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Re: Elastic energy

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• From: Janek Kozicki <janek_listy@xxxxx>
• Date: Sun, 4 Jul 2010 21:31:33 +0200

```Janek Kozicki said:     (by the date of Sun, 4 Jul 2010 20:47:40 +0200)

> I'm not sure if I worded myself clearly.... I hope that you can
> understand what I mean?

to rephrase in even different words: in plastic regime the Fs remains
constant at the limiting value. While the sphere slipping distance
sᵢ-sᵢ₋₁ can be actually bigger, that what has been contributed into
the value of Fs, and is 's'.

hmm, I see that ElasticContactLaw.cpp lines 96-104 are actually
implementing a simlar thing to what I'm saying, by actually
calculating the difference between s and sᵢ-sᵢ₋₁.

So apparently there is just a bug inside here. Is maxFs correct? is
trialForce the right value? Is the formula really similar?

Let's rewrite this:

plasticDissipation +=	((1/currentContactPhysics->ks)*(trialForce-shearForce))//plastic disp.
.dot(shearForce);//active force

into:

Ep += (Fs-Fsmax)⋅Fsmax/ks

And compare this with:

Es+=(Fs²⁄ks)⁄2

hmmm is there a "⁄2" missing?
I mean:

Ep += ((Fs-Fsmax)⋅Fsmax/ks)⁄2

I'm not sure in fact :)

Next I am trying to check this again:

Es+=(Fs²⁄ks)⁄2

where in fact:  Fs²⁄ks = (ks*s)²/ks = ks*s²

And compare again with this:

Ep += ((Fs-Fsmax)⋅Fsmax/ks)⁄2

and try to rewrite it:

(Fs-Fsmax)⋅Fsmax/ks = Fs⋅Fsmax/ks-Fsmax²/ks =
= (ks*s)*(ks*∑(sᵢ-sᵢ₋₁))/ks - (ks*∑(sᵢ-sᵢ₋₁)²/ks =
= s*ks*∑(sᵢ-sᵢ₋₁) - ks*∑(sᵢ-sᵢ₋₁)²

but I must leave now. You could start derivation from this place
where I stopped, and try to continue verifying if Ep is correct or
not.

Also I wanted to check how Ep could be expressed if using the same
terms as in (or in terms of)

Ws =∑ (Fsᵢ+Fsᵢ₋₁)*(sᵢ-sᵢ₋₁)⁄2

but unfortunately not today, I must go. But maybe you will be able to
figure this out?

best regards
--
Janek Kozicki                               http://janek.kozicki.pl/  |

```

References

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