Drucker Stability Check in Abaqus

[cthen]

Dear Jorgen,
Dear all,

I have employed the Drucker check (for Hyperfoam material model) into an own optimization routine. It all works ok and puts out the limits given in the .dat file for the employed sets of material parameters. A material paramter set though, where mu is negative does not produce any stability warning in the .dat file. A warning, however should be issued as mu is a factor to be multiplied with the material stiffness matrix |D as indicated in the Theory manual (4.6.2 Fitting of hyperelastic and hyperfoam constants, Drucker Stability Check, Hyperfoam). Therefore every matrix component is multiplied with a negative mu and then the 1.st inequality (D11+D22+D33>0) is not satisfied that needs to be satisfied though to assure material stability.
Can you think of any reason that might be and maybe give me a hint where I might be wrong?

I do thank you very much in advance and would be very grateful for any information I get!

Kind Regards,
Christophe Then

[Jorgen]

It seems reasonable that mu should be larger than 0. Are you saying that ABAQUS does not give you a warning if mu < 0?

- Jorgen

[cthen]

Dear Jorgen,

thank you very much for your help & time! E.g. taking the order of the series expansion N=2 and then using both mu<0.0, I would think that the inequality D11+D22+D33 would become negative and a warning message should be given in the .dat-file, but this is not the case.
If you could tell me why this may be or where I could be wrong, I would very much appreaciate it!

Kind Regards,
Christophe

[Jorgen]

It seems to me that your are probably right. Perhaps you should issue a bug report with ABAQUS.

- Jorgen

[cthen]

Dear Jorgen,

thank you so much for your reply! I may go ahead and check with Abaqus.

I found, using the Hyperfoam material model, that if checking the Drucker stability with N>1, the roots of a rather ugly polynom in lamda2 and lamda1, respectively, have to be found in order to provide the lamda2 value that corresponds to a given lamda1. That works in many cases but in some special situations, root finding appears to be difficult. If those special cases (certain combination of parameters) happen to violate one of the 3 inequations, Abaqus sometimes does not provide any stability limits at all. So, all in all, it seems quite intransparent to me.

Thank you again so much for help & time!
Christophe