Stress and Strain

Hooke’s law describes how a particular spring or wire responds to force, but the spring constant k depends on the dimensions of the object — a thicker wire is stiffer than a thinner one of the same material. To compare the mechanical properties of materials rather than specific objects, we need quantities that are independent of the object’s size and shape. That is where stress and strain come in.

Tensile Stress

Tensile stress is the force per unit cross-sectional area experienced by a material under tension:

$\sigma = \frac{F}{A}$

where:

σ (Greek letter sigma) is stress, measured in Pa (pascals) or N m⁻²
F is the applied force in N
A is the cross-sectional area in m²

Stress tells you how concentrated the force is. A 1000 N force applied to a wire of cross-sectional area 1.0 × 10⁻⁶ m² produces a stress of 1.0 × 10⁹ Pa (1 GPa) — the same force distributed over a 1 m² area produces a stress of only 1000 Pa.

Worked example 1: A steel wire has a diameter of 0.80 mm and supports a mass of 5.0 kg. What is the tensile stress in the wire?

Area = π(d/2)² = π(0.40 × 10⁻³)² = π × 1.6 × 10⁻⁷ = 5.03 × 10⁻⁷ m²
Force = mg = 5.0 × 9.81 = 49.05 N
Stress = F/A = 49.05 / 5.03 × 10⁻⁷ = 9.75 × 10⁷ Pa ≈ 97.5 MPa

Worked example 2: A lift cable has a safe working stress of 150 MPa and a diameter of 12 mm. What is the maximum safe load?

Area = π(6.0 × 10⁻³)² = 1.131 × 10⁻⁴ m²
Maximum force = σ × A = 150 × 10⁶ × 1.131 × 10⁻⁴ = 16 970 N ≈ 17.0 kN
Maximum mass = 17 000 / 9.81 = 1730 kg

Tensile Strain

Tensile strain is the fractional change in length of a material:

$\varepsilon = \frac{\Delta L}{L}$

where:

ε (Greek letter epsilon) is strain — it is dimensionless (no units) because it is a ratio of two lengths
ΔL is the extension (change in length) in m
L is the original length in m

Strain is often expressed as a percentage. A strain of 0.005 means the material has extended by 0.5% of its original length. Metals typically fracture at strains of a few percent, while rubber can sustain strains of several hundred percent.

Worked example 3: A wire of original length 2.50 m extends by 1.8 mm under load. What is the strain?

ε = ΔL/L = 1.8 × 10⁻³ / 2.50 = 7.2 × 10⁻⁴ (or 0.072%)

Stress–Strain Graphs

A stress-strain graph plots stress (σ) on the y-axis against strain (ε) on the x-axis. Because stress and strain are independent of the sample dimensions, a stress-strain graph is a property of the material itself — all samples of the same material give the same curve, regardless of their size or shape.

The key features of a stress-strain graph for a typical ductile metal (such as copper or mild steel) are:

Linear (Elastic) Region

The initial straight-line portion where stress is directly proportional to strain. The gradient of this region is the Young modulus (discussed in the next lesson). In this region, deformation is elastic — the material returns to its original length when the stress is removed.

Limit of Proportionality (P)

The point where the graph begins to curve. Beyond P, stress and strain are no longer proportional, though deformation may still be elastic.

Elastic Limit (E)

The maximum stress the material can withstand and still return to its original length. Beyond E, permanent (plastic) deformation occurs.

Yield Point (Y)

The stress at which a large increase in strain occurs with little or no increase in stress. This is where the material begins to “flow” plastically. In mild steel, there is a distinct upper and lower yield point where the graph shows a noticeable dip.

Ultimate Tensile Strength (UTS)

The maximum stress the material can withstand. On the graph, this is the highest point. Beyond this, the material begins to neck — a localised region thins rapidly.

Breaking Stress / Fracture Point (B)

The stress at which the material fractures. The actual stress at the fracture surface is higher than the breaking stress shown on the graph because the cross-sectional area has reduced due to necking, but the stress on the graph is calculated using the original cross-sectional area (this is called engineering stress).

Stress–Strain Data for Common Materials

Material	Young Modulus (GPa)	Yield Stress (MPa)	UTS (MPa)	Strain at fracture
Mild steel	200	250	400–550	0.15–0.25
Cast iron	170	N/A (brittle)	200–400	0.003–0.005
Copper	130	70	220	0.30–0.50
Aluminium	70	95	110–200	0.10–0.40
Glass	70	N/A (brittle)	50	0.001
Rubber	0.01–0.1	N/A	15–25	5.0–8.0
CFRP	150	N/A	600	0.01–0.02

Elastic and Plastic Regions

Property	Elastic Region	Plastic Region
Deformation	Reversible	Permanent
Return to original length?	Yes	No
Energy	Stored and recoverable	Dissipated (converted to internal energy)
Atomic explanation	Bonds stretch but atoms return to equilibrium positions	Atoms slip past each other to new permanent positions

Comparing Materials on Stress–Strain Graphs

Different materials produce very different stress-strain curves:

Mild steel: Linear elastic region, distinct yield point, large plastic region, necking before fracture. High UTS, moderate strain to fracture.
Cast iron: Linear elastic region, very little plastic deformation, fractures suddenly. A brittle material.
Copper: Linear elastic region, no distinct yield point, very large plastic region. Highly ductile.
Rubber: Non-linear elastic behaviour, very large strains (up to 500%), returns to original length but with hysteresis.

These differences determine how and where each material is used in engineering.

Worked Calculation: Full Stress–Strain Analysis

Problem: A copper wire of diameter 0.60 mm and original length 3.00 m supports a load of 25 N. The Young modulus of copper is 130 GPa.

(a) Calculate stress: A = π(0.30 × 10⁻³)² = 2.827 × 10⁻⁷ m² σ = F/A = 25 / 2.827 × 10⁻⁷ = 88.4 MPa

(b) Calculate strain: ε = σ/E = 88.4 × 10⁶ / 130 × 10⁹ = 6.80 × 10⁻⁴

(c) Calculate extension: ΔL = εL = 6.80 × 10⁻⁴ × 3.00 = 2.04 × 10⁻³ m = 2.04 mm

(d) Is the wire within its elastic limit? The yield stress of copper is about 70 MPa. Since σ = 88.4 MPa > 70 MPa, the wire has exceeded its yield stress and is undergoing plastic deformation. The Young modulus calculation is not strictly valid here — the actual extension will be larger than 2.04 mm.

Units and Common Prefixes

Stress values in physics problems often involve large numbers:

1 kPa = 10³ Pa
1 MPa = 10⁶ Pa (common for many engineering materials)
1 GPa = 10⁹ Pa (used for stiff materials like steel: Young modulus ≈ 200 GPa)

Always check your units carefully — a common error is failing to convert diameters from mm to m before calculating area.

Common Exam Mistakes

Using diameter instead of radius for area: A = π(d/2)² = πd²/4, NOT πd².
Forgetting to convert mm to m: 0.80 mm = 0.80 × 10⁻³ m, not 0.80 m.
Thinking stress-strain curves depend on sample size: They do not. That is the whole point of using stress and strain rather than force and extension.
Confusing yield point with UTS: The yield point is where plastic flow begins; the UTS is the maximum stress the material can withstand.
Saying strain has units: Strain is a dimensionless ratio — it has no units.

A-Level Deep Dive: Stress and Strain

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, sub-topic on stress, strain and the stress–strain graph requires students to define tensile stress as $\sigma = F/A$ and tensile strain as $\varepsilon = \Delta L/L$ , to interpret stress–strain curves for ductile, brittle and polymeric materials, and to identify limit of proportionality, elastic limit, yield point, ultimate tensile strength and breaking stress (refer to the official Pearson Edexcel specification document for exact wording). Stress and strain are the bridge between Hooke's law (object-specific) and the Young modulus (material-specific): every Topic 4 question on extension or stiffness ultimately routes through them. The content is examined in 9PH0 Paper 1 (Mechanics and Materials, often as a multi-part calculation), in Paper 3 (the Practical Skills paper, where CP2 — determination of the Young modulus of a metal wire — is the direct experimental application) and synoptically in Paper 2 through energy stored per unit volume in stretched samples. The Edexcel formula booklet provides $\sigma = F/A$ , $\varepsilon = \Delta L/L$ and $E = \sigma/\varepsilon$ , but the unit pascal (Pa $\equiv$ N m⁻²) and the dimensionless nature of strain must be quoted from memory.

Worked example with full mark scheme

Question (8 marks):

A copper wire of original length $L = 1.80$ m and uniform diameter $d = 0.45$ mm hangs vertically from a clamp. A force $F = 28$ N is applied to its lower end. The extension is measured as $\Delta L = 0.66$ mm.

(a) Calculate the tensile stress in the wire. (2)

(b) Calculate the tensile strain. (1)

(c) Determine the Young modulus and comment on whether it is consistent with copper ( $E_{\text{copper}} \approx 130$ GPa). (3)

(d) Explain whether the wire has remained within its elastic limit, given that the yield stress of copper is approximately 70 MPa. (2)

Solution with mark scheme:

(a) Convert and compute the cross-sectional area: $d = 4.5 \times 10^{-4}$ m, so $r = 2.25 \times 10^{-4}$ m, giving $A = \pi r^2 = 1.59 \times 10^{-7}$ m².

M1 — correct $A = \pi r^2$ with $r = d/2$ in metres. The near-universal slip is to use $\pi d^2$ (forgetting the factor of 4) or leave $d$ in mm and obtain area in mm² uncorrected — either error misplaces stress by a power of ten.

$\sigma = \dfrac{F}{A} = \dfrac{28}{1.59 \times 10^{-7}} = 1.76 \times 10^{8}\ \text{Pa} = 176\ \text{MPa}$

A1 — value in the range 170–180 MPa, with Pa or MPa stated.

(b) With $\Delta L = 6.6 \times 10^{-4}$ m and $L = 1.80$ m:

$\varepsilon = \dfrac{\Delta L}{L} = \dfrac{6.6 \times 10^{-4}}{1.80} = 3.67 \times 10^{-4}$

B1 — dimensionless answer in the range $3.6$ – $3.7 \times 10^{-4}$ , no unit.

$E = \dfrac{1.76 \times 10^{8}}{3.67 \times 10^{-4}} = 4.79 \times 10^{11}\ \text{Pa} = 479\ \text{GPa}$

M1 — correct substitution. A1 — value 470–490 GPa.

B1 (AO3.2a) — comment that 479 GPa is far higher than the accepted 130 GPa for copper. Plausible explanations: the extension was mis-measured (slack not removed, zero-error, or stage too small to resolve); the wire is not pure copper; or — see (d) — the wire is in the plastic region, where $E = \sigma/\varepsilon$ does not apply.

(d) Applied stress $\sigma = 176$ MPa exceeds the yield stress $\sigma_y \approx 70$ MPa, so the wire has been loaded beyond its yield point.

M1 — explicit numerical comparison. A1 — conclusion that the wire is undergoing plastic deformation and the linear $E$ -formula is not strictly valid; this is consistent with the anomalous (c) result.

Total: 8 marks (M3 A4 B1).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): The stress–strain graph for two different metal wires, X and Y, is shown schematically. Both wires have the same dimensions. Wire X is linear up to a stress of 240 MPa at a strain of $1.2 \times 10^{-3}$ , then yields and continues to extend at roughly constant stress until fracture at a strain of 0.18. Wire Y is linear up to a stress of 80 MPa at a strain of $1.0 \times 10^{-3}$ , then fractures almost immediately at a strain of $1.1 \times 10^{-3}$ .

(a) Calculate the Young modulus of each wire from the linear region. (2)

(b) State, with a reason, which wire is brittle and which is ductile. (2)

(c) The wires are to be used as load-bearing supports where a small permanent deformation under overload is preferable to sudden failure. State, with reasoning that refers to the graph, which wire should be chosen. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1b) — applying $E = \sigma/\varepsilon$ to each wire: $E_X = 2.40 \times 10^8 / 1.2 \times 10^{-3} = 2.0 \times 10^{11}$ Pa $= 200$ GPa.
A1 (AO1.1b) — $E_Y = 8.0 \times 10^7 / 1.0 \times 10^{-3} = 8.0 \times 10^{10}$ Pa $= 80$ GPa, with both values quoted in GPa.

(b)

B1 (AO2.1) — Y is brittle: it fractures at a strain of $1.1 \times 10^{-3}$ , almost identical to the strain at the limit of proportionality, so it shows essentially no plastic region.
B1 (AO2.1) — X is ductile: it sustains a strain of 0.18 (about 150 times its yield strain) before fracture, indicating an extensive plastic region.

(c)

M1 (AO3.1a) — choosing X.
A1 (AO3.2b) — reasoning that the long plastic region in X gives visible warning of impending failure (sagging, necking) before fracture, whereas Y would shatter without warning; this is a property linked to area-under-graph (toughness) rather than peak stress alone.

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. This is a graph-interpretation question of the kind 9PH0 Paper 1 favours: numerical recall ( $E = \sigma/\varepsilon$ ) is worth one-third, the rest is qualitative materials-selection reasoning.

Synoptic links

Stress and strain knit Topic 4 to four other strands of A-Level Physics:

Topic 4 — Hooke's law and Young modulus: $F = k\Delta x$ is the object version of the linear stress–strain region. Dividing by $A$ gives $\sigma = (kL/A)\varepsilon$ , identifying $E = kL/A$ , hence $k = EA/L$ — a thicker or shorter wire of the same material has a larger spring constant. This is a frequent Paper 1 derivation.
Topic 4 — elastic energy stored: the area under a force–extension graph is $W = \tfrac{1}{2}F\Delta x$ ; dividing by volume $V = AL$ gives energy density $u = \tfrac{1}{2}\sigma\varepsilon$ in J m⁻³ — a per-unit-volume quantity comparable across samples of different sizes.
Topic 6 — oscillations: the restoring force in a spring–mass oscillator is $F = -kx$ , giving $\omega = \sqrt{k/m}$ . Substituting $k = EA/L$ shows the natural frequency of a longitudinal wire oscillator depends on $\sqrt{E/\rho}$ — exactly the speed of longitudinal sound waves through the wire.
Topic 5 — fluid flow: shear stress $\tau = F/A$ and shear strain rate $d\gamma/dt$ define viscosity $\eta = \tau / (d\gamma/dt)$ for a Newtonian fluid. The same stress definition adapts to fluids by replacing strain with strain rate.