Tension test of polyurethane in Cosmos

[C2H5OH]

Hello,

I'm a self-educated user of FEM softwares. I use Cosmos/DS or Cosmos/Works. I usually use FEM for linear static analyses, but now I need to simulate hyperelastic polyurethane. I am beginner in nonlinear. I made a simple tension tests on real sample piece of polyurethane.

Maximal stress is 1,625 MPa and elongation is cca 30%

I calculated stress (force/cross-section area) and stretch ratio (epsilon+1) and I got table with 8 parameters. These parameters I used for calculation of two mooney-rivlin coefficients. For poisson's ratio I used 0,495 like for nearly incompressible materials.

But when I simulate tension test - I use force for example 50N, what gave to me 1,25MPa for 40 square mm of cross-section area - I never get the same stress (SX) and maximal displacement (UX) that are in my table from uniaxial tension test.

I know, that my question is probably stupid and I didn't write all necessary parateters to this forum, but can anybody tell me where is problem with testing of mooney-rivlin model and if the normal stress is f.e. cauchy's stress or something like this?

Thanks to everybody, who will spent a time to explain me where is problem

[Jorgen]

Let's see if I understand. You have experimental data for your material. You saved this data in a table with 8 parameters. You then used that table to calculate the Mooney-Rivlin parameters.

That sounds great.

Then you tried to the software to predict one data point in the table using the calibrated Mooney-Rivlin model, and you got results that were not identical with the experimental data that you started with.

Well, in general you cannot expect that the predictions from a material model (such as the Mooney-Rivlin model) will be identical to a set of experimental data. The model is hopefully relatively close to the experimental data, if not it is likely not an appropriate material model.

- Jorgen

[C2H5OH]

Thanks J_B for answer.

I introduce an example:

Table with parameters that I got from simple tension:

Stretch ratio Stress (N/m^2)
1...................0
1.05..............250000
1.082.............500000
1.11..............750000
1.144............1000000
1.184............1250000
1.25..............1500000
1.292............1625000

F.e. the sixth row is calculated from force 50N and elongation 9,2 mm between the two marks with distance 50 mm. For cross-section area 10*4 mm was calculated stress = 1250000 and stretch ratio = 1 + dL/L = 1,184.

When I put this parameters to Cosmos/Works, it calculate two Mooney-Rivlin coefficients. With poisson ratio = 0.495 (it's just a tip) I startet to elongate the 1/8 part of the symetrical model (3 symetrical planes) to final length 4,6 mm (half of the length because of symetry).

I expected the same stress as in my table - software have to calculate with input values. But instead of 1,25 MPa I got 1,36 MPa. Reaction forces are 11.57 * 4 = 46,28N (because of 1/4 of area because of symetry) instead of 50N.

And I realize that there are causchy (real) stresses in results and reaction force is divided by deformed cross-section area (33,86 square mm instead of 40 square mm of unloaded model).


When I load the model by force 12,5 N (what is 50N for full area), I got stress 1,5 MPa (instead of 1,25 MPa) and displacement 5*2 = 10 mm (instead of 9,2 mm).

Are these values acceptable deviations or have I use another model than MR for simulate or is there any other mistake?
When I will simulate the bending, what is a combination of pulling and pressing, need I necessarily have a compression test?

Thaks

[Grant]

C2H5OH, are you testing solid polyurethane (not foam)? If so, can I ask how you made a tensile specimen of solid polyurethane? I had a desperate need for one about a year ago and several chemists and experimenters I consulted with assured me that it could not be done.

[C2H5OH]

Hello Grant,

Polyurethane (solid, not foam), that I use is two-part resin. I casted it to the form with specified dimensions (STN ISO 527-2). The specimen looks like a "bone". The polyuretane has low viscosity, but I rather use vacuum chamber for exhaustion all air bubbles.
This specimen is good enough for my usage. The final product, that I will make will be produced with same technologic process.

[Jorgen]

I think that your results seem reasonable. I think that the Mooney-Rivin (MR) is good enough for your case.

You do not need to perform more experiments, but if you do you might be able to improve the accuracy of your predictions.