NX Nastran Strain

[NikonF6]

This is about Strain in the case of large deformation. Material annealed Steel.
I do not need stress but ONLY STRAIN. Deformstions are larger than at the yield point, so there are plastic deformations.
Do I need non-linear analysis? I work with NX-Nastran.

 

[rb1957]

you'd think, wouldn't you ? (strains > yield = plasticity = non-linearity ... no?)

Quando Omni Flunkus Moritati

 

[NikonF6]

Thanks for your time.
As i understand, you say that either linear and non-linear analysis will yield the same result for the strain in cases when the strain is much higher than strain at yield.
CORRECT?

 

[rb1957]

100% INCORRECT !

Linear and NL will have the same results BELOW yield.

Beyond yield NL (which understands the reduced stiffness beyond yield) will be less than Linear (which just extends E).

Quando Omni Flunkus Moritati

 

[NikonF6]

Thnks again, and I hope you are talking about STRAIN (I have no interest for STRESS). Just to make sure...

 

[rb1957]

strain and stress are reasonably equivalent terms, related by the stiffness curve of the material.
if you're above yield, you need to use NL to get sensible results.

Quando Omni Flunkus Moritati

 

[NikonF6]

This is clear now. Thanks a lot for useful comments.
Now is something why I asked this.
Strean is linearly depend on Modulus of Elasticity. how will works this:
find the Strain at the point by linear analysis, say it is 0.3
find slope of the strain-stress curve (I have the curve) where strain is 0.3
correct strain by the ration (tan of slope below yield - which is E) / (tan of slope beyond yield)
the resultant strain at the point is now larger than strain found by linear analysis, by the above factor

Hope you got it, and find again time for a nice answer

 

[rb1957]

not quite !
you can't use the strain for a linear analysis to infer much of anything about the non-linear state.

the reason is the change in stiffness as the material goes plastic, which redistributes load internally.

to follow your proposal, if you mean to replace the linear stress with a stress deduced from the stress/strain curve for the linear strain, then you've just removed a lot of the internal strain energy of the model, and this has to be added back somewhere, somehow. NL FEA works by figuring out the point of plasticity, then incremently adding load changing the stiffness of yield elements.

Quando Omni Flunkus Moritati

 

[shaun8567]

To add to rb1957 post, a large deformation analysis is not tied to JUST large strains. You'll also need to do a large deformation analysis if the displacements/rotations are large(typically if the displacements are large enough to be visible at scale of 1, or the rotations are greater than 15 deg), even if the strains are well below yield.