yade-dev team mailing list archive

[chiara modenese <c.modenese@xxxxxxxxx>]

Hello Bruno,
I send again to the list a couple of mails I addressed to you.
Chiara

On 16 May 2010 01:23, chiara modenese <c.modenese@xxxxxxxxx> wrote:

>
>
> On 6 May 2010 23:11, chiara modenese <c.modenese@xxxxxxxxx> wrote:
>
>> Hi Bruno,
>>
>> could you explain me in formula how did you get the plastic dissipation
>> for the friction component? I see
>>
>> plasticDissipation +=
>> (shearDisp+(1/currentContactPhysics->ks)*(shearForce-prevForce)).dot(shearForce)
>>
>> Is shearDisp the total shear displacement, right? I would compute the
>> dissipation incrementally, something like
>>
>> plasticDissipation += [delta_us + (Fs_max_current - Fs_max_prev)/ks] *
>> Fs_max_current
>>
>> Is this the same as you did? If not where I am eventually wrong? Sorry if
>> I ask..
>>
>> Thanks a lot,
>> Chiara
>>
>
>
> Hi Bruno,
> I am checking the formulation for the energy friction dissipation. Here it
> is what you proposed (possible option, not committed yet I think):
>
> if( shearForce.squaredNorm() > maxFs ){
>           Real ratio = Mathr::Sqrt(maxFs) / shearForce.norm();
>           Vector3r trialForce=shearForce;//store prev force for definition
> of plastic slip
>           shearForce *= ratio;
>           plasticDissipation +=
>           ((1/currentContactPhysics->ks)
> *(trialForce-shearForce))//plastic disp.
>           .dot(shearForce);//active force           }
>
> I have one big guess. How can this formulation being incremental using
> trialForce? Let's take a simple case: normal force is set to a constant
> value. Then say that trialForce will increase. In this case shearForce is
> the same since normal force is as well (like maintaining constant the
> maximum elastic shear displacement). If any time you get the work done as
> above, I would say that
> 1/ks*(trialForce-shearForce)
> is the total plastic displacement, is it? Why do you think it is
> incremental? In which way it is?
>
> thanks, Chiara
>
>
>