Force ouput of 2D asymmetric compression model less than experimental value

1

Hey all,

I am doing the simulation of compression of an O-Ring using the Neo Hooke compression model.

The O-Ring is compressed for 2.4 mm between 2 parallel flat plates.
Because of revolutionary geometrical characteristics and symmetry, compression model is a 2D axisymmetric model:

Screenshot
Screenshot711×530 41 KB

The yellow nodes at diameter are the reaction force nodes.

The problem is that the experiment result of the compression show a max force of 1400? but the fem model shows just a force of around 10N

I am not being able to figure out what I am missing out, so that the forces from simulation are so much lesser than the actual experiment. Would be great if someone could help with that.
SInce the model is a 2D asymmetric model, do the forces have to be integrated along the circumference after the simulation ?

( I set the displacement in INP file to be half of the actual compression in experiment, since the fem geometry is half cross section of the actual.)

This is the input file:


*****************************************************
** Material properties definitions
*****************************************************
*MATERIAL, NAME = metal
*ELASTIC
**youngs modulus in N/m2
2.1e11, 0.3

*MATERIAL, NAME=rubber
*HYPERELASTIC, Neo Hooke
1.6273, 0


*****************************************************
** Geometry-material assignments
*****************************************************
*SOLID SECTION, ELSET = Eoring_com, MATERIAL = rubber
*SOLID SECTION, ELSET = Eplate, MATERIAL = metal

*****************************************************
** Constraints and homogenous boundary conditions
*****************************************************
*BOUNDARY 
Nplate, 1
*INCLUDE, INPUT = symm_y_2.bou

*****************************************************
** Contact definitions
*****************************************************
*SURFACE INTERACTION, NAME = compression
*SURFACE BEHAVIOR, PRESSURE-OVERCLOSURE = linear
200,0.00639,1e-3
*CONTACT PAIR, INTERACTION = compression, TYPE = node to surface
Scontact_oring, Scontact_plate
*FRICTION
0.9, 0.055

*****************************************************
** Time sequence definitions
*****************************************************
*TIME POINTS, NAME = sampling, GENERATE
0,1, 0.02

*****************************************************
** STEPS
*****************************************************
*STEP, NLGEOM, INC =200
*STATIC
0.1,1,,1e30
*BOUNDARY
Nplate, 2,,-0.934

*NODE FILE, OUTPUT = 2D
U
*EL FILE
S

*node print, nset=Nreaction_com,time points=sampling, totals = yes
**node print, nset=Nreaction_com, totals = yes
RF
*CONTACT FILE
CDIS, cstr
*END STEP
2

CalculiX expands axisymmetric elements to one layer of 3D elements in a 2 degree sector. So you have to multiply the reaction forces by 180.

How is the sample constrained in the experiment ? Where did you get the hyperelastic material data from ?

3

the hyperlastic model looks like in MPa but metal in Pa.

4

Thanks for the quick reply @Calc_em. The values are now much better after multiplying it by 180.

The sampled is just constrained vertically in the experiment, compressed between two flat parallel plates.
The C10 value of Neo Hooke from another study, I still need to calibrate it to the test data but I first wanted to test the model with a value.

I had one more doubt :
The actual compression in the test is 2.41, but since I have just half cross section of the cord of the ORing, I need half the compression in the fem model or does it have to be 2.41 in the model as well ?

5

Yeah that’s a mistake I did, but I tried it in MPa and it didnt have any effect on the force outcome

6

of course, it just has to balance the applied load. In that respect is isostatic.

7

I also have 1/8th model for the tensile test of the O Ring.
The displacement in the actual test is 40 mm and I also gave 40 disp in the FEM model, so my question is:
Do the reaction forces after the simulation need to be multiplied by 8 and the displacement needs to be half of test (20mm) ? or it doesn’t need to be multiplied and the displacement in model can be 40

Screensho
Screensho576×661 91.3 KB

[Red nodes are for reaction forces]

8

So it looks like this in 3D ?

ring
ring756×691 42.6 KB

9

Correct ! And displacement is only on the top pin, the bottom pin is fixed in the test

10

@Calc_em, do you have any suggestions regarding my doubt ?
Would be really helpful for me.

11

Then you can’t use the axisymmetric model here since it assumes contact on the whole circumference. You would have to use planar symmetry.

Regarding your questions about 3D models with symmetry - you don’t have to reduce the applied displacement but you will get lower reaction forces.

Same reaction force (RF) here:

symm 1
symm 1830×499 19 KB

Half (RF/2) here:

symm 2

Quarter (RF/4) here:

symm 3

12

Thanks for the detailed Reply.
So also for the compression model I don’t need to half the actual test displacement in my model, but only need to multiply force by 180, if my test looks like this,:

Capture
Capture775×466 44 KB

and my FEM model is this :

Screenshot
Screenshot711×530 41 KB

13

In this case, you have to use half of the displacement. But you will get the expected reaction force after multiplying the result by 180.

1 Like
14

Thanks a lot for the help :pray: @Calc_em