Texture evolution in rolled magnesium during uniaxial tension

This page compares two Taylor-model predictions for rolled magnesium pulled along its rolling direction. The starting sheet has a basal fibre texture, but the two temperatures use different slip resistances and reach different strains. The calculation shows how those choices change both the texture and the activity assigned to each deformation family.

The motivating tension experiments reached approximately 30 percent strain at room temperature and 70 percent at 250 degrees Celsius.

Set the crystal and specimen frames

The sheet is orthotropic, so its specimen symmetry is the orthorhombic group 222. Expressing it in a rolling frame names the specimen axes rolling direction (RD), transverse direction (TD), and normal direction (ND), and every pole figure below is drawn and annotated in those axes. Here the tension axis is RD.

ss = specimenSymmetry('222');
ss.frame = specimenFrame.rolling

ss = orthorhombic specimenSymmetry (TD←RD↑)

Load the lattice parameters and hexagonal symmetry from the magnesium crystal-information file. properGroup keeps the rotational part of the point group used to generate oriented slip and twinning systems.

cs = crystalSymmetry.load('Mg-Magnesium.cif')
cs = cs.properGroup;

cs = crystalSymmetry (⊙c→a)
 
  mineral        : Mg               
  symmetry       : 6/mmm            
  elements       : 24               
  a, b, c        : 3.2, 3.2, 5.2    
  reference frame: X||a*, Y||b, Z||c

Build the initial basal fibre texture

In an ideal rolled magnesium sheet, the crystal c-axes are parallel to ND while rotations about that axis are random. A fibre ODF represents that basal fibre texture directly.

odf = fibreODF(cs.cAxis,vector3d.Z,ss);
odf = FourierODF(odf);

Plot the basal pole, a prismatic pole, and a pyramidal pole. Pole figures are antipodal here, so opposite directions are drawn as the same pole.

h = Miller({0,0,0,1},{1,0,-1,0},{1,0,-1,1},cs);
plotPDF(odf,h,'contourf','complete','upper')
mtexColorbar

The basal poles form one maximum at ND. The prismatic poles form a ring because the initial model contains every rotation about the c-axis with equal probability. RD is also the tension direction in all three plots.

Choose temperature-dependent families

The Slip Systems page defines critical resolved shear stress (CRSS) and explains why hexagonal materials have no universal slip set. CRSS depends on material, temperature, and experiment, and published values vary widely. The dimensionless values below are illustrative ratios normalized to basal slip, not a universal magnesium parameter set.

At room temperature, basal slip commonly dominates magnesium deformation and extension twins can also have a low CRSS. For this basal texture, tension perpendicular to the c-axis does not activate extension twinning. The model therefore includes only compression twins and assigns them the largest CRSS.

sScold = [slipSystem.basal(cs,1),...
  slipSystem.prismatic2A(cs,66),...
  slipSystem.pyramidalCA(cs,80),...
  slipSystem.twinC1(cs,100)];

% Generate all symmetry-related systems and remember their family ids.
[sScold,slipId] = sScold.symmetrise;

At higher temperature, the assumed CRSS of both non-basal slip families decreases. The compression-twin CRSS remains high in this comparison.

sSwarm = [slipSystem.basal(cs,1),...
  slipSystem.prismatic2A(cs,15),...
  slipSystem.pyramidalCA(cs,10),...
  slipSystem.twinC1(cs,100)];
sSwarm = sSwarm.symmetrise;

Define the two strain states

Plastic incompressibility requires each infinitesimal strain tensor to have zero trace. The room-temperature case assumes unequal contraction along TD and ND. The 250-degree case assumes that this transverse anisotropy is negligible, so both directions contract equally.

epsCold = 0.3 * strainTensor(diag([1 -0.6 -0.4]))
epsWarm = 0.7 * strainTensor(diag([1 -0.5 -0.5]))

epsCold = strainTensor (y↓→x)
  type: Lagrange 
  rank: 2 (3 × 3)
 
 *10^-2
  30   0   0
   0 -18   0
   0   0 -12
 
epsWarm = strainTensor (y↓→x)
  type: Lagrange 
  rank: 2 (3 × 3)
 
 *10^-2
  70   0   0
   0 -35   0
   0   0 -35

Solve the Taylor model for the starting texture

Draw a synthetic polycrystal from the initial ODF. optimalSample places the orientations so that they reproduce the ODF as closely as possible, which needs far fewer of them than a random draw. Both simulations start from this same polycrystal.

ori = odf.optimalSample(5000)

ori = orientation (Mg → TD←RD↑ (222))
  size: 5280 × 1

Express each strain in each crystal frame and solve the Taylor problem. The columns of bCold and bWarm are slip or twin amounts for the symmetrized systems. The spin tensors describe the corresponding lattice rotations.

[~,bCold,Wcold] = calcTaylor(inv(ori) .* epsCold,sScold);
[~,bWarm,Wwarm] = calcTaylor(inv(ori) .* epsWarm,sSwarm);

Apply each crystallographic spin to the initial orientations.

oriCold = ori .* orientation(-Wcold);
oriWarm = ori .* orientation(-Wwarm);

meanRotation = [mean(angle(ori,oriCold)),...
  mean(angle(ori,oriWarm))] ./ degree

meanRotation =
    8.3179   15.4162

The mean orientation changes are 8.3179 degrees at room temperature and 15.4162 degrees at 250 degrees Celsius. The larger warm value reflects both its larger imposed strain and its different CRSS ratios.

One-step approximation

This page evaluates the spin only at the starting orientations and applies the entire 30 or 70 percent strain in one update. It is therefore an illustrative one-step approximation, especially at the larger strain. Texture Evolution shows the more accurate incremental calculation in which orientations are updated repeatedly.

Compare the pole figures

Add the room-temperature and 250-degree results beneath the initial pole figures, then arrange the three states as rows on common specimen axes.

newMtexFigure('layout',[3,3])
plotPDF(odf,h,'contourf','complete','upper','grid','grid_res',30*degree)

nextAxis
plotPDF(oriCold,h,'contourf','upper','complete',...
  'grid','grid_res',30*degree,'noLabel','noTitle')

nextAxis
plotPDF(oriWarm,h,'contourf','upper','complete',...
  'grid','grid_res',30*degree,'noLabel','noTitle')
mtexColorbar

Compare each column from top to bottom: initial, room temperature, then 250 degrees Celsius. Both deformed rows depart from the ideal basal fibre, and they differ from one another because their CRSS ratios, strain shapes, and total strains are different. The plot cannot attribute a difference to temperature alone because all three inputs change together.

Summarize deformation-family activity

Sum the Taylor coefficients of symmetry-related systems using slipId. Dividing by the number of sampled crystals gives the mean amount assigned to each family per crystal.

slipId = repmat(slipId.',length(ori),1);
statSsCold = accumarray(slipId(:),bCold(:)) ./ length(ori);
statSsWarm = accumarray(slipId(:),bWarm(:)) ./ length(ori);

familyActivity = array2table([statSsCold.';statSsWarm.'],...
  'VariableNames',{'Basal','Prismatic','Pyramidal','CompressionTwin'},...
  'RowNames',{'RoomTemperature','250DegreesC'})

familyActivity =
  2×4 table
                        Basal      Prismatic    Pyramidal    CompressionTwin
                       ________    _________    _________    _______________
    RoomTemperature    0.021981     0.37224      0.29477       2.3338e-10   
    250DegreesC        0.044826     0.72982      0.84239       4.9066e-11

Use a logarithmic scale because the active family totals span several orders of magnitude.

figure(2)
bar([statSsCold.';statSsWarm.'])
set(gca,'YScale','log','XTickLabel',{'RT','250 degrees C'})
ylabel('Mean deformation amount per crystal')
legend({'Basal slip','Prismatic slip','Pyramidal slip','Comp. twin'},...
  'Location','eastoutside')
legend('boxoff')

Prismatic and pyramidal slip carry almost all deformation in this example. The warm CRSS ratios favour pyramidal slip over prismatic slip, whereas the order is reversed at room temperature. Compression-twin activity is below \(3 \times 10^{-10}\) per crystal, nine to ten orders of magnitude under prismatic and pyramidal slip. Because the imposed total strains differ, compare the family ranking within a row rather than absolute bar heights between temperatures.

References

Next

Lankford uses Taylor factors to predict the plastic strain ratio of a sheet as its loading direction changes. It turns the same orientation-dependent deformation model into a measure of sheet anisotropy.