yade-users team mailing list archive

[chiara modenese <c.modenese@xxxxxxxxx>]

On 10 May 2010 18:03, chiara modenese <c.modenese@xxxxxxxxx> wrote:

>
>
> On 10 May 2010 17:43, Bruno Chareyre <bruno.chareyre@xxxxxxxxxxx> wrote:
>
>>
>>
>>> Hi Bruno, please see attached file. I am not completely sure but I think
>>> it would be correct to distinguish between loading and unloading phase. I am
>>> trying to prove it analytically, I am not yet totally confident.. What do
>>> you think?
>>>
>>>  There is no trialFs in your figures, this is the key factor.
>>
>> us + (current-prev)/ks could make sense in some formulations if *=ratio
>> was applied _before,_ but it is applied _after, _so that "current" is in
>> fact the "trial" value (as you pointed out).
>>
>> OTOH, us is exactly (trialFs-prevFs)/ks, by definition of trialFs,
>> (trialFs-currentFs)/ks is exactly the plastic part ,
>> and (currentFs-prevFs)/ks is the elastic part.
>>
>> I don't see any difference between loading and unloading.
>
>
> I agree that the plastic displacement is (trialFs-currentFs)/ks. However
> your formulation is not incremental because you are using the trial value.
> To me this would be the right procedure only in the case in which shearForce
> is maintained constant. But shearForce is actually varying (right? Where do
> you account for this variation in your formula?) and as I sketched in my
> picture there is an elastic contribution that turns into plastic
> dissipation. Can you see my point? I will find out the analytical
> explanation.
>
>
When I say shearForce is varying I mean the maximum friction value. Maybe I
am a little confused now, perhaps I am wrong. Could you just tell me if you
see any effect due to the variation of the maximum friction? Because that is
why I was making distinction between loading and unloading and for the same
reason I am not convinced that using Trial force would be the best choice.
thanks, chiara


> Chiara
>
>>
>>
>> Bruno
>>
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>