The big idea: SF = ultimate load (or stress) ÷ allowable load (or stress).
Rearranged, allowable = ultimate ÷ SF — which is how a rating is set. The margin covers material variation, manufacturing defects, overloading, degradation over time, and the assumptions in the calculation.
The margin at SF 1, 2 and 4 — then everything that lives inside it.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
From a load to a section
Find the maximum intended load
Not the average user: the heaviest plausible one, plus whatever they carry, plus any dynamic peak. A step-ladder is rated for a 150 kg user, not a 70 kg one.
Choose the safety factor
From the consequence of failure and the certainty of the loads. Furniture 1.5-2, general structures 2-3, lifting gear and play equipment 4-6.
Work out the required ultimate load
Ultimate = allowable × SF. A 1,500 N working load at SF 4 means the part must not fail below 6,000 N.
Size the section for that
Required area = required load ÷ the material's ultimate stress. Then round UP to a standard size, because a calculated section is a minimum.
A worked sizing: A steel tie must carry 2,000 N in service with a safety factor of 3. The steel's ultimate tensile stress is 400 MPa.
Ultimate load = 2,000 × 3 = 6,000 N. Required area = 6,000 ÷ 400 × 10⁶ = 15 × 10⁻⁶ m² = 15 mm². A 5 mm diameter rod is 19.6 mm², so specify that — rounding up, to a size that is actually stocked.
Feeling unprepared for exams?
Get a clear study plan, practice with real questions, and know exactly where you stand before exam day. No more guessing.
| Typical SF | Used when | Examples |
|---|---|---|
| 1.5 to 2 | Loads are well known, the material is consistent, and a failure is not dangerous | Furniture, brackets, product housings, general machine parts |
| 2 to 3 | Loads vary, or the material is less predictable | Structural timber, general steelwork, vehicle components |
| 4 to 6 | A failure injures somebody, loads are uncertain, or the part cannot be inspected | Lifting gear, ropes and slings, play equipment, pressure vessels, lifts |
| 8 upwards | The consequence is fatal and the loads are genuinely unpredictable | Climbing ropes, some crane and rigging components |
A bigger factor is not automatically better: Every increase means more material, more mass, more cost and more embodied impact — and sometimes a product that is worse in use because it is heavy.
A large factor can also hide sloppy analysis. It is often cheaper and safer to reduce the uncertainty, by testing and inspecting, than to cover it with material.
How this is tested — calculating safety factors and maximum intended loads, and designing with one. It comes up two ways:
Paper 1 — multiple choice
- Calculate a safety factor from two loads.
- Find an allowable load from an ultimate load and a factor.
Paper 2 — analysing a product
- Size a member for a stated load and safety factor.
- Justify the safety factor chosen for a named product.
The trap: Multiplying where you should divide. Allowable = ultimate ÷ SF, and the required ultimate = allowable × SF. Write the formula down before substituting.
A playground swing seat hangs from two chains. The heaviest intended user is 100 kg, the swing is used dynamically, and the chain's ultimate tensile load is 12,000 N per chain. Apply a suitable safety factor and check the design.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.