# Is Uniaxial Compression the Same as Biaxial Tension?

## Introduction

I frequently meet people that have opinions about whether uniaxial compression can be used as a substitute for a biaxial tension test. If you think about, it would certainly be great if it was true, since biaxial tension experiments are both difficult to perform and expensive. In this article I will try to answer, once and for all, if this is true or an urban legend. ### Uniaxial Compression

• Easy to perform
• Friction issues

Assume incompressible:

$$\mathbf{F} = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & 1/\sqrt{\lambda} & 0 \\ 0 & 0 & 1/\sqrt{\lambda} \end{bmatrix}$$

Take $$\lambda=0.5$$, giving:

$$\mathbf{F} = \begin{bmatrix} 0.5 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \end{bmatrix}$$

### Biaxial Tension

• Difficult to perform
• No friction
• Useful for calibrating an I2-based hyperelastic model
Assume incompressible:
$$\mathbf{F} = \begin{bmatrix} 1/\lambda^2 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}$$

Take $$\lambda = \sqrt{2}$$, giving:

$$\mathbf{F} = \begin{bmatrix} 0.5 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \end{bmatrix}$$

$$\boldsymbol{\sigma} = \begin{bmatrix} \sigma_u & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

Incompressible biaxial tension:

$$\boldsymbol{\sigma} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & \sigma_b & 0 \\ 0 & 0 & \sigma_b \end{bmatrix} = \begin{bmatrix} -\sigma_b & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + \sigma_b \mathbf{I}$$

The following table is from my book. It shows that one can convert from uniaxial compression data to biaxial tension data with good accuracy if the Poisson’s ratio is close to 0.5. That is, if the material is a rubber. For thermoplastics, which have a Poisson’s ratio around 0.4, you cannot accurately convert uniaxial compression data to biaxial tension data.   