Say I have two different specimens of a material, one is 10mm thick and the other is 1mm thick. Assume that this material has properties which vary depending on the rate of deformation (I would call this behavior non-Newtonian in a fluid, I'm not sure if this has a different name in solids). I want to test their mechanical properties under compression (e.g. modulus, plastic deformation, etc). Should I run the compression tests for these materials using a constant crosshead speed (e.g. 5mm/min, which would be 50%/min for the thick sample and 500%/min for the thin sample), or using a constant strain rate (e.g. 50%/min, which would be 5mm/min for the thick sample and 0.5mm/min for the thin sample)? If both samples are made from identical materials, which control method will show them as having identical material properties?

  • $\begingroup$ I have decided to go with constant crosshead speed. I did a test to try to find a difference, and it was minimal across the range of thicknesses I'm likely to see, so I've gone with crosshead speed since it's a bit simpler to report. $\endgroup$ Jun 12, 2020 at 18:55

1 Answer 1


It depends on many factors.

Let's say your material's young modulus is quadratically related to the strain versus linearly.

So when you compress it to shrink its thickness the stress wave traveling from the outer surface to the core can stratify the specimen into deferent layers starting from high stress but relatively lower strain near the exterior layers because they were immediately strained to jump to the higher ratio. But the middle layers are still at lower stress versus high strain threshold. So even tho the stress can settle to equilibrium the strain can vary across the depth.

However, the thinner sample may not stratify as much and behave totally differently. Here lies the nightmare of calculating the properties of the specimen.

That may have been the reason for assuming stress/strain proportionality linear in engineering.


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