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 <[email protected]> wrote: > > > On 6 May 2010 23:11, chiara modenese <[email protected]> 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 > > >
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