Contact Mechanics - Friction Test
Posted: 10 Jul 2019, 12:13
Hello ELMER-users,
I am trying to simulate a simple friction test, where I pull a cube over the floor. I want to validate this simulation by using COULOMBS friction law: F_R = mu * F_N, where F_R is the dynamic Friction load (in opposite direction of the direction in which the cube is pulled), mu the dynamic friction coefficient and F_N the normal load of the cube. I use displacement as a BC and save the displacement loads in the direction in which I pull. Usually after a few timesteps, when the speed of the cube becomes constant, these loads should correspond to the dynamic friction load, which is easy to calculate by hand.
I used displacement as a BC at the y-side of the cube. Unfortunately, the contact was only recognized on the floor in the area of this boundary. Then I tried many things and it turned out that it works best if I put displacement as a BC on EVERY boundary of the cube and prohibit the displacement of the top floor boundary (Master boundary) (pretty sure this is not physically correct anymore).
The cube and the floor consist of austenitic steel, so I chose the dynamic friction coefficient mu=0,2. However, the solution looks good and the contact is well-recognized. The displacement loads I receive are also in the correct magnitude, but when I change the friction coefficient for testing-reasons, the loads do not change at all.
Is simulation of friction contact in ELMER even possible yet? Why is the friction coefficient not taken into account? Or did I miss anything or make a mistake?
I attached my sif and a picture that illustrates my problem.
Thanks in advance
Maike
I am trying to simulate a simple friction test, where I pull a cube over the floor. I want to validate this simulation by using COULOMBS friction law: F_R = mu * F_N, where F_R is the dynamic Friction load (in opposite direction of the direction in which the cube is pulled), mu the dynamic friction coefficient and F_N the normal load of the cube. I use displacement as a BC and save the displacement loads in the direction in which I pull. Usually after a few timesteps, when the speed of the cube becomes constant, these loads should correspond to the dynamic friction load, which is easy to calculate by hand.
I used displacement as a BC at the y-side of the cube. Unfortunately, the contact was only recognized on the floor in the area of this boundary. Then I tried many things and it turned out that it works best if I put displacement as a BC on EVERY boundary of the cube and prohibit the displacement of the top floor boundary (Master boundary) (pretty sure this is not physically correct anymore).
The cube and the floor consist of austenitic steel, so I chose the dynamic friction coefficient mu=0,2. However, the solution looks good and the contact is well-recognized. The displacement loads I receive are also in the correct magnitude, but when I change the friction coefficient for testing-reasons, the loads do not change at all.
Is simulation of friction contact in ELMER even possible yet? Why is the friction coefficient not taken into account? Or did I miss anything or make a mistake?
I attached my sif and a picture that illustrates my problem.
Thanks in advance
Maike