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ES/S6 – Prelim 41082 P0021884

Engineering StudiesPreliminary CourseStage 6

Braking systems

AcknowledgmentsThis publication is copyright Learning Materials Production, Open Training and Education Network –Distance Education, NSW Department of Education and Training, however it may contain material fromother sources which is not owned by Learning Materials Production. Learning Materials Productionwould 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 ingood 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 theCopyright 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

i

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 andengineering materials application – 1 ..........................1–49

Part 2: Development of braking systems andengineering materials application – 2 ..........................1–33

Part 3: Engineering mechanics, hydraulicsand communications in braking systems – 1..............1–57

Part 4: Engineering mechanics, hydraulicsand communications in braking systems – 2..............1–47

Part 5: Engineering report for braking systems.......................1–27

Bibliography........................................................................................29

Module evaluation .............................................................................31

iii

Subject overview

Stage 6 Engineering Studies Preliminary Course and HSC Course eachhave five modules.

Engineering Studies Preliminary CourseHousehold appliances examines common appliancesfound in the home. Simple appliances are analysedto identify materials and their applications.Electrical principles, researching methods andtechniques to communicate technical information areintroduced. The first student engineering report iscompleted undertaking an investigation of materialsused in a household appliance.

Landscape products investigates engineeringprinciples by focusing on common products, such aslawnmowers and clothes hoists. The historicaldevelopment of these types of products demonstratesthe effect materials development and technologicaladvancements have on the design of products.Engineering techniques of force analysis aredescribed. Orthogonal drawing methods areexplained. An engineering report is completed thatanalyses lawnmower components.

Braking systems uses braking components andsystems to describe engineering principles. Thehistorical changes in materials and design areinvestigated. The relationship between internalstructure of iron and steel and the resultingengineering properties of those materials is detailed.Hydraulic principles are described and examplesprovided in braking systems. Orthogonal drawingtechniques are further developed. An engineeringreport is completed that requires an analysis of abraking system component.

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Bio-engineering examines both engineeringprinciples and also the scope of the bio-engineeringprofession. Careers and current issues in this fieldare explored. Engineers as managers and ethicalissues confronted by the bio-engineer areconsidered. An engineering report is completed thatinvestigates a current bio- engineered product anddescribes the related issues that the bio-engineerwould need to consider before, during and after thisproduct development.

Irrigation systems is the elective topic for thepreliminary modules. The historical development ofirrigation systems is described and the impact ofthese systems on society discussed. Hydraulicanalysis of irrigation systems is explained. Theeffect on irrigation product range that has occurredwith the introduction of is detailed. An engineeringreport on an irrigation system is completed.

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HSC Engineering Studies modules

Civil structures examines engineering principles asthey relate to civil structures, such as bridges andbuildings. The historical influences of engineering,the impact of engineering innovation, andenvironmental implications are discussed withreference to bridges. Mechanical analysis of bridgesis used to introduce concepts of truss analysis andstress/strain. Material properties and application areexplained with reference to a variety of civilstructures. Technical communication skillsdescribed in this module include assembly drawing.The engineering report requires a comparison of twoengineering solutions to solve the same engineeringsituation.

Personal and public transport uses bicycles, motorvehicles and trains as examples to explainengineering concepts. The historical development ofcars is used to demonstrate the developing materiallist available to the engineer. The impact on societyof these developments is discussed. The mechanicalanalysis of mechanisms involves the effect offriction. Energy and power relationships areexplained. Methods of testing materials, andmodifying material properties are examined. Aseries of industrial manufacturing processes isdescribed. Electrical concepts, such as powerdistribution, are detailed are introduced. The use offreehand technical sketches.

Lifting devices investigates the social impact thatdevices raging from complex cranes to simple carjacks, have had on our society. The mechanicalconcepts are explained, including the hydraulicconcepts often used in lifting apparatus. Theindustrial processes used to form metals and themethods used to control physical properties areexplained. Electrical requirements for many devicesare detailed. The technical rules for sectionedorthogonal drawings are demonstrated. Theengineering report is based on a comparison of twolifting devices.

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Aeronautical engineering explores the scope of theaeronautical engineering profession. Careeropportunities are considered, as well as ethicalissues related to the profession. Technologies uniqueto this engineering field are described. Mechanicalanalysis includes aeronautical flight principles andfluid mechanics. Materials and material processesconcentrate on their application to aeronautics.The corrosion process is explained and preventativetechniques listed. Communicating technicalinformation using both freehand and computer-aideddrawing is required. The engineering report is basedon the aeronautical profession, current projects andissues.

Telecommunications engineering examines thehistory and impact on society of this field. Ethicalissues and current technologies are described.The materials section concentrates on specialisedtesting, copper and its alloys, semiconductors andfibre optics. Electronic systems, such as analogueand digital, are explained and an overview of avariety of other technologies in this field ispresented. Analysis, related to telecommunicationproducts, is used to reinforce mechanical concepts.Communicating technical information using bothfreehand and computer-aided drawing is required.The engineering report is based on thetelecommunication profession, current projects andissues.

Figure 0.1 Modules

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Module overview

This module will build upon the material covered in Householdappliances and introduce new concept relating to Braking systems.

Historical developments provides an overview of developments intechnology and society over time.

Engineering mechanics and hydraulics analyses friction in brakingsystems, stress, strain, the modulus of elasticity, work, energy, power,fluid mechanics, including Pascal’s Principle, pressure, hydraulics andArchimedes’ Principle.

Engineering materials examines; wrought iron, steel and composites,analyses the structure, properties, manufacturing methods, modificationof properties and uses as applied to braking systems, and investigatesmaterials testing of tensile strength, compressive strength and hardness.

Communications concentrates on visualisation and sketching of objectsusing pictorial drawing. Isometric projection is extensively covered andisometric 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. Detaildrawings of components that give a full shape and size description areprovided.

The last part of this module involves an engineering report. This reportrequires a detailed investigation of one braking component.

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Module components

Each module contains three components, the preliminary pages, theteaching/learning section and additional resources.

• The preliminary pages include:

– module contents

– subject overview

– module overview

– icons

– glossary

– directive terms.

Figure 0.2 Preliminary pages

Figure 0.3 Teaching/learning section

• The teaching/learning parts mayinclude:

– part contents

– introduction

– teaching/learning text and tasks

– exercises

– check list.

• The additional information mayinclude:

– module appendix

– bibliography

– module evaluation.

Additional resources

Figure 0.4 Additional materials

Support materials such as audiotapes, video cassettes and computer diskswill sometimes accompany a module.

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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 ofmaterials in engineering (P2.1)

• use mathematical, scientific and graphical methods to solveproblems of engineering practice (P3.1)

• develop written, oral and presentation skills and apply these toengineering reports (P3.2)

• apply graphics as a communication tool (P3.3)

• describe developments in technology and their impact on engineeringproducts (P4.1)

• describe the influence of technological change on engineering and itseffect on people (P4.2)

• demonstrate the ability to work both individually and in teams (P5.1)

• apply skills in analysis, synthesis and experimentation related toengineering (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.

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Indicative time

The Preliminary course is 120 hours (indicative time) and the HSCcourse is 120 hours (indicative time).

The following table shows the approximate amount of time you shouldspend 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 fivehours of work. You should aim to complete the module within 20 to 25 hours.

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Resource requirementsYou will need the following equipment for this module:

• technical drawing equipment

– rule, 0.5 mm pencil with B lead, protractor, set of compasses, drawingboard, tee-square, 60º-30º and 45º set squares, eraser, circle templateand 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 isquestionable. If you access information from sites that are reputable, theinformation can be used confidently and quoted.

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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 indicateparticular types of tasks you need to complete in this module.

The list below shows the icons and outlines the types of tasks for Stage 6Engineering studies.

ComputerThis icon indicates tasks such as researching using anelectronic database or calculating using a spreadsheet.

DangerThis icon indicates tasks which may present a danger andto proceed with care.

DiscussThis icon indicates tasks such as discussing a point ordebating an issue.

ExamineThis icon indicates tasks such as reading an article orwatching a video.

Hands onThis icon indicates tasks such as collecting data orconducting experiments.

RespondThis icon indicates the need to write a response or drawan object.

ThinkThis icon indicates tasks such, as reflecting on yourexperience or picturing yourself in a situation.

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ResearchThis icon indicates you will need to do someinvestigative work.

ReturnThis icon indicates exercises for you to return to yourteacher when you have completed the part. (OTEN OLPstudents will need to refer to their Learner's Guide forinstructions on which exercises to return).

xv

Glossary

As you work through the module you will encounter a range of terms thathave specific meanings. The first time a term occurs in the text it willappear in bold.

The list below explains the terms you will encounter in this module.

anti-lock brakingsystem

abbreviated to ABS prevents wheels from lockingduring emergency braking situations

absolutecoordinates

coordinates, used in CAD, that take allmeasurements along the x and y axes from theorigin

alloy the addition of another element or elements to ametal used to change the properties of that metal

angle of friction the angle that the resultant makes with the normalwhen the friction force and the normal reaction arereplaced by a single force

Archimedes’Principle

when a body is wholly or partially immersed in afluid, it is acted upon by an upthrust which is equalto the weight of the fluid displaced

AS1100 the drawing standards used in Australia for alltechnical drawings, such as mechanical and civilengineering, survey and architectural drawings

back pedal brake a common braking system for bicycles used in the1950s requiring a freewheel system that enabled thepedals to be pushed backwards to apply the brakingforce to the rear wheel

batching the combining or premixing of materials inpreparation for forming or manufacturingcomponents – used for composite or polymer basedcomponents

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Brinell hardnesstest

a hardness test that uses a hardened steel ortungsten carbide ball indentor pressed into thesurface of a material for 10 to15 seconds – the loadsused are 500, 1 500 and 3 000 kg

buoyancy for a body to float in a fluid, the upward thrust dueto the weight of the displaced fluid, must be equalto the weight of the floating body, this upwardthrust is buoyancy

cable brakes a braking system introduced by Daimler in 1899,using a cable anchored to the chassis, and woundaround a drum

cast iron cast iron is a ferrous metal generally containing1.8% carbon to 4.0% carbon

cementite a phase in the microstructure of steel consisting of6.67% carbon dissolved in BCC iron – it is aninterstitial compound, Fe3C, that is extremely hardand brittle

coefficient offriction

the ratio of the limiting frictional resistance to thenormal reaction

compositematerial

a composite material consists of two or morematerials combined to utilise the individualproperties of those materials to give distinctlydifferent service properties to the manufacturedcomposite

compressionmoulding

used in the manufacture of components that aremade from thermosetting polymers or fromcomposites based upon thermosetting polymers; itconsists of compressing raw material into a mouldor cavity of the desired shape, and then applyingheat and pressure

compressive test a test conducted on a prepared specimen, held in agripping device and a gradually increasing axialload applied which shortens the specimen; theapplied load is plotted against the compression, toproduce a load-compression graph

compressive stress the internal resistance of a body to a deformingforce that is tending to shorten the body

continuousprecipitation

a precipitation of a new phase that completelysurrounds the existing equiaxed grains, forming acontinuous phase throughout the structure, anexample is the continuous precipitation ofcementite around the pearlite grains in a 1.2%carbon steel

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contracting bandbrake

a braking system developed in the 1890’s inresponse to the introduction of pneumatic tyres –the main type operated on the principle of a steelband acting externally on a hub or drum

deformed grains the grains, visible in a microstructure, that havebeen squashed and deformed as a result of coldworking

dendrites the skeleton shaped grains formed during thesolidification of many metals; also a microstructuralfeature, formed only when cooling an alloy from aliquid, and consist of skeleton shaped grains whichare drawn using curved lines

detail drawing an orthogonal drawing which gives a full size andshape description of the component, it also includesthe material from which the component is to bemanufactured

drum brakes a braking system introduced in 1902 by LouisRenault that operated on the principle of two hingedshoes being forced apart onto the inside of arotating drum

enlarging scale a drawing scale that is used to enable small objectsto be drawn to a suitable scale on a piece ofdrawing paper (a scale of 2:1 means that you drawthe object twice full size, whilst a scale of 10:1means 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 thegrain

equilibriumstructure

the structure formed in a material as a result of aslow rate of cooling which enables all reactions totake place

eutectoid steel a steel having a composition of 0.8% carbon

explodedisometric

an exploded isometric drawing is a pictorialdrawing of an assembly in which the componentsare drawn separated so that details of eachcomponent can be seen

ferrite a phase in the microstructure of steel, consisting ofcarbon 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 interiordetails as visible outline

xviii

friction the resistance to motion that occurs when twosurfaces slide or tend to slide over each other

grey cast iron a cast iron produced when molten iron, containing2.8% to 4.0% carbon, is slowly or moderatelycooled in a mould; the resultant structure hasgraphite flakes in a pearlite or ferrite matrix – it isvery strong in compression, but weak in tension

half-section a standard method of drawing used only withsymmetrical components, to show the interiordetails on one side of the symmetry line as visibleoutline, and the exterior details on the other side ofthe symmetry line also as visible outline

hidden outline lines that represent the edges of an object thatcannot be seen as visible outline when viewed fromthe required direction; they are represented as thindark dashed lines, usually 0.25 mm thickness whenusing A4 size paper

Hooke’s Law extension is proportional to the applied load in atensile test

hydraulic system a brake operating system using fluids to transferpressure throughout the system by the applicationof Pascal’s Principle

inertia the amount of matter in a body; it is also describedas the tendency of a body to remain at rest or, ifmoving, remain in motion in a straight line

isometricprojection

a three dimensional pictorial drawing that usesangles 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 onthe point of moving

leading shoe the shoe in a drum brake that tends to be pulledagainst the drum surface due to the rotation of thedrum

malleable castiron

a cast iron produced when white cast iron isreheated to 800∞C and soaked for 30 to 50 hours; amoderate cooling rate produces graphite rosettes ina pearlite matrix while a slow cooling rate producesgraphite rosettes in a ferrite matrix – it has highertensile properties than gray cast iron, and is alsotougher

xix

matrix the continuous phase in a material that holds theother constituents together

mechanicalenergy

a body’s capacity to do work

mechanical work the work done when a force acts upon a body andproduces a displacement is mechanical work; it isdetermined by the product of the force and thedisplacement of the point of application of thatforce

nodules Carbon is deposited in nodular or spherical forms

orthogonaldrawing

a method of drawing utilising two dimensionalviews and dimensions to give a shape and sizedescription of components – orthogonal drawingmust follow AS1100 Drawing Standards

part-section a standard method of drawing used to show therelevant interior details of part of the component asvisible outline

Pascal’s Principle if the pressure at any point in a liquid that isenclosed and at rest, is changed, then the pressure atall points in the liquid is changes by the sameamount

pearlite a microstructural constituent consisting of twophases, ferrite and cementite, pearlite has a lamellaor plate like structure, alternating between plates offerrite and plates of cementite; it is drawn in amicrostructure to give the appearance of a fingerprint

phase a physically distinct, chemically homogeneous partof a material

pictorial drawing a three dimensional drawing used to show theshape, and sometimes size description of acomponent; isometric projection is one method ofdrawing pictorials

pneumatic tyres vulcanised rubber tubular tyres that use air to inflatethe tyre or inner tube

polar coordinates coordinates used in CAD that take radialmeasurements from the last point entered, using theangle measured in a counterclockwise directionfrom the positive x axis

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potential energy the energy a body possesses due to its position; it isdetermined by the amount of work required to lift abody through a vertical height

power power is the time rate of doing work, and isdetermined by the ratio of work done to the timetaken to do the work

pressure pressure is force per unit area

reactive force a force that acts as a response to an applied force orapplied forces; Newton, in his third law said that toevery action there is an equal and opposite reaction

reducing scale a drawing scale that is used to enable large objectsto be drawn to scale on a piece of drawing paper (ascale of 1:2 means that you draw the object half fullsize, whilst a scale of 1:10 means that you draw theobject one tenth full size

relativecoordinates

coordinates used in CAD that take actualmeasurements along the x and y directions from thelast point entered – negative values are frequentlyused

Rockwellhardness test

a hardness test that uses a variety of indentors,including an industrial diamond cone, and a 1.5 mmand 3 mm hardened steel ball, the indentor isinitially pressed into the surface of the material by aminor load of 10 kg and the major load is thenapplied

rosettes Carbon is deposited around a central core withradiating arms

service properties the performance properties of a manufacturedcomponent when being used for its designedpurpose

servo-assisted the assistance in a drum brake of the rotating drumthat tends to pull the brake shoe against the rotatingsurface of the drum

servo-assistedbrake

drum brakes that are designed so the leading shoeor shoes are pulled in towards the braking surfaceand thus increase the braking force

shape description a full definition of the shape of a component intechnical drawing, using a drawing or a number ofviews of that component

xxi

shear stress the internal resistance of a body to a deformingforce that is tending to slide one part of the bodyacross another part of the body

size description a full definition of the size of a component intechnical drawing, showing all the dimensions ofthat component

solid solution(substitutional)

an alloy system in which the atoms of one elementreplace the atoms of the other element in the latticestructure of the metal

spheroidalgraphite cast iron

abbreviated to SGCI, is a cast iron alloyed withmagnesium to produce nodules of graphite in thecooling process; a moderate cooling rate producesgraphite nodules or spheroids in a pearlite matrixwhile a slow cooling rate produces graphite nodulesor spheroids in a ferrite matrix

steel ferrous metal that contains carbon of varyingamounts generally from 0.05% to 1.4%

strain the ratio of change in length of a body with respectto its original length;: it is calculated as deformationper unit length

strain energy the energy a body possesses due to its deformation;it is determined by the amount of work done indeforming the body

stress a body’s internal resistance to an externally appliedforce that tends to deform a body; it is calculated asload per unit area

tensile stress the internal resistance of a body to a deformingforce that is tending to stretch the body

tensile test a test conducted on a prepared specimen, held in agripping device and a gradually increasing axialload applied which stretches the specimen – theapplied load is plotted against the extension, toproduce a load-extension graph

trailing shoe shoe in a drum brake that tends to be pushed awayfrom the drum surface due to the rotation of thedrum

Vickers hardnesstest

a hardness test that uses an industrial diamondindentor in the shape of an inverted square pyramidwhich is pressed into the surface of a material for15 seconds

xxii

visible outline lines that represent the edges of an object in atechnical drawing, they are represented as thickdark continuous lines, usually of 0.5 mm thicknesswhen using A4 size paper

vulcanisation a mechanism used to strengthen the mechanicalproperties of rubber by forming sulphur cross-linksbetween the polymer chains

white cast iron a cast iron produced when molten iron, containing2.8% to 4.0% carbon, is rapidly cooled in a mould,the resultant structure has dendrites of pearlite in acementite matrix; it is extremely hard and brittle

wrought iron a ferrous metal containing little or no carbon; itusually has slag inclusions which align in thedirection of working

Young’s Modulus also known as the modulus of elasticity wherestress is proportional to strain within the elasticlimit

xxiii

Directive terms

The list below explains key words you will encounter in assessment tasksand examination questions.

account account for: state reasons for, report on;give an account of: narrate a series of events ortransactions

analyse identify components and the relationship betweenthem, 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 orinformation

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, knowledgeand understanding, logic, questioning, reflectionand quality to (analysis/evaluation)

deduce draw conclusions

define state meaning and identify essential qualities

demonstrate show by example

xxiv

describe provide characteristics and features

discuss identify issues and provide points for and/or against

distinguish recognise or note/indicate as being distinct ordifferent from; to note differences between

evaluate make a judgement based on criteria; determine thevalue of

examine inquire into

explain relate cause and effect; make the relationshipsbetween 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 mainfeatures of

predict suggest what may happen based on availableinformation

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.


Braking systems

Part 1: Development of braking systems andmaterials application – 1

Part 1: Development of braking systems and materials application – 1 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

2 Braking systems

Introduction

Think of all the different types of braking systems, or methods, that youcould use to stop a bicycle – there are front and rear calliper brakes, andback 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 inbraking systems

• discuss the social implications of technological change in braking systems

• investigate the structure and properties of appropriate materials usedin braking systems.




Developing of braking systems

A brake is a device used to slow down or stop a moving object. It is alsoused to hold a stationary vehicle or object at rest. It operates as a resultof friction by converting the energy of motion, kinetic energy, intosome 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, alarge tree branch was used to slow his carriage as it descended a steepincline at Mount York.

The design of the first bicycle, the Draisine, patented in 1818 by FreiherrDrais, 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 brakeintroduced on horse-drawn wagons in the 18th century. It consisted of acurved wooden block or shoe, designed to press against the wroughtiron rim of the wheels when a force was applied through a system oflevers and linkages. This system was effective as it was used inconjunction with the horse when stopping the wagon. It was mainly usedas a parking brake.

4 Braking systems

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 ironbrake shoes rubbing against cast iron wheels.

Up until the 1870s hand-operated brakes were used on the tender andvans of steam-driven railway carriages. In 1875 Westinghouse developeda compressed air brake, which operated automatically if the trainseparated. It was made compulsory on all trains in Britain in 1889.

It was the advent of the motor vehicle that caused braking technology todevelop. Initially hand-operated lever brakes were used, operatingdirectly onto the solid tyre tread, similar to the contemporary horsedrawn carriages. They were quite effective at low speeds but were not


effective in wet weather and would damage the tyres. Karl FriedrickBenz 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 shoebrake obsolete. The contracting band brake was developed. It operatedon the principle of a band acting on a hub. The brake was less effective inwet weather. Dirt often became trapped between the lining and the hubreducing the braking effectiveness. The band brake would not operatewhen the vehicle was in reverse.

The drum brake

The next development in braking systems was the introduction of thedrum brake. A mechanically operated drum brake was first used by LouisRenault in 1902. The drum brake was unaffected by dirt and weathersince the brake shoes were enclosed in the brake drum. The mechanicaldrum brake was inefficient due to frictional losses in the joints. Severewear of the moving parts required constant maintenance, and the systemhad to be meticulously balanced to deliver equal, safe braking forces tothe brake shoes.

Slave cylinder

Dust seal

Brake shoe and lining

Wheel hubReturn spring

Figure 1.3 Hydraulic drum brake assembly

Courtesy: Newgas Automotive Taren Point© LMP

6 Braking systems

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 technologywhen a hydraulic system was introduced to operate the rear brakes.

By 1910 most motor vehicles were using two independent and separatebrake operating mechanisms on the rear wheels; the first a hand operatedlever system, the second either a pedal operated mechanical system or apedal operated hydraulic system.

Front wheel brakes

Around this time front wheel brakes also began to appear. The advantagesof having brakes on all four wheels was that the stopping distance could bereduced. When brakes are applied on a motor vehicle, much of the weightforce of the vehicle is thrown forward onto the front wheels, leaving therear brakes relatively ineffective.

Figure 1.5 Front wheel brakes required


The introduction of hydraulically assisted ‘servo-brakes’, and, in 1924the ‘vacuum servo’, led to power assisted braking systems. During the1930s hydraulic systems were gradually introduced to all brakingsystems in vehicles.

Disc brakes

The next major development in braking systems was the use of discbrakes. Although originally developed in the early 1900s, it wasregarded as a ‘new invention’ at the London Motor Show in 1951.Previously disc brakes had only been used on motorcycles, aeroplanesand trucks but not motor cars. This development revolutionised thebraking industry, so much so, that by the 1960s the use of disc brakeswas widespread in British and European cars.

Figure 1.6 Disc brake

Brake linings and pads

Along with the development of brake mechanisms, frictional material forbrake linings also developed. Early brake pad liners were made byweaving an asbestos yarn into the desired shape. Today the brake padlinings are produced from a combination of many materials: fibres, suchas glass fibre, kevlar, steel wool and carbon fibres; fillers, such as clay,calcium carbonate, barytes, fine metal and schist; binders, such as phenolformaldehyde; 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 brakingsystem, (ABS). This system prevents wheels from locking during

8 Braking systems

emergency braking situations, enabling drivers to steer the vehicle whilestopping.

ABS uses wheel speed sensors to detect rapid deceleration. Anelectronic control unit constantly monitors the wheel speed information,and when an emergency situation is detected, it activates an hydraulicunit with solenoid valves which build up and release pressure, ‘pumping’the brakes much more effectively than a driver can, to prevent the wheelsfrom locking.

The following table will be supplemented with more specific historicalperspective throughout the module, along with related developments inareas of materials and technology.

Brake system Operation Advantages Disadvantages

External shoebrake

18th –20th century

Used on horse-drawn carts

∑ hand operatedby lever

∑ uses linkages

∑ pressureapplied toshoe, forcedagainst metalrim

∑ mainly aparking brake

∑ supplementedthe horse

∑ appropriate forhorse-drawnvehicle

∑ cheap toproduce

∑ simpletechnologyavailable

∑ materialscheap andeasy to obtain

∑ needed a largeforce tooperate

∑ worked only asa supplementto the horses

∑ not effective inwet and dustyconditions

∑ safety problemdue to exposedlinkage

Contracting bandbrakes

Late 19th century

Early model cars

∑ contractingband acting ona hub

∑ hand operated

∑ worked only inforward motion

∑ appropriate forearly modelcars withrubber tyres

∑ new technologyneeded

∑ steel industrydeveloping

∑ would notoperate inreverse

∑ not effective inwet and dustyconditions

∑ not effective asparking brakes

Drum brakes

From 1902

Cars and trucks

∑ internalexpanding-shoes

∑ mechanically/hydraulicallyoperated

∑ operated in alltypes ofweather

∑ servo-assisted

∑ twoindependentsystems

∑ brake fade

∑ heat dissipationproblems


Disc brakes

1930s in trucks

From 1952 in cars

∑ calipers forcepads againstthe rotatingdisc

∑ hydraulicallyoperated withpowerassistance

∑ special designrequired tooperate thedisc brake as ahand brake

∑ more efficient

∑ improved heatdissipation

∑ lighter weight

∑ easier paddesign

∑ little or no fade

∑ special designneeded forparking brake

∑ powerassistancerequired

∑ moreexpensive

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 brakingsystem, the contracting band system. You will see why there was a needto develop a system to replace the hand operated lever brake, and look atthe different engineering designed systems that were developed.

References: to complete this case study, the Historical Development ofBraking Systems and the history from Materials for Braking Systemsfrom this module were used, along with the Repco-PBR Sound Filmstripfrom the ‘Stop – Braking Systems for Cars’.

In the mid-nineteenth century simple hand operated lever brakes were used onhorse drawn coaches, steam carriages and railway locomotives. They were quiteeffective at low speeds, were excellent as parking brakes but were not aseffective in wet weather.

From1800–1880, wooden wheels with wrought iron rims were used on horsedrawn carriages. The lever brake used a wooden shoe and leather liner. Fromthe 1830s, steam carriages, both rail and road, used cast iron wheels withwrought iron brake shoes. Both systems used an external shoe brake.

In 1841, Goodyear patented the vulcanisation of rubber which enabled the usein 1871 of solid rubber tyres on wheels. In 1888 Dunlop patented pneumatictyres, which meant the eventual end of the external shoe brake.

In 1895 the Michelin brothers had begun the move towards replacing steel-rimmed wheels with pneumatic rubber tyres and found that the old technologyof applying a brake shoe directly to the tyre was unsatisfactory.

As a direct result, contracting band brakes were developed. These brakesoperated on the principle of a band acting externally on a hub or drum. Twoearly devices attempted to apply the force of friction to the axle and to a drum

10 Braking systems

on the axle. One used wooden blocks inside an external, flexible, contractingsteel band.

In 1899, Daimler used a cable anchored to the chassis and wound around adrum. When the cable tightened while the car was moving forward, the rotationof the drum increased the tightness and grip of the cable, thus increasingbraking efficiency. This was called servo-assistance, and is still an importantfactor in the design of expanding shoe drum brakes.

Both band brakes and cable brakes proved ineffective. With band brakes, dirtoften became trapped in-between the lining and the hub, reducing the brakingeffectiveness. It was also considerably less effective in wet weather. Neithersystem 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 brakingsystem, the drum brake. You will look at the different engineeringdesigns that were developed, along with the materials used.

The introduction of the vulcanisation of rubber and the subsequentdevelopment of pneumatic tyres led to the demise of the externallyapplied shoe brake.

The use of band brakes and cable brakes also proved ineffective as carsbecame heavier and faster. They were considerably less effective in wetweather, 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 in1902. The design used two hinged shoes which were forced apart by aninterposed arm pushing each shoe against the inside of a rotating drum.The brake was unaffected by dirt and weather since the brake shoes wereenclosed inside the brake drum.

Initially, with the two shoes pivoted separately at their lower end, one shoewas self-energising and the other was not. If the drum is considered to berotating clockwise, the right hand shoe is tending to be pulled against thedrum surface, so that the braking effect is increased. This shoe is called theleading shoe . The other shoe is pushed off by the effect of the rotatingdrum and its braking effect is reduced. This shoe is called the trailingshoe. The leading shoe wears more quickly as it does more work.

Design advancement saw the introduction of a brake with the two shoeslinked together thus giving the effect of two leading shoes. This is knownas a servo-assisted brake and is the basis for the drum brakes used today.


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 toprovide equal, safe braking forces to the brake shoes.

Early drums were made from pressed, medium carbon steel, however, theywere not strong enough to maintain their shape, they were easily scoredand were poor conductors of heat.

A nickel-iron alloy was used in the 1920s; it had greater rigidity and betterfriction properties. Cast aluminium alloys with cast iron liners were alsoused but were considered too expensive. Grey cast iron was found to bethe best material for use in drums, but this was replaced in the 1970s withspheroidal graphite cast iron, SGCI, providing greater toughness.

Disc brakes have now replaced drum brakes on the front wheels in all newcars and on all four wheels in many models.

The effect of engineering innovation

In this section of work we will examine the effect of engineeringinnovations on personal transport since 1940, and compare systems inplace then, with the systems in place today. We will especially look atthe effect that improved braking system technology had on the lives ofpeople who have lived through this era.

During the 1940s not many families were not able to afford a car forpersonal transport. Transport around the towns was by bicycle. Deliveries ofbread, milk, fruit and vegetables, and of ice for the ice-chest, were madedoor to door by vendors using a horse and cart. Personal transport around thecities was also available by tram or train.

Brakes on bicycles were either a ‘back-pedal’ brake, or the conventionalcaliper brake on the rear wheel, similar to today’s bicycles. Many bicyclesdid not have brakes. Bikes were stopped using the fixed wheel drive throughthe 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 cartwas stationary, the hand brake, consisting of a lever, linkages and a brakeshoe with a leather liner, was applied. The brake shoe applied a force to themild steel rim of the wooden wheel. Like the cars of today, there were twosystems in use.

The trams and trains used a braking system that is still in use today – metalshoe brakes applied by an air-operated system.

Cars had mechanical brake systems, lever and cable operated for the handbrake, and hydraulically operated for the foot brake.

12 Braking systems

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 manypeople for recreation, sport and fitness, and for travel to and from school andwork. Horse and sulkies are only seen at country shows, at the Royal EasterShow, 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 acar. The cars travel at much greater velocity than the cars of the 1940s andrequire far greater stopping power.

Social and economical conditions have changed dramatically since the1940s. Ask your grandparents or people that you know, or that your familyknows, to describe some of these changes and the effect that the changes hadon their lives.



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 importantselection criteria when considering the material for the product orcomponent.

Steels and cast irons for brakingsystems

Steels and cast irons have been used in braking systems for many years.They are ferrous metals that contain varying amounts of carbon alongwith other alloying elements. For this course you will focus on plaincarbon steels and cast irons.

Steels contain 0.05% – 1.4% carbon.

Commercial cast irons contain from 1.8% – 4% carbon.

14 Braking systems

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 supplementto wood and cast iron. In London, 1839, wrought iron was used for smalltrusses to span the roof of Euston station. In Paris, 1889, 7417 tonne ofwrought iron was used in the construction of the 300 m high EiffelTower.

The use of wrought iron braking systems

From 1800 and 1880, wooden wheels with wrought iron rims were usedon horse-drawn carriages. The lever brake had a wooden shoe, andleather liner. From the 1830’s, steam carriages used cast iron wheels withwrought iron brake shoes. In 1841, Goodyear patented the vulcanisationof rubber which enabled the use in 1871 of solid rubber tyres on wheels.In 1888 Dunlop patented pneumatic tyres, which meant the end of theexternal shoe brake, and the end of wrought iron rims and shoes.

The summaries detail composition and structure, including theappropriate microstructures. They incorporate properties of the material,then specify manufacturing and service properties. Manufacturingtechnologies 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 directionof rolling.

ii Properties

Very soft, malleable, ductile, tough, (due to the iron matrix).

• Availability

– Readily available, (produced in puddling furnace).

• Manufacturing technology

– Hot rolled into strips, shaped and hot welded by blacksmith,heat shrunk onto rim.


• Manufacturing properties

– Easily formed, good thermal expansion, soft, malleable, ductile.

• Service Properties

– Adequate hardness and toughness.

• Modification of properties

– Can be work hardened or alloyed.

• Microstructure phases

– Iron and slag inclusions.

Slag

Ferrite

Figure 1.7 Microstructure, Wrought Iron

A microstructure

A microstructure is a magnified view of portion of the material as seen under areflecting light microscope. Magnification is usually between 150x and 500x.

When viewed using the reflecting light microscope, the grain structure isvisible. The method of determining the structure is outside the scope of thesyllabus. Interpretation of the structure and the drawing of the structure arevital for interpretation of the syllabus in terms of the properties of thatmaterial.

Steels


Steel has been used for 2000 years but it was not until the1850s that thesteel industry began to develop with the availability of cheaper steel. In1856 Henry Bessemer announced the development of his BessemerConverter, a tilting furnace that allowed the air to be blasted throughliquid pig iron to decarburise the molten metal to produce steel. In the1860s the Seimens open-hearth furnace was introduced.

16 Braking systems

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 bridgeconstruction.

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-hearthfurnace by adding a large quantity of scrap metal to the pig iron. Thisenabled the recycling of scrap and better quality control of the steelproduced. In 1880 Carnegie built the first big furnace in the UnitedStates.

Developments in mechanisation, and technology enabled the US toproduce three times the quantity of steel than England did by the end ofthe century. The world output of steel rose from 500 000 tonne in 1870 to28 000 000 tonne in 1899.

Property/structure relationships

Property/structure relationships is very important in understanding steeland its use by engineers. The microstructure of steels and how thestructure affects the properties of the various steels must be known

Equilibrium structure

In steels, the equilibrium structure is very similar to the annealedstructure and can be considered the same for this course. The structure ofsteel can be modified by heat treatment, therefore the type of structureshown must be specified.

The microstructure of steel

The microstructures show only two phases, ferrite and cementite. It isthe amount of each phase and the distribution of the phases throughoutthe microstructure that determine the properties of the steel.


A phase

A phase is a chemically distinct, homogeneous part of a material. Ferrite isone 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 verysmall amount of carbon dissolved in the iron. The amount of carbondissolved varies with the temperature, ranging from 0.008% at roomtemperature to 0.025% at 723ºC.

Cementite

Cementite is a very hard, brittle phase comprising of Body Centred CubicStructure (BBC) iron with 6.67% carbon dissolved in the iron. It is acompound and thus has a chemical formula, Fe3C.

Pearlite

Pearlite is a mixture of the two phases, ferrite and cementite and istherefore not a phase. It appears in most steel equilibriummicrostructures. Pearlite is a lamella or plate-like structure withalternating thin plates of ferrite and cementite. It is a micro-constituentas 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%)

– Equiaxed grains of ferrite, small grains of pearlite(approximately 12%). (Pearlite is a lamella structure, alternatethin plates of ferrite and cementite)

• Properties

– Very soft, malleable, ductile, tough (due to ferrite matrix).

• Availability

– Readily available, produced in Bessemer or Open-hearthfurnace.

• Manufacturing technology, 1880

– Hot rolled into strips, shaped and forge welded by blacksmith,heat shrunk onto rim.

18 Braking systems


– Easily formed, soft, malleable, ductile.

• Service properties

– Adequate hardness and toughness.


– Can be work–hardened or alloyed.

• Microstructure

– Phases, ferrite and cementite (12% of the structure is in the formof grains of pearlite, a lamella structure, thin alternating platesof ferrite and cementite).

Pearlite

Ferrite

Figure 1.8 Microstructure, Steel, 0.1% C

Mild steel – used in brake nuts and bolts, 1920s

• Composition

– Iron, 0.15% to 0.3% carbon.

• Equilibrium Structure (0.3% C)

– Equiaxed grains of ferrite, small grains of pearlite,(approximately 30%).

• Properties

– Soft, malleable, ductile, tough (due to ferrite matrix).

• Availability

– Readily available, high steel production.

• Manufacturing technology 1920

– Hot rolled into bars, hot forged to shape, thread, machineformed.


– Easily formed by hot working, good machinability.

• Service Properties

– Good shear and tensile strength, tough.



– Can be work hardened or alloyed.

• Microstructure

– Phases, ferrite and cementite (30% of the structure in the formof grains of pearlite, a lamella structure, thin alternating platesof ferrite and cementite).

Ferrite

Pearlite


Medium carbon steel – used in brake springs, 1950s

• Composition

– Iron, 0.3% to 0.6% carbon.


– Small equiaxed grains of ferrite, and large grains of pearlite,(approximately 75%).

• Properties

– Tough, heat treatable, hard, good machinability.

• Availability

– Readily available, very high steel production.


– Hot rolled into rods, hot drawn to wire shape. Methods ofproducing springs; draw into wire, form the helical shape,harden and temper the spring.


– Good formability by hot working, heat treatable to produce‘spring’ properties.


– Resilient, high elasticity, not corroded by brake fluid.


– Can be heat treated to produce different properties, or alloyed.

20 Braking systems

• Microstructure

– Phases, ferrite and cementite (75% of structure grains ofpearlite).

Pearlite

Ferrite


Eutectoid steel – used in brake cable wire, 1950s

Note eutectoid steel is steel that contains 0.8% carbon

• Composition

– Iron, 0.8 % carbon.


– Grains of pearlite, (100%).

• Properties

– Heat treatable.

• Availability


• Manufacturing technology

– 1950: hot rolled into rods, hot drawn to wire.


– Good formability by hot working.


– Good toughness and high tensile strength.



• Microstructure

– Phases, ferrite and cementite in the form of grains of pearlite.


Pearlite


High carbon steel – used in brake cable wire, 1950s

• Composition

– Iron, 0.6% to 0.9 % carbon.


– Grains of pearlite surrounded by a continuous precipitation ofcementite at the grain boundaries.

• Properties

– Heat treatable, hard with low ductility, (brittle) high tensilestrength, poor machinability.

• Availability



Hot rolled into rods, hot drawn to wire.


– Good formability by hot working, heat treatable.


– Good toughness and high tensile strength.



• Microstructure

– Phases, ferrite and cementite in the form of grains of pearlite,surrounded by a continuous precipitation of cementite at thegrain boundaries.

CementitePearlite


22 Braking systems

Tool steel – used in cutting tools, 1950s

• Composition

– Iron, 0.9% to 1.4 % carbon.

i Equilibrium Structure (1.3% C)

Grains of pearlite surrounded by a greater continuousprecipitation of cementite at the grain boundaries.

ii Properties

Heat treatable, very hard with very low ductility, (brittle) lowertensile strength, very poor machinability.

• Availability



– Hot worked into shape, heat treated to obtain desired properties.


– Good formability by hot working, heat treatable.


– Heat treatable for good toughness and tensile strength, very hardwearing surface.



• Microstructure

– Phases, ferrite and cementite in the form of grains of pearlite,surrounded by a greater continuous precipitation of cementite atthe grain boundaries.

Cementite

Pearlite



Brakes, steels and engineersLet’s now consider how a materials engineer involved in thedevelopment of braking systems for a car manufacturer would use thestudy of steels and the relationships between the structure and propertiesof those steels.

A reportA materials engineer is required to prepare a report on the selection of plaincarbon steels for use in the production of various components for a brakemanufacturing 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 ofsteel: 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.

0.1% C 0.3% C 0.6% C 0.8% C 1.2% C

perlite cementite

ferrite

Figure 1.14 Steel microstructures

Backgroundi The micro-constituents.

There are only two phases present, ferrite and cementite. Pearlite, alamella structure of alternating plates of ferrite and cementite phasesis 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 andbrittle.

24 Braking systems

iii The structure of pearlite and the phases that are present in pearlite.

Pearlite, is a lamella structure of alternating plates of ferrite andcementite phases. Pearlite is a micro-constituent of all of themicrostructures, 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 possessesthese 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 offerrite grains, with a small amount of pearlite in the form of plates offerrite and cementite. The ferrite is soft, ductile and malleable, andas the ferrite is the continuous phase, and the predominate phase, themechanical properties are those of ferrite.

c Modification of properties – the properties can be modified by coldworking or alloying. The steel cannot be hardened by heattreatment.

d Use in brake systems – backing plate for discs.

0.3% carbon steela Two mechanical properties – ductile and tough.

b Microstructure/properties – the microstructure consists mainly offerrite grains, with approximately 30% of pearlite in the form ofplates of ferrite and cementite. The ferrite is soft, ductile andmalleable, and is still the continuous phase. Due to the increasedamount of cementite the UTS and hardness is higher than that of the0.1% C steel.

c Modification of properties – the properties can be modified by coldworking or alloying. The steel cannot be hardened by heat treatment.

d Use in brake systems – nuts and bolts.


0.6% carbon steel

a Two mechanical properties – hard and tough.

b Microstructure/properties – the microstructure consists of ferritegrains, with approximately 75% of pearlite in the form of plates offerrite and cementite. The ferrite is soft, ductile and malleable, and isstill the continuous phase. Due to the increased amount of cementitethe UTS and hardness are higher than that of the 0.3% C steel.

c Modification of properties – the properties can be modified by usingheat treatment to harden and temper the steel. Alloying can also beused.

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 inthe form of plates of ferrite and cementite. Due to the increasedamount of cementite and its distribution throughout themicrostructure, the UTS and hardness are very high.


d Use in brake systems – brake cable wire.

1.2% carbon steel

a Two mechanical properties – very hard with very low ductility, thatis, it is very brittle.

b Microstructure/properties – the microstructure consists of pearlitegrains in the form of plates of ferrite and cementite, surrounded by acontinuous phase of cementite. Due to the increased amount ofcementite and its distribution throughout the microstructure as acontinuous phase, the hardness is very high, and ductility very low.


d Use in brake systems – cutting tools.

Conclusion and recommendations

The five steel all have applications for the braking systems beingdeveloped. The various applications are listed for each steel. It isrecommended that the steels continue to be used for these applications.

26 Braking systems

Cast irons

Another ferrous metal that has a wide ranging use in braking systems iscast iron. A materials engineer must be familiar with the structure of thecast irons and the relationship between the structure and the properties.


Until the introduction of the blast furnace in the middle ages there was nomeans of producing molten iron in quantity for casting. By the fifteenthcentury the casting of iron, made possible by higher furnace temperaturesand 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 thetemperature to be raised sufficiently to allow the metal to be cast intopigs. The addition of silicon to the re-melted pig iron produces cast iron.

Timeline

1700 Coke-smelting iron developed and horse-drawn railway linesused 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 wroughtiron 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 1875Westinghouse developed a compressed air brake, which operatedautomatically if the train separated. In 1889, automatic, continuouspower braking systems were made compulsory on all trains in Britain.


Property/structure relationships

The syllabus requirement of property/structure relationships is also veryimportant in understanding cast iron. The microstructure of cast irons andhow the structure affects the properties of the various cast irons must beknown.

The microstructure of cast iron

Interpretation of the structure of cast iron and the drawing of themicrostructure is vital to the interpretation of the syllabus in terms of theproperties of that material.

Except for white cast iron, the microstructures show three phases,ferrite, cementite and graphite. The amount of each phase, thedistribution of the phases throughout the microstructure and the shape ofthe graphite phase determines the properties of the cast iron.

You have been given definitions of a phase, ferrite, cementite andpearlite in the previous notes on steel. A reminder that ferrite is a verysoft, ductile phase and cementite is a very hard, brittle phase.

Graphite

Graphite has little mechanical strength. The microstructural shape of thegraphite determines many of the properties of the cast iron.

• Graphite flakes –Graphite exists as flakes in the microstructure of grey cast iron. Theflakes have sharp ends. It is the sharp ends of the graphite flakesthat are responsible for the grey cast iron having a very low tensilestrength. Under tensile loading stress concentration occurs at thesharp ends. This stress concentration causes the cast iron to fractureat a low tensile loading.

• Graphite nodules and rosettes –Graphite exists as nodules in spheroidal graphite cast iron and asrosettes in malleable cast iron. These shapes do not cause stressconcentration.

Steel matrix

The matrix surrounding the graphite can be ferrite, pearlite or acombination of each. The matrix is often referred to as a ‘steel’ matrix todescribe this occurrence.

28 Braking systems

White cast iron – used in dies and wearing plates, 1950s

• Composition

– Iron: 1.8% to 3.6% carbon; 0.5% to 2.0% silicon.

• Structure

– Dendrites of pearlite in a matrix of cementite.

• Properties

– Very hard, zero ductility, (extremely brittle), not machinable.

• Availability

– Readily available, high production.

• Production technology

– Cast to shape, rapid cooling.


– Excellent castability.


– Extremely hard, strong in compression.


– Can be heat treated to produce pearlitic or ferritic malleable castiron.

• Microstructure

– Phases, ferrite and cementite in the form of dendrites of pearlite,surrounded by a matrix of cementite.

PearliteCementite

Figure 1.15 Microstructure, White Cast Iron


Grey cast iron – used in brake master cylinders, 1970s

• Composition

– Iron; 2.4% to 3.6% carbon; 1.0% to 3.0% silicon.

• Structure

– Graphite flakes in a ‘steel’ matrix of either pearlite or ferrite, ora combination of both.

• Properties

– Relatively soft and machinable. Strong in compression butweak in tension.

• Availability

– Readily available, high production.


– Cast to shape, moderate cooling produces pearlitic grey castiron, slow cooling produces ferritic grey cast iron.


– Excellent castability, excellent machinability


– Not corroded by brake fluid, strong in compression.


– Pearlitic grey cast iron can be heat treated to produce a ferriticmatrix.

• Microstructure

– Pearlitic grey cast iron; phases, graphite flakes in a matrix offerrite and cementite in the form of pearlite.

– Ferritic grey cast iron; phases, graphite flakes in a matrix offerrite.

Graphiteflakes

Pearlitematrix

Ferritematrix

Pearlitic greycast iron

Ferretic greycast iron

Figure 1.16 Pearlitic grey cast iron, ferritic grey cast iron

30 Braking systems

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

– Readily available.


– White cast iron reheated to 800º C and soaked for 30 to 50 hours.Moderate cooling produces pearlitic malleable cast iron, slowcooling produces ferritic malleable cast iron.


– Good ductility, excellent machinability


– Tough, strong in tension and compression.


– Pearlitic malleable cast iron can be heat treated to produce aferritic matrix.

• Microstructure

– Pearlitic malleable cast iron; phases, graphite rosettes in amatrix of ferrite and cementite in the form of pearlite.

– Ferritic malleable cast iron; phases, graphite rosettes in a matrixof ferrite.

Graphiterosettes

Pearlitematrix

Ferritematrix

Pearlitic malleablecast iron

Ferretic malleablecast iron

Figure 1.17 Pearlitic malleable cast iron, ferritic malleable cast iron


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 orferrite, 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 graphiteCI, slow cooling produces ferritic spheroidal graphite cast iron.


– Good ductility, excellent machinability


– Tough, strong in tension and compression.


– Pearlitic spheroidal graphite cast iron can be heat treated toproduce a ferritic matrix.

• Microstructure

– Pearlitic spheroidal graphite cast iron; phases, graphite nodulesor spheres in a matrix of ferrite and cementite in the form ofpearlite.

– Ferritic spheroidal graphite cast iron; phases, graphite nodulesor spheres in a matrix of ferrite.

Graphitespheroidsor nodules

Pearlitematrix

Ferritematrix

Pearlitic spheroidal graphitecast iron

Ferretic spheroidal graphitecast iron

Figure 1.18 Microstructure of cast irons

32 Braking systems

Brakes, cast irons and engineers

Now consider how a materials engineer involved in the development ofbraking systems for a car manufacturer would use the study of cast ironsand the relationships between the structure and properties of those castirons.

A report

A materials engineer is required to prepare a report on the selection ofvarious cast irons for use in the production of discs, brake drums andwheel cylinders for a brake manufacturing company. Portions of thatreport are given in this example.

The three cast irons to be considered are white cast iron, grey cast ironand 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:

Pearlite

Cementite

Ferritematrix

Graphite flakes Graphite nodules

White cast iron Grey cast iron Spheroidal graphitecast iron

Figure 1.19 White, Grey and Spheroidal graphite cast iron

Background

i The micro-constituents.

There are possibly three phases present, ferrite, cementite andgraphite. Pearlite, a lamella structure of alternating plates offerrite and cementite phases may possibly be a micro-constituent of all of the microstructures. If the matrix ispearlitic it will be a part of the structure, however, if the ‘steel’matrix is ferritic, then pearlite will not be part of the structure.


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 ironpossesses 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 ofpearlite grains in the form of dendrites surrounded by a matrixof cementite. The cementite is the continuous phase, and thepredominate phase. The mechanical properties are thereforethose of cementite.

c Modification of properties – the properties can be modified byheat treatment to produce a malleable cast iron.

d Recommendation – cannot be used in brake systems as whitecast iron, but can be used to produce malleable cast iron for usein discs, drums or wheel cylinders.

Grey cast iron

a Two mechanical properties – very high compressive strength butpoor tensile strength.

b Microstructure/properties – the microstructure consists ofgraphite flakes surrounded by a ‘steel’ matrix consisting ofeither pearlite (cementite and ferrite plates), or of ferrite.Usually the ‘steel’ matrix is a combination of both.

34 Braking systems

The shape of the graphite flakes, with points at each end, causesstress concentration to occur when the material is placed undertension. This results in the material having very poor tensilestrength properties.

c Modification of properties – pearlitic grey cast iron can bemodified by heat treatment to produce ferritic grey cast iron.

d Recommendation – can be used in brake systems for drums orwheel cylinders. Previously used in discs but SGCI is nowpreferred, due to better toughness.

Spheroidal graphite cast iron

a Two mechanical properties – very high compressive strengthand excellent toughness.

b Microstructure/properties – the microstructure consists ofgraphite spheroids surrounded by a ‘steel’ matrix consisting ofeither 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 materialhaving good tensile properties, while the ‘steel’ matrix, whetherpearlitic or ferritic gives good compressive strength andexcellent toughness.

c Modification of properties – pearlitic spheroidal graphite castiron can be modified by heat treatment to produce ferriticspheroidal graphite cast iron.

d Recommendation – can be used in brake systems in drums orwheel cylinders and is excellent in discs, due to its toughness.

Conclusion and recommendations

The three cast irons all have applications for the braking systemsbeing developed. The various applications are listed for each castiron. It is recommended that the cast irons continue to be used forthese applications.

Turn to the exercise sheets and complete exercises 1.3 to 1.6.


Exercises

Exercise 1.1

a Name four devices in which brakes are used.

_______________________________________________________

_______________________________________________________

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b Describe the braking device the first horse-drawn carriage to crossthe Blue Mountains used to descend the very rough and steep trackdown Mount York.

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c Describe the braking system used to stop the first bicycle.

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d Describe the earliest known type of mechanical braking system – thelever brake – used on horse-drawn wagons.

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36 Braking systems

e Name the materials used in brakes for steam carriages from the1830s for the:

i external shoes

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ii carriage wheels

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f State two advantages of simple hand-operated lever brakes used onhorse-drawn coaches, steam carriages and railway locomotives in themid-nineteenth century.

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g Name the two developments, one in 1841, the other in 1888, whichgreatly affected the design of tyres and led to the subsequent demiseof the externally applied shoe brake.

i 1841 _______________________________________________

ii 1888 _______________________________________________

h Describe the contracting band brake, a direct result of Dunlop’spatent.

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_______________________________________________________

_______________________________________________________

_______________________________________________________

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i In 1899, a cable anchored to the chassis and wound around a drumwas used as a braking system.

i Name the person who developed the cable brake.

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ii State the main disadvantage of this brake.

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___________________________________________________

j Identify the first person to introduce mechanically operated drumbrakes first used in cars in 1902.

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k List two reasons why the introduction of front wheel brakes was animportant development.

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_______________________________________________________

l What do the letters ABS stand for in braking systems?

_______________________________________________________

Exercise 1.2

Social and economical conditions have changed dramatically since the1940s. You should talk to people who lived through these changes.

In the space below, list some of the changes and describe the effect thatthe changes had on their lives. Make reference to the development ofcars and in particular, the resulting development in braking systems.

You may submit this exercise as a computer generated word processeddocument and attach your work to this page.

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38 Braking systems

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Exercise 1.3

a List three reasons pressed medium carbon steel, used in early drumbrakes, was not satisfactory.

i ___________________________________________________

___________________________________________________

ii ___________________________________________________

___________________________________________________

iii ___________________________________________________

___________________________________________________

b Name three materials used for brake drums prior to the 1970s.

i ___________________________________________________

ii ___________________________________________________

iii ___________________________________________________

c Until the 1970s grey cast iron was the main material used for brakedrums 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 afterthe 1970s.

_______________________________________________________


Exercise 1.4

A materials engineer has to prepare a report on the selection of plaincarbon steels for use in the production of various components for a brakemanufacturing company.

Assume that you are the engineer, complete the unfinished sections of thereport. 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 0.35% C 0.8% C 1.1% C

c Outline the properties of both of the phases listed below.

Ferrite: _________________________________________________

Cementite: ______________________________________________

d Describe the structure of pearlite, and name the phases that arepresent 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 possessesthese properties

_______________________________________________________

_______________________________________________________

_______________________________________________________

iii write two methods that may be used to modify these properties

_______________________________________________________

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40 Braking systems

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


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________


0.8% carbon steel (Eutectoid steel)


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________

1.1% carbon steel


_______________________________________________________

_______________________________________________________


______________________________________________________________________________________________________________

______________________________________________________________________________________________________________


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________

42 Braking systems

Exercise 1.5

A materials engineer has to prepare a report on the selection of variouscast irons for use in the production of brake discs, brake drums and wheelcylinders for a brake manufacturing company.

Assume that you are the engineer, complete the unfinished sections of thereport. The report must be able to be interpreted by all of the directors.

a Draw the microstructures for white cast iron, grey cast iron andspheroidal graphite cast iron.

White cast iron Grey cast iron(ferritic)

Spheroidal graphitecast iron (pearlitic)

b Name the micro-constituents for each given microstructure bylabelling 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 ironpossesses these properties;

iii describe how the properties may be modified

iv write your recommendation for use of the cast iron in theproduction of brake discs, brake drums and wheel cylinders.


White cast iron


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________

iv Recommendation:

_______________________________________________________

_______________________________________________________

Grey cast iron (ferritic)


_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________

_______________________________________________________


_______________________________________________________

_______________________________________________________

iv Recommendation:

_______________________________________________________

_______________________________________________________

44 Braking systems

Spheroidal graphite cast iron (Pearlitic)

a Two mechanical properties:

_______________________________________________________

_______________________________________________________

b Microstructure/properties:

_______________________________________________________

_______________________________________________________

_______________________________________________________

c Modification of properties:

_______________________________________________________

_______________________________________________________

d Recommendation:

_______________________________________________________

_______________________________________________________

Exercise 1.6

Visit a variety of web sites then explain why ABS are used on heavyvehicles, how ABS work and their application to current model cars.

You may submit this exercise as a computer generated word processeddocument and attach your work to the back of this page. State, in thebibliography, at least two sources of information you located.

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Bibliography

1 ___________________________________________________

2 ___________________________________________________

46 Braking systems


Progress check

In this part you explored the early history of brakes, and the relationshipbetween properties and applications of materials.

Take a few moments to reflect on your learning then tick the box that bestrepresents your level of achievement.

❏✓ Agree – well done

❏✓ Disagree – revise your work

❏✓ Uncertain – contact your teacher

Ag

ree

Dis

agre

e

Un

cert

ain

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.

I have learnt to

• examine the changing applications of materials tocomponents in braking systems

• discuss the social implications of technological change inbraking systems

• investigate the structure and properties of appropriatematerials used in braking systems.



During the next part you will continue to explore the history of brakes, andthe relationship between properties and applications of materials.

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 yourresponses to the cover sheet.

If you study Stage 6 Engineering Studies through a Distance EducationSchool/Centre (DEC) you will need to return the exercise sheets and yourresponses at the completion of each part of a module.

If you study Stage 6 Engineering Studies through the OTEN OpenLearning Program (OLP) refer to the Learner’s Guide to determine whichexercises you need to return to your teacher along with the Mark RecordSlip.

Braking systems

Part 2: Development of braking systems andmaterials 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

2 Braking systems

Introduction

In this part of the module you will explore the early history of discbrakes, the developments in Britain and Europe and the reason why theUnited States was so slow in adopting and developing this ‘newtechnology’.

You will also explore the relationship between properties andapplications of materials in engineering.



• historical and societal influences


– 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.

Students learn to:

• examine the changing applications of materials to components inbraking systems

• discuss the social implications of technological change in brakingsystems

• investigate the structure and properties of appropriate materials usedin braking systems

• conduct relevant mechanical tests on materials.




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, wereregarded as a new invention. This development ‘revolutionised’ theautomobile braking industry, so that by the 1960s the use of disc brakeswas widespread in British and European cars.

Early history of disc brakes – a timeline1902 F W Lanchester, England, first

patent of automotive disc brake.

1904 W H Barrett, England, patented abrake that pressed two externaldiscs onto a revolving inner disc,(giving twice as much brakingforce).

1905 Lanchester installed disc brakes ona car. They was used as atransmission brake, not attachedto the wheels.

1911 Metz, America, used multiple discsinside the hub of each wheel.

1914 Autocarriers Ltd, A C cars, askedHenry Ford to design a brake toreplace the ineffective band brake. He designed an auxiliary system.

1919 AC used the auxiliary disc brakebehind the transmission.

1930s Girling in England and Lockheedin America developed caliper typedisc brakes for wheel installation.

1940s WW II disc brakes were used onthe wheels of the Daimler armouredcar. Goodyear developed a discbrake with a ventilated disc for useon aircrafts, while Sikorskyproduced a sandwich disc of monelmetal.

1951 Girling bought a licence to producecar disc brakes under Dunlop

patents. The prototype wasexhibited at the London MotorShow.

1952 Jaguar won at Le Mans usingGirling disc brakes.

1955 All British racing cars used discbrakes.

1956 Citroen, in France, used slidingcaliper disc brakes on its cars.

1961 Ford, England, fitted mass-produced disc brakes to all models.

America, with its very big cars, wasslow to adopt the disc brake,although a small West Coastcompany, Airheart, supplied theracing industry.

1958 All Indy 500 cars were fitted withdisc brakes.

1965 Ford America fitted disc brakes toits Galaxy.

1966 Bendix supplied disc brakes toChrysler and Buick.

During the 1960s, front wheel discbrakes were gradually used in somemass produced cars, and becamestandard during the 1970s. Duringthe 1980s four wheel disc brakingsystems became more common,while today most cars use discbrakes on all four wheels.

4 Braking systems

The effects on society

Society is often affected by technology developments. Consider somesocial implications and effects of various developments of the discbrakes. Two samples with written solutions are given below. Thesolutions have been researched and contain more detailed information ondisc brakes. Again, study methods can be used to highlight key phrasesor techniques used.

Sample 1

Briefly examine the contribution that the car racing industry had on thedevelopments of the disc brake, in England and in America.

Prior to 1951 the use of disc brakes was restricted to the aviationindustry and military vehicles. In 1951, Girling bought a licence toproduce car disc brakes under Dunlop patents, and exhibited theprototype 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 C-type roadster racing team and easily defeated the Ferrari V12 coupeswhich had a much faster top speed. This victory encouraged the use ofdisc brakes and in 1955 all British racing cars were fitted with discbrakes. Even Ferrari fitted them in 1958.

American cars were much bigger and heavier and travelled at a highervelocity. For these reasons the adoption of the disc brakes was notused by the big companies, however, a small west coast company,Airheart supplied the racing industry. In 1958 all Indy 500 cars werefitted with disc brakes. In 1965, Ford America commenced using themon their Galaxy model, and the next year Chrysler and Buick usedBendix brakes.

The developments in both England and America were very muchinfluence by the results of the racing cars using disc brakes, andresulted in the adoption of this system to the mass produced familycars during the 1960s.


Sample 2

One reason that disc brake technology was developed was the need toslow and stop cars that were becoming much faster and more powerfuleach year. Another reason was the changing attitude of society to carsafety.

Discuss the social implications that led to the development of improvedbraking technology and improved safety of cars from the 1960s.

Car safety became a big issue during the 1960s. Each year the numberof fatalities and injuries increased as a result of more accidents.

Poor braking, along with bad roads, the use of plain window glass inwindscreens, and poor safety awareness all contributed to the increaseinjury and death by car accidents.

A gradual increase in public awareness lead to many changes thatactually slowed the rate of increase in accidents and in the past fewyears dramatically reduced the road toll.

Design engineers were involved in improved road design, improvedbrake technology, and in the introduction of laminated glasswindscreens.

The government introduced restrictions on the permissible alcoholblood level that a driver had to obey.

The improvements in brake system technology included the use ofdisc brakes, initially on the front wheels and later the rear wheels, aswell as the use of ABS, anti-lock braking systems.

All these improvements came about because of increased publicawareness for the need to improve car safety and to decrease the everrising fatality rate from car accidents.

Turn to the exercise sheets and complete exercises 2.1.

6 Braking systems

Materials in braking systems

In this section of work you will learn about composite materials andhow 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 brakingsystems

A composite material consists of two or more materials combinedto form the composite. The composite utilises the properties of theindividual materials to give distinctly different service properties tothe manufactured composite product.

When selecting materials to form a composite, the following must beconsidered:

• 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, highlyeffective, maintains its effectiveness under extreme conditions, requireslittle or no maintenance, can last the life of the vehicle and isenvironmental friendly.


Specific service properties for brake pads

• Uniform friction properties. The coefficient of friction should notvary appreciably with variation of load, temperature, and velocity.The brakes must stop the vehicle.

• High thermal stability; resistance to thermal shock and fatigue. Thematerial should not break down due to temperature variations.

• Low noise generation. Noise and brake-squealing must be kept to aminimum.

• Adequate compressive and shear strength. The composite should notshear 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 centurybecause of its friction properties, strength, low cost and good thermalqualities. During the past 25 years, requirements for braking materialshave dramatically changed. There is a tendency towards smaller frictionelements operating at higher temperatures and pressure. Asbestos doesnot fulfil the requirements needed for heat resistance at much highertemperatures, higher coefficient of friction, nor the extended durabilityrequired in today’s brakes.

Health and safety risk

As you may know, asbestos also represents a serious occupational healthand safety risk. It has been proved that exposure to asbestos can lead toasbestosis, a form of lung cancer. For this reason many countries havebanned the use of asbestos.

Fibres – a replacement for asbestos

Research to find a replacement for asbestos fibres has includedinvestigation of steel wool, glass fibre, wool fibre, aramid (aromaticpolyamide polymer) fibre, kevlar and carbon fibre.

Required properties of fibres for brake pads includes:

• good friction properties

• good processing performance

8 Braking systems

• 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, andthus requires care when mixing into the friction compound to preventbreakage. It softens at high temperatures, acting as a lubricant,producing a sudden loss of friction. It has excellent binding propertieswith the matrix, is able to be produced in long and short fibres givinggood dimensional stability, rigidity and strength to the final product.

Carbon fibre

Carbon fibre has been around for more than a century, with ThomasEdison utilised carbon filament in his newly invented electrical lightbulb. Up until the late seventies it was used in composits for brakes inthe aerospace industry, racing cars and high performance militaryaircraft. In the early eighties it was used in the brakes of the Concorde.

In 1975 research was accelerated to find a relatively cheap multipurposecarbon fibre similar to the fibre used in the aerospace and sporting goodsindustries.

Properties of carbon fibre for brake pad materials

Carbon fibre properties as related to braking materials include:

• 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


• 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 otheringredients, be tough and strong in shear and compression, and havegood thermal shock resistance.

The most common matrix is phenol-formaldehyde or a modifiedphenolic.

Fillers for brake pads

Fillers are generally low cost materials, such as clay or calciumcarbonate, that are added to extend the material in the composite, occupyspace and reduce costs. They usually influence wear properties of thecomposite.

Friction modifiers

Friction modifiers are many and varied. Some are listed below.

• Metal chips, used to modify friction properties, and to controlcleanliness 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 tosuppress noise and provide protection against disc wear.

• Friction ‘dust’ or powder, such as cross linked phenolic andmodified phenolic based polymers, are used to provide thermalstability, reduce the wear factor and to provide even frictionproperties under extreme conditions.

• Barium sulphate, used to improve the wear resistance of the frictionmaterial at low temperatures, accelerate the curing of the binder andimprove compressive and shear strength.

10 Braking systems

Manufacture of brake pads

Brake pads are manufactured by compression moulding. Compressionmoulding consists of compressing raw material into a cavity or mould ofthe desired shape and applying heat and pressure.

Batch formulation

Batching is the combining or premixing of the materials in preparationfor forming.

There are many combinations of these materials that may be used to formbraking materials. Batch formulas contain up to ten or even fifteenmaterials, combined together. Generally the matrix or binder is 10–25%,the fibre, 15–30%, non-organic fillers, 10–20%, metals 1–8%, frictiondust 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 minimisefracture in the mixing process.

Pre-forming

Pre-forming is used to economise in the use of materials. Apredetermined amount of the batched material is cold pressed into a pre-form mould at a pressure of 7–15 MPa. This pre-formed shape is thenplaced 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 thepolymer. The pre-formed batched material is then placed into the hotmould and put under pressure of 14–50 MPa. The material softens and iscompressed into the shape of the mould cavity.

Post-cured

The finished product is then post-cured in an air-forced oven, at150–200º C for several hours. The cure time depends upon the thicknessof the product, the polymer used, its state of polymerisation whencharged into the mould, the mould temperature, and the mouldingpressure used


Finishing

The brake pads then undergo thickness grinding.

Brake liners require edge grinding to width, inside and outside grindingto thickness, squaring and cutting to length, and drilling whereappropriate.


12 Braking systems


Testing of materials

In this section of work you will learn about mechanical testing ofmaterials, including hardness testing, and tensile and compressivetesting. In previous modules you studied the properties of materials andthe 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 indentationas a basis for hardness testing.

Hardness tests are non-destructive tests. They are used in industry toverify that the required properties have been produced following the heattreatment of components during production. They are used for control orproduction line testing, as well as for research, and comparison testing

You will learn about three standardised hardness tests, Brinell, Vickersand Rockwell. All three use machines which apply a specified load to anindentor. The indentation is then measured to give the tested material ahardness number.

1 Brinell hardness test

The Brinell hardness test was Introduced in 1900 by Swedishmetallurgist J. A. Brinell.

Method

A hardened steel or tungsten carbide ball indentor is pressed into thesurface of a material for 10–15 seconds. The loads used are 500, 1 500and 3 000 kg.

14 Braking systems

Measured

The diameter of the indentation is measured using a low-poweredgraduated microscope, and the Hardness Brinell number determined froma prepared table.

Recorded

The hardness number is given, followed by the letters HB, then bynumbers indicating the diameter of the ball and the load used for the test.For example 250 HB 10/3000, indicates that a Brinell Hardness testnumber of 250 was obtained using a 10 mm diameter ball and a load of3 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, orfor plated or hardened surfaces.

2 Vickers hardness test

Vickers hardness test was introduced in 1922 in England by R. Smithand G. Sunderland.

Method

An industrial diamond indentor in the shape of an inverted squarepyramid is pressed into the surface of a material for 15 seconds.

Measured

The surface area of the indentation is determined, and the HardnessVickers number read from prepared tables. The numbers have beencalculated by dividing the load by the surface area of the indentation.

Recorded

The hardness number is given, followed by the letters HV, then by anumber indicating the load used for the test. For example, 650 HV 30,indicates a Vickers Hardness test number of 650 was obtained using aload of 30 kg.


Application

Used for a full range of materials with a wide range of hardness. It isused for sheet metal, for very hard material, and for case hardenedsurfaces.

3 Rockwell hardness test

The Rockwell hardness test was introduced in 1922 by Americanmetallurgist 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 initiallypressed into the surface of the material by a minor load of 10 kg and thedial indicator is set to zero. The major load is then applied. When thedial indicator is steady, the major load is removed.

Nine scales of hardness are available from A to K, having variousindentors used with different major loads for various materials. The mostcommon Rockwell tests are B and C.

16 Braking systems

The Rockwell B test uses a 1.5 mm hardened steel ball, with a major loadof 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 loadof 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 majorloadings is used as the measure of hardness. The hardness number is readdirectly from the dial using the appropriate scale for that test. This directreading enables the Rockwell testing to be done quickly and accuratelyduring the actual production of the component. The test can also beautomated.

Recorded

A number indicating the related hardness of the material for that scale,followed by HR and the appropriate letter for the Rockwell Hardness testused, is given. For example, 60 HRC indicates a test hardness number of60 was obtained using the appropriate load and indentor for the RockwellC 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 quiteoften 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 plasticdeformation.

In all of these processes the material is subjected to tensile, compressiveand shear forces. The relationship between a force and the deformation itproduces is required knowledge for the engineer in manufacturing. Twoof the most important mechanical property tests are the tensile test andcompressive test.

Analysis of the curve produced during a load-deformation test canprovide information essential to the mechanical engineer. Yield stress,ultimate tensile stress (UTS), modulus of elasticity, percentage


elongation, and percentage reduction in area can be determined alongwith interpretation of properties such as ductility and toughness.

Tensile tests

Tensile tests are conducted using a tensometer or a universal testingmachine. A prepared specimen, usually of standardised size, is held in agripping device and a gradually increasing axial load applied to thespecimen. The load is usually applied until fracture occurs, alternatively,it may only be applied within the elastic limit, or up to the yield point insome 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 specialadaptor, a universal testing machine, or a specialised compression testingmachine. The test is similar to the above except that a compressive load isapplied.

In ductile materials barreling usually occurs in the specimen. This is dueto the frictional forces existing between the ends of the specimen and thesurface 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 expandedwhile the ends have not.

18 Braking systems

Load-deformation verses stress/strain curves

To be able to compare different materials and similar or the samematerials, a standardised specimen is used, or better still the load-deformation diagram is converted to a stress-strain diagram. This allowscomparisons to be validly made.

You need to conduct relevant mechanical tests on materials. This is verydifficult for you to be able to do. However, there are ways that you maybe able to experience tensile, compression and hardness testing.

Your School of Distance Education or the associated TAFE collage maybe able to organise a practical workshop day where the testing machinesare available. Machines are available at many secondary schools, a visitmay be able to be arranged. University Engineering faculties havetesting equipment, TAFE at Ultimo also has a testing laboratory. Manyindustries have testing laboratories as part of their manufacturing. AIS inWollongong has a very comprehensive testing lab, while in theSutherland Shire, Dowell Industries and Davies Kent do mechanicaltesting 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 theircomparative properties.



Investigation of a braking system:materials analysis

In this section you will consider how a materials engineer would analysethe materials used in components for a braking system. Also, for each ofthe components you will look at the analysis of an alternative materialthat could be used for that component. A recommendation will then begiven based upon the analysis.

Analysis of structure and properties

The materials engineer will identify the main service properties of the selectedcomponent, and then proceed to analyse the structure of the materials. Bothmicrostructure and lattice or molecular structure will be analysed whereappropriate, and the properties of the materials determined. A recommendationas 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 2.3will be used for the analysis. The components to be considered by thematerials engineer are: the rivets used to secure the liner to the brake shoe, thepiston, spring, slave cylinder and the dust seal.

Piston sealDust seal

Lining

Spring

Slave cylinder

Brake shoe

Figure 2.3 Drum brake slave cylinder assembly

Courtesy: Trinity College Auburn© LMP

20 Braking systems

An exploded isometric drawing

An exploded isometric drawing is a pictorial drawing of the separatedcomponents in their relative position to each other. To assemble thecomponents, all of the components would be moved along their line ofcentres, into the slave cylinder. The exploded drawing allows you to seethe 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 annealedstate; deformed grains when cold worked.

Properties: high shear strength in the cold worked condition, workhardened, adequate toughness, able to withstand impact forces, doesnot corrode in the ‘braking environment’.

Suitability: very suitable for the rivets.

Alternative material: 70–30 brass, an alloy of copper and 30% zinc. Itis a non-ferrous metal.

Lattice structure: 70–30 brass has a FCC structure. The zinc atomstake the place of some of the copper atoms in the original copperlattice structure. A structure such as this is called a substitutionalsolid solution.

Microstructure: equiaxed grains of the solid solution when in theannealed state, deformed grains when cold worked.

Properties: high shear strength, adequate toughness, able to withstandimpact forces. Could corrode in the ‘braking environment’.

Suitability: suitable except in adverse corrosive conditions.

Recommendation: the recommendation is to retain the copper rivets.


2 Component: piston

Service properties: adequate hardness, good compressive strength,able to absorb impact loads, tough corrosion resistant in ‘brakingenvironment’.

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 pearlitewhen in the annealed state, deformed grains when cold worked.

Properties: adequate hardness, good compressive strength, adequatetoughness, 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 annealedstate, deformed grains when cold worked.

Properties: inadequate hardness, the aluminium is too soft.Inadequate compressive strength, excellent toughness, able towithstand impact forces and does not corrode in the ‘brakingenvironment’.

Suitability: not suitable due to its mechanical properties.

Recommendation: the recommendation is to retain the mild steelpistons.

3 Component: spring

Service properties: good ‘spring’ properties, able to absorb impactloads, 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,able to absorb impact loads, tough corrosion resistant in ‘brakingenvironment’.

22 Braking systems

Suitability: suitable for the piston. Must be in the cold workedcondition

Alternative material: high carbon steel, 0.7% carbon. A ferrous metal.


Microstructure: almost all pearlite, with a small amount of ferritegrains.

Properties: good ‘spring’ properties in the cold worked condition,able to absorb impact loads, tough corrosion resistant in ‘brakingenvironment’.

Suitability: suitable for the spring, however it must be shaped by hotworking, then be heat treated to obtain the required properties.

Recommendation: The recommendation is to retain the mediumcarbon steel springs due to the greater cost in forming the high carbonsteel springs.

4 Component: cylinder

Service properties: adequate tensile strength (the ability to withstandthe 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 acrystalline 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 equiaxedgrains of ferrite with possibly some areas of pearlite.

Properties: adequate tensile strength, corrosion resistant in ‘brakingenvironment’.

Suitability: suitable for the slave cylinder.

Alternative material: aluminium alloy, containing silicon andmagnesium, a non-ferrous alloy.

Lattice structure: aluminium alloy has a FCC structure.

Microstructure: equiaxed grains of aluminium (with areas of lamellamagnesium silicide Mg2Si, which is outside the scope of the course);deformed grains when cold worked.

Properties: good tensile strength, does not corrode in the ‘brakingenvironment’.


Suitability: very suitable due to its properties and to its low weight.

Recommendation: the recommendation is to retain the spheroidalgraphite cast iron for the slave cylinder, but the comparative cost ofchanging to aluminium alloy should be further investigated.

5 Component: dust seal

Service properties: flexibility. Elastomer, able to be compressedrepeatedly 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 anetwork 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. Corrosionresistant 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.

24 Braking systems


Exercises

Exercise 2.1

a Outline the contribution that the car racing industry had on thedevelopments of the disc brake, in England and in America.

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

b One reason that disc brake technology was developed was the needto slow and stop cars which were becoming much faster and morepowerful each year. Discuss the social implications of thisstatement.

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

26 Braking systems

Exercise 2.2

a Define the term ‘composite material’.

_______________________________________________________

_______________________________________________________

_______________________________________________________

b When studying a composite material it is important to consider theindividual materials that are combined to form that composite.Explain why this is important.

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________

c Service properties are the first consideration for the materialsengineer when designing a composite for brake pads. List four ofthese service properties.

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________

d Asbestos is now considered unsuitable to be used as a fibre in brakepad composites. List three reasons why asbestos is no longersuitable for use in brake pads.

_______________________________________________________

_______________________________________________________

_______________________________________________________

e Carbon fibre is used in braking material in aircraft and high poweredmotor vehicles. List three properties of carbon fibre that makes itsuitable for use in brake composites.

_______________________________________________________

_______________________________________________________

_______________________________________________________

f Metal chips, such as zinc, are used in brake pad composites. Statethe specific reasons for using zinc.

_______________________________________________________

_______________________________________________________

_______________________________________________________


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 shapebrake pads.

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

Exercise 2.3

a Define the term hardness.

______________________________________________________

______________________________________________________

______________________________________________________

b Hardness testing is described as a non-destructive test. Explain themeaning of non-destructive test.

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

c Name two areas where hardness tests are used in industry.

i ___________________________________________________

ii ___________________________________________________

d Name and briefly describe three standardised hardness tests used inindustry.

i ___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

28 Braking systems

ii ___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

iii ___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

e Briefly describe the methods used for a tensile test and for acompressive test.

i tensile ______________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

ii compressive _________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________


Exercise 2.4

In the drum brake slave cylinder sample given in the previous notes, thestructure and properties of ten materials were analysed to determine thesuitability 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, non-ferrous alloys, ferrous metals or polymers.

copper 70-30 brass

0.2% carbon steel aluminium

0.4% carbon steel 0.7% carbon steel

spheroidal graphitecast iron

aluminiumalloy

neoprene PVC

b Draw and label the microstructures of 0.2% C steel and 0.7% C steel

0.2% C steel 0.7% C steel

c The two steels, 0.2% C steel and 0.7% C steel have differenthardness properties and different ductility. In terms of theirmicrostructures explain the reasons for the difference in theseproperties of the two steels.

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

30 Braking systems

d Neoprene and PVC have different molecular structures and differentbonding. For each material describe the molecular structure andname the bonding.

i Neoprene _____________________________________

_____________________________________

_____________________________________

ii PVC _____________________________________

_____________________________________

_____________________________________


Progress check

During this part you explored the early history of disc brakes and therelationship between properties, uses and applications of materials inengineering.

Take a few moments to reflect on your learning then tick the box whichbest represents your level of achievement.




Ag

ree

Dis

agre

e

Un

cert

ain

I have learnt about

• Historical and societal influences


– the effect of engineering innovations on people’slives

– environmental implications from the use of materialsin braking systems

• Engineering materials

– materials for braking systems

– testing of materials.

I have learnt to

• examine the changing applications of materials tocomponents in braking systems

• discuss the social implications of technological changein braking systems

• investigate the structure and properties of appropriatematerials used in braking systems

• conduct relevant mechanical tests on materials.

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.

32 Braking systems

During the next part you will employ mathematical and graphicalmethods used to solve problems of engineering practice and also furtherdevelop your skills in isometric and orthogonal drawing.



Exercises 2.1 to 2.4 Name: _____________________________

Check!


❐ Exercise 2.1

❐ Exercise 2.2

❐ Exercise 2.3

❐ Exercise 2.4

Locate and complete any outstanding exercises then attach yourresponses to this sheet.

If you study Stage 6 Engineering Studies through a Distance EducationCentre/School (DEC) you will need to return the exercise sheet and yourresponses at the completion of each part of a module.


Braking systems

Part 3: Engineering mechanics, hydraulics andcommunication – 1

Part 3: Engineering mechanics, hydraulics and communication – 1 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


2 Braking systems

Introduction

In this part you will explore mathematical and graphical methods used tosolve problems of engineering practice and also learn more aboutisometric 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 andbraking components applying appropriate Australian Standards(AS 1100).


Refer to <http//ww.boardofstudies.nsw.edu.au> for original and current documents.


Engineering mechanics and hydraulics

In this section we will examine the nature of friction, as well as lookingat stress and strain.

Friction

When you walk on a rough surface, such as a footpath, then try to walkon an ice rink there are two very different results.

To be able to walk on a rough surface such as the footpath, friction isused to allow you to make progress. To try to walk on an ice rink, wherethere 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 surfacesslide, 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 frictionforce increased? Did you also notice that your hand got hotter as theforce increased? Were you able to apply a big enough downward forcethat 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 asyour desk top, a book cover, your computer keyboard, a TD set squareand an eraser.

4 Braking systems

Did you notice that the same results occurred as before? Did you alsonotice that the frictional resistance was different for the differentmaterials used in the activity? Were you again able to apply a brakingforce? 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 twotouching surfaces. The friction force varies directly with the appliedforce that pushes the bearing surfaces together, and therefore with thenormal reaction.

Frictional force(reactive)

Opposing motionNormal reaction

Applied forcePushing surfacestogether

F

P Applied forcetending to movethe body

Figure 3.1 Analysis of forces

Friction forces in braking systems

Friction forces occur in braking systems due to the reaction between thespecially compounded materials of the brake linings/pads, and the castiron 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 rotationof the wheels through developing a ‘controlled’ friction that convertskinetic energy of the vehicle into heat (energy). The heat energy is thendissipated into the surrounding air through the brake drums or brakediscs.

KE = 12

mv2

From this formula, when brakes are applied a combination of the velocityof the vehicle and mass of the vehicle determine the amount of kineticenergy that must be converted to heat energy. However, it is actually the


rolling friction between the tyres and the road surface that eventuallybrings the vehicle to a stop.

Coefficient of friction

Coefficient of friction µ (Mu) is the ratio of the limiting frictionalresistance to the normal reaction. The coefficient of friction is a constantfor any two materials in contact.

m = FN

R

Coefficient of friction in braking materials

If the coefficient of friction between the materials of the brakeliners/pads and the materials of the brake drums/discs is excessive, theabrasion would quickly wear down both the liners/pads and thedrums/discs. A high coefficient would also cause the brakes to lock.Brake materials are therefore manufactured with a range of coefficientsfrom low friction, 0.25–0.3 through to high friction, 0.4–0.45.

6 Braking systems

Analysis of friction problems

Friction force

Friction force is a reaction or a reactive force that opposes motion orpending motion that occurs due to an applied force.

F = NR m

Force analysis

Analysis should commence with the determination of the direction of thefriction force. The friction force should then be drawn on the forceanalysis diagram. All other forces acting on the body should then bedrawn on the analysis diagram. These forces should then be drawn as afreebody diagram that shows only the previously analysed forces.

Note the friction force always opposes motion or pending motion.

P = Applied forcem = mass of bodyg = 9.8 m/s2

N = Normal reactionFR = Frictional resistance

mg

P

NFR

Force analysis Free body diagram

FR

N

mg

P

Figure 3.2 Force analysis, free body diagram

Solve, using the linear equilibrium equations;

SS

H =

V = 0

0

Two important formulae which must be known.

Limiting friction

Limiting friction is the frictional resistance acting when a body is on thepoint of moving.


Coefficient of friction µ (Mu) and friction formula

Coefficient of friction µ (Mu) is the ratio of the limiting frictionalresistance to the normal reaction.

m = FN

R

Alternative method of writing formula

F = NR m

Note: this formula only applies when limiting friction is involved.

Worked example 1

In the following five examples different forces are being applied to abody 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 theanalysis of the problem and will refer to the diagram as an analysisdiagram.

To solve the problem:

1 Analysis diagram –

The analysis diagram is used to determine and show all of the forcesacting on the body. As the friction force opposes motion or pendingmotion the first step must be to determine and show the friction forceon the analysis diagram. Having shown the friction force, show allof the other forces acting on the body. Remember that if the body isto 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 thesystem. It is easier for analysis if the forces are drawn with thearrows pointing away from the point of intersection as shown in thesolutions. This method uses the principle of transmissibility, andthus does not alter the question.

3 Equation for FR –

Write an equilibrium equation for FR using the two linear equilibriumequations, S H = 0, S V = 0 and F =µN.

Note: the friction force always opposes motion or pending motion.

8 Braking systems

mg

P

NFR

Body at point of moving � force horizontal

mg (given)

NFR P FR = mN

= mmg

mg

P

NFR

Body at rest � force horizontal

N (given)

mg (given)

FR P FR = P

(ÂH = 0)

mg

N

Body at rest

FR = 0

No horizontal forceis acting

Analysis diagram Free body diagram Equation for FR

mg

P

NFR

Body at point of moving � force downward, 30∞

P cos 30∞FR

P sin 30∞

mgN

FR = mN= m(mg + P sin 30∞)

30∞

mgP cos 30∞ FR

P sin 30∞

N

30∞

mg

NFR

Body at point of moving � force upward, 30∞

PFR = mN

= m(mg - P sin 30∞)

N

mg

Figure 3.3 Analysis diagrams


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 theforces acting on the body. The first step is to determine and showthe friction force on the analysis diagram and then show all of theother 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 thesystem, mg, P and the resultant force R.

The solution is now found using a force diagram. You solve the problemusing a graphical method or using trigonometry.

Once you learn this method of analysis you will find it much easier andquicker to use than the previous method when solving limiting frictionproblems.

Sample solution

Pending motion

Force analysis Free body diagram

mg

Pmg

R

P

mg

P

N

FR

Rf

R

Figure 3.4 Angle of friction

The solution using a force triangle as shown is a much quicker method.

10 Braking systems

Worked example 2

Repeat the force analysis for parts iii, iv and v, using the angle of frictionmethod.

Body at point of moving � force horizontal

Body at point of moving � force downward, 30∞

Body at point of moving � force upward, 30∞

mg

R f

FR

N

P

mgP

N

FR

30∞

mg

P

N

FRmg (given)

PR f

FR

N

mg

Rf

FR

NP

mgP

N

FR

30∞

R

f

P

mg

mg

R

P

f

mg

R

P

f

Figure 3.5 Inclined Plane 1 Inclined Plane 2


Friction on an inclined plane

Basic introduction. In Landscape products you were introduced to theanalysis of forces on an inclined plane. You should revise this workbefore commencing this basic introduction to friction on an inclinedplane.

Limiting friction on an inclined plane

When a body is at rest on an inclined plane, and is on the point ofmoving, (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 inmagnitude, but opposite in sense to the weight component down theplane.

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 tothe 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, orsheet of garnet paper, to one side and another material to the other.

Method 1

Determine the horizontal force required to move the body across asurface. You may be able to secure a spring balance to the body.

12 Braking systems

Alternately you could use a fixed pulley and attached masses. Now,using the friction formula F = µN determine the coefficient of friction.

Sand could be added to thebucket until movement occurs

Spring balance

or

Figure 3.7 Coefficient of friction on a horizontal plane

Method 2

Place the body on an inclined plane. Determine the angle of repose ofthe inclined plane, that is, the angle of inclination of the plane when thebody just commences to move. Now, determine the coefficient offriction 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



Stress and strain

When a force is applied to an object, if it doesn’t move, then there mustbe some force opposing it.

Opposing ForceOpposing Force

What is happening to the object?

It’s under stress.

An engineer must be aware of these stresses as they could cause thestructure 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 extensionwas proportional to the applied load – Hooke’s Law. His contribution tothe study of the strength of materials and the resulting effect on thedesign of components was enormous.

Young’s Modulus

Thomas Young, in 1807, showed that Hooke’s Law was only effectiveup to a certain limit, was a characteristic of the material and appliedequally to the compression of a body as well as to tension. His name isgiven to the Modulus of Elasticity, or Young’s Modulus.

When a body has a load or force applied to it, some deformation, eitherextension or contraction will occur. Depending upon the size of the loadand the mechanical strength of the body, the deformation will be eitherelastic or plastic.

Elastic deformation

Elastic deformation means that the body will return to its original shapeand size when the deforming force is removed.

14 Braking systems

Plastic deformation

Plastic deformation means that the body will not return to its originalshape and size when the deforming force is removed. The body is said tohave taken a permanent set.

In both of the above cases, when the deforming force is being applied, aninternal resistance is tending to prevent the body from deforming. Thisinternal resistance to deformation is called stress.

Stress

Stress is a body’s internal resistance to an externally applied force thattends 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 =

LoadArea

s =

PA

Symbol: s (sigma)

Units

Pascals, (Pa). Other permitted units include:

kilo Pascal kPa, 103 Pa

Mega Pascal MPa, 106 Pa

Giga Pascal GPa, 109 Pa

Note: you must always convert to basic units when doing calculations.

Basic units: Load = newtons (N)

Area = square metres (m2)

Stress = Pascals (Pa)


Tensile stress

Tensile stress occurs when the externally applied load tends to stretchthe body.

120 N120 N120 N

Figure 3.9 Tensile stress

Compressive stress

Compressive stress occurs when the externally applied load tends toshorten the body.

120 N120 N120 N

Figure 3.10 Compressive loads

Shear stress

Shear stress occurs when the externally applied load tends to slide onepart 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 practiceto be able to analyse and solve problems. (See examples 1 and 2 whichfollow).

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 thecross sectional area. In shear loads, the area being stressed is the sheararea, that is the area that would have to break if the component were tofail under the applied load.

16 Braking systems

Method

All problems should be set out and presented as follows.

i Summarise the given data, using the related symbols from theformula, 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 sketchof this area.

iv Select and write the appropriate formula.

v Substitute the data into the formula, ensuring that you use the basicunits.

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 atensile force of 5 kN. Determine the tensile stress in the rod.

d = 12 mm

= 12 x 10-3m

P = 5 kN

= 5 x 103 N

s = ?

CSA

5 kN

5 kN

Figure 3.12 Analysis diagram

Area being stressed is the cross sectional area (CSA).

Area = pr2 or

pd2

4

A =pd2

4

=

p ¥ ¥ -( )12 104

3 2

=

p ¥ ¥ -144 104

6

= 113 10 6 3¥ - m

Now s =

PA


P =5 10

113 10

3

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 abrake liner of thickness 5 mm.

i If, during the punching operation, the compressive stress in thepunch is 120 MPa, determine the force used to punch out the hole.

d = 8 mm

= 8 ¥ 10-3 m

= 120 Mpa

= 120 ¥ 106 Pa

P = ?

compressive area (CSA)


Area being stressed is the cross sectional area.

A =pd2

4

=p ¥ ¥ -( )8 10

4

3 2

=p ¥ ¥ -64 10

4

6

= 50 27 10 6 2. ¥ - m

Now s =PA

P = s ¥ A

= 120 ¥ 106 ¥ 50.27 ¥ 10-6

= 6032.4 N

= 6.032 kN

ii Using the previous data as well as the calculated force in the punchfrom part i, determine the shear stress in the lining material.

Area being sheared is the curved surface area of the cylindricalshape beig punched out of the liner.

18 Braking systems

Circumference = 2p pr or d

Shear area = Circumference ¥ thickness

= pd k¥

= p ¥ ¥ ¥ ¥- -8 10 5 103 3

= 125.67 ¥ 10-6m2

shear area


Now s =PA

P =6 032 10

125 67 10

3

6

.

.

¥¥ -

= 48.2 ¥ 106 Pa

= 48 MPa

You can see from these two worked examples that it is very important toanalyse each question. It is especially important to determine the areabeing stressed, so that errors do not occur.

Strain

Strain is the ratio of the change in length of a body with respect to itsoriginal length. It is calculated as deformation per unit length.

Formula

Strain =Change in lengthOriginallength

e = e/L

Symbol: e (eta)

Units

Strain is a ratio. It is sometimes expressed as a percentage, an example ispercentage elongation, that is, strain expressed as a percentage.

Now we will consider the contribution of Robert Hooke and ThomasYoung to the scientific design of engineering structures.

The ratio of stress to strain, within the elastic limit is a constant for agiven material. It is a measure of the elasticity or stiffness of the body.


Formula

Modulas of Elasticity = StressStrain

E = se (within the elastic limit)

Derived formula

E = PLAe

Units: Pascals (Pa), and engineering multiples; kPa, MPa and GPa.

Worked example 3

A cylindrical braking rod made from 15 mm diameter medium carbonsteel, is subjected to a compressive load of 25 kN. If the original lengthof the rod is 800 mm and the modulus of elasticity is 210 GPa, determinethe contraction of the rod.

d = 15 mm

= 15 ¥ 10-3m

P = 25 kN

= 25 ¥ 103N

L = 800 mm

= 800 ¥ 10-3m

E = 210 Gpa

= 210 ¥ 109 Pa

e = ?

CSA

25 kN

25 kN


Area being stressed is the cross sectional area.

A =pd2

4

=p ¥ ¥ -( )15 10

4

3 2

= 176.71 ¥ 10-6 m2

Now E =PLAe

Eae = PL

20 Braking systems

\ e =PLEA

=25 10 800 10

210 10 176 71 10

3 3

9 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 wireduring an experiment to confirm Hooke’s Law. If the modulus ofelasticity of the mild steel wire is 210 GPa, and the diameter of the wireis 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 convertedto basic units, kilograms, by multiplying by 103, and then to the weightforce, in Newtons, by multiplying by 10.

m = 1.2 t

= 1.2 ¥ 103kg

P = mg

= 1.2 ¥ 103 ¥ 10N

= 12 ¥ 103

L = 12 m

E = 210 Gpa

= 210 ¥ 109 Pa

e = ?

CSA


i Area being stressed is the cross sectional area.

A =pd2

4

=p ¥ ¥ -( )5 10

4

3 2

= 19.64 ¥ 10-6m2

Now E =PLAe


=120 10 12

210 10 19 64 10

3

9 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 50mm in gauge length and a cross sectional area of 160 mm2.

Extension(mm)

0.40 0.80 1.20 1.40 2.0 3.0 3.5 4.0 5.0

Load(kN)

20 40 60 62 70 80 82 80 70

i On the given axes below, plot the load extension diagram.

0

20

40

60

80

Load

(kN

)

1 2 3 4 5Extension (mm)

Figure 3.17 Load extension diagram

ii Determine the ultimate tensile strength of the material.

Ultimate tensile strength (UTS) occurs where the load is amaximum.

22 Braking systems

\ P = 82 kN

= 82 ¥ 103 N

A = 160 mm2

s = ?

s =PA

=82 10

160 10

3

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 stressto strain within the elastic limit. The straight line portion of thegraph 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 =PLAe

=60 10 50 10

160 10 1 20 10

3 3

6 3

¥ ¥ ¥¥ ¥ ¥

-

- -.

= 15.625 ¥ 109Pa

= 15.6 Gpa


Part 3: Engineering mechanics and hydraulics – communications 23

Communication

In this section of work you are going to build upon the freehand drawingof three dimensional objects that you did in Household appliances. Youwill cover, in detail, pictorial drawings using isometric projection.

Pictorial drawing

Pictorial drawing is very important to engineers as it enables thevisualisation of components. Freehand pictorial is used extensively ininitial design work. It has been used to design the communicationsections of this module. Pictorial drawing includes isometric, oblique,axonometric, perspective and dimetric projection. In this section of workyou will learn to draw one of these methods of pictorial, isometricprojection. You should then be able to interpret the shapes of otherpictorial drawings.

Isometric projection

You will learn about isometric projection and in particular how to drawisometric circles. If you have done technical drawing in earlier years youshould find this section relatively easy, although revision may berequired. If you have not covered this work before, you will needextensive practice on the topic, especially in the visualisation ofcomponents.

Isometric projection, visualisation

Worked exercise 1

A stepped block is manufactured from a rectangular prism, length 18mm, width 10 mm and height 9 mm. The length is evenly divided intothree and the height is also evenly divided into three.

24 Braking systems

The above paragraph gives details of the size of the block. Size detailsmay also be given as dimensions on a drawing.

The top view, front view and left side view of the block are given in thirdangle projection, drawn to a scale of 2:1 figure 3.18.

To assist you with your pictorial drawing, a basic shape of the originalblock has been given. The block has been divided into a grid pattern toassist 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 thestepped block.

FRONT VIEW

TOP VIEW

SCALE 2:1

PICTORIALLEFT SIDE

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. Thefront view provides the details needed. That is the block has been cut into astepped 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 outlinedto complete the pictorial drawing.


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 eightdifferent shapes are cut from rectangular prisms having the dimensionsgiven in Worked Exercise 1. Three orthogonal views of each block havealso been drawn, along with a pictorial grid. The drawings are to a scaleof 1:1. In the exercise you are to complete freehand, the pictorialdrawing 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, bothfreehand 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, useisometric 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 coulddraw four quadrants to form a circle. A circle in isometric projection,using the four centre method, is represented by combining four separateisometric quadrants. The two figures below show a circle and anisometric ‘circle’; (actually an ellipse).

26 Braking systems

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 tolocate 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.

Radius

Rad

ius

90∞

90∞

CornerContact

point

Contact point

90∞

90∞

Corner

Radius

Radius

Contact pointCentre

Centre

Quadrant 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 theisometric circle.

Four quadrants are combined to form the isometric circle.

Method:

1 draw the isometric square having sides equal in length to thediameter of the required circle using very light construction lines


2 locate and mark the middle of each side of the square – these middlepoints represent the contact points for each quadrant

3 draw lines at 90º to the sides of the square from these middle orcontact points – where these lines intersect are the centres for each ofthe quadrants

4 set your compasses at a radius equal to the distance from the centreto 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 25mm 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 theisometric circle.

Four quadrants are combined to form the isometric circle.

28 Braking systems

Method:

1 draw the isometric square having sides equal in length to thediameter of the required circle using very light construction lines

2 locate and mark the middle of each side of the square – these middlepoints 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 centreto 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 20mm in a vertical face.

Figure 3.25 Circle in vertical face


Projecting an isometric quadrant to another face

The following drawings show you a method of projecting the quadrant toanother face. You could fully construct another quadrant, but it isquicker 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 pointstowards the new surface using very light construction lines

2 set your dividers to the given thickness and accurately mark off thedistance to the new surface, down from the centre point and the twocontact points

3 check accuracy then draw the quadrant for the new surface.

project downfor new centre

project down fornew 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 inthe lower face. You therefore do not have to draw the whole quadrant inthis lower face; only half of the quadrant will be visible.

Method:

1 draw the quadrant in the top face, then project downward, asdescribed below, to the lower face

2 project downward from the centre point and the contact pointtowards the new surface using very light construction lines

3 project downward the profile edge (note the profile edge is a linetangential to the quadrants that represents the outside edge of theobject)

4 set to the required thickness of the object, mark off the distances tolocate the new centre point and contact point for the lower surfaceusing dividers

30 Braking systems

5 check accuracy then draw the part quadrant for the new surface.

6 darken the profile edge.

Quadrant radius = 40 mmThickness = 15 mm

Profile edge

Projectedcontact point

Projected centre

Figure 3.27 Profile edge

From a vertical plane, no profile edge

The following drawings show the method of projecting a quadrant from aleft and a right vertical face. The method is similar to the one describedfor 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 aleft and a right vertical face. The method is similar to the one describedfor the horizontal plane, and as such the method will not be described.


Figure 3.29 Profile edge

Projecting an isometric quadrant (with a profile edge)

You now have sufficient information to draw isometric circles, eitherfreehand or by using instruments, but you will need a great deal ofpractice to be able to complete drawings quickly and accurately. Keepthese notes as a reference, and use them as often as possible.

Orthogonal drawing

In this section of work you will build upon the freehand orthogonaldrawing introduced in Household appliances and the orthogonaldrawing from Landscape products.

You will be shown two orthogonal drawings as worked examples. If youare inexperienced at technical drawing you may wish to attempt thesetwo drawings, following the given steps, as practice.

Worked example 1

Draw, in orthogonal projection using a scale of 1: 2, a front view of thehand brake lever, shown in figure 3.30, when viewed from the directionof the arrow.

The front view of the 12 mm diameter lower hole has been given as astarting point for the drawing.

32 Braking systems

90

130

70

Ø 20

30

5520

14

6

14

Ø 4

40

R 10: INSIDE R 8.5

Ø 12

Ø 12R 20

R 20

Figure 3.30 Pictorial – hand brake lever

Steps and method

Note the scale of 1:2 means that you will use half size measurements forthe drawing. This is a reducing scale, used so that the drawing can fitonto the drawing page. You must divide all dimensions by two.

1 Locate the centreline position of, and draw the higher 12 mmdiameter hole. The vertical dimension is 40 mm, therefore measureto scale 20 mm above the given centreline. The horizontaldimension is 20 mm therefore measure 10 mm to the left of thegiven centreline. The circle is diameter 12mm therefore draw thecircle using a measurement of diameter 6 mm.

Note the circle should be drawn using a circle template.

2 Locate the top of the handle; project up from the located centrelineand measure the required distance. The dimension is 30 mm(R20 + R10) therefore measure 15 mm above the located centreline.


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 markoff distances of 35 mm, 65 mm and 45 mm.

4 Draw the left hand end of the lever. The dimension is diameter 20mm therefore draw down a distance of 10 mm.

5 From this left hand end draw the parallel portion of the bottom of thelever. The dimension is 90 mm therefore measure 45 mm to theright.

6 Draw the left hand sloping section of the bottom of the lever. It slopesdownward to a point 30 mm below the top edge, therefore measuredownward a distance of 15 mm to locate the end point of the slopingline.

7 Draw the middle sloping section of the bottom of the lever. It slopesdownward to a point 55 mm below the top edge, therefore measuredownward a distance of 27.5 mm to locate the end point.

8 Now you have to complete the right hand end of the lever. Lightlydraw the two R20 mm radius curves on the two centrelines. Thedimension is 20 mm radius, therefore use a measurement of 10 mmradius. You should use your compasses to do this construction. Usevery 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 thepreviously 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 thecentreline of the previously drawn top 12 mm diameter hole.Dimensions are 6 mm to the left, and 14 mm above, the centrelineand the diameter is 4 mm; therefore use measurements of 3 mm, 7mm and 2 mm. Again you must use your circle template to draw thecircle.

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 thindark chain line for the long centreline of the barrel of the lever, andthin continuous lines for the circle centrelines. Note that thincontinuous dark lines are used to indicate short centrelines.

If you have decided to attempt this drawing, you have now completed thefront view of the hand brake lever, in orthogonal projection.Congratulations.

Note that you were not requested to show any dimensions so do not showany.

The completed drawing is shown in figure 3.31.

34 Braking systems

Figure 3.31 Front view of hand brake lever

Worked example 2

Draw a front view of the piston from a hydraulic brake cylinder, shownin figure 3.32, using on enlarging scale 2:1 and the drill hole positionedto show the 6mm diameter hole using a part-section, a standard methodto 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 alongwith dimensioning of the hole.

Ø 25 Ø 12 Ø 25 Ø 25Ø 12

R 4

HOLE Ø 6DEPTH 12

6 6 62032

Figure 3.32 Piston from on hydraulic brake cylinder

The quickest method:

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 curvesusing 0.5 mm, thick, dark lines for the quadrants.


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 diameterof 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 theright hand end

2 mark off the diameter of the hole, 6 mm above and below thecentreline

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 isdrawn using two lines, each of 60º, from the left hand end of thepreviously drawn rectangle. The full drill hole is now shown as visibleoutline 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 theof the drill hole to indicate the limit of the part-section.

36 Braking systems

Hatching the sectioned area

The part-sectioned area is hatched, using thin dark lines, equally spacedat 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 theAS1100. 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 AS1100standards that you should be aware of include:

• All lines are drawn as dark lines.

• There are two different thicknesses of dark lines used on thedrawing.

• 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 indicatediameter and depth of the drill hole.

You must use current AS1100 standards in your drawings.


60

Ø 6

1

2

Figure 3.35 Part sectioned front view

Turn to the exercise sheets and complete exercise 3.8 and 3.9.

38 Braking systems


Exercises

Exercise 3.1

Determine the coefficient of friction between two materials and write a 1

2page report on the experiment and state your conclusions.

You should submit this exercise as a word processed document andattach your document to this page.

40 Braking systems

Exercise 3.2

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 ofthe disc is 0.35, determine the total braking force.

b A drum braking system has a force of 12 kN applied by the brakeshoe to the drum surface. If the coefficient of friction between thematerials of the shoe liner and of the drum is 0.3, determine thebraking force.


c A family sedan is parked with its hand brake on. The hand brakeoperates only on the rear wheels. A truck, attempting to park, bumpsthe sedan with a horizontal force of 2 kN. If each of the rear wheelsof the sedan supports a mass of 300 kg, and the coefficient of frictionbetween 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.

42 Braking systems

Exercise 3.3

Select the alternative, A, B, C. or D that best answers the question.

1 Braking systems are effective as a result of :

a an extremely high coefficient of friction between the brakingmaterials

b an extremely low coefficient of friction between the brakingmaterials

c a range of coefficients of friction between the braking materialsfrom 0.25 through to 0.45

d having no coefficient of friction between the braking materials.

2 Effective dissipation of heat energy is important in braking systemsto:

a keep the driver warm in winter

b allow the conversion of kinetic energy to heat energy tocontinue during heavy braking operations.

c allow the drums/discs to stay hot during braking.

d allow fade when braking.

3 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 suddenlyto avoid a collision. If the front wheels are fitted with disc brakes:

i draw a force analysis diagram showing all of the forces actingbetween one of the front discs and the brake pads

ii draw a free body diagram of the braking area of the discshowing all of the forces

iii write an equilibrium equation that would be used to determinethe magnitude of the braking force



Pads

Disc

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 diagramshowing all of the forces acting between one of the front wheelsand the road surface.

ii Draw a free body diagram of the braking area of the front wheeland the road surface showing all of the forces.

iii Replace the friction force and normal reaction with a singleforce. Clearly label the angle of friction.

iv Draw a force triangle that would be used to determine themagnitude of the braking force.

Analysis diagram Free body diagram Force triangle

Wheel rotates clockwise

Figure 3.37

c A truck is parked on the side of a gravel road. The angle ofinclination of the road is 6º. Due to the loose gravel surface, thetruck is on the point of sliding down the hill.

Determine the coefficient of friction between the truck tyres and thegravel surface.

6∞

Figure 3.38

44 Braking systems

Exercise 3.5

Complete the analysis of forces being applied to a body on a horizontalplane by showing all of the forces acting on the body, then completingthe free body diagram, (showing only the forces acting), then writing anequilibrium equation for FR.

The first example, a body at rest, is completed for you. Note: the frictionforce opposes motion or pending motion.


mg

P

N

Body at point of moving – force downward, 30∞

mg

FR =30∞

30∞

mg

N

Body at point of moving – force upward, 30∞

P

mg

N

Body at rest

FR = 0

No horizontal forceis acting

N

mg

mg

P

N

Body at rest – force horizontal

N (given)

mg (given)

FR =

mg

P

N

Body at point of moving – force horizontal

mg (given)

FR =

mgFR =

Figure 3.39 Friction analysis


Exercise 3.6

a Define the term mechanical stress.

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________

b A cylindrical braking rod of mild steel, diameter 10 mm, is subjectedto a tensile force of 15 kN. Determine the tensile stress in the rod.

46 Braking systems

c A cylindrical braking rod made from 12 mm diameter mediumcarbon steel, is subjected to a tensile load of 30 kN. If the originallength of the rod is 900 mm and the modulus of elasticity is 210GPa, determine the extension of the rod.

d A cylindrical punch, of diameter 9 mm is used to punch out the holesof a brake liner of thickness 6 mm.

i If, during the punching operation, the compressive stress in thepunch is 120 MPa, determine the force used to punch out thehole.


ii Using the previous data as well as the calculated force in the punchfrom part i above, determine the shear stress in the lining material.

48 Braking systems

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 apictorial grid. The drawings are to a scale of 1:1. Complete freehand,the pictorial drawing of each shaped block.

Figure 3.40 Isometric exercises


Exercise 3.8

Shape and size details of a washer from a master cylinder assembly aregiven in the dimensioned orthogonal drawing in figure 3.41.

i Sketch, freehand, in isometric projection, a pictorial drawing of thewasher. The scale used may be selected by you. The starting pointfor the centrelines of the washer has been given.

ii Draw, using instruments, in isometric projection, a pictorial drawingof the washer. Use a scale of 2:1. The starting point for thecentrelines of the washer has been given.

TOP VIEW

Ø 10

Ø 30

FRONT VIEW

3

Figure 3.41 Washer

50 Braking systems

iii FREEHAND PICTORIAL

iv INSTRUMENT DRAWING OF WASHER

Figure 3.42 Washer


Exercise 3.9

Shape and size details of a disc brake rotor are given in figure 3.43. Onthe grid paper attached:

i Draw freehand, in orthogonal projection using a scale of 1: 3, a frontview of the disc brake rotor, when viewed from the direction of thearrow. Include the principle dimensions.

ii Project freehand, using third angle projection, a left side view of thebrake pad.

3060

20

Ø 50

Ø 30

Ø 150Ø 300

Ø 12 X 10 HOLES

EQUALLY SPACED

Figure 3.43 Disk brake rotor

52 Braking systems

Figure 3.44 Grid


Exercise 3.10

The front and rear disc brake from a modern motorbike are shown infigure 3.45. The rotors are considerably different in size, at the frontthere 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.

___________________________________________________________

___________________________________________________________

___________________________________________________________

___________________________________________________________

___________________________________________________________

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___________________________________________________________

___________________________________________________________

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54 Braking systems


Progress check

In this part you explored mathematical and graphical methods to solveproblems of engineering practice.





Ag

ree

Dis

agre

e

Un

cert

ain

I have learnt about

• engineering mechanics and hydraulics– friction (without calculations)– stress and strain

stress (tensile and compression)load/extension diagramstrain (tensile and compression)

• communication– pictorial and orthogonal drawings

– Australian Standard AS1100, includingdimensioning.

I have learnt to

• distinguish between force, stress and strain

• produce pictorial, and orthogonal drawings of brakingsystems and braking components applying appropriateAustralian Standard (AS 1100).



During the next part you will continue to explore mathematical andgraphical methods used to solve problems of engineering practice andalso learn more about isometric and orthogonal drawing.

56 Braking systems



Exercises 3.1 to 3.10 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

Locate and complete any outstanding exercises then attach yourresponses to this sheet.

If you study Stage 6 Engineering Studies through a Distance EducationCentre/School (DEC) you will need to return the exercise sheet and yourresponses at the completion of each part of a module.


Braking systems

Part 4: Engineering mechanics, hydraulics andcommunication – 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


Introduction

In this part you will continue to explore mathematical and graphical methods usedto solve problems of engineering practice and also develop your skills in isometricand orthogonal drawing.

What you will learn?You will learn about:

• Engineering mechanics and hydraulics

– 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 simplebraking systems

• detail drawings of braking systems and braking components applyingappropriate Australian Standard (AS 1100)

• produce simple computer assisted drawing(s).




Engineering mechanics and hydraulics

In this section of work you will learn about the meaning of mechanicalwork, energy and power.

Work, power, energyYou may have studied these topics before, but you will now require anunderstanding of each term, how each affects engineering problems andspecifically, how each relates to braking systems.

Mechanical work

Mechanical work is done when a force acts upon a body and produces adisplacement.

The work done by a force is determined by the product of the force and thedisplacement of the point of application of that force.

Work = Force x Displacement

W = 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, adistance, s, against a frictional resistance, the work done by the appliedforce in overcoming this resistance is:

Work = Frictional resistance x Displacement

W = FR x s


Against gravity

When the centre of mass of a body is raised through a vertical distance (h), thework 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 theplane, the work done is:

Work = mg sin q ¥ s

But sin q = HeightDisplacement along the plane

=hs

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


Example 1

Consider the work done by a motor car:

1 Travelling at constant velocity on a horizontal surface.

Work done = Total resistance x Displacement

W = Rs

Note if there is no resistance, work done is zero

2 Accelerating uniformly on a horizontal surface.

Work done = Total resistance x Displacement + mas

W = Rs + mas

3 Accelerating uniformly up an inclined surface.

Work done = Total resistance x Displacement + mas + mgh

Where h is the vertical displacement of the car.

W = Rs + mas + mgh

4 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 isin operation, but in (3) and (4), ineffectively.

1 Car stationary on a horizontal surface.

• Work done by the hand brake = 0 as displacement is zero.

2 Car stationary on an inclined surface.

• Work done by the hand brake = 0 as displacement is zero.

3 Car on an inclined surface, moving downward at constant velocity.

• Work done by the hand brake = mgh - Rs

4 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 isdifferent from the work done by the braking system.


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 ofenergy is determined by the quantity of work it can do. Units are therefore thesame as for work, Joule, J.

Kinetic energy

Kinetic energy is the energy a body possesses due to its motion. It is determinedby the amount of work done in bringing the body to rest.

KE =12

mv2

Potential energy

Potential energy is the energy possesses due to its position. It is determined bythe 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 isdetermined by the amount of work done in deforming the body, such as a spring.

SE =12

Fs

It may also be determined from the load-extension diagram, following a tensiletest. Strain energy is found by determining the area under the graph, up to theelastic 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 andbraking systems. More importantly is the rate at which a car or brake can do thework 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.


Power

Power is the time rate of doing work, and is determined by the ratio ofthe work done over time taken to do the work.

Power =Work doneTime taken

P =Wt

Derived from this formula are two very important formulas.

P =Wt

but W = Fs

\ P =Fst

andst

= v (velocity)

\ P = Fv

P =Wt

= Fst

= Fv

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 =12

mv2

= Work done by the brakes in stopping the car

2 Accelerating uniformly on a horizontal surface.

KE =12

mv2 – 12

mu2

= Work done by the car in accelerating


Mechanical efficiency of a machine

Another important consideration for engineers is the efficiency of a machine suchas a car or braking system.

Mechanical Efficiency =Power outputPower input

Also in cars Mechanical Efficiency =BrakePower

IndicatedPower

When dealing with hydro-electricity the relationship between the volume and themass 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 mechanicsIn this section of work you will study the basic principles of the hydraulics used inbraking systems. If you need to refresh your memory of the developments ofbraking 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, causingmany service problems in maintaining linkages and providing equalisation ofbraking pressure to the brakes.

Hydraulic systems were developed based on the simple principle that pressureexerted at any point on a confined fluid will be transmitted throughout the fluidequally and undiminished in all directions.

Liquid flows freely, assumes the shape of the container, cannot be appreciablycompressed, and, if contained in a sealed system, allows pressure to be equallyand evenly distributed throughout the system. The hydraulic-brake actuatingsystem 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 brakesare that it gave completely uniform pressure throughout, greatly reduced adjustmentproblems, and provided even braking on all wheels at all times. Later developmentsof front brakes and ABS braking systems modified the last ‘advantage’.


Pressure

To understand the basic principles of the hydraulic braking system you mustfirstly understand what is meant by pressure and be able to do simple calculationsinvolving pressure.

Pressure is force per unit area.

p =FA

Thus total force or thrust on a surface is the area of the surface, times thepressure on that surface.

F = p x A

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

= 1.162 kPa

ii F = mg

= 3 ¥ 10

= 30 N

A = b ¥ t

= 110 ¥ 10-3 ¥ 75 ¥ 10-3

= 8250 x 10-6 mm2

P = F ∏ A


= 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 differentweight at various depths.

1 Drill, or pierce, three small holes along the side of a large PET drinkbottle 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 greaterthe amount of water above the hole, the greater the pressure.

ii pressure exerted by the liquid is always perpendicular to the surface itcontacts.

iii pressures are the same at all points on the same horizontal level in a liquid atrest.


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 canexert an equal and undiminished pressure to all other portions of the liquid and tothe walls of the container.

1 Attach the PET drink bottle used in the previous activity to anotherintact PET drink bottle so the two join at the neck.

2 Fill the bottles with water, connect and squeeze the container withoutholes 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 closedcontainer 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 principlenamed after him.

Pascal’s PrinciplePascal’s Principle states that if the pressure at any point in a liquid that isenclosed and at rest, is changed, then the pressure at all points in the liquid ischanges 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 istransmitted equally throughout the whole of the enclosed fluid.

If pressure is applied to a liquid in a sealed container or system, through theapplication of a force (F1) in a cylinder of cross sectional area (A1), an equalpressure will be transmitted to a larger piston and cylinder, of area (A2), causing athrust or force in this piston, of magnitude F2.


FA

1

1

=F

A2

2

F2 =F A

A1 2

1

¥

If A2 is very large compared to A1 a comparatively smaller force applied to thesmaller piston can overcome a large resistance acting on the larger piston.Additionally, this can apply to a number of different cylinders and pistonsattached to the sealed system.

Braking systems

This principle forms the basis of hydraulic machines, including the hydraulicpress, 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 isapplied 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.

rear wheel cylinder

100 N

A1 = 600 mm2

A2 = 300 mm2

A3 = 900 mm2

250

50

front wheel cylinder

Figure 4.4 Hydraulic braking system


Solution

The solution is based on Pascal’s Principle, that any pressure applied to aliquid in a confined container or system is transmitted equally andundiminished to all parts of the container or system.

i Determine by moments the resultant force on the master cylinder caused bythe 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

P =FA

=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

= 250 N

iv 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 thepressure as needed. This is particularly applicable to the different stoppingforces needed at the front and back wheels.

When the brakes are applies the reaction at the front wheels is greatlyincreased due to the tendency of the vehicle to continue its forwardmotion. The front brakes therefore need a greater applied force than therear wheels. One way that this can be done is by using larger wheelcylinders on the front brakes.


Archimedes’ PrincipleArchimedes 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 byan upthrust which is equal to the weight of the fluid displaced. Thisupthrust, or buoyancy, acts through the centre of mass of the displacedfluid. The centre of mass is therefore referred to as the centre ofbuoyancy.

Buoyancy

From your previous reading you will remember that fluid exerts an equalpressure 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 thedisplaced fluid, must be equal to the weight of the floating body. Thisupward 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 ofthin cotton around the brick page and attempt to lift the object. You mayhave to us a wooden handle on the string to prevent the cotton cutting intoyour fingers.

Normally the cotton will break.

Now attempt the same experiment with the brick immersed in water, either in abucket or washing tub. The lift should be successful while the brick remainsimmersed.

Is this because the brick has less mass in the water, or is it due to thebuoyancy?

I think that you can agree with Archimedes on that question.

Turn to the exercise sheets and complete exercises 4.2 and 4.3.


Communication

In this section of work you will learn more about AS1100 standards, what ismeant by a detail drawing, and the standard sectioning techniques that maybe used. As examples you will be shown how to design the best solutionsfor, and complete, two detail drawings on brake components.

Detail drawingA detail drawing is a specialised type of orthogonal drawing used tocommunicate information from the designer or engineer to themanufacturing 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 detaildrawing must provide sufficient information for the manufacture of thatcomponent.

The shape description

The shape description is usually given in an orthogonal drawing. Adecision must be made as to which views, and how many views arenecessary to show the full shape description. Sometimes three views, a topview, front view and left or right side view are required, whilst other, morecomplicated components may need up to five views.

The simple components from a brake master cylinder used in the followingexamples can use as few as two views, or even a single view thatincorporates dimensioning to provide the full shape description.


Sectioned views

Where the component has interior details that need to be shown, sectioningmust be used to show these details as visible outline. Hidden outlineshould be avoided where possible. You will learn about the different typesof 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 thecomponents, using AS 1100 dimensioning standards.

The material

The material to be used in the manufacture of the component is given on thedrawing 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 designdecisions have to be made. The best approach when designing a detaildrawing, is to complete a number of freehand drawings showing variousoptions with regard to the number of views, the sectioning methods, andthen, the placement of the dimensions.

In the following two worked examples you will be shown how this designtechnique is applied.

Worked examples, the master cylinder

A master cylinder for a hydraulic braking system is shown below infigure 4.5. You will be required to draw detail drawings of some of thecomponents in the exercise section of this module part.


Figure 4.5 Master cylinder

Courtesy: Trinity College Auburn© LMP

Worked example 1

Design a detail drawing of the piston seal from the master cylindercomponents. The psiton seal in shown in figure 4.6.

a Show the designs for four possible detail drawings, using freehand drawingtechniques:

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 badpoints of the designed detail drawing.

c Fully dimension each drawing, using different placements for the dimensions.


Material: neopreneScale 2:1

Ø 22

RIGHT SIDE VIEWFRONT VIEW

10

Ø 2

5

Ø 1

2

10

Ø 12

Ø 22SEAL

Ø 25

Figure 4.6 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 showthe 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.

Ø 22

RIGHT SIDE VIEWFRONT VIEW

10Ø 25

Ø 12


Figure 4.7 Possible solution 2 (using a full-section)


• Front View – As full sectioning is used, no hidden outline is shown. Thedetails of the hole are shown as visible outline; these are correct standards.


• Right Side View – The drawing is clear and gives good shape description.

• Dimensioning – Dimensioning is poorly designed with too many dimensionsshown 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.

FRONT VIEW

10


Ø 2

2

Ø 2

5

RIGHT SIDE VIEW

Ø 12

Figure 4.8 Possible solution 3 (using a half-section)


• Front View – As half-sectioning is used, no hidden outline is shown. Thedetails of the hole are shown on one side of the centre line and the exteriordetails on the other side of the centreline; correct standards. Note that a half-section 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 dimensionsshown on the Front View.

• New Methods – The use of circle templates, the use of a half-section.

• Decision – Good solution, could be used.


FRONT VIEW

10


Ø 2

2

Ø 2

5

Ø 1

2

Figure 4.9 Possible solution 4 (using a part-section)


• Front View – As part-sectioning is used, no hidden outline is shown. Thedetails 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 theRight Side View to be omitted. The circular shapes have been defined by theuse 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 darkcontinuous freehand line.

• Decision – Good solution, will use.

Possible solution 4, using a part-section, is the quickest and preferredmethod. You will be required to draw this detail drawing of the piston sealusing instruments, as exercise 4.1 in your exercises.


Varying the design process

In the next worked exercise you will be shown how the design process may bevaried to suit the component being drawn. The right side view would show onlythree concentric circles. These are rather difficult to draw freehand, and reallyserve no purpose in repeating them in the design process. The right side view willtherefore be omitted. Again, the design of the dimensioning can be completedlater.

Worked example 2

Design a detail drawing of the valve from the master cylinder components. Youare to show four possible detail drawings, using freehand drawing techniques.The valve is to be made from 70-30 brass.

i without the use of a section



iv using a part-section.

Ø 16Ø6

Ø 30

12

2

Figure 4.10 Valve

As the right side view is only three concentric circles, you can omit the RSV inthe 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 thedimensions later.

The four possible front views of the valve.

FRONT VIEW(part-section)

FRONT VIEW(half-section)

FRONT VIEW(full section)

FRONT VIEW(no section)

Figure 4.11 Four possible solutions


The four possible solutions are given above. When you have designed thesolutions you should check that the solutions are correct, and that the correctAS1100 standards have been used.

Again, as with worked example 1, Possible solution 4, using a part-section, is thequickest and preferred method, and again the right side view may be omitted if thedimensions 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.

When the design is completed the drawing is done using instruments or a CADsystem. The completed detail drawing is shown below.

PART-SECTIONED FRONT VIEW Material: 70–30 brass

Ø 3

0

Ø 6

Ø 1

6

2

Figure 4.12 Detail drawing of valve

AS 1100 standards

When you have completed a drawing you should check to see that you havecorrectly used AS1100 standards.

All lines, other than construction lines should be the same darkness. During theHSC marking the darkness of lines is always checked. The lines should be darkenough so that if photocopied they would give a good dark outline.

Line thickness is also important. Visible outline should be drawn with thick darklines. All lines other than visible outline are thin dark.

The dimensioning standards should also be checked. There are two smalldimensions, 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 thereis 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 tobe neatly lettered. The 2 mm dimension is written outside the extension lines asshown, and in line with the leader line.


Computer aided drawingIn 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 aideddrawing. You will learn to produce simple computer assisted drawings usingtools and coordinates.

Coordinates

All objects drawn using CAD are defined by the positioning of points. A line maybe defined by the positioning of its two end points. A circle may be defined bythe positioning of its centre point and a point on the circle, that is, the radius of thecircle.

Graphing points in mathematics

You have already used x and y coordinates to plot points, lines and curves whendrawing 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.


+y

0

(2,3)

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 theend points, (2, 3) and (6, 7).

Solution

Mark off the horizontal distance 6 units to the right of the origin. Mark off thevertical 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 inspace by using the x , y and z axes radiating from a fixed, or predetermined pointcalled the origin. To position a point on a flat surface, such as on drawing paperor 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 thuslines and objects. As with mathematics the coordinates use both positive andnegative 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 ydirections from the last point entered. Negative values are frequently used.

• polar coordinates – you measure the radial distance from the last pointentered, and the angle, measured in a counter clockwise direction from thepositive x axis. Note, we could also use absolute polar coordinates.


Example of absolute coordinates

When using absolute coordinates in CAD, the x and y values are measured fromthe origin, (0, 0). The positive x value is measured horizontally to the right of theorigin. The positive y value is measured vertically upward from the origin.

Thus the Absolute Cartesian Coordinates (2, 3) of a point, P, indicate that thepoint to be plotted is 2 units to the right and 3 units above the origin. This isshown in the previous diagram, and also applies to CAD.

Absolute coordinates are not widely used as it is difficult to calculate the values ofall points in a complicated drawing.

Example 2

40

40

20

20

A

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 drawingon a computer screen. Point A, having coordinates of (25, 20) is plotted on theaxes.

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. Besure 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 sideview of the ratchet block using the absolute coordinates method. If not, readthrough and study the method.


+y

0

A (25,20)

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 horizontalline. 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 thetop of this vertical line, horizontally, 20 mm to the right, to meet the slopingline.

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. Thesolutions 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 thepoint, 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>.




Note that AutoCAD remembers the last point specified.

Example of relative coordinates

When using relative coordinates the actual dimensions are measured along the xand 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 relativecoordinates describe the actual distance from the first point to the second point.

Remember that the relative coordinate distances are measured along the axes fromthe last point entered, not from the origin. This enables the dimensions of theobject to be used without having to calculate the absolute coordinates for eachpoint from the origin. This is a quicker and easier method.

Example 3

The front view of a ratchet block follows.

40

20

2010

40

B

Figure 4.16 Front view of ratchet block

The origin, the x axis and the y axis drawn below represent a CAD drawing on acomputer 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 thedrawing. Assume that you draw the 40 mm square first, commencing frompoint B and drawing in a counter clockwise direction. Remember, thecoordinates are relative to the previous point plotted. Assume that you drawthe 20 mm square next, commencing at the bottom right hand corner.

Note: record the relative coordinates using the AutoCAD system, ie therelative 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 ofthe ratchet block using the relative coordinates method.


If not, read through and study the methods.

+y

0

B (25,20)

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 horizontalline. Draw a horizontal line from the top of the vertical line, 40 mm to theleft. Draw a vertical line 40 mm downward from the left hand end of thishorizontal line to point B.

Find a point on the bottom line 30 mm to the right of point B then draw avertical line 20 mm upward from this point. Draw a horizontal line from thetop of this vertical line, 20 mm to the left. Draw a vertical line 20 mmdownward from the left hand end of the previously drawn line.

ii Determining the relative coordinator for using AutoCAD. Absolutecoordinates 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

change in x value, 0, change in y value, 40 @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 the20 mm square, 55,20.


The relative coordinates of each of the points for the 20 mm square, listed incyclic order from the first point, in a clockwise direction are:

Determining relative coordinates AutoCAD method

change in x value, 0, change in y value, 20 @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 theright 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 theEnter key. This tells the computer that the line you wish to draw begins at thepoint, 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. Therelative coordinates are the most commonly used, so we will only consider them.

When using relative polar coordinates the actual dimensions are measured in aradial direction from the last point entered, the angle is measured in a counterclockwise direction from the positive x axis.

The first point is entered using absolute coordinates, as described previously.


The relative polar coordinates of the second point are then entered. These relativecoordinates describe the angle of rotation and the actual distance from the firstpoint to the second point.

4020

201040

B

Figure 4.18 Top view of ratchet block

The origin, the x axis and the y axis drawn below represent a CAD drawing on acomputer 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 thedrawing. Assume that you draw the 40 mm square first, commencing frompoint B then drawing in a counter clockwise direction. Remember, thecoordinates are relative to the previous point plotted. Assume that you drawthe 20 mm square next, commencing at the bottom right hand corner.

AutoCAD system; the relative polar coordinate for the right hand end of thefirst line is @40<0. This means the required point is a distance of 40 mmfrom 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 ofthe ratchet block using the relative polar coordinates method. If not, readthrough and study the methods.

+y

0

B (25,20)

Figure 4.19 Plotting point B (25,20)


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 mmsquare, 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




Absolute coordinates of the first point, the bottom right hand corner, for the20 mm square, 55, 20.

The relative coordinates of each of the points for the 20 mm square, listed incyclic order in a clockwise direction from the first point as follows.

Determining relative polar coordinates AutoCAD method




You should read the notes a number of times, highlighting the areas of importanceor concern. Complete all of the problems, including the computer portion. If youdo not have access to a CAD system, work through the exercise using freehandsketching methods.



Exercises

Exercise 4.1

a Define the term ‘mechanical work’.

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b Analyse each of the following three problems, and determine an equation forthe work done in moving the car.

i A car travelling at constant velocity on a horizontal plane, against africtional 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:


c Determine the work done by a braking system in the following. In each casethe 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 forcalculating that energy.

i Kinetic energy

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ii Potential energy

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iii Strain energy

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e Define the term ‘power’ when referring to mechanics.

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f Two very important formulas may be derived from the basic formula, powerequals 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.

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b Define the term ‘pressure’.

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c State Pascal’s Principle.

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d State Archimedes’ Principle.

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e A concrete brick of mass 5 kg and dimensions 430 x 150 x 100 rests on ahorizontal surface, flat on its largest face. Determine the pressure applied tothe horizontal surface.

f The front wheel brakes provide approximately 67% of the braking forces dueto ‘dipping’ of the car when braking. Explain how the hydraulic system isable to provide for this need for greater braking forces at the front wheels.

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Exercise 4.3

Figure 4.20 represents a sealed hydraulic braking system. A force of 200 N isapplied to the brake pedal as shown. Size details of the pedal, master cylinder, andfront 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.


rear wheel cylinder

100 N

A1 = 600 mm2

A2 = 300 mm2

A3 = 900 mm2

250

50

front wheel cylinder

Figure 4.20 Hydraulic braking system

i Determine by moments the resultant force on the master cylinder caused bythe 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.


Exercise 4.4

Draw, using instruments, to a scale of 2:1, a detail drawing of the piston seal infigure 4.21. Note: it is sufficient to draw only one view, a part-sectioned frontview, 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 yourinstruments.


10

Ø 12

Ø 22SEAL

Ø 25

Figure 4.21 Piston seal


Exercise 4.5

a Design a detail drawing of the spring seal from the master cylinder components infigure 4.22. You are to show four possible detail drawings, using freehanddrawing techniques:i without the use of a section



iv using a part-section.

The spring seal is to be made from neoprene.

2 10

Ø 22

Ø 25

Figure 4.22 Spring seal

As the right side view consists of only four concentric circles, two visible and twohidden outline, you can omit the Right Side View and show only the four frontviews. Do not show the dimensions on these design sketches.

Front View Front View Front View Front View(No section) (Full section) (Half-section) (Part-section)

Methods iii and iv are the best solutions. You should now complete thehalf-section solution to give you experience with this standard.

First design the position of the dimensions.


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 thenfully dimension the valve.


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 dimensionalisometric drawing. The size details are given using dimensioning.

B

A10

20

40

20

2040

Figure 4.23 Ratchet block

The origin, the x axis and the y axis, drawn on the next page represent a CADdrawing 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 viewof the spacing block.

ii From this front view, project using third angle projection a right side view ofthe spacing block. Note: the absolute coordinates of the right side view ofPoint 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: theabsolute coordinates of the top view of Point B, the starting point for the topview, are (40,130). Name the TOP VIEW.

iv Determine and neatly label the absolute coordinates of each of the points onthe drawing.

v Commencing at Point B, (40,60), fully describe how you would draw theorthogonal drawing of the spacing block using a AutoCAD package. Youmay use either relative or polar coordinate methods or a combination of thetwo methods. List every step in sequence, giving the coordinate entries foreach point plotted.

Space for this exercise is provided after figure 4.24.


+x AXIS0 ORIGIN

+y AXIS

B (40,60)

Figure 4.24 CAD drawing


Describe how you would draw the orthogonal drawing of the spacing block,commencing at Point B, (40,60), using AutoCAD. Underline the method youwill use: relative coordinates, polar coordinates, combination of the two.

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Progress check

In this part you used mathematical and graphical methods to solve problems ofengineering practice and applied graphics as a communication tool.





Ag

ree

Dis

agre

e

Un

cert

ain

I have learnt about

• engineering mechanics and hydraulics

– 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 fluidmechanics to simple braking systems

• produce detail drawings of braking systems andbraking components applying appropriate AustralianStandard (AS 1100)

• produce simple computer assisted drawing(s).



During the next part you will generate an engineering report.



Exercises 4.1 to 4.6 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 thissheet.

If you study Stage 6 Engineering Studies through a Distance EducationCentre/School (DEC). You will need to return the exercise sheet and yourresponses at the completion of each part of a module.

If you study Stage 6 Engineering Studies through the OTEN Open LearningProgram (OLP) refer to the Learner’s Guide to determine which exercises youneed to return to your teacher along with the Mark Record Slip.

Braking systems

Part 5: Engineering report

Braking systems 1

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

2 Part 5: Engineering report

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 ofbrake or component of a braking system, integrating computersoftware.

• conduct research using appropriate computer technologies

• work with others and appreciate the value of collaborative working.



Braking systems 3

Engineering report

An engineering report is a formal, considered document which drawstogether information gained about a product or filed, through research andanalysis, to arrive at a conclusion or present recommendations based oninvestigation.

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 additionalaspects 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 andcommunication skills related to the content

• include data relevant to the area under investigation


• present information clearly and concisely so that it is easily understoodby the reader through the use of tables, graphs and diagrams toillustrate 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 andgives the date when the report was completed.

Abstract

The abstract is a concise statement that describes the content of theengineering report. It covers the scope of the report (what it is about) andthe approaches used to complete the analysis (how the information wasassembled).

The purpose of the abstract is to allow a reader to decide if the engineeringreport contains relevant information.

The abstract should be no more than two or three paragraphs – shorter ifpossible.

Braking systems 5

Introduction

The introduction provides an overview of the subject, purpose and scope of theengineering report and may contain background information regarding thetopic.

It also outlines the sections of the engineering report including why theinvestigation was undertaken, what research occurred, how data was collectedand what anaylsis was conducted.

Analysis

The analysis is the body of the engineering report and should show evidence ofresearch and experimentation. Information about materials and the mechanicsof products should be collected or calculated for all engineering reports. Thissection must contain information required to satisfy the aim and purpose of thereport.

Tables and graphs, used to summarise detailed data in a concise form, arecommon features of an engineering report.

Result summary

The result summary should present the results concisely and note anylimitations on the investigation.

The results inform and support the conclusions and recommendations.

Conclusions/recommendations

The conclusions/recommendations summarises major points or issues in earliersections of the engineering report.

This section requires the writer to draw conclusions or make recommendationsbased 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’swork that has contributed to the engineering report.

Bibliography

The bibliography demonstrates that the report is well researched – allreferences need to be included. Bibliographic entries should follow establishedguidelines.


A standard approach is the Harvard systems of referencing requiring theauthors surname initials, date of publication, title of reference, publisher andplace of publication.

For example:

Kalpakjian, H. R. and Wrighton, H. 1967, Practical MicroscopicalMetallurgy, Addison Westley, USA.

This information allows the reader to source the information for confirmationof the details or conduct further research.

Appendices

The appendices should contain detail that has been separated from the mainbody of the engineering report. The information in this section is not essentialbut enhances the other data. Examples could be engineering drawings of theproducts being compared where the overall dimensions of the product may nothave been part of the comparison in the report, but may be relevant to somereaders.

During the engineering course this section may contain a technical drawing andcould include information collected from organisations.

Developing an engineering report

Research and collaboration are the keys to developing an accurate andinformative engineering report.

Research methods

In addition to popular research methods, like individual input and electronicmedia, traditional reference materials remain a valuable source ofinformation and include:

• textbooks

• booklets, brochures and pamphlets

• newspapers, journals and magazines

• videos.

Collaborative work practices

Discussion with colleagues constitutes collaboration and can providevaluable information.

Braking systems 7

These approaches to research and collaboration can be used by you, alongwith 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 previousmodules so you will be expected to present a more comprehensive reportthis time.

The engineering report for this module must be based upon theinvestigation 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 theanalysis of an individual component, such as the caliper braking systemused on bicycles.

You should communicate the selected topic for your engineeringreport to your teacher before commencing the report. You maynegotiate with the teacher for an alternative topic based upon a brakingsystem if you have a particular interest or resource available.

You must be aware of the need for all safety precautions to be followedduring research and experimentation. Do not tamper with the brakingsystem of a registered vehicle. Tampering may make the vehicleunroadworthy which could have fatal consequences.

The following section contains a sample engineering report that you mayuse as a guide when presenting your work.

The sample of engineering report focuses on the investigation of acomponent of a braking system – the brake shoe used in the rear drumbrakes of a current model car.

To assist you the sample engineering report will include notes explainingthe reasons for the selection or use of the information in the report. Thesenotes have been boxed to separate comments from the report.

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, analysesthe 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 ifa 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 beanalysed 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 pictorialdrawings. Another method is to sketch the pictorial from looking at theactual 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 thansketching freehand pictorials. A drawing involving largecircles such as in a brake shoe requires more skill, especiallywith concentric circles. In cases like this an engineer wouldprobably use an aid such as a radius curve.

You have to decide how many views are required to give acomplete shape description for the report. Refer back to Part 4where you did detail drawings, then decide on the most simplemethod to show the brake shoe.

The drawing completed above uses two views, a front view anda left side view that fully describes the shape of the brake shoe.Well maybe it does if the person reading the report caninterpret an orthogonal drawing. Don’t worry too much, theycan also refer to the pictorial which is easier for the untrainedperson to understand. Add all of the dimensions.

If a problem still exists, provide a model of the component orinclude the actual component in an attachment.

If the component is too big to be included, attach a video. Beinnovative, and design a solution to a perceived problem.

A detail drawing of the product

4

Ø 10

R 10

5024

R 104

R 88

R 120

15∞

30∞

42

4

LEFT SIDE VIEW FRONT VIEW Scale 1:2Material 0.2% C Steel

Ø 4

Ø 6

Figure 5.04 A detail of a brake show

It is good practice to include a completed drawing of the componentor 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 andthe drilled web.

The shape of the curved plate is stamped from 4 mm mild steel strap thencurved to the required radius.

The shape of the web is stamped from 4 mm mild steel strap and thenstamped 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 thatservice properties refer only to the properties that the material needswhen in use, not the properties needed during manufacture.

Material

The metal components of the brake shoe are manufactured using mildsteel (0.2% carbon). Two service properties of mild steel that make it asuitable 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 underapplied loads.

Environmental effects that mild steel might have:

a during production of the material

The mining of the iron ore causes environmental problems with thesurrounding area, and also affects the mining area, the flora andfauna in the mine area and near the mine.

The smelting and production of steel has a very adverse effect uponthe surrounding area near the steelworks.

b during manufacture of the product

As the brake shoe components are produced by stamping andpressing, the main environmental effect would be the noisepollution.

The welding causes fumes and produces welding light both of whichcan be detrimental to the operator or personnel near the weldingarea.

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. Anevaluation of the advantages and disadvantages of each materialcompared 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 stresscalculations must be conducted to determine the force required tostamp out the shape of each component.

Compressive stress and strain calculations are also required todetermine the forces in the punch or punches used to stamp out theshapes.

2 During service

The web of the brake shoe undergoes compressive stress duringservice due to the forces applied by the slave cylinder. Thecompressive forces must be determined to ensure that the yieldstrength of the steel is adequate for the required service condition.

Friction


Friction is involved during the stamping process and must beconsidered during the design of the punches.

2 During service

Friction is involved during operation of the braking system. Thebraking force produces shearing stresses in the welded web, andmust be considered during design calculations.

Hydraulics

The forces applied by the slave cylinder to the brake shoe must bedetermined using moment calculations, and hydraulic pressurecalculation involving Pascal’s Principle.


The stamping machine uses a hydraulic press, which could involvecalculations 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 bothmanufacture and service.

During manufacture

The energy used to stamp out the shapes could be determined as acomparison between the different materials being investigated for use.

During service

The energy generated by the friction forces needs to be dissipated as heatenergy. Calculations of the mechanical energy generated at the brakelining surface could be made.

Note that it would be sufficient for you to cover only two situationsinvolving 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 areused to punch out the holes of the web of a brake shoe of thickness 4mm in the one simultaneous operation.

i If, during the punching operation, the total compressive stress inthe punches is 720 MPa, determine the total force required topunch 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 = 720 ¥ 106 Pa

P = ?


Area being stressed is the total cross sectional area.

A1 =p ¥ d 2

4

=p ( )10 10

4

3 2¥ -

= 78.54 ¥ 10-6

A2 =p ¥ d 2

4

=p ( )6 10

4

3 2¥ -

= 28.27 ¥ 10-6

A3 =p ¥ d 2

4

=p ( )4 10

4

3 2¥ -

= 12.57 ¥ 10-6

Total Area = A1 + A2 + A3

= 119.38 x 10-6m2

Now s =

PA

P = s ¥ A

= 720 ¥ 106 ¥ 119.38 ¥ 10-6

= 85.953.6

= 85.95 kN

ii Determine the shear stress in the 4 mm thick material.

Area being sheared is the total surface area of the threecylindrical shapes being punched out of the material for the webof the brake shoe.

Total Shear Area = circumference of three holes x thickness

SA = p ¥ (d1 + d2+ d3) ¥ t

= p ¥ (10 + 6+ 4) ¥10–3¥ 5¥ 10-3

= p ¥ 100 ¥ 10-6m2 Figure 5.06 Analysis diagram

Now s = P ∏ A

=85 954

100 10 6p ¥ ¥ -

= 273 6 106. ¥ Pa

= 273.6 MPa

2 A pressure of 50 MPa is produced in the slave cylinder. If theinternal area of the cylinder is 30 mm:

i determine the force applied to the brake shoe.

Pressure =

FA

Force = Pressure ¥ Area

= 50 ¥ 106 ¥ 30 ¥ 10-6

= 1 500 N

= 1.5 kN

ii If the force applied to the front disc brake by the samehydraulic system is to be twice the size of the force applied tothe rear drum brake, determine the internal area of the frontwheel 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 washoping to conduct tensile and compressive test using a Hounsfieldtensometer.

I wanted to carry out comparative testing of the three materials, 0.2% Csteel, 0.35% C steel and gray cast iron.

I decided to conduct compressive tests and impact tests on the threematerials and to research the tensile properties of the three materials.

The tests are comparative only. Two pieces of each of the three materialswere 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 appliedaxially to each sample.

The impact test was conducted by holding the sample in a vice andrepeatedly 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 theresults for the steels. The compressive tests were complete failures. Ineeded to gain access to testing machines but was unable to do so at thelocal high school.

Collected data

Material 0.2% carbonsteel

0.35% carbonsteel

Gray cast iron

Yield stress MPa 345 375 NA

UTS Mpa 440 580 170

Izod impact test 117 65 10

Yield UTS Impact

NA

Gray cast iron

0.35% C steel

0.2% C steel

Figure 5.07 Graph

Note: you may also carry out an experiment that fails to provide thedesired results. If this occurs, you still need to provide data that isrelevant to the report. Research data, provided that it is clearlyidentified as such, may be used

Health and safety issues

1 The performance of braking systems for cars and trucks are regularlytested. The vehicle, to be registered as roadworthy, must pass aninspection each year. Heavy trucks are also randomly tested by theRTA inspectors throughout the year.

Safety issues are thus extremely important when associated with thebrake shoe performance and design.

2 The poor performance of braking systems is still responsible formany vehicle accidents and as such contributes greatly to thehospitalisation of victims. The number of fatal accidents has beenreduced over the past ten years. The improvement in design andmaintenance 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 twomaterials 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 productionof 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 beaddressed by the Workplace Health and Safety Committee.

Glossary

auxiliarybrakes

are additional brakes that are fitted to avehicle and are used to assist the majorbraking system in the vehicle.

detaildrawing

a detail drawing is an orthogonal drawingwhich gives a full size and shapedescription of the component. It alsoincludes the material from which thecomponent is to be manufactured.

explodedisometric

an exploded isometric drawing is apictorial drawing of an assembly in whichthe components are drawn separated sothat 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, Introduction to Materials Science,Jacaranda Wiley, Australia.

Greaves, H.R. & Wrighton, H. 1967, Practical MicroscopicalMetallurgy, Chapman and Hall, England.

Moffatt, W. Pearsall, G. and Wulff, J. 1964, The Structure & Propertiesof Materials, John Wily and Sons, New York.

Kalpakjian, S. 1985, Manufacturing Processes for EngineeringMaterials, Addison Wesley, USA.

Appendices

Historical development of braking and brake shoes

Shoe brakes were used extensively in the 18th and 19th centuries as part ofthe hand-operated lever brakes used on horse drawn vehicles. Thesebrakes were essentially used as parking brakes to hold the vehiclestationary. The main braking system was provided by the horse (orbullock) slowing down and then stopping. The shoe brake could beconsidered as an auxiliary brake that assisted the braking operation.

These brakes were still in use on delivery carts during the 1940s and maystill be seen in carriages and horse-drawn sulkies at shows.

The externally applied brake shoe used initially was made from wood andoperated by applying external pressure to the wrought iron rim of thevehicles’ wheel. Wood proved inadequate as vehicles became faster andheavier, so liners were introduced.

Mild steel shoes were introduced in the late 19th century, having woodattachments and leather liners.

When pneumatic tyres were patented in 1888, the use of external shoebrakes became limited. Various brake systems were used, including bandand cable brakes. In 1902, Louis Renault introduced the drum brake,incorporating internal brake shoes.

The mechanical design of the drum brake systems has varied anddeveloped during the past century. The basis design of the brake shoehas not altered except for the required shape designed for individualvehicles.

The internal brake shoe consists basically of two parts, the curved plateand the drilled web. The material used is mild steel, and the manufactureinvolves the stamping out of the two shapes, bending of the plate into therequired curved shape, then welding the two components together.

External brake shoes are still in use on railway carriages. These involvethe use of medium carbon steel or gray cast iron shoes acting directlyonto spheroidal graphite cast iron wheels.

Braking systems 23

Exercises

Exercise 5.1

Select a braking system/component and complete an engineering reportstructured under the headings used in the sample report.

You may obtain a component or components from a garage, awrecking 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 graphicspackage to aide in the generation of your engineering report.

24 Part 5 – Engineering report

Braking systems 25

Progress check

In this part you completed an engineering report.





Agr

ee

Dis

agre

e

Unc

erta

in

I have learnt about

• engineering report writing

• communication

– research methods including Internet.

I have learnt to

• complete an engineering report based on the analysisof one type of brake or component of braking system,integrating computer software

• conduct research using appropriate computertechnologies.



Congratulations! You have now completed Braking systems.

Braking systems 27


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.

If you study Stage 6 Engineering Studies through a Distance EducationCentre/School (DEC) you will need to return the exercise pages with yourresponses.

Return the exercise pages with the Title Page cover attached. Do not return allthe notes, they should be filed for future reference.

If you study Stage 6 Engineering Studies through the OTEN Open LearningProgram (OLP) refer to the Learner’s Guide to determine which exercises youneed to return to your teacher along with the Mark Record Slip.

Please complete and return the module evaluation that follows.

29

Bibliography

Board of Studies, 1999, The New Higher School Certificate AssessmentSupport Document, Board of Studies NSW, Sydney.

Board of Studies, 1999, Stage 6 Engineering Studies Examination, Assessmentand Reporting, Board of Studies NSW, Sydney.

Board of Studies, 1999, Stage 6 Engineering Studies Support Document, Boardof 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 HSCExamination 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 ofMaterials, 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.

31

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