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Message #21770
Re: [Question #688134]: Not rational result in the Triaxial code
Question #688134 on Yade changed:
https://answers.launchpad.net/yade/+question/688134
Status: Open => Answered
Jan Stránský proposed the following answer:
0)
you have no running in the deviatoric part, so "at the end of deviatoric part" = "at the end of reaching target porosity"
1+3) (same problem)
sigma2 = axialS, q = axialS-sigmaIso
because sigma2=axialS has no relation to sigmaIso (see below)
> if unb<stabilityThreshold and abs(sigmaIso-
triax.meanStress)/sigmaIso<0.001:
since sigmaIso is negative, the fraction in the second condition is always negative and therefore the second condition is always met.
So the loop is stopped based only on unbalanced force= with arbitrary mean stress.
In your case triax.meanStress = 0 (according to the output)
2)
> Why axial deformation ... is zero?
confining - because internalCompaction=True (default value)
deviatoric - you have no running, the strain remains 0
cheers
Jan
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