Young Modulus

The Young modulus is the single most important quantity for describing the stiffness of a material. Named after Thomas Young, who formalised the concept in the early 19th century, it tells you how much a material resists elastic deformation under stress.

Defining Young Modulus

The Young modulus (E) is defined as the ratio of stress to strain in the linear (elastic) region:

$E = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L / L} = \frac{FL}{A \Delta L}$

where:

E is the Young modulus in Pa (or more commonly GPa)
F is the applied force in N
A is the cross-sectional area in m²
L is the original length in m
ΔL is the extension in m

The Young modulus has units of pressure (Pa) because strain is dimensionless, so E has the same units as stress.

Reference Table: Young Modulus Values

Material	Young Modulus / GPa	Category
Rubber	0.01 – 0.1	Elastomer
Polythene	0.2 – 0.5	Polymer
Bone	10 – 20	Biological
Concrete	30	Ceramic composite
Aluminium	70	Metal
Glass	70	Ceramic
Titanium	110	Metal
Copper	130	Metal
Steel	200	Metal
Tungsten	400	Metal
Diamond	1220	Ceramic

Notice that stiffness is not the same as strength. Glass and aluminium have similar Young moduli (both about 70 GPa), meaning they resist elastic deformation equally well. However, glass is brittle (it fractures with almost no plastic deformation) while aluminium is ductile (it can deform plastically by a large amount before fracturing).

Calculating Young Modulus from Data

Worked example 1: A copper wire of length 2.00 m and diameter 0.50 mm supports a mass of 3.0 kg. The wire extends by 0.72 mm. Calculate the Young modulus of copper.

Step 1: Calculate the cross-sectional area. A = π(d/2)² = π(0.25 × 10⁻³)² = 1.963 × 10⁻⁷ m²

Step 2: Calculate the force. F = mg = 3.0 × 9.81 = 29.43 N

Step 3: Apply the Young modulus formula. E = FL/(AΔL) = (29.43 × 2.00) / (1.963 × 10⁻⁷ × 0.72 × 10⁻³) E = 58.86 / (1.414 × 10⁻¹⁰) E = 4.16 × 10¹¹ Pa ≈ 416 GPa

This is higher than the accepted value of 130 GPa, which suggests measurement errors — most likely in the small extension or the wire diameter.

Worked example 2: A steel rod of original length 0.50 m and cross-sectional area 2.0 × 10⁻⁴ m² is subjected to a compressive force of 100 kN. The Young modulus of steel is 200 GPa. By how much does the rod shorten?

Rearranging E = FL/(AΔL): ΔL = FL/(AE) = (100 × 10³ × 0.50) / (2.0 × 10⁻⁴ × 200 × 10⁹) ΔL = 50 000 / (4.0 × 10⁷) = 1.25 × 10⁻³ m = 1.25 mm

Worked example 3: Two wires of the same material (E = 200 GPa) but different dimensions are loaded with the same force of 50 N. Wire A: length 1.0 m, diameter 0.40 mm. Wire B: length 3.0 m, diameter 0.80 mm.

Wire A: A = π(0.20 × 10⁻³)² = 1.257 × 10⁻⁷ m² ΔL_A = FL/(AE) = (50 × 1.0) / (1.257 × 10⁻⁷ × 200 × 10⁹) = 50 / 25 140 = 1.99 × 10⁻³ m ≈ 2.0 mm

Wire B: A = π(0.40 × 10⁻³)² = 5.027 × 10⁻⁷ m² ΔL_B = (50 × 3.0) / (5.027 × 10⁻⁷ × 200 × 10⁹) = 150 / 100 540 = 1.49 × 10⁻³ m ≈ 1.5 mm

Despite being three times longer, wire B extends less because its cross-sectional area is four times larger (diameter doubled → area quadrupled).

Young Modulus from a Stress–Strain Graph

On a stress-strain graph, the Young modulus is the gradient of the linear (proportional) region:

$E = \frac{\Delta \sigma}{\Delta \varepsilon} = \text{gradient of linear region}$

To find the Young modulus from a graph:

Identify the straight-line portion of the graph (from the origin to the limit of proportionality).
Choose two points well apart on this straight line.
Calculate the gradient: E = (σ₂ − σ₁) / (ε₂ − ε₁).

The advantage of using the gradient (rather than a single point) is that it averages out small measurement errors. Always draw a line of best fit rather than relying on a single data point.

Young Modulus from a Force–Extension Graph

If you have a force-extension graph instead of a stress-strain graph, you can still find the Young modulus. The gradient of a force-extension graph is F/x, which is the spring constant k. To convert to Young modulus:

$E = \frac{\text{gradient} \times L}{A} = \frac{kL}{A}$

This relationship shows why the spring constant depends on dimensions (length and area) while the Young modulus does not.

Specific Stiffness: Comparing Materials for Lightweight Design

The Young modulus allows meaningful comparison between materials regardless of sample dimensions. For applications where weight matters (aerospace, automotive, sports equipment), engineers use the specific stiffness — the ratio E/ρ:

Material	E (GPa)	ρ (kg m⁻³)	E/ρ (MN m kg⁻¹)	Application
Steel	200	7800	25.6	Bridges, buildings
Aluminium	70	2700	25.9	Aircraft fuselages
Titanium	110	4500	24.4	Jet engine blades
CFRP	150	1600	93.8	Racing cars, aircraft wings
Wood (along grain)	12	600	20.0	Traditional structures

This analysis shows that while steel has the highest absolute stiffness, CFRP has by far the best stiffness-to-weight ratio, which is why it dominates modern aircraft and racing car design. Interestingly, steel, aluminium, and titanium all have very similar specific stiffnesses (≈25 MN m kg⁻¹).

Elastic Strain Energy from a Stress–Strain Graph

The area under a stress-strain graph (up to the elastic limit) gives the elastic strain energy per unit volume of the material:

$\text{Energy per unit volume} = \frac{1}{2} \sigma \varepsilon = \frac{\sigma^2}{2E}$

This is analogous to E = ½kx² for a spring, but expressed per unit volume so it is a material property rather than a property of a specific sample.

Worked example 4: A steel wire is stressed to 200 MPa (within its elastic limit). E = 200 GPa. What is the elastic strain energy stored per unit volume?

Energy/volume = σ²/(2E) = (200 × 10⁶)² / (2 × 200 × 10⁹) = 4 × 10¹⁶ / (4 × 10¹¹) = 1.0 × 10⁵ J m⁻³ = 100 kJ m⁻³

Worked Example: Finding Extension in a Real Structure

Bridge cable problem: A suspension bridge uses steel cables with total cross-sectional area 0.050 m² and length 200 m. The bridge deck has a mass of 2.0 × 10⁶ kg. Eₛₜₑₑₗ = 200 GPa.

Step 1 — Force: F = mg = 2.0 × 10⁶ × 9.81 = 1.96 × 10⁷ N

Step 2 — Stress: σ = F/A = 1.96 × 10⁷ / 0.050 = 3.92 × 10⁸ Pa = 392 MPa

Step 3 — Check: UTS of steel ≈ 500 MPa, so σ < UTS — the cables are safe but under significant load.

Step 4 — Strain: ε = σ/E = 3.92 × 10⁸ / 200 × 10⁹ = 1.96 × 10⁻³

Step 5 — Extension: ΔL = εL = 1.96 × 10⁻³ × 200 = 0.392 m ≈ 39 cm

This means each cable stretches by about 39 cm under the full load of the bridge deck — a surprisingly large extension, but the strain is still very small (0.2%).

Relationship Between Spring Constant and Young Modulus

The spring constant k of a wire depends on its dimensions AND the material:

$k = \frac{AE}{L}$

This means:

A thicker wire (larger A) is stiffer (larger k)
A shorter wire (smaller L) is stiffer (larger k)
A wire made from a stiffer material (larger E) has a larger k

Change	Effect on k	Effect on E
Double the diameter	k × 4 (area quadruples)	No change
Double the length	k ÷ 2	No change
Change material	k changes	E changes
Same material, different dimensions	k changes	No change

This table highlights the key distinction: k is a property of the specific sample, while E is a property of the material.

Common Mistakes

Forgetting to halve the diameter when calculating cross-sectional area. If given a diameter d, the area is π(d/2)² = πd²/4, not πd².
Mixing units — extension in mm but length in m, diameter in mm but area needed in m². Convert everything to SI base units first.
Using the gradient of the whole graph rather than just the linear region. The Young modulus is only valid in the proportional region.
Confusing stiffness with strength — a material can be very stiff (high E) but also very brittle (low strain to fracture).
Thinking the Young modulus changes with sample dimensions — it does not. E is a material constant. Only the spring constant k = AE/L changes with dimensions.

A-Level Deep Dive: Young Modulus

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, sub-topic on the Young modulus requires students to define $E = \sigma/\varepsilon$ , recall and use $E = FL/(A\Delta L)$ , extract $E$ from the gradient of the linear region of a stress–strain graph, and apply $E$ to engineering and biological problems (refer to the official Pearson Edexcel specification document for exact wording). The Young modulus is the load-bearing concept of Topic 4: it is the synthesis of stress, strain and Hooke's law into a single material constant. It is examined directly in 9PH0 Paper 1, in core practical questions on Paper 3 (CP2 — determination of the Young modulus of a metal wire), and synoptically in mechanics (oscillations of a loaded wire treated as a vertical spring with $k = AE/L$ ). The Edexcel formula booklet provides the stress and strain definitions, and lists $E = \sigma/\varepsilon$ , but the rearranged working form $E = FL/(A\Delta L)$ — and the strain-energy density $\sigma^2/(2E)$ — must be derived in the moment.

Worked example with full mark scheme

Question (8 marks):

A student performs CP2 to determine the Young modulus of a steel wire. The wire has an unstretched length $L = 1.800 \pm 0.001$ m and a diameter measured at three points along its length giving a mean of $d = 0.380 \pm 0.005$ mm. The wire is loaded with masses in 0.5 kg increments and the extension $\Delta L$ is measured with a travelling microscope. A graph of stress $\sigma$ against strain $\varepsilon$ is plotted; the linear region passes through the origin and a chosen point at $\sigma = 1.60 \times 10^8$ Pa, $\varepsilon = 8.00 \times 10^{-4}$ .

(a) Calculate the cross-sectional area of the wire and quote the percentage uncertainty in $A$ . (2)

(b) Calculate the Young modulus from the gradient of the stress–strain graph. (2)

(c) Estimate the absolute uncertainty in $E$ given that the percentage uncertainty in $\Delta L$ over the linear region is $\pm 4\%$ . Quote the final answer to an appropriate precision. (4)

Solution with mark scheme:

(a) Step 1 — compute area from diameter.

Convert: $d = 3.80 \times 10^{-4}$ m, so $r = 1.90 \times 10^{-4}$ m.

$A = \pi r^2 = \pi (1.90 \times 10^{-4})^2 = 1.134 \times 10^{-7}\ \text{m}^2$

M1 — correct conversion of mm to m and use of radius (not diameter) in $\pi r^2$ . The dominant slip is using $\pi d^2$ instead of $\pi (d/2)^2$ , quadrupling the area and halving the deduced $E$ .

A1 — area to 3 sf with percentage uncertainty $2 \Delta d/d = 2 \times (0.005/0.380) = 2.63\%$ (the diameter is squared, so its fractional uncertainty doubles).

(b) Step 1 — gradient of linear region.

$E = \dfrac{\Delta \sigma}{\Delta \varepsilon} = \dfrac{1.60 \times 10^8}{8.00 \times 10^{-4}} = 2.00 \times 10^{11}\ \text{Pa} = 200\ \text{GPa}$

M1 — gradient calculated from two well-separated points on the linear region (not from a single data pair).

A1 — answer in the range 195–205 GPa, consistent with the accepted value for steel ( $\approx 200$ GPa).

$E = FL/(A\Delta L)$ — the dominant contributors are $\Delta L$ (4%) and $A$ (2.63% from above). $L$ contributes $0.001/1.800 \approx 0.06\%$ — negligible. Force from gravity carries percentage uncertainty $\Delta m/m$ , also small if 0.5 kg masses are calibrated. Add fractional uncertainties:

$\dfrac{\Delta E}{E} = \dfrac{\Delta L_{\text{ext}}}{\Delta L} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta L_0}{L_0} + \dfrac{\Delta F}{F} \approx 4.0\% + 2.6\% + 0.06\% + \text{small} \approx 6.7\%$

M1 — recognising that percentage uncertainties add in a product/quotient.

M1 — explicitly doubling $\Delta d/d$ .

Step 2 — convert to absolute uncertainty.

$\Delta E = 6.7\% \times 200 = 13.4$ GPa, sensibly rounded to 1 sf for the uncertainty: $\pm 10$ GPa.

A1 — absolute uncertainty in the range 10–15 GPa.

Step 3 — quote sensibly.

$E = (2.0 \pm 0.1) \times 10^{11}$ Pa, or equivalently $E = 200 \pm 10$ GPa.

A1 — value and uncertainty quoted with consistent precision; the absolute uncertainty governs the precision of the mean.

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A copper wire of length 2.50 m and diameter 0.45 mm hangs vertically. A mass of 4.0 kg is attached to its lower end. Take $E_{\text{Cu}} = 130$ GPa, $g = 9.81$ N kg⁻¹.

(a) Calculate the extension of the wire. (3)

(b) The wire is removed and replaced with a wire of the same material but twice the diameter and half the length. The same 4.0 kg mass is attached. Without recalculating from scratch, determine the new extension and justify your reasoning. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2.1) — area $A = \pi (0.225 \times 10^{-3})^2 = 1.59 \times 10^{-7}$ m² and force $F = mg = 39.24$ N.
M1 (AO1.1b) — substitute into $\Delta L = FL/(AE)$ : $\Delta L = (39.24 \times 2.50)/(1.59 \times 10^{-7} \times 1.30 \times 10^{11})$ .
A1 (AO1.1b) — $\Delta L = 4.74 \times 10^{-3}$ m $\approx 4.7$ mm.

(b)

M1 (AO2.1) — recognise that $\Delta L \propto L/A$ and that doubling the diameter quadruples $A$ , while halving $L$ halves the numerator. So new extension $= \Delta L_{\text{old}} \times (1/2)/(4) = \Delta L_{\text{old}}/8$ .
M1 (AO2.5) — explicit reasoning chain stated, not just the numerical factor.
A1 (AO1.1b) — new extension $= 4.74/8 = 0.59$ mm.

Total: 6 marks split AO1 = 4, AO2 = 2. Part (b) is the AO2 reasoning question Edexcel uses to separate B from A — most candidates can recompute, but the marks are awarded for the proportional argument.

Synoptic links

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