I'm trying to compute the stress-strain curve for an elastic material with cylindrical geometry subject to an increasing uniaxial load. I understand that this requires:

stress = $F/A_0$ where $A_0$, the initial cross-sectional area of the cylinder, is a constant whereas $F$ is increasing.

strain = $\Delta{L}/L_0$ where $\Delta{L}$ varies but $L_0$, the inital length of the cylinder, is constant.

Is my understanding correct? My confusion stems from the fact that the wikipedia page on Young's modulus says that $A_0$ is the actual cross-sectional area rather than the original cross-sectional area. Subject to an increasing uniaxial load the actual cross-sectional area would be decreasing so $A_0$ wouldn't be a constant.

  • $\begingroup$ There are different measures of stress and strain. If you use the nominal force with respect to the initial area you are using de first Piola-Kirchhoff stress: en.wikipedia.org/wiki/…. With respecto to strain you are using the infinitessimal strain $\varepsilon = (L_f/L_0) -1 = \Delta L /L_0$ $\endgroup$
    – Davius
    Commented Jul 5, 2022 at 12:36
  • $\begingroup$ Related: physics.stackexchange.com/q/265100/226902 $\endgroup$
    – Quillo
    Commented Jul 22, 2023 at 12:37

1 Answer 1


Technically WikiPedia is correct but we usually use apparent stress during stress strain curve (Mostly) and not actual stress . Apparent stress(or engineering stress) is the one you mentioned and the one mentioned by Wikipedia is actual strees(or true stress)enter image description here

As you can see in this picture A is engineering stress strain curve whereas b is true stress strain curve.

  • $\begingroup$ Ok. But, do we use the actual(i.e. instantaneous) cross-sectional area or the original cross-sectional area? $\endgroup$
    – user29305
    Commented Jun 28, 2016 at 8:24
  • $\begingroup$ @AidanRocke If you use the actual cross-sectional area, you obtain true stress-strain curve (the blue curve in the above figure) and if you use the original cross-sectional area, you obtain engineering stress-strain curve (red curve). $\endgroup$
    – lucas
    Commented Jun 28, 2016 at 12:09
  • $\begingroup$ @AidanRocke - keep in mind that the concept of engineering stress/strain was developed more than a century ago, when rapid, in situ methods for measuring actual dimensions were not available. Hence, the engineering stress/strain is basically all based on measuring the force, and change of length in one dimension (the one you are pulling on). As KashishGarg points out, they can be related. $\endgroup$
    – Jon Custer
    Commented Jun 28, 2016 at 12:50
  • $\begingroup$ @JonCuster If I'll be using actual stress, should I not also be using actual strain? $\endgroup$
    – user29305
    Commented Jun 29, 2016 at 6:04
  • 1
    $\begingroup$ @AidanRocke if you are using true stress you should use true strain , true strain is given by ln((ΔL+L0)/L0) (It is almost same as engineering strain if ΔL/L0 is less than .2) $\endgroup$ Commented Jun 29, 2016 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.