I was conducting a FEA analysis of a symmetric isotropic structure under a loading condition. I will not mention the material I was using. Under that loading, one part of the structure was in tension and other in compression. When I reversed the loading condition, the part that was in compression previously was now in tension and vice versa. However, the displacements and the Von-Mises stresses that were seen on the part of the structure that was in tension in the initial loading condition, and again which was in tension in the reversed loading condition, I saw a difference in it and I couldn't understand why it was happening. So I thought maybe it is possible for the material to have different elastic modulus and Poisson's ratio under compression and tension. Is this true? If yes, then in most of the FEA softwares, there is option to add only one elastic modulus. Why is that? Is there any specific class of materials where this happens, or we have to conduct experiments to individually check each and every material if their elastic modulus is the same of different under tension and compression?

  • $\begingroup$ The short answer is that it can, but it rarely occurs (most usually in material systems) or it has a noticeable difference. However, the question is what exactly are you noticing in the simulation because it could have many sources. Could you post more details on your FEA analysis/results? $\endgroup$
    – NMech
    Jun 15, 2021 at 5:54
  • $\begingroup$ @NMech, yeah so it just cleared up that the elastic modulus for general metals should be the same under both, tension and compression. Then it appears as if the geometry is not exactly symmetric, thats why I am seeing a difference in the Von-Mises stresses. $\endgroup$ Jun 15, 2021 at 11:55
  • $\begingroup$ Again, it would be difficult to assess without looking at your model. $\endgroup$
    – NMech
    Jun 15, 2021 at 12:29
  • 1
    $\begingroup$ Isotropic material properties, nonlinear material properties, and structural symmetry, are independent of each other. Since you don't tell us the material, and the question implies there was time-dependent loading, we don't have enough information to give an useful answer. $\endgroup$
    – alephzero
    Jun 15, 2021 at 12:37

1 Answer 1


For small strains of stable materials, the tensile and compressive elastic moduli are equal. This is equivalent to saying that a smooth energy minimum looks like a (symmetric) parabola up close; an energy well in the shape of a parabola characterizes an ideal spring with equal elongation and contraction spring constants.

This approximation works well for metals, ceramics, and crosslinked polymers, for example, whose elastic strain is small (1%, say). This is why you have the option to enter only one Young's modulus, for example. However, the approximation typically does not work for large elastic strains of hyperelastic materials such as elastomers, which may stretch their own length (100% strain) and much more. As shown below (source), the stress–strain slopes are visually identical for slight positive and negative strains but differ for larger strains.

Since the same argument holds for the bulk and shear modulus, it must hold for Poisson's ratio, which is not independent if Young's modulus and either the bulk or shear modulus are specified.

  • $\begingroup$ So why is this the case that for general metals, some of the metals have difference in yield/ultimate strength for tension and compression? Although, as you have said, that the elastic modulus should be the same for small strains? And I just want to confirm this, so you mean to say that if I stretch a metal, like Aluminum, with a 100 N force, and imagine it displaces about, say, 1 mm, then if I would compress it with the same force, the compressive displacement should again be equal to 1 mm, right? $\endgroup$ Jun 15, 2021 at 11:51
  • $\begingroup$ @RameezUlHaq yield/ultimate stresses are a lot easier to explain (and there are many mechanisms). For example In tension, when a crack forms, in tension it tends to increase, however, in compression effectively the edges of the crack are brought together. Another effect is due to the fact that although in tesnion you can have theoretically more than 100% strain (at least in some materials), in compression that is not achievable. $\endgroup$
    – NMech
    Jun 15, 2021 at 12:28
  • $\begingroup$ @Rameez See this post for the usual (idealized) Elastic Modulus for steel of different grades. engineering.stackexchange.com/questions/44686/… $\endgroup$
    – r13
    Jun 15, 2021 at 17:03
  • $\begingroup$ @NMech, ofcourse I am aware of the crack formation dependence on the type of loading applied. But I am still unsure why can the compressive yield/ultimate be different than the tensile one, for small strains, when both the same elastic modulus. I want to know the reason behind it. Like what is it dependent upon? $\endgroup$ Jun 16, 2021 at 12:51
  • $\begingroup$ @r13, no that post was not at all useful for what I have asked here. $\endgroup$ Jun 16, 2021 at 12:53

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