Merging grains that satisfy a set of conditions

[Tuomo Nyyssönen]

Thank you for making this toolbox. It is very nice not to have to rely on commercial software all the time and to apply the computational powers of MATLAB to crystallographic data.

My problem concerns the merging of ferritic bcc grains that originate from a single parent austenitic fcc grain. That is, the bcc orientations of the neighboring grains can be duplicated with a particular rotation performed on different crystallographic solutions of the original parent austenite orientation.

One way to find the common parent austenite orientation is to perform the inverse of the fcc to bcc rotation on all crystallographically equivalent solutions of two separate bcc orientations. The two sets of resulting orientations are then compared with one another; if there is a matching orientation in the sets (the misorientation angle between them being suitably low), this is the parent austenite orientation for both bcc orientations.

I have attached some experimental data and a Matlab script that hopefully clarify the problem. Figure 1 shows a martensitic packet, the data points of which all satisfy the condition. Figure 2 shows the 001 pole figures for all bcc orientations, as well as the discovered parent austenite orientation. Figure 3 shows all recalculated martensite variants on top of experimental data. There is at least some kind of match.

I would like to be able to identify all the bcc grains that satisfy the common parent austenite orientation condition and transform them to a merged fcc grain with the parent orientation.

This is a big problem and by any means, I don't expect a complete solution. However, any advice on how to proceed would be greatly appreciated. I am using MTEX 4.0.9.

BR,

Tuomo

[Ralf Hielscher]

Hi Tuomo,

this is indeed an interesting question. I have looked through your code and maybe I can give some hints:

general hint: Whenever possible try to work with orientations and misorientations in MTEX (no convertion to rotation) as MTEX takes care about crystal symmetries.

Assume you have two Ferrit orientations ori1 and ori2, both originating from the same Austenite orientation ori. Then we have

ori1 * cs_bcc1 = ori * cs_fcc1 *  bcc2fcc

ori2 * cs_bcc2 = ori * cs_fcc2 *  bcc2fcc

where bcc2fcc is a bcc2fcc pase transition misorientation and cs_bcc1, cs_bcc2 are some bcc symmetry operations. Then we obtain

ori1 * cs_bcc1  * inv(bcc2fcc) * inv(cs_fcc1) = ori2 * cs_bcc2 * inv(bcc2fcc) * inv(cs_fcc2)

and hence

inv(cs_bcc2) * inv(ori2) * ori1 * cs_bcc1 = inv(bcc2fcc) * inv(cs_fcc2) * cs_fcc1 * bcc2fcc

The left hand side is simply the misorientation between two grains modulo symmetry which we can extract from grain boundaries by

gB = grains.boundary('ferrit','ferrit')
gB.misorientation

The right hand side is a list of ferrite to ferrite misorientations which we can define by

fcc2bcc = orientation('Euler',4.8*degree,46.1*degree,4.8*degree,cs_fcc,cs_bcc);
fcc_trans = fcc2bcc * cs_fcc.properGroup * inv(fcc2bcc)

Now we can compute the angle between all boundary misorientations and all misorientations from the above list and take the minimum with respect to the this list

omega = min(angle_outer(gB.misorientation,fcc_trans),[],2);

One can visualize this deviation by

plot(gB,omega./degree)
colorbar

or merge all those grains having a common boundary misorientation close to fcc_trans 

grains_austenite = merge(grains,gB(omega < 2*degree))

I hope this helps. I have also attached the whole code. Which gives this result where the magenta grain boundaries are those of the bcc phase.

Ralf.

[Tuomo Nyyssönen]

Thank you very much for your reply and explanations! The code you posted works as desired and is certainly enough to proceed.

Best regards,
Tuomo