Materials & Properties
Stress, strain and Young's modulus describe how engineering materials respond to forces — the foundation of every structural and mechanical design decision, from choosing a bolt to sizing a bridge beam.
2Topics
8Examples
4Calculators
12Practice Qs
🔧Stress & Strainσ = F ÷ A, ε = x ÷ L
📈Young's ModulusE = σ ÷ ε
⚙️Applications4 workshop calculators
🎮Game12 randomised questions
📝Practice12 graded questions
Why this matters in engineeringReal world
When a structural engineer chooses a steel bolt, they must ensure the stress in the bolt under load stays below the material's yield strength. Young's modulus tells them how much the bolt will stretch — critical for joint clamping force. These calculations apply to every structural component: shafts, beams, columns and fasteners.
🔧
Stress & Strain
Stress measures the internal force per unit area inside a material. Strain measures how much it deforms relative to its original length.
🔧Stress & Strain CalculatorChoose what to find, enter values, press Calculate
Choose what to find, enter the other values, then press Calculate.
📚 Key vocabulary
- Stress (σ)
- force per unit cross-sectional area inside a material, measured in pascals (Pa) or N/mm².
- Strain (ε)
- the fractional change in length — extension divided by original length. A dimensionless ratio.
- Tensile stress
- stress that tries to stretch or pull apart a material (positive by convention).
- Compressive stress
- stress that squashes or shortens a material (negative by convention).
- Cross-sectional area (A)
- the area of the face perpendicular to the applied force — for a circular bar: A = πr².
- Pascal (Pa)
- 1 N/m². Engineering stresses are typically millions of pascals — quoted in MPa or N/mm².
👁️ Visual aid — stress in a tensile bar: force arrows pull the bar, which stretches by extension x
✍️Worked example 1 — stress in a bar
QuestionA steel bar with cross-sectional area 0.001 m² carries a tensile force of 50 000 N. Calculate the stress.
Formula
σ = F ÷ ASubstitute
σ = 50 000 ÷ 0.001Divide
50 000 ÷ 0.001 = 50 000 000Convert
50 000 000 Pa = 50 MPaAnswerσ = 50 MPa
✍️Worked example 2 — strain in a component
QuestionA component of original length 2.0 m extends by 4 mm under load. Calculate the strain.
Convert
4 mm = 0.004 mFormula
ε = x ÷ LSubstitute
ε = 0.004 ÷ 2.0Divide
0.004 ÷ 2.0 = 0.002Answerε = 0.002 (no units — it is a ratio)
✍️Worked example 3 — finding force from stress
QuestionA bolt has a cross-sectional area of 200 mm² and the allowable stress is 150 N/mm². What is the maximum safe force?
Formula
F = σ × ASubstitute
F = 150 × 200Multiply
150 × 200 = 30 000AnswerF = 30 000 N = 30 kN
✍️Worked example 4 — finding extension from strain
QuestionA shaft of original length 0.5 m has a strain of 0.003 under axial load. How much does it extend?
Formula
x = ε × LSubstitute
x = 0.003 × 0.5Multiply
0.003 × 0.5 = 0.0015 mConvert
0.0015 m = 1.5 mmAnswerx = 0.0015 m (1.5 mm)
📈
Young's Modulus
The ratio of stress to strain — a measure of stiffness. A high Young's modulus means a material resists deformation; it barely stretches under load.
📈Young's Modulus CalculatorChoose what to find, enter values, press Calculate
Choose what to find, enter the other values, then press Calculate.
📚 Key vocabulary
- Young's modulus (E)
- stress divided by strain — the stiffness of a material. Measured in GPa for engineering metals.
- Elastic region
- the range of stress within which the material returns to its original shape when the load is removed.
- Elastic limit
- the maximum stress a material can withstand and still return fully to its original shape.
- Yield point
- the stress at which a material starts to deform permanently — design stress must stay below this.
- Stiffness
- resistance to deformation under load. A stiffer material has a higher Young's modulus.
- Gigapascal (GPa)
- 1×10&sup9; Pa. Steel's Young's modulus is approximately 200 GPa.
👁️ Visual aid — stress-strain graph: the slope of the straight elastic section equals Young's modulus E
✍️Worked example 5 — Young's modulus from test data
QuestionA tensile test on a steel rod gives a stress of 200 MPa and a corresponding strain of 0.001. Calculate Young's modulus.
Convert
200 MPa = 200 × 10&sup6; Pa = 2 × 10&sup8; PaFormula
E = σ ÷ εSubstitute
E = 200 000 000 ÷ 0.001Divide
200 000 000 ÷ 0.001 = 200 000 000 000Convert
2 × 10¹¹ Pa = 200 GPaAnswerE = 200 GPa (typical for steel)
✍️Worked example 6 — stress from E and strain
QuestionAn aluminium alloy has Young's modulus 70 GPa. If the strain in a component is 0.0015, find the stress in MPa.
Convert
70 GPa = 70 × 10&sup9; PaFormula
σ = E × εSubstitute
σ = 70 × 10&sup9; × 0.0015Multiply
70 × 10&sup9; × 0.0015 = 105 000 000 PaConvert
105 000 000 ÷ 10&sup6; = 105 MPaAnswerσ = 105 MPa
✍️Worked example 7 — strain from E and stress
QuestionA steel component (E = 200 GPa) is stressed to 160 MPa. Calculate the strain.
Formula
ε = σ ÷ EConvert
160 MPa = 160 × 10&sup6; Pa, E = 200 × 10&sup9; PaSubstitute
ε = 160 × 10&sup6; ÷ 200 × 10&sup9;Divide
160 ÷ 200 000 = 0.0008Answerε = 0.0008
✍️Worked example 8 — extension using E (combined)
QuestionA steel tie rod (E = 200 GPa, A = 500 mm², L = 3 m) carries a load of 300 kN. Find the extension.
Step 1: stress
σ = F ÷ A = 300 000 ÷ (500 × 10−&sup6;) = 600 × 10&sup6; Pa = 600 MPaStep 2: strain
ε = σ ÷ E = 600 × 10&sup6; ÷ 200 × 10&sup9; = 0.003Step 3: ext
x = ε × L = 0.003 × 3 = 0.009 mAnswerx = 0.009 m = 9 mm
⚙️
Engineering Applications
Four structural and mechanical calculators applying stress, strain and Young's modulus.
🔧 Bolt stress check
Check whether the stress in a fastener stays below the allowable design stress for the material.
σ = F ÷ A
🧮 Calculator
Enter force and area, then press Calculate.
📈 Beam extension (E given)
Find the extension of a loaded member when Young's modulus, dimensions and load are known.
x = (F × L) ÷ (A × E)
🧮 Calculator
Enter all values, then press Calculate.
🔨 Material identification (E from test)
Identify a material by calculating its Young's modulus from tensile test results.
E = σ ÷ ε = (F ÷ A) ÷ (x ÷ L)
🧮 Calculator
Enter stress and strain, then press Calculate.
🛠️ Safe load from allowable stress
Find the maximum load a cross-section can carry without exceeding its design stress limit.
F = σallow × A
🧮 Calculator
Enter allowable stress and area, then press Calculate.
🎮
Materials Challenge
12 randomised questions per round. New numbers every time.
📈 Quick-fire round
Question 1 / 12
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Press Start to begin!
📝
Practice Questions
12 graded questions. Attempt on paper first, then reveal the full worked solution.
Use SI units throughout unless stated. 1 MPa = 10&sup6; Pa. 1 GPa = 10&sup9; Pa. Show formula, substitution and every step.
Question 1 · Stress1 mark
A force of 20 000 N acts on a cross-section of area 0.002 m². Calculate the stress.
σ = F ÷ A = 20 000 ÷ 0.002 = 10 000 000 Pa = 10 MPa
Question 2 · Stress2 marks
A bolt has a circular cross-section with area 314 mm² and carries a tensile load of 47 100 N. Find the stress in N/mm².
σ = F ÷ A
σ = 47 100 ÷ 314 = 150 N/mm² (= 150 MPa)
σ = 47 100 ÷ 314 = 150 N/mm² (= 150 MPa)
Question 3 · Stress2 marks
A structural column has an allowable stress of 80 MPa and a cross-sectional area of 600 mm². What is the maximum safe load in kN?
F = σ × A
F = 80 × 600 = 48 000 N = 48 kN
F = 80 × 600 = 48 000 N = 48 kN
Question 4 · Strain1 mark
A wire of original length 3.0 m extends by 6 mm under tension. Calculate the strain.
ε = x ÷ L = 0.006 ÷ 3.0 = 0.002
Question 5 · Strain2 marks
A component has a strain of 0.0025 and an original length of 0.8 m. Calculate the extension in mm.
x = ε × L
x = 0.0025 × 0.8 = 0.002 m = 2 mm
x = 0.0025 × 0.8 = 0.002 m = 2 mm
Question 6 · Young's modulus2 marks
A test sample shows a stress of 140 MPa and a strain of 0.002. Calculate Young's modulus in GPa.
E = σ ÷ ε
E = 140 × 10&sup6; ÷ 0.002 = 70 000 000 000 Pa = 70 GPa
E = 140 × 10&sup6; ÷ 0.002 = 70 000 000 000 Pa = 70 GPa
Question 7 · Young's modulus2 marks
Steel has a Young's modulus of 200 GPa. A steel component has a strain of 0.0006. Find the stress in MPa.
σ = E × ε
σ = 200 × 10&sup9; × 0.0006 = 120 000 000 Pa = 120 MPa
σ = 200 × 10&sup9; × 0.0006 = 120 000 000 Pa = 120 MPa
Question 8 · Young's modulus2 marks
An aluminium component (E = 70 GPa) has a stress of 105 MPa. Calculate the strain.
ε = σ ÷ E
ε = 105 × 10&sup6; ÷ 70 × 10&sup9; = 0.0015
ε = 105 × 10&sup6; ÷ 70 × 10&sup9; = 0.0015
Question 9 · Combined2 marks
A rod (A = 0.0005 m², L = 1.5 m, E = 200 GPa) carries 150 kN. Find the stress and then the strain.
σ = F ÷ A = 150 000 ÷ 0.0005 = 300 000 000 Pa = 300 MPa
ε = σ ÷ E = 300 × 10&sup6; ÷ 200 × 10&sup9; = 0.0015
ε = σ ÷ E = 300 × 10&sup6; ÷ 200 × 10&sup9; = 0.0015
Question 10 · Combined2 marks
A component has strain 0.0015 and Young's modulus 200 GPa. Calculate the stress, then find the force if A = 400 mm².
σ = E × ε = 200 × 10&sup9; × 0.0015 = 300 MPa
F = σ × A = 300 × 400 = 120 000 N = 120 kN
F = σ × A = 300 × 400 = 120 000 N = 120 kN
Question 11 · Extension3 marks
A steel tie rod (E = 200 GPa, A = 250 mm², L = 2.0 m) carries 100 kN. Find the stress, strain and extension.
σ = F ÷ A = 100 000 ÷ (250×10−&sup6;) = 400 MPa
ε = σ ÷ E = 400×10&sup6; ÷ 200×10&sup9; = 0.002
x = ε × L = 0.002 × 2.0 = 0.004 m = 4 mm
ε = σ ÷ E = 400×10&sup6; ÷ 200×10&sup9; = 0.002
x = ε × L = 0.002 × 2.0 = 0.004 m = 4 mm
Question 12 · Extension3 marks
An aluminium strut (E = 70 GPa, A = 600 mm², L = 1.2 m) carries a compressive load of 126 kN. Find the stress, strain and shortening.
σ = 126 000 ÷ (600×10−&sup6;) = 210 MPa
ε = 210×10&sup6; ÷ 70×10&sup9; = 0.003
x = 0.003 × 1.2 = 0.0036 m = 3.6 mm shortening
ε = 210×10&sup6; ÷ 70×10&sup9; = 0.003
x = 0.003 × 1.2 = 0.0036 m = 3.6 mm shortening