# Understanding stress-strain calculations

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.

• 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 $\varepsilon = (L_f/L_0) -1 = \Delta L /L_0$ Jul 5, 2022 at 12:36
• Jul 22 at 12:37

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)

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

• Ok. But, do we use the actual(i.e. instantaneous) cross-sectional area or the original cross-sectional area?
– user29305
Jun 28, 2016 at 8:24
• @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). Jun 28, 2016 at 12:09
• @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. Jun 28, 2016 at 12:50
• @JonCuster If I'll be using actual stress, should I not also be using actual strain?
– user29305
Jun 29, 2016 at 6:04
• @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) Jun 29, 2016 at 9:36