Engineering Science
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🧰 Materials Science — Properties, Testing & Selection

Engineers choose materials by matching their mechanical properties to the demands of the application. A shaft that needs to resist fatigue requires a different material from a bracket that needs to be lightweight, or a cutting tool that needs hardness.

σ = F ÷ A ε = ΔL ÷ L E = σ ÷ ε UTSHardnessToughness
2Topics
8Examples
4Calculators
12Practice Qs
🧰 Level 3 context: Stress and strain are the two most fundamental quantities in solid mechanics. Young's modulus tells you how stiff a material is — not how strong. These three quantities underpin every structural and mechanical design decision.

🧰Properties & Stressσ, ε, E, UTS
🔬Testing & SelectionTensile, hardness, impact
⚙️Applications4 materials calculators
🎮Game12 randomised questions
📝Practice12 graded questions
🧰

Mechanical Properties & Stress–Strain

Stress is force per unit area; strain is the fractional change in length. Their ratio gives Young's modulus — the stiffness of a material, independent of its size.

σ = F ÷ Aε = ΔL ÷ L₀E = σ ÷ εPaGPa

📜 Key rules

Stress: σ = F ÷ A — force (N) divided by cross-sectional area (m²). Unit: Pa (or N/m²; engineering: MPa, GPa).
Strain: ε = ΔL ÷ L₀ — extension divided by original length. Dimensionless (no unit).
Young's modulus: E = σ ÷ ε — stiffness of the material. Unit: Pa (GPa for metals).
Extension from E: ΔL = FL ÷ (AE) — rearranged from the above three equations combined.
1 MPa = 10&sup6; Pa  |  1 GPa = 10&sup9; Pa  |  Steel E ≈ 200 GPa  |  Aluminium E ≈ 70 GPa.
🎮Live explorer — stress, strain & Young's modulusDrag sliders — diagram updates
Stress σ
100 MPa
σ = F ÷ A
Strain ε (steel E=200 GPa)
0.000500
ε = σ ÷ E
Extension ΔL
0.50 mm
ΔL = ε × L₀
F F A = 500 mm² | L₀ = 1000 mm σ = 100 MPa | ε = 0.000500 | ΔL = 0.50 mm
💡 Stress depends on force and area. Strain depends on stress and material stiffness (E). Extension = strain × original length.
📋Try it — Stress, Strain & E CalculatorUpdates as you type
Select what to find and enter values above.

📚 Key vocabulary

Stress (σ)
force per unit cross-sectional area. Unit: Pa (N/m²). Engineering uses MPa or N/mm².
Strain (ε)
fractional change in length — dimensionless. Always a very small number for metals in the elastic range.
Young's modulus (E)
stress divided by strain within the elastic region — a material constant. Unit: Pa (GPa for metals).
Elastic deformation
reversible deformation — the material returns to its original shape when the load is removed.
Plastic deformation
permanent deformation — occurs beyond the elastic limit (yield point) of the material.
Ultimate tensile strength (UTS)
the maximum stress a material can withstand before fracture. Unit: MPa.
👁️ Visual aid — stress–strain curve: elastic region, yield point, UTS and fracture for a typical steel
Strain ε Stress σ (MPa) Elastic region Yield point UTS Fracture Slope = E σ ÷ ε = E
✍️Worked example 1 — stress in a steel bar  step through it
QuestionA steel bar of cross-sectional area 400 mm² carries a tensile load of 80 kN. Calculate the stress in MPa.
ConvertF = 80 kN = 80 000 N   A = 400 mm² = 400 × 10&sup6; m² = 4 × 10&sup4; m²
Formulaσ = F ÷ A
Substituteσ = 80 000 ÷ (400 × 10&sup6;)   or   80 000 N ÷ 0.0004 m²
Calculate80 000 ÷ 0.0004 = 200 000 000 Pa = 200 MPa
Answerσ = 200 MPa ✓
Step 1 of 6
✍️Worked example 2 — Young's modulus from test data
QuestionA 250 mm gauge-length specimen extends by 0.125 mm under a stress of 100 MPa. Calculate Young's modulus.
Strainε = ΔL ÷ L₀ = 0.125 ÷ 250 = 0.000 500
EE = σ ÷ ε = 100 × 10&sup6; ÷ 0.000 500 = 200 × 10&sup9;
AnswerE = 200 GPa (consistent with steel)
✍️Worked example 3 — extension of an aluminium rod
QuestionAn aluminium rod (E = 70 GPa), length 2.0 m, cross-section 600 mm², carries 42 kN. Find the extension.
FormulaΔL = FL ÷ (AE)
SubstituteΔL = (42 000 × 2.0) ÷ (600×10&sup6; × 70×10&sup9;)
SimplifyΔL = 84 000 ÷ (0.0006 × 70×10&sup9;) = 84 000 ÷ 42 000 000 = 0.002 m
AnswerΔL = 2.0 mm
✍️Worked example 4 — minimum area to limit stress
QuestionA steel tie bar must carry 120 kN without exceeding a stress of 150 MPa. Find the minimum cross-sectional area.
RearrangeA = F ÷ σ = 120 000 ÷ 150 × 10&sup6;
CalculateA = 120 000 ÷ 150 000 000 = 0.0008 m² = 800 mm²
AnswerA = 800 mm²
🔬

Testing Methods & Material Selection

Engineers test materials to measure properties before selecting them for a design. The key tests are tensile, hardness and impact — each measuring a different property.

Tensile testBrinell / VickersCharpy / IzodFOSFactor of safety

📜 Key rules

Factor of safety: FOS = UTS ÷ Working stress — always >1; typically 2–5 for structural engineering.
Tensile test gives: E (elastic region slope), yield strength, UTS, ductility (% elongation), toughness (area under curve).
Hardness tests: Brinell (HB) uses steel ball — good for castings. Vickers (HV) uses diamond pyramid — accurate across all materials.
Impact tests: Charpy & Izod measure toughness (energy absorbed before fracture). Unit: Joules.
Selection criteria: strength, stiffness, toughness, hardness, corrosion resistance, machinability, cost and density — balance for the application.
🎮Live explorer — factor of safetyDrag sliders to explore safe working stress
UTS
400 MPa
FOS
2.5
Safe working stress
160 MPa
UTS = 400 MPa Safe = 160 MPa (FOS 2.5)
💡 Higher FOS = safer but heavier/more expensive. FOS = 1 means working at the limit — never acceptable in practice.
📋Try it — FOS & Working Stress CalculatorUpdates as you type
Select what to find and enter values above.

📚 Key vocabulary

Factor of safety (FOS)
ratio of material strength to working stress — accounts for uncertainty, overloads and imperfections.
Toughness
energy absorbed before fracture — area under the stress–strain curve. Measured in J by impact tests.
Hardness
resistance to surface indentation. Brinell (HB) and Vickers (HV) are the standard engineering scales.
Ductility
ability to deform plastically before fracture — expressed as percentage elongation or reduction in area.
Brittleness
fracture with little or no plastic deformation — high strength but low toughness (e.g. cast iron, glass).
Fatigue
failure under repeated cyclic loading at stresses below UTS — the most common cause of engineering failures.
👁️ Visual aid — tensile test: specimen gripped in jaws, load applied, extensometer measures ΔL live
Gauge length specimen Tensile load F ↓ Reaction R ↑ Ext. From test data: σ = F/A   ε = ΔL/L₀ E = σ/ε = slope
✍️Worked example 5 — factor of safety  step through it
QuestionA steel component has UTS = 460 MPa. The design requires a factor of safety of 3.0. Calculate the maximum allowable working stress.
FormulaWorking stress = UTS ÷ FOS
SubstituteWorking stress = 460 ÷ 3.0
Divide460 ÷ 3.0 = 153.3
AnswerWorking stress = 153.3 MPa ✓
Step 1 of 5
✍️Worked example 6 — calculating FOS from measured stress
QuestionA component is under a working stress of 120 MPa. Its material has UTS = 480 MPa. Calculate the actual FOS.
FormulaFOS = UTS ÷ Working stress
SubstituteFOS = 480 ÷ 120 = 4
AnswerFOS = 4.0
✍️Worked example 7 — selecting material for a shaft
QuestionA shaft experiences a working stress of 200 MPa. FOS required = 2.5. What minimum UTS must the material have?
FormulaUTS = Working stress × FOS
SubstituteUTS = 200 × 2.5 = 500
AnswerMinimum UTS = 500 MPa — medium-carbon steel would be suitable.
✍️Worked example 8 — ductility and % elongation
QuestionA tensile test specimen has original gauge length 50 mm. After fracture, the gauge length is 64 mm. Calculate the percentage elongation.
Formula% elongation = (Lᵢ − L₀) ÷ L₀ × 100
Substitute= (64 − 50) ÷ 50 × 100 = 14 ÷ 50 × 100
Answer% elongation = 28% — a ductile material.
⚙️

Engineering Applications

Four materials calculators for structural and design scenarios.

🧰 Tensile stress in a bolt

Calculate the tensile stress in a bolt from the clamping force and bolt root area, then check against the material yield strength.

σ = F ÷ A   (compare with yield strength)
🧮 Calculator
Enter clamping force and bolt root area.

🔨 Extension of a tie bar

Find how much a structural tie bar extends under load using the elastic extension formula.

ΔL = FL ÷ (AE)
🧮 Calculator
Enter force, length, area and E.

📈 Safe working stress from FOS

Calculate the maximum allowable stress for a component given its material's UTS and the required factor of safety.

Working stress = UTS ÷ FOS
🧮 Calculator
Enter UTS and factor of safety.

🔋 Minimum section area

Find the minimum cross-sectional area required to keep a member's stress within its safe working stress limit.

A = F ÷ σ𝑤𝑜𝑟𝑞
🧮 Calculator
Enter force and safe working stress.
🎮

Materials Challenge

12 randomised questions — stress, strain, E and factor of safety.

🧰 Quick-fire round

Question 1 / 12 Score: 0 Streak: 0 🔥
Press Start to begin!
📝

Practice Questions

12 graded questions — type your answer to check, then reveal full working.

🧰 Show formula, unit conversions and every step. 1 MPa = 10&sup6; Pa. 1 GPa = 10&sup9; Pa. 1 mm² = 10&sup6; m².
Q1 · Stress1 mark
A bar of cross-sectional area 200 mm² carries a 60 kN tensile load. Calculate the stress in MPa.
σ = F÷A = 60000÷(200×10⁻&sup6;) = 60000÷0.0002 = 300 MPa
Q2 · Stress2 marks
A 25 kN compressive force acts on a column of cross-section 500 mm². Find the compressive stress in MPa.
σ = 25000 ÷ 0.0005 = 50 MPa
Q3 · Strain2 marks
A 500 mm bar extends by 0.3 mm under load. Calculate the strain.
ε = ΔL ÷ L₀ = 0.3 ÷ 500 = 0.0006
Q4 · Young's modulus2 marks
A material under 250 MPa stress shows a strain of 0.00125. Calculate Young's modulus in GPa.
E = σ÷ε = 250×10&sup6;÷0.00125 = 200×10&sup9; = 200 GPa
Q5 · Extension2 marks
A steel rod (E = 200 GPa, A = 300 mm², L = 1.5 m) carries 45 kN. Find the extension in mm.
ΔL = FL÷(AE) = 45000×1.5÷(0.0003×200×10&sup9;) = 67500÷60000000 = 0.001125 m = 1.125 mm
Q6 · Area from stress2 marks
A tie bar must carry 90 kN without exceeding 180 MPa stress. Find the minimum cross-sectional area in mm².
A = F÷σ = 90000÷(180×10&sup6;) = 0.0005 m² = 500 mm²
Q7 · Factor of safety1 mark
A component made from steel with UTS = 540 MPa operates at 180 MPa. Calculate the FOS.
FOS = UTS ÷ working stress = 540 ÷ 180 = 3
Q8 · Working stress2 marks
A material has UTS = 350 MPa. A FOS of 2.5 is required. Find the maximum safe working stress.
Working stress = UTS ÷ FOS = 350 ÷ 2.5 = 140 MPa
Q9 · Percentage elongation2 marks
A specimen with original gauge length 80 mm fractures at 104 mm gauge length. Find the percentage elongation.
% elongation = (104−80)÷80×100 = 24÷80×100 = 30%
Q10 · Aluminium extension2 marks
An aluminium rod (E = 70 GPa, L = 3.0 m, A = 800 mm²) carries 56 kN. Find the extension in mm.
ΔL = FL÷(AE) = 56000×3.0÷(0.0008×70×10&sup9;) = 168000÷56000000 = 0.003 m = 3.0 mm
Q11 · Combined3 marks
A 30 kN load is applied to a steel rod (E = 200 GPa, L = 2.0 m, A = 150 mm²). Find (a) stress, (b) strain, (c) extension.
σ = 30000÷0.00015 = 200 MPa
ε = 200×10&sup6;÷(200×10&sup9;) = 0.001
ΔL = 0.001×2000 mm = 2.0 mm
Q12 · Combined3 marks
A shaft operates at 160 MPa. The required FOS is 3.5. (a) Find the minimum UTS. (b) Would a steel with UTS 500 MPa be suitable? (c) What actual FOS does that give?
Min UTS = 160×3.5 = 560 MPa → 500 MPa steel not suitable (560>500)
Actual FOS = 500÷160 = 3.125 (below required 3.5)
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