Strain & Poisson''s Ratio

Calculate axial strain, shear strain, and transverse deformation via Poisson''s ratio.

Inputs

l

Original Length

l'

Deformation

Formula Interpretation

Axial Strain — Formula ①

\varepsilon = \dfrac{l'}{l} \quad \text{(dimensionless)}

$l$ is the original length (mm); $l'$ is the absolute deformation (mm); $\varepsilon$ is the resulting dimensionless strain.

Shear Strain — Formula ②

\gamma = \dfrac{\lambda}{l} = \tan\varphi \approx \varphi \quad \text{(rad)}

$\lambda$ is the lateral displacement (mm); $l$ is the element height (mm); $\gamma$ ≈ $\varphi$ for small angles (radians).

Poisson''s Ratio — Formula ③

\nu = \dfrac{\varepsilon_1}{\varepsilon} = \dfrac{1}{m}

$\varepsilon_1$ is the transverse strain; $\varepsilon$ is the axial strain; $\nu$ is Poisson's ratio; $m$ is Poisson's number.

Poisson''s Number — Formula ④

m = \dfrac{1}{\nu} = \dfrac{\varepsilon}{\varepsilon_1}

$m$ = 1/ $\nu$ is the reciprocal of Poisson's ratio.

Knowledge Points

Axial Strain

Axial strain ε is the ratio of deformation to original length. Tensile strain is positive (elongation); compressive strain is negative (shortening). If represented by l'', ε = l''/l.

Shear Strain

Shear strain γ is the angular deformation of an element due to shear loading. For small angles, γ = λ/l ≈ tan φ ≈ φ (in radians).

Poisson''s Ratio & Number

Poisson''s ratio ν is the ratio of transverse strain to axial strain: ν = ε₁/ε. Its reciprocal m = 1/ν is called Poisson''s number. Within the elastic limit, ν is constant for a given material.

Worked Example

An aluminum rod with length $l = 100\,\text{mm}$ and a square cross-section of side $b = 30\,\text{mm}$ is subjected to compressive loading. The length decreases by $l' = 0.78\,\text{mm}$ . The Poisson's ratio of aluminum is $\nu = 0.34$ . Find the cross-sectional area after compression.

Step 1 — Compute axial strain ε

\varepsilon = \dfrac{l'}{l} = \dfrac{0.78}{100} = 0.0078

Step 2 — Compute transverse strain ε₁ using Poisson''s ratio

\varepsilon_1 = \dfrac{1}{m} \cdot \varepsilon = \nu \cdot \varepsilon = 0.34 \times 0.0078 = 0.002652

Step 3 — Find new side length after lateral expansion

b' = b(1 + \varepsilon_1) = 30 \times (1 + 0.002652) \approx 30.08 \text{ (mm)}

Step 4 — Compute new cross-sectional area

A = b'^2 = 30.08^2 \approx 905 \text{ (mm}^2\text{)}

The cross-sectional area after compression is $\approx 905\,\text{mm}^2$ .

Extended Knowledge

•Poisson''s ratio is a fundamental material property used in finite element analysis (FEA) to model three-dimensional deformation under complex loading.
•For incompressible materials (e.g. rubber, biological tissues), ν ≈ 0.5 meaning volume is conserved under loading.
•Shear strain is directly related to shear stress through the shear modulus G: τ = G·γ, analogous to Hooke''s Law for normal stress.
•In composite materials, Poisson''s ratio can vary with fibre orientation, making multi-axis strain analysis critical in aerospace and automotive design.
•Strain gauges measure surface strain directly and are widely used in structural health monitoring, load cells and experimental stress analysis.