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Attached is a figure to illustrate the difference in the value of yield strength as observed from the true stress/strain curve and engineering stress/strain curve. While conducting FEA in ANSYS for a isotropic ductile material, I need to compare the maximum stress regions with that of the yield strength to know if that structure is undergoing plastic deformation or not. My question is which value should be used for yield strength; the one obtained from true or engineering? enter image description here

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    $\begingroup$ IMO, it depends on your task on hand, if it is for real-world design, then stick to the engineering properties as they are upheld by the codes. Also, you are subjecting to liability and scrutiny for deviations shall any bad thing occurs. If it is research, then the true properties may lead to better interpretation and understanding, and the result could have great significance though limited applicability. $\endgroup$
    – r13
    Jun 24 at 2:18
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UPDATE Based on comments

The question was:

After I obtain the maximum stress results from my simulation, should I compare the maximum stress that I observe in the simulation to the engineering or the true yield stress of the material?

IMHO, it doesn't really matter, at least for steel (there are other materials that will exhibit significant differences in yield stress, but steel is not amongst them IMHO).

The reason is that at small strains (i.e. in the linear region below the yield stress) the true stress and the engineering strain are almost identical.

The relationship between true stress ($\sigma_t$) and engineering stress ($\sigma$) and strain ($\varepsilon$) can be approximated by:

$$\sigma_t = \sigma \cdot (1 + \varepsilon)$$

E.g. For example, for steel, the yield strain is about 0.2%, therefore the change between true and engineering strain is about 0.2%.

In most cases of real life problems, - IMHO- if the error in the maximum stress is within 5% of the actual stress you've done a superb job. So checking a value that is 0.2% when your error is about 5% doesn't make sense IMHO

Again, what I wrote above is applicable to steel, or other materials that exhibit yield strain at really low values (less than 5%).

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  • $\begingroup$ I diagree. It depends on the how the material model is formulated in the FE code. Almost all FE codes have the option to use material properties based on engineering stress and strain, but some also have the option for true stress and logarithmic strain, intended for use in large-strain nonlinear analysis. For a material where the "yield point" is at 30% or 50% strain, the difference is definitely not small. $\endgroup$
    – alephzero
    Jun 23 at 13:05
  • $\begingroup$ You are correct in the general case (for the difference in the yield stress at materials with yield at 50% strain). Before, I edit my answer to make it more correct, in your experience, would you consider correct what I wrote for most steels and their yield point (I thought I had put that asterisk in there along with the small strains but upon a more careful reading I must have missed it). $\endgroup$
    – NMech
    Jun 23 at 13:35
  • $\begingroup$ Also, regarding the option for engineering and true stress and strain, I must confess that I've never seen the option (until about 20 years ago) that I used to read the solver codes more diligently, but I know you are more experienced, so I take your word for it. I can't help thinking though, that if someone started to develop from scratch a basic version of a linear FE solvers, then the true properties would be more easily applied (instead of the engineering). (I hope what I am trying to say makes sense). $\endgroup$
    – NMech
    Jun 23 at 13:44
  • $\begingroup$ @alephzero I forgot to tag you in my replies to your comment. I'd really appreciate your insight and feedback (positive or negative). $\endgroup$
    – NMech
    Jun 23 at 17:01
  • $\begingroup$ @alephzero, first of all, I don't think the FE code at all uses the yield/ultimate strength at all neither it knows what engineering or true stress/strain curve is. We just need to input the basic istoropic properties like Elastic Modulus and poisson's ratio and thats about it, if we are in the elastic range. Although I believe that the FE codes actually uses the 'true' meaning to find out the stresses and strains within the nodes/elements. $\endgroup$ Jun 24 at 8:09
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The choice is up to your task/interest and its application.

The ultimate strength is completely obscured in a true stress-strain curve. However, the engineering stress-strain curve hides the true effect of strain hardening. The true stress-strain curve is ideal for showing the actual strain (and strength) of the material.

The engineering stress-strain curve is ideal for performance applications (meant design). The true stress-strain curve is ideal for material property analysis (meant research or study). For everyone except (some) materials scientists, the engineering stress-strain curve is simply more useful than the true stress-strain curve.

However, the true stress-strain curve is ideal for showing the actual strain (and strength) of the material.

Some materials scientists may be interested in fundamental properties of the material. In this case, the true stress-strain curve is better. This curve tells the actual state of stress in the material at any point. It also shows strain hardening without being affected by the changing area of the sample.

For example, many metals show strain-hardening behavior that can be modeled as:

enter image description here

Where K is a constant and n is the strain-hardening exponent. n is always less than 1.

https://msestudent.com/true-stress-strain-vs-engineering-stress-strain/#:~:text=The%20ultimate%20strength%20is%20completely%20obscured%20in%20a,the%20actual%20strain%20%28and%20strength%29%20of%20the%20material.

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