amquen
Mechanical
- Mar 6, 2018
- 3
Dear all,
I am performing a large strain analysis and had some difficult in understanding some results. The issue was with an hyperelastic Marlow-material model, but I tried to develop a simpler one to show what is happening. Hope you can help me to make it clear.
-The simplified model is 3D analisys of a rectangular bar.
-Dimensions are 4.00 x 2.25 and 60.00 mm length (length in U3 direction).
-Material is linear elastic isotropic, with Young Modulus 5000 MPa and Poisson 0.44.
-***Non-linear geometrics is enabled***.
-The boundary conditions are: one 4.00x2.25 face fixed for U3 displacements, and the opposite face displaced by 100 mm also in U3 direction. I know values may be a bit strange, but focus on the results obtained.
- Mesh is all of C3D8R bricks, reduced integration, reasonably fine size.
For LE results (standard abaqus output), I obtained:
LE33= 0,9808; LE11 and LE22= -0,420.
Deformed geometry: 2.6281 x 1.4783 x 160 mm
Ok, first I can use the Logarithm Strains (LE) to obtain the Engineering Strains (E), using E = exp(LE)-1 (derived from the classic LE=ln(E+1))
Hence, E33= 1,667; E11 and E22= -0,5220
However, I can also calculate these Engineering Strains using the final dimensions, where:
E33= (160-60)/60= 1,667; E22= (4.00-2.6281)/4.00= 0,3430; E33= (2.25-1.4783)/2.25= 0,3430
Here it is possible to verify that the engineering strains are OK for the 33 direction, but are not as expected for 22 and 11 directions.
Also, I was in doubt about how abaqus uses the poisson information. In principle, I believed that the 0.44 coefficient could be checked using the Engineering Strains, once in tests it is obtained with extensometers. What I found was:
poisson (LE values)= 0,420/0,9808= 0,428
poisson (E values, calculated from LE)= 0,5220/1,667= 0,313
poisson (E values, calculated from dimensions)= 0,3430/1,667= 0,206
So, it seems that Abaqus used poisson to calculate the transverse LE values, which was not what I expected. Even though, in a model simple like that I couldn´t find out why it was not exactly 0.44.
For information, I ran this same model using NLGeom=off. For this case, abaqus output was the Engineering Strain. Results are below:
E11= 1.667; E22 and E33= -0.7333;
Deformed geometry= 1.0667 x 0.60 x 160 mm which leaded to the same E11, E22 and E33 shown in the line above.
The calculated LE would be: LE11= 0.9808; LE22 and LE33= -0.55;
From this, poisson values would be:
poisson (LE values, calculated from E)= 0,55/0,9808= 0,56
poisson (E values)= 0,7333/1,667= 0,44
Here, Abaqus used the poisson to calculte the transverse Engineering Strains, and all values are "as expected".
Concluding, what I would like to understand:
- Why transverse LE strains for the first model are different when calculated from final dimensions and from abaqus output? The same formula worked for the 11 direction...
- For large strain abaqus uses the poisson for calculating Logarith Strains? Does it make sense? I have not seen any information regarding "converting" poisson anywhere, and it doesn´t seems logical to me.
I am performing a large strain analysis and had some difficult in understanding some results. The issue was with an hyperelastic Marlow-material model, but I tried to develop a simpler one to show what is happening. Hope you can help me to make it clear.
-The simplified model is 3D analisys of a rectangular bar.
-Dimensions are 4.00 x 2.25 and 60.00 mm length (length in U3 direction).
-Material is linear elastic isotropic, with Young Modulus 5000 MPa and Poisson 0.44.
-***Non-linear geometrics is enabled***.
-The boundary conditions are: one 4.00x2.25 face fixed for U3 displacements, and the opposite face displaced by 100 mm also in U3 direction. I know values may be a bit strange, but focus on the results obtained.
- Mesh is all of C3D8R bricks, reduced integration, reasonably fine size.
For LE results (standard abaqus output), I obtained:
LE33= 0,9808; LE11 and LE22= -0,420.
Deformed geometry: 2.6281 x 1.4783 x 160 mm
Ok, first I can use the Logarithm Strains (LE) to obtain the Engineering Strains (E), using E = exp(LE)-1 (derived from the classic LE=ln(E+1))
Hence, E33= 1,667; E11 and E22= -0,5220
However, I can also calculate these Engineering Strains using the final dimensions, where:
E33= (160-60)/60= 1,667; E22= (4.00-2.6281)/4.00= 0,3430; E33= (2.25-1.4783)/2.25= 0,3430
Here it is possible to verify that the engineering strains are OK for the 33 direction, but are not as expected for 22 and 11 directions.
Also, I was in doubt about how abaqus uses the poisson information. In principle, I believed that the 0.44 coefficient could be checked using the Engineering Strains, once in tests it is obtained with extensometers. What I found was:
poisson (LE values)= 0,420/0,9808= 0,428
poisson (E values, calculated from LE)= 0,5220/1,667= 0,313
poisson (E values, calculated from dimensions)= 0,3430/1,667= 0,206
So, it seems that Abaqus used poisson to calculate the transverse LE values, which was not what I expected. Even though, in a model simple like that I couldn´t find out why it was not exactly 0.44.
For information, I ran this same model using NLGeom=off. For this case, abaqus output was the Engineering Strain. Results are below:
E11= 1.667; E22 and E33= -0.7333;
Deformed geometry= 1.0667 x 0.60 x 160 mm which leaded to the same E11, E22 and E33 shown in the line above.
The calculated LE would be: LE11= 0.9808; LE22 and LE33= -0.55;
From this, poisson values would be:
poisson (LE values, calculated from E)= 0,55/0,9808= 0,56
poisson (E values)= 0,7333/1,667= 0,44
Here, Abaqus used the poisson to calculte the transverse Engineering Strains, and all values are "as expected".
Concluding, what I would like to understand:
- Why transverse LE strains for the first model are different when calculated from final dimensions and from abaqus output? The same formula worked for the 11 direction...
- For large strain abaqus uses the poisson for calculating Logarith Strains? Does it make sense? I have not seen any information regarding "converting" poisson anywhere, and it doesn´t seems logical to me.