Designing textured polycrystals 165
For brevity, let us simply write R for R
1
. Then aggregate
✁
1
is of type D
2
R O. If we
write R ψ
, θ , φ
= α, β, α, then R
3
ψ , θ
, φ = π2 − α, β, π2 − α. Construct
an aggregate e
✁
by rearranging the grains in
✁
1
so that R is replaced by R
T
, which has Euler angles π
− α, β, π − α and is equivalent to π − α, β, π2 − α for a D
2
R
T
O aggregate. If we rotate e
✁
by Re
3
, −π2, we obtain aggregate
✁
3
because Re
3
, −π2D
2
= D
2
Re
3
, −π2.
Hence
✁
3
is of type D
2
R
T
O.
4.2. Grains of other crystal symmetries
In Eq. 29 we obtain an aggregate
✁
1
of type D
2
R O, which has its elasticity tensor
✂
isotropic. From this solution we can construct, for crystallites of any
cr
⊂ , an aggregate with an isotropic
✂
. The method is as follows: Let R
1
be any rotation and
2
be any finite subgroup of which satisfies the crystallographic restriction. If we rotate the aggregate
✁
1
by R
1
, the rotated aggregate
✁
1 R
still has its
✂
isotropic. Now append grains to
✁
1 R
to obtain an aggregate of type
2
R
1
D
2
R O, which is simply an assembly of N
2
the order of
2
rotated copies of
✁
1 R
. Clearly the new assembly has an isotropic
✂
. By Remark 2, we conclude that the aggregate of type O R
T
D
2
R
T 1
2
, which consists of 24 × 4 = 96 grains with
cr
=
2
, also has an isotropic
✂
. In other words, for crystallites with its
cr
being a finite rotation group, including triclinic crystallites with
cr
= C
1
, we can always design an aggregate with 96 identical grains which has an isotropic elasticity tensor.
The appearance of an arbitrary rotation R
1
in the preceding scheme suggests that this recipe generally will not lead to a solution with the least possible number of grains. Indeed for many
crystal symmetries we can achieve our goal using less grains. Let us now present one other solution for each
cr
⊂ other than C
1
.
cr
= D
2
, D
4
, D
6
By Remark 2, O R
T
D
2
is a solution with 24 orthorhombic grains. Moreover, if
1
con- tains D
2
as a subgroup, then the 24-grain aggregate of type O R
T 1
also has an isotropic
✂
. Indeed, let q
= |
1
||D
2
| and 34
1
=
q
[
i=1
g
i
D
2
, disjoint union
where {g
i
: i = 1, ..., q} is a set of left coset representatives of D
2
in
1
. An aggregate of type
1
R O can be taken as a “super-aggregate” of q rotated copies of the aggregate of type
D
2
R O, where g
i
i = 1, ..., q describe the rotations in question. Since each rotated copy has
an isotropic
✂
, so does the super-aggregate. It follows from Remark 2 that an aggregate of type O R
T 1
also has an isotropic
✂
. The same argument in fact proves a general assertion, which we put as the next remark.
R
EMARK
3. Let
a
and
b
be point groups such that
a
⊂
b
⊂ = SO3. If an
aggregate of type
p
R
p−1 p−1
...
1
R
0 a
has its material tensors
1
, ...,
p
isotropic, so does an aggregate of type
p
R
p−1 p−1
...
1
R
0 b
.
166 C.-S. Man - L. Noble
By the preceding remark, cubic aggregates of 24 tetragonal and hexagonal crystallites which are of type O R
T
D
4
and O R
T
D
6
, respectively, have their elasticity tensor isotropic; here we take rotations Re
2
, π and Re
3
, π 2 as the two generators of group D
4
and rotations Re
2
, π and
Re
3
, π 3 as the two generators of group D
6
.
cr
= C
2
, C
4
, C
6
Let C
1 2
= {I , Re
3
, π } and C
2 2
= {I , Re
2
, π }, where I is the identity in . The
solution of type D
2
R O can be looked upon as of type C
1 2
I C
2 2
R O. By Remark 2, we
obtain a solution of type O R
T
C
2 2
I C
1 2
, which consists of 24 × 2 = 48 C
2
-grains of equal volume.
Let Re
3
, π 2 and Re
3
, π 3 be the generator of group C
4
and C
6
, respectively. Since both C
4
and C
6
include C
1 2
as a subgroup, by Remark 3 we conclude that aggregates of type O R
T
C
2 2
I C
4
and O R
T
C
2 2
I C
6
are also solutions. These aggregates are made up of 48 C
4
- and C
6
-grains, respectively.
cr
= C
3
First we present a solution of hexagonal grains which exhibits C
3
texture symmetry. To start with, we arrange an aggregate of 8 identical hexagonal grains so that it has tetragonal texture
symmetry i.e.
1
= D
4
, where Re
2
, π and Re
3
, π 2 are taken as generators. Then,
by determining the independent coefficients for l = 4 and solving the resulting equations with
texture coefficients of form shown in Eq. 14, we find that the orientation 35
R ψ
, θ , φ
= 0.39269908, 1.22389959, 0 generates an aggregate of type D
4
R D
6
which has all its c
4 mn
coefficients zero. By placing three copies of this aggregate in such a way that the super-aggregate has C
3
texture i.e.
2
= C
3
, with Re
3
, 2π3 as generator, we are able to determine that among
the c
2 mn
coefficients of the super-aggregate only the coefficient c
2 00
is independent, and c
2 00
= 0 renders all c
2 mn
coefficients zero. Moreover, we find that R
1
ψ
1
, θ
1
, φ
1
= 0, 0.95531662, 0 is a solution of c
2 00
= 0, where the texture coefficient is of form Eq. 16 with R given by
Eq. 35. Thus we obtain an aggregate of type C
3
R
1
D
4
R D
6
, which has an isotropic elasticity tensor
✂
. By Remark 2, aggregates of type D
6
R
T
D
4
R
T 1
C
3
, which consist of 12 × 8 = 96 C
3
-grains of equal volume, have their elasticity tensor isotropic.
cr
= D
3
We found an arrangement of 24 D
3
-grains, for which the elasticity tensor
✂
of the aggregate is isotropic. The arrangement is of type O R
D
3
, where R
ψ , θ
, φ = 0.55357436, π2, 0.
Designing textured polycrystals 167
cr
= T In paper [1] Bertram et al. have presented a solution of type T R
O, where 36
R ψ
, θ , φ
= 0.24002358, 2.67480609, 2.90156907. Hence there is a solution of type O R
T
T with 24 tetrahedral grains. In fact, R
T
= R in this
case. In summary, we have presented at least one solution for each
cr
which is a finite rotation group that satisfies the crystallographic restriction. For each
cr
, our best solution at present where using a smaller number of grains means better requires 4 grains for
cr
= O; 24 grains for
cr
= D
2
, D
3
, D
4
, D
6
, or T ; 48 grains for
cr
= C
2
, C
4
, or C
6
; 96 grains for
cr
= C
1
, or C
3
. Except for the case of cubic grains, where a proof was given by Bertram et al. [1], it remains unclear whether the solution we presented would be a minimal solution, i.e., one with the least
possible number of identical grains for the
cr
in question. In fact, we believe that many of our present “best solutions” can be improved upon.
5. Example: acoustoelastic tensor
