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[Solved] True Stress-Strain vs Engineering Stress-Strain

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Posts: 5
Topic starter
(@hossein-shf)
Active Member
Joined: 4 years ago

Hi Mr.Jorgen,

There are some Polyurethane compression test specimens (According to ASTM-D575 standard - 12mm thickness) in 3 hardness 55, 65 and 75 ShoreA. The test results are reported in the form of time-force-displacement and now I'm trying to plot true stress-strain curve of compression behavior of these specimens. If the strain will be calculated using "delta l / l" formula (engineering strain), the strain range will be 0-100% that is physically sensible but using the strain integration (integral of "dl/l") (true strain), the strain is from 0 to near 300% (the specimens compressed about 11.8 mm). Now, my question is that, this range for true strain is correct? and what does physically means?

Thanks in advance, best regards.

1 Reply
Posts: 3993
(@jorgen)
Member
Joined: 4 years ago

Good question. The \(\Delta L/L_0\) definition of strain gives strain values that cannot ever be less than -1 (that is -100%). This is the classical definition of engineering strain.

The integral strain integration equation is consistent with the Logarithmic strain definition (also called true strain). In this case the strain is given by \(\varepsilon = \log(1 + \Delta L / L_0)\). With this definition the compressive strain can be any negative number.

It is no problem to have a 300% compressive strain (which can also be written as -3.0), as long as it is expressed as a true strain.

-Jorgen

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