Hooke's Law & the Elastic Moduli

For small deformations, stress is proportional to strain.

12 minUpdated 21 October 2026Hooke's lawYoung's modulusshear modulusbulk moduluscompressibilityPoisson's ratio

Lesson at a glanceShow the key points

Hooke's law: for small deformations stress ∝ strain; the constant is a modulus of elasticity.
Young's modulus Y = (F/A)/(ΔL/L) = FL/(AΔL) (14.6); the same for tension and compression. Steel ≈ 2.0 × 10¹¹ N/m².
Shear (rigidity) modulus η = (F/A)/(x/h) (14.7).
Bulk modulus B = −P/(ΔV/V) = −V dP/dV (14.8); compressibility k = 1/B (14.9). Water: k ≈ 4.9 × 10⁻¹⁰ m²/N.
Poisson's ratio σ = −(Δd/d)/(ΔL/L) (14.10): a stretched rod gets thinner.

Double the load on a wire and, as long as it’s a modest load, it stretches twice as far. Robert Hooke spotted this in 1660. The constants of proportionality tell you how stiff a material is.

Hooke’s law

Each kind of stress–strain pair gives its own constant, a modulus of elasticity.

Young’s modulus

A rod of length L and cross-section A is stretched by forces F at its ends and lengthens by ΔL:

Young's modulus(14.5, 14.6)

Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}

the same value applies to compression

Interactive

Stretching a wire

Hang a load on a wire. Hooke's law: strain = stress/Y, so the extension is ΔL = FL/(AY). The wire also gets slightly thinner: Poisson's ratio σ links the two fractional changes. (The picture exaggerates the stretch enormously.)

F/A

9.95 × 10⁷N/m²

\Delta L/L

4.97 × 10⁻⁴

\Delta L = FL/AY

1.492mm

diameter shrinks by0.106µm

Material

Load

M

10.0kg

Length

L

3.0m

Radius

r

0.56mm

Worked example

Measuring Y for steel

A steel wire 3.0 m long with cross-section 1.0 mm² stretches by 1.5 mm when a 10 kg load hangs from it. Find Young’s modulus.

Shear modulus (modulus of rigidity)

Shear modulus(14.7)

\eta = \frac{\text{shearing stress}}{\text{shearing strain}} = \frac{F/A}{x/h} = \frac{Fh}{Ax}

Worked example

Shearing a steel plate

A steel block 1.0 cm thick has a face of area 50 cm². A tangential force of 1.0 × 10⁵ N acts on that face, the opposite face being fixed. How far does it shift? (η = 8.4 × 10¹⁰ N/m²)

Bulk modulus and compressibility

A volume stress (pressure) P changes the volume by ΔV (a decrease, so ΔV is negative):

Bulk modulus(14.8, 14.9)

B = -\frac{P}{\Delta V/V} = -V\frac{dP}{dV}, \qquad k = \frac1B\ \text{(compressibility)}

the minus sign makes B positive

Worked example

Squeezing water

By what fraction does water’s volume change at the bottom of a deep ocean trench, where the extra pressure is about 100 atm (1.0 × 10⁷ Pa)? (k = 4.9 × 10⁻¹⁰ m²/N)

Poisson’s ratio

Stretch a rod and it gets longer and also thinner. The fractional change in width is proportional to the fractional change in length:

Poisson's ratio(14.10)

\sigma = -\frac{\Delta d/d}{\Delta L/L}

the minus sign makes σ positive; typical metals 0.2–0.35

\text{rod of diameter }1\ \text{cm, strain }10^{-3},\ \sigma = 0.3:\quad \Delta d = -0.3\times10^{-3}\times1\ \text{cm} = -3\ \mu\text{m}

Material	Y (10¹¹ N/m²)	η (10¹¹ N/m²)	B (10¹¹ N/m²)	σ
Aluminium	0.70	0.30	0.70	0.16
Brass	0.91	0.36	0.61	0.26
Copper	1.1	0.42	1.4	0.32
Iron	1.9	0.70	1.0	0.27
Steel	2.0	0.84	1.6	0.19
Tungsten	3.6	1.5	2.0	0.20

What to remember

1Hooke's law: for small deformations stress ∝ strain.
2 $Y = FL/(A\Delta L)$ , $\eta = Fh/(Ax)$ , $B = -V\,dP/dV$ , $k = 1/B$ .
3Poisson’s ratio $\sigma = -(\Delta d/d)/(\Delta L/L)$ .

Check yourself

Two wires of the same material and length; one has twice the diameter. Under the same load, its extension is…

Which material in the table is stiffest in tension?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchAn introduction to stress and strainThe Efficient Engineer ReadStress, strain and elastic modulusOpenStax University Physics §12.3 ReadElastic propertiesHyperPhysics ReadYoung's modulusWikipedia ReadShear modulusWikipedia ReadBulk modulusWikipedia ReadPoisson's ratioWikipedia

PreviousElasticity, Stress & Strain Next Beyond Hooke: Stress–Strain Curves

Hooke's Law & the Elastic Moduli

For small deformations, stress is proportional to strain.

12 minUpdated 21 October 2026Hooke's lawYoung's modulusshear modulusbulk moduluscompressibilityPoisson's ratio

Lesson at a glanceShow the key points

Hooke's law: for small deformations stress ∝ strain; the constant is a modulus of elasticity.
Young's modulus Y = (F/A)/(ΔL/L) = FL/(AΔL) (14.6); the same for tension and compression. Steel ≈ 2.0 × 10¹¹ N/m².
Shear (rigidity) modulus η = (F/A)/(x/h) (14.7).
Bulk modulus B = −P/(ΔV/V) = −V dP/dV (14.8); compressibility k = 1/B (14.9). Water: k ≈ 4.9 × 10⁻¹⁰ m²/N.
Poisson's ratio σ = −(Δd/d)/(ΔL/L) (14.10): a stretched rod gets thinner.

Double the load on a wire and, as long as it’s a modest load, it stretches twice as far. Robert Hooke spotted this in 1660. The constants of proportionality tell you how stiff a material is.

Hooke’s law

Each kind of stress–strain pair gives its own constant, a modulus of elasticity.

Young’s modulus

A rod of length L and cross-section A is stretched by forces F at its ends and lengthens by ΔL:

Young's modulus(14.5, 14.6)

Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}

the same value applies to compression

Interactive

Stretching a wire

F/A

9.95 × 10⁷N/m²

\Delta L/L

4.97 × 10⁻⁴

\Delta L = FL/AY

1.492mm

diameter shrinks by0.106µm

Material

Load

M

10.0kg

Length

L

3.0m

Radius

r

0.56mm

Worked example

Measuring Y for steel

A steel wire 3.0 m long with cross-section 1.0 mm² stretches by 1.5 mm when a 10 kg load hangs from it. Find Young’s modulus.

Shear modulus (modulus of rigidity)

Shear modulus(14.7)

\eta = \frac{\text{shearing stress}}{\text{shearing strain}} = \frac{F/A}{x/h} = \frac{Fh}{Ax}

Worked example

Shearing a steel plate

A steel block 1.0 cm thick has a face of area 50 cm². A tangential force of 1.0 × 10⁵ N acts on that face, the opposite face being fixed. How far does it shift? (η = 8.4 × 10¹⁰ N/m²)

Bulk modulus and compressibility

A volume stress (pressure) P changes the volume by ΔV (a decrease, so ΔV is negative):

Bulk modulus(14.8, 14.9)

B = -\frac{P}{\Delta V/V} = -V\frac{dP}{dV}, \qquad k = \frac1B\ \text{(compressibility)}

the minus sign makes B positive

Worked example

Squeezing water

By what fraction does water’s volume change at the bottom of a deep ocean trench, where the extra pressure is about 100 atm (1.0 × 10⁷ Pa)? (k = 4.9 × 10⁻¹⁰ m²/N)

Poisson’s ratio

Stretch a rod and it gets longer and also thinner. The fractional change in width is proportional to the fractional change in length:

Poisson's ratio(14.10)

\sigma = -\frac{\Delta d/d}{\Delta L/L}

the minus sign makes σ positive; typical metals 0.2–0.35

\text{rod of diameter }1\ \text{cm, strain }10^{-3},\ \sigma = 0.3:\quad \Delta d = -0.3\times10^{-3}\times1\ \text{cm} = -3\ \mu\text{m}

Material	Y (10¹¹ N/m²)	η (10¹¹ N/m²)	B (10¹¹ N/m²)	σ
Aluminium	0.70	0.30	0.70	0.16
Brass	0.91	0.36	0.61	0.26
Copper	1.1	0.42	1.4	0.32
Iron	1.9	0.70	1.0	0.27
Steel	2.0	0.84	1.6	0.19
Tungsten	3.6	1.5	2.0	0.20

What to remember

1Hooke's law: for small deformations stress ∝ strain.
2 $Y = FL/(A\Delta L)$ , $\eta = Fh/(Ax)$ , $B = -V\,dP/dV$ , $k = 1/B$ .
3Poisson’s ratio $\sigma = -(\Delta d/d)/(\Delta L/L)$ .

Check yourself

Two wires of the same material and length; one has twice the diameter. Under the same load, its extension is…

Which material in the table is stiffest in tension?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousElasticity, Stress & Strain Next Beyond Hooke: Stress–Strain Curves

Hooke’s law#

Young’s modulus

Shear modulus (modulus of rigidity)

Bulk modulus and compressibility

Poisson’s ratio

Check yourself

Keep exploring

Hooke’s law#

Young’s modulus

Shear modulus (modulus of rigidity)

Bulk modulus and compressibility

Poisson’s ratio

Check yourself

Keep exploring

Hooke’s law

Hooke’s law