The Young Modulus

Quantifying material stiffness and the required practical for measuring it

SPEC 3.4.2.2 • MATERIALS

What is Young Modulus?

The Young modulus (E) is a measure of how stiff a material is. It tells us how much a material resists being stretched or compressed.

Definition

Young modulus is the ratio of tensile stress to tensile strain:

E = \frac{σ}{ε} = \frac{stress}{strain}

Key Concept

Young modulus is a material property - it depends only on what the material is made of, not on the shape or size of the object. A thin steel wire and a thick steel beam have the same Young modulus.

This is different from spring constant (k), which is a property of the specific object.

Formula and Units

Key Formulas

E = \frac{σ}{ε} = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}

Wire under tension: Original length L, cross-sectional area A, extends by ΔL when force F applied

Units

E is measured in pascals (Pa)

Or equivalently: N m⁻²

Typical values are in GPa (10⁹ Pa)

Why Pa?

Since strain is dimensionless:

[E] = \frac{[ σ ]}{[ ε ]} = \frac{Pa}{1} = Pa

MaterialYoung Modulus (GPa)

Steel200

Copper120

Aluminium70

Glass70

Rubber0.01 - 0.1

Material vs Object Property

Young Modulus (E)

Material property

Same for all objects made of that material

Does not depend on dimensions

Steel wire and steel beam: same E

Spring Constant (k)

Object property

Depends on material AND dimensions

Changes with length, area, shape

Thick wire vs thin wire: different k

Relationship

For a wire of length L and cross-sectional area A:

k = \frac{E A}{L}

This shows how the spring constant depends on both the material (E) and the geometry (A/L).

Worked Example01

A steel wire has length 2.0 m and diameter 0.50 mm. When a force of 40 N is applied, the wire extends by 0.80 mm. Calculate the Young modulus of steel.

1

Calculate cross-sectional area:

A = π r^{2} = π \times (0.25 \times 1 0^{- 3})^{2} = 1.96 \times 1 0^{- 7} m^{2}

2

Calculate stress:

σ = \frac{F}{A} = \frac{40}{1.96 \times 1 0 ^{- 7}} = 2.04 \times 1 0^{8} Pa

3

Calculate strain:

ε = \frac{Δ L}{L} = \frac{0.80 \times 1 0 ^{- 3}}{2.0} = 4.0 \times 1 0^{- 4}

4

Calculate Young modulus:

E = \frac{σ}{ε} = \frac{2.04 \times 1 0 ^{8}}{4.0 \times 1 0 ^{- 4}} = 5.1 \times 1 0^{11} Pa

Answer: E = 5.1 × 10¹¹ Pa = 510 GPa

(Note: This is higher than typical steel ~200 GPa, suggesting measurement uncertainty in the question data)

Worked Example02

An aluminium rod has Young modulus 70 GPa, length 1.5 m, and cross-sectional area 2.0 × 10⁻⁴ m². Calculate: (a) the extension when a 50 kN force is applied, (b) the spring constant of the rod.

a)

Rearrange

E = \frac{F L}{A Δ L}

for extension:

Δ L = \frac{F L}{E A}

•

Δ L = \frac{50 \times 1 0 ^{3} \times 1.5}{70 \times 1 0 ^{9} \times 2.0 \times 1 0 ^{- 4}} = \frac{75000}{14 \times 1 0 ^{6}} = 5.4 \times 1 0^{- 3} m

Answer (a): ΔL = 5.4 mm

b)

Spring constant:

k = \frac{E A}{L} = \frac{70 \times 1 0 ^{9} \times 2.0 \times 1 0 ^{- 4}}{1.5} = 9.3 \times 1 0^{6} N m^{- 1}

Answer (b): k = 9.3 MN m⁻¹

Worked Example03

Two wires P and Q are made of the same material. Wire P has length L and diameter d. Wire Q has length 2L and diameter 2d. Compare: (a) their Young modulus values, (b) their spring constants.

a)

Young modulus is a material property. Since both wires are made of the same material, they have the same Young modulus.

Answer (a): E_P = E_Q (same material)

b)

Spring constant

k = \frac{E A}{L}

•

For P:

k_{P} = \frac{E \times π d ^{2} /4}{L}

•

For Q:

k_{Q} = \frac{E \times π ( 2 d ) ^{2} /4}{2 L} = \frac{E \times 4 π d ^{2} /4}{2 L} = \frac{4 E π d ^{2} /4}{2 L} = 2 k_{P}

Answer (b): k_Q = 2k_P

(Area increases by factor 4, length by factor 2 → k increases by factor 4/2 = 2)

Young Modulus from Graphs

The Young modulus can be found from the gradient of a stress-strain graph in the linear (elastic) region.

Stress-strain graph

Gradient = E

Finding E from Graph

E = gradient = \frac{Δ σ}{Δ ε}

Only valid in the linear region (up to the elastic limit)

Use a large triangle for accuracy when reading gradient.

Common Mistake

Don't confuse a stress-strain graph (gradient = E) with a force-extension graph (gradient = k = EA/L). You must convert force-extension to stress-strain first, or use the relationship carefully.

Worked Example04

A stress-strain graph for a metal wire shows the following data points in the linear region: at strain 0.001, stress = 120 MPa; at strain 0.003, stress = 360 MPa. Calculate the Young modulus.

1

Identify the two points: (0.001, 120 MPa) and (0.003, 360 MPa)

2

Calculate the gradient:

E = \frac{Δ σ}{Δ ε} = \frac{360 - 120}{0.003 - 0.001} \times 1 0^{6}

3

E = \frac{240 \times 1 0 ^{6}}{0.002} = 1.2 \times 1 0^{11} Pa = 120 GPa

Answer: E = 120 GPa

(This is consistent with copper - the graph tells us the material)

Worked Example05

In a Young modulus experiment, a student plots Force (N) against Extension (m) and finds the gradient is 2.5 × 10⁴ N m⁻¹. The wire has length 1.80 m and diameter 0.40 mm. Calculate the Young modulus.

1

The gradient of F-ΔL graph = k = EA/L, so E = (gradient × L)/A

2

Calculate area:

A = \frac{π d ^{2}}{4} = \frac{π \times ( 0.40 \times 1 0 ^{- 3} ) ^{2}}{4} = 1.26 \times 1 0^{- 7} m^{2}

3

Calculate E:

E = \frac{gradient \times L}{A} = \frac{2.5 \times 1 0 ^{4} \times 1.80}{1.26 \times 1 0 ^{- 7}}

4

E = \frac{4.5 \times 1 0 ^{4}}{1.26 \times 1 0 ^{- 7}} = 3.6 \times 1 0^{11} Pa = 360 GPa

Answer: E = 360 GPa

Worked Example06

A student measures the diameter of a wire as 0.48 mm with uncertainty ±0.02 mm. Calculate: (a) the cross-sectional area, (b) the percentage uncertainty in the area.

a)

Area:

A = \frac{π d ^{2}}{4} = \frac{π \times ( 0.48 \times 1 0 ^{- 3} ) ^{2}}{4} = 1.81 \times 1 0^{- 7} m^{2}

Answer (a): A = 1.81 × 10⁻⁷ m²

b)

Percentage uncertainty in d:

\frac{0.02}{0.48} \times 100% = 4.2%

•

Since A ∝ d², the percentage uncertainty in A is doubled: 2 × 4.2% = 8.4%

Answer (b): 8.4%

Worked Example07

A copper wire (E = 120 GPa) and a steel wire (E = 200 GPa) have the same length and diameter. They are joined end-to-end and a tensile force is applied. Which wire experiences: (a) greater stress, (b) greater strain, (c) greater extension?

a)

Stress σ = F/A. Both wires carry the same force F and have the same area A.

Answer (a): Same stress in both wires

b)

Strain ε = σ/E. Same stress, but copper has lower E.

•

ε_{C u} = \frac{σ}{120}

and

ε_{St} = \frac{σ}{200}

Answer (b): Copper has greater strain (lower E means more strain for same stress)

c)

Extension ΔL = εL. Same length, but copper has greater strain.

Answer (c): Copper has greater extension

Practice

Question 1: A steel cable has length 25 m, diameter 8.0 mm, and Young modulus 200 GPa. Calculate the extension when a force of 15 kN is applied.

1

Area:

A = \frac{π \times ( 8.0 \times 1 0 ^{- 3} ) ^{2}}{4} = 5.03 \times 1 0^{- 5} m^{2}

2

Extension:

Δ L = \frac{F L}{E A} = \frac{15 \times 1 0 ^{3} \times 25}{200 \times 1 0 ^{9} \times 5.03 \times 1 0 ^{- 5}}

3

Δ L = \frac{375000}{1.006 \times 1 0 ^{7}} = 3.7 \times 1 0^{- 2} m = 37 mm

Answer: ΔL = 37 mm

Practice

Question 2: Two wires X and Y are made of materials with Young modulus 200 GPa and 100 GPa respectively. Both have the same length and diameter. If the same force is applied to each, calculate the ratio of their extensions (extension of Y : extension of X).

1

Extension formula:

Δ L = \frac{F L}{E A}

2

For same F, L, and A:

Δ L \propto \frac{1}{E}

3

Ratio:

\frac{Δ L _{Y}}{Δ L _{X}} = \frac{E _{X}}{E _{Y}} = \frac{200}{100} = 2

Answer: Extension of Y : Extension of X = 2 : 1

(Lower E means larger extension for same stress)

Practice

Question 3: A wire of length 2.5 m and cross-sectional area 1.2 × 10⁻⁶ m² is stretched. When the stress is 150 MPa, the strain is 7.5 × 10⁻⁴. Calculate: (a) the Young modulus, (b) the extension, (c) the elastic strain energy stored.

a)

E = \frac{σ}{ε} = \frac{150 \times 1 0 ^{6}}{7.5 \times 1 0 ^{- 4}} = 2.0 \times 1 0^{11} Pa = 200 GPa

Answer (a): E = 200 GPa

b)

Δ L = ε \times L = 7.5 \times 1 0^{- 4} \times 2.5 = 1.875 \times 1 0^{- 3} m = 1.9 mm

Answer (b): ΔL = 1.9 mm

c)

Force:

F = σ A = 150 \times 1 0^{6} \times 1.2 \times 1 0^{- 6} = 180 N

•

Energy:

E_{s t r ain} = \frac{1}{2} F Δ L = \frac{1}{2} \times 180 \times 1.875 \times 1 0^{- 3} = 0.17 J

Answer (c): Strain energy = 0.17 J

Summary

Key Formulas

FormulaDescription

E = \frac{σ}{ε}

Young modulus definition

E = \frac{F L}{A Δ L}

Combined formula

k = \frac{E A}{L}

Spring constant from E

Gradient of σ-ε graphYoung modulus from graph

Key Exam Points

Young modulus is a material property - same for all objects of that material

Spring constant is an object property - depends on E, A, and L

Gradient of stress-strain graph = E (only in linear region)

Higher E = stiffer material (small strain for large stress)

Use

k = \frac{E A}{L}

to relate spring constant to Young modulus