Taylor Model

Schmid analysis asks which single slip system is best aligned with an applied stress. The Taylor model asks a different question: which combination of slip systems can produce an imposed strain? It assumes that every grain undergoes the specimen strain, and therefore represents the equal-strain limit of a polycrystal model.

Set the specimen frame and strain

A rolling frame is a named specimen frame with rolling direction (RD), transverse direction (TD), and normal direction (ND). Its plotting convention places RD north, TD west, and ND out of the page.

specimenFrame.rolling.makeDefault

Begin with the three predefined bcc slip families. Symmetrization expands them to 96 systems because both signs of each Burgers vector are needed to represent signed slip with nonnegative coefficients.

cs = crystalSymmetry('432');
sSRepresentative = slipSystem.bcc(cs)
sS = sSRepresentative.symmetrise;
length(sS)

sSRepresentative = slipSystem (432)
 size: 1 × 3
 
   u    v    w  | h    k    l CRSS
   1   -1    1    0    1    1    1
  -1    1    1    2    1    1    1
  -1    1    1    3    2    1    1
ans =
    96

The strain below extends RD, leaves TD unchanged, and compresses ND by the same amount. More generally, \(q\) partitions the transverse contraction between TD and ND while keeping the strain trace zero.

q = 0;
epsilon = strainTensor(diag([1 -q -(1-q)]))

epsilon = strainTensor (TD←RD↑)
  type: Lagrange 
  rank: 2 (3 × 3)
 
  1  0  0
  0  0  0
  0  0 -1

Solve one crystal

An orientation maps the crystal frame into the specimen frame. The inverse therefore expresses the imposed specimen strain in this crystal's frame, where the slip systems are defined.

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

[M,gamma,W] = calcTaylor(epsilonCrystal,sS);
M
W

ori = orientation (432 → TD←RD↑)
 
  Bunge Euler angles in degree
  phi1  Phi phi2
     0   30   15
 
M =
    2.1208
 
W = spinTensor (crystal)
  rank: 2 (3 × 3)
 
 *10^-2
      0     51  65.65
    -51      0 -23.04
 -65.65  23.04      0

For this orientation the Taylor factor is 2.1208. The vector gamma contains one nonnegative slip amount per system, while W is the crystallographic spin required by that combination. It is important not to confuse gamma with sS.b, which stores the Burgers vectors.

active = gamma > 1e-8;
bar(find(active),gamma(active))
xlabel('slip-system index')
ylabel('slip amount')

Several systems share the imposed deformation. Their summed activity, normalized by the strain magnitude, gives the Taylor factor.

Map the Taylor factor over orientation space

If the strain remains in the specimen frame, calcTaylor returns the Taylor factor and spin as orientation-dependent functions. They can then be evaluated at any crystal orientation.

[MFun,~,WFun] = calcTaylor(epsilon,sS)

MFun.eval(ori)
WFun.eval(ori)

MFun = SO3FunHarmonic (432 → TD←RD↑)
  bandwidth: 32
  weight: 3.1
 
 
WFun = SO3VectorFieldHarmonic (1 → TD←RD↑)
  bandwidth: 32
  tangent space: rightSpinTensor
  intern symmetries: 432 → TD←RD↑
  intern tangent space: leftVector
 
ans =
    2.1009
 
ans = spinTensor (crystal)
  rank: 2 (3 × 3)
 
 *10^-2
      0  38.89  49.68
 -38.89      0 -14.97
 -49.68  14.97      0

The harmonic functions approximate the direct solution above. Their main advantage is that the expensive Taylor solve is performed once rather than separately for every later orientation.

sP = phi1Sections(cs);
sP.phi1 = (0:10:90)*degree;

plot(MFun,'smooth',sP)
mtexColorbar

hold on
plot(WFun,'color','black')
hold off

The colours reproduce the orientation dependence of the Taylor factor in Fig. 5 of Bunge (1970). Darker and lighter regions require different total slip for the same strain. The black vectors show the accompanying spin direction and magnitude in orientation space.

Map the spin magnitude

For an infinitesimal strain step, the norm of the spin tensor is the angle through which the corresponding crystal rotates in the Taylor model. The values below are angles for the strain step stored in epsilon and scale with the size of that step.

plot(norm(WFun)/degree,'smooth',sP,'resolution',0.5*degree)
mtexColorbar

The symmetry-related peaks mark orientations that rotate most during this plane-strain increment. This plot reproduces the construction in Fig. 8 of Bunge (1970).

sP = sigmaSections(cs);
plot(norm(WFun)./degree,'smooth',sP)
mtexColorbar

Sigma sections slice the same spin-magnitude function along a different family of paths through orientation space. The extrema are unchanged.

Apply the model grain by grain

Return to the ordinary specimen frame and adopt the plotting convention of the CSL map. The data supply one mean orientation for each grain.

specimenFrame.specimen.makeDefault
plottingConvention.default('y↓→x');
mtexdata csl

grains = calcGrains(ebsd,'minPixel',3);
grains = smoothBoundary(grains,5);

ebsd = EBSD (y↓→x)
 
 Phase   Orientations     Mineral         Color  Symmetry  Crystal reference frame
     0    5 (0.0032%)  notIndexed          none                                   
    -1  154107 (100%)        iron  LightSkyBlue      m-3m                         
 
 Properties: ci, error, iq
 Scan unit : um
 X × Y × Z : [0 → 511] × [0 → 300] × [0 → 0]
 Normal vector: (0,0,1)

Apply the same plane strain to every grain. The inverse mean orientations express it in the individual crystal frames. The fcc family contains both Burgers-vector signs for the nonnegative slip amounts used by the solver.

q = 0;
epsilon = strainTensor(diag([1 -q -(1-q)]));
sS = symmetrise(slipSystem.fcc(grains.CS));

[MGrain,gammaGrain,WGrain] = calcTaylor( ...
  inv(grains.meanOrientation) * epsilon,sS);

plot(grains,MGrain)
mtexColorMap white2black
mtexColorbar

Grain colour now measures the total slip needed to impose the common strain. A high Taylor factor identifies an orientation that is hard to deform under the equal-strain assumption.

[~,gammaMaxId] = max(gammaGrain,[],2);
sSGrains = grains.meanOrientation .* sS(gammaMaxId);

hold on
quiver(grains,sSGrains.b,'autoScaleFactor',0.7, ...
  'displayName','Burgers vector','project2plane')
quiver(grains,sSGrains.trace,'autoScaleFactor',0.7, ...
  'displayName','slip plane trace')
hold off

The arrows identify the system with the largest slip amount in each grain. One arrow is the surface projection of its Burgers vector, and the other is the trace of its slip plane. The full combination still contains the other systems with nonzero entries in gammaGrain.

newMtexFigure
plot(sSGrains.b)
text([xvector,yvector,zvector],'labeled','BackGroundcolor','w')

All but one of the selected directed Burgers vectors fall in the upper hemisphere for this \(q=0\) strain. The solver uses nonnegative slip amounts and chooses between the two stored Burgers-vector signs, so this is the selected shear sense rather than antipodal plotting. Changing \(q\) changes the strain path and can change both the selected systems and their signs.

How the Taylor solve works

For every system \(\alpha\), calcTaylor forms the symmetric unit-shear tensor \(\mathbf P^\alpha\). It finds nonnegative slip amounts \(\gamma^\alpha\) that reproduce the five independent components of the deviatoric strain:

\[\epsilon = \sum_\alpha \gamma^\alpha\mathbf P^\alpha.\]

Among feasible combinations, the solver minimizes the CRSS-weighted slip activity \(\sum_\alpha \mathrm{CRSS}^\alpha\gamma^\alpha\). With the equal CRSS used here, the Taylor factor is

\[M = \frac{\sum_\alpha\gamma^\alpha} {\|\epsilon\|}.\]

The antisymmetric parts of the same shears give the spin tensor W. This construction is why the Taylor assumption needs enough independent slip systems to span all five deviatoric strain components.

Texture evolution during rolling

Iterating the Taylor spin over small strain increments simulates how the orientation distribution of a polycrystal evolves during deformation. Texture Evolution develops that calculation and explains the required step-size approximation.

close all

References

Next

Continue with Sachs Model for the complementary equal-stress limit and its comparison with the Taylor upper bound.