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Consider a 1x1x1 cube. It has a volume of 1. If the material is isotropic and the poissons ratio is 0.5 the volume should be conserved.

consider then I stretch one side so it has a length of 2. The remaining dimensions are then (2Xsqrt(2)/2)Xsqrt(2)/2 , if the volume is still to equal 1.

Applying the formula for poissons ratio:

Poissons ratio = lateralstrain1/(longitudinal*strain)

Lateral strain = (sqrt(2)/2 - 1)/1

Longitudinal strain = (2 - 1)/1

poissons ratio = (sqrt(2)/2 - 1)/(2 - 1) = 0.29

Which is not 0,5. What is happening?

Thank you

ps sorry the equations are not in "proper" format. Don't know how to do that here yet

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1 Answer 1

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The equivalence of a Poisson ratio of 1/2 and constant volume (i.e., incompressibility) holds only in the limit of infinitesimal strains. You posited a very large strain—100%—and as a result, the relation is strongly inaccurate.

You can prove this to yourself by calculating the volume change expected from a tensile strain in one direction and the corresponding Poisson-mediated contractile strains in the other two directions.

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  • $\begingroup$ hmm yea I just tried with a strain elongation of 1.01 instead of 2. and it equaled almost 0.5. Thanks for the answer $\endgroup$
    – Jan F. S
    Nov 27, 2022 at 19:00
  • $\begingroup$ Good check! You’re welcome. $\endgroup$ Nov 27, 2022 at 20:12

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