Grain Boundary character: 5parameter descriptions

1
Grain Boundary character 5-parameter
descriptions

• Advanced Characterization Microstructural
  Analysis
• 27-750, A.D. Rollett, 2007

2
Motivation

• Why do we need 5 parameters to describe the
  macroscopic character of a grain boundary
  (neglecting the microscopic parameters that
  account for the atomic structure)?
• Reason once you have described the difference in
  lattice orientation, which is equivalent to a
  proper rotation with 3 parameters, there are two
  more parameters that are required to describe a
  unit vector that locates the normal to the
  boundary. This normal can be described in the
  sample frame, or either of the two crystal frames.

3
Notation

• How do we write down a mathematical description
  of a 5-parameter grain boundary? Answer there
  are 3 useful methods.
• Disorientation plane
• Plane of 1st grain plane of 2nd grain twist
  angle
• Matrix (4 x 4)

4
Tilt versus Twist

• Definitions a tilt boundary is one in which the
  rotation axis lies in the boundary plane.
• A twist boundary is one in which the rotation
  axis is perpendicular to the boundary plane.
• Example a coherent twin boundary (in fcc) is a
  pure twist boundary, 60 lt111gt.

5
Properties

• Relevance? For low angle boundaries, twist
  boundaries have screw dislocation structures
  tilts have edge dislocations.
• The properties of twist and tilt boundaries are
  sometimes different evidence in fcc metals
  suggests that only 40 lt111gt tilt boundaries are
  highly mobile, but not the twist 40 lt111gt type
  (in recrystallization).

6
Orientations vs. Boundaries

• A useful approach to dealing with parameterizing
  grain boundaries is to return to the idea of
  orientations specified by a direction and normal,
  using Miller indices.
• Write down the indices for the normal of the
  boundary on each side and the direction that lies
  in the boundary plane that is parallel to a
  corresponding direction on the other side.
• The reference frame is then the same for the two
  sides of the boundaries.

7
Miller Index Definition of a Crystal Orientation

• Use a set of three orthogonal directions as the
  reference frame. Mathematicians set up a set of
  unit vectors called e1, e2 and e3.
• In many cases we use the names Rolling Direction
  (RD) // e1, Transverse Direction (TD) // e2, and
  Normal Direction (ND) // e3.
• We then identify a crystal (or plane normal)
  parallel to 3rd axis (ND) and a crystal direction
  parallel to the 1st axis (RD), written as
  (hkl)uvw.

8
Geometry of hklltuvwgt
Transformation of Axes fromSample to Crystal
(primed)


e3
Miller indexnotation oftexture
componentspecifies directioncosines of
crystaldirections tosample axes.
e3 (hkl)
001
010

e2


e2 t

e1 uvw

e1
t hkl x uvw
100
9
Form matrix from Miller Indices
The matrix g represents a transformation from
Sample to Crystal axes.
10
Example (110)112
(110)

• This happens to be the specification for the
  Brass texture component. They key here is to
  note the specification of a reference coordinate
  frame.

11
Two Crystals

• Example of (110)001 with (-3-2-1)-103 -
  normalized to unit vectors
• Note how we have to be careful to specify the
  negative of the face normal for the second
  crystal (outward-pointing normals) so that, when
  the crystals are joined at a boundary, the
  reference axes coincide

12
Two Crystals Joined

• Imagine that we have flipped the second crystal
  over and laid it on top of the first one to make
  the grain boundary
• Then note the coincidence in alignment of the
  axes
• Remember that the misorientation axis is not
  necessarily related in any special way to the
  crystal directions identified here.
• The twist axis, however, must be parallel (or
  anti-parallel) to the boundary normal and the
  tilt axis must lie somewhere in the plane of the
  boundary

13
Output from convert_hkl2rfh

• Enter RF-vector 0 or planesdirections 1?
• (Any other value will stop the program)
• 1
• Now we enter the specification on the A side of
  the boundary integer or real values can be
  input. Spaces are required between the values,
  as h k l u v w.
• Enter GB normaldirection (in plane) for xtal A
• 1 1 0 0 0 1
• The program shows you what it calculated for
  Bunge Euler angles, quaternion and orientation
  matrix.
• HKLUVW2QUAT Bunge angles 90. 90. 45.
• HKLUVW2QUAT Quat 0.653281152 0.270597667
  0.653281629 0.270598829
• HKLUWV2QUAT Orientation matrix
• 0.000 0.707 0.707
• 0.000 -0.707 0.707
• 1.000 0.000 0.000
• Now we enter the specification on the B side of
  the boundary integer or real values can be
  input. Spaces are required between the values,
  as h k l u v w.
• Enter GB normaldirection (in plane) for xtal B
• 3 2 1 -1 0 3
• HKLUVW2QUAT Bunge angles 100.102615
  74.498642 56.3099327
• HKLUVW2QUAT Quat 0.561619043 0.225727558
  0.779205382 0.162696317
• HKLUWV2QUAT Orientation matrix

• The program displays the misorientation matrix
  that effects transformation of axes from A to B
  (to go from B to A, use the transpose of this
  matrix)
• Misorientation matrix from A to B
• 0.926 0.208 -0.316
• -0.220 0.976 0.000
• 0.309 0.069 0.949
• Angle 22.3484268
• misorientation axis based on matrix -0.091
  0.822 0.563
• misorientation axis based on quat -0.091
  0.822 0.563
• The program also writes out the misorientation
  angle and the rotation axis derived from both the
  matrix and the (equivalent) quaternion
  description of the misorientation.
• Then it takes the uvw in A (normalized to a
  unit vector) and applies the misorientation to
  this to check that we obtain the uvw in B,
  again performed with both the matrix and the
  quaternion descriptions.
• Direction in A 0.000 0.000 1.000
• Rotated Direction in B -0.316 0.000
  0.949
• Compare with original Direction in B -0.316
  0.000 0.949
• Rotated Direction in B from matrix -0.316
  0.000 0.949
• Similarly for the normals it takes the (hkl) in
  A (normalized to a unit vector) and applies the
  misorientation to this to check that we obtain
  the (hkl) in B, again performed with both the
  matrix and the quaternion descriptions.
• normal in A 0.707 0.707 0.000
• Rotated normal in B 0.802 0.535 0.267
• Compare with original normal in B 0.802
  0.535 0.267

14
Fundamental Zone

• What is the smallest possible range of parameters
  needed to describe a 5-parameter grain boundary?
• Disorientation based use the same FZ for
  disorientation hemisphere for the plane
• Boundary plane based use SST for the first plane
  double-SST for the second plane (0,2p for
  the twist angle
• Matrix no FZ established.

15
Disorientation based RF-space
Disorientations can be described in the
Rodrigues-Frank spacewhose FZ for cubic
materials is a truncated pyramid. The boundary
plane requires a full hemisphere in order to be
described.
Misorientation axis, e.g. 111
16
Disorientation based boundary plane based
2nd plane normal requires double SST
1st plane normal requires single SST

f 0
f 2p

17
Matrix description

• Construct the matrix from the 3x3 orthogonal
  rotation matrix that describes the misorientation
  and add a columnrow to describe the plane
  normals of each side of the boundary Morawiec,
  A. (1998). Symmetries of the grain boundary
  distributions. Int. Conf. on Grain Growth,
  Pittsburgh, PA, TMS. The determinant of the
  resulting matrix, B, is 1, as it is for the
  misorientation matrix, ?g. Additionally, 4x4
  symmetry matrices can be constructed from the
  conventional 3x3 matrices.

18
The Number of Grain Boundary Types
19
Measurement

• We can measure the orientation of the grains on
  either side of a boundary as gA and gB, as well
  as the boundary normal, n.
• Compare the misorientation axis in specimen
  axes with the boundary normal by forming the
  scalar product.
• If b0, we have a tilt boundary if b/-1, then
  it is a pure twist boundary.

20
Tilt-twist character

• If cos-1(b)0, boundary is pure twist if
  cos-1(b)90, boundary is pure tilt.

n
b
21
Ambiguities

• Caution! Although the misorientation axis is
  unambiguous for low angle boundaries, it is not
  necessarily so for high angle boundaries thanks
  to the crystal symmetry.
• Finding the disorientation and locating the
  misorientation axis in the SST, makes the choice
  unambiguous, but other selections may be
  physically reasonable.

22
Grain Boundary Plane

• The planes that make up a boundary are readily
  identified.
• Transform the boundary normal to crystal axes in
  either crystal.
• We can specify the boundary completely by
  identifying indices for both sides of the
  boundary and a twist angle between the lattices.

23
Example of G.B. plane

• If we measure the boundary normal as the vector
  (1,0,0), and the orientation of the grain on the
  A side is given by

ns
B
A
x1
Then the boundary normal in crystal(A)
coordinates is gAnS 1/v3(1,1,1)reasonable
gA112lt111gt!
x2
24
GB planes, disorientation

• Calculation of the disorientation with particular
  choices of symmetry operators affects the
  location of the rotation axis.

25
GB planes, disorientation, contd.

• Must re-calculate the planes and tilt/twist
  characterbut, we can calculate the
  tilt/twist character in the sample axes, if we
  choose.

If cos-1(b)0, boundary is pure twist if
cos-1(b)90, boundary is pure tilt.
26
(hkl)1(hkl)2Twist

• A standard representation of a grain boundary
  that is particularly useful for symmetric tilt
  boundaries is the (hkl)1(hkl)2Twist definition.
  This specifies the two planes the comprise the
  boundary, and the twist angle between the two
  lattices. The difference between (hkl)1 and
  (hkl)2 defines the tilt angle between the
  lattices.

27
Example(s) of GB plane

• Reminder full description of g.b. type requires
  plane as well as misorientation.
• Example strong lt111gt fiber leads to boundaries
  that are pure tilt boundaries with lt111gt
  misorientation axes.
• Similarly, strong lt100gt fiber leads to pure lt100gt
  tilt boundaries.
• Thus, in a drawn wire (e.g. fcc metal) with a
  mixture of lt111gt and lt100gt fiber components, the
  grain boundaries within each component will be
  predominantly lt111gt and lt100gt tilt boundaries,
  respectively.

28
Microstructure

• Euler angles1. (10,55,45)2. (87,55,45)3.
  (32,55,45)4. (54,55,45)...

q1213q2355 q3422 etc.
2
3
1
4
2
3
cube- on- corner
1
29
Grain Boundary descriptions
2
Specimen axes- misorientation axes (0,0,1)-
g.b. normals (x,y,0)/v(x2y2)Crystal Axes -
misorientation axes 1/v3(1,1,1)- g.b. normals ?
1/v3(1,1,1), i.e., in the zone
110-112Boundary Planes are limited tothe
zone of (111).Misorientations include
S3,7,13b,19b.
3
1
2
3
_
_
1
4
30
lt111gt Fiber RF-space
Misorientations lie on R1R2R3 line
Boundary planes lie on 111 zoneand are pure tilt
lt111gt boundaries
Misorientation axis
31
Viewing Five Parameter Grain Boundary
Distributions
(MRD)
Planes for all boundaries, 40 rotation around
lt111gt
View point by point
(MRD)
Planes for all boundaries
Average over Dg
Average over n (conventional MDF)
or
32
Definitions
(hkl)1 (hkl)2
Twist angle
33
Experimental Information

• In a typical experiment, we measure (a) the
  orientations of the two grains adjacent to a
  boundary, gA and gB. In addition, we measure the
  boundary normal, ns, in specimen axes (outward
  pointing with respect to, e.g., grain A).
• Objective is to provide a unique, 5-parameter
  description, based on the experimental
  information.

34
Boundary Normal

• Define the normal to a boundary plane as the
  outward pointing normal based on the grain to
  which the normal is referred.

ns(A)
B
Example n(A) (1,0,0) n(B) (-1,0,0)
A
ns(B)
x1
x2
graduate
35
Locate Plane Normal in SST

• The interface-plane method seeks to emphasize the
  crystallographic surfaces that are joined at the
  boundary. The obvious choice of fundamental zone
  would appear to be the 001-101-111 unit triangle
  for each plane as will become apparent, however,
  a combination is needed of a single unit triangle
  for one plane and a double unit triangle,
  001-101-111-011 for the plane on the other side
  of the boundary, combined with the fifth (twist)
  angle.

graduate
36
Equivalent Descriptions of a 5-parameter
Boundaryswitching symmetries
graduate
37
Equivalencies

• Geometry the misorientation carries the boundary
  normal from one crystal into the other ?gAB
  gBgAT nB ?gAB nA.
• Rule 1 if you apply symmetry to one orientation
  (crystal), then you must also apply it to the
  boundary plane. Thus if we have some property,
  f, of a grain boundary, f(?g,nA) f(?gOAT,
  OAnA) f(OB?g, nA).
• Rule 2 centrosymmetry has this effect f(?g,nA)
  f(?g,-nA).
• Rule 3 switching symmetry applies f(?g,nA)
  f(?gT,-?gTnA).

38
Locate Normals in SST

• Apply symmetry operators to locate the first
  boundary normal in the SST, possibly (switching
  symmetry) relabeling A as B and vice versa.

Then repeat the process for the second plane
normal, except this time, the result falls in a
double triangle
39
Unit triangles for plane normals
1st triangle
2nd triangle
40
Coincidence of triangles?
Look down the misorientationaxis at the grain
boundary planeof a twist boundary
Case A coincidence not possible by rotation
(hkl)
Case B coincidence is possible by rotation
Conclusion it is possible to have the same
rotation axis in the two crystals and yet for
there to be no twist angle that yields zero
misorientation.
41
Demonstration

• To demonstrate the difference between (hkl)(hkl)
  and (hkl)(khl), we can draw the crystal axes.
  Clearly in case 1, the axes can be made to
  coincide at some twist angle (? no boundary). In
  case 2, however, this cannot be accomplished at
  any twist angle.

nA hkl
nA hkl
- - -
nB hkl
nB hkl
2 (hkl)(khl)
1 (hkl)(hkl)
42
Examples planes inside the SST
43
Example dissimilar planes inside the SST
44
Rodrigues space
(123)(213)
(123)(123)
The twist angle is notated at each plotted point.
Note how the misorientation axis varies between
the two sequences of grain boundaries.
45
Special Cases

• The exception to the above analysis occurs when
  one of the planes lies on a mirror plane, i.e. in
  the zone of either lt100gt or lt110gt. That is to
  say, there is always a zero-misorientation twist
  angle when you combine (hhl) or (h00) with
  itself.
• Why? The mirror plane means that (hkl) and (khl)
  are equivalent.
• A more subtle point is that (hhl)(mno) is
  equivalent to (hhl)(nmo) except for a reversal
  in the sense of twist. Specifically,
  (hhl)(mno)(f) ? (hhl)(nmo)(-f).
• How to demonstrate this? Calculate the full
  disorientationplane.

46
Special Cases

• No boundary (hkl)1 -1(hkl)2 twist0.
• Pure Twist (hkl)1 -1(hkl)2 twist ? 0.
• Symmetrical Tilt (h,k,0)1 -1(-h,k,0)2
  twist0.
• Asymmetric tilt (hkl)1 ? -1(hkl)2 twist0.
• Note only proper rotations O(432) permitted as
  symmetry operators to permute the indices of the
  planes.

47
More Special Cases

• If (hkl)1 (hkl)2 twist0, then we have a
  twin relationship that is equivalent to (hkl)1
  -1(hkl)2 twist180. Note that matching (hkl)
  does not necessarily allow a zero-boundary
  condition to be found.

48
Conversions

• It is useful to be able to convert from one type
  of description to another.
• First we describe conversion from ?gn to
  (hkl)A(hkl)BTwist.
• Then we describe the inverse process to convert
  from (hkl)A(hkl)BTwist to ?gn.
• See Takashima, M., A. D. Rollett and P.
  Wynblatt (2000). "A representation method for
  grain boundary character." Philosophical Magazine
  A 80 2457-2465..

49
Calculation of Tilt, Twist parameters, given the
grain orientations and boundary plane

• Given the orientations of the two grains we can
  compute the misorientation and decompose it into
  a pair of tilt and a twist rotations

The tilt angle, y, is obtained as (the twist
rotation does not alter the position of the plane
normal)
50
Tilt-Twist Decomposition
Note it is not possible to obtain the twist
angle from this relationship based on a knowledge
of the misorientation and tilt angles because the
inverse trigonometric functions are limited to a
range of 0xp.
RTWIST n
f
q
2(1 cosq) (1 cosf)(1 cosy)
RTILT
y
51
Tilt Rotation

• The tilt axis is obtained from the cross product
  of the normals
• Rotation matrix for the (active) tilt rotation,
  R(rtilt,y)A-gtB

52
Twist Angle

• Form the tilt rotation matrix, and calculate the
  twist rotation matrix from

Then obtain the twist angle thus
Note x and nB may be parallel or anti-parallel,
which is importantbecause the range of the twist
angle is 0f2p.
53
5-parameter conversions nAnBtwist??gn

• Given (hkl)A(hkl)Btwist, with the normals
  defined as outward-pointing normals
• Obtain the tilt angle from rotation required to
  bring the two normals (with inversion of nB) into
  coincidence
• Vector product gives the rotation axis

54
nAnBtwist??gn, contd.

• Rotation matrix for the (active) tilt rotation,
  R(rtilt,y)A-gtB
• Twist rotation axis nB angle twist angle, f,
  giving R(rtwist,f)A-gtB

55
nAnBtwist??gn, no. 3

• Misorientation (axis transformation) product of
  tilt, twist rotations not a disorientation! ?
  gTAB R(rtwist,f) R(rtilt,y)
• Apply symmetry to identify the disorientation
  (i.e. minimum angle, axis in the SST) ?g(gBOc)
  -1(gAOc) Oc?gABOc-1

56
nAnBtwist??gn, no. 4

• Choose whether to use nA or nB to characterize
  the plane.
• Apply the (same) symmetry operator as used to
  identify the disorientation to the boundary
  plane
• nA,nB,twist lt-gt ?gAB,nA
• Other conversions given elsewhere.

57
Examples

• In this next section, we give some examples of
  graphical representations of various sets of
  grain boundaries.
• One example is a series of symmetrical tilt
  boundaries.
• Another example is a set of experimentally
  measured grain boundaries in Al.
• A third example is a set of boundaries measured
  in an Al foil.

58
Symmetric Tilt Boundaries lt110gt
Symmetric tilt boundaries arebased on the
concept of a tiltaxis lying in the plane of
theboundary and with boundaryplanes
symmetrically disposedabout a low index
(symmetry)plane. Example lt110gt tilt
boundariesare based on rotations abouta 110
axis. It is convenient andaccurate to think
about rotatingeach crystal by equal and
oppositeamounts (through half the specified
misorientation angle).
nA
nB
59
Symmetric Tilt Boundaries lt100gt
Symmetric tilt boundaries arebased on the
concept of a tiltaxis lying in the plane of
theboundary and with boundaryplanes
symmetrically disposedabout a low index
(symmetry)plane. Example lt100gt tilt
boundariesare based on rotations abouta 100
axis. It is convenient andaccurate to think
about rotatingeach crystal by equal and
oppositeamounts (through half the specified
misorientation angle).
nA
nB
60
lt110gt symmetric tilts disorientations
Plot of the misorientations for a series of
symmetric tilt grain boundaries based on
rotations about lt110gt. For tilt angles between 0
and 60, and between 120 and 180, the
misorientations lie on the lt110gt line in the
first section (R30). For tilt angles between
60 and 120, the misorientations follow a
complex path through Rodrigues-Frank space that
includes the S3 and S17 positions.
61
lt110gt symmetric tilts (hkl)twist
nA (111)
Boundary plane normals for a series of lt110gt
symmetric tilt boundaries. The normal for the B
crystal is plotted in a quadrant which is a
double unit triangle. The range of normals for
the A crystal is delineated by a box in each
triangle.
nA (011)
nA (001)
62
lt110gt symmetric tilts ?gplane
near twin (S3)
Boundary plane normals for the same set of
symmetric tilt boundaries as shown in the
previous figure. Poles on the upper hemisphere
are plotted as solid points and poles on the
lower hemisphere as gray points. The
misorientation axis, indicated by a circle on the
projected sphere, has been placed in a single
unit triangle, and is also plotted in a Rodrigues
space triangle, as in fig. 2.
lt110gt axes
low angle boundaries
63
Example of Annealed Al bulk spec.interface
plane description
64
Example of Annealed Al (bulk)disorientation
plane description
lt110gt axes
low angle boundaries
65
Example of Annealed Al (foil)disorientation
plane description
Strong 001lt100gtcube texture presentin the
foil.
66
Boundary Plane Analysis from Single Plane Section

• At first sight, it is not possible to measure the
  full 5-parameter nature of a boundary from a
  single section plane. Serial sectioning is
  required in order to accomplish the
  characterization.
• Statistical stereology can mitigate this problem
  we perform measurement of boundary tangents,
  coupled with analysis in the crystal boundary
  plane space.

67
Boundary Tangents

• Measure the (local) boundary tangent the normal
  must lie in its zone.

gB
ns(A)
B
ts(A)
A
x1
gA
x2
68
G.B. tangent disorientation

• Select the pair of symmetry operators that
  identifies the disorientation, i.e. minimum angle
  and the axis in the SST.

69
Tangent ? Boundary space

• Next we apply the same symmetry operator to the
  tangent so that we can plot it on the same axes
  as the disorientation axis.
• We transform the zone of the tangent into a great
  circle.

Boundary planes lie on zone of the boundary
tangent in this examplethe tangent happens to
be coincident withthe disorientation axis.
Disorientation axis
70
Tangent Zone

• The tangent transforms thus tA OAgAtS(A)
• This puts the tangent into the boundary plane (A)
  space.
• To be able to plot the great circle that
  represent its great circle, consider spherical
  angles for the tangent, ct,ft, and for the zone
  (on which the normal must lie), cn,fn.

71
Spherical angles
chi declination phi azimuth
f
c
72
Tangent Zone, parameterized

• The scalar product of the (unit) vectors
  representing the tangent and its zone must be
  zero

73
Integration

• In order to obtain a distribution of the grain
  boundary character, an integration must be
  performed in the space of the grain boundary
  plane.
• Each boundary (segment) contributes uniform
  intensity over the length of its associated
  tangent zone.
• Wherever many of the tangent zones intersect,
  there must be a high frequency of that boundary
  type.

74
Example coherent twins

• If there is a high density of coherent twin
  boundaries, these correspond to the pure twist
  boundary on 60lt111gt.
• Such a condition will be evident in many tangent
  zones for 60lt111gtdisorientation types
  crossing at the location of the disorientation
  axis.

75
Another Illustration
The probability that the correct plane is in the
zone is 1. The probability that all planes are
sampled is lt 1.
Poles of possible planes
The grain boundary surface trace is the zone axis
of the possible boundary planes.
Trace pole
76
Recovering the Distribution from Planar Sections
The correct planes are observed more frequently
than the incorrect planes
3.0
add 2
add 1
2.0
1.0
Subtract background
add 3
77
Illustration of Stereology Approach
Randomly sample two types of planes. Movie
shows evolution of the distribution as
observations are accumulated.
78
Summary

• In general, five parameters needed to describe
  crystallographic grain boundary character (the
  macroscopic degrees of freedom).
• Tilt versus twist character can be quantified by
  the relationship between the boundary normal and
  the disorientation axis.
• Two standard parameterizations for 5-parameter
  boundaries are (a) disorientationplane (b)
  plane in A plane in B twist angle.
• Conversion from one to the other parameterization
  requires considerable care.
• Statistical stereology offers a way to obtain the
  full 5-parameter grain boundary character
  distribution and avoid serial sectioning.