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Logarithmic Strain 1
- Thread starter Capko
- Start date
Hello,
I think LE refers to the true strain and not engineering strain. You see true stress, you input true stress and true strain. It seems good to see true strain.
21.1.2 Material data definition
"When giving material properties for finite-strain calculations, “stress” means “true” (Cauchy) stress (force per current area) and “strain” means logarithmic strain. For example, unless otherwise indicated, for uniaxial behavior epsilon = ln(L/L0)"
10.2.2 Stress and strain measures for finite deformation
"where L is the current length, L0 is the original length, and epsilon is the true strain or logarithmic strain."
E is "All strain components. For geometrically nonlinear analysis using element formulations that support finite strains, E is not available for output to the output database (.odb) file."
I think LE refers to the true strain and not engineering strain. You see true stress, you input true stress and true strain. It seems good to see true strain.
21.1.2 Material data definition
"When giving material properties for finite-strain calculations, “stress” means “true” (Cauchy) stress (force per current area) and “strain” means logarithmic strain. For example, unless otherwise indicated, for uniaxial behavior epsilon = ln(L/L0)"
10.2.2 Stress and strain measures for finite deformation
"where L is the current length, L0 is the original length, and epsilon is the true strain or logarithmic strain."
E is "All strain components. For geometrically nonlinear analysis using element formulations that support finite strains, E is not available for output to the output database (.odb) file."
- Thread starter
- #3
I think so. But if you recall a stress-strain diagram with engineering and true values plotted, the difference between these curves is small at smaller strains. Perhaps this isn't really an issue in most cases?
LS-Dyna support page
"From engineering to true strain, true stress
First of all, you may check that your experimental data from a uniaxial tension test is expressed in terms of true stress vs. true strain, not engineering stress or strain.
True strain = ln(1 + engineering strain) where ln designates the natural log"
So for Rp0.2 the true strain is
et = ln(1+0.002) = 0.00199800
which is very close to 0.002.
What do you think?
LS-Dyna support page
"From engineering to true strain, true stress
First of all, you may check that your experimental data from a uniaxial tension test is expressed in terms of true stress vs. true strain, not engineering stress or strain.
True strain = ln(1 + engineering strain) where ln designates the natural log"
So for Rp0.2 the true strain is
et = ln(1+0.002) = 0.00199800
which is very close to 0.002.
What do you think?
- Thread starter
- #5
This is discussed in the Abaqus Users Guide:
Section 1.2.2 Conventions - Stress and Strain Measures
Section 4.2.1 Abaqus Standard Output Variable Identifiers - Strain Output
Total strain (E) is only available in geometrically linear analyses. For geometrically nonlinear analyses, you can output logarithmic (LE) and nominal strain (NE).
Section 1.2.2 Conventions - Stress and Strain Measures
Section 4.2.1 Abaqus Standard Output Variable Identifiers - Strain Output
Total strain (E) is only available in geometrically linear analyses. For geometrically nonlinear analyses, you can output logarithmic (LE) and nominal strain (NE).
- Thread starter
- #7
LE is the true strain, the strain you would measure with a strain gauge. For measuring strain it should be advantageous.
Same thing with stress, it is the actual stress and not a "fake", constructed, stress.
You are comparing two models, correct? I agree that comparing two logarithmic values can be questionable but even if you could compare engineering strains it would be strange since your geometry is not linear (otherwise you would have E).
Same thing with stress, it is the actual stress and not a "fake", constructed, stress.
You are comparing two models, correct? I agree that comparing two logarithmic values can be questionable but even if you could compare engineering strains it would be strange since your geometry is not linear (otherwise you would have E).
- Thread starter
- #9
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1
- #10
How large are your strains? If they are low, it doesn't matter if they are logarithmic or not in my oppinion (I could be wrong).
If the strains are high or you have some nonlinear geometry behaviour then the model assuming linear behaviour could be incorrect.
What you could do to compare is to manually calculate strain by looking at displacement and original length of some chosen part of the geometry, or from section forces maybe.
If the strains are high or you have some nonlinear geometry behaviour then the model assuming linear behaviour could be incorrect.
What you could do to compare is to manually calculate strain by looking at displacement and original length of some chosen part of the geometry, or from section forces maybe.
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