Section 1: Material Categories
Engineers must choose the most appropriate material for every part of a design. The choice depends on what the part needs to do, what forces it will experience, and the environment it will operate in. Materials are grouped into four main categories:
- Steel — structural beams, car bodies, tools
- Aluminium — aircraft, packaging, bike frames
- Copper — electrical wiring, plumbing
- Titanium — aerospace, medical implants
- Polyethylene (PE) — plastic bags, bottles
- Polypropylene (PP) — food containers, rope
- Nylon — gears, bearings, textiles
- Acrylic — signs, windows, displays
- Glass — windows, lenses, optical fibre
- Concrete — foundations, roads, bridges
- Porcelain — electrical insulators, tiles
- Silicon carbide — cutting tools, brake discs
- Carbon fibre reinforced polymer (CFRP) — aircraft, F1 cars, bikes
- Glass fibre (GRP/fibreglass) — boat hulls, helmets
- Reinforced concrete — buildings, bridges
Section 2: Material Properties
To choose the right material, engineers consider a range of mechanical, physical and other properties. Understanding these properties allows you to justify a material choice in the exam.
| Property | Definition | Example where important |
|---|---|---|
| Strength | Ability to withstand a force without breaking | Structural beams, lifting cables |
| Hardness | Resistance to scratching or indentation | Cutting tools, drill bits, gear teeth |
| Toughness | Ability to absorb impact energy without fracturing | Helmets, crash barriers, tool handles |
| Ductility | Ability to be drawn into wires or bent without breaking | Electrical wiring, bent brackets |
| Malleability | Ability to be hammered or pressed into shape | Aluminium car body panels, coins |
| Stiffness (rigidity) | Resistance to bending or deformation under load | Machine frames, structural columns |
| Elasticity | Ability to return to original shape after a force is removed | Springs, rubber seals |
| Density | Mass per unit volume — how heavy the material is | Aircraft components — low density preferred |
| Thermal conductivity | How well the material conducts heat | Heat sinks, cooking pans, insulators |
| Electrical conductivity | How well the material conducts electricity | Wiring (conductors), casings (insulators) |
| Corrosion resistance | Resistance to chemical attack or rusting | Outdoor structures, marine applications |
| Cost | Material and processing cost per unit | Always a factor in real engineering decisions |
Section 3: Selecting a Material
In the SQA exam you will be given a scenario and asked to select a suitable material and justify your choice. A good justification always links the properties of the material directly to the demands of the application.
How to approach a material selection question
- Read the scenario carefully — identify what forces or conditions the part must withstand
- List the key requirements (e.g. must be lightweight, must conduct electricity, must resist corrosion)
- Match a material to those requirements
- Justify each property you mention — say why it matters for this specific application
Question: A bicycle frame must be lightweight and strong. Select a suitable material and justify your choice.
Aluminium alloy would be a suitable material. It has a high strength-to-weight ratio, meaning the frame can withstand the forces from the rider without being unnecessarily heavy. Aluminium also has good corrosion resistance, which is important as bicycle frames are used outdoors in wet conditions. Its ductility allows it to be formed into the required tube shapes during manufacture.
Question: Select a suitable material for the plastic insulation surrounding electrical cables and justify your choice.
PVC (polyvinyl chloride) is commonly used. It is an electrical insulator, which prevents the user from receiving an electric shock by stopping current flowing through the outer casing. It is also flexible, so the cable can be bent without cracking, and it is low cost making it economical to produce in large quantities.
Question: A suspension bridge cable must carry very large tensile forces and operate outdoors for many years. Select a suitable material and justify your choice.
High-tensile steel wire is used. Steel has very high tensile strength, meaning it can carry the enormous pulling forces in the cables without breaking. It is also stiff, which limits the amount of sag in the bridge. Galvanised or stainless steel variants are used to provide corrosion resistance for long outdoor service life.
Section 4: Stress
When a force is applied to a structural member, the material inside it is being squeezed or stretched. The intensity of this internal force — the force per unit area — is called stress. Stress is what determines whether a material will fail: if the stress in a member exceeds the material's strength, it will break.
Rearrangements: Force = Stress × Area | Area = Force ÷ Stress
The cross-sectional area is the area of the cut face of the member — perpendicular to the direction of the force. A smaller cross-section experiences greater stress for the same force. This is why load-bearing columns and cables are sized carefully.
Calculating cross-sectional areas
The cross-section shape determines how you calculate the area:
- Rectangle: A = width × height (mm × mm = mm²)
- Circle: A = πr² = π(d/2)² — where d is the diameter in mm
- Square: A = side × side
A square bar of 20 mm × 20 mm cross-section is subjected to a tensile load of 500 N. Calculate the stress in the bar.
A column of cross-section 0.25 m² has a maximum allowable stress of 50 N/mm². Calculate the maximum compressive load.
A steel wire supports a load of 8 kN. The stress must not exceed 200 N/mm². Find the minimum wire diameter required.
d² = 4A ÷ π = (4 × 40) ÷ π = 50.93
d = √50.93 = 7.14 mm
Show all working. Check units carefully — forces in N, areas in mm², stress in N/mm².
Section 5: Strain
Every material deforms slightly when a force is applied to it — it stretches under tension or shortens under compression. This change in length, expressed as a proportion of the original length, is called strain. Strain tells us how much a material has deformed relative to its size.
Both ΔL and L must be in the same units before dividing.
Rearrangements: ΔL = ε × L | L = ΔL ÷ ε
Tensile and compressive strain
If a member is under tension (being stretched), ΔL is an increase in length — the strain is positive. If a member is under compression (being shortened), ΔL is a decrease in length — the strain is sometimes described as compressive.
A steel wire of length 5 m stretches 2.5 mm when a load is applied. Calculate the strain.
The allowable strain on a concrete column must not exceed 5 × 10−4. The column is 3 m tall. Find the maximum reduction in height.
A mass of 2 500 kg is hung from a vertical steel bar. The bar has a rectangular cross-section of 75 mm × 50 mm and an original length of 0.5 m. The bar stretches by 2.5 mm. Calculate both the stress and the strain. (Use g = 9.81 m/s².)
ε = 2.5 ÷ 500 = 0.005
Show all working. Remember to convert both lengths to the same unit before dividing. State clearly that strain has no units.