Engineer's Notebook §1.2

Poisson's ratio

Stretch a bar and it gets thinner; squash it and it bulges. Poisson's ratio is the lateral strain per unit axial strain:

ν = −εlatεax
ν
Poisson's ratio
—
εax
axial strain
—
εlat
lateral strain
—
d
width
m
E
Young's modulus
Pa
G
shear modulus
Pa
K
bulk modulus
Pa
V
volume
m³
A rectangular bar, outlined dashed at its original size and drawn solid after loading: longer and narrower in tension, shorter and wider in compression. d = 20 mm → 19.94 mm strains drawn ×10
Fig. 1.2-1Dashed: original shape. Strains exaggerated by the scale noted.
Δd = εlatd−0.06 mm
ΔV/V = εax(1 − 2ν)0.4%
εlat = −νεax−0.3%

Worked example

Uses the values in Fig. 1.2-1. Change them and the working updates.

Given
  • νsteel0.3
  • εaxaxial strain1%
  • doriginal width20 mm
Find

The lateral strain and the change in width.

Solution
εlat = −νεax = −0.3 × 1% = −0.3% Δd = εlatd = −3 × 10⁻³ × 20 mm = −0.06 mm
Answer
εlat = −0.3%

The width changes by −0.06 mm, and the volume by 0.4%.

Related elastic constants

  • Shear modulus

    E = 2G(1 + ν)

  • Bulk modulus

    K = E3(1 − 2ν)

  • Volume change, uniaxial

    ΔVV = εax(1 − 2ν)

Assumptions
  • The material is isotropic and linear-elastic, and strains are small.
  • The stress is uniaxial: the sides are free to move.
  • For isotropic materials −1 < ν ≤ 0.5. At 0.5 the volume never changes, which rubber nearly achieves.
  • Values are typical. Cork's ratio is close to zero, which is why a cork goes into a bottle without bulging.

Derivation
  1. Apply axial strain ε. Each lateral direction then strains by −νε.
  2. A block of volume V becomes V(1 + ε)(1 − νε)².
  3. Expand and drop the squared terms, which are tiny for small strains: V(1 + ε − 2νε).
  4. So ΔV/V = ε(1 − 2ν).

With ν = 0.5 the volume stays constant, which is why rubber is nearly incompressible yet easy to stretch.