Stress, Strain and Young's Modulus

This lesson covers AQA Spec 3.4.2.1 (Bulk properties of solids) in detail, including definitions, stress–strain graphs, experimental determination of Young's modulus, and comparison of materials.

Spec mapping (AQA 7408 §3.4.2.1–§3.4.2.2): This lesson develops tensile stress σ = F/A, tensile strain ε = ΔL/L, Hooke's law within the limit of proportionality, and Young's modulus E = σ/ε as a material property. The treatment includes interpretation of stress-strain graphs (limit of proportionality, elastic limit, yield point, ultimate tensile stress, fracture point) and the elastic energy stored per unit volume as ½σε. Required Practical 2 (determination of Young's modulus, RP2) sits inside this strand and is treated in depth in Lesson 10. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Young's modulus connects to several other A-Level topics. (i) Hooke's law for springs (Lesson 8, §3.4.2.1) is the discrete-spring analogue of Young's modulus for a continuous wire — both express linear restoring force, with k = EA/L for a uniform wire treated as an effective spring. (ii) Elastic potential energy (Lesson 5, §3.4.1.7) is the same ½kx² result expressed in the language of materials as the area under a stress-strain curve up to the elastic limit. (iii) Simple harmonic motion of a mass on a spring (Year 13, §3.6.1.2) uses the spring-constant form of Hooke's law as the restoring force — wires under tension exhibit SHM with effective k = EA/L, giving angular frequency ω = √(EA/(mL)). (iv) Material selection in engineering uses Young's modulus as the primary stiffness criterion: aerospace components prefer high E/ρ ratios (carbon-fibre composites, titanium alloys); the same data table you see at A-Level is the entry point to engineering design.

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1 Tensile Stress

Key Definition: Tensile stress (σ) is the force per unit cross-sectional area.

σ = F / A

SI unit: pascal (Pa). 1 Pa = 1 N m⁻².

Typical values: steel wire under tension might experience stresses of 10⁸ Pa = 100 MPa.

Worked Example 1

A steel wire has a cross-sectional area of 2.0 × 10⁻⁶ m² and supports a load of 500 N. Find the stress.

Solution:

σ = F/A = 500 / (2.0 × 10⁻⁶) = 2.5 × 10⁸ Pa = 250 MPa

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2 Tensile Strain

Key Definition: Tensile strain (ε) is the fractional change in length (extension divided by original length).

ε = ΔL / L

Strain has no units — it is a ratio (dimensionless). It is often expressed as a percentage.

Worked Example 2

A wire of original length 2.5 m extends by 1.5 mm under a load. Find the strain.

Solution:

ε = ΔL/L = 1.5 × 10⁻³ / 2.5 = 6.0 × 10⁻⁴ = 0.060%

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3 Young's Modulus

Key Definition: Young's modulus (E) is the ratio of tensile stress to tensile strain, within the limit of proportionality.

E = σ / ε = (F/A) / (ΔL/L) = FL / (AΔL)

SI unit: Pa (or N m⁻²). Young's modulus is a property of the material, not the object.

Material	Young's Modulus (GPa)
Steel	200
Copper	130
Aluminium	70
Glass	70
Rubber	0.01–0.1
Bone	15–20

Worked Example 3

A copper wire of length 3.0 m and diameter 0.80 mm supports a 50 N load. Young's modulus for copper is 130 GPa. Find the extension.

Solution:

Cross-sectional area: A = π(d/2)² = π(0.40 × 10⁻³)² = π × 1.6 × 10⁻⁷ = 5.03 × 10⁻⁷ m²

From E = FL/(AΔL): ΔL = FL/(AE)

ΔL = 50 × 3.0 / (5.03 × 10⁻⁷ × 130 × 10⁹) = 150 / 65 390 = 2.29 × 10⁻³ m = 2.3 mm

Worked Example 4

A nylon guitar string is 0.64 m long, has a diameter of 1.0 mm, and is under a tension of 80 N. It extends by 0.50 mm. Find Young's modulus for nylon.

Solution:

A = π(0.50 × 10⁻³)² = 7.854 × 10⁻⁷ m²

E = FL/(AΔL) = 80 × 0.64 / (7.854 × 10⁻⁷ × 0.50 × 10⁻³)

E = 51.2 / (3.927 × 10⁻¹⁰) = 1.30 × 10¹¹ Pa ≈ 130 GPa

Common Misconception: Students often forget to convert diameter to radius and then to area. Always use A = πr² = π(d/2)². Watch the powers of 10 carefully.

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4 Stress–Strain Graphs

A stress–strain graph reveals the mechanical properties of a material.

Described diagram — Stress–strain graph for a typical ductile metal (e.g., mild steel):

• O to P (Limit of Proportionality): Straight line — stress is proportional to strain (Hooke's law region). Young's modulus = gradient.
• P to E (Elastic Limit): Slightly curved — still elastic (returns to original length on unloading) but no longer proportional.
• E to Y (Yield Point): The material begins to deform plastically. For mild steel, there is a distinct upper and lower yield point where strain increases with little increase in stress.
• Y to UTS (Work Hardening): Stress increases with strain as the material work-hardens. The cross-section begins to reduce.
• UTS (Ultimate Tensile Stress): The maximum stress the material can withstand. Beyond this, necking occurs.
• UTS to Fracture: Stress appears to decrease (because we use original cross-section) as the neck narrows rapidly until fracture.

Key Points on the Graph

Point/Region	Significance
Gradient (linear region)	Young's modulus
Limit of proportionality	End of Hooke's law behaviour
Elastic limit	Beyond this, permanent deformation occurs
Yield point	Onset of significant plastic flow
UTS	Maximum stress before necking
Area under curve	Energy per unit volume (energy density)

Exam Tip: The area under a stress–strain graph represents the energy stored per unit volume (in J m⁻³). For the elastic region, this equals ½σε = ½Eε².

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5 Experimental Determination of Young's Modulus

The Searle's Apparatus Method

Equipment: Two long, identical wires suspended from the same support; one is the test wire, the other is a reference wire (to compensate for thermal expansion). A micrometer measures the relative extension.

Procedure:

1. Measure the original length L of the test wire with a metre rule.
2. Measure the diameter d at several points along the wire using a micrometer, and calculate the mean. Find A = π(d/2)².
3. Add masses in increments (e.g., 100 g), recording the extension ΔL for each load using the micrometer or vernier scale.
4. Plot a graph of force F (y-axis) against extension ΔL (x-axis).
5. The gradient = F/ΔL.
6. Calculate: E = (F/ΔL) × (L/A) = gradient × L/A

Alternative Method: Long Wire and Ruler

Hang a long wire (2–3 m) from the ceiling. Attach a ruler and marker. Add masses and measure extension against the ruler (or use a travelling microscope for precision).

Sources of Uncertainty:

Source	How to Minimise
Measuring diameter	Take multiple readings, rotate the micrometer
Measuring extension	Use a vernier scale; take readings loading and unloading
Wire not straight initially	Apply a small initial load to straighten the wire
Temperature changes	Use a reference wire (Searle's method)

AQA Required Practical: Determination of Young's modulus — see Lesson 10 for full practical details.

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6 Comparing Materials Using Stress–Strain Graphs

Property	Steel	Copper	Glass	Rubber
Young's modulus	Very high (200 GPa)	High (130 GPa)	High (70 GPa)	Very low (~0.01 GPa)
Yield stress	High	Moderate	N/A (brittle)	N/A
UTS	Very high	Moderate	Low in tension	Low
Behaviour	Ductile	Ductile	Brittle	Elastic (large strains)
Breaking strain	~0.3	~0.5	~0.001	~5.0

Worked Example 5 — Comparing Two Wires

Wire A (steel): length 2.0 m, diameter 1.0 mm, E = 200 GPa. Wire B (copper): length 2.0 m, diameter 1.0 mm, E = 130 GPa.

Both support a 100 N load. Which extends more, and by how much?

Solution:

Both have the same A = π(0.50 × 10⁻³)² = 7.854 × 10⁻⁷ m²

ΔL_A = FL/(AE) = 100 × 2.0 / (7.854 × 10⁻⁷ × 200 × 10⁹) = 200 / 157 080 = 1.27 × 10⁻³ m = 1.27 mm

ΔL_B = FL/(AE) = 100 × 2.0 / (7.854 × 10⁻⁷ × 130 × 10⁹) = 200 / 102 102 = 1.96 × 10⁻³ m = 1.96 mm

The copper wire extends more, by 1.96 − 1.27 = 0.69 mm.

Worked Example 6 — Energy Stored in a Loaded Steel Wire

A steel wire (E = 200 GPa) has length 1.50 m, diameter 0.80 mm, and supports a 300 N load. Calculate (a) the extension, (b) the strain, (c) the elastic energy stored per unit volume, and (d) the total elastic energy stored.

Solution:

(a) A = π × (0.40 × 10⁻³)² = π × 1.6 × 10⁻⁷ = 5.027 × 10⁻⁷ m².

ΔL = FL/(AE) = 300 × 1.50 / (5.027 × 10⁻⁷ × 200 × 10⁹) = 450 / (1.005 × 10⁵) = 4.476 × 10⁻³ m = 4.48 mm.

(b) ε = ΔL/L = 4.476 × 10⁻³ / 1.50 = 2.98 × 10⁻³ (about 0.30%).

(c) σ = F/A = 300 / 5.027 × 10⁻⁷ = 5.968 × 10⁸ Pa = 597 MPa.

Energy density u = ½σε = ½ × 5.968 × 10⁸ × 2.98 × 10⁻³ = 8.89 × 10⁵ J m⁻³.

(d) Total elastic energy E_tot = u × V where V = A × L = 5.027 × 10⁻⁷ × 1.50 = 7.54 × 10⁻⁷ m³.

E_tot = 8.89 × 10⁵ × 7.54 × 10⁻⁷ = 0.670 J.

Check via ½kx² where k = EA/L = 200 × 10⁹ × 5.027 × 10⁻⁷ / 1.50 = 6.703 × 10⁴ N m⁻¹:

E_tot = ½ × 6.703 × 10⁴ × (4.476 × 10⁻³)² = ½ × 6.703 × 10⁴ × 2.003 × 10⁻⁵ = 0.671 J ✓ (within rounding).

This worked example confirms the equivalence of the energy-density formulation (½σε integrated over volume) with the spring-energy formulation (½kx²).

AQA Specification Reference: Section 3.4.2.1 (Bulk properties of solids — Hooke's law, Young's modulus, stress–strain graphs).

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Specimen Exam Question

Specimen question modelled on the AQA paper format — not a verbatim past-paper item.

A copper wire of original length 2.00 m and diameter 0.50 mm is suspended vertically and used to support an increasing load. The accompanying force-extension data are recorded. The wire begins to show non-linear behaviour above an extension of 1.50 mm; the wire fractures at a load of 80 N.

Force F (N)	Extension ΔL (mm)
0	0
10	0.50
20	1.00
30	1.50
40	2.20
50	3.20
60	5.00

(a) Use the linear portion of the data to calculate Young's modulus for copper. [3 marks]

(b) Calculate the stress in the wire at the point of fracture. [2 marks]

(c) Sketch the stress-strain curve for this wire, labelling the limit of proportionality, the region of plastic deformation, and the fracture point. [2 marks]

(d) Calculate the elastic energy stored per unit volume in the wire at the limit of proportionality. [2 marks]

AO Breakdown