Hooke's Law and the Elastic Constants
Four constants, and only two of them are independent — plus the reason Poisson's ratio can never exceed a half.
Skip to the animationWithin the elastic region stress is proportional to strain, σ = Eε — and an isotropic material has four elastic constants of which only two are independent, tied together by E = 2G(1+ν) and E = 3K(1−2ν).
Why stress and strain rather than load and extension
A thick bar stretches less than a thin one under the same load, and a long bar stretches more than a short one. Dividing force by area and extension by original length removes both effects, so what is left describes the material rather than this particular specimen.
The gradient of the resulting straight line is Young's modulus E. For steel it is about 200 GPa regardless of grade, which turns out to matter enormously when columns are discussed.
Poisson's ratio
Pulling a bar lengthways makes it contract sideways. The ratio of lateral strain to longitudinal strain is Poisson's ratio ν, defined with a minus sign so that it comes out positive for ordinary materials.
| Material | ν | Consequence |
|---|---|---|
| Cork | ≈ 0 | Pushes into a bottleneck without bulging |
| Concrete | 0.2 | Modest lateral expansion under load |
| Steel | 0.3 | The standard assumption |
| Rubber | ≈ 0.4999 | Effectively incompressible |
Four constants, two of them independent
- E — Young's modulus
- Direct stress divided by direct strain. From a tensile test.
- G — modulus of rigidity
- Shear stress divided by shear strain. From a torsion test.
- K — bulk modulus
- Hydrostatic pressure divided by volumetric strain.
- ν — Poisson's ratio
- Lateral strain divided by longitudinal strain, negated.
For an isotropic material these are linked by E = 2G(1+ν) and E = 3K(1−2ν). Measure any two and the rest follow. Substituting steel's E = 200 GPa and ν = 0.3 returns G = 77 GPa and K = 167 GPa — exactly the handbook values, which is worth checking once.
Why ν cannot exceed 0.5
In K = E/3(1−2ν), the denominator vanishes as ν approaches 0.5, sending the bulk modulus to infinity — the material becomes incompressible. Rubber sits just below that, which is why it changes shape freely but cannot be squeezed in a sealed cavity.
Above 0.5 the bulk modulus would go negative: the material would expand in volume when compressed. So 0.5 is a genuine physical ceiling rather than an empirical observation.
When two constants are not enough
Everything above assumes isotropy — identical properties in every direction. Steel, aluminium and glass qualify. Timber, composites and heavily rolled sheet do not.
- Timber's modulus along the grain exceeds that across it by more than twenty times.
- Timber is usually modelled as orthotropic, with nine independent constants.
- A fully anisotropic material requires twenty-one.
- Which is why a timber beam is loaded along the grain, and why joints across the grain govern the design.
Where the law stops
Hooke's law holds only to the proportional limit. Past it the curve bends over, E stops describing anything, and beyond yield the deformation is permanent — the bar does not come back.
Every elastic calculation in this subject — bending stress, deflection, torsion, buckling — assumes the material is still on the straight portion. Keeping it there is what the factor of safety is for.
The numbers you will be asked for
- Hooke's law
σ = E · ε
- Extension of a bar
δ = PL / AE
- Poisson's ratio
ν = −ε_lateral / ε_longitudinal
- Shear and direct
E = 2G(1 + ν)
- Bulk and direct
E = 3K(1 − 2ν)
and this is what caps ν at 0.5
- Volumetric strain
ε_v = (1 − 2ν)·(σx + σy + σz)/E
Watch it work
Check yourself
question 1 / 4
One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.