yade-dev team mailing list archive

[Bruno Chareyre <bruno.chareyre@xxxxxxxxxxx>]

>  written by
> my boss, so that page is certainly right ;-) )
>
>   
Hehe. There is a french saying like "the world is small" (le monde est 
petit).
Yes, this page looks serious.
> Current Matrix3r strain:
>
> Δx/x₀ δx/y₀ δx/z₀
> δy/x₀ Δy/y₀ δy/z₀
> δz/x₀ δz/y₀ Δz/z₀
>
>   
Ok, this is the usual gradient of displacement (alternatively, gradient 
of velocity if you replace δx by dx/dt - in answer to your last email - 
the same theories hold).
And this is really F - Id.
> I think we can really decribe any transformation by a Matrix3, since we
> always map box to a parallelepiped, i.e. 3 points to 3 points (9
> coordinates, exactly number of elements of a 3x3 matrix): e.g.
>
>   
I agree. Another way to explain why we "can" is we assume a homogeneous 
cell deformation and, therefore, a linear expression for disp. u(x,y,z).

>   0,y₀,0 → 0+y₀*εyx,y₀+y₀*εyy,0+z₀*εyz
>
> (similar for x₀,0,0 and 0,0,z₀). (This is not bound to small strains in
> any way, as far as I see). It seems then that Hsize is the product of
>
>   x₀,0,0   
>   0,y₀,0 * strain
>   0,0,z₀
>
> right?
>
>   
Yes for the product. The tricky part is defining "strain" is not 
straightforward, it is easier to consider it as a result of the integration.
In order to keep a complex case in mind, you can imagine a rotational 
"strain" (in fact disp/vel gradient) with exy=-eyx=pi/100 rad/second. 
After 100 seconds, Hsize will be

0,y₀,0   
-x₀,0,0 
0,0,z₀

after 50 seconds it would be :

x₀/sqrt2,x₀/sqrt2,0   
-y₀/sqrt2,y₀/sqrt2,0 
0,0,z₀

If the code can handle this, then we can consider it general enough 
IMHO. The diagonal multiplication cannot do that probably.


> The strain matrix can be "decomposed" in its diagonal and non-diagonal.
>
> The diagonal (Δx/x₀, Δy/y₀, Δz/z₀) is normal (extensional) strain.
> Multiplying refSize by normal strain (by components) gives _size.
>
>   
Only if you assume no rotation?
> shear matrix (_shearTrsfMatrix) is strain matrix non-diagonal terms, but
> with 1s on the diagonal:
>
> 1     δx/y₀ δx/z₀
> δy/x₀ 1     δy/z₀
> δz/x₀ δz/y₀ 1
>
> This matrix is used frequently, since collisions etc are detected in
> scaled but UNsheared coordinates (positions are in scaled and sheared
> coordinates).
>
>   
I think it can be correct to decompose the transformation as the product 
of rotation and strain (polar decomposition in the wiki page) but 
spliting shear and stretch into a sum is weird, and I'm afraid it could 
mislead us to wrong maths.

> Relative volume difference is det(strain); if it is upper-triangular or
> lower-triangular, this determinant is product of the diagonal terms;
> therefore applying pure shear doesn't change volume (since it doesn't
> change _size).
>
>   
Yes, detF is V/V0.


> 5. Way to directly prescribe cell normal&shear strain (like what the
> strain matrix has now).
>
>   
One way could be to add members like this :

void setStrainRate(int i, Real value) {gradVel[i][i]=value;}
void setSize(int i, Real value) {Hsize[i][i]=value;} //Fixes the problem 
of Hsize initialisation at the same time

It would make the interface simple for users using no complex 
deformations, others could just access data gradVel/Hsize directly, or 
with additional members.

> Ugh, I am writing in a little confusing way. Better to continue
> tomorrow. Hope this doesn't make more mess than there was.
>
>   
That's ok, I think we are very close to a good solution now. :-)

Bruno

--

_______________
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
________________