How to analyze slip transmission at grain boundaries

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Burgers vector and normal directions |

Rotating slip systems |

Consider hexagonal symmetry of alpha-Titanium

cs = crystalSymmetry('622',[3,3,4.7],'x||a','mineral','Titanium (Alpha)')

cs = crystalSymmetry mineral : Titanium (Alpha) symmetry : 622 a, b, c : 3, 3, 4.7 reference frame: X||a, Y||b*, Z||c

Then basal slip is defined by the Burgers vector (or slip direction)

`b = Miller(2,-1,-1,0,cs,'UVTW')`

b = Miller size: 1 x 1 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c) U 2 V -1 T -1 W 0

and the slip plane normal

`n = Miller(0,1,-1,0,cs,'HKIL')`

n = Miller size: 1 x 1 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c) h 0 k 1 i -1 l 0

Accordingly we can define a slip system in MTEX by

sSBasal = slipSystem(b,n)

sSBasal = slipSystem mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c) CRSS: 1 size: 1 x 1 U V T W | H K I L 2 -1 -1 0 0 1 -1 0

The most important slip systems for cubic, hexagonal and trigonal crystal lattices are already implemented into MTEX. Those can be accessed by

sSBasal = slipSystem.basal(cs)

sSBasal = slipSystem mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c) CRSS: 1 size: 1 x 1 U V T W | H K I L 1 1 -2 0 0 0 0 1

Obviously, this is not the only basal slip system in hexagonal lattices. There are also symmetrically equivalent ones, which can be computed by

`sSBasalSym = sSBasal.symmetrise('antipodal')`

sSBasalSym = slipSystem mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c) CRSS: 1 size: 3 x 1 U V T W | H K I L 1 1 -2 0 0 0 0 1 1 -2 1 0 0 0 0 1 -2 1 1 0 0 0 0 1

The length of the burgers vector, i.e., the amount of displacment is

sSBasalSym.b.norm

ans = 3.0000 3.0000 3.0000

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