# yade-users team mailing list archive

## Re: How to keep a constant stress boundary?

```kan a écrit :
```
```Thanks, Bruno,

```
On Wed, Sep 23, 2009 at 11:04 AM, Bruno Chareyre <bruno.chareyre@xxxxxxxxxxx <mailto:bruno.chareyre@xxxxxxxxxxx>> wrote:
```
> The algorithm is : displacement = (stress offset) / (total
box-spheres stiffness)

hmmmm..... for the unit...
stress offset unit is  N/m^2
stiffness unit is   N/m  ?am I right?
```
then (N/m^2) / (N/m) =(1/m), this is not the unit of displacement of m, I think if we use force may be better.
```then Force/stiffness=(N)/(N/m)=m.
```
Yes. Or consider that the stifness is a stress/meter (increment of stress on the boundary associated to a unit displacement).
```

```
```    On one hand you compute the total stiffness on boundary N = the
sum of all sphere-box(N) contact stiffnesses.

```
ok, here the stiffness is due to the sphere-box are parallel connected along the boundary.
```

On the other hand you have the current stress (or, say, the
current force if you prefer) applied on boundary N. From this you
can define the "offset", which is (target stress - current stress).

```
so from here, we can get the stress offset, if the applied force (stress) is not the target force (stress)
```
Then, the displacement computed in the equation above is the one
"so that the new force will be exactly the one we want after the
displacement". (assuming spheres are not moving in the meantime,
which is not true, but it will be corrected again on the next time
step, so it finally converge relatively well)

```
well, if the sphere move fast (such as compressing loose sand to a solid condition), then this may not sufficient to be corrected in the next timestep (I am not very sure , but I feel it may not sufficient to...) . but for static condition, I believe it is correct. One more question: Assume the boundary spheres will not change (always those spheres nomatter they are moving or not), how about I just always apply a constant force on each sphere (if the stress on this boundary is S, then assume the area of the sphere is A, then we can get the force F=S*A, so that since the stress S is a constant, so the force is also a constant ) and then nomatter it is static or it is dynamic, can we say it is also under a constant stress boundary? what do you think?
It sounds complex. How will you define which spheres are on the boundary? What will happen if one sphere is going out of the packing even with this constant force on it? You need a stiffness here, not a force.
```
Bruno

```
```Thanks .
Yongfeng
```
```    This algorithm has no name that I know.

Bruno

kan a écrit :

Thanks, Bruno,

I checked the code, but I still did not understand the
physical meaning of the code.
May I ask : what is the physical meaning behind the code? or
what is the mathematic of the algorithm?
how is it called ?
Thanks a lot.
yongfeng

On Fri, Sep 11, 2009 at 10:22 AM, Bruno Chareyre
<bruno.chareyre@xxxxxxxxxxx
<mailto:bruno.chareyre@xxxxxxxxxxx>
<mailto:bruno.chareyre@xxxxxxxxxxx
<mailto:bruno.chareyre@xxxxxxxxxxx>>> wrote:

More precisely, it is in TriaxialStressController. This engine
needs to be assigned 6 boxes (one for each boundary). The boxes
are used to define the surfaces (since you need a surface to
define a stress).
If you don't have 6 boxes, you need to find a workaround.

The algorithm is : displacement = (stress offset) / (total
box-spheres stiffness)

Bruno

```
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```    _______________
Chareyre Bruno
Maître de Conférences

Grenoble INP
Laboratoire 3SR - bureau E145
BP 53 - 38041, Grenoble cedex 9 - France
Tél : 33 4 56 52 86 21
Fax : 33 4 76 82 70 43
________________

_______________________________________________

```
```

--

_______________
Chareyre Bruno
Maître de Conférences

Grenoble INP
Laboratoire 3SR - bureau E145
BP 53 - 38041, Grenoble cedex 9 - France
Tél : 33 4 56 52 86 21
Fax : 33 4 76 82 70 43
________________

```