© D.J.Dunn www.freestudy.co.uk 1

UNIT 21: MATERIALS ENGINEERING

Unit code: F/601/1626

QCF level: 4

Credit value: 15

LEARNING OUTCOME 3

TUTORIAL 1

On successful completion of this unit a learner will:

3 Be able to select suitable materials and processing methods for a specific product

Design constraints: e.g. working conditions such as applied forces, environment electrical/magnetic

requirements, shape, form and function of the product

Materials, properties and processing: inter-relationship between product design, material selection and

processing methods; merit index/index of suitability; ability to be re-used

Processing limitations: effects of the manufacturing processing capabilities on the structure of materials

and preventing or facilitating product design, effect on environment (such as sustainability, emissions,

energy conservation)

MATERIALS AND PROCESS SELECTION

CONTENTS

1. INTRODUCTION

2. PRODUCT ANALYSIS

3. DESIGN CONSTRAINTS

Properties

Processability

Performance Index

Materials Selection Chart

4. COST RESTRAINTS

Cost Performance Index

Material cost chart

Raw Material

Storage

Availability

Quantity

Form of supply

5. DATA SOURCES

Textbooks

Data books

Manufacturer’s literature

Internet

Standards

Data Base

6. CHOICE OF PROCESS

Material removal

No Material removal

© D.J.Dunn www.freestudy.co.uk 2

1. INTRODUCTION

The properties of materials and their processing methods have already been covered in previous tutorials.

The purpose of this tutorial is to get you to use the knowledge to choose appropriate materials and

manufacturing processes for specific cases. As there are so many examples you could study and so many

different areas of engineering and technology it is impractical to think that this tutorial could teach everyone

examples appropriate to their particular field of study. The designer of any product must get involved with

material selection. Only occasionally will the exact grade of material be specified by the customer. Even

then the designer must understand the material to be able to design the product. Often this will be decided in

consultation with the intended manufacturer. There are so many materials and so much information

available to a designer that it is quite bewildering. The question is how to decide? In the process of deciding

you will find there are constraints that are vital and produce a narrow selection (rigid requirements). Other

constraints may not be so vital (soft requirements) and can be left to the final stages of selection. One way to

summarise the process is like this.

PRODUCT ANALYSIS CONSTRAINT PERFORMANCE INDEX

What is its function? Identify specific constraints. Maximise or minimise

properties. A component has more than one

function such as strength to

carry the loads acting on it,

hardness to resist wear, conduct

heat, look good and so on.

The component may be constrained to carry

a maximum force, work within a certain

range of temperatures, function in certain

environments and so on. The manufacturing

process may be a constraint.

You might want to minimise cost,

minimise weight, maximise

strength, hardness, conductivity,

magnetic strength and so on.

2. PRODUCT ANALYSIS

It is usual to start by just analysing the product. Here are some of the decisions you might have to make.

Is it performance or cost? Perhaps the first thing to ask is if the product is performance driven or cost

driven? This makes a huge difference when choosing materials (e.g. cheap steel for mass production,

carbon fibre for top of the range). Is the product intended to be a small production quality item or a

mass production regular item?

What must the product do? (Function) In many cases like cars, bicycles, TV sets and so on this is

obvious. In other cases it might not be obvious and if you are asked to design a specific part without

knowing what it is for you should find out. It might be part of a more complex system and its best to

understand its function in that system.

What is the intended lifespan of the product? Obviously some materials will last longer than others.

What level of reliability is required? A material may be more reliable than others if chosen so that a

system needs no or little maintenance thus eliminating human failing in keeping a system going.

Where will the product be used? The environment must be understood so that the product does not fail

prematurely.

Who uses it? This has safety implications such as being used by children who may abuse the system.

What should it cost? You need to have an idea what the customer is expecting to pay.

What should it look like (colours etc.)? Clearly the finishing process may have many purposes and

visual appearance is one of them.

Does ergonomics feature? (e.g. comfort, ease of use of product)

© D.J.Dunn www.freestudy.co.uk 3

3. DESIGN CONSTRAINTS

Design engineers have a very large say in the choice of materials. The material selection will depend mainly

on the constraints imposed by function and performance expectations. Some of the things they should

consider in the design process are given here.

Properties

Mechanical properties

Strength - The component/structure/product must not fail under the action of expected stresses and forces during its intended life span. Strength can mean many things, tensile, compression,

shear or torsional. The strength can be weakened due to service factors such as stress corrosion

and fatigue so there are many things to consider.

Elasticity - The elasticity of a component depends on its modulus and explanations of Elastic,

Shear and Bulk modulus along with the relationship with Poisson’s ratio will be found in other

modules that you should be studying.

Degradation - The material must not become degraded due to service or environmental factors. This will reduce its intended life span. There are many things that cause a material to degrade

such as corrosion, wear, chemical attack and radiation.

Wear - Wear is a form of degradation due to surfaces rubbing together. The designer needs to understand Tribology (friction and wear of rubbing surfaces). He must select materials with

suitable compatibility and wear resistance.

Impact resistance - This can be a form of degradation but also affects the strength. It occurs when the surface becomes damaged due to being struck. This could lead to fatigue failure or to

sudden cracking in brittle materials. It also affects the visual appearance and may be important

in house hold goods such as work tops and cooker surfaces.

Surface finish - The final treatment of manufactured parts is called the finishing process and

materials must be suitable for the process. This is conducted in order to do the following.

Protect the part from corrosion and other chemical attacks.

Produce enhanced physical surface properties.

Produce an attractive appearance.

Aesthetic – While a gear wheel needs not be aesthetic maybe the casing in which it will be placed need to be attractive in form, texture, colour etc. This will also be considered in any

finish used for protection purposes.

Here is a list of finishing processes that might be used.

Galvanising

Sheradising

Calorising

Chromising

Chromating

Phosphating

Metal Spraying

Cladding

Anodising

Electroplating

Plastic Coating

Paint Coating

Plasma coating

Surface peening

© D.J.Dunn www.freestudy.co.uk 4

Thermal and other physical properties

Melting point

Solidus

Latent heat of fusion

Thermal conductivity

Thermal expansion

Temperature coefficient of resistance

Brittle transition temperature

Glass Temperature

Maximum service temperature

Melt flow

Processing Temperature

Vicat softening Temperature

Emissivity

Reflection Coefficient

Refractive Index

Water Absorption

Oxygen Index

Solubility

Electrical and Magnetic

Resistivity/conductivity

Temperature coefficient of

resistance

Permittivity

Permeability

Break down voltage

Hysteresis characteristic

Processability The materials to be used must be suited to the intended process of manufacture and manufacturing

properties may be a deciding factor such as:

Machineability

Weldability

Arc Resistance

Ability to be hot and cold

rolled

Ability to be drawn

Ability to be forging

Ability to be Extruded

Ability to be Cast

Mould Shrinkage

Melt flow rate

Performance Index

This is a way to choose a material based on the largest or smallest value of one or more of its properties. For

example we might be looking for a material with maximum strength, maximum stiffness, maximum

hardness, maximum thermal conductivity and so on. Further on you are introduced to property charts and

really it’s a case of picking the material with the right combination of properties. There are many learned

papers about “performance index” and its use but in the author’s opinion, this only complicates an otherwise

simple process. Never-the-less it is covered in the following work which mainly applies to mechanical

engineering but the same principles work for other areas of study.

If we wanted a material that was as stiff as possible we would simply choose one with the largest modulus E

for direct stress or G for torsion. If we wanted a material that was as strong as possible we choose one with

the maximum stress criterion σ for direct stress or τ for torsion. If we wanted a material with maximum heat

conduction we look for the maximum conductivity and so on. Often we need to combine more than one

property. For example you might want the stiffest and lightest material for a beam or strut. In this case we

could use a performance index defined simply as P = E/ρ or P = G/ρ where ρ is the density. The maximum

value of P is the material that is stiffest and lightest or strongest and lightest.

Whilst there is nothing wrong with this method, for reasons that are not clear many articles on the subject

use a performance index derived as shown in the following work. It is usual to define the performance index

as P = K/m. K is a constant that results from the applied forces and dimensions and m is the mass. If we use

P = K/m then the smallest mass yields the largest P and it is this definition used here.

Let’s start by deriving the one based on strength for a tie or a strut with a uniform cross section.

© D.J.Dunn www.freestudy.co.uk 5

Strength per unit mass for a simple strut or tie

If a material is subjected to an applied force F the maximum allowable direct stress is σ = Force/ Area = F/A

For maximum strength we need to know the maximum allowable stress either in tension (tie) or

compression (strut). A = F/σ

The mass : m = Volume x density m = ρAL L is the length and ρ is the density.

Substitute for A m = FLρ/σ

F and L are fixed properties regardless of the material FL = K m = K (ρ/σ)

ρ/σ is the value that varies with material based on strength and density.

P = σ/ρ

If we have values of P for different materials we can see that the largest values are the strongest and lightest

and the smallest values are the weakest and heaviest.

Stiffness per unit mass for simple strut or tie

For the same situation as before the stiffness (elasticity) depends on the modulus of elasticity E.

E = σ/ = F/A is the strain in the material. Rearrange A= F/E

m = ρAL= ρL F/E = (FL/)(ρ/E) = K(ρ/E)

F, L and are the constraints that do not depend on the material.

K = (FL/) This will be the same for any material for the same loads and dimensions.

P = E/ρ

We can see that the largest values are the lightest and stiffest and the smallest values are the most elastic and

heaviest.

Strength per unit mass for simple torsion of a round shaft.

This applies to a uniform round shaft in torsion (being twisted).

If you have studied shear stress and torsion you will know that the strength is based on the ultimate shear

stress τ. The shaft twists an angle θ when a toque T is applied. The torsion formula is

32

πDJ

4

D is the diameter.

The mass is m = ρAL = ρLπD2/4

K is the constant part that is the same for all materials with the same loading and dimensions.

If we have values of P for different materials we can see that the largest values are the lightest and strongest

and the smallest values are the weakest and heaviest.

© D.J.Dunn www.freestudy.co.uk 6

Strength per unit mass for simple torsion of a round shaft.

This applies to a uniform round shaft in torsion as before. The torsional stiffness depends on the modulus of

rigidity G. The formula that that links it to the torque is

where θ is the angle twisted by the torque.

This is the same for materials with the same load and dimensions.

This tells us which materials are stiff and light or elastic and heavy.

Stiffness per unit mass for a beam in bending

A typical beam is shown. Due to a combination of loads and end

fixings the bending moment M may be found and the stress and

deflection calculated. There are various ways to arrive at the

performance index. This method assumes you understand beam

theory. The usual formula relating all the parameters is:

I is the second moment of area of the section depending on its

shape and dimensions. R is the radius of curvature at a point and relates the deflection and hence stiffness of

the beam. y is normally the dimension from the neutral (unstressed) layer to the extreme edge.

For stiffness we use

For a uniform round section I = πD4/64 and y = D/2 A = πD

2/4 so I = A

2/4π

The mass of a uniform length L is m = ρAL

K contains all the constants placed by the dimensions and forces. The performance index is

Strength per unit mass for a beam in bending

For strength we use

Many other cases could be examined but here is a list of the main performance indexes used in Mechanical

Engineering.

STIFFNESS P

Tie (tension) ρ/E

Strut (compression) ρ/E

Solid shaft in Torsion ρ /G½

Hollow Shaft in Torsion ρ/G1/3

Beam Bending ρ/E½

Flat Plate Bending ρ/E1/3

Pressurised hollow thin walled cylinder ρ/E

Pressurised hollow thin walled sphere

STRENGTH P

Tie (tension) ρ/σ

Strut (compression) ρ/σ

Shaft in Torsion ρ/τ2/3

Hollow Shaft in Torsion ρ/τ1/2

Beam Bending ρ/σ2/3

Flat Plate Bending σ1/2

/ρ

Pressurised hollow thin walled cylinder ρ /σ/ρ

Pressurised hollow thin walled sphere ρ /σ/ρ

© D.J.Dunn www.freestudy.co.uk 7

Using Performance Indexes

In the design stage the forces, stresses and deflections are calculated and so an index can be determined for

the particular case and property. In the cases examined previously we would need the values of Modulus,

design stress (e.g. yield stress in tension) and density of all possible materials and compile a list of the

performance factor. This is a bit impractical but software can be purchased with a comprehensive list of

properties. With such a list the materials need to be narrowed down. This can be done with property charts

covered further on in this tutorial. Here is a list of indexes for many materials. The values given are typical

or average and should not be used seriously for design purposes. The table was simply created with a

data base. This can be used to sort the contents in order to suit your search. Note the performance ratio

figure depends on the units used and these must be the same for all materials being compared.

Material property Performance Index

Material ρ kg/m3 σ MPa E GPa

Tensile (strut) Bending (Beam)

σ/ρ E/ρ σ2/3

/ρ E1/2

/ρ

Alumina, ceramic 3800 150

345 0.0395 0.0908 0.0074 0.0049

Aluminium 2710 80 0.34 71 0.0295 0.0262 0.0069 0.0031

Aluminium strong alloy 2800 600 0.34 71 0.2143 0.0254 0.0254 0.0030

Brass (70Cu/30Zn) 8500 550 0.35 100 0.0647 0.0118 0.0079 0.0012

Carbon Fibre 1750 4300 0.2 181 2.4571 0.1034 0.1511 0.0077

Cast Iron 7000 1000 0.22 211 0.1429 0.0301 0.0143 0.0021

Concrete 2400 4 0.15 14 0.0017 0.0058 0.0010 0.0016

Copper 8930 150 0.35 117 0.0168 0.0131 0.0032 0.0012

Diamond (approx) 3300 150000 0.2 1200 45.4545 0.3636 0.8555 0.0105

Epoxy resin 1120 50 0.37 4.5 0.0446 0.0040 0.0121 0.0019

Fluon (PTFE) 2200 22 0.46 0.34 0.0100 0.0002 0.0036 0.0003

Glass (crown) 2600 100 0.21 71 0.0385 0.0273 0.0083 0.0032

Glass Fibre 2600 3400 0.23 42 1.3077 0.0162 0.0870 0.0025

Iron, cast grey 7156 100 0.27 110 0.0140 0.0154 0.0030 0.0015

Lead 11340 15 0.44 18 0.0013 0.0016 0.0005 0.0004

Magnesium 1740 190 0.29 44 0.1092 0.0253 0.0190 0.0038

Melamine Formaldehyde 1500 70 0.23 9 0.0467 0.0060 0.0113 0.0020

Nickel 8900 300 0.36 207 0.0337 0.0233 0.0050 0.0016

Nickel Iron Alloy 8900 490 0.38 200 0.0551 0.0225 0.0070 0.0016

Nickel, strong alloy 8500 1300 0.38 110 0.1529 0.0129 0.0140 0.0012

Nylon (6) 1150 70 0.35 3 0.0609 0.0026 0.0148 0.0015

Perspex 1190 50 0.39 3 0.0420 0.0025 0.0114 0.0015

Phenol formaldehyde 1300 50 0.46 6.9 0.0385 0.0053 0.0104 0.0020

Polyethylene (high den) 955 260 0.46 0.43 0.2723 0.0005 0.0427 0.0007

Polyethylene (low den) 920 13 0.46 0.18 0.0141 0.0002 0.0060 0.0005

Polypropylene 900 35 0.45 1.2 0.0389 0.0013 0.0119 0.0012

Polystyrene 1050 50 0.35 3.1 0.0476 0.0030 0.0129 0.0017

PVC (non-rigid) 1250 15 0.35 0.01 0.0120 0.0000 0.0049 0.0001

PVC (rigid) 1700 60 0.35 2.8 0.0353 0.0016 0.0090 0.0010

Rubber (polyisoprene) 910 17 0.48 0.02 0.0187 0.0000 0.0073 0.0002

Silicon Carbide 3200 3340 0.19 450 1.0438 0.1406 0.0698 0.0066

Spruce (with grain) 600 84

14 0.1400 0.0233 0.0320 0.0062

Stainless Steel (18Cr/8Ni) 7930 600 0.29 210 0.0757 0.0265 0.0090 0.0018

Steel, mild 7860 460 0.29 210 0.0585 0.0267 0.0076 0.0018

Titanium Alloy (Grade 5) 4420 1000 0.36 110 0.2262 0.0249 0.0226 0.0024

Titanium carbide 4.9 258 0.188 300 52.6531 61.2245 8.2709 3.5348

Tungstun Carbide 15.8 334.8 0.24 534.4 21.1899 33.8228 3.0516 1.4631

Wood, oak (with grain) 650 60

12 0.0923 0.0185 0.0236 0.0053

Zinc 7140 150 0.25 110 0.0210 0.0154 0.0040 0.0015

© D.J.Dunn www.freestudy.co.uk 8

Materials selection charts

You will find an interactive tutorial at this link.

You can download a useful tutorial on this link.

You can download some worked examples on this link.

This link is another full tutorial on Ashby’s methods.

This link allows you to download a range of charts.

Performance indicators and selection charts are the result of work done by Ashby. For example if we plotted

Modulus E against density ρ we would get a scatter chart with a point on it for every material. It is then

possible to draw an envelope around the points for all similar materials (metals, polymers and so on) and the

result is as shown. We could do the same for other properties (e.g. σ and ρ). These are simple to use but

should only be used to make a rough selection. We can see immediately which class of materials has the

two values that you require. The usefulness can be increased by plotting lines representing a given value of

performance factor P so it is then possible to see which materials meet the performance criteria. As these

charts are copyrighted they cannot be shown fully here but click on the thumbnail chart to download them

free.

Software can be purchased to create these charts at this link.

Click on the chart to download the full size version.

MODULUS – DENSITY CHART

TENSILE STRENGTH – DENSITY CHART

FRACTURE TOUGHNESS – MODULUS CHART

THERMAL EXPANSION – THERMAL CONDUCTIVITY CHART

Consider the Modulus - Density Chart. Click on it to see it. This has a logarithmic scale in order to cope

with a range from very small to very large. If we want a light and stiff material we need to choose materials

near the top left corner of the chart and composites look a possibility. If we want a heavy elastic material we

need the region in the lower right corner and this looks like rubber. There are charts for many combinations

of properties, e.g. 'strength - toughness' and 'electrical resistivity - cost'.

If we plotted lines of constant P on any of these charts they would be straight lines. For example if take the

case of P = ρ/E½

Rearrange and E = ρ2/ P

2 and taking logs log E = 2 log ρ – 2 log P and this is a straight line on

the chart with gradient 2.

© D.J.Dunn www.freestudy.co.uk 9

4. COST RESTRAINTS

You may be restricted in choice by the cost you have to achieve. There are some tools that link cost to

materials in the form of an index. The cost of the product depends on the design and manufacture

(processing) as covered in the preceding work.

Cost Index

Various cost indexes have been devised to assist in selection. These are basically the same as the

performance index but modified as follows. P was defined as P = K/m so m = P/K

The material cost is c per unit mass then the cost is c M = c P/K

Because material prices fluctuate usually the cost is based on relative cost to a reference material

such as steel plate. Relative cost is cr = c/cref

Because performance indexes are selected for maximum values and cost indexes for minimum

values we define the cost index as C = cr /P The cheapest material has the smallest value of C.

Consider the one based on strength in tension (Tie). P = σ/ρ Cost Index = crρ/σ

Consider the one based on stiffness in tension (Tie). P = E/ρ Cost Index = crρ/E

Cost chart

These are similar to the performance indicator charts but based on cost instead of density. Click on

the thumbnails below to see the various charts on www-materials.eng.cam.ac.uk Click on the chart to download the full size version.

STRENGTH – COST CHART

ALL MATERIALS

METALS AND ALLOYS

CERAMICS

POLYMERS

WOOD PRODUCTS

COMPOSITES

© D.J.Dunn www.freestudy.co.uk 10

Raw material

When more than one material meets the required specifications, the cheapest material would be

logically chosen and these depend largely on the price of the raw material. For example when copper

is expensive, there is a tendency to make electrical conductors from aluminium even though the

cable diameter has to be increased to meet the resistance criteria.

Storage

The material to be used and the end product may have to be stored and transported so the material

must not degrade whilst in storage. Steel stock, for example should not be stored in the open where

rain will accelerate rust. If a supplier can reliably supply stock quickly, then you need not bear the

cost of storage.

Availability

Again if more than one material meets the required specification, the final choice of material may

depend on the availability and the one most readily available would be chosen.

Quantity The price of materials may well depend on contractual arrangements with discount for quantities and

regularity of orders. The choice of material, all other being equal, is not so important for small

quantities. The manufacturing cost is reduced by large scale production so an investment in mass

production moulds and dies would be worth while. For small quantities, modern machine tools can

produce items fairly cheaply especially if the component design has been produced in a form that

can be downloaded direct to the machine.

Forms of raw material supply The choice of material may depend on the form of supply. The manufacturing process governs this

to a large extent. For example a cylindrical component might be made from stock tube or made from

flat sheet. The latter is cheaper but extra costs are involved in forming it. The material may be

supplied in a form close to required finished product so cost can be saved in the manufacture.

Standard extrusions and standard sections may be cheaper to buy and modify than fabricating or

machining the shape from cheaper forms such as bar stock. Other forms of material might include

ingots, castings and mouldings, forgings and pressings, granules and powders and liquids.

Here are some of the many forms of material

Plate Coil Bar Ingots

Tube and Hollow sections Extruded Sections Polymer Granules

© D.J.Dunn www.freestudy.co.uk 11

WORKED EXAMPLE 1

A designer has determined that the material for a certain component should have a tensile strength of at least

100 MPa and should be as light as possible. It should also be tough. What are the possibilities from the

chart? Using the cost chart, determine the cheapest raw material available.

SOLUTION

The chart indicates that a small chance of using parallel grain wood or polymer but composites and

magnesium alloy are also very possible. We can rule out stone because it is strong in compression and weak

and brittle in tension. On the cost chart it seems that metals and alloys are the cheaper but you would need a

more detailed chart to see if magnesium alloy is in this range because the alloy region spreads from low to

very high cost.

© D.J.Dunn www.freestudy.co.uk 12

WORKED EXAMPLE 2

A solid circular section beam is put into bending. From the list of materials given below which is the

strongest for the same size beam?

Given the relative material cost shown below which material would be the cheapest?

Material cr

Carbon fibre composite 80

Glass fibre composite 40

Aluminium alloy 15

Titanium alloy 110

Mild Steel 5

SOLUTION

Select the properties from the table given earlier but note more realistically you would use a precise material

specification and manufacturer’s data. The data below is typical but can vary widely. Select on the basis of

the strongest possible per unit mass.

Material Density ρ Max tensile stress σ cr

Carbon fibre composite 1750 kg/m3

4 300 MPa 80

Glass fibre composite 2600 kg/m3 3 400 MPa 40

Aluminium alloy 2800 kg/m3

600 MPa 15

Titanium alloy 4420 kg/m3

1 000 MPa 110

Mild Steel 7860 kg/m3

460 MPa 5

Calculate the performance index for simple bending.

Material P = σ2/3

/ρ cr C = cr/P

Carbon fibre composite 0.155 80 516

Glass fibre composite 0.087 40 458

Aluminium alloy 0.0254 15 472

Titanium alloy 0.0226 110 4867

Mild Steel 0.0076 5 658

For maximised strength Carbon Fibre is the best. (Remember this is based on the same dimensions and load

for each material).

The cheapest material will be Glass fibre composite.

© D.J.Dunn www.freestudy.co.uk 13

5. DATA SOURCES

The main sources of data are textbooks, data books, internet, manufacturer’s literature and appropriate

standards. These days, all this is available in electronic form to be accessed on you computer.

TEXT BOOKS

Good for general information and always on hand to refer to. Some have tables of properties but these are

not usually detailed and only general.

DATA BOOKS

These are one of the quickest sources of detailed information and often contain grades and specifications as

well as properties. Often pocket sized data books are obtained from manufacturers and professional

institutes and provide a handy source of data.

MANUFACTURERS AND STOCKISTS and DATA BASES

These vary in quality and usefulness but can be very useful. They are likely only to have their own materials

in it and so cannot be easily used to compare materials. They are good for final selection before ordering.

INTERNET

The internet is a handy source of information but the information available is vast and often hard to narrow

down as a lot of useless information comes back in the search. The information is sometimes wrong and

data should be used with care.

Finding actual properties on the web is difficult. Here are some you might find useful.

Mechanical data -

Steel stock at Corus -

Properties of a range of copper, nickel and beryllium alloys -

Aluminium alloy properties -

Plastics and information about polymers -

Comprehensive materials data base -

Search tool for polymers -

Material search including compatibility -

Data base for noble metals -

General material properties -

Useful information on Aluminium-

Downloads on properties of metals -

Properties and calculator tools for material failures -

International standards and materials –

DATA BASES

These are probably the best way to check out a material. A data base might be on the internet (e.g. Matweb

www.matweb.com) or it may be purchased and stored on your computer. Usually a purchased data base is

updated regularly for a subscription and it may even be linked into a design package used to construct your

2D or 3D model.

This link has a powerful software package that runs on your desktop PC. It combines information on

materials and process properties with a series of powerful tools for browsing and searching that information,

and for using it to compare and select materials.

© D.J.Dunn www.freestudy.co.uk 14

STANDARDS ORGANISATIONS

A designer should always adhere to standards set down for the use, safety, legal requirements etc. by

various organisations, they contain much useful information. Here are some of them.

British standards specifications (BS) - This specifies sizes and properties to which manufacturers

should conform. For example BS4 gives the dimensions of standard rolled steel universal beams.

These standards are available for purchase.

International standards organisation (ISO) - Most British Standards now conform to ISO and

when viewing BS the equivalent ISO is given.

American national standards institution (ANSI) - As the voice of the U.S. standards and

conformity assessment system, the American National Standards Institute (ANSI)

empowers its members and constituents to strengthen the U.S. marketplace position in the

global economy while helping to assure the safety and health of consumers and the

protection of the environment.

American society for testing and materials (ASTM) - ASTM International, formerly known as the

American Society for Testing and Materials (ASTM), is a globally recognized leader in the

development and delivery of international voluntary consensus standards. Today, some 12,000

ASTM standards are used around the world to improve product quality, enhance safety, facilitate

market access and trade, and build consumer confidence.

6. CHOICE OF PROCESS

Although we usually choose materials first, sometimes it is the shape and process which is the limiting

factor. For each component we need to decide how it will be made. This is a key question which has a

massive influence on materials selection.

The cost of producing the component is always a big factor and the designer must consider how the

component/structure/product is manufactured. To a large extent, the manufacturing process is governed by

the material but the material is also influenced by the method of manufacture. The shape, size and quantities

of the component are a major factor governing the manufacturing process.

The mechanical properties of the finished component are affected by the manufacturing method. For

example forging a crankshaft is better than turning one on a lathe because it produces a grain flow that

makes it stronger and more resistant to fatigue failure. Grinding and polishing also produces better fatigue

strength.

The tolerance on the finished size also governs the method. Casting and moulding does not produce a high

tolerance and generally material removal is the best way to produce an accurate size or fit. (e.g. grinding the

outer and inner ring of a bearing race). If a mass produced component with a high tolerance is to be made,

special machine tools such as broaches might be best.

Here is a list of manufacturing processes. It is not a complete list. The processes are described fully in

earlier tutorials.

MATERIAL REMOVAL WITHOUT REMOVING MATERIAL

Turning (lathes)

Milling

Drilling

Shaping

Broaching

Blanking

Electrical erosion

Chemical erosion

Grinding

Casting

Moulding

Forging

Drawing

Bending

Pressing

© D.J.Dunn www.freestudy.co.uk 15

WORKED EXAMPLE 3

Choose a suitable material for a telescopic antenna pole. The length and wall thickness

of the section is specified.

SOLUTION

Function

Each part is a hollow long and slender tube that must support a vertical compressive

load so it is strut and it will be prone to buckling. It will also have to survive outside

in a variety of atmospheric conditions.

Constraints The length is specified. It must not buckle or corrode.

We will need to minimise the mass and maximise the stiffness to prevent buckling.

Free Variables

Diameter and choice of materials

From the list of performance indicators for a strut we must maximise P = E½/ρ

The weight is minimized by selecting materials with the greatest value P

On the modulus - density chart a line of constant P = 0.00316 passes through the group of materials in the

top right corner where we have the stiffest and lightest materials.

Candidate materials include some ceramics, but rule these out as they are brittle. Other possibilities are

metal alloys or even carbon and glass composites.

Durability and Manufacture

Composites do not corrode or deteriorate in the atmosphere but are difficult to form into tubes cheaply.

Aluminium alloys are readily available in tubular form and have good resistance to degradation in the

atmosphere. Steel is prone to degradation and would need surface treatment. Other alloys are expensive.

Cost

Examining the cost charts available on the internet there is not a great deal to choose between these

materials but manufacturing would make aluminium alloy the most likely candidate.

© D.J.Dunn www.freestudy.co.uk 16

SELF ASSESSMENT EXERCISE No. 1

1. Go to the Website Matweb.com and search for the materials listed below. Print off a copy of what you

find. For each material find out the following:

The exact composition

The tensile strength

The yield strength

The elongation %

Materials to be found are as follows.

Aluminium 1050-0

Copper UNS C10200 soft

Grey cast iron.

Magnesium alloy AZ 31B-0

Aluminium Alloy 1201 20 mm Plate

For each material, find an example of what it is used to make.

2. The resistance of an electric conductor is Calculated by the formula R = ρeL/A where ρe is the electrical

resistivity.

The mass is given by M = ρd LA where ρd is the density.

A is the cross sectional area and L the length.

What is the ratio of the price per kg of aluminium and copper when used to make electric wires of the

same length and resistance? Show your strategy for working this out.

3. Find a suitable polymer material for making carbonated (fizzy) drink bottles. Decide the properties that

are needed and show how you arrive at the decision.

4. A large heat exchanger shell will contain radioactive hot carbon dioxide gas at a fairly high pressure.

Discuss the things to be considered in the material selection and method of manufacture.

5. A flywheel basically as shown in the diagram is to be used to for high energy density storage. Show

how to select a suitable material taking into account the energy stored and the possibility of the wheel

bursting under the stress.