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.

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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