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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.

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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.

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  • $\begingroup$ Ok. But, do we use the actual(i.e. instantaneous) cross-sectional area or the original cross-sectional area? $\endgroup$ – user29305 Jun 28 '16 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 Jun 28 '16 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 Jun 28 '16 at 12:50
  • $\begingroup$ @JonCuster If I'll be using actual stress, should I not also be using actual strain? $\endgroup$ – user29305 Jun 29 '16 at 6:04
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    $\begingroup$ @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) $\endgroup$ – Kashish Garg Jun 29 '16 at 9:36

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