I've been somewhat confused over the concepts of true and engineering strain, and I just want to see if I am understanding this correctly. Let us denote true and engineering strain as $\epsilon_t$ and $\epsilon_e$.

I know that $\epsilon_t = \ln(1+\epsilon_e)$.

For small deformations, we have the following relationship between strain and displacement:

$\epsilon = 0.5[\nabla u + (\nabla u)^T]$

This relationship is for the engineering strain, and not the true strain right?

And I also understand that strain gauges measure the engineering strain.

If all of this is true, then what is the point of true strain?


3 Answers 3


As described here, some advantages of true strain are:

It’s the exact value, not an approximation predicated on small values, which makes it useful when characterizing materials that deform by large amounts (considerable fractions of their length up to many times their length). Examples include hydrogels, elastomers, biological tissue, and fabric.

Sequential strains can be added, which is not the case with engineering strain.

It’s geometrically symmetric: that is, if the strain associated with being stretched to n times the original length is e, then the strain associated with being compressed to 1/n the original length is −e.

  • $\begingroup$ Thanks! If the true strain is "exact," then why do strain gauges measure the engineering strain? Shouldn't an experimental apparatus seek to measure what is true? $\endgroup$
    – anonuser01
    Nov 28, 2018 at 1:30
  • $\begingroup$ You seem to imply that the true (logarithmic) strain is fundamentally better than other large strain measures (e.g., Cauchy strain tensor) for characterizing the behavior of real materials at large deformations. Do you have experimental evidence to back this up? $\endgroup$ Nov 28, 2018 at 1:32
  • $\begingroup$ @Iamanon It's not that a strain gauge inherently measures engineering strain, but through a series of idealizations and assumptions, we obtain a functional form that conveniently matches that of engineering strain. Note that the true and engineering strains are essentially identical at the working ranges generally used with strain gauges. $\endgroup$ Nov 28, 2018 at 16:30
  • $\begingroup$ @ChesterMiller I did not mean to imply this. $\endgroup$ Nov 28, 2018 at 16:31

The second formula approximates the true strain, but is not equal to it. On the one hand the true strain is:

$$\varepsilon_t = \frac{\text{d}L_f}{\text{d}L_0} - 1$$

On the other hand the second formula proceeds from defining the right Cauchy–Green deformation tensor as:

$$\mathbf{C} = (\mathbf{1} + \nabla\boldsymbol{u})^T(\mathbf{1} + \nabla\boldsymbol{u})$$

This tensor is related to the true strain by:

$$(1+\varepsilon_t)^2 = \left(\frac{\text{d}L_f}{\text{d}L_0}\right)^2 = \boldsymbol{n}\cdot (\mathbf{C}\boldsymbol{n})$$

where $\boldsymbol{n}$ is a unit vector in the direction in which you are calculating the strain. The above formula implies that:

$$1+\varepsilon_t = \sqrt{\boldsymbol{n}\cdot \mathbf{C}\cdot \boldsymbol{n}} = \sqrt{ [(\mathbf{1} + \nabla\boldsymbol{u})\boldsymbol{n}]^T[(\mathbf{1} + \nabla\boldsymbol{u})\boldsymbol{n}]}$$

Using a Taylor series development to linearize the above expression, we obtain:

$$1+\varepsilon_t \approx 1 + \boldsymbol{n}\cdot \left(\frac{(\nabla\boldsymbol{u})^T + \nabla\boldsymbol{u}}{2}\right) \cdot \boldsymbol{n}$$

Therefore, the infinitesimal tensor $[(\nabla\boldsymbol{u})^T + \nabla\boldsymbol{u}]/2$ does not coincide with either the true strain or the engineering strain. If we had not done this last linerization we would have obtained the true strain, but computationally the formula obtained is disadvantageous.


The second formula you show is for the true strain, this is the physical strain in the material. It is is different in every location of the material, and is practically impossible to measure. In order to have something more practical to work with, the engineering strain has been invented. This is nothing more than some measure for the average strain in a standard tensile test specimen. It has very little physical meaning, but because everybody uses it the same way, it is very useful for comparing material properties.

Strain gauges cant really measure strain, they can only measure their own average elongation. This gives information about the true strain of the material it is attached to, in one direction only.

This link gives a good explanation:

  • $\begingroup$ Wait, so strain gauges measure the engineering strain right? I was told that the small deformation equation that I wrote for the strain is the engineering strain, which didn't make sense, but now that I see it is actually the true strain. $\endgroup$
    – anonuser01
    Nov 28, 2018 at 15:36
  • $\begingroup$ Strain gauges measure true strain, but for small deformations there is no difference. $\endgroup$
    – Orbit
    Nov 28, 2018 at 20:24
  • $\begingroup$ Now I am confused again. researchgate.net/post/… In this post, the answers are saying strain gauges measure engineering strain? Yes I understand engineering strain is approximately the same as true strain for small deformations, which is evident from their log relationship. Are you saying that the strain gauge measures the true strain because it's about the same as engineering strain for small deformations? Or are you saying that strain gauge measures true strain regardless of how large the deformation? $\endgroup$
    – anonuser01
    Nov 29, 2018 at 1:06
  • $\begingroup$ Please look at the last large post from Jean-Philippe Mathieu in your link, he explains it quite well. Engineering strain only applies to standard tensile tests. Engineering strain is defined as the extension of the tensile test machine clams, divided by the original length of the test specimen. This is usually measured by an extensometer, but you could try to use a strain gauge on a piece of rubber instead. True strain is a real strain that exists for every situation, but it can usually not be measured because you can only glue a gauge on the surface. $\endgroup$
    – Orbit
    Nov 29, 2018 at 17:34
  • $\begingroup$ Ah. Hmm. So when people are modeling a structural experiment using FEM, the FEM model gives strain values. And then the experimental data from strain gauges can be directly compared to the FEM results without any conversion? $\endgroup$
    – anonuser01
    Nov 30, 2018 at 16:03

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