Mechanisms & structures · National 5
Materials, stress & strain
3 cards · 3 quiz · 0 exam
Engineering Science · National 5 · mrkellydt.co.uk
What you're learning in this topic
Learning intention WALT
We are learning about materials, stress & strain, specifically to:
understand
- a material from a data table
be able to
- calculate direct stress and strain
Success criteria WILF
- I can choose and justify a material from a data table
- I can calculate direct stress and strain
Choosing materials
Select from a table of properties and justify with properties linked to the job: "Material C, because it is corrosion resistant, so it won't degrade in sea water, and has high strength, so it resists the forces on the mast." "Strong" alone doesn't score.
Stress and strain
- Direct stress σ = F ÷ A, in N/mm² (force in N, area in mm²). E.g. 26 000 N on a 710 mm² cable: σ = 37 N/mm².
- Strain ε = Δl ÷ l: change in length ÷ original length. It has no units. E.g. 0.00036 = Δl ÷ 120 gives Δl = 0.043 mm.
- A bigger force on the same area gives a bigger stress; a bigger area reduces stress.
Stress and strain worked examples
- Rectangular bar 180 × 120 mm with 500 N: A = 21 600 mm², σ = 500 ÷ 21 600 = 0.023 N/mm².
- Wire diameter to carry 8 kN at no more than 200 N/mm²: A = 8000 ÷ 200 = 40 mm²; d = √(4 × 40 ÷ π) = 7.1 mm (round up for safety).
- Strain: a 4.0 m wire stretches 2.5 mm: ε = 2.5 ÷ 4000 = 0.000 63 (no units). Put both lengths in the same units first.
- Change in length: ΔL = ε × L = 0.0075 × 2500 mm = 19 mm.
Design challenge: climbing wall anchor
The brief. An outdoor centre in Aviemore needs a steel tie rod that will carry 12 kN without the stress going above 150 N/mm².
- Calculate the minimum cross-sectional area.
- Calculate the minimum diameter of the rod.
- The rod is 400 mm long and stretches 0.3 mm. Calculate the strain.
- Justify the choice of material.
What a good answer includes
- A = F ÷ σ = 12 000 ÷ 150 = 80 mm².
- A = πd² ÷ 4 → d = √(4 × 80 ÷ π) = 10.1 mm, so use the next size up.
- ε = 0.3 ÷ 400 = 0.00075 (no units).
- Steel is strong in tension and tough; galvanise it to stop it corroding outdoors.
Key terms
| Stress | σ = F ÷ A (N/mm²) |
|---|---|
| Strain | ε = Δl ÷ l (no units) |
| Justifying a material | Name the property and link it to the job. |
Quick check
1. A 1152 N force on 36 mm² gives a stress of…
- 41 472 N/mm²
- 32 N/mm²
- 0.031 N/mm²
- 1188 N/mm²
Show answer
B: 32 N/mm². F ÷ A.
2. Strain has units of…
- N/mm²
- mm
- None
- N
Show answer
C: None. It's a ratio.
3. Which justification scores?
- It is strong, so it is a good choice for the boat
- It is low cost and easy for the builder to get hold of
- It is a good material that will last a long time
- It is corrosion resistant, so it won't degrade in sea water
Show answer
D: It is corrosion resistant, so it won't degrade in sea water. Property + link to job.
Printable worksheet
An I do, we do, you do worksheet for this topic, with a separate answer sheet.