Texture Evolution edit page

Plastic deformation does not only change the shape of a polycrystal, it also rotates its crystals. The Taylor model tells us, for a given strain and a given family of slip systems, by how much a crystal in a given orientation has to rotate. Applying that rotation to every orientation of a sample and iterating over small strain increments is the simplest possible simulation of texture evolution, and that is what this page does.

Two related pages take different routes to the same question - the single slip model solves the continuity equation for the ODF analytically, and VPSC results computed outside of MTEX can be imported and analyzed here.

The crystallographic spin

We consider fcc slip and plane strain, i.e. rolling.

cs = crystalSymmetry('432');
sS = symmetrise(slipSystem.fcc(cs))
sS = slipSystem (432)
 size: 24 x 1
 
   u    v    w  | h    k    l CRSS
   0    1   -1    1    1    1    1
  -1    0    1    1    1    1    1
   1   -1    0    1    1    1    1
   0   -1    1    1    1    1    1
   1    0   -1    1    1    1    1
  -1    1    0    1    1    1    1
   1   -1    0    1    1   -1    1
   1    0    1    1    1   -1    1
   0    1    1    1    1   -1    1
  -1    1    0    1    1   -1    1
  -1    0   -1    1    1   -1    1
   0   -1   -1    1    1   -1    1
   0    1   -1   -1    1    1    1
   1    0    1   -1    1    1    1
   1    1    0   -1    1    1    1
   0   -1    1   -1    1    1    1
  -1    0   -1   -1    1    1    1
  -1   -1    0   -1    1    1    1
  -1    0    1    1   -1    1    1
   1    1    0    1   -1    1    1
   0    1    1    1   -1    1    1
   1    0   -1    1   -1    1    1
  -1   -1    0    1   -1    1    1
   0   -1   -1    1   -1    1    1
q = 0;
epsTotal = 0.6 * strainTensor(diag([1 -q -(1-q)]))
epsTotal = strainTensor (y↓→x)
  type: Lagrange 
  rank: 2 (3 x 3)
 
 *10^-2
  60   0   0
   0   0   0
   0   0 -60

For a single crystal calcTaylor returns, besides the Taylor factor, the spin tensor W that the crystal experiences

ori = orientation.byEuler(0,30*degree,15*degree,cs);

eps1 = 0.01 * strainTensor(diag([1 -q -(1-q)]));

[M,~,W] = calcTaylor(inv(ori) * eps1, sS);

M, W
M =
    2.3153
 
W = spinTensor (crystal)
  rank: 2 (3 x 3)
 
 *10^-4
      0 -47.11 -68.65
  47.11      0  24.65
  68.65 -24.65      0

and the updated orientation is obtained by applying it

oriNew = exp(ori,-W);

angle(ori,oriNew) ./ degree
ans =
    0.4975

One percent of strain rotates this crystal by about half a degree. Since the spin depends on the orientation, the crystals of a polycrystal drift apart at different rates and the texture sharpens.

Iterating over a polycrystal

Evaluating calcTaylor separately for every orientation would be far too slow. Called without an orientation it instead returns the Taylor factor and the spin as orientation dependent functions, which are then cheap to evaluate on a whole list of orientations. We compute this spin field once, for a single strain increment.

numIter = 60;

[~,~,spin] = calcTaylor(epsTotal ./ numIter, sS)
spin = SO3VectorFieldHarmonic (1 → y↓→x)
  bandwidth: 32
  tangent space: rightSpinTensor
  intern symmetries: 432 → y↓→x
  intern tangent space: leftVector

Note that this step takes the bulk of the computing time, and its cost depends only on the bandwidth of the harmonic representation, not on the number of orientations. Passing 'bandwidth',16 makes it noticeably faster at the price of a relative error of a few percent in the spin.

We start from a uniform texture

rng(0)
ori = orientation.rand(2e4,cs);

odf0 = calcDensity(ori,'halfwidth',10*degree);

norm(odf0)^2
ans =
    1.0010

and step through the deformation

pC = progressCounter(numIter);
for k = 1:numIter

  % the spin experienced by each individual orientation, in crystal
  % coordinates - the field was set to return a right sided spin tensor
  W = spin.eval(ori);

  % update the orientations
  ori = exp(ori,-W);

  pC.show(k);
end

The resulting rolling texture

% the resulting texture lives in the rolling frame - RD to the north, TD
% to the west and ND out of the page; the frame also annotates the pole
% figures with RD, TD, ND
specimenFrame.rolling.makeDefault

plotPDF(ori,Miller({0,0,1},{1,1,1},cs),'contourf')
mtexColorbar

The initially uniform texture has developed the familiar fcc rolling components. How strong the texture has become is best measured by the texture index \(\lVert f \rVert^2\)

odf = calcDensity(ori,'halfwidth',10*degree);

norm(odf)^2
ans =
    1.9160

For reference, the same computation stopped at 20 and at 40 percent strain gives 1.15 and 1.50 - the texture index grows steadily but the texture is still far from being sharp at 60 percent.

plotSection(odf,'phi2',[0 45 65]*degree,'contourf')
mtexColorbar

Things worth knowing

  • The step size matters. The spin field is computed for one strain increment and then applied numIter times, which is an explicit Euler scheme - too few steps and the trajectories are wrong, too many and the computation is needlessly slow.
  • The model deforms every crystal by exactly the same strain, which is the defining Taylor assumption. It over predicts the sharpness of real textures, because in a real material grains accommodate each other.
  • Nothing here depends on the strain being plane strain. Changing epsTotal to strainTensor(diag([-0.5 -0.5 1])) gives axisymmetric tension, and the same loop produces the corresponding fibre texture.
  • The slip systems enter only through sS. Replacing slipSystem.fcc by slipSystem.bcc or by a hexagonal family, see Taylor Model for Hexagonal Materials, changes the predicted texture completely.