DEM Application

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WARNING: This page is still under construction.

The DEM Kratos Team


Contents

Theory

This application solve the the equations.... Mathematical approach to the problems.

Nothing numerical

Integration Schemes

Forward Euler Scheme

Contact Laws

Concept of indentation HMD, LSD

Normal Force Laws
Linear Repulsive Force

The most simple representation of a repulsive contact force between a sphere and a wall is given by a linear law, where the force acting on the sphere when contacting a plane is a linear function of the indentation, which in turn would bring a quadratic dependence with the contact radius. The next figure shows this simple law:

Jkr cohesion linear force.png


Hertzian Repulsive Force

On the other hand, Hertz solved in 1882 the non-cohesive normal contact between a sphere and a plane. In 1971 Johnson, Kendall and Roberts presented the solution (JKR-Theory) for the same problem in this case adding cohesive behaviour. Not much later, Derjaguin, Müller and Toporov published similar results (DMT-Theory).

Both theories are very close and correct, the main difference being that the JKR theory is oriented to the study of flexible, large spheres, while the DMT theory is specially suited to represent the behaviour of rigid, small ones.

Jkr cohesion hertz.jpeg

The previous figure shows the standard representation of a Linear or Hertzian contact between a sphere and a wall. The distribution of contact pressures between both bodies follow a parabolic law.

JKR Cohesive Force

Jkr cohesion jkr.jpeg

The preceding capture shows the a representation of a JKR contact between a sphere and a wall. In this case, the distribution of pressures between both bodies is more complex due to the formation of a neck at the boundaries of the contact. A later figure shows a detailed view of the pressures involved.

Jkr cohesion forces.png

In the previous graphic, it is very interesting to note the existence of two singular values of contact radius: one for which the total forces acting on the contacting sphere is zero, and another for which the maximum value of adhesion is achieved.


Jkr cohesion pressures.png

In the previous figure, the blue area represents the distribution of pressures acting on the sphere when contacting a wall if a Hertzian Law is followed. On the other hand, the green area represents the JKR distribution. Note the larger values and the existence of adhesive behaviour at both sides of the pressures distribution.


An example of granular simulation without cohesive forces:

JKR no cohesion 1 33.png JKR no cohesion 2 33.png JKR no cohesion 3 33.png JKR no cohesion 4 33.png

The same simulation as before, this time with cohesive forces in both sphere-sphere and sphere-plane contacts.

JKR cohesion 1 33.png JKR cohesion 2 33.png JKR cohesion 3 33.png JKR cohesion 4 33.png


References:

V. L. Popov. Contact Mechanics and Friction (2010).


Tangential Force Laws
Damping Force Laws

(restit. coef)

Numerical approach (implementation)

Structure of the code (Strategy, Scheme, Element, Node, Utilities, functions frequently used like FastGet,...)


DEM strategies

Non-cohesive materials Strategy
Continuum Strategy

DEM schemes

DEM elements

Spheric Particle
Spheric Continuum Particle
Spheric Swimming Particle

Benchmarks

The DEM Benchmarks consist of a set of 9 simple tests which are run every night and whose object is to make sure both that the application performs correctly and that the code did not break after the daily changes. They are the following:

Test1: Elastic normal impact of two identical spheres

Check the evolution of the elastic normal contact force between two spheres with time.

Benchmark1.jpeg Benchmark1 graph.png

If the coefficient of restitution is 1, the module of the initial and final velocities should be the same. Also, by symmetry, velocities should be equal for both spheres.

_

Test2: Elastic normal impact of a sphere with a rigid plane

Check the evolution of the elastic normal contact force between a sphere and a plane.

Benchmark2.jpeg Benchmark2 graph.png

If the coefficient of restitution is equal to 1, the module of the initial and final velocity should remain unchanged.

_

Test3: Normal contact with different restitution coefficients

Check the effect of different restitution coefficients on the damping ratio.

Benchmark3.jpeg Benchmark3 graph.png

If total energy is conserved, the restitution coefficient and the damping ratio values should be identical.

_

Test4: Oblique impact of a sphere with a rigid plane with a constant resultant velocity but at different incident angles

Check the tangential restitution coefficient, final angular velocity and rebound angle of the sphere.

Benchmark4.jpeg Benchmark4 graph1.png Benchmark4 graph2.png

_

Test5: Oblique impact of a sphere with a rigid plane with a constant normal velocity but at different tangential velocities

Check the final linear and angular velocities of the sphere.

Benchmark5.jpeg Benchmark5 graph1.png _

Test6: Impact of a sphere with a rigid plane with a constant normal velocity but at different angular velocities

Check the final linear and angular velocities of the sphere.

Benchmark6.jpeg Benchmark6 graph1.png _

Test7: Impact of two identical spheres with a constant normal velocity and varying angular velocities

Check the final linear and angular velocities of both spheres.

Benchmark7.jpeg Benchmark7 graph1.png

By symmetry, the tangential final velocity of both spheres should be zero. Additionally, for a coefficient of restitution of 1, there should be no changes in the modules of both linear and angular velocities and their values should conserve symmetry.

_

Test8: Impact of two differently sized spheres with a constant normal velocity and varying angular velocities

Check the final linear and angular velocities of both spheres.

Benchmark8.jpeg Benchmark8 graph1.png

In this case, it is interesting to note that, the bigger and/or denser sphere 2 is, the more this test resembles the sphere versus plane simulation.

_

Test9: Impact of two identical spheres with a constant normal velocity and different restitution coefficients

Check the effect of different restitution coefficients on the damping ratio.

Benchmark9.jpeg Benchmark9 graph1.png

If total energy is conserved, the restitution coefficient and the damping ratio values should be identical.


References:

Y.C.Chung, J.Y.Ooi. Benchmark tests for verifying discrete element modelling codes at particle impact level (2011).

How to analyse using the current application

Pre-Process

GUI's & GiD

D-DEMPack

D-DEMPack is the package that allows a user to create, run and analyze results of a DEM simulation for discontinuum / granular / little-cohesive materials. It is written for GiD. So in order to use this package, you should install GiD first.

You can read the D-DEMPack manual or follow the D-DEMPack Tutorials for fast learning on how to use the GUI.

C-DEMPack

Continuum / Cohesive

F-DEMPack

Fluid coupling

Post-Process

Application Dependencies

The Swimming DEM Application depends on the DEM application

Other Kratos Applications used in current Application

FEM-DEM


Programming Documentation

The source code is accessible through this site.

Problems!

What to do if the Discrete Elements behave strangely

In the case you notice that some discrete elements cross walls, penetrate in them or simply fly away of the domain at high velocity, check the following points:


In the case of excessive penetration:

  • Check that the Young Modulus is big enough. A small Young Modulus makes the Elements and the walls behave in a very smooth way. Sometimes they are so soft that total penetration and trespass is possible.
  • Check the Density of the material. An excessive density means a big weight and inertia that cannot be stopped by the walls.
  • Check the Time Step. If the time step is too big, the Elements can go from one side of the wall to the other with no appearence of a reaction.
  • Check the frequency of neighbour search. If the search is not done frequently enough, the new contacts with the walls may not be detected soon enough.


In the case of excessive bounce:

  • Check that the Young Modulus is not extremely big. An exaggerated Young Modulus yields extremely large reactions that can make the Elements bounce too fast in just one time step. Also take into account that the stability of explicit methods depends on the Young Modulus (the higher the modulus, the closer to instability).
  • Check the Density of the material. A very low density means a very small weight and inertia, so any force exerted by other elements or the walls can provoque big accelerations on the element.
  • Check the Time Step. If the time step is too big, the method gains more energy, and gets closer to instability.
  • Check the restitution coefficient of the material. Explicit integration schemes gain energy noticeably, unless you use a really small time step. In case the time step is chosen to be big (but still stable), use the restitution coefficient to compensate the gain of energy and get more realistic results.

Contact

Contact us for any question regarding this application:


-Miguel Angel Celigueta: maceli@cimne.upc.edu

-Guillermo Casas: gcasas@cimne.upc.edu

-Salva Latorre: latorre@cimne.upc.edu

-Miquel Santasusana: msantasusana@cimne.upc.edu

-Ferran Arrufat: farrufat@cimne.upc.edu

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