the internal resistance of a body to a deforming
force that is tending to slide one part of the body
across another part of the body
size description
a full definition of the size of a component in
technical drawing, showing all the dimensions of
that component
solid solution
(substitutional)
an alloy system in which the atoms of one element
replace the atoms of the other element in the lattice
structure of the metal
spheroidal
graphite cast iron
abbreviated to SGCI, is a cast iron alloyed with
magnesium to produce nodules of graphite in the
cooling process; a moderate cooling rate produces
graphite nodules or spheroids in a pearlite matrix
while a slow cooling rate produces graphite nodules
or spheroids in a ferrite matrix
steel
ferrous metal that contains carbon of varying
amounts generally from 0.05% to 1.4%
strain
the ratio of change in length of a body with respect
to its original length;: it is calculated as deformation
per unit length
strain energy
the energy a body possesses due to its deformation;
it is determined by the amount of work done in
deforming the body
stress
a body’s internal resistance to an externally applied
force that tends to deform a body; it is calculated as
load per unit area
tensile stress
the internal resistance of a body to a deforming
force that is tending to stretch the body
tensile test
a test conducted on a prepared specimen, held in a
gripping device and a gradually increasing axial
load applied which stretches the specimen – the
applied load is plotted against the extension, to
produce a load-extension graph
trailing shoe
shoe in a drum brake that tends to be pushed away
from the drum surface due to the rotation of the
drum
Vickers hardness
test
a hardness test that uses an industrial diamond
indentor in the shape of an inverted square pyramid
which is pressed into the surface of a material for
15 seconds
xxi
visible outline
lines that represent the edges of an object in a
technical drawing, they are represented as thick
dark continuous lines, usually of 0.5 mm thickness
when using A4 size paper
vulcanisation
a mechanism used to strengthen the mechanical
properties of rubber by forming sulphur cross-links
between the polymer chains
white cast iron
a cast iron produced when molten iron, containing
2.8% to 4.0% carbon, is rapidly cooled in a mould,
the resultant structure has dendrites of pearlite in a
cementite matrix; it is extremely hard and brittle
wrought iron
a ferrous metal containing little or no carbon; it
usually has slag inclusions which align in the
direction of working
Young’s Modulus also known as the modulus of elasticity where
stress is proportional to strain within the elastic
limit
xxii
Directive terms
The list below explains key words you will encounter in assessment tasks
and examination questions.
account
account for: state reasons for, report on;
give an account of: narrate a series of events or
transactions
analyse
identify components and the relationship between
them, draw out and relate implications
apply
use, utilise, employ in a particular situation
appreciate
make a judgement about the value of
assess
make a judgement of value, quality, outcomes,
results or size
calculate
ascertain/determine from given facts, figures or
information
clarify
make clear or plain
classify
arrange or include in classes/categories
compare
show how things are similar or different
construct
make, build, put together items or arguments
contrast
show how things are different or opposite
critically
(analyse/evaluate)
add a degree or level of accuracy depth, knowledge
and understanding, logic, questioning, reflection
and quality to (analysis/evaluation)
deduce
draw conclusions
define
state meaning and identify essential qualities
demonstrate
show by example
xxiii
describe
provide characteristics and features
discuss
identify issues and provide points for and/or against
distinguish
recognise or note/indicate as being distinct or
different from; to note differences between
evaluate
make a judgement based on criteria; determine the
value of
examine
inquire into
explain
relate cause and effect; make the relationships
between things evident; provide why and/or how
extract
choose relevant and/or appropriate details
extrapolate
infer from what is known
identify
recognise and name
interpret
draw meaning from
investigate
plan, inquire into and draw conclusions about
justify
support an argument or conclusion
outline
sketch in general terms; indicate the main
features of
predict
suggest what may happen based on available
information
propose
put forward (for example a point of view, idea,
argument, suggestion) for consideration or action
recall
present remembered ideas, facts or experiences
recommend
provide reasons in favour
recount
retell a series of events
summarise
express, concisely, the relevant details
synthesise
putting together various elements to make a whole
Extract from The New Higher School Certificate Assessment Support Document,
© Board of Studies, NSW, 1999.
xxiv
Braking systems
materials application – 1
Part 1 contents
Introduction.......................................................................................... 2
What will you learn?................................................................... 2
Development of braking systems..................................................... 3
Early history of brakes ............................................................... 3
The effect of engineering innovation ..........................................11
Investigating materials......................................................................13
Steels and cast iron for braking systems ....................................13
Brakes, steels and engineers ....................................................23
Brakes, cast irons and engineers...............................................32
Exercises ............................................................................................35
Progress check ..................................................................................47
Exercise cover sheet.........................................................................49
Part 1: Development of braking systems and materials application – 1
1
Introduction
Think of all the different types of braking systems, or methods, that you
could use to stop a bicycle – there are front and rear calliper brakes, and
back pedal brakes, in an emergency using your foot on the back wheel,
sliding the bike or ‘laying it down’ are also effective.
In this part you will examine the development of braking systems.
What will you learn?
You will learn about:
•
•
historical and societal influences
–
historical developments of braking systems
–
the effect of engineering innovations on people’s lives
–
environmental implications from the use of materials in braking systems
engineering materials
–
materials for braking systems.
You will learn to:
•
examine the changing applications of materials to components in
braking systems
•
discuss the social implications of technological change in braking systems
•
investigate the structure and properties of appropriate materials used
in braking systems.
2
Braking systems
Developing of braking systems
A brake is a device used to slow down or stop a moving object. It is also
used to hold a stationary vehicle or object at rest. It operates as a result
of friction by converting the energy of motion, kinetic energy, into
some other form of energy, usually heat energy.
Today brakes are used in motor vehicles, trains, lifts, aircraft, cranes,
bicycles and many other machines or vehicles.
Early history of brakes
In 1815, when Governor Macquarie was crossing the Blue Mountains, a
large tree branch was used to slow his carriage as it descended a steep
incline at Mount York.
The design of the first bicycle, the Draisine, patented in 1818 by Freiherr
Drais, used the rider's feet to stop the bike. You have probably used this
‘braking system’ to stop your bike.
External shoe brake
The earliest known type of mechanical braking system was a lever brake
introduced on horse-drawn wagons in the 18th century. It consisted of a
curved wooden block or shoe, designed to press against the wrought
iron rim of the wheels when a force was applied through a system of
levers and linkages. This system was effective as it was used in
conjunction with the horse when stopping the wagon. It was mainly used
as a parking brake.
3
Figure 1.1 1850s Brake
Figure 1.2 Early model braking system
From the 1830s, steam carriages used a hand operated braking system,
the application still being through linkages and levers to wrought iron
brake shoes rubbing against cast iron wheels.
Up until the 1870s hand-operated brakes were used on the tender and
vans of steam-driven railway carriages. In 1875 Westinghouse developed
a compressed air brake, which operated automatically if the train
separated. It was made compulsory on all trains in Britain in 1889.
It was the advent of the motor vehicle that caused braking technology to
develop. Initially hand-operated lever brakes were used, operating
directly onto the solid tyre tread, similar to the contemporary horse
drawn carriages. They were quite effective at low speeds but were not
4
Braking systems
effective in wet weather and would damage the tyres. Karl Friedrick
Benz applied this system to his first internal combustion vehicle in 1885.
Contracting band brake
In the late 1890s the use of pneumatic tyres made the external shoe
brake obsolete. The contracting band brake was developed. It operated
on the principle of a band acting on a hub. The brake was less effective in
wet weather. Dirt often became trapped between the lining and the hub
reducing the braking effectiveness. The band brake would not operate
when the vehicle was in reverse.
The drum brake
The next development in braking systems was the introduction of the
drum brake. A mechanically operated drum brake was first used by Louis
Renault in 1902. The drum brake was unaffected by dirt and weather
since the brake shoes were enclosed in the brake drum. The mechanical
drum brake was inefficient due to frictional losses in the joints. Severe
wear of the moving parts required constant maintenance, and the system
had to be meticulously balanced to deliver equal, safe braking forces to
the brake shoes.
Dust seal
Slave cylinder
Brake shoe and lining
Return spring
Wheel hub
Figure 1.3 Hydraulic drum brake assembly
Courtesy: Newgas Automotive Taren Point
© LMP
5
Figure 1.4 The brake drum fits over the brake assembly
Courtesy: Newgas Automotive Taren Point
© LMP
Hydraulic braking systems
In 1904, the Mutton Car Company heralded a revolution in braking technology
when a hydraulic system was introduced to operate the rear brakes.
By 1910 most motor vehicles were using two independent and separate
brake operating mechanisms on the rear wheels; the first a hand operated
lever system, the second either a pedal operated mechanical system or a
pedal operated hydraulic system.
Front wheel brakes
Around this time front wheel brakes also began to appear. The advantages
of having brakes on all four wheels was that the stopping distance could be
reduced. When brakes are applied on a motor vehicle, much of the weight
force of the vehicle is thrown forward onto the front wheels, leaving the
rear brakes relatively ineffective.
Figure 1.5
6
Front wheel brakes required
Braking systems
The introduction of hydraulically assisted ‘servo-brakes’, and, in 1924
the ‘vacuum servo’, led to power assisted braking systems. During the
1930s hydraulic systems were gradually introduced to all braking
systems in vehicles.
Disc brakes
The next major development in braking systems was the use of disc
brakes. Although originally developed in the early 1900s, it was
regarded as a ‘new invention’ at the London Motor Show in 1951.
Previously disc brakes had only been used on motorcycles, aeroplanes
and trucks but not motor cars. This development revolutionised the
braking industry, so much so, that by the 1960s the use of disc brakes
was widespread in British and European cars.
Figure 1.6 Disc brake
Brake linings and pads
Along with the development of brake mechanisms, frictional material for
brake linings also developed. Early brake pad liners were made by
weaving an asbestos yarn into the desired shape. Today the brake pad
linings are produced from a combination of many materials: fibres, such
as glass fibre, kevlar, steel wool and carbon fibres; fillers, such as clay,
calcium carbonate, barytes, fine metal and schist; binders, such as phenol
formaldehyde; and friction modifiers, such as elastomers, brass and zinc,
which are moulded into shape and cured.
Anti-lock braking systems
Another major development in recent years is the anti-lock braking
system, (ABS). This system prevents wheels from locking during
7
emergency braking situations, enabling drivers to steer the vehicle while
stopping.
ABS uses wheel speed sensors to detect rapid deceleration. An
electronic control unit constantly monitors the wheel speed information,
and when an emergency situation is detected, it activates an hydraulic
unit with solenoid valves which build up and release pressure, ‘pumping’
the brakes much more effectively than a driver can, to prevent the wheels
from locking.
The following table will be supplemented with more specific historical
perspective throughout the module, along with related developments in
areas of materials and technology.
Brake system
Operation
Advantages
Disadvantages
External shoe
brake
∑ hand operated
by lever
18th –20th century
∑ uses linkages
∑ appropriate for
horse-drawn
vehicle
∑ needed a large
force to
operate
∑ cheap to
produce
∑ worked only as
a supplement
to the horses
Used on horsedrawn carts
∑ pressure
applied to
shoe, forced
against metal
rim
∑ mainly a
parking brake
∑ supplemented
the horse
Contracting band ∑ contracting
brakes
band acting on
Late 19th century
Early model cars
a hub
∑ hand operated
From 1902
Cars and trucks
8
∑ materials
cheap and
easy to obtain
∑ appropriate for
early model
cars with
rubber tyres
∑ not effective in
wet and dusty
conditions
∑ safety problem
due to exposed
linkage
∑ would not
operate in
reverse
∑ new technology
needed
∑ not effective in
wet and dusty
conditions
∑ steel industry
developing
∑ not effective as
parking brakes
∑ internal
expandingshoes
∑ operated in all
types of
weather
∑ brake fade
∑ mechanically/
hydraulically
operated
∑ servo-assisted
∑ worked only in
forward motion
Drum brakes
∑ simple
technology
available
∑ heat dissipation
problems
∑ two
independent
systems
Braking systems
Disc brakes
1930s in trucks
From 1952 in cars
∑ calipers force
pads against
the rotating
disc
∑ more efficient
∑ hydraulically
operated with
power
assistance
∑ lighter weight
∑ special design
required to
operate the
disc brake as a
hand brake
∑ little or no fade
∑ improved heat
dissipation
∑ easier pad
design
∑ special design
needed for
parking brake
∑ power
assistance
required
∑ more
expensive
Turn to the exercise sheets and complete exercise 1.1.
The contracting band system – a case study
In this section of work you will follow a case study of one braking
system, the contracting band system. You will see why there was a need
to develop a system to replace the hand operated lever brake, and look at
the different engineering designed systems that were developed.
References: to complete this case study, the Historical Development of
Braking Systems and the history from Materials for Braking Systems
from this module were used, along with the Repco-PBR Sound Filmstrip
from the ‘Stop – Braking Systems for Cars’.
In the mid-nineteenth century simple hand operated lever brakes were used on
horse drawn coaches, steam carriages and railway locomotives. They were quite
effective at low speeds, were excellent as parking brakes but were not as
effective in wet weather.
From1800–1880, wooden wheels with wrought iron rims were used on horse
drawn carriages. The lever brake used a wooden shoe and leather liner. From
the 1830s, steam carriages, both rail and road, used cast iron wheels with
wrought iron brake shoes. Both systems used an external shoe brake.
In 1841, Goodyear patented the vulcanisation of rubber which enabled the use
in 1871 of solid rubber tyres on wheels. In 1888 Dunlop patented pneumatic
tyres, which meant the eventual end of the external shoe brake.
In 1895 the Michelin brothers had begun the move towards replacing steelrimmed wheels with pneumatic rubber tyres and found that the old technology
of applying a brake shoe directly to the tyre was unsatisfactory.
As a direct result, contracting band brakes were developed. These brakes
operated on the principle of a band acting externally on a hub or drum. Two
early devices attempted to apply the force of friction to the axle and to a drum
9
on the axle. One used wooden blocks inside an external, flexible, contracting
steel band.
In 1899, Daimler used a cable anchored to the chassis and wound around a
drum. When the cable tightened while the car was moving forward, the rotation
of the drum increased the tightness and grip of the cable, thus increasing
braking efficiency. This was called servo-assistance, and is still an important
factor in the design of expanding shoe drum brakes.
Both band brakes and cable brakes proved ineffective. With band brakes, dirt
often became trapped in-between the lining and the hub, reducing the braking
effectiveness. It was also considerably less effective in wet weather. Neither
system would operate when the vehicle was in reverse.
The design solution was the development of the expanding shoe drum brake.
The drum brake – a case study
In this section of work you will follow a case study of another braking
system, the drum brake. You will look at the different engineering
designs that were developed, along with the materials used.
The introduction of the vulcanisation of rubber and the subsequent
development of pneumatic tyres led to the demise of the externally
applied shoe brake.
The use of band brakes and cable brakes also proved ineffective as cars
became heavier and faster. They were considerably less effective in wet
weather, dirt often became trapped in-between the lining and the hub,
and neither system would operate when the vehicle was in reverse.
Mechanically operated drum brakes were first used by Louis Renault in
1902. The design used two hinged shoes which were forced apart by an
interposed arm pushing each shoe against the inside of a rotating drum.
The brake was unaffected by dirt and weather since the brake shoes were
enclosed inside the brake drum.
Initially, with the two shoes pivoted separately at their lower end, one shoe
was self-energising and the other was not. If the drum is considered to be
rotating clockwise, the right hand shoe is tending to be pulled against the
drum surface, so that the braking effect is increased. This shoe is called the
leading shoe. The other shoe is pushed off by the effect of the rotating
drum and its braking effect is reduced. This shoe is called the trailing
shoe. The leading shoe wears more quickly as it does more work.
Design advancement saw the introduction of a brake with the two shoes
linked together thus giving the effect of two leading shoes. This is known
as a servo-assisted brake and is the basis for the drum brakes used today.
10
Braking systems
Mechanical operation of the drum brake was through a series of levers,
rods and Bowden cables to a cam which pushed the shoes apart.
Hydraulics were introduced to improve the operation of the systems, and to
provide equal, safe braking forces to the brake shoes.
Early drums were made from pressed, medium carbon steel, however, they
were not strong enough to maintain their shape, they were easily scored
and were poor conductors of heat.
A nickel-iron alloy was used in the 1920s; it had greater rigidity and better
friction properties. Cast aluminium alloys with cast iron liners were also
used but were considered too expensive. Grey cast iron was found to be
the best material for use in drums, but this was replaced in the 1970s with
spheroidal graphite cast iron, SGCI, providing greater toughness.
Disc brakes have now replaced drum brakes on the front wheels in all new
cars and on all four wheels in many models.
The effect of engineering innovation
In this section of work we will examine the effect of engineering
innovations on personal transport since 1940, and compare systems in
place then, with the systems in place today. We will especially look at
the effect that improved braking system technology had on the lives of
people who have lived through this era.
During the 1940s not many families were not able to afford a car for
personal transport. Transport around the towns was by bicycle. Deliveries of
bread, milk, fruit and vegetables, and of ice for the ice-chest, were made
door to door by vendors using a horse and cart. Personal transport around the
cities was also available by tram or train.
Brakes on bicycles were either a ‘back-pedal’ brake, or the conventional
caliper brake on the rear wheel, similar to today’s bicycles. Many bicycles
did not have brakes. Bikes were stopped using the fixed wheel drive through
the pedals, or by applying a foot to the tyre, if a freewheel drive was used.
The braking system on a moving cart was always the horse. When the cart
was stationary, the hand brake, consisting of a lever, linkages and a brake
shoe with a leather liner, was applied. The brake shoe applied a force to the
mild steel rim of the wooden wheel. Like the cars of today, there were two
systems in use.
The trams and trains used a braking system that is still in use today – metal
shoe brakes applied by an air-operated system.
Cars had mechanical brake systems, lever and cable operated for the hand
brake, and hydraulically operated for the foot brake.
11
Today personal transport is by car, train or bus, plane, mono-rail, light-rail
(tram), and sometimes even by skate board. Bicycles are used by many
people for recreation, sport and fitness, and for travel to and from school and
work. Horse and sulkies are only seen at country shows, at the Royal Easter
Show, or special events and are seldom used for personal transport.
The braking systems for trains, trams and bikes remain basically the same,
however, cars have seen tremendous improvements. Most families have a
car. The cars travel at much greater velocity than the cars of the 1940s and
require far greater stopping power.
Social and economical conditions have changed dramatically since the
1940s. Ask your grandparents or people that you know, or that your family
knows, to describe some of these changes and the effect that the changes had
on their lives.
12
Braking systems
Investigating materials
The selection of materials for braking systems is influence by:
•
mechanical properties –
ductility, hardness, hardenability, elasticity and toughness
•
physical properties –
density, thermal expansion and conductivity
•
chemical properties –
oxidation and corrosion
•
comparative cost and availability of materials
•
manufacturing properties –
critical when selecting methods of forming, machining, casting,
welding, surface treatment and heat treatment to be used
•
service properties –
such as wear resistance, strength, hardness, toughness, fatigue,
corrosion resistance, environmental effects and safety are important
selection criteria when considering the material for the product or
component.
Steels and cast irons for braking
systems
Steels and cast irons have been used in braking systems for many years.
They are ferrous metals that contain varying amounts of carbon along
with other alloying elements. For this course you will focus on plain
carbon steels and cast irons.
Steels contain 0.05% – 1.4% carbon.
Commercial cast irons contain from 1.8% – 4% carbon.
13
Wrought iron
Historical perspective
Between 1850 and 1870 the use of wrought iron produced by the
‘puddling’ process, increased. Prior to this it was used as a supplement
to wood and cast iron. In London, 1839, wrought iron was used for small
trusses to span the roof of Euston station. In Paris, 1889, 7417 tonne of
wrought iron was used in the construction of the 300 m high Eiffel
Tower.
The use of wrought iron braking systems
From 1800 and 1880, wooden wheels with wrought iron rims were used
on horse-drawn carriages. The lever brake had a wooden shoe, and
leather liner. From the 1830’s, steam carriages used cast iron wheels with
wrought iron brake shoes. In 1841, Goodyear patented the vulcanisation
of rubber which enabled the use in 1871 of solid rubber tyres on wheels.
In 1888 Dunlop patented pneumatic tyres, which meant the end of the
external shoe brake, and the end of wrought iron rims and shoes.
The summaries detail composition and structure, including the
appropriate microstructures. They incorporate properties of the material,
then specify manufacturing and service properties. Manufacturing
technologies and the modification of properties are also included.
Wrought Iron – used in wheel rims of carriages, 1860s
•
Composition
Iron, with slag inclusions.
i
Structure
Equiaxed grains of iron, slag inclusions aligned in the direction
of rolling.
ii
Properties
Very soft, malleable, ductile, tough, (due to the iron matrix).
•
Availability
–
•
Manufacturing technology
–
14
Readily available, (produced in puddling furnace).
Hot rolled into strips, shaped and hot welded by blacksmith,
heat shrunk onto rim.
Braking systems
•
Manufacturing properties
–
•
Service Properties
–
•
Adequate hardness and toughness.
Modification of properties
–
•
Easily formed, good thermal expansion, soft, malleable, ductile.
Can be work hardened or alloyed.
Microstructure phases
–
Iron and slag inclusions.
Ferrite
Slag
Figure 1.7 Microstructure, Wrought Iron
A microstructure
A microstructure is a magnified view of portion of the material as seen under a
reflecting light microscope. Magnification is usually between 150x and 500x.
When viewed using the reflecting light microscope, the grain structure is
visible. The method of determining the structure is outside the scope of the
syllabus. Interpretation of the structure and the drawing of the structure are
vital for interpretation of the syllabus in terms of the properties of that
material.
Steels
Steel has been used for 2000 years but it was not until the1850s that the
steel industry began to develop with the availability of cheaper steel. In
1856 Henry Bessemer announced the development of his Bessemer
Converter, a tilting furnace that allowed the air to be blasted through
liquid pig iron to decarburise the molten metal to produce steel. In the
1860s the Seimens open-hearth furnace was introduced.
15
Time line
1869
first transcontinental railway in US.
1875
Westinghouse Brake developed for railways. (adopted 1889)
1877
Reinforced concrete patented by Monier.
1877
British Board of Trade authorised the use of steel in bridge
construction.
1883
Brooklyn Bridge completed.
1885
Rover ‘safety’ bicycle produced.
1893
Benz produced his first four wheeled ‘car’.
1893
Henry Ford’s first automobile.
1903
Henry Ford established the mass production technique.
Cheaper steel and better quality control
Pierre Martin, in 1864 was able to produce steel in the open-hearth
furnace by adding a large quantity of scrap metal to the pig iron. This
enabled the recycling of scrap and better quality control of the steel
produced. In 1880 Carnegie built the first big furnace in the United
States.
Developments in mechanisation, and technology enabled the US to
produce three times the quantity of steel than England did by the end of
the century. The world output of steel rose from 500 000 tonne in 1870 to
28 000 000 tonne in 1899.
Property/structure relationships
Property/structure relationships is very important in understanding steel
and its use by engineers. The microstructure of steels and how the
structure affects the properties of the various steels must be known
Equilibrium structure
In steels, the equilibrium structure is very similar to the annealed
structure and can be considered the same for this course. The structure of
steel can be modified by heat treatment, therefore the type of structure
shown must be specified.
The microstructure of steel
The microstructures show only two phases, ferrite and cementite. It is
the amount of each phase and the distribution of the phases throughout
the microstructure that determine the properties of the steel.
16
Braking systems
A phase
A phase is a chemically distinct, homogeneous part of a material. Ferrite is
one phase seen in the microstructure of steel, cementite is the other phase.
Ferrite
Ferrite is a very soft, ductile phase comprising of BCC iron with a very
small amount of carbon dissolved in the iron. The amount of carbon
dissolved varies with the temperature, ranging from 0.008% at room
temperature to 0.025% at 723ºC.
Cementite
Cementite is a very hard, brittle phase comprising of Body Centred Cubic
Structure (BBC) iron with 6.67% carbon dissolved in the iron. It is a
compound and thus has a chemical formula, Fe 3C.
Pearlite
Pearlite is a mixture of the two phases, ferrite and cementite and is
therefore not a phase. It appears in most steel equilibrium
microstructures. Pearlite is a lamella or plate-like structure with
alternating thin plates of ferrite and cementite. It is a micro-constituent
as it is a feature in the microstructure.
Now consider the property/structure relationships of various steels.
Dead mild steel – used in wheel rims, 1880s
•
Composition
Iron, 0.05% to 0.15% carbon.
•
Equilibrium Structure (0.15%)
–
•
Properties
–
•
Very soft, malleable, ductile, tough (due to ferrite matrix).
Availability
–
•
Equiaxed grains of ferrite, small grains of pearlite
(approximately 12%). (Pearlite is a lamella structure, alternate
thin plates of ferrite and cementite)
Readily available, produced in Bessemer or Open-hearth
furnace.
Manufacturing technology, 1880
–
Hot rolled into strips, shaped and forge welded by blacksmith,
heat shrunk onto rim.
17
•
–
•
Service properties
–
•
Adequate hardness and toughness.
–
•
Easily formed, soft, malleable, ductile.
Can be work–hardened or alloyed.
Microstructure
–
Phases, ferrite and cementite (12% of the structure is in the form
of grains of pearlite, a lamella structure, thin alternating plates
of ferrite and cementite).
Ferrite
Pearlite
Figure 1.8
Microstructure, Steel, 0.1% C
Mild steel – used in brake nuts and bolts, 1920s
•
Composition
–
•
Equilibrium Structure (0.3% C)
–
•
Easily formed by hot working, good machinability.
Service Properties
–
18
Hot rolled into bars, hot forged to shape, thread, machine
formed.
–
•
Readily available, high steel production.
Manufacturing technology 1920
–
•
Soft, malleable, ductile, tough (due to ferrite matrix).
Availability
–
•
Equiaxed grains of ferrite, small grains of pearlite,
(approximately 30%).
Properties
–
•
Good shear and tensile strength, tough.
Braking systems
•
–
•
Can be work hardened or alloyed.
Microstructure
–
Phases, ferrite and cementite (30% of the structure in the form
of grains of pearlite, a lamella structure, thin alternating plates
of ferrite and cementite).
Pearlite
Ferrite
Figure 1.9 Microstructure, Steel, 0.3% C
Medium carbon steel – used in brake springs, 1950s
•
Composition
–
•
–
•
Good formability by hot working, heat treatable to produce
‘spring’ properties.
Service properties
–
•
Hot rolled into rods, hot drawn to wire shape. Methods of
producing springs; draw into wire, form the helical shape,
harden and temper the spring.
–
•
Readily available, very high steel production.
–
•
Tough, heat treatable, hard, good machinability.
Availability
–
•
Small equiaxed grains of ferrite, and large grains of pearlite,
(approximately 75%).
Properties
–
•
Resilient, high elasticity, not corroded by brake fluid.
–
Can be heat treated to produce different properties, or alloyed.
19
•
Microstructure
–
Phases, ferrite and cementite (75% of structure grains of
pearlite).
Ferrite
Pearlite
Eutectoid steel – used in brake cable wire, 1950s
Note eutectoid steel is steel that contains 0.8% carbon
•
Composition
–
•
–
•
Microstructure
–
20
Good toughness and high tensile strength.
–
•
Good formability by hot working.
Service properties
–
•
1950: hot rolled into rods, hot drawn to wire.
–
•
Manufacturing technology
–
•
Heat treatable.
Availability
–
•
Grains of pearlite, (100%).
Properties
–
•
Iron, 0.8 % carbon.
Phases, ferrite and cementite in the form of grains of pearlite.
Braking systems
Pearlite
High carbon steel – used in brake cable wire, 1950s
•
Composition
–
•
–
•
Heat treatable, hard with low ductility, (brittle) high tensile
strength, poor machinability.
Availability
–
•
Grains of pearlite surrounded by a continuous precipitation of
cementite at the grain boundaries.
Properties
–
•
Iron, 0.6% to 0.9 % carbon.
Hot rolled into rods, hot drawn to wire.
•
–
•
Service properties
–
•
Good toughness and high tensile strength.
–
•
Good formability by hot working, heat treatable.
Microstructure
–
Pearlite
Phases, ferrite and cementite in the form of grains of pearlite,
surrounded by a continuous precipitation of cementite at the
grain boundaries.
Cementite
21
Tool steel – used in cutting tools, 1950s
•
Composition
–
Iron, 0.9% to 1.4 % carbon.
i
Grains of pearlite surrounded by a greater continuous
precipitation of cementite at the grain boundaries.
ii
Properties
Heat treatable, very hard with very low ductility, (brittle) lower
tensile strength, very poor machinability.
•
Availability
–
•
–
•
Heat treatable for good toughness and tensile strength, very hard
wearing surface.
–
•
Good formability by hot working, heat treatable.
Service properties
–
•
Hot worked into shape, heat treated to obtain desired properties.
–
•
Microstructure
–
Phases, ferrite and cementite in the form of grains of pearlite,
surrounded by a greater continuous precipitation of cementite at
the grain boundaries.
Pearlite
Cementite
22
Braking systems
Brakes, steels and engineers
Let’s now consider how a materials engineer involved in the
development of braking systems for a car manufacturer would use the
study of steels and the relationships between the structure and properties
of those steels.
A report
A materials engineer is required to prepare a report on the selection of plain
carbon steels for use in the production of various components for a brake
manufacturing company. Portions of that report are given in this example.
The five steels to be considered are; 0.1%, 0.3%, 0.6%, 0.8% and 1.2%
carbon steels.
A report by a materials engineer
Abstract:
Steel used in braking systems.
Topic:
Investigate the affect of micro-constituents on the properties of
steel: Five steels are to be considered. They are; 0.1%, 0.3%, 0.6%,
0.8% and 1.2% carbon steels.
The microstructures are shown for each of these five steels.
perlite
cementite
ferrite
0.1% C
Figure 1.14
0.3% C
0.6% C
0.8% C
1.2% C
Steel microstructures
Background
i
The micro-constituents.
There are only two phases present, ferrite and cementite. Pearlite, a
lamella structure of alternating plates of ferrite and cementite phases
is a micro-constituent of all of the microstructures.
ii
The properties of each of the phases.
Ferrite: is soft, malleable and ductile. Cementite: is very hard and
brittle.
23
iii
The structure of pearlite and the phases that are present in pearlite.
Pearlite, is a lamella structure of alternating plates of ferrite and
cementite phases. Pearlite is a micro-constituent of all of the
microstructures, and appears in the microstructures as a ‘finger print’
pattern.
The steels
Each of the five steels will be compared by:
a
listing two mechanical properties of the steel
b
explaining in terms of the microstructure why the steel possesses
these properties
c
stating two methods that may be used to modify these properties
d
identifying one example of where the steel could be used in brakes.
0.1% carbon steel
a
Two mechanical properties – malleable and ductile.
b
Microstructure/properties – the microstructure consists mainly of
ferrite grains, with a small amount of pearlite in the form of plates of
ferrite and cementite. The ferrite is soft, ductile and malleable, and
as the ferrite is the continuous phase, and the predominate phase, the
mechanical properties are those of ferrite.
c
Modification of properties – the properties can be modified by cold
working or alloying. The steel cannot be hardened by heat
treatment.
d
Use in brake systems – backing plate for discs.
0.3% carbon steel
24
a
Two mechanical properties – ductile and tough.
b
Microstructure/properties – the microstructure consists mainly of
ferrite grains, with approximately 30% of pearlite in the form of
plates of ferrite and cementite. The ferrite is soft, ductile and
malleable, and is still the continuous phase. Due to the increased
amount of cementite the UTS and hardness is higher than that of the
0.1% C steel.
c
Modification of properties – the properties can be modified by cold
working or alloying. The steel cannot be hardened by heat treatment.
d
Use in brake systems – nuts and bolts.
Braking systems
0.6% carbon steel
a
Two mechanical properties – hard and tough.
b
Microstructure/properties – the microstructure consists of ferrite
grains, with approximately 75% of pearlite in the form of plates of
ferrite and cementite. The ferrite is soft, ductile and malleable, and is
still the continuous phase. Due to the increased amount of cementite
the UTS and hardness are higher than that of the 0.3% C steel.
c
Modification of properties – the properties can be modified by using
heat treatment to harden and temper the steel. Alloying can also be
used.
d
Use in brake systems – brake springs.
0.8% carbon steel (Eutectoid steel)
a
Two mechanical properties – very hard with a high tensile strength.
b
Microstructure/properties – the microstructure consists of pearlite in
the form of plates of ferrite and cementite. Due to the increased
amount of cementite and its distribution throughout the
microstructure, the UTS and hardness are very high.
c
used.
d
Use in brake systems – brake cable wire.
1.2% carbon steel
a
Two mechanical properties – very hard with very low ductility, that
is, it is very brittle.
b
Microstructure/properties – the microstructure consists of pearlite
grains in the form of plates of ferrite and cementite, surrounded by a
continuous phase of cementite. Due to the increased amount of
cementite and its distribution throughout the microstructure as a
continuous phase, the hardness is very high, and ductility very low.
c
used.
d
Use in brake systems – cutting tools.
Conclusion and recommendations
The five steel all have applications for the braking systems being
developed. The various applications are listed for each steel. It is
recommended that the steels continue to be used for these applications.
25
Cast irons
Another ferrous metal that has a wide ranging use in braking systems is
cast iron. A materials engineer must be familiar with the structure of the
cast irons and the relationship between the structure and the properties.
Until the introduction of the blast furnace in the middle ages there was no
means of producing molten iron in quantity for casting. By the fifteenth
century the casting of iron, made possible by higher furnace temperatures
and the production of an iron having a relatively high carbon content,
enabled cast iron to be used, especially in the development of artillery.
By 1700 the blast furnace had been progressively developed enabling the
temperature to be raised sufficiently to allow the metal to be cast into
pigs. The addition of silicon to the re-melted pig iron produces cast iron.
Timeline
1700
Coke-smelting iron developed and horse-drawn railway lines
used in mining and canal transport.
1767
Rails cast at Coalbrookdale.
1776
Watts Steam Engine invented.
1779
Iron Bridge at Coalbrookdale opened.
1801
Trevithick produced a steam road carriage.
1804
Trevithick produced a steam railway locomotive.
1805
Surrey railway opened.
1819
McAdams published A practical Essay on Roads.
1829
Stephenson produced the Rocket steam driven locomotive.
1830
Liverpool to Manchester railway.
1851
Great Exibition, Crystal Palace built of cast iron, with wrought
iron used for tension components.
1865
Red Flag Act, (limited the development of road steamers)
enacted.
1893
Red Flag Act repealed.
1950s
Spheroidal graphite cast iron developed.
Up until the 1870s hand-operated brakes were used on rail carriages.
Wrought iron shoe brakes were used on the cast iron wheels. In 1875
Westinghouse developed a compressed air brake, which operated
automatically if the train separated. In 1889, automatic, continuous
power braking systems were made compulsory on all trains in Britain.
26
Braking systems
Property/structure relationships
The syllabus requirement of property/structure relationships is also very
important in understanding cast iron. The microstructure of cast irons and
how the structure affects the properties of the various cast irons must be
known.
The microstructure of cast iron
Interpretation of the structure of cast iron and the drawing of the
microstructure is vital to the interpretation of the syllabus in terms of the
properties of that material.
Except for white cast iron, the microstructures show three phases,
ferrite, cementite and graphite. The amount of each phase, the
distribution of the phases throughout the microstructure and the shape of
the graphite phase determines the properties of the cast iron.
You have been given definitions of a phase, ferrite, cementite and
pearlite in the previous notes on steel. A reminder that ferrite is a very
soft, ductile phase and cementite is a very hard, brittle phase.
Graphite
Graphite has little mechanical strength. The microstructural shape of the
graphite determines many of the properties of the cast iron.
•
Graphite flakes –
Graphite exists as flakes in the microstructure of grey cast iron. The
flakes have sharp ends. It is the sharp ends of the graphite flakes
that are responsible for the grey cast iron having a very low tensile
strength. Under tensile loading stress concentration occurs at the
sharp ends. This stress concentration causes the cast iron to fracture
at a low tensile loading.
•
Graphite nodules and rosettes –
Graphite exists as nodules in spheroidal graphite cast iron and as
rosettes in malleable cast iron. These shapes do not cause stress
concentration.
Steel matrix
The matrix surrounding the graphite can be ferrite, pearlite or a
combination of each. The matrix is often referred to as a ‘steel’ matrix to
describe this occurrence.
27
White cast iron – used in dies and wearing plates, 1950s
•
Composition
–
•
Structure
–
•
Extremely hard, strong in compression.
–
•
Excellent castability.
Service properties
–
•
Cast to shape, rapid cooling.
–
•
Readily available, high production.
Production technology
–
•
Very hard, zero ductility, (extremely brittle), not machinable.
Availability
–
•
Dendrites of pearlite in a matrix of cementite.
Properties
–
•
Iron: 1.8% to 3.6% carbon; 0.5% to 2.0% silicon.
Can be heat treated to produce pearlitic or ferritic malleable cast
iron.
Microstructure
–
Pearlite
Phases, ferrite and cementite in the form of dendrites of pearlite,
surrounded by a matrix of cementite.
Cementite
Figure 1.15 Microstructure, White Cast Iron
28
Braking systems
Grey cast iron – used in brake master cylinders, 1970s
•
Composition
–
•
Structure
–
•
Not corroded by brake fluid, strong in compression.
–
•
Excellent castability, excellent machinability
Service properties
–
•
Cast to shape, moderate cooling produces pearlitic grey cast
iron, slow cooling produces ferritic grey cast iron.
–
•
Readily available, high production.
–
•
Relatively soft and machinable. Strong in compression but
weak in tension.
Availability
–
•
Graphite flakes in a ‘steel’ matrix of either pearlite or ferrite, or
a combination of both.
Properties
–
•
Iron; 2.4% to 3.6% carbon; 1.0% to 3.0% silicon.
Pearlitic grey cast iron can be heat treated to produce a ferritic
matrix.
Microstructure
–
Pearlitic grey cast iron; phases, graphite flakes in a matrix of
ferrite and cementite in the form of pearlite.
–
Ferritic grey cast iron; phases, graphite flakes in a matrix of
ferrite.
Pearlite
matrix
Graphite
flakes
Pearlitic grey
cast iron
Ferrite
matrix
Ferretic grey
cast iron
Figure 1.16 Pearlitic grey cast iron, ferritic grey cast iron
29
Malleable cast iron – used in brake shoes, 1970s
•
Composition
–
Iron; 1.8% to 3.6% carbon, 1.0% to 3.0% silicon.
i
Structure
Graphite ‘rosettes’ in a ‘steel matrix’ of either pearlite or ferrite,
or a combination of both.
ii
Properties
Soft and ductile, malleable, tough, machinable.
•
Availability
–
•
–
•
Tough, strong in tension and compression.
–
•
Good ductility, excellent machinability
Service properties
–
•
White cast iron reheated to 800º C and soaked for 30 to 50 hours.
Moderate cooling produces pearlitic malleable cast iron, slow
cooling produces ferritic malleable cast iron.
–
•
Readily available.
Pearlitic malleable cast iron can be heat treated to produce a
ferritic matrix.
Microstructure
–
Pearlitic malleable cast iron; phases, graphite rosettes in a
matrix of ferrite and cementite in the form of pearlite.
–
Ferritic malleable cast iron; phases, graphite rosettes in a matrix
of ferrite.
Pearlite
matrix
Graphite
rosettes
Pearlitic malleable
cast iron
Ferrite
matrix
Ferretic malleable
cast iron
Figure 1.17 Pearlitic malleable cast iron, ferritic malleable cast iron
30
Braking systems
Spheroidal graphite cast iron – used in brake discs, 1980s
•
Composition
–
•
Iron; 3.0% to 4.0% carbon, 1.8% to 3.0% silicon.
Structure
–
•
Graphite ‘spheroids’ in a ‘steel matrix’ of either pearlite or
ferrite, or a combination of both.
Properties
–
•
Soft and ductile, malleable, tough, machinable.
Availability
–
•
Readily available since the 1960s.
–
•
Addition of magnesium produces nodules of graphite in a ‘steel’
matrix. Moderate cooling produces pearlitic spheroidal graphite
CI, slow cooling produces ferritic spheroidal graphite cast iron.
–
•
Good ductility, excellent machinability
Service properties
–
•
Tough, strong in tension and compression.
–
•
Pearlitic spheroidal graphite cast iron can be heat treated to
produce a ferritic matrix.
Microstructure
–
Pearlitic spheroidal graphite cast iron; phases, graphite nodules
or spheres in a matrix of ferrite and cementite in the form of
pearlite.
–
Ferritic spheroidal graphite cast iron; phases, graphite nodules
or spheres in a matrix of ferrite.
Pearlite
matrix
Graphite
spheroids
or nodules
Pearlitic spheroidal graphite
cast iron
Ferrite
matrix
Ferretic spheroidal graphite
cast iron
Figure 1.18 Microstructure of cast irons
31
Brakes, cast irons and engineers
Now consider how a materials engineer involved in the development of
braking systems for a car manufacturer would use the study of cast irons
and the relationships between the structure and properties of those cast
irons.
A report
A materials engineer is required to prepare a report on the selection of
various cast irons for use in the production of discs, brake drums and
wheel cylinders for a brake manufacturing company. Portions of that
report are given in this example.
The three cast irons to be considered are white cast iron, grey cast iron
and spheroidal graphite cast iron.
A report by a materials engineer
Abstract:
Cast iron in braking systems.
Topic:
The affect of micro-constituents on the properties of cast iron:
Cementite
Pearlite
White cast iron
Figure 1.19
Graphite flakes
Graphite nodules
Ferrite
matrix
Grey cast iron
Spheroidal graphite
cast iron
White, Grey and Spheroidal graphite cast iron
Background
i
The micro-constituents.
There are possibly three phases present, ferrite, cementite and
graphite. Pearlite, a lamella structure of alternating plates of
ferrite and cementite phases may possibly be a microconstituent of all of the microstructures. If the matrix is
pearlitic it will be a part of the structure, however, if the ‘steel’
matrix is ferritic, then pearlite will not be part of the structure.
32
Braking systems
ii
The properties of each of the phases.
Graphite: has little or no mechanical strength.
Ferrite: is soft, malleable and ductile.
Cementite: is very hard and brittle.
The cast irons
Each of the three cast irons will be compared by;
a
listing two mechanical properties of the cast iron
b
explaining in terms of the microstructure why the cast iron
possesses these properties
c
stating two methods that may be used to modify these properties
d
recommending where the cast iron could be used in brakes.
White cast iron
a
Two mechanical properties – extremely hard and brittle.
b
Microstructure/properties – the microstructure consists of
pearlite grains in the form of dendrites surrounded by a matrix
of cementite. The cementite is the continuous phase, and the
predominate phase. The mechanical properties are therefore
those of cementite.
c
Modification of properties – the properties can be modified by
heat treatment to produce a malleable cast iron.
d
Recommendation – cannot be used in brake systems as white
cast iron, but can be used to produce malleable cast iron for use
in discs, drums or wheel cylinders.
Grey cast iron
a
Two mechanical properties – very high compressive strength but
poor tensile strength.
b
graphite flakes surrounded by a ‘steel’ matrix consisting of
either pearlite (cementite and ferrite plates), or of ferrite.
Usually the ‘steel’ matrix is a combination of both.
33
The shape of the graphite flakes, with points at each end, causes
stress concentration to occur when the material is placed under
tension. This results in the material having very poor tensile
strength properties.
c
Modification of properties – pearlitic grey cast iron can be
modified by heat treatment to produce ferritic grey cast iron.
d
Recommendation – can be used in brake systems for drums or
wheel cylinders. Previously used in discs but SGCI is now
preferred, due to better toughness.
Spheroidal graphite cast iron
a
Two mechanical properties – very high compressive strength
and excellent toughness.
b
graphite spheroids surrounded by a ‘steel’ matrix consisting of
either pearlite (cementite and ferrite plates), or of ferrite.
Usually the ‘steel’ matrix is a combination of both.
The shape of the graphite spheroids results in the material
having good tensile properties, while the ‘steel’ matrix, whether
pearlitic or ferritic gives good compressive strength and
excellent toughness.
c
Modification of properties – pearlitic spheroidal graphite cast
iron can be modified by heat treatment to produce ferritic
spheroidal graphite cast iron.
d
Recommendation – can be used in brake systems in drums or
wheel cylinders and is excellent in discs, due to its toughness.
Conclusion and recommendations
The three cast irons all have applications for the braking systems
being developed. The various applications are listed for each cast
iron. It is recommended that the cast irons continue to be used for
these applications.
Turn to the exercise sheets and complete exercises 1.3 to 1.6.
34
Braking systems
Exercises
Exercise 1.1
a
Name four devices in which brakes are used.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
b
Describe the braking device the first horse-drawn carriage to cross
the Blue Mountains used to descend the very rough and steep track
down Mount York.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
c
Describe the braking system used to stop the first bicycle.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
d
Describe the earliest known type of mechanical braking system – the
lever brake – used on horse-drawn wagons.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
35
e
Name the materials used in brakes for steam carriages from the
1830s for the:
i
external shoes
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
ii
carriage wheels
___________________________________________________
___________________________________________________
___________________________________________________
f
State two advantages of simple hand-operated lever brakes used on
horse-drawn coaches, steam carriages and railway locomotives in the
mid-nineteenth century.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
g
h
Name the two developments, one in 1841, the other in 1888, which
greatly affected the design of tyres and led to the subsequent demise
of the externally applied shoe brake.
i
1841 _______________________________________________
ii
1888 _______________________________________________
Describe the contracting band brake, a direct result of Dunlop’s
patent.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
36
Braking systems
i
In 1899, a cable anchored to the chassis and wound around a drum
was used as a braking system.
i
Name the person who developed the cable brake.
___________________________________________________
ii
State the main disadvantage of this brake.
___________________________________________________
___________________________________________________
___________________________________________________
j
Identify the first person to introduce mechanically operated drum
brakes first used in cars in 1902.
_______________________________________________________
k
List two reasons why the introduction of front wheel brakes was an
important development.
_______________________________________________________
_______________________________________________________
l
What do the letters ABS stand for in braking systems?
_______________________________________________________
Exercise 1.2
Social and economical conditions have changed dramatically since the
1940s. You should talk to people who lived through these changes.
In the space below, list some of the changes and describe the effect that
the changes had on their lives. Make reference to the development of
cars and in particular, the resulting development in braking systems.
You may submit this exercise as a computer generated word processed
document and attach your work to this page.
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
37
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
Exercise 1.3
a
List three reasons pressed medium carbon steel, used in early drum
brakes, was not satisfactory.
i
___________________________________________________
___________________________________________________
ii
___________________________________________________
___________________________________________________
iii
___________________________________________________
___________________________________________________
b
c
Name three materials used for brake drums prior to the 1970s.
i
___________________________________________________
ii
___________________________________________________
iii
___________________________________________________
Until the 1970s grey cast iron was the main material used for brake
drums and brake discs.
i
Draw the microstructure of grey cast iron
ii
List the reasons for suitability as brake drum material:
___________________________________________________
___________________________________________________
___________________________________________________
d
Name the material that was used to manufacture brake drums after
the 1970s.
_______________________________________________________
38
Braking systems
Exercise 1.4
A materials engineer has to prepare a report on the selection of plain
carbon steels for use in the production of various components for a brake
manufacturing company.
Assume that you are the engineer, complete the unfinished sections of the
report. The report must be able to be interpreted by all of the directors.
a
Draw the microstructures for the following steels; 0.15%; 0.35%;
0.8% and; 1.1% carbon steels.
b
Label the phases present in each microstructure.
0.15% C
c
0.35% C
0.8% C
1.1% C
Outline the properties of both of the phases listed below.
Ferrite: _________________________________________________
Cementite: ______________________________________________
d
Describe the structure of pearlite, and name the phases that are
present in pearlite.
_______________________________________________________
_______________________________________________________
_______________________________________________________
e
For each of the four steels nominated:
i
list two mechanical properties of the steel
_______________________________________________________
_______________________________________________________
ii
explain in terms of the microstructure, why the steel possesses
these properties
_______________________________________________________
_______________________________________________________
_______________________________________________________
iii write two methods that may be used to modify these properties
_______________________________________________________
_______________________________________________________
39
iv
give one example where the steel could be used in brakes.
_______________________________________________________
0.15% carbon steel
i
Two mechanical properties:
_______________________________________________________
_______________________________________________________
ii
Microstructure/properties:
_______________________________________________________
_______________________________________________________
iii Modification of properties:
_______________________________________________________
_______________________________________________________
iv Use in brake systems:
_______________________________________________________
_______________________________________________________
0.35% carbon steel
i
_______________________________________________________
_______________________________________________________
ii
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
40
Braking systems
0.8% carbon steel (Eutectoid steel)
i
_______________________________________________________
_______________________________________________________
ii
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
1.1% carbon steel
i
_______________________________________________________
_______________________________________________________
ii
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
41
Exercise 1.5
A materials engineer has to prepare a report on the selection of various
cast irons for use in the production of brake discs, brake drums and wheel
cylinders for a brake manufacturing company.
Assume that you are the engineer, complete the unfinished sections of the
report. The report must be able to be interpreted by all of the directors.
a
Draw the microstructures for white cast iron, grey cast iron and
spheroidal graphite cast iron.
White cast iron
Grey cast iron
(ferritic)
Spheroidal graphite
cast iron (pearlitic)
b
Name the micro-constituents for each given microstructure by
labelling the phases present in each.
c
Outline the properties of each of the following phases:
Graphite – ______________________________________________
Ferrite – _______________________________________________
Cementite – _____________________________________________
d
For each of the three cast irons listed:
i
name two mechanical properties of the cast iron
ii
explain in terms of the microstructure, why the cast iron
possesses these properties;
iii describe how the properties may be modified
iv
42
write your recommendation for use of the cast iron in the
production of brake discs, brake drums and wheel cylinders.
Braking systems
White cast iron
i
_______________________________________________________
_______________________________________________________
ii
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
iv Recommendation:
_______________________________________________________
_______________________________________________________
Grey cast iron (ferritic)
i
_______________________________________________________
_______________________________________________________
ii
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
iv Recommendation:
_______________________________________________________
_______________________________________________________
43
Spheroidal graphite cast iron (Pearlitic)
a
_______________________________________________________
_______________________________________________________
b
_______________________________________________________
_______________________________________________________
_______________________________________________________
c
Modification of properties:
_______________________________________________________
_______________________________________________________
d
Recommendation:
_______________________________________________________
_______________________________________________________
Exercise 1.6
Visit a variety of web sites then explain why ABS are used on heavy
vehicles, how ABS work and their application to current model cars.
You may submit this exercise as a computer generated word processed
document and attach your work to the back of this page. State, in the
bibliography, at least two sources of information you located.
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
44
Braking systems
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
Bibliography
1
___________________________________________________
2
___________________________________________________
45
46
Braking systems
Progress check
In this part you explored the early history of brakes, and the relationship
between properties and applications of materials.
✓
❏
Disagree – revise your work
✓
❏
Uncertain – contact your teacher
Uncertain
Agree – well done
Disagree
✓
❏
Agree
Take a few moments to reflect on your learning then tick the box that best
represents your level of achievement.
I have learnt about
•
•
– historical developments of braking systems
– the effect of engineering innovations on people’s lives
– environmental implications from the use of materials
in braking systems
– materials for braking systems.
I have learnt to
•
•
•
examine the changing applications of materials to
components in braking systems
discuss the social implications of technological change in
braking systems
investigate the structure and properties of appropriate
materials used in braking systems.
During the next part you will continue to explore the history of brakes, and
the relationship between properties and applications of materials.
47
48
Braking systems
Exercise cover sheet
Exercises 1.1 to 1.6
Name:
_____________________________
Check!
Have you have completed the following exercises?
❐ Exercise 1.1
❐ Exercise 1.2
❐ Exercise 1.3
❐ Exercise 1.4
❐ Exercise 1.5
❐ Exercise 1.6
Locate and complete any outstanding exercises then attach your
responses to the cover sheet.
If you study Stage 6 Engineering Studies through a Distance Education
School/Centre (DEC) you will need to return the exercise sheets and your
responses at the completion of each part of a module.
If you study Stage 6 Engineering Studies through the OTEN Open
Learning Program (OLP) refer to the Learner’s Guide to determine which
exercises you need to return to your teacher along with the Mark Record
Slip.
49
Braking systems
materials application – 2
Part 2 contents
Introduction .......................................................................................... 2
What will you learn?...................................................................... 2
Development of disc brakes.............................................................. 3
Early history of disc brakes – a time line...................................... 3
The effects on society ............................................................... 4
Materials in braking systems............................................................. 6
Composite materials for braking systems .................................... 6
Testing of materials .......................................................................... 13
Hardness testing...................................................................... 13
Compression and tension testing .............................................. 16
Investigation of a braking system: materials analysis.................. 19
Exercises............................................................................................ 25
Progress check ................................................................................. 31
Exercise cover sheet........................................................................ 33
1
Introduction
In this part of the module you will explore the early history of disc
brakes, the developments in Britain and Europe and the reason why the
United States was so slow in adopting and developing this ‘new
technology’.
You will also explore the relationship between properties and
applications of materials in engineering.
•
•
–
historical developments of braking systems
–
the effect of engineering innovations on people’s lives
–
environmental implications from the use of materials in braking systems
–
materials for braking systems
–
testing of materials.
Students learn to:
•
examine the changing applications of materials to components in
braking systems
•
discuss the social implications of technological change in braking
systems
•
investigate the structure and properties of appropriate materials used
in braking systems
•
conduct relevant mechanical tests on materials.
2
Braking systems
Development of disc brakes
Although originally developed in the early 1900s, disc brakes for cars,
displayed at the London Motor Show, in Earls Court, in 1951, were
regarded as a new invention. This development ‘revolutionised’ the
automobile braking industry, so that by the 1960s the use of disc brakes
was widespread in British and European cars.
Early history of disc brakes – a timeline
1902
F W Lanchester, England, first
patent of automotive disc brake.
1904
W H Barrett, England, patented a
brake that pressed two external
discs onto a revolving inner disc,
(giving twice as much braking
force).
1905
Lanchester installed disc brakes on
a car. They was used as a
transmission brake, not attached
to the wheels.
1911
Metz, America, used multiple discs
inside the hub of each wheel.
1914
Autocarriers Ltd, A C cars, asked
Henry Ford to design a brake to
replace the ineffective band brake.
He designed an auxiliary system.
1919
AC used the auxiliary disc brake
behind the transmission.
1930s
Girling in England and Lockheed
in America developed caliper type
disc brakes for wheel installation.
1940s
WW II disc brakes were used on
the wheels of the Daimler armoured
car. Goodyear developed a disc
brake with a ventilated disc for use
on aircrafts, while Sikorsky
produced a sandwich disc of monel
metal.
1951
Girling bought a licence to produce
car disc brakes under Dunlop
patents. The prototype was
exhibited at the London Motor
Show.
1952
Jaguar won at Le Mans using
Girling disc brakes.
1955
All British racing cars used disc
brakes.
1956
Citroen, in France, used sliding
caliper disc brakes on its cars.
1961
Ford, England, fitted massproduced disc brakes to all models.
America, with its very big cars, was
slow to adopt the disc brake,
although a small West Coast
company, Airheart, supplied the
racing industry.
1958
All Indy 500 cars were fitted with
disc brakes.
1965
Ford America fitted disc brakes to
its Galaxy.
1966
Bendix supplied disc brakes to
Chrysler and Buick.
During the 1960s, front wheel disc
brakes were gradually used in some
mass produced cars, and became
standard during the 1970s. During
the 1980s four wheel disc braking
systems became more common,
while today most cars use disc
brakes on all four wheels.
3
The effects on society
Society is often affected by technology developments. Consider some
social implications and effects of various developments of the disc
brakes. Two samples with written solutions are given below. The
solutions have been researched and contain more detailed information on
disc brakes. Again, study methods can be used to highlight key phrases
or techniques used.
Sample 1
Briefly examine the contribution that the car racing industry had on the
developments of the disc brake, in England and in America.
Prior to 1951 the use of disc brakes was restricted to the aviation
industry and military vehicles. In 1951, Girling bought a licence to
produce car disc brakes under Dunlop patents, and exhibited the
prototype at the London Motor Show, in Earls Court, in the same year.
At the Le Mans race track in 1952 Jaguar fitted disc brakes to its Ctype roadster racing team and easily defeated the Ferrari V12 coupes
which had a much faster top speed. This victory encouraged the use of
disc brakes and in 1955 all British racing cars were fitted with disc
brakes. Even Ferrari fitted them in 1958.
American cars were much bigger and heavier and travelled at a higher
velocity. For these reasons the adoption of the disc brakes was not
used by the big companies, however, a small west coast company,
Airheart supplied the racing industry. In 1958 all Indy 500 cars were
fitted with disc brakes. In 1965, Ford America commenced using them
on their Galaxy model, and the next year Chrysler and Buick used
Bendix brakes.
The developments in both England and America were very much
influence by the results of the racing cars using disc brakes, and
resulted in the adoption of this system to the mass produced family
cars during the 1960s.
4
Braking systems
Sample 2
One reason that disc brake technology was developed was the need to
slow and stop cars that were becoming much faster and more powerful
each year. Another reason was the changing attitude of society to car
safety.
Discuss the social implications that led to the development of improved
braking technology and improved safety of cars from the 1960s.
Car safety became a big issue during the 1960s. Each year the number
of fatalities and injuries increased as a result of more accidents.
Poor braking, along with bad roads, the use of plain window glass in
windscreens, and poor safety awareness all contributed to the increase
injury and death by car accidents.
A gradual increase in public awareness lead to many changes that
actually slowed the rate of increase in accidents and in the past few
years dramatically reduced the road toll.
Design engineers were involved in improved road design, improved
brake technology, and in the introduction of laminated glass
windscreens.
The government introduced restrictions on the permissible alcohol
blood level that a driver had to obey.
The improvements in brake system technology included the use of
disc brakes, initially on the front wheels and later the rear wheels, as
well as the use of ABS, anti-lock braking systems.
All these improvements came about because of increased public
awareness for the need to improve car safety and to decrease the ever
rising fatality rate from car accidents.
Turn to the exercise sheets and complete exercises 2.1.
5
Materials in braking systems
In this section of work you will learn about composite materials and
how they can be used in braking systems.
Other composites are used in engineering, including cermets, asphalt,
glass reinforced polymers, timber laminates and plywood.
Composite materials for braking
systems
A composite material consists of two or more materials combined
to form the composite. The composite utilises the properties of the
individual materials to give distinctly different service properties to
the manufactured composite product.
When selecting materials to form a composite, the following must be
considered:
•
properties of the individual materials
•
cost of the materials
•
manufacturing properties
•
cost of production
•
macrostructure or microstructure of the final composite
•
service properties required of the component.
Composite material for a brake pad
The ultimate brake pad composite material is light, inexpensive, highly
effective, maintains its effectiveness under extreme conditions, requires
little or no maintenance, can last the life of the vehicle and is
environmental friendly.
6
Braking systems
Specific service properties for brake pads
•
Uniform friction properties. The coefficient of friction should not
vary appreciably with variation of load, temperature, and velocity.
The brakes must stop the vehicle.
•
High thermal stability; resistance to thermal shock and fatigue. The
material should not break down due to temperature variations.
•
Low noise generation. Noise and brake-squealing must be kept to a
minimum.
•
Adequate compressive and shear strength. The composite should not
shear or fail through compressive stress.
•
Suitable hardness. Only minimal wearing or scoring of surfaces,
including the disc and the pads should occur.
•
Suitable toughness. The pad must be able to withstand impact loads.
Materials used in brake liners and pads
Asbestos
Asbestos has been used in braking materials for most of this century
because of its friction properties, strength, low cost and good thermal
qualities. During the past 25 years, requirements for braking materials
have dramatically changed. There is a tendency towards smaller friction
elements operating at higher temperatures and pressure. Asbestos does
not fulfil the requirements needed for heat resistance at much higher
temperatures, higher coefficient of friction, nor the extended durability
required in today’s brakes.
Health and safety risk
As you may know, asbestos also represents a serious occupational health
and safety risk. It has been proved that exposure to asbestos can lead to
asbestosis, a form of lung cancer. For this reason many countries have
banned the use of asbestos.
Fibres – a replacement for asbestos
Research to find a replacement for asbestos fibres has included
investigation of steel wool, glass fibre, wool fibre, aramid (aromatic
polyamide polymer) fibre, kevlar and carbon fibre.
Required properties of fibres for brake pads includes:
•
good friction properties
•
good processing performance
7
•
high reinforcing effectiveness
•
high shear and compressive strength
•
good adhesion to binding matrix
•
adequate heat resistance
•
low specific gravity.
Research for replacement fibre
Let us consider two replacements; glass fibre and carbon fibre.
Glass fibre
Glass fibre is very hard and can abrade the drum or disc. It is brittle, and
thus requires care when mixing into the friction compound to prevent
breakage. It softens at high temperatures, acting as a lubricant,
producing a sudden loss of friction. It has excellent binding properties
with the matrix, is able to be produced in long and short fibres giving
good dimensional stability, rigidity and strength to the final product.
Carbon fibre
Carbon fibre has been around for more than a century, with Thomas
Edison utilised carbon filament in his newly invented electrical light
bulb. Up until the late seventies it was used in composits for brakes in
the aerospace industry, racing cars and high performance military
aircraft. In the early eighties it was used in the brakes of the Concorde.
In 1975 research was accelerated to find a relatively cheap multipurpose
carbon fibre similar to the fibre used in the aerospace and sporting goods
industries.
Properties of carbon fibre for brake pad materials
Carbon fibre properties as related to braking materials include:
8
•
high strength, equal to or better than steel
•
light weight, 20% that of steel
•
high temperature resistance, MP greater than 3000º C
•
resistance to oxidation, even at high temperature
•
low thermal expansion, maintaining dimensional stability
•
self lubricating
•
good wear resistance
•
excellent reinforcing properties, long and short fibres
Braking systems
•
good coefficient of friction with cast iron
•
still expensive.
Matrix for brake pads
The matrix is the continuous phase in the braking composite that holds or
‘glues’ the materials and fibres together. It must bind with the other
ingredients, be tough and strong in shear and compression, and have
good thermal shock resistance.
The most common matrix is phenol-formaldehyde or a modified
phenolic.
Fillers for brake pads
Fillers are generally low cost materials, such as clay or calcium
carbonate, that are added to extend the material in the composite, occupy
space and reduce costs. They usually influence wear properties of the
composite.
Friction modifiers
Friction modifiers are many and varied. Some are listed below.
•
Metal chips, used to modify friction properties, and to control
cleanliness on the brake interface.
•
Lead and zinc, low temperature frictional properties.
•
Copper and brass, high temperature frictional properties.
•
Lubricants, such as graphite and carbon black powder are added to
suppress noise and provide protection against disc wear.
•
Friction ‘dust’ or powder, such as cross linked phenolic and
modified phenolic based polymers, are used to provide thermal
stability, reduce the wear factor and to provide even friction
properties under extreme conditions.
•
Barium sulphate, used to improve the wear resistance of the friction
material at low temperatures, accelerate the curing of the binder and
improve compressive and shear strength.
9
Manufacture of brake pads
Brake pads are manufactured by compression moulding. Compression
moulding consists of compressing raw material into a cavity or mould of
the desired shape and applying heat and pressure.
Batch formulation
Batching is the combining or premixing of the materials in preparation
for forming.
There are many combinations of these materials that may be used to form
braking materials. Batch formulas contain up to ten or even fifteen
materials, combined together. Generally the matrix or binder is 10–25%,
the fibre, 15–30%, non-organic fillers, 10–20%, metals 1–8%, friction
dust 3–10%, and other modifiers 3–15%.
Batching
During batching, the powdered phenolic and fillers are first blended,
modifiers are added and mixed. The fibres are added last to minimise
fracture in the mixing process.
Pre-forming
Pre-forming is used to economise in the use of materials. A
predetermined amount of the batched material is cold pressed into a preform mould at a pressure of 7–15 MPa. This pre-formed shape is then
placed into the cavity of the compression moulding machine,
Compression moulding
The cavity and plunger of the mould are attached to a compression press.
The mould is heated to a temperature of 130–190º C, depending upon the
polymer. The pre-formed batched material is then placed into the hot
mould and put under pressure of 14–50 MPa. The material softens and is
compressed into the shape of the mould cavity.
Post-cured
The finished product is then post-cured in an air-forced oven, at
150–200º C for several hours. The cure time depends upon the thickness
of the product, the polymer used, its state of polymerisation when
charged into the mould, the mould temperature, and the moulding
pressure used
10
Braking systems
Finishing
The brake pads then undergo thickness grinding.
Brake liners require edge grinding to width, inside and outside grinding
to thickness, squaring and cutting to length, and drilling where
appropriate.
11
12
Braking systems
Testing of materials
In this section of work you will learn about mechanical testing of
materials, including hardness testing, and tensile and compressive
testing. In previous modules you studied the properties of materials and
the modification of the properties.
Hardness testing
Hardness is a measure of a material’s resistance to indentation, abrasion,
machining or scratching. Engineering tests use resistance to indentation
as a basis for hardness testing.
Hardness tests are non-destructive tests. They are used in industry to
verify that the required properties have been produced following the heat
treatment of components during production. They are used for control or
production line testing, as well as for research, and comparison testing
You will learn about three standardised hardness tests, Brinell, Vickers
and Rockwell. All three use machines which apply a specified load to an
indentor. The indentation is then measured to give the tested material a
hardness number.
1
Brinell hardness test
The Brinell hardness test was Introduced in 1900 by Swedish
metallurgist J. A. Brinell.
Method
A hardened steel or tungsten carbide ball indentor is pressed into the
surface of a material for 10–15 seconds. The loads used are 500, 1 500
and 3 000 kg.
13
Measured
The diameter of the indentation is measured using a low-powered
graduated microscope, and the Hardness Brinell number determined from
a prepared table.
Recorded
The hardness number is given, followed by the letters HB, then by
numbers indicating the diameter of the ball and the load used for the test.
For example 250 HB 10/3000, indicates that a Brinell Hardness test
number of 250 was obtained using a 10 mm diameter ball and a load of
3 000 kg.
Application
Used for materials such as cast iron, due to the large indentor giving an
‘average’ hardness. Unsuitable for sheet metal, for very hard material, or
for plated or hardened surfaces.
2
Vickers hardness test
Vickers hardness test was introduced in 1922 in England by R. Smith
and G. Sunderland.
Method
An industrial diamond indentor in the shape of an inverted square
pyramid is pressed into the surface of a material for 15 seconds.
Measured
The surface area of the indentation is determined, and the Hardness
Vickers number read from prepared tables. The numbers have been
calculated by dividing the load by the surface area of the indentation.
Recorded
The hardness number is given, followed by the letters HV, then by a
number indicating the load used for the test. For example, 650 HV 30,
indicates a Vickers Hardness test number of 650 was obtained using a
load of 30 kg.
14
Braking systems
Application
Used for a full range of materials with a wide range of hardness. It is
used for sheet metal, for very hard material, and for case hardened
surfaces.
3
Rockwell hardness test
The Rockwell hardness test was introduced in 1922 by American
metallurgist S. P. Rockwell.
Figure 2.1 Rockwell hardness tester
Courtesy Picnic Point High School
© LMP
Method
A variety of indentors are used, including an industrial diamond cone,
and a 1.5 mm and 3 mm hardened steel ball. The indentor is initially
pressed into the surface of the material by a minor load of 10 kg and the
dial indicator is set to zero. The major load is then applied. When the
dial indicator is steady, the major load is removed.
Nine scales of hardness are available from A to K, having various
indentors used with different major loads for various materials. The most
common Rockwell tests are B and C.
15
The Rockwell B test uses a 1.5 mm hardened steel ball, with a major load
of 100 kg. It is used for testing softer metals such as copper, brass,
aluminium, malleable cast iron and grey cast iron.
The Rockwell C test uses an industrial diamond cone, with a major load
of 150 kg. It is used for testing harder metals such as white cast iron,
hardened and case hardened steel.
Measured
The difference in depth of the indentation caused by the minor and major
loadings is used as the measure of hardness. The hardness number is read
directly from the dial using the appropriate scale for that test. This direct
reading enables the Rockwell testing to be done quickly and accurately
during the actual production of the component. The test can also be
automated.
Recorded
A number indicating the related hardness of the material for that scale,
followed by HR and the appropriate letter for the Rockwell Hardness test
used, is given. For example, 60 HRC indicates a test hardness number of
60 was obtained using the appropriate load and indentor for the Rockwell
C hardness test.
Application
Used for a full range of materials with a wide range of hardness.
Compression and tension testing
The manufacturing methods and techniques used to shape materials quite
often depend upon plastic deformation. These processes include forging,
rolling, extrusion, and wire and rod drawing. Sheetmetal processing,
folding, pressing, deep drawing and spinning also rely upon plastic
deformation.
In all of these processes the material is subjected to tensile, compressive
and shear forces. The relationship between a force and the deformation it
produces is required knowledge for the engineer in manufacturing. Two
of the most important mechanical property tests are the tensile test and
compressive test.
Analysis of the curve produced during a load-deformation test can
provide information essential to the mechanical engineer. Yield stress,
ultimate tensile stress (UTS), modulus of elasticity, percentage
16
Braking systems
elongation, and percentage reduction in area can be determined along
with interpretation of properties such as ductility and toughness.
Tensile tests
Tensile tests are conducted using a tensometer or a universal testing
machine. A prepared specimen, usually of standardised size, is held in a
gripping device and a gradually increasing axial load applied to the
specimen. The load is usually applied until fracture occurs, alternatively,
it may only be applied within the elastic limit, or up to the yield point in
some tests.
The applied load is plotted, usually automatically, against the extension,
to produce a load-extension graph or curve.
Figure 2.2 Tensometer
Courtesy Picnic Point High School
© LMP
Compressive tests
Compressive tests can also be conducted using a tensometer, with a special
adaptor, a universal testing machine, or a specialised compression testing
machine. The test is similar to the above except that a compressive load is
applied.
In ductile materials barreling usually occurs in the specimen. This is due
to the frictional forces existing between the ends of the specimen and the
surface of the dies that retard the free flow or expansion of the material.
The resulting shape is similar to that of a barrel, the middle has expanded
while the ends have not.
17
Load-deformation verses stress/strain curves
To be able to compare different materials and similar or the same
materials, a standardised specimen is used, or better still the loaddeformation diagram is converted to a stress-strain diagram. This allows
comparisons to be validly made.
You need to conduct relevant mechanical tests on materials. This is very
difficult for you to be able to do. However, there are ways that you may
be able to experience tensile, compression and hardness testing.
Your School of Distance Education or the associated TAFE collage may
be able to organise a practical workshop day where the testing machines
are available. Machines are available at many secondary schools, a visit
may be able to be arranged. University Engineering faculties have
testing equipment, TAFE at Ultimo also has a testing laboratory. Many
industries have testing laboratories as part of their manufacturing. AIS in
Wollongong has a very comprehensive testing lab, while in the
Sutherland Shire, Dowell Industries and Davies Kent do mechanical
testing on aluminium alloys, and polymers respectively.
If you have access to a tensometer, or hardness testing equipment,
conduct a series of tests on a variety of specimens to assertain their
comparative properties.
18
Braking systems
Investigation of a braking system:
materials analysis
In this section you will consider how a materials engineer would analyse
the materials used in components for a braking system. Also, for each of
the components you will look at the analysis of an alternative material
that could be used for that component. A recommendation will then be
given based upon the analysis.
Analysis of structure and properties
The materials engineer will identify the main service properties of the selected
component, and then proceed to analyse the structure of the materials. Both
microstructure and lattice or molecular structure will be analysed where
appropriate, and the properties of the materials determined. A recommendation
as to the suitability of the materials for the component will then be given.
Drum brake, slave cylinder assembly
Components from the drum brake slave cylinder assembly shown in figure 2.3
will be used for the analysis. The components to be considered by the
materials engineer are: the rivets used to secure the liner to the brake shoe, the
piston, spring, slave cylinder and the dust seal.
Dust seal
Piston seal
Slave cylinder
Spring
Brake shoe
Lining
Figure 2.3 Drum brake slave cylinder assembly
Courtesy: Trinity College Auburn
© LMP
19
An exploded isometric drawing
An exploded isometric drawing is a pictorial drawing of the separated
components in their relative position to each other. To assemble the
components, all of the components would be moved along their line of
centres, into the slave cylinder. The exploded drawing allows you to see
the size and shape details of each of the components.
Analysis of slave cylinder assembly components
1
Component: rivets
Service properties: high shear strength, able to absorb impact loads,
tough, corrosion resistant in ‘braking environment’.
Material used: copper, a pure, non-ferrous metal.
Lattice structure: copper has a FCC structure.
Microstructure: equiaxed grains of copper, when in the annealed
state; deformed grains when cold worked.
Properties: high shear strength in the cold worked condition, work
hardened, adequate toughness, able to withstand impact forces, does
not corrode in the ‘braking environment’.
Suitability: very suitable for the rivets.
Alternative material: 70–30 brass, an alloy of copper and 30% zinc. It
is a non-ferrous metal.
Lattice structure: 70–30 brass has a FCC structure. The zinc atoms
take the place of some of the copper atoms in the original copper
lattice structure. A structure such as this is called a substitutional
solid solution.
Microstructure: equiaxed grains of the solid solution when in the
annealed state, deformed grains when cold worked.
Properties: high shear strength, adequate toughness, able to withstand
impact forces. Could corrode in the ‘braking environment’.
Suitability: suitable except in adverse corrosive conditions.
Recommendation: the recommendation is to retain the copper rivets.
20
Braking systems
2
Component: piston
Service properties: adequate hardness, good compressive strength,
able to absorb impact loads, tough corrosion resistant in ‘braking
environment’.
Material used: mild steel, 0.2% carbon, a ferrous metal.
Lattice structure: ferrite has a BCC structure.
Microstructure: equiaxed grains of ferrite with small areas of pearlite
when in the annealed state, deformed grains when cold worked.
Properties: adequate hardness, good compressive strength, adequate
toughness, able to withstand impact forces. Does not corrode in the
‘braking environment.
Suitability: very suitable for the piston.
Alternative material: aluminium. It is a pure, non-ferrous metal.
Lattice structure: aluminium has a FCC structure.
Microstructure: equiaxed grains of aluminium when in the annealed
state, deformed grains when cold worked.
Properties: inadequate hardness, the aluminium is too soft.
Inadequate compressive strength, excellent toughness, able to
withstand impact forces and does not corrode in the ‘braking
environment’.
Suitability: not suitable due to its mechanical properties.
Recommendation: the recommendation is to retain the mild steel
pistons.
3
Component: spring
Service properties: good ‘spring’ properties, able to absorb impact
loads, tough, corrosion resistant in “braking environment”
Material used: medium carbon steel, 0.4% carbon, a ferrous metal.
Microstructure: equiaxed grains of ferrite with areas of pearlite (50%)
when in the annealed state, deformed grains when cold worked.
Properties: good ‘spring’ properties in the cold worked condition,
environment’.
21
Suitability: suitable for the piston. Must be in the cold worked
condition
Alternative material: high carbon steel, 0.7% carbon. A ferrous metal.
Microstructure: almost all pearlite, with a small amount of ferrite
grains.
Properties: good ‘spring’ properties in the cold worked condition,
environment’.
Suitability: suitable for the spring, however it must be shaped by hot
working, then be heat treated to obtain the required properties.
Recommendation: The recommendation is to retain the medium
carbon steel springs due to the greater cost in forming the high carbon
steel springs.
4
Component: cylinder
Service properties: adequate tensile strength (the ability to withstand
the internal pressure, known as hoop tension). Corrosion resistant in
‘braking environment’.
Material used: spheroidal graphite cast iron, 3.0% carbon and 2%
silicon, a ferrous alloy.
Lattice structure: the ferrite has a BCC structure. The graphite is a
crystalline form of carbon that has a layered or plate like structure,
making it a good lubricant.
Microstructure: nodules of graphite, in a ‘steel’ matrix of equiaxed
grains of ferrite with possibly some areas of pearlite.
Properties: adequate tensile strength, corrosion resistant in ‘braking
environment’.
Suitability: suitable for the slave cylinder.
Alternative material: aluminium alloy, containing silicon and
magnesium, a non-ferrous alloy.
Lattice structure: aluminium alloy has a FCC structure.
Microstructure: equiaxed grains of aluminium (with areas of lamella
magnesium silicide Mg2Si, which is outside the scope of the course);
deformed grains when cold worked.
Properties: good tensile strength, does not corrode in the ‘braking
environment’.
22
Braking systems
Suitability: very suitable due to its properties and to its low weight.
Recommendation: the recommendation is to retain the spheroidal
graphite cast iron for the slave cylinder, but the comparative cost of
changing to aluminium alloy should be further investigated.
5
Component: dust seal
Service properties: flexibility. Elastomer, able to be compressed
repeatedly and return to its original shape, corrosion resistant in
‘braking environment’.
Material used: neoprene, a synthetic rubber.
Molecular structure: chains of chloroprene, cross linked to form a
network structure, which has covalent bonding.
Microstructure: is not applicable in polymers and rubbers.
Properties: flexible and an elastomer, adequate tensile strength,
excellent corrosion resistant in ‘braking environment’.
Suitability: suitable for the dust seal.
Alternative material: PVC, polyvinyl chloride.
Molecular structure: chain structure having covalent bonding.
Secondary bonds between the chains.
Microstructure: is not applicable in polymers and rubbers.
Properties: soft PVC is flexible but not an elastomer. Corrosion
resistant in ‘braking environment’ not good as hardening occurs
Suitability: not suitable for the dust seal.
Turn to the exercise sheets and complete exercises 2.4.
23
24
Braking systems
Exercises
Exercise 2.1
a
Outline the contribution that the car racing industry had on the
developments of the disc brake, in England and in America.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
b
One reason that disc brake technology was developed was the need
to slow and stop cars which were becoming much faster and more
powerful each year. Discuss the social implications of this
statement.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
25
Exercise 2.2
a
Define the term ‘composite material’.
_______________________________________________________
_______________________________________________________
_______________________________________________________
b
When studying a composite material it is important to consider the
individual materials that are combined to form that composite.
Explain why this is important.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
c
Service properties are the first consideration for the materials
engineer when designing a composite for brake pads. List four of
these service properties.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
d
Asbestos is now considered unsuitable to be used as a fibre in brake
pad composites. List three reasons why asbestos is no longer
suitable for use in brake pads.
_______________________________________________________
_______________________________________________________
_______________________________________________________
e
Carbon fibre is used in braking material in aircraft and high powered
motor vehicles. List three properties of carbon fibre that makes it
suitable for use in brake composites.
_______________________________________________________
_______________________________________________________
_______________________________________________________
f
Metal chips, such as zinc, are used in brake pad composites. State
the specific reasons for using zinc.
_______________________________________________________
_______________________________________________________
_______________________________________________________
26
Braking systems
g
The term batching is used when referring to composite materials.
Explain the meaning of batching.
______________________________________________________
______________________________________________________
______________________________________________________
h
Name and briefly describe the method of manufacture used to shape
brake pads.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
Exercise 2.3
a
Define the term hardness.
______________________________________________________
______________________________________________________
______________________________________________________
b
Hardness testing is described as a non-destructive test. Explain the
meaning of non-destructive test.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
c
d
Name two areas where hardness tests are used in industry.
i
___________________________________________________
ii
___________________________________________________
Name and briefly describe three standardised hardness tests used in
industry.
i
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
27
ii
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
iii
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
e
Briefly describe the methods used for a tensile test and for a
compressive test.
i
tensile ______________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
ii
compressive _________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
28
Braking systems
Exercise 2.4
In the drum brake slave cylinder sample given in the previous notes, the
structure and properties of ten materials were analysed to determine the
suitability of the materials for various components of the slave cylinder.
The materials considered are listed below.
a
Classify each of the listed materials as pure non-ferrous metals, nonferrous alloys, ferrous metals or polymers.
copper
70-30 brass
0.2% carbon steel
aluminium
0.4% carbon steel
0.7% carbon steel
spheroidal graphite
cast iron
aluminium
alloy
neoprene
PVC
b
Draw and label the microstructures of 0.2% C steel and 0.7% C steel
0.2% C steel
c
0.7% C steel
The two steels, 0.2% C steel and 0.7% C steel have different
hardness properties and different ductility. In terms of their
microstructures explain the reasons for the difference in these
properties of the two steels.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
29
d
Neoprene and PVC have different molecular structures and different
bonding. For each material describe the molecular structure and
name the bonding.
i
Neoprene _____________________________________
_____________________________________
_____________________________________
ii
PVC
_____________________________________
_____________________________________
_____________________________________
30
Braking systems
Progress check
During this part you explored the early history of disc brakes and the
relationship between properties, uses and applications of materials in
engineering.
✓
❏
✓
❏
Uncertain
Agree – well done
Disagree
✓
❏
Agree
Take a few moments to reflect on your learning then tick the box which
best represents your level of achievement.
I have learnt about
•
•
Historical and societal influences
– historical developments of braking systems
– the effect of engineering innovations on people’s
lives
– environmental implications from the use of materials
in braking systems
Engineering materials
– materials for braking systems
– testing of materials.
I have learnt to
•
•
•
•
examine the changing applications of materials to
components in braking systems
discuss the social implications of technological change
in braking systems
investigate the structure and properties of appropriate
materials used in braking systems
conduct relevant mechanical tests on materials.
31
During the next part you will employ mathematical and graphical
methods used to solve problems of engineering practice and also further
develop your skills in isometric and orthogonal drawing.
32
Braking systems
Name: _____________________________
Check!
❐ Exercise 2.1
❐ Exercise 2.2
❐ Exercise 2.3
❐ Exercise 2.4
responses to this sheet.
Centre/School (DEC) you will need to return the exercise sheet and your
Slip.
33
Braking systems
Part 3: Engineering mechanics, hydraulics and
communication – 1
Part 3 contents
Introduction.......................................................................................... 2
What you will learn?................................................................... 2
Engineering mechanics and hydraulics .......................................... 3
Friction .................................................................................... 3
Stress and strain ......................................................................13
Communication ..................................................................................23
Pictorial drawing.......................................................................23
Orthogonal drawing ..................................................................31
AS1100 standards....................................................................36
Exercises ............................................................................................39
Progress check ..................................................................................55
Part 3: Engineering mechanics, hydraulics and communication – 1
1
Introduction
In this part you will explore mathematical and graphical methods used to
solve problems of engineering practice and also learn more about
isometric and orthogonal drawing.
What you will learn?
•
•
engineering mechanics and hydraulics
–
friction (without calculations)
–
stress and strain
–
stress (tensile and compression)
–
load/extension diagram
–
strain (tensile and compression)
communication
–
pictorial and orthogonal drawings
–
Australian Standard AS1100, including dimensioning.
You will learn to:
•
distinguish between force, stress and strain
•
produce pictorial and orthogonal drawings of braking systems and
braking components applying appropriate Australian Standards
(AS 1100).
Refer to <http//ww.boardofstudies.nsw.edu.au> for original and current documents.
2
Braking systems
Engineering mechanics and hydraulics
In this section we will examine the nature of friction, as well as looking
at stress and strain.
Friction
When you walk on a rough surface, such as a footpath, then try to walk
on an ice rink there are two very different results.
To be able to walk on a rough surface such as the footpath, friction is
used to allow you to make progress. To try to walk on an ice rink, where
there is little or no friction, is almost impossible.
Friction allows you to walk, the lack of friction allows you to iceskate,
but not to walk on ice.
Friction is the resistance to motion that occurs when two bearing surfaces
slide, or tend to slide, over each other.
1
Place your right index finger, against the palm of your left hand.
2
Very lightly push your finger across your palm.
3
Repeat, but slightly increase the downward force.
4
Now really increase the downward force and try moving your finger.
Did you notice that as you increased the downward force, that the friction
force increased? Did you also notice that your hand got hotter as the
force increased? Were you able to apply a big enough downward force
that prevented you moving your finger across your palm?
Your palm has just applied a braking force to your finger.
Repeat the activity, pushing your finger against various surfaces such as
your desk top, a book cover, your computer keyboard, a TD set square
and an eraser.
3
Did you notice that the same results occurred as before? Did you also
notice that the frictional resistance was different for the different
materials used in the activity? Were you again able to apply a braking
force? Did the braking force vary with the different material?
Now you will see how these activities apply to friction forces.
A friction force is the reactive force opposing the movement of two
touching surfaces. The friction force varies directly with the applied
force that pushes the bearing surfaces together, and therefore with the
normal reaction.
F
Applied force
Pushing surfaces
together
P
Frictional force
(reactive)
Opposing motion
Figure 3.1
Applied force
tending to move
the body
Normal reaction
Analysis of forces
Friction forces in braking systems
Friction forces occur in braking systems due to the reaction between the
specially compounded materials of the brake linings/pads, and the cast
iron or aluminium alloy metals of the brake drums/discs.
How brakes work
Brakes are able to slow or stop a moving vehicle by retarding the rotation
of the wheels through developing a ‘controlled’ friction that converts
kinetic energy of the vehicle into heat (energy). The heat energy is then
dissipated into the surrounding air through the brake drums or brake
discs.
KE
=
1
mv 2
2
From this formula, when brakes are applied a combination of the velocity
of the vehicle and mass of the vehicle determine the amount of kinetic
energy that must be converted to heat energy. However, it is actually the
4
Braking systems
rolling friction between the tyres and the road surface that eventually
brings the vehicle to a stop.
Coefficient of friction
Coefficient of friction µ (Mu) is the ratio of the limiting frictional
resistance to the normal reaction. The coefficient of friction is a constant
for any two materials in contact.
m
=
FR
N
Coefficient of friction in braking materials
If the coefficient of friction between the materials of the brake
liners/pads and the materials of the brake drums/discs is excessive, the
abrasion would quickly wear down both the liners/pads and the
drums/discs. A high coefficient would also cause the brakes to lock.
Brake materials are therefore manufactured with a range of coefficients
from low friction, 0.25–0.3 through to high friction, 0.4–0.45.
5
Analysis of friction problems
Friction force
Friction force is a reaction or a reactive force that opposes motion or
pending motion that occurs due to an applied force.
FR
=
mN
Force analysis
Analysis should commence with the determination of the direction of the
friction force. The friction force should then be drawn on the force
analysis diagram. All other forces acting on the body should then be
drawn on the analysis diagram. These forces should then be drawn as a
freebody diagram that shows only the previously analysed forces.
Note the friction force always opposes motion or pending motion.
N
mg
P
FR
N
Force analysis
FR
P
P =
m =
g =
N =
FR =
Applied force
mass of body
9.8 m/s2
Normal reaction
Frictional resistance
mg
Free body diagram
Figure 3.2 Force analysis, free body diagram
Solve, using the linear equilibrium equations;
SH
SV
= 0
= 0
Two important formulae which must be known.
Limiting friction
Limiting friction is the frictional resistance acting when a body is on the
point of moving.
6
Braking systems
Coefficient of friction µ (Mu) and friction formula
Coefficient of friction µ (Mu) is the ratio of the limiting frictional
resistance to the normal reaction.
m
=
FR
N
Alternative method of writing formula
FR
=
mN
Note: this formula only applies when limiting friction is involved.
Worked example 1
In the following five examples different forces are being applied to a
body on a horizontal plane.
The problem is usually described in words accompanied by a diagram.
The diagram is called a space diagram. You will use this diagram for the
analysis of the problem and will refer to the diagram as an analysis
diagram.
To solve the problem:
1
Analysis diagram –
The analysis diagram is used to determine and show all of the forces
acting on the body. As the friction force opposes motion or pending
motion the first step must be to determine and show the friction force
on the analysis diagram. Having shown the friction force, show all
of the other forces acting on the body. Remember that if the body is
to be in equilibrium the two linear equilibrium equations, S H = 0,
S V = 0, must be satisfied.
2
Free body diagram –
The free body diagram is used to show only the forces acting on the
system. It is easier for analysis if the forces are drawn with the
arrows pointing away from the point of intersection as shown in the
solutions. This method uses the principle of transmissibility, and
thus does not alter the question.
3
Equation for FR –
Write an equilibrium equation for FR using the two linear equilibrium
equations, S H = 0, S V = 0 and F =µN.
Note: the friction force always opposes motion or pending motion.
7
Analysis diagram
Free body diagram
Equation for FR
Body at rest
mg
N
FR = 0
mg
N
No horizontal force
is acting
Body at rest – force horizontal
mg
P
N (given)
P
FR
mg (given)
N
FR = P
(ÂH = 0)
FR
Body at point of moving – force horizontal
mg
N
P
P
mg (given)
FR
N
FR = mN
= mmg
FR
Body at point of moving – force downward, 30∞
mg
N
P
FR
30∞
P cos 30∞
mg
N
FR
FR = mN
= m(mg + P sin 30∞)
P sin 30∞
Body at point of moving – force upward, 30∞
mg
N
P
P sin 30∞
30∞
P cos 30∞
FR
FR
mg
FR = mN
= m(mg - P sin 30∞)
N
Figure 3.3 Analysis diagrams
8
Braking systems
An alternative method, the angle of friction
Angle of friction f, (phi) also only applies to limiting friction.
If the friction force and normal reaction are replaced by a resultant force,
R, the angle that R makes with the normal is f, the angle of friction, and
tan f = µ.
To solve the problem:
1
Analysis diagram –
The analysis diagram is again used to determine and show all of the
forces acting on the body. The first step is to determine and show
the friction force on the analysis diagram and then show all of the
other forces acting on the body.
2
Angle of friction method –
Replace the friction force and normal reaction with a resultant force,
R. Indicate the angle that R makes with the normal as f, the angle of
friction
3
Free body diagram –
On the free body diagram show the three forces acting on the
system, mg, P and the resultant force R.
The solution is now found using a force diagram. You solve the problem
using a graphical method or using trigonometry.
Once you learn this method of analysis you will find it much easier and
quicker to use than the previous method when solving limiting friction
problems.
Sample solution
Pending motion
mg
R
P
P
FR
N
f
P
R
Force analysis
Figure 3.4
mg
R
mg
Free body diagram
Angle of friction
The solution using a force triangle as shown is a much quicker method.
9
Worked example 2
Repeat the force analysis for parts iii, iv and v, using the angle of friction
method.
mg
f
RfN
P
mg
P
FR
FR
R
mg (given)
N
P
f
mg
P
30∞
RfN
FR
FR
mg
R
P
mg
P
N
mg
P
30∞
P
P Nf R
mg
mg
FR
FR
f
R
N
Figure 3.5 Inclined Plane 1
10
Inclined Plane 2
Braking systems
Friction on an inclined plane
Basic introduction. In Landscape products you were introduced to the
analysis of forces on an inclined plane. You should revise this work
before commencing this basic introduction to friction on an inclined
plane.
Limiting friction on an inclined plane
When a body is at rest on an inclined plane, and is on the point of
moving, (ie pending motion), the angle of inclination, q, is equal to the
angle of friction, f, and tan f = µ.
tan f = µ
This angle, q, is sometimes called the angle of repose.
q = f, and tan f = µ
Inclined plane 1
Inclined plane 2
Figure 3.6 Inclined planes
A body at rest on Plane 1 is in equilibrium. The friction force is equal in
magnitude, but opposite in sense to the weight component down the
plane.
A body on Plane 2 is on the point of slipping (that is, pending motion).
Again the friction force is equal in magnitude, but opposite in sense to
the weight component down the plane. As it is on the point of moving,
(that is, pending motion), the angle of inclination, q, is equal to the angle
of friction, f, and tan f = µ.
Determine the coefficient of friction between two selected materials.
Using a ream of paper, or a brick, glue a material, such as a fabric, or
sheet of garnet paper, to one side and another material to the other.
Method 1
Determine the horizontal force required to move the body across a
surface. You may be able to secure a spring balance to the body.
11
Alternately you could use a fixed pulley and attached masses. Now,
using the friction formula F = µN determine the coefficient of friction.
Spring balance
or
Sand could be added to the
bucket until movement occurs
Figure 3.7 Coefficient of friction on a horizontal plane
Method 2
Place the body on an inclined plane. Determine the angle of repose of
the inclined plane, that is, the angle of inclination of the plane when the
body just commences to move. Now, determine the coefficient of
friction knowing that the angle of inclination, q, is equal to the angle of
friction, f, and tan f = µ.
Figure 3.8 Determining the angle of limiting friction
12
Braking systems
Stress and strain
When a force is applied to an object, if it doesn’t move, then there must
be some force opposing it.
Opposing Force
Opposing Force
What is happening to the object?
It’s under stress.
An engineer must be aware of these stresses as they could cause the
structure to deform and subsequently collapse.
Hooke’s Law
Robert Hooke, in 1662, as the Curator of the Royal Society,
demonstrated, using the tensile loading of a piece of wire that extension
was proportional to the applied load – Hooke’s Law. His contribution to
the study of the strength of materials and the resulting effect on the
design of components was enormous.
Young’s Modulus
Thomas Young, in 1807, showed that Hooke’s Law was only effective
up to a certain limit, was a characteristic of the material and applied
equally to the compression of a body as well as to tension. His name is
given to the Modulus of Elasticity, or Young’s Modulus.
When a body has a load or force applied to it, some deformation, either
extension or contraction will occur. Depending upon the size of the load
and the mechanical strength of the body, the deformation will be either
elastic or plastic.
Elastic deformation
Elastic deformation means that the body will return to its original shape
and size when the deforming force is removed.
13
Plastic deformation
Plastic deformation means that the body will not return to its original
shape and size when the deforming force is removed. The body is said to
have taken a permanent set.
In both of the above cases, when the deforming force is being applied, an
internal resistance is tending to prevent the body from deforming. This
internal resistance to deformation is called stress.
Stress
Stress is a body’s internal resistance to an externally applied force that
tends to deform the body. It may be tensile, compressive or shear,
depending upon the applied load.
Calculation
Stress is calculated as load per unit area.
Formula
Stress =
s =
Load
Area
P
A
Symbol: s (sigma)
Units
Pascals, (Pa). Other permitted units include:
kilo Pascal
3
kPa, 10 Pa
Mega Pascal
6
MPa, 10 Pa
Giga Pascal
9
GPa, 10 Pa
Note: you must always convert to basic units when doing calculations.
Basic units:
Load = newtons (N)
Area = square metres (m2)
Stress = Pascals (Pa)
14
Braking systems
Tensile stress
Tensile stress occurs when the externally applied load tends to stretch
the body.
120 N
Figure 3.9
120 N
120 N
Tensile stress
Compressive stress
Compressive stress occurs when the externally applied load tends to
shorten the body.
120 N
120 N
120 N
Figure 3.10 Compressive loads
Shear stress
Shear stress occurs when the externally applied load tends to slide one
part of the body across another part of the body.
120 N
120 N
Figure 3.11 Shear load
Problem solving
There is only one very simple formula to learn, but it takes much practice
to be able to analyse and solve problems. (See examples 1 and 2 which
follow).
Area being stressed
The area being stressed varies with the different application of the load.
In tensile and compressive loads the area being stressed is usually the
cross sectional area. In shear loads, the area being stressed is the shear
area, that is the area that would have to break if the component were to
fail under the applied load.
15
Method
All problems should be set out and presented as follows.
i
Summarise the given data, using the related symbols from the
formula, and the given units.
ii
Convert all units to basic units where appropriate.
iii Determine the area under stress. It sometimes helps to use a sketch
of this area.
iv Select and write the appropriate formula.
v
Substitute the data into the formula, ensuring that you use the basic
units.
vi Complete the necessary calculations.
vii Write the solution to the problem using correct engineering units.
Worked example 1
A cylindrical braking rod of mild steel, diameter 12 mm, is subjected to a
tensile force of 5 kN. Determine the tensile stress in the rod.
d
P
s
5 kN
=
12 mm
=
12 x 10-3m
=
5 kN
CSA
3
=
5 x 10 N
=
?
5 kN
Figure 3.12 Analysis diagram
Area being stressed is the cross sectional area (CSA).
Area
A
pr 2 or
=
pd 2
4
=
p ¥ (12 ¥ 10 -3 )2
4
=
p ¥ 144 ¥ 10 -6
4
=
113 ¥ 10 -6 m 3
Now
16
pd 2
4
=
s =
P
A
Braking systems
P
=
5 ¥ 103
113 ¥ 10 -6
=
44.2 ¥ 106
=
44.2 Mpa
Worked example 2
A cylindrical punch, of diameter 8 mm is used to punch out the holes of a
brake liner of thickness 5 mm.
i
If, during the punching operation, the compressive stress in the
punch is 120 MPa, determine the force used to punch out the hole.
d = 8 mm
compressive area (CSA)
= 8 ¥ 10-3 m
= 120 Mpa
Figure 3.13
Analysis diagram
6
= 120 ¥ 10 Pa
P = ?
Area being stressed is the cross sectional area.
A =
Now
s
pd 2
4
=
p ¥ (8 ¥ 10 -3 )2
4
=
p ¥ 64 ¥ 10 -6
4
=
50.27 ¥ 10 -6 m 2
=
P
A
P =
ii
s¥A
=
120 ¥ 106 ¥ 50.27 ¥ 10-6
=
6032.4 N
=
6.032 kN
Using the previous data as well as the calculated force in the punch
from part i, determine the shear stress in the lining material.
Area being sheared is the curved surface area of the cylindrical
shape beig punched out of the liner.
17
Circumference =
2p r or pd
Shear area = Circumference ¥ thickness
=
pd ¥ k
=
p ¥ 8 ¥ 10 -3 ¥ 5 ¥ 10 -3
= 125.67 ¥ 10-6m2
Now s
=
P
A
P
=
6.032 ¥ 103
125.67 ¥ 10 -6
shear area
= 48.2 ¥ 106 Pa
= 48 MPa
You can see from these two worked examples that it is very important to
analyse each question. It is especially important to determine the area
being stressed, so that errors do not occur.
Strain
Strain is the ratio of the change in length of a body with respect to its
original length. It is calculated as deformation per unit length.
Formula
Strain
=
Change in length
Originallength
e
=
e/L
Symbol: e (eta)
Units
Strain is a ratio. It is sometimes expressed as a percentage, an example is
percentage elongation, that is, strain expressed as a percentage.
Now we will consider the contribution of Robert Hooke and Thomas
Young to the scientific design of engineering structures.
The ratio of stress to strain, within the elastic limit is a constant for a
given material. It is a measure of the elasticity or stiffness of the body.
18
Braking systems
Formula
Modulas of Elasticity
=
Stress
Strain
E
=
s
e
(within the elastic limit)
Derived formula
E
PL
Ae
=
Units: Pascals (Pa), and engineering multiples; kPa, MPa and GPa.
Worked example 3
A cylindrical braking rod made from 15 mm diameter medium carbon
steel, is subjected to a compressive load of 25 kN. If the original length
of the rod is 800 mm and the modulus of elasticity is 210 GPa, determine
the contraction of the rod.
d =
=
P =
=
L =
=
E =
=
e =
15 mm
25 kN
-3
15 ¥ 10 m
25 kN
CSA
25 ¥ 103N
800 mm
800 ¥ 10-3m
25 kN
210 Gpa
210 ¥ 109 Pa
?
A =
Now
pd 2
4
=
p ¥ (15 ¥ 10 -3 )2
4
=
176.71 ¥ 10-6 m2
E =
PL
Ae
Eae =
PL
19
\ e =
=
PL
EA
25 ¥ 103 ¥ 800 ¥ 10 -3
210 ¥ 109 ¥ 176.71 ¥ 10 -6
= 0.539 ¥ 10-3
= 0.54 mm
Worked example 4
A mass of 1.2 tonne is suspended from a 12 m length of fencing wire
during an experiment to confirm Hooke’s Law. If the modulus of
elasticity of the mild steel wire is 210 GPa, and the diameter of the wire
is 5 mm:
i
determine the extension of the wire
ii
determine the extension from a load of 0.6 tonne.
Note in this example mass is given as 1.2 tonne. This must be converted
to basic units, kilograms, by multiplying by 103, and then to the weight
force, in Newtons, by multiplying by 10.
m
P
=
1.2 ¥ 103kg
=
mg
CSA
3
1.2 ¥ 10 ¥ 10N
=
12 ¥ 103
L
=
12 m
E
=
210 Gpa
=
210 ¥ 109 Pa
=
?
A
Now
20
1.2 t
=
e
i
=
pd 2
4
=
=
p ¥ (5 ¥ 10 -3 )2
4
=
19.64 ¥ 10-6m2
E =
PL
Ae
Braking systems
120 ¥ 103 ¥ 12
210 ¥ 109 ¥ 19.64 ¥ 10 -6
=
= 0.349 m
= 350 mm
ii
Determine the extension from a load of 0.6 tonne
From the above calculations, the only value to change is the mass,
which is halved. The extension must also be halved.
\
Extension =
175 mm
Load-extension diagram
When a tensile test is conducted, a graph is produced during the test,
plotting load on the vertical axis and extension on the horizontal axis.
Worked example 5
The following results were obtained in a tensile test with a test piece 50
mm in gauge length and a cross sectional area of 160 mm2.
Extension
(mm)
Load
(kN)
i
0.40
0.80 1.20 1.40
2.0
3.0
3.5
4.0
5.0
20
40
70
80
82
80
70
60
62
On the given axes below, plot the load extension diagram.
80
Load (kN)
60
40
20
0
Figure 3.17
ii
1
2
3
Extension (mm)
4
5
Load extension diagram
Determine the ultimate tensile strength of the material.
Ultimate tensile strength (UTS) occurs where the load is a
maximum.
21
\ P = 82 kN
= 82 ¥ 103 N
A = 160 mm2
s = ?
s
=
P
A
=
82 ¥ 103
160 ¥ 10 -6
=
512 ¥ 106 Pa
=
512.5 Mpa
iii Determine the Young’s modulus.
Young’s modulus or the modulus of elasticity, E, is the ratio of stress
to strain within the elastic limit. The straight line portion of the
graph is from the origin to the point having coordinates, load = 60,
extension = 1.20. These values are therefore used to determine E.
P = 60 kN
= 60 ¥ 103N
e = 1.20 mm
= 1.20 ¥ 10-3m
L = 50 mm
= 50 ¥ 10-3m
A = 160 mm2
= 160 ¥ 10-3m2
E = ?
E =
=
PL
Ae
60 ¥ 103 ¥ 50 ¥ 10 -3
160 ¥ 10 -6 ¥ 1.20 ¥ 10 -3
= 15.625 ¥ 109Pa
= 15.6 Gpa
22
Braking systems
Communication
In this section of work you are going to build upon the freehand drawing
of three dimensional objects that you did in Household appliances. You
will cover, in detail, pictorial drawings using isometric projection.
Pictorial drawing
Pictorial drawing is very important to engineers as it enables the
visualisation of components. Freehand pictorial is used extensively in
initial design work. It has been used to design the communication
sections of this module. Pictorial drawing includes isometric, oblique,
axonometric, perspective and dimetric projection. In this section of work
you will learn to draw one of these methods of pictorial, isometric
projection. You should then be able to interpret the shapes of other
pictorial drawings.
Isometric projection
You will learn about isometric projection and in particular how to draw
isometric circles. If you have done technical drawing in earlier years you
should find this section relatively easy, although revision may be
required. If you have not covered this work before, you will need
extensive practice on the topic, especially in the visualisation of
components.
Isometric projection, visualisation
Worked exercise 1
A stepped block is manufactured from a rectangular prism, length 18
mm, width 10 mm and height 9 mm. The length is evenly divided into
three and the height is also evenly divided into three.
Part 3: Engineering mechanics and hydraulics – communications
23
The above paragraph gives details of the size of the block. Size details
may also be given as dimensions on a drawing.
The top view, front view and left side view of the block are given in third
angle projection, drawn to a scale of 2:1 figure 3.18.
To assist you with your pictorial drawing, a basic shape of the original
block has been given. The block has been divided into a grid pattern to
assist you with your freehand work when approximating sizes.
Remember to use 30º lines and vertical lines only.
Using this given shape, complete, freehand, the pictorial drawing of the
stepped block.
SCALE 2:1
TOP VIEW
PICTORIAL
LEFT SIDE
VIEW
FRONT VIEW
Figure 3.18 Orthogonal and isometric
Did you answer?
Method
The top view and the left side view show only the outside shape and two edges.
The visualisation of the shape of the block requires more information. The
front view provides the details needed. That is the block has been cut into a
stepped shape.
This stepped shape is drawn on the front face of the isometric block.
The steps are projected back towards the left at 30º. Each step is then outlined
to complete the pictorial drawing.
24
Braking systems
PICTORIAL
Figure 3.19 Stepped block
So that you will have practice at visualisation and freehand sketching,
eight different blocks have been given in Exercise 3.7. The eight
different shapes are cut from rectangular prisms having the dimensions
given in Worked Exercise 1. Three orthogonal views of each block have
also been drawn, along with a pictorial grid. The drawings are to a scale
of 1:1. In the exercise you are to complete freehand, the pictorial
drawing of each shaped block.
Turn to the exercise section and complete exercise 3.7.
Isometric circles
In this section you will learn how to draw isometric circles, both
freehand and by using the following instruments; a 60∞–30∞ set square
and a set of compasses.
Most engineers would use freehand methods, or if necessary, use
isometric ellipse templates. CAD systems could also be used. However,
the freehand method for quick visualisation is the most useful.
Four centre method to construct an isometric circle
A circle can be divided into four quadrants, or conversely, you could
draw four quadrants to form a circle. A circle in isometric projection,
using the four centre method, is represented by combining four separate
isometric quadrants. The two figures below show a circle and an
isometric ‘circle’; (actually an ellipse).
25
Circle
Isometric circle
Figure 3.20 Quadrants
Drawing a quadrant in isometric projection
Method:
1
draw the corner that contains the quadrant
2
accurately mark off the radius from the corner, along each side to
locate the contact points
3
draw lines at 90º to the sides from these contact points
4
where these lines meet is the centre for the quadrant
5
check accuracy to each contact point and draw the quadrant.
The two figures below show the method of constructing a true quadrant,
and its application to an isometric quadrant in a horizontal surface.
Corner
Radius
Radius
90∞
Contact
point
Corner
90∞
Contact point
Ra
diu
s
Centre
Quadrant
s
diu 90∞
Ra
Centre
90∞
Contact point
Isometric quadrant
Figure 3.21 Quadrant method
Drawing an isometric circle in a horizontal face
The quadrant method given on the previous page is used to construct the
isometric circle.
Four quadrants are combined to form the isometric circle.
Method:
1
26
draw the isometric square having sides equal in length to the
diameter of the required circle using very light construction lines
Braking systems
2
locate and mark the middle of each side of the square – these middle
points represent the contact points for each quadrant
3
draw lines at 90º to the sides of the square from these middle or
contact points – where these lines intersect are the centres for each of
the quadrants
4
set your compasses at a radius equal to the distance from the centre
to the contact points (note that this radius will not be 25 mm)
5
check your accuracy and draw the quadrant
6
complete the other three quadrants to form a full circle.
Two of the four corners are represented below, the quadrants are shown.
Corner 1
Corner 2
Figure 3.22 Quadrant horizontal face
The four centre method is used to draw an isometric circle of radius 25
mm in a horizontal face.
Figure 3.23 Circle in horizontal face
Drawing an isometric circle in a vertical face
The quadrant method given on the previous pages is used to construct the
isometric circle.
Four quadrants are combined to form the isometric circle.
27
Method:
1
draw the isometric square having sides equal in length to the
diameter of the required circle using very light construction lines
2
locate and mark the middle of each side of the square – these middle
points represent the contact points for each quadrant
3
draw lines at 90º to the sides of the square from these middle or contact points
– where these lines intersect are the centres for each of the quadrants
4
set your compasses at a radius equal to the distance from the centre
to the contact points (note that this radius will not be 25 mm)
5
check your accuracy and draw the quadrant
6
complete the other three quadrants to form a full circle.
Two of the four corners are represented below, the quadrants are shown.
Corner 1
Corner 2
Figure 3.24 Quadrants in vertical face
The four centre method is used to draw an isometric circle of radius 20
mm in a vertical face.
Figure 3.25 Circle in vertical face
28
Braking systems
Projecting an isometric quadrant to another face
The following drawings show you a method of projecting the quadrant to
another face. You could fully construct another quadrant, but it is
quicker to use the methods shown below.
From a horizontal plane, no profile edge
Method:
1
project downward from the centre point and the two contact points
towards the new surface using very light construction lines
2
set your dividers to the given thickness and accurately mark off the
distance to the new surface, down from the centre point and the two
contact points
3
check accuracy then draw the quadrant for the new surface.
project down
for new centre
project down for
new contact points
Figure 3.26 Projecting quadrant in horizontal face
From a horizontal plane, with a profile edge
A profile or outer edge of a solid object will hide part of the quadrant in
the lower face. You therefore do not have to draw the whole quadrant in
this lower face; only half of the quadrant will be visible.
Method:
1
draw the quadrant in the top face, then project downward, as
described below, to the lower face
2
project downward from the centre point and the contact point
towards the new surface using very light construction lines
3
project downward the profile edge (note the profile edge is a line
tangential to the quadrants that represents the outside edge of the
object)
4
set to the required thickness of the object, mark off the distances to
locate the new centre point and contact point for the lower surface
using dividers
29
5
check accuracy then draw the part quadrant for the new surface.
6
darken the profile edge.
Quadrant radius = 40 mm
Thickness
= 15 mm
Profile edge
Projected centre
Projected
contact point
Figure 3.27 Profile edge
From a vertical plane, no profile edge
The following drawings show the method of projecting a quadrant from a
left and a right vertical face. The method is similar to the one described
for the horizontal plane, and as such the method will not be described.
Figure 3.28 Projected quadrant in vertical face
From a vertical plane, with a profile edge
The following drawings show the method of projecting a quadrant from a
left and a right vertical face. The method is similar to the one described
for the horizontal plane, and as such the method will not be described.
30
Braking systems
Figure 3.29 Profile edge
Projecting an isometric quadrant (with a profile edge)
You now have sufficient information to draw isometric circles, either
freehand or by using instruments, but you will need a great deal of
practice to be able to complete drawings quickly and accurately. Keep
these notes as a reference, and use them as often as possible.
Orthogonal drawing
In this section of work you will build upon the freehand orthogonal
drawing introduced in Household appliances and the orthogonal
drawing from Landscape products.
You will be shown two orthogonal drawings as worked examples. If you
are inexperienced at technical drawing you may wish to attempt these
two drawings, following the given steps, as practice.
Worked example 1
Draw, in orthogonal projection using a scale of 1: 2, a front view of the
hand brake lever, shown in figure 3.30, when viewed from the direction
of the arrow.
The front view of the 12 mm diameter lower hole has been given as a
starting point for the drawing.
31
90
Ø
20
13
0
Ø
14
70
4
30
R1
0
Ø : INS
12 IDE
R8
R2
.5
0
40
55
6
14
20
Ø
12
R2
0
Figure 3.30 Pictorial – hand brake lever
Steps and method
Note the scale of 1:2 means that you will use half size measurements for
the drawing. This is a reducing scale, used so that the drawing can fit
onto the drawing page. You must divide all dimensions by two.
1
Locate the centreline position of, and draw the higher 12 mm
diameter hole. The vertical dimension is 40 mm, therefore measure
to scale 20 mm above the given centreline. The horizontal
dimension is 20 mm therefore measure 10 mm to the left of the
given centreline. The circle is diameter 12mm therefore draw the
circle using a measurement of diameter 6 mm.
Note the circle should be drawn using a circle template.
2
32
Locate the top of the handle; project up from the located centreline
and measure the required distance. The dimension is 30 mm
(R20 + R10) therefore measure 15 mm above the located centreline.
Braking systems
3
Draw the top of the lever, drawing from the located position in part
(2). The dimension are 70 mm, 130 mm and 90 mm therefore mark
off distances of 35 mm, 65 mm and 45 mm.
4
Draw the left hand end of the lever. The dimension is diameter 20
mm therefore draw down a distance of 10 mm.
5
From this left hand end draw the parallel portion of the bottom of the
lever. The dimension is 90 mm therefore measure 45 mm to the
right.
6
Draw the left hand sloping section of the bottom of the lever. It slopes
downward to a point 30 mm below the top edge, therefore measure
downward a distance of 15 mm to locate the end point of the sloping
line.
7
Draw the middle sloping section of the bottom of the lever. It slopes
downward to a point 55 mm below the top edge, therefore measure
downward a distance of 27.5 mm to locate the end point.
8
Now you have to complete the right hand end of the lever. Lightly
draw the two R20 mm radius curves on the two centrelines. The
dimension is 20 mm radius, therefore use a measurement of 10 mm
radius. You should use your compasses to do this construction. Use
very light construction lines.
9
Use your set square to join the tops of the two R20 arcs.
10 Use your set square to join the bottom of the R20 arc to the
previously drawn sloping line at the bottom of the lever.
11 Use radius curves to darken the curves drawn in part (8) above.
12 Locate and draw the diameter 4 mm hole. Measure from the
centreline of the previously drawn top 12 mm diameter hole.
Dimensions are 6 mm to the left, and 14 mm above, the centreline
and the diameter is 4 mm; therefore use measurements of 3 mm, 7
mm and 2 mm. Again you must use your circle template to draw the
circle.
13 Darken in all visible outline, using thick, 0.5 mm, dark lines.
Darken all centrelines, using thin, (0.25 mm), dark lines. Use a thin
dark chain line for the long centreline of the barrel of the lever, and
thin continuous lines for the circle centrelines. Note that thin
continuous dark lines are used to indicate short centrelines.
If you have decided to attempt this drawing, you have now completed the
front view of the hand brake lever, in orthogonal projection.
Congratulations.
Note that you were not requested to show any dimensions so do not show
any.
The completed drawing is shown in figure 3.31.
33
Figure 3.31 Front view of hand brake lever
Worked example 2
Draw a front view of the piston from a hydraulic brake cylinder, shown
in figure 3.32, using on enlarging scale 2:1 and the drill hole positioned
to show the 6mm diameter hole using a part-section, a standard method
to show interior details as visible outline.
Dimension the overall length of the piston and the drill hole.
The method of drawing the shape of the drill hole will be covered along
with dimensioning of the hole.
Ø 25
Ø 12
Ø 25 Ø 12 Ø 25
R4
HOLE Ø 6
DEPTH 12
6
32
6
20
6
Figure 3.32 Piston from on hydraulic brake cylinder
The quickest method:
34
1
mark off distances along the given centreline, from the right hand end,
12 mm, 20 mm, 12 mm, 64 mm and 12 mm using the scale of 2:1
2
mark off distances either side of the centreline, of 12 mm and 25 mm –
all lines should be light construction lines
3
draw the four R2 quadrants, using the 4mm size on your radius curves
using 0.5 mm, thick, dark lines for the quadrants.
Braking systems
Drawing the drill hole
The following steps describe how to draw the shape of the drill hole.
Note that the hole takes the pointed shape of the drill.
The depth of the drill hole
The depth of the drill hole is measured as the distance of the full diameter
of the hole. The depth does not include the distance to the point.
1
mark off the depth of the 12 mm hole, measuring 24 mm from the
right hand end
2
mark off the diameter of the hole, 6 mm above and below the
centreline
3
outline the rectangular shape of the hole.
Figure 3.33 Depth of drill hole
The pointed end of the drill hole
The pointed end of the drill hole has an included angle of 120º. It is
drawn using two lines, each of 60º, from the left hand end of the
previously drawn rectangle. The full drill hole is now shown as visible
outline using thick dark lines.
60
∞
DEPTH
Figure 3.34 Shape of drill hole
The part-section
A thin dark continuous freehand line is now drawn just to the left of the
of the drill hole to indicate the limit of the part-section.
35
Hatching the sectioned area
The part-sectioned area is hatched, using thin dark lines, equally spaced
at an angle of 45º. The area of the drill hole is not hatched.
Dimensioning the drill hole
The drill hole is fully dimension to show diameter and depth using the
AS1100. 1992 symbols for diameter and depth as shown in figure 3.35.
Then dimension the overall length of the piston.
AS 1100 standards
The completed drawing is shown in figure 3.35. Some of the AS1100
standards that you should be aware of include:
•
All lines are drawn as dark lines.
•
There are two different thicknesses of dark lines used on the
drawing.
•
Thick dark lines are used to draw the visible outline.
•
Thin dark lines are used to indicate
•
–
the part-section line
–
the hatching lines
–
the centre line
–
the extension lines for the dimensions and
–
the dimension lines.
The dimensioning uses the current standard symbols to indicate
diameter and depth of the drill hole.
You must use current AS1100 standards in your drawings.
36
Braking systems
Ø6
12
60
Figure 3.35 Part sectioned front view
Turn to the exercise sheets and complete exercise 3.8 and 3.9.
37
38
Braking systems
Exercises
Exercise 3.1
Determine the coefficient of friction between two materials and write a
page report on the experiment and state your conclusions.
1
2
You should submit this exercise as a word processed document and
attach your document to this page.
39
Exercise 3.2
40
a
A disc brake system has a force of 8 kN applied to each of the pads.
If the coefficient of friction between the materials of the pad and of
the disc is 0.35, determine the total braking force.
b
A drum braking system has a force of 12 kN applied by the brake
shoe to the drum surface. If the coefficient of friction between the
materials of the shoe liner and of the drum is 0.3, determine the
braking force.
Braking systems
c
A family sedan is parked with its hand brake on. The hand brake
operates only on the rear wheels. A truck, attempting to park, bumps
the sedan with a horizontal force of 2 kN. If each of the rear wheels
of the sedan supports a mass of 300 kg, and the coefficient of friction
between the tyres and the surface of the parking area is 0.2,
determine if the sedan will move forward as a result of the collision.
41
Exercise 3.3
Select the alternative, A, B, C. or D that best answers the question.
1
2
3
Braking systems are effective as a result of :
a
an extremely high coefficient of friction between the braking
materials
b
an extremely low coefficient of friction between the braking
materials
c
a range of coefficients of friction between the braking materials
from 0.25 through to 0.45
d
having no coefficient of friction between the braking materials.
Effective dissipation of heat energy is important in braking systems
to:
a
keep the driver warm in winter
b
allow the conversion of kinetic energy to heat energy to
continue during heavy braking operations.
c
allow the drums/discs to stay hot during braking.
d
allow fade when braking.
The angle of friction is:
a
equal to the coefficient of friction
b
used during calculations only when limiting friction applies
c
equal to the normal reaction
d
equal to the friction force.
Exercise 3.4
a
A family sedan, moving at a velocity of 100 km/h, brakes suddenly
to avoid a collision. If the front wheels are fitted with disc brakes:
i
draw a force analysis diagram showing all of the forces acting
between one of the front discs and the brake pads
ii
draw a free body diagram of the braking area of the disc
showing all of the forces
iii write an equilibrium equation that would be used to determine
the magnitude of the braking force
42
Braking systems
Pads
Disc
Analysis diagram
Free body diagram
Equation for FR
Figure 3.36
b
The front wheels equally support 2/3 of the total mass of the sedan.
i
Draw on the following diagram a force analysis diagram
showing all of the forces acting between one of the front wheels
and the road surface.
ii
Draw a free body diagram of the braking area of the front wheel
and the road surface showing all of the forces.
iii Replace the friction force and normal reaction with a single
force. Clearly label the angle of friction.
iv
Draw a force triangle that would be used to determine the
magnitude of the braking force.
Wheel rotates clockwise
Analysis diagram
Free body diagram
Force triangle
Figure 3.37
c
A truck is parked on the side of a gravel road. The angle of
inclination of the road is 6º. Due to the loose gravel surface, the
truck is on the point of sliding down the hill.
Determine the coefficient of friction between the truck tyres and the
gravel surface.
6∞
Figure 3.38
43
Exercise 3.5
Complete the analysis of forces being applied to a body on a horizontal
plane by showing all of the forces acting on the body, then completing
the free body diagram, (showing only the forces acting), then writing an
equilibrium equation for FR.
The first example, a body at rest, is completed for you. Note: the friction
force opposes motion or pending motion.
Analysis diagram
Body at rest
Free body diagram
Equation for FR
mg
N
FR = 0
mg
N
No horizontal force
is acting
Body at rest – force horizontal
mg
N (given)
P
FR =
mg (given)
N
mg
P
FR =
mg (given)
N
mg
P
30∞
FR =
mg
N
mg
P
30∞
FR =
mg
N
Figure 3.39 Friction analysis
44
Braking systems
Exercise 3.6
a
Define the term mechanical stress.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
b
A cylindrical braking rod of mild steel, diameter 10 mm, is subjected
to a tensile force of 15 kN. Determine the tensile stress in the rod.
45
c
A cylindrical braking rod made from 12 mm diameter medium
carbon steel, is subjected to a tensile load of 30 kN. If the original
length of the rod is 900 mm and the modulus of elasticity is 210
GPa, determine the extension of the rod.
d
A cylindrical punch, of diameter 9 mm is used to punch out the holes
of a brake liner of thickness 6 mm.
i
46
If, during the punching operation, the compressive stress in the
punch is 120 MPa, determine the force used to punch out the
hole.
Braking systems
ii
Using the previous data as well as the calculated force in the punch
from part i above, determine the shear stress in the lining material.
47
Exercise 3.7
Eight different shapes are cut from rectangular prisms in figure 3.40.
Three orthogonal views of each block have been drawn, along with a
pictorial grid. The drawings are to a scale of 1:1. Complete freehand,
the pictorial drawing of each shaped block.
Figure 3.40 Isometric exercises
48
Braking systems
Exercise 3.8
Shape and size details of a washer from a master cylinder assembly are
given in the dimensioned orthogonal drawing in figure 3.41.
Sketch, freehand, in isometric projection, a pictorial drawing of the
washer. The scale used may be selected by you. The starting point
for the centrelines of the washer has been given.
ii
Draw, using instruments, in isometric projection, a pictorial drawing
of the washer. Use a scale of 2:1. The starting point for the
centrelines of the washer has been given.
3
i
TOP VIEW
Ø 30
Ø 10
FRONT VIEW
Figure 3.41 Washer
49
iii
iv
FREEHAND PICTORIAL
INSTRUMENT DRAWING OF WASHER
Figure 3.42 Washer
50
Braking systems
Exercise 3.9
Shape and size details of a disc brake rotor are given in figure 3.43. On
the grid paper attached:
i
Draw freehand, in orthogonal projection using a scale of 1: 3, a front
view of the disc brake rotor, when viewed from the direction of the
arrow. Include the principle dimensions.
ii
Project freehand, using third angle projection, a left side view of the
brake pad.
30
60
20
0
Ø5
0
Ø3
ES
HOLCED
0
1
A
SP
2X
Ø 1 ALLY
U
Q
E
Figure 3.43
50
Ø1
00
Ø3
Disk brake rotor
51
Figure 3.44 Grid
52
Braking systems
Exercise 3.10
The front and rear disc brake from a modern motorbike are shown in
figure 3.45. The rotors are considerably different in size, at the front
there are two large rotors whilst at the back there is a single smaller rotor.
Figure 3.45 Front and rear disc brakes on a modern motorbike
Explain the dynamics of why this is an effective brake set-up.
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
53
54
Braking systems
Progress check
In this part you explored mathematical and graphical methods to solve
problems of engineering practice.
✓
❏
✓
❏
Uncertain
Agree – well done
Disagree
✓
❏
Agree
I have learnt about
•
•
– friction (without calculations)
– stress and strain
stress (tensile and compression)
load/extension diagram
strain (tensile and compression)
communication
– pictorial and orthogonal drawings
– Australian Standard AS1100, including
dimensioning.
I have learnt to
•
•
distinguish between force, stress and strain
produce pictorial, and orthogonal drawings of braking
systems and braking components applying appropriate
Australian Standard (AS 1100).
During the next part you will continue to explore mathematical and
graphical methods used to solve problems of engineering practice and
also learn more about isometric and orthogonal drawing.
55
56
Braking systems
Name: _______________________________
Check!
❐ Exercise 3.1
❐ Exercise 3.2
❐ Exercise 3.3
❐ Exercise 3.4
❐ Exercise 3.5
❐ Exercise 3.6
❐ Exercise 3.7
❐ Exercise 3.8
❐ Exercise 3.9
❐ Exercise 3.10
responses to this sheet.
Centre/School (DEC) you will need to return the exercise sheet and your
Slip.
57
Braking systems
Part 4: Engineering mechanics, hydraulics and
communication – 2
Part 4 contents
Introduction.................................................................................2
What will you learn? ............................................................................ 2
Engineering mechanics and hydraulics ......................................3
Work, power, energy ........................................................................... 3
Fluid mechanics................................................................................... 8
Pascal’s Principle............................................................................... 11
Archimedes’ Principle........................................................................ 14
Communication......................................................................... 15
Detail drawing .................................................................................... 15
Computer aided drawing ................................................................... 23
Exercises.................................................................................. 33
Progress check......................................................................... 45
Exercise cover sheet ................................................................ 47
1
Introduction
In this part you will continue to explore mathematical and graphical methods used
to solve problems of engineering practice and also develop your skills in isometric
and orthogonal drawing.
What you will learn?
•
–
work, power, energy (without calculations)
–
fluid mechanics
Pascal’s and Archimedes’ Principles
hydrostatic pressure
applications to braking systems
•
Communication
–
detail drawing
–
computer graphics, computer assisted drawing (CAD).
You will learn to:
•
experiment with and apply the basic principles of fluid mechanics to simple
braking systems
•
detail drawings of braking systems and braking components applying
appropriate Australian Standard (AS 1100)
•
produce simple computer assisted drawing(s).
2
In this section of work you will learn about the meaning of mechanical
work, energy and power.
Work, power, energy
You may have studied these topics before, but you will now require an
understanding of each term, how each affects engineering problems and
specifically, how each relates to braking systems.
Mechanical work
Mechanical work is done when a force acts upon a body and produces a
displacement.
The work done by a force is determined by the product of the force and the
displacement of the point of application of that force.
Work =
W =
Force x Displacement
Fs
Units of work: Joule, J
Work done by a force against a resistance
Against a frictional resistance
When a body, on an horizontal plane, moves with uniform velocity, a
distance, s, against a frictional resistance, the work done by the applied
force in overcoming this resistance is:
Work
=
W =
Frictional resistance x Displacement
FR x s
3
Against gravity
When the centre of mass of a body is raised through a vertical distance (h), the
work done against gravity is:
Work =
=
Mass x Acceleration due to gravity x Height
mgh
Up a smooth inclined plane (no friction)
When a body on an inclined plane of angle q to the horizontal, is moved at
uniform velocity a distance (s), up the incline, by a force applied parallel to the
plane, the work done is:
Work = mg sin q ¥ s
But sin q =
=
Height
Displacement along the plane
h
s
W = mgh
Against inertia
When a force acting on a body causes that body to accelerate, the applied force
(F), is opposed by inertia, ma and
F = ma
If the displacement during the acceleation is s, then:
Work = FS
= mas
4
Example 1
Consider the work done by a motor car:
1
Travelling at constant velocity on a horizontal surface.
Work done =
W =
Total resistance x Displacement
Rs
Note if there is no resistance, work done is zero
2
Accelerating uniformly on a horizontal surface.
Work done =
W =
3
Total resistance x Displacement + mas
Rs + mas
Accelerating uniformly up an inclined surface.
Work done =
Total resistance x Displacement + mas + mgh
Where h is the vertical displacement of the car.
W =
4
Rs + mas + mgh
Accelerating uniformly down an inclined surface.
Work done =
Total resistance x Displacement + mas - mgh
Where h is the vertical displacement of the car.
W =
Rs + mas - mgh
Worked example 1
Let us consider the work done by a braking system. In each case the hand brake is
in operation, but in (3) and (4), ineffectively.
1
Car stationary on a horizontal surface.
•
2
Car stationary on an inclined surface.
•
3
Work done by the hand brake = 0 as displacement is zero.
Car on an inclined surface, moving downward at constant velocity.
•
4
Work done by the hand brake = 0 as displacement is zero.
Work done by the hand brake = mgh - Rs
Car on an inclined surface, accelerating uniformly downward.
•
Work done by the hand brake = mgh – Rs – mas
From these examples, you can see that the work done by the car engine is
different from the work done by the braking system.
5
Mechanical energy
Mechanical energy is a body’s capacity to do work.
A body that has a capacity to do work is said to possess energy; the amount of
energy is determined by the quantity of work it can do. Units are therefore the
same as for work, Joule, J.
Kinetic energy
Kinetic energy is the energy a body possesses due to its motion. It is determined
by the amount of work done in bringing the body to rest.
KE =
1
mv2
2
Potential energy
Potential energy is the energy possesses due to its position. It is determined by
the amount of work done in lifting the body through a vertical height.
PE = mgh
Strain energy
Strain energy is the energy a body possesses due to its deformation. It is
determined by the amount of work done in deforming the body, such as a spring.
SE =
1
Fs
2
It may also be determined from the load-extension diagram, following a tensile
test. Strain energy is found by determining the area under the graph, up to the
elastic limit.
Conservation of mechanical energy
When considering the vertical movement of a body.
Loss of PE = Gain in KE
and conversely
Loss of KE = Gain in PE
Work and energy are very important to engineers when designing cars and
braking systems. More importantly is the rate at which a car or brake can do the
work or transfer energy. This rate of doing work is called power.
A car having an engine with a high power rating, can do work more quickly.
6
Power
Power is the time rate of doing work, and is determined by the ratio of
the work done over time taken to do the work.
Power =
P =
Work done
Time taken
W
t
Derived from this formula are two very important formulas.
P =
W
t
but
\ P =
Fs
t
and
W = Fs
s
t
= v (velocity)
\ P = Fv
P =
W
t
Fs
= Fv
t
=
Units of Power: Watt (W)
Worked example 2
Let us consider the energy of a motor car and the braking system:
1
Travelling velocity on a horizontal surface
KE =
1
mv2
2
= Work done by the brakes in stopping the car
2
Accelerating uniformly on a horizontal surface.
KE =
1
mv2
2
–
1
mu2
2
= Work done by the car in accelerating
7
Mechanical efficiency of a machine
Another important consideration for engineers is the efficiency of a machine such
as a car or braking system.
Also in cars
Mechanical Efficiency =
Power output
Power input
Mechanical Efficiency =
BrakePower
Indicated Power
When dealing with hydro-electricity the relationship between the volume and the
mass of water is also very important to engineers:
1 000 l of water has a volume of 1 m3 and a mass of 1 000 kg.
Fluid mechanics
In this section of work you will study the basic principles of the hydraulics used in
braking systems. If you need to refresh your memory of the developments of
braking systems revisit Part 2 and Part 3 of this module.
Basic hydraulic braking systems
Originally, automobile brakes were operated by mechanical means using levers,
cables and rods as linkages. These braking systems were very unreliable, causing
many service problems in maintaining linkages and providing equalisation of
braking pressure to the brakes.
Hydraulic systems were developed based on the simple principle that pressure
exerted at any point on a confined fluid will be transmitted throughout the fluid
equally and undiminished in all directions.
Liquid flows freely, assumes the shape of the container, cannot be appreciably
compressed, and, if contained in a sealed system, allows pressure to be equally
and evenly distributed throughout the system. The hydraulic-brake actuating
system thus provides equalised transfer of pressure from the applied force,
through the system, to the brake shoes or discs.
The advantages of this basic hydraulic system over the original mechanical brakes
are that it gave completely uniform pressure throughout, greatly reduced adjustment
problems, and provided even braking on all wheels at all times. Later developments
of front brakes and ABS braking systems modified the last ‘advantage’.
8
Pressure
To understand the basic principles of the hydraulic braking system you must
firstly understand what is meant by pressure and be able to do simple calculations
involving pressure.
Pressure is force per unit area.
p =
F
A
Thus total force or thrust on a surface is the area of the surface, times the
pressure on that surface.
F = pxA
Basic units of pressure: Pascal (Pa).
Worked example 3
A brick of mass 3 kg and dimensions 230 x 110 x 75 rests on a horizontal surface:
i
flat on its largest face
ii
on its end.
Determine the pressure applied to the horizontal surface in each case.
i
F = mg
= 3 ¥ 10
= 30 N
A = 1¥b
= 230 ¥ 10-3 ¥ 110 ¥ 10-3
= 25 300 ¥ 10-6 mm2
p = F∏ A
= 30 ∏ 25300 x 10-6
ii
= 1.162 kPa
F = mg
= 3 ¥ 10
= 30 N
A = b¥ t
= 110 ¥ 10-3 ¥ 75 ¥ 10-3
= 8250 x 10-6 mm2
P = F∏A
9
= 30 ∏ 8250 ¥ 10-6
= 3.564 kPa
As you can see the pressure is increased as the area is decreased.
Note, the area was determined in mm2, so the answer was in kPa.
Pressure in liquids
Open container
A liquid at rest in an open container exerts a pressure due to its different
weight at various depths.
1
Drill, or pierce, three small holes along the side of a large PET drink
bottle at various heights – one near the bottom, middle and top.
2
Fill the container with water and observe the result.
Figure 4.1 Water overflow from an open container
You should observe that the:
i
pressure on the water in the open container varies with the depth; the greater
the amount of water above the hole, the greater the pressure.
ii
pressure exerted by the liquid is always perpendicular to the surface it
contacts.
iii pressures are the same at all points on the same horizontal level in a liquid at
rest.
10
Closed container
Now consider the pressures in a closed or sealed container.
Any pressure that is applied from outside a sealed container full of liquid can
exert an equal and undiminished pressure to all other portions of the liquid and to
the walls of the container.
1
Attach the PET drink bottle used in the previous activity to another
intact PET drink bottle so the two join at the neck.
2
Fill the bottles with water, connect and squeeze the container without
holes and observe the result.
Figure 4.2 Water flow from a closed container
You should observe that the pressure on the water in the sealed or closed
container is the same for each of the holes.
This observation would also apply to connected sealed containers.
A French scientist, Pascal, made similar observations in 1650, in his publication,
Principles of Transmission of Pressure in Fluids and today has a law or principle
named after him.
Pascal’s Principle
Pascal’s Principle states that if the pressure at any point in a liquid that is
enclosed and at rest, is changed, then the pressure at all points in the liquid is
changes by the same amount.
Thus when a fluid completely fills a sealed container, or connected containers,
and pressure is applied by means of a cylinder and piston, that pressure is
transmitted equally throughout the whole of the enclosed fluid.
If pressure is applied to a liquid in a sealed container or system, through the
application of a force (F1) in a cylinder of cross sectional area (A1), an equal
pressure will be transmitted to a larger piston and cylinder, of area (A2), causing a
thrust or force in this piston, of magnitude F2.
11
F1
A1
=
F2 =
F2
A2
F1 ¥ A2
A1
If A2 is very large compared to A1 a comparatively smaller force applied to the
smaller piston can overcome a large resistance acting on the larger piston.
Additionally, this can apply to a number of different cylinders and pistons
attached to the sealed system.
Braking systems
This principle forms the basis of hydraulic machines, including the hydraulic
press, hoist, jack and hydraulic braking systems.
Figure 4.3 Hydraulic Braking System
Worked example 4
Figure 4.4 represents a sealed hydraulic braking system. A force of 100 N is
applied to the brake pedal as shown. Size details of the pedal, master cylinder,
and front and back wheel cylinders are given on the diagram.
Determine the thrust (force) delivered by each of the wheel cylinders.
100 N
250
A2 = 300 mm2 rear wheel cylinder
A1 = 600 mm2
50
A3 = 900 mm2 front wheel cylinder
Figure 4.4 Hydraulic braking system
12
Solution
The solution is based on Pascal’s Principle, that any pressure applied to a
liquid in a confined container or system is transmitted equally and
undiminished to all parts of the container or system.
i
Determine by moments the resultant force on the master cylinder caused by
the applied force of 100 N exerted on the brake pedal.
∑M about pivot:
100 x 250 = 50 ¥ R
Resultant force on master cylinder = 500 N
Note: the lever system has provided a mechanical advantage of 5!
ii
Determine the pressure generated in the system by this resultant force.
Pressure generated at master cylinder = Force/area
F
P =
A
=
500
600 ¥ 10 -6
= 833 kPa
iii Determine the thrust at the rear wheel cylinder.
Since the pressure of 833 kPa is equal in all directions
Force exerted by the piston = pressure x area
= 833 ¥ 103 x 300 ¥ 10-6
iv
= 250 N
Determine the thrust at the front wheel cylinder.
Since the pressure of 833 kPa is equal in all directions,
Force exerted by the piston = pressure x area
= 833 ¥ 103 ¥ 900 ¥ 10-6
= 750 N
Application to braking systems
By varying the diameter of the cylinders it is possible to distribute the
pressure as needed. This is particularly applicable to the different stopping
forces needed at the front and back wheels.
When the brakes are applies the reaction at the front wheels is greatly
increased due to the tendency of the vehicle to continue its forward
motion. The front brakes therefore need a greater applied force than the
rear wheels. One way that this can be done is by using larger wheel
cylinders on the front brakes.
13
Archimedes’ Principle
Archimedes was a Greek philosopher who lived in the third century BC.
Archimedes’ Principle, still has wide application today.
When a body is wholly or partially immersed in a fluid, it is acted upon by
an upthrust which is equal to the weight of the fluid displaced. This
upthrust, or buoyancy, acts through the centre of mass of the displaced
fluid. The centre of mass is therefore referred to as the centre of
buoyancy.
Buoyancy
From your previous reading you will remember that fluid exerts an equal
pressure to all parts of a body in contact with, or immersed in the fluid.
For a body to float in a fluid, the upward thrust due to the weight of the
displaced fluid, must be equal to the weight of the floating body. This
upward thrust is buoyancy.
Buoyancy force = mass of fluid displaced ¥ g
= density of fluid x volume ¥ 10
Use the brick, or ream paper, from a previous experiment. Tie a length of
thin cotton around the brick page and attempt to lift the object. You may
have to us a wooden handle on the string to prevent the cotton cutting into
your fingers.
Normally the cotton will break.
Now attempt the same experiment with the brick immersed in water, either in a
bucket or washing tub. The lift should be successful while the brick remains
immersed.
Is this because the brick has less mass in the water, or is it due to the
buoyancy?
I think that you can agree with Archimedes on that question.
Turn to the exercise sheets and complete exercises 4.2 and 4.3.
14
Communication
In this section of work you will learn more about AS1100 standards, what is
meant by a detail drawing, and the standard sectioning techniques that may
be used. As examples you will be shown how to design the best solutions
for, and complete, two detail drawings on brake components.
Detail drawing
A detail drawing is a specialised type of orthogonal drawing used to
communicate information from the designer or engineer to the
manufacturing personnel.
A detail drawing gives a full shape and size description of the component.
It also gives the material that the component is to be made from. The detail
drawing must provide sufficient information for the manufacture of that
component.
The shape description
The shape description is usually given in an orthogonal drawing. A
decision must be made as to which views, and how many views are
necessary to show the full shape description. Sometimes three views, a top
view, front view and left or right side view are required, whilst other, more
complicated components may need up to five views.
The simple components from a brake master cylinder used in the following
examples can use as few as two views, or even a single view that
incorporates dimensioning to provide the full shape description.
15
Sectioned views
Where the component has interior details that need to be shown, sectioning
must be used to show these details as visible outline. Hidden outline
should be avoided where possible. You will learn about the different types
of sectioning that may be used; full-section, half-section and part-section.
A part-section was used in the orthogonal drawing of the piston in figure 3.35.
The size description
The size description is given by fully dimensioning the drawing of the
components, using AS 1100 dimensioning standards.
The material
The material to be used in the manufacture of the component is given on the
drawing or in a materials list if the component is part of a larger drawing.
Designing a detail drawing
With all of these requirements and options for the drawing, many design
decisions have to be made. The best approach when designing a detail
drawing, is to complete a number of freehand drawings showing various
options with regard to the number of views, the sectioning methods, and
then, the placement of the dimensions.
In the following two worked examples you will be shown how this design
technique is applied.
Worked examples, the master cylinder
A master cylinder for a hydraulic braking system is shown below in
figure 4.5. You will be required to draw detail drawings of some of the
components in the exercise section of this module part.
16
Figure 4.5
Master cylinder
Courtesy: Trinity College Auburn
© LMP
Worked example 1
Design a detail drawing of the piston seal from the master cylinder
components. The psiton seal in shown in figure 4.6.
a
Show the designs for four possible detail drawings, using freehand drawing
techniques:
i
without the use of a section
ii
using a full-section
iii using a half-section
v
using a part-section.
b
Comments should be made as to the standards used and the good and bad
points of the designed detail drawing.
c
Fully dimension each drawing, using different placements for the dimensions.
17
10
10
Ø
25
Ø 12
Ø 25
Ø 22
S
FRONT VIEW
Figure 4.6
RIGHT SIDE VIEW
L
EA
Ø
Ø
12
22
Material: neoprene
Scale 2:1
Possible solution 1 (without the use of a section)
Comments on solution 1
•
Front View – As no sectioning is used, hidden outline must be used to show
the details of the hole. Hidden outline should be avoided where possible.
•
Right Side View – The drawing is clear and gives a good shape description.
•
Dimensioning – Dimensioning is clear and easily interpreted.
•
New Methods – The use of circle templates.
•
Decision – will not use this drawing due to hidden outline.
10
Ø 25
Ø 22
Ø 12
FRONT VIEW
RIGHT SIDE VIEW
Material: neoprene
Scale 2:1
Figure 4.7 Possible solution 2 (using a full-section)
•
Front View – As full sectioning is used, no hidden outline is shown. The
details of the hole are shown as visible outline; these are correct standards.
18
•
Right Side View – The drawing is clear and gives good shape description.
•
Dimensioning – Dimensioning is poorly designed with too many dimensions
shown on the Right Side View.
•
New Methods – The use of circle templates, the use of a full section.
•
Decision – Good solution but will not use as better solutions can be found.
10
Ø 22
Ø 25
Ø 12
FRONT VIEW
RIGHT SIDE VIEW
Material: neoprene
Scale 2:1
Figure 4.8 Possible solution 3 (using a half-section)
•
Front View – As half-sectioning is used, no hidden outline is shown. The
details of the hole are shown on one side of the centre line and the exterior
details on the other side of the centreline; correct standards. Note that a halfsection may only be used when drawing a symmetrical component.
•
Right Side View – The drawing is clear and gives a good shape description.
•
Dimensioning – Dimensioning is poorly designed with too many dimensions
shown on the Front View.
•
New Methods – The use of circle templates, the use of a half-section.
•
Decision – Good solution, could be used.
19
Ø 22
Ø 12
Ø 25
10
FRONT VIEW
Material: neoprene
Scale 2:1
Figure 4.9 Possible solution 4 (using a part-section)
•
Front View – As part-sectioning is used, no hidden outline is shown. The
details of the hole are shown as visible outline; correct standards.
•
Right Side View – No view is needed, shape is defined by the dimensions.
Note that the dimensioning of the diameters in the Front View allows the
Right Side View to be omitted. The circular shapes have been defined by the
use of these diameter dimensions.
•
Dimensioning – dimensioning is good, correct AS1100 standards.
•
New Methods – the use of a part-section. The part-section line is a thin dark
continuous freehand line.
•
Decision – Good solution, will use.
Possible solution 4, using a part-section, is the quickest and preferred
method. You will be required to draw this detail drawing of the piston seal
using instruments, as exercise 4.1 in your exercises.
20
Varying the design process
In the next worked exercise you will be shown how the design process may be
varied to suit the component being drawn. The right side view would show only
three concentric circles. These are rather difficult to draw freehand, and really
serve no purpose in repeating them in the design process. The right side view will
therefore be omitted. Again, the design of the dimensioning can be completed
later.
Worked example 2
Design a detail drawing of the valve from the master cylinder components. You
are to show four possible detail drawings, using freehand drawing techniques.
The valve is to be made from 70-30 brass.
i
ii
iv
2
Ø
30
12
Ø6
16
Ø
Figure 4.10 Valve
As the right side view is only three concentric circles, you can omit the RSV in
the design process and show only the four methods for the various front views.
Do not show the dimensions on these design sketches, you can design the
dimensions later.
The four possible front views of the valve.
FRONT VIEW
(no section)
FRONT VIEW
(full section)
FRONT VIEW
(half-section)
FRONT VIEW
(part-section)
Figure 4.11 Four possible solutions
21
The four possible solutions are given above. When you have designed the
solutions you should check that the solutions are correct, and that the correct
AS1100 standards have been used.
Again, as with worked example 1, Possible solution 4, using a part-section, is the
quickest and preferred method, and again the right side view may be omitted if the
dimensions are able to be clearly placed on the front view.
Now the design for the position of the dimensions should be done on this solution.
Ø 16
Ø6
Ø 30
When the design is completed the drawing is done using instruments or a CAD
system. The completed detail drawing is shown below.
2
PART-SECTIONED FRONT VIEW
Figure 4.12
Material: 70–30 brass
Detail drawing of valve
AS 1100 standards
When you have completed a drawing you should check to see that you have
correctly used AS1100 standards.
All lines, other than construction lines should be the same darkness. During the
HSC marking the darkness of lines is always checked. The lines should be dark
enough so that if photocopied they would give a good dark outline.
Line thickness is also important. Visible outline should be drawn with thick dark
lines. All lines other than visible outline are thin dark.
The dimensioning standards should also be checked. There are two small
dimensions, the diameter 6 mm and the thickness 2 mm, that should be checked.
In both cases the arrows have been positioned outside the extension lines as there
is insufficient space to neatly draw the arrows inside these extension lines.
Similarly the space for the 2 mm dimension is too small to enable the number to
be neatly lettered. The 2 mm dimension is written outside the extension lines as
shown, and in line with the leader line.
22
Computer aided drawing
In Landscape products you were introduced to Computer aided drawing (CAD).
Before commencing this section you should take time to review that work.
In this module you will learn more about computer graphics and computer aided
drawing. You will learn to produce simple computer assisted drawings using
tools and coordinates.
Coordinates
All objects drawn using CAD are defined by the positioning of points. A line may
be defined by the positioning of its two end points. A circle may be defined by
the positioning of its centre point and a point on the circle, that is, the radius of the
circle.
Graphing points in mathematics
You have already used x and y coordinates to plot points, lines and curves when
drawing graphs in mathematics.
•
The x axis is the horizontal axis.
•
The y axis is the vertical axis.
•
The coordinates, (2, 3) represent the x value, 2, and the y value, 3.
•
To plot the point defined by the coordinates (2, 3);
–
mark off a distance of 2 units from the origin, along the horizontal x axis,
–
mark off a distance of 3 units along the vertical y axis, from the origin,
–
locate and label the plotted point (2, 3).
The axes have both positive and negative values, measured from the origin:
•
x axis: positive to the right, negative to the left from the origin
•
y axis: positive above, negative below the origin.
23
+y
(2,3)
0
Figure 4.13 Graph of point 2, 3
Example 1
On the axes given above, plot the point (6, 7), then draw the line defined by the
end points, (2, 3) and (6, 7).
Solution
Mark off the horizontal distance 6 units to the right of the origin. Mark off the
vertical distance 7 units above the origin. Locate and label point (6, 7).
Draw a line to join points (2, 3) and (6, 7).
Cartesian coordinate system in CAD
In CAD the Cartesian coordinate system is used to define the position of a point in
space by using the x , y and z axes radiating from a fixed, or predetermined point
called the origin. To position a point on a flat surface, such as on drawing paper
or a computer screen, you need only use two coordinates, (x,y) as in mathematics.
CAD uses the x and y values to precisely specify the location of points and thus
lines and objects. As with mathematics the coordinates use both positive and
negative values.
There are three different methods you can use in CAD systems to locate points:
•
absolute coordinates – you measure the x and y values from the origin, (0,0)
as you do in mathematics.
•
relative coordinates – you measure the actual sizes along the x and y
directions from the last point entered. Negative values are frequently used.
•
polar coordinates – you measure the radial distance from the last point
entered, and the angle, measured in a counter clockwise direction from the
positive x axis. Note, we could also use absolute polar coordinates.
24
Example of absolute coordinates
When using absolute coordinates in CAD, the x and y values are measured from
the origin, (0, 0). The positive x value is measured horizontally to the right of the
origin. The positive y value is measured vertically upward from the origin.
Thus the Absolute Cartesian Coordinates (2, 3) of a point, P, indicate that the
point to be plotted is 2 units to the right and 3 units above the origin. This is
shown in the previous diagram, and also applies to CAD.
Absolute coordinates are not widely used as it is difficult to calculate the values of
all points in a complicated drawing.
Example 2
20
40
20
A
40
Figure 4.14 Right side view of ratchet block
The origin, the x axis and the y axis are drawn below to represent a CAD drawing
on a computer screen. Point A, having coordinates of (25, 20) is plotted on the
axes.
i
On the given axes, sketch to scale the right side view of the ratchet block.
ii
Determine and label the coordinates of each of the points on the drawing. Be
sure to write the x value first, then the y value.
iii If you have access to a computer with a CAD package, draw the right side
view of the ratchet block using the absolute coordinates method. If not, read
through and study the method.
25
+y
A (25,20)
0
Figure 4.15
Plotting point (25,20)
Solution (sketching)
i
To sketch the shape, draw a horizontal line from point A, 40 mm to the right.
Draw a vertical line 40 mm upward from the right hand end of this horizontal
line. Draw a line from the top of this vertical line to point A.
Now draw a vertical line 20 mm upward from point A. Draw a line from the
top of this vertical line, horizontally, 20 mm to the right, to meet the sloping
line.
ii
The absolute coordinates of each of the points, listed in cyclic order from A,
in a clockwise direction are: (65, 20); (65, 60); (45, 40) and (25, 40).
iii The method used will vary with the CAD package that you are using. The
solutions uses AutoCAD.
The following solution assumes that you can create a new drawing.
We will use a line tool in each exercise, other tools could have been used.
Solution (CAD)
Click the Line button in the toolbox.
Type the absolute coordinates 25, 20 at the <From> point prompt, then press the
<Enter> key. This tells the computer that the line you wish to draw begins at the
point, 25, 20.
Type the absolute coordinates of the end point, 65, 20; <Enter>.
Type the absolute coordinates of the next end point, 65, 60; <Enter>.
Type the absolute coordinates of the next point, 25, 20; <Enter>.
26
Note that AutoCAD remembers the last point specified.
Example of relative coordinates
When using relative coordinates the actual dimensions are measured along the x
and y directions from the last point entered.
The first point is entered using absolute coordinates, as described previously.
The relative coordinates of the second point are then entered. These relative
coordinates describe the actual distance from the first point to the second point.
Remember that the relative coordinate distances are measured along the axes from
the last point entered, not from the origin. This enables the dimensions of the
object to be used without having to calculate the absolute coordinates for each
point from the origin. This is a quicker and easier method.
Example 3
The front view of a ratchet block follows.
20
40
40
B
10
Figure 4.16
20
Front view of ratchet block
The origin, the x axis and the y axis drawn below represent a CAD drawing on a
computer screen. Point B, having coordinates of (25, 20) is plotted.
i
On the given axes, sketch to scale the front view of the ratchet block.
ii
Determine and label the relative coordinates of each of the points on the
drawing. Assume that you draw the 40 mm square first, commencing from
point B and drawing in a counter clockwise direction. Remember, the
coordinates are relative to the previous point plotted. Assume that you draw
the 20 mm square next, commencing at the bottom right hand corner.
Note: record the relative coordinates using the AutoCAD system, ie the
relative coordinate for the right hand end of the first line is @40, 0.
iii If you have access to a computer with a CAD package, draw the front view of
the ratchet block using the relative coordinates method.
27
If not, read through and study the methods.
+y
B (25,20)
0
Figure 4.17 Plotting B (25,20)
Solution
i
To sketch the shape, draw a horizontal line from point B, 40 mm to the right.
Draw a vertical line 40 mm upward from the right hand end of this horizontal
line. Draw a horizontal line from the top of the vertical line, 40 mm to the
left. Draw a vertical line 40 mm downward from the left hand end of this
horizontal line to point B.
Find a point on the bottom line 30 mm to the right of point B then draw a
vertical line 20 mm upward from this point. Draw a horizontal line from the
top of this vertical line, 20 mm to the left. Draw a vertical line 20 mm
downward from the left hand end of the previously drawn line.
ii
Determining the relative coordinator for using AutoCAD. Absolute
coordinates of point B, 25, 20.
The relative coordinates of each of the other points for the 40mm square,
listed in cyclic order from B, in a clockwise direction are:
Determining relative coordinates
AutoCAD method
change in x value, 40, change in y value, 0
@40, 0
@0,4 0
change in x value, -40, change in y value, 0
@-40, 0
change in x value, 0, change in y value, -40
@0, -40
Absolute coordinates of the first point, the bottom right hand corner, for the
20 mm square, 55,20.
28
The relative coordinates of each of the points for the 20 mm square, listed in
cyclic order from the first point, in a clockwise direction are:
Determining relative coordinates
AutoCAD method
@0, 20
change in x value, -20, change in y value, 0
@-20, 0
change in x value, 0, change in y value, -20
@0, -20
Note the use of negative coordinates
iii Method of drawing using AutoCAD.
Click the Line button in the toolbox.
To draw the 40 mm square, type the absolute coordinates 25, 20 at the
<From> point prompt, then press the <Enter> key.
Now select the next point at the known distance of 40 mm horizontally to the
right of point B.
(Note: to enter relative coordinates, 40,0 in AutoCAD, type @40, 0.)
Now; type @40, 0 press <Enter>.
type @0, 40 press <Enter>.
type @-40, 0 press <Enter>.
type @0, -40 press <Enter>.
To draw the 20 mm square, start at the bottom right hand corner.
Type the absolute coordinates 55, 20 at the From point prompt., then press the
Enter key. This tells the computer that the line you wish to draw begins at the
point, 55,20. Now enter the relative coordinates.
type @0, 20 press <Enter>.
type @-20, 0 press <Enter>.
type @0, -20 press <Enter>.
Example of relative polar coordinates
Polar coordinates can be either absolute coordinates or relative coordinates. The
relative coordinates are the most commonly used, so we will only consider them.
When using relative polar coordinates the actual dimensions are measured in a
radial direction from the last point entered, the angle is measured in a counter
clockwise direction from the positive x axis.
The first point is entered using absolute coordinates, as described previously.
29
The relative polar coordinates of the second point are then entered. These relative
coordinates describe the angle of rotation and the actual distance from the first
point to the second point.
20
40
40
B
10
Figure 4.18
20
Top view of ratchet block
The origin, the x axis and the y axis drawn below represent a CAD drawing on a
computer screen. Point B, having coordinates of (25, 20) is plotted.
i
On the given axes, sketch to scale the top view of the ratchet block.
ii
Determine and label the relative polar coordinates of each of the points on the
drawing. Assume that you draw the 40 mm square first, commencing from
point B then drawing in a counter clockwise direction. Remember, the
coordinates are relative to the previous point plotted. Assume that you draw
the 20 mm square next, commencing at the bottom right hand corner.
AutoCAD system; the relative polar coordinate for the right hand end of the
first line is @40<0. This means the required point is a distance of 40 mm
from B at an angle of 0º from the x axis.
iii If you have access to a computer with a CAD package, draw the front view of
the ratchet block using the relative polar coordinates method. If not, read
through and study the methods.
+y
B (25,20)
0
Figure 4.19 Plotting point B (25,20)
30
Solution
i
The solution is the same as for the front view.
ii
Absolute coordinates of point B, 25, 20.
The relative polar coordinates of each of the other points for the 40 mm
square, listed in cyclic order from B, in a clockwise direction are as follows.
Determining relative polar coordinates
AutoCAD method
Radial distance 40. Angle from x axis 0º
@40<0
@40<90
@40<180
@40<270
Absolute coordinates of the first point, the bottom right hand corner, for the
20 mm square, 55, 20.
The relative coordinates of each of the points for the 20 mm square, listed in
cyclic order in a clockwise direction from the first point as follows.
Determining relative polar coordinates
AutoCAD method
@40<90
@40<180
@40<270
You should read the notes a number of times, highlighting the areas of importance
or concern. Complete all of the problems, including the computer portion. If you
do not have access to a CAD system, work through the exercise using freehand
sketching methods.
31
32
Exercises
Exercise 4.1
a
Define the term ‘mechanical work’.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
b
Analyse each of the following three problems, and determine an equation for
the work done in moving the car.
i
A car travelling at constant velocity on a horizontal plane, against a
frictional resistance (FR) a distance (s).
Work done in overcoming this resistance is:
ii
Against gravity when raising a car of mass (m), through a height (h):
Work done against gravity is:
iii A car travelling at constant velocity, up an inclined plane of angle q to the
horizontal, through a distance (s).
Work done is:
33
c
Determine the work done by a braking system in the following. In each case
the hand brake is in operation, but in iii ineffectively.
i
Car stationary on a horizontal surface.
Work done by the hand brake =
ii
Car stationary on an inclined surface.
iii Car on an inclined surface, moving downward at constant velocity.
d
Define the following types of energy and give the formula for
calculating that energy.
i
Kinetic energy
___________________________________________________
___________________________________________________
___________________________________________________
ii
Potential energy
___________________________________________________
___________________________________________________
___________________________________________________
iii Strain energy
___________________________________________________
___________________________________________________
___________________________________________________
34
e
Define the term ‘power’ when referring to mechanics.
______________________________________________________
______________________________________________________
______________________________________________________
f
Two very important formulas may be derived from the basic formula, power
equals work divided by time. Show how these two formulas are derived.
Exercise 4.2
a
Explain what is meant by a hydraulic system used in brakes.
______________________________________________________
______________________________________________________
______________________________________________________
b
Define the term ‘pressure’.
______________________________________________________
______________________________________________________
c
State Pascal’s Principle.
______________________________________________________
______________________________________________________
______________________________________________________
d
State Archimedes’ Principle.
______________________________________________________
______________________________________________________
______________________________________________________
35
e
A concrete brick of mass 5 kg and dimensions 430 x 150 x 100 rests on a
horizontal surface, flat on its largest face. Determine the pressure applied to
the horizontal surface.
f
The front wheel brakes provide approximately 67% of the braking forces due
to ‘dipping’ of the car when braking. Explain how the hydraulic system is
able to provide for this need for greater braking forces at the front wheels.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
Exercise 4.3
Figure 4.20 represents a sealed hydraulic braking system. A force of 200 N is
applied to the brake pedal as shown. Size details of the pedal, master cylinder, and
front and back wheel cylinders are given on the diagram.
Determine the thrust (force) delivered by each of the wheel cylinders.
Note the four steps needed to complete this question are set out in parts i, ii, iii,
and iv.
36
100 N
250
A2 = 300 mm2 rear wheel cylinder
A1 = 600 mm2
50
A3 = 900 mm2 front wheel cylinder
Figure 4.20
Hydraulic braking system
i
Determine by moments the resultant force on the master cylinder caused by
the applied force of 200 N exerted on the brake pedal.
ii
Determine the pressure generated in the system by this resultant force.
iii Determine the thrust at the rear wheel cylinder.
iv
Determine the thrust at the front wheel cylinder.
37
Exercise 4.4
Draw, using instruments, to a scale of 2:1, a detail drawing of the piston seal in
figure 4.21. Note: it is sufficient to draw only one view, a part-sectioned front
view, then fully dimension the piston seal.
The design of the drawing has already been completed in Worked Example 1.
You should now draw the best solution to AS1100 standards, using your
instruments.
10
Ø
25
AL
SE
Ø
Ø
12
22
Material: neoprene
Scale 2:1
Figure 4.21 Piston seal
38
Exercise 4.5
Design a detail drawing of the spring seal from the master cylinder components in
figure 4.22. You are to show four possible detail drawings, using freehand
drawing techniques:
i
10
2
a
ii
iv
The spring seal is to be made from neoprene.
Ø
25
Ø
22
Figure 4.22 Spring seal
As the right side view consists of only four concentric circles, two visible and two
hidden outline, you can omit the Right Side View and show only the four front
views. Do not show the dimensions on these design sketches.
Front View
(No section)
Front View
(Full section)
Front View
(Half-section)
Front View
(Part-section)
Methods iii and iv are the best solutions. You should now complete the
half-section solution to give you experience with this standard.
First design the position of the dimensions.
39
b
Draw, using instruments, to a scale of 2:1, a detail drawing of the valve.
Note; it is sufficient to draw only one view, a half-sectional front view and then
fully dimension the valve.
40
Exercise 4.6
Shape and size details of a ratchet block are given figure 4.23.
Note: the shape details are fully shown as the illustration is a three dimensional
isometric drawing. The size details are given using dimensioning.
20
40
20
40
B
20
10
A
Figure 4.23 Ratchet block
The origin, the x axis and the y axis, drawn on the next page represent a CAD
drawing on a computer screen. The scale of the drawing is 1:1.
Point B, having coordinates of (40,60) is plotted on the axes.
i
On the given axes, commencing at Point B, sketch to scale 1:1 the front view
of the spacing block.
ii
From this front view, project using third angle projection a right side view of
the spacing block. Note: the absolute coordinates of the right side view of
Point A, the starting point for the right side view, are (110,60).
iii Also from the front view, project a top view of the spacing block. Note: the
absolute coordinates of the top view of Point B, the starting point for the top
view, are (40,130). Name the TOP VIEW.
iv
Determine and neatly label the absolute coordinates of each of the points on
the drawing.
v
Commencing at Point B, (40,60), fully describe how you would draw the
orthogonal drawing of the spacing block using a AutoCAD package. You
may use either relative or polar coordinate methods or a combination of the
two methods. List every step in sequence, giving the coordinate entries for
each point plotted.
Space for this exercise is provided after figure 4.24.
41
+y AXIS
B (40,60)
0 ORIGIN
Figure 4.24
+x AXIS
CAD drawing
42
Describe how you would draw the orthogonal drawing of the spacing block,
commencing at Point B, (40,60), using AutoCAD. Underline the method you
will use: relative coordinates, polar coordinates, combination of the two.
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
__________________________________________________________
43
44
Progress check
In this part you used mathematical and graphical methods to solve problems of
engineering practice and applied graphics as a communication tool.
✓
❏
✓
❏
Uncertain
Agree – well done
Disagree
✓
❏
Agree
I have learnt about
•
– work, power, energy (without calculations)
– fluid mechanics
Pascal’s and Archimedes’ Principles
hydrostatic pressure
applications to braking systems
•
communication
– detail drawing
– computer graphics, computer assisted drawing
(CAD).
I have learnt to
•
experiment with and apply the basic principles of fluid
mechanics to simple braking systems
•
produce detail drawings of braking systems and
braking components applying appropriate Australian
Standard (AS 1100)
•
produce simple computer assisted drawing(s).
During the next part you will generate an engineering report.
45
46
Name: ________________________
Check!
❐
Exercise 4.1
❐
Exercise 4.2
❐
Exercise 4.3
❐
Exercise 4.4
❐ Exercise 4.5
❐ Exercise 4.6
Locate and complete any outstanding exercises then attach your responses to this
sheet.
Centre/School (DEC). You will need to return the exercise sheet and your
If you study Stage 6 Engineering Studies through the OTEN Open Learning
Program (OLP) refer to the Learner’s Guide to determine which exercises you
need to return to your teacher along with the Mark Record Slip.
47
Braking systems
Part 5:
Engineering report
Part 5 contents
Introduction.......................................................................................... 2
What will you learn?................................................................... 2
Engineering report.............................................................................. 3
Aims of an engineering report..................................................... 3
Structure of the engineering report.............................................. 4
Developing an engineering report ............................................... 6
Sample engineering report ......................................................... 6
Exercise ..............................................................................................23
Progress check ..................................................................................25
Bibliography........................................................................................29
Module evaluation .............................................................................31
Braking systems
1
Introduction
In this part you will:
•
examine the components of an engineering report
•
read through a sample engineering report
•
report on a braking system or brake component.
•
engineering report writing
•
communication
–
research methods including … libraries
–
collaborative work practices.
You will learn to:
•
complete an engineering report based on the analysis of one type of
brake or component of a braking system, integrating computer
software.
•
conduct research using appropriate computer technologies
•
work with others and appreciate the value of collaborative working.
2
Part 5: Engineering report
Engineering report
An engineering report is a formal, considered document which draws
together information gained about a product or filed, through research and
analysis, to arrive at a conclusion or present recommendations based on
investigation.
Engineers do not communicate with words alone. In an engineering report,
technical information is presented using a combination of text, tables,
graphs and diagrams.
An engineering report for an application module involves:
•
outlining the area under investigation
•
collecting and analysing available data
•
drawing conclusions and/or proposing recommendations
•
acknowledging contributions form individuals or groups
•
recording sources of information
•
including any relevant additional support material.
An engineering report for a focus module involves covering additional
aspects such as:
•
examining the nature of the work done by the profession
•
discussing issues related to the field.
Aims of an engineering report
A well structured engineering report aims to:
•
demonstrate effective management, research, analysis and
communication skills related to the content
•
include data relevant to the area under investigation
Braking systems
3
•
present information clearly and concisely so that it is easily understood
by the reader through the use of tables, graphs and diagrams to
illustrate mathematical and scientific facts
•
justify the purpose using observations, calculations, or other evidence,
to support a conclusion or recommendations.
•
document contributions and sources of information.
Structure of the engineering report
An engineering report generally includes the following sections:
•
title page
•
abstract
•
introduction
•
analysis
•
result summary
•
conclusions/recommendations
•
acknowledgments
•
bibliography
•
appendices.
Title page
The title page gives the title of the report, identifies its writer or writers and
gives the date when the report was completed.
Abstract
The abstract is a concise statement that describes the content of the
engineering report. It covers the scope of the report (what it is about) and
the approaches used to complete the analysis (how the information was
assembled).
The purpose of the abstract is to allow a reader to decide if the engineering
report contains relevant information.
The abstract should be no more than two or three paragraphs – shorter if
possible.
4
Introduction
The introduction provides an overview of the subject, purpose and scope of the
engineering report and may contain background information regarding the
topic.
It also outlines the sections of the engineering report including why the
investigation was undertaken, what research occurred, how data was collected
and what anaylsis was conducted.
Analysis
The analysis is the body of the engineering report and should show evidence of
research and experimentation. Information about materials and the mechanics
of products should be collected or calculated for all engineering reports. This
section must contain information required to satisfy the aim and purpose of the
report.
Tables and graphs, used to summarise detailed data in a concise form, are
common features of an engineering report.
Result summary
The result summary should present the results concisely and note any
limitations on the investigation.
The results inform and support the conclusions and recommendations.
Conclusions/recommendations
The conclusions/recommendations summarises major points or issues in earlier
sections of the engineering report.
This section requires the writer to draw conclusions or make recommendations
based on data collected. If the purpose of the report was to ‘select the best…..’,
then the selection should be stated and the reason for the selection explained.
Acknowledgments
The acknowledgment section provides the opportunity to credit other people’s
work that has contributed to the engineering report.
Bibliography
The bibliography demonstrates that the report is well researched – all
references need to be included. Bibliographic entries should follow established
guidelines.
Braking systems
5
A standard approach is the Harvard systems of referencing requiring the
authors surname initials, date of publication, title of reference, publisher and
place of publication.
For example:
Kalpakjian, H. R. and Wrighton, H. 1967, Practical Microscopical
Metallurgy, Addison Westley, USA.
This information allows the reader to source the information for confirmation
of the details or conduct further research.
Appendices
The appendices should contain detail that has been separated from the main
body of the engineering report. The information in this section is not essential
but enhances the other data. Examples could be engineering drawings of the
products being compared where the overall dimensions of the product may not
have been part of the comparison in the report, but may be relevant to some
readers.
During the engineering course this section may contain a technical drawing and
could include information collected from organisations.
Developing an engineering report
Research and collaboration are the keys to developing an accurate and
informative engineering report.
Research methods
In addition to popular research methods, like individual input and electronic
media, traditional reference materials remain a valuable source of
information and include:
•
textbooks
•
booklets, brochures and pamphlets
•
newspapers, journals and magazines
•
videos.
Collaborative work practices
Discussion with colleagues constitutes collaboration and can provide
valuable information.
6
These approaches to research and collaboration can be used by you, along
with the Information Technology (IT), to develop your engineering report,
as well as for any other research you may need to undertake.
Sample engineering report
You have already completed two engineering reports in the previous
modules so you will be expected to present a more comprehensive report
this time.
The engineering report for this module must be based upon the
investigation of a braking system such as:
•
band brake
•
drum brake
•
disc brake
•
multiple disc brakes systems
•
regenerative braking systems.
Alternately you may wish to complete your engineering report on the
analysis of an individual component, such as the caliper braking system
used on bicycles.
You should communicate the selected topic for your engineering
report to your teacher before commencing the report. You may
negotiate with the teacher for an alternative topic based upon a braking
system if you have a particular interest or resource available.
You must be aware of the need for all safety precautions to be followed
during research and experimentation. Do not tamper with the braking
system of a registered vehicle. Tampering may make the vehicle
unroadworthy which could have fatal consequences.
The following section contains a sample engineering report that you may
use as a guide when presenting your work.
The sample of engineering report focuses on the investigation of a
component of a braking system – the brake shoe used in the rear drum
brakes of a current model car.
To assist you the sample engineering report will include notes explaining
the reasons for the selection or use of the information in the report. These
notes have been boxed to separate comments from the report.
Braking systems
7
8
Braking systems
Report title:
A brake shoe
Module:
Braking systems – Module 3
Authors name:
F. Riction
Date:
February 2000
Abstract
The report provides a brief history of braking and brake shoes, analyses
the materials used and the mechanical situations involved.
Introduction
This report will investigate a braking system components- a brake shoe.
The report aims to:
‘analyse the rear brake shoes used in a current model car and determine if
a better product could be produced’.
Function of the product
The function of a brake shoe may be summarised as follows:
•
to provide a braking force to the brake drum
•
to adequately support a brake lining
•
to transmit the applied force from the hydraulic system.
The function of the product or system selected by you must be
analysed and fully described in this section.
Figure 5.1
A brake shoe with lining material attached
Courtesy: Trinity Catholic College
© LMP
A freehand pictorial sketch
Figure 5.02 A freehand pictorial sketch
Engineers frequently use freehand sketching, particularly for pictorial
drawings. Another method is to sketch the pictorial from looking at the
actual component. Either method helps you with your presentation.
A sketched detail drawing
Figure 5.03 A sketched detail drawing of the brake shoe (without dimensions)
Sketching freehand orthogonal drawings is easier than
sketching freehand pictorials. A drawing involving large
circles such as in a brake shoe requires more skill, especially
with concentric circles. In cases like this an engineer would
probably use an aid such as a radius curve.
You have to decide how many views are required to give a
complete shape description for the report. Refer back to Part 4
where you did detail drawings, then decide on the most simple
method to show the brake shoe.
The drawing completed above uses two views, a front view and
a left side view that fully describes the shape of the brake shoe.
Well maybe it does if the person reading the report can
interpret an orthogonal drawing. Don’t worry too much, they
can also refer to the pictorial which is easier for the untrained
person to understand. Add all of the dimensions.
If a problem still exists, provide a model of the component or
include the actual component in an attachment.
If the component is too big to be included, attach a video. Be
innovative, and design a solution to a perceived problem.
A detail drawing of the product
42
50
24
4
Ø4
30 ∞
8
R8
R
R
Ø6
0
12
15∞
10
4
Ø 10
R 10
4
LEFT SIDE VIEW
FRONT VIEW
Scale 1:2
Material 0.2% C Steel
Figure 5.04 A detail of a brake show
It is good practice to include a completed drawing of the component
or product in the report. It makes the report look more professional.
It is also another opportunity to practice orthogonal drawing.
Analysis
The main components of the product
There are only two components in the brake shoe, the curved plate and
the drilled web.
The shape of the curved plate is stamped from 4 mm mild steel strap then
curved to the required radius.
The shape of the web is stamped from 4 mm mild steel strap and then
stamped a second time to produce the required holes.
The two components are welded together.
Note: that the components required only one material, mild steel and that
service properties refer only to the properties that the material needs
when in use, not the properties needed during manufacture.
Material
The metal components of the brake shoe are manufactured using mild
steel (0.2% carbon). Two service properties of mild steel that make it a
suitable material for the brake shoe are:
1
adequate toughness, able to absorb impact forces without fracture
2
adequate tensile and compressive strength to retain shape under
applied loads.
Environmental effects that mild steel might have:
a
during production of the material
The mining of the iron ore causes environmental problems with the
surrounding area, and also affects the mining area, the flora and
fauna in the mine area and near the mine.
The smelting and production of steel has a very adverse effect upon
the surrounding area near the steelworks.
b
during manufacture of the product
As the brake shoe components are produced by stamping and
pressing, the main environmental effect would be the noise
pollution.
The welding causes fumes and produces welding light both of which
can be detrimental to the operator or personnel near the welding
area.
c
during service in the product
The material causes no environmental problem during service.
The material can be reused to produce steel for other products.
Alternative materials that could be used
Two possibilities are: medium carbon steel and gray cast iron. An
evaluation of the advantages and disadvantages of each material
compared to the use of mild steel, 0.2% C follow.
Medium carbon steel, 0.35% C
Advantages
•
Higher carbon content, gives greater strength.
•
Increased toughness.
•
Increased tensile and compressive strength.
•
Increased shear strength.
•
Able to be hardened by heat treatment.
Disadvantages
•
Higher carbon content, requires larger forces.
•
Increased cost of production, larger forces needed.
•
Product would be overdesigned.
Gray cast iron
Advantages
•
Brake shoe is able to be produced as a single component.
Disadvantages
•
Decreased toughness.
•
Decreased tensile and compressive strength.
•
Increased cost of production.
•
Heavier component.
Mechanic and hydraulic situations involving the brake shoe
Stress and strain
1
During manufacture
The components are stamped from steel strap. Shear stress
calculations must be conducted to determine the force required to
stamp out the shape of each component.
Compressive stress and strain calculations are also required to
determine the forces in the punch or punches used to stamp out the
shapes.
2
During service
The web of the brake shoe undergoes compressive stress during
service due to the forces applied by the slave cylinder. The
compressive forces must be determined to ensure that the yield
strength of the steel is adequate for the required service condition.
Friction
1
During manufacture
Friction is involved during the stamping process and must be
considered during the design of the punches.
2
During service
Friction is involved during operation of the braking system. The
braking force produces shearing stresses in the welded web, and
must be considered during design calculations.
Hydraulics
The forces applied by the slave cylinder to the brake shoe must be
determined using moment calculations, and hydraulic pressure
calculation involving Pascal’s Principle.
1
During manufacture
The stamping machine uses a hydraulic press, which could involve
calculations of force and pressure.
2
During service
The forces applied by the slave cylinder would have to be determined.
Energy and power
Energy and power calculations affect the brake shoe during both
manufacture and service.
During manufacture
The energy used to stamp out the shapes could be determined as a
comparison between the different materials being investigated for use.
During service
The energy generated by the friction forces needs to be dissipated as heat
energy. Calculations of the mechanical energy generated at the brake
lining surface could be made.
Note that it would be sufficient for you to cover only two situations
involving the mechanics and/or hydraulics for your selected product.
1
Stress and strain:
Three cylindrical punches, of diameters 10 mm, 6 mm and 4 mm are
used to punch out the holes of the web of a brake shoe of thickness 4
mm in the one simultaneous operation.
i If, during the punching operation, the total compressive stress in
the punches is 720 MPa, determine the total force required to
punch out the holes.
d1
=
10 mm
=
10 ¥ 10–3 m
d2
=
6 mm
=
6 ¥ 10–3 m
d3
=
4 mm
=
4 ¥ 10–3 m
s
=
720 Mpa =
P
=
?
720 ¥ 106 Pa
Area being stressed is the total cross sectional area.
A1
=
p¥d
2
4
-3 2
=
=
A2
=
p (10 ¥ 10 )
4
78.54 ¥ 10-6
p¥d
2
4
-3 2
=
=
A3
=
p (6 ¥ 10 )
4
28.27 ¥ 10-6
p¥d
2
4
-3 2
=
p (4 ¥ 10 )
4
=
12.57 ¥ 10-6
=
A1 + A2 + A3
=
119.38 x 10-6m 2
Now s
=
P
A
P
=
s¥A
=
720 ¥ 10 6 ¥ 119.38 ¥ 10-6
=
85.953.6
=
85.95 kN
Total Area
ii
Determine the shear stress in the 4 mm thick material.
Area being sheared is the total surface area of the three
cylindrical shapes being punched out of the material for the web
of the brake shoe.
Total Shear Area = circumference of three holes x thickness
SA = p ¥ (d1 + d 2 + d3) ¥ t
–3
= p ¥ (10 + 6+ 4) ¥10 ¥ 5¥ 10-3
= p ¥ 100 ¥ 10-6m 2
Now s = P ∏ A
=
85 954
p ¥ 100 ¥ 10
-6
= 273.6 ¥ 10 6 Pa
= 273.6 MPa
2
A pressure of 50 MPa is produced in the slave cylinder. If the
internal area of the cylinder is 30 mm:
i
determine the force applied to the brake shoe.
Pressure =
F
A
Force = Pressure ¥ Area
= 50 ¥ 10 6 ¥ 30 ¥ 10 -6
= 1 500 N
= 1.5 kN
ii
If the force applied to the front disc brake by the same
hydraulic system is to be twice the size of the force applied to
the rear drum brake, determine the internal area of the front
wheel cylinder.
Area required would be twice the area of the rear cylinder
\ Area = 60 mm.
Experiment to test alternative materials for the brake shoe
Note I was not able to gain access to materials testing machines. I was
hoping to conduct tensile and compressive test using a Hounsfield
tensometer.
I wanted to carry out comparative testing of the three materials, 0.2% C
steel, 0.35% C steel and gray cast iron.
I decided to conduct compressive tests and impact tests on the three
materials and to research the tensile properties of the three materials.
The tests are comparative only. Two pieces of each of the three materials
were cut and shaped to size; 20 mm long, 6 mm wide and 4 mm thick.
The compression tests were conducted in a vice, the force being applied
axially to each sample.
The impact test was conducted by holding the sample in a vice and
repeatedly striking the sample with a dumpy hammer.
The tests failed to give comparative results, the impact tests worked well,
especially with the gray cast iron, but I was unable to differentiate the
results for the steels. The compressive tests were complete failures. I
needed to gain access to testing machines but was unable to do so at the
local high school.
Collected data
Material
0.2% carbon
steel
0.35% carbon Gray cast iron
steel
Yield stress MPa 345
375
NA
UTS Mpa
440
580
170
Izod impact test
117
65
10
0.2% C steel
0.35% C steel
Gray cast iron
NA
Yield
Figure 5.07
UTS
Impact
Graph
Note: you may also carry out an experiment that fails to provide the
desired results. If this occurs, you still need to provide data that is
relevant to the report. Research data, provided that it is clearly
identified as such, may be used
Health and safety issues
1
The performance of braking systems for cars and trucks are regularly
tested. The vehicle, to be registered as roadworthy, must pass an
inspection each year. Heavy trucks are also randomly tested by the
RTA inspectors throughout the year.
Safety issues are thus extremely important when associated with the
brake shoe performance and design.
2
The poor performance of braking systems is still responsible for
many vehicle accidents and as such contributes greatly to the
hospitalisation of victims. The number of fatal accidents has been
reduced over the past ten years. The improvement in design and
maintenance of braking systems has contributed to this reduction.
Result summary
List of strong points
•
Cost effective.
•
Ease of production.
•
Transfers force effectively from wheel cylinder to brake drum.
•
Adequate strength properties when in use.
List of weak points
•
Some distortion under high and sustained temperatures.
•
Material may corrode in adverse environment.
Recommendations and conclusions
Conclusions
•
The material used is adequate and far superior to the other two
materials investigated for this report.
•
The shape of the components is the best for the designed purpose.
•
The manufacturing method is the most cost effective.
•
Health and environmental problems that occur during the production
of the components should be considered.
Recommendations
•
The material, 0.2% carbon steel is retained.
•
The design is retained.
•
The noise problem that occurs during the stamping operation be
addressed by the Workplace Health and Safety Committee.
Glossary
auxiliary
brakes
are additional brakes that are fitted to a
vehicle and are used to assist the major
braking system in the vehicle.
detail
drawing
a detail drawing is an orthogonal drawing
which gives a full size and shape
description of the component. It also
includes the material from which the
component is to be manufactured.
exploded
isometric
an exploded isometric drawing is a
pictorial drawing of an assembly in which
the components are drawn separated so
that details of each component can be seen
Acknowledgements
Gary Smith, Garage owner, Brakes and Brake Shoes
George Michaels, Physics Teacher, Hydraulics
Graeme Hamer, RTA Inspector, Brake Regulations
Tom Livingston, Manager, Metal Pressings Manufacturing Methods
Video Stop Better Brakes Audio-visual Production History
Bibliography
Schlenker, B.R. 1983,
Jacaranda Wiley, Australia.
Introduction to Materials Science,
Greaves, H.R. & Wrighton, H. 1967, Practical Microscopical
Metallurgy, Chapman and Hall, England.
Moffatt, W. Pearsall, G. and Wulff, J. 1964, The Structure & Properties
of Materials, John Wily and Sons, New York.
Kalpakjian, S. 1985, Manufacturing Processes for Engineering
Materials, Addison Wesley, USA.
Appendices
Historical development of braking and brake shoes
Shoe brakes were used extensively in the 18th and 19th centuries as part of
the hand-operated lever brakes used on horse drawn vehicles. These
brakes were essentially used as parking brakes to hold the vehicle
stationary. The main braking system was provided by the horse (or
bullock) slowing down and then stopping. The shoe brake could be
considered as an auxiliary brake that assisted the braking operation.
These brakes were still in use on delivery carts during the 1940s and may
still be seen in carriages and horse-drawn sulkies at shows.
The externally applied brake shoe used initially was made from wood and
operated by applying external pressure to the wrought iron rim of the
vehicles’ wheel. Wood proved inadequate as vehicles became faster and
heavier, so liners were introduced.
Mild steel shoes were introduced in the late 19th century, having wood
attachments and leather liners.
When pneumatic tyres were patented in 1888, the use of external shoe
brakes became limited. Various brake systems were used, including band
and cable brakes. In 1902, Louis Renault introduced the drum brake,
incorporating internal brake shoes.
The mechanical design of the drum brake systems has varied and
developed during the past century. The basis design of the brake shoe
has not altered except for the required shape designed for individual
vehicles.
The internal brake shoe consists basically of two parts, the curved plate
and the drilled web. The material used is mild steel, and the manufacture
involves the stamping out of the two shapes, bending of the plate into the
required curved shape, then welding the two components together.
External brake shoes are still in use on railway carriages. These involve
the use of medium carbon steel or gray cast iron shoes acting directly
onto spheroidal graphite cast iron wheels.
Exercises
Exercise 5.1
Select a braking system/component and complete an engineering report
structured under the headings used in the sample report.
You may obtain a component or components from a garage, a
wrecking yard, or from a vehicle that is to be scraped.
Alternatively use a bicycle part that you can see and measure.
Use computer software such as a word processing program or graphics
package to aide in the generation of your engineering report.
Braking systems
23
24
Part 5 – Engineering report
Progress check
In this part you completed an engineering report.
✓
❏
✓
❏
Uncertain
Agree – well done
Disagree
✓
❏
Agree
I have learnt about
•
engineering report writing
•
communication
– research methods including Internet.
I have learnt to
•
complete an engineering report based on the analysis
of one type of brake or component of braking system,
integrating computer software
•
conduct research using appropriate computer
technologies.
Congratulations! You have now completed Braking systems.
Braking systems
25
26
Exercises 5.1
Name: ______________________
Check!
Have you completed the following exercise and included all the sections?
❐ Exercise 5.1
•
title page
•
abstract
•
introduction
•
analysis
•
result summary
•
conclusions/recommendations
•
acknowledgments
•
bibliography
•
appendices.
Centre/School (DEC) you will need to return the exercise pages with your
responses.
Return the exercise pages with the Title Page cover attached. Do not return all
the notes, they should be filed for future reference.
If you study Stage 6 Engineering Studies through the OTEN Open Learning
Program (OLP) refer to the Learner’s Guide to determine which exercises you
need to return to your teacher along with the Mark Record Slip.
Please complete and return the module evaluation that follows.
Braking systems
27
28
Bibliography
Board of Studies, 1999, The New Higher School Certificate Assessment
Support Document, Board of Studies NSW, Sydney.
Board of Studies, 1999, Stage 6 Engineering Studies Examination, Assessment
and Reporting, Board of Studies NSW, Sydney.
Board of Studies, 1999, Stage 6 Engineering Studies Support Document, Board
of Studies NSW, Sydney.
Board of Studies, 1999, Stage 6 Engineering Studies Syllabus,
Board of Studies NSW, Sydney.
Board of Senior School Studies, 1972–1998, Engineering Science HSC
Examination Papers, Board of Senior school Studies NSW, Sydney.
Greaves, H.R. and Wrighton, H. 1967, Practical Microscopical Metallurgy,
Chapman and Hall, England.
Kalpakjian, S. 1985, Manufacturing Processes for Engineering Materials,
Addison Wesley, USA.
Moffatt, W. Pearsall, G. and Wulff, J. 1964, The Structure & Properties of
Materials, John Wily and Sons, New York.
Schlenker, B.R. 1974, Introduction to materials Science, John Wiley & Sons,
Sydney.
Warren, N.G. 1990, Physics Outlines, Pergoman Press,
New York.
Better Brakes, Hydraulic Disc and Drum Brakes.
Mitsubishi, Maintenance and Repair Manual for Mitsubishi Magna.
Matteucci, M. 1971, History of the Motor Car, Crown Publisher,
New York.
29
30
Module evaluation
To help us make improvements to future learning materials we would
like your comments on this material.
Tick the box which best describes you.
Gender
male
female
Study through
DEC
OTEN - OLP
Other
Age group
under 20 years
20 – 30 years
over 30
Circle the number that best represents your rating of this material.
The number 1 indicates a low rating and the number 5 indicates a high
rating.
There is room to make comment if you would like.
1
Rate your enjoyment of the
material.
___________________________
1
___________________________
2
3
4
5
___________________________
2
Rate your understanding of the
content.
___________________________
1
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2
3
4
5
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3
Rate the usefulness of the
activities.
___________________________
1
___________________________
2
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31
4
Rate the relevance of the
exercises.
____________________________
1
____________________________
2
3
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5
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5
Rate the accuracy of the
indicative time given.
____________________________
1
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2
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6
Rate the ease of obtaining the
resources.
____________________________
1
____________________________
2
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7
Rate the helpfulness of any
support material.
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8
Rate your achievement of the
outcomes for the material.
____________________________
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____________________________
Finally!
Which were the most challenging parts of the material?
__________________________________________________________
__________________________________________________________
Please return this form to your teacher to forward on to OTEN – LMP.
Thank you for this valuable information.
32
Learning Materials Production
Training and Education Network – Distance Education
NSW Department of Education and Training

Braking systems - Westfields Sports High School

Transcription

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