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Re: [Question #434659]: the displacement diagram using periodic boundary

 

Question #434659 on Yade changed:
https://answers.launchpad.net/yade/+question/434659

    Status: Open => Answered

Bruno Chareyre proposed the following answer:
>coordinate as (0,0,0) and use the maximum coordinate as (1,1,1), even
if the values of minimum coordiante and maximum >coordinate in the
absolute coordinate system are changing.

Unfortunately no, that's not what I meant, but I understand it was unclear. The minimum is always (0,0,0) but the maximum can be anything and it is changing with time (even in the negative quadran after some time if the velocity gradient is rotational).
Some implementations are indeed setting all coordinates between 0 and 1 as you suggest, but yade is not.
When writing (1,1,1) I was mentally thinking (1*v1+1*v2+1*v3)  where the vi's are the vectors defining the periodic box (and they are changing with time), it is not represented in this way in the code.

>if I want to display the characteristic of displacement tendency under
a certain load(such as pure shear), I should look >symmetric in the
window from (-0.5,-0.5,-0.5) to (0.5,0.5,0.5).

I'm not sure what "characteristic of displacement tendency" means, but if you ask how to see it symmetric, yes, you need to look through a window between -0.5*(v+v2+v3) and +0.5*(v+v2+v3).
Which is technically impossible with yade without considerable changes in the source code, by the way.
So, by default, simple shear will look like [1].
If you want to see it (skew)symmetric you superimpose a constant rigidbody velocity, in the post processing as Jan suggested (by far the simplest) or even in the simulation itself by assigning the same initial velocity to all particles.

Bruno

http://www.flowillustrator.com/wp-
content/uploads/2015/02/ViscosityShearFlow.png

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