Dear Prithivi,

D is a representation of the stretch of material fibre in a given direction while W represents the angular velocity of the fibre about a material point.

The formulation F = FeFp, L=F'F-1 , Le=Fe'Fe-1 and Lp =Fp'Fp-1 is not consistent with the decomposition L = Le+Lp, D=De +Dp and W = We +Wp (In general). It only becomes consistent if you make some assumption for instance small elastic strains in case of metals. So basically, when you decompose D (or L) (or W) in general, it will have at least one term which has an elasto-plastic coupling.

Secondly, the formulation  L = F'eFe -1 + Fe F'p Fp -1 Fe -1  is arbitrary. If you take the intermediate configuration as only plastic i.e. F=FpFe then L will give the formulation L = Fp'Fp-1+FpFe'Fe-1Fp-1.  So it is really hard to identify what is actually happening in the intermediate configuration because it is not possible to decouple every physical quantity.

2) W (skew symmetric) represents spin (angular velocity) 

F = RU

L = F'F-1

L = R'RT+RU'U-1RT=W+D

where R'RT is the rate of rotation tensor and is not necessarily equal to W. It will only be equal if U'U-1 is symmetric. So in general, W is not zero in the intermediate configuration even if the rotation in intermediate configuration is zero.

Also see the section 'Model formulation' and 'single crystal plasticity model' in the attached article. If you pay good attention, you will see that they make many assumptions (which might be valid for their case). So you can always make some assumptions to make the model consistent for your specific case.

Dr. Niazi,

Thanks for your comments, they were very helpful. I have a couple more queries -

1) I understand that W represents the angular velocity but how do we understand the difference between the elastic spin and plastic spin ? how are they different ? I can easily visualize the difference between the elastic strain and plastic strain .

2)  By the definition of Fp, Rp = I

Fp = RpUp = Up

As Lp = F'p Fp-1

Lp = U'p Up-1

Up is definitely a symmetric tensor by definition, I believe that U'p Up-1

is also symmetric

Would Lp be just Dp in such an scenario and does this formulation make any sense to you ?

Thanks,

Prithivi

Dear Prithivi,

1) Imagine a block of material under simple shear loading. In this case you have stretch as well as rotation. The coordinate system is continuously rotating while shearing. Now you unload at a certain point then the block will rotate back slightly due to elastic unloading. The change in coordinate system before and after unloading is the elastic rotation at that configuration while the rest is plastic rotation. Note that plastic strain/rotation is an internal variable which is not directly observable. But elastic strain/rotation is a state variable which can be observed directly. That is why the internal variables are evolved based on an arbitrary dissipation potentials and you can use (develop) any model for these under certain rules/principles. On the other hand the state (observable) variables are evolved based on state potential which are based on stored energy in the system at a given state.

2) I am not a crystal plasticity expert so i can not advise on the assumptions you make. If you see the article I have attached in the previous comment, they assumed plastic spin for the YSDH model to be zero but for single crystal plasticity they consider plastic spin. I would suggest to look carefully into the equations they have described.

For you formulation:

Uij = Uji  agree

(U.U)ij = (U.U)ji   also agree

But

(U'.U-1)ij = (U'.U-1)ji   not always i.e. not true in general. Multiplying two symmetric but diffetent second order tensors do not yield a symmetric second order tensor.

If Rp = I then Lp= Up'Up-1 agree but the L= Le+FeLpFe-1 so you still have the velcity gradient not decoupled. You must think if you really need to decouple the velocity gradient (for instance are you using Eulerian formulation).

Since Up'Up-1 is not symmetric in general, the pladtic spin is not zero even if Rp=I. So Lp = Dp + Wp in this formulation.

Thanks Dr. Niazi.

That was a silly mistake, I got that

(AB)' = B'A' = BA ( not necessarily AB)

See the following articles. Plastic Spin is explained much clearly in the first article. They make assumptions / formulations to tackle the issue of plastic spin. Might be interesting for your work.

Dr.. Niazi ,

Thank you very much for taking time to answer our questions, its really helping me a lot !

I think, when you consider the continuum mechanical framework in crystal plasticity, the plastic velocity gradient in the current configuration is FeLpFe-1 where Lp is in intermediate configuration. Now this Lp is related to crystallographic slip which doesn't have any lattice rotation in it.  So when we solve the constitutive equations, Lp is assumed and from that Fp, Fe and S is evaluated , where S is the 2nd PK stress and with that S Lp is again evaluated until the error is less that a tolerance.

Moreover, the skew-symm tensors represent angular momentum about a point, which isn't zero but conserved and is a little different from rotation which is associated with the polar decomposition of a tensor. In other words, we go from reference to intermediate configuration by Fp( or Lp) due to plastic slip in specific crystallographic planes. Now due to this there is a change of coordinate frames from the lab coordinate (aligned with crystal coordinate in ref frame) and crystal coordinates in intermediate configuration. So the rotation tensor associated with Fp fixes helps align the lattice coordinate axis with lab coordinate axis in intermediate configuration. I guess this also answers your second question. And Dp which is the rate of deformation tensor is related to crystallographic slip deformation in some constitutive models.

Thanks a lot Arita for joining the discussion.

I have always thought that W( skew symmetric spin tensor) represents the angular velocity about the 3 coordinate axes but not angular momentum, Could you provide any reference for that ?

Yeah, its the angular velocity, which a few people relate to the angular momentum equation but mostly in fluid dynamics, http://journals.cambridge.org/download.php?file=%2F25059_D0377079D3D977305A3E8A984044CF9D_journals__PSP_PSP74_02_S0305004100048131a.pdf&cover=Y&code=c641db34c190fc82a239898b8d35813, sorry for the confusion.

What I was trying to say was, the W part is different than Rotation R, and Rp =1 is a matter of choice , and when it is done other quantities have to be rotated accordingly because the lattice frame of reference and the lab frame of reference is no longer same in intermediate configuration as was in reference configuration.

hope this helps.

A few additional comments:

1) We represents lattice rotation. Unfortunately, this contribution is often termed as elastic rotation. This is however inaccurate since, in contrast with De, the We contribution cannot be calculated from the stress rate.

Dp is the plastic strain rate. It tells you how lengths and angles are affected by plastic flow. Wp is the plastic rotation rate. It tells you how a fiber is rotated during a infintesimal time step.

2) No, you cannot say that Lp is zero. In the intermediate configuration, the plastic rotation rate is generally not zero. For instance, if you consider a pure shear transformation (activation of a single slip system), you should see that Wp is not zero.