Poisson’s ratio describes how a material changes sideways when it is stretched or compressed lengthwise.
If a rod is stretched, many ordinary materials become slightly thinner. If the rod is compressed, they become
slightly wider.
1. Longitudinal strain
Longitudinal strain is the fractional change in length:
\[
\varepsilon_{\mathrm{long}}=\frac{\Delta L}{L_0}.
\]
In the sign convention used here:
- Tension gives \(\varepsilon_{\mathrm{long}}>0\).
- Compression gives \(\varepsilon_{\mathrm{long}}<0\).
2. Lateral strain
Lateral strain is the fractional change in a sideways dimension such as diameter or width:
\[
\varepsilon_{\mathrm{lat}}=\frac{\Delta d}{d_0}.
\]
Here, \(d_0\) is the original diameter or width, and \(\Delta d=d_f-d_0\).
3. Poisson’s ratio definition
Poisson’s ratio is defined as
\[
\boxed{
\nu=-\frac{\varepsilon_{\mathrm{lat}}}{\varepsilon_{\mathrm{long}}}
}.
\]
The negative sign makes \(\nu\) positive for ordinary materials in tension, because tension has
\(\varepsilon_{\mathrm{long}}>0\) and lateral contraction has \(\varepsilon_{\mathrm{lat}}<0\).
4. Lateral strain from Poisson’s ratio
Rearranging the definition gives
\[
\boxed{
\varepsilon_{\mathrm{lat}}=-\nu\varepsilon_{\mathrm{long}}
}.
\]
The diameter or width change is then
\[
\boxed{
\Delta d=\varepsilon_{\mathrm{lat}}d_0
}.
\]
The final diameter or width is
\[
\boxed{
d_f=d_0+\Delta d
}.
\]
5. Volume change from Poisson’s effect
For small strains in a rod with equal lateral strain in two transverse directions, the approximate volume strain is
\[
\boxed{
\frac{\Delta V}{V_0}
\approx
\varepsilon_{\mathrm{long}}+2\varepsilon_{\mathrm{lat}}
}.
\]
Substitute \(\varepsilon_{\mathrm{lat}}=-\nu\varepsilon_{\mathrm{long}}\):
\[
\frac{\Delta V}{V_0}
\approx
\varepsilon_{\mathrm{long}}(1-2\nu).
\]
For a nearly incompressible material, \(\nu\) is close to \(0.5\), so \(1-2\nu\) is close to zero and volume changes
very little.
6. Exact finite-strain volume estimate
A simple finite-strain estimate for a rectangular or cylindrical sample is
\[
\frac{V_f}{V_0}
=
(1+\varepsilon_{\mathrm{long}})
(1+\varepsilon_{\mathrm{lat}})^2.
\]
Therefore,
\[
\boxed{
\frac{\Delta V}{V_0}
=
(1+\varepsilon_{\mathrm{long}})
(1+\varepsilon_{\mathrm{lat}})^2-1
}.
\]
7. Typical values
| Material type |
Typical Poisson’s ratio |
Behavior |
| Cork-like materials |
\(\nu\approx0\) |
Very little lateral contraction or expansion |
| Metals |
\(\nu\approx0.25{-}0.35\) |
Moderate lateral contraction in tension |
| Rubber-like materials |
\(\nu\approx0.49\) |
Nearly incompressible; volume changes very little |
| Auxetic materials |
\(\nu<0\) |
Expand laterally when stretched |
8. Isotropic modulus relations
For isotropic linear elastic materials, Young’s modulus \(E\), shear modulus \(G\), bulk modulus \(K\), and
Poisson’s ratio \(\nu\) are connected.
Young’s modulus and shear modulus:
\[
\boxed{
E=2G(1+\nu)
}.
\]
Therefore,
\[
\boxed{
G=\frac{E}{2(1+\nu)}
}.
\]
Young’s modulus and bulk modulus:
\[
\boxed{
K=\frac{E}{3(1-2\nu)}
}.
\]
This expression requires \(\nu<0.5\). As \(\nu\) approaches \(0.5\), the bulk modulus becomes very large, which is
the mathematical sign of near incompressibility.
9. Solving Poisson’s ratio from modulus values
From \(E=2G(1+\nu)\),
\[
\boxed{
\nu=\frac{E}{2G}-1
}.
\]
From \(K=E/[3(1-2\nu)]\),
\[
\boxed{
\nu=\frac{3K-E}{6K}
}.
\]
10. Worked example
A steel rod is stretched with
\[
\varepsilon_{\mathrm{long}}=0.0012,
\qquad
\nu=0.29.
\]
The lateral strain is
\[
\varepsilon_{\mathrm{lat}}
=
-\nu\varepsilon_{\mathrm{long}}.
\]
\[
\varepsilon_{\mathrm{lat}}
=
-(0.29)(0.0012)
=
-0.000348.
\]
If the original diameter is \(12\ \mathrm{mm}\), then
\[
\Delta d=\varepsilon_{\mathrm{lat}}d_0.
\]
\[
\Delta d=(-0.000348)(12\ \mathrm{mm})
=
-0.00418\ \mathrm{mm}.
\]
So the rod becomes slightly thinner while it is stretched.
The small-strain volume change estimate is
\[
\frac{\Delta V}{V_0}
\approx
\varepsilon_{\mathrm{long}}+2\varepsilon_{\mathrm{lat}}.
\]
\[
\frac{\Delta V}{V_0}
\approx
0.0012+2(-0.000348)
=
0.000504.
\]
The volume increases slightly because steel is not incompressible.
Key idea: Poisson’s ratio compares sideways strain with lengthwise strain and explains why rods usually become
thinner when stretched and wider when compressed.
