I have already read: Stress- Strain curve but this does not answer my question completely.

My problem is to understand the decrease of stress ($\sigma$) while the strain ($\epsilon$) is increasing, after the maximum on the red curve:

enter image description here


For that, there are two points:

(1) Normally, we put on the $x$ axis "what we control/decide to do", and we consider on the $y$ axis "what we obtain from the conditions that we have chosen".

I assume that "what we control/decide to do" is the stress ($\sigma$) and what we measure if the strain ($\epsilon$). So why do we put this in a non logical orientation? (why aren't we putting the stress on the x axis and why aren't we putting the strain on the y axis ?)

(2) at the maximum in stress of the red curve, there is permanent deteriation of the material. ok. But in any case, if, starting from this point, we reduce the stress, whatever deteriorated is the material, I would assume that the strain should reduce, while it is the opposite: it increase further. If the material would be so damaged, then I would assume that after the maximum of $\sigma$, the line would no more continue, or if you prefer, the line would be a sudden vertical line towards $\sigma=0$.

Could somebody explain?

Remark: I have assumed, maybe wrongly, that the strain is a kind of measurement of the length of the object. Maybe this is the key point of my misunderstanding?

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    $\begingroup$ In a real mechanical test (e.g. Instron), the strain is actually what is applied - a giant screw is turned at a given rate, making the sample strain. A load cell measures the stress applied to the sample as the strain is applied. Past the peak of the red curve, what is actually happening is that the material is necking down, reducing the cross section, so it takes less stress to allow it to continue to elongate. $\endgroup$
    – Jon Custer
    Aug 26, 2020 at 21:19
  • $\begingroup$ @Jon Custer "so it takes less stress to allow it to continue to elongate" Less nominal stress, that is tension divided by original area, surely? Because of the necking (and thinning prior to pre-necking) the true stress (tension divided by actual area) continues to rise. $\endgroup$ Aug 26, 2020 at 21:34
  • $\begingroup$ @JonCuster, that looks like an answer. $\endgroup$ Aug 26, 2020 at 21:34
  • $\begingroup$ @JonCuster Looks like an answer to me too (and a good one). Post it. $\endgroup$
    – Bob D
    Aug 26, 2020 at 21:38
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    $\begingroup$ I found this YT video quite helpful. $\endgroup$
    – Shub
    Jan 13, 2022 at 5:53

2 Answers 2


For question (1), the plot is actually 'normal' - a tensile test machine (e.g. Instron, no affiliation, just used them) varies the strain in the material directly, often using a giant screw, sometimes hydraulics. The resulting stress on the material is then measured by a load cell. Strain is applied, stress is measured, so the independent and dependent variables in the plot are as you would expect.

Now, what happens as you apply strain? The blue line is the (nominally) elastic response - you elongate the sample, creating stress, then let it recover. I say nominally because for real materials there are usually plastic deformations even for low strains or stresses. The engineering definition that plastic response starts at 0.2% elongation (deformation) after cycling kind of indicates there can be less than 0.2% changes without being "plastic"

For the first part of the red curve, plastic deformation has kicked in. Often for metals this is the result of dislocation formation, allowing slip systems to activate, so not much more stress is generated as the strain keeps increasing.

At the peak of the red curve (ultimate yield stress), basically you are seeing all that slip deformation (often concentrated in slip bands) resulting in a decrease in the actual cross section of the sample, while the measured strain is still based on the original cross section.

If it helps, and you like caramel candy, you see the same thing as you pull on a caramel - it takes a certain level of applied strain, then the caramel starts to neck down, and continuing to pull the caramel apart takes less and less applied stress for the increase in strain.

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    $\begingroup$ Your explanation is crystal clear, and everything works perfectly if the strain is what we apply. But on this link : physics.stackexchange.com/questions/480717/stress-strain-curve one guy says that what is applied is the "stress" : in this case, it is very difficult to understand the curve ? To the sentence : "So I've read that stress is basically the result of strain acting on a body and that strain is applied first, then comes the stress.", one guy replied : "Actually, you have it reversed. Strain is the result of stress, not the other way around." : so meaning "stress is applied" $\endgroup$ Aug 27, 2020 at 6:59
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    $\begingroup$ @MathieuKrisztian, maybe it would help if you said, "controlled" rather than "applied." We could argue about what the testing machines that Jon talks about apply to the sample, but what they control is displacement. The machine stretches the sample at a controlled, constant rate, and it measures the force. So, the X-axis in the plots represents the controlled/independent variable, and the Y-axis represents the measured/dependent variable. $\endgroup$ Aug 27, 2020 at 9:50
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    $\begingroup$ Also note: The plots show force as a function of displacement (also, of time if anyone cares.) If you swap X and Y, the plot no longer looks like a function, which makes it somewhat more difficult to explain. $\endgroup$ Aug 27, 2020 at 9:54
  • $\begingroup$ @besmirched : ok, I made the change of the word $\endgroup$ Aug 27, 2020 at 10:15

First, what you're looking at is a plot of engineering stress $\sigma_e=F/A_0$ vs. engineering strain $\epsilon_e=\Delta L / L_0$, where $A_0$ and $L_0$ are the original cross-sectional area and length of the material. This is not, in fact, a plot of true stress $\sigma_t=F/A$ and true strain $\epsilon_t = \int \delta L/L$, which use the current cross-sectional area and length of the material.

This is important because, once the material reaches its ultimate strength (the maximum of the red curve), it begins "necking". At this point, there is enough stress on the material that its cross-sectional area begins to shrink (i.e. it forms a "neck"). Thus, the true stress is greater than the engineering stress, even if the same force is applied. It turns out that this downward slope is entirely an artifact of using engineering stress and engineering strain, and it goes away if you use true stress and true strain, as you can see below:

enter image description here

Also, a slight misconception in your question: permanent deformation occurs anywhere on the red curve, not just at the top.

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    $\begingroup$ thank you for your kind explanations $\endgroup$ Aug 27, 2020 at 7:00
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    $\begingroup$ Note that this explaantion is not universally true: For many materials (in particular rocks), there is indeed a peak in true stress with subsequent decline when the material is damaged. This is often quite brittle in tensile tests, but more ductile in compression. $\endgroup$
    – Toffomat
    Aug 27, 2020 at 10:52

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