Kroner

Derivation and role of Kroner’s
cubic in the calculation of shear
moduli of an aggregate of cubic
crystals
Armand Attia
April 12, 2019
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Contents
1 Abstract 3
2 Introduction 4
3 Properties of fourth-order tensors 4
4 Equilibrium 6
5 Integral equation for strains 6
6 Role of Eshelby’s tensor in integral equation for strains 10
7 Symmetry properties and model of tensor D 11
8 Derivation of tensor A for isotropic matrix and isotropic inhomogeneity
12
9 Self consistent scheme for cubic polycrystals 13
10 Elastic moduli of a cubic polycrystal 15
11 Rotational averages 17
12 Derivation of tensor A for anisotropic inhomogeneity and Kroner’s
cubic 19
13 Comparison with experimental data 22
14 Major symmetries of tensor D 25

15 Iterative Newton’s method solution of Kroner’s cubic 28

1 Abstract
The pioneering work of Hershey [1], Kroner [2], and Eshelby [3] has laid the
foundations for modeling polycrystals. The focus here is on Kroner’s cubic
which provides a remarkable estimate of the shear modulus of an aggregate of
cubic crystals, arranged in such random orientations that a rotational average
provides a good isotropic estimate. A detailed derivation of Kroner’s cubic is
provided beginning with the Green tensor calculation by Weinberger [4], leading
to an integral equation for strains by Gubernatis [5], from which Markov [6]
derives Kroner’s cubic. Comparison with experimental data shows good agreement
except for the Carbon family. Markov’s approach opens the potential for
modeling non-cubic crystals.

2 Introduction
This is a detailed review of the derivation of Kroner’s cubic [2], which is used
to calculate the macroscopic shear modulus of an aggregate of cubic crystals.
The macroscopic modulus is assumed to apply to an isotropic ensemble of an
aggregate of cubic crystals in perfectly randomized orientations.
The first major part of this derivation is based on the treatment of equilibrium
by Gubernatis [5] and the calculation of the Green tensor function as
derived by Weinberger et al [4], all leading to an integral equation for strains
by Gubernatis [5].
The second major part is based on Markov’s [6] treatment for deriving Kroner’s
cubic from this integral equation. The third major part compares measured
shear moduli with the values predicted by Kroner’s cubic with Hashin’s bounds.
But first, a few words about fourth-order tensors.
3 Properties of fourth-order tensors
Fourth-order tensors are designated with such script (C), while second-order
tensors are designated without such script (C). Markov [6] defines the composition
of two fourth-order tensors C : B as:
(C : B) ijkl = C ijrs B srkl (1)
For a fourth-order tensor B with only minor symmetry:
the identity tensor I
B ijkl = B jikl (2)
B ijkl = B jilk (3)
has the property:
Since:
I ijkl = 1 2 (δ ikδ jl + δ il δ jk ) (4)
B : I = I : B = B (5)
B ijrs I srkl = 1 2 B ijrs(δ sk δ rl + δ sl δ rk ) = 1 2 (B ijlk + B ijkl ) = B ijkl (6)
I ijrs B srkl = 1 2 (δ irδ js + δ is δ jr )B srkl = 1 2 (B jilk + B ijkl ) = B ijkl (7)
Consider isotropic fourth-order tensors B with both major and minor symmetries:
B ijkl = B jikl = B ijlk = B klij (8)

Such a tensor can be decomposed as a linear combination of the following
orthogonal set of fourth-order tensors I ′ and I ′′
where
with the properties:
so that:
B = b ′ I ′ + b ′′ I ′′ (9)
I ′ ijkl = 1 3 δ ijδ kl (10)
I” ijkl = I ijkl − I ′ ijkl (11)
I ′ : I ′ = I ′ (12)
I ′′ : I ′′ = I ′′ (13)
I ′ : I ′′ = I ′ : (I − I ′ ) = I ′ : I − I ′ : I ′ = 0 (14)
Note, that because of the above properties of I ′ and I ′′ , we find that:
B : I ′ = b ′ I ′ (15)
B : I ′′ = b ′′ I ′′ (16)
so that it is clear that b ′ and b ′′ are eigenvalues of B with corresponding eigentensors
I ′ and I ′′ and that with respect to these eigentensors, B is in diagonal
form. Therefore, its inverse is a tensor A given in terms of the reciprocals of
the eigenvalues (if they do not vanish):
A = B −1 = 1 b ′ I′ + 1
b ′′ I′′ (17)
The scalar product of two fourth-order tensors is defined as:
A :: B = A ijkl B ijkl (18)
Notice that while the tensors I ′ and I ′′ are orthogonal, only I ′ is normalized,
i.e.:
I ′ :: I ′ = 1 (19)
I ′′ :: I ′′ = 5 (20)
Therefore, given an isotropic fourth-order tensor B with major and minor symmetries,
if we write
Then to solve for b ′ and b ′′ , we have:
since I ′ is normalized, but:
B = b ′ I ′ + b ′′ I ′′ (21)
b ′ = B :: I ′ (22)
since I ′′ is not normalized.
b ′′ =
B :: I′′
I ′′ :: I ′′ (23)

4 Equilibrium
Consider a linear elastic solid. At equilibrium with no body forces, the stress
satisfies:
∂σ ij (x)
∂x j
= ∂
∂x j
(C ijkl (x)ɛ kl (x)) = 0 (24)
where the strain is related to the displacement via:
and following Gubernatis [5]:
(25)
ɛ kl = 1 2 (u k,l + u l,k ) (26)
C(x) = C 0 + [C](x) (27)
[C](x) = C 2 (x) − C 1 (x) (28)
in which the elastic modulus C 0 is constant, and the elastic moduli C 1 (x) and
C 2 (x) may vary over an infinite region of volume V but differ only over an
inclusion of finite volume W . C 1 is associated with the ”matrix” region, and
C 2 is associated with an inhomogeneity in the inclusion. We will focus on a
spherical inclusion and on constant moduli C 1 and C 2 .
Therefore, the equilibrium requirement Eq. 24 may be written as:
−C 0 klmnɛ mn,l = ([C] klmn ɛ mn ) ,l
(29)
5 Integral equation for strains
This concerns the determination of displacement fields in a solid matrix (medium
1) containing an inhomogeneity (medium 2) whose elastic properties differ from
those of the surrounding matrix. The goal is to relate the strain in the inhomogeneity
to the known strain when there is no inhomogeneity. Only the
matrix is assumed to be isotropic. Since the equilibrium equation is based on
linear elasticity, Green’s (tensor) function can be used to construct the general
solution for the displacement. The construction of the Green tensor function
follows Weinberger et al [4]. The Green tensor function G mr (x, x ′ ) is defined as
the displacement at x in the m th direction in response to a concentrated unit
force applied in the r th direction at x ′ . In an infinite homogeneous body, the
Green’s function depends only on the relative displacement between the points
because the underlying differential operator here has constant coefficients and
thus is invariant under translation (See Stakgold [7] p. 49). Therefore, Green’s
function may be written as:
G mr (x, x ′ ) = G mr (x − x ′ ) (30)

For a constant point force F acting at x ′ , the displacement at x is given by:
u m (x) = G mr (x − x ′ )F r (x ′ ) (31)
So that the displacement gradient and the strain take the form:
u m,n = G mr,n (x − x ′ )F r (32)
ɛ mn = 1 2 (u m,n + u n,m ) (33)
and from the constitutive relation, the stress is given by:
σ kl = C 0 klmnɛ mn (34)
because of the symmetry of the linear elastic coefficients:
σ kl = C 0 klmnu m,n (35)
σ kl = C 0 klmnG mr,n (x − x ′ )F r (36)
Now, following Weinberger et al [4], for a given volume V , from the definition
of the Dirac delta function, the k component of the force F acting at x ′ can be
described by:
∫
F k δ(x − x ′ )dV (37)
V
Equilibrium requires that the force F be balanced by tractions acting on the
surface S enclosing this volume V , so that:
∫
∫
F k δ(x − x ′ )dV + σ kl n l dS = 0 (38)
V
S
∫
∫
F k δ(x − x ′ )dV + CklmnG 0 mr,n (x − x ′ )n l F r dS = 0 (39)
V
S
Applying Gauss’s theorem:
∫
∫
F k δ(x − x ′ )dV + CklmnG 0 mr,nl (x − x ′ )F r dV = 0 (40)
V
V
∫
[F r δ kr δ(x − x ′ ) + CklmnG 0 mr,nl (x − x ′ )F r ]dV = 0 (41)
from which:
V
−C 0 klmnG mr,nl (x − x ′ ) = δ kr δ(x − x ′ ) (42)
For an unbounded isotropic matrix, Weinberger et al [4] derive that Green
tensor function for x ′ = 0 in their Eq. (1.79) as:
G mr (x) =
1
(
(3 − 4ν)δ mr + x )
mx r
16πµ(1 − ν)x
x 2
(43)

which clearly shows that G mr vanishes as x = |x| approaches infinity, and where
µ, ν are the shear modulus and Poisson’s ratio of the matrix material. Following
Gubernatis [5], the solution for the displacement is decomposed into:
u i (x) = u 0 i (x) + u 1 i (x) (44)
where u 0 i (x) is any solution of the homogeneous equation:
C 0 klmnɛ mn,l = 0 (45)
and u 1 i (x) satisfies the inhomogeous equation Eq. 29. The solution for the
displacement u 1 i (x) is given by applying Stakgold [7] Eq. (5.100) to Eqs. 29
and 42 to find the displacement response at x due to the equilibrating ”force”
at x ′ given by the right hand side of Eq. 29:
∫
u 1 i (x) = dx ′ G ik (x − x ′ )σkl,l ∗ ′(x′ ) (46)
or,
∫
u 1 i (x) =
σ ∗ kl,l ′(x′ ) =
V
V
∂
∂x ′ l
([C] klmn (x ′ )ɛ mn (x ′ )) (47)
dx ′ G ik (x − x ′ ) ∂
∂x ′ ([C] klmn (x ′ )ɛ mn (x ′ )) (48)
l
so that the total displacement is given by:
∫
u i (x) = u 0 i (x) + dx ′ G ik (x − x ′ ) ∂ (
)
V
∂x ′ [C] klmn (x ′ )ɛ mn (x ′ )
l
(49)
Integrating by parts:
∫
u i (x) = u 0 i (x) + dx ′ ∂
(
)
V ∂x ′ G ik (x − x ′ )[C] klmn (x ′ )ɛ mn (x ′ ) (50)
l
∫
− dx ′ ∂ (
Gik (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (51)
V
∂x ′ l
Applying the divergence theorem and using the fact that the Green function
vanishes on the boundary at infinity:
∫
u i (x) = u 0 i (x) − dx ′ ∂ (
Gik (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (52)
Since:
∂
∂x ′ l
V
∂x ′ l
G ik (x − x ′ ) = − ∂
∂x l
G ik (x − x ′ ) (53)
The displacement solution becomes:
∫
u i (x) = u 0 i (x) + dx ′ ∂ (
Gik (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (54)
∂x l
V

or, defining:
we can write:
∫
u i (x) = u 0 i (x) +
Now calculate the strains.
G ik,l (x − x ′ ) =
V
∂
∂x l
(
Gik (x − x ′ ) ) (55)
dx ′( G ik,l (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (56)
ɛ ij = 1 2 (u i,j + u j,i ) (57)
Since only the Green tensor function depends on x:
∫
u i,j = u 0 i,j + dx ′( G ik,lj (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (58)
V
∫
u j,i = u 0 j,i + dx ′( G jk,li (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (59)
V
∫
ɛ ij (x) = ɛ 0 ij(x) + dx ′ G ijkl (x − x ′ )[C] lkmn (x ′ )ɛ mn (x ′ ) (60)
where:
V
G ijkl (x) = 1 2 (G ik,lj(x) + G jk,li (x)) (61)
Since the moduli differ only over the inclusion W , Eq. 60 reduces to:
∫
ɛ ij (x) = ɛ 0 ij(x) + dx ′ G ijkl (x − x ′ )[C] lkmn (x ′ )ɛ mn (x ′ ) (62)
W
Markov’s [6] treatment of this integral equation begins here. Markov proceeds
as follows. ”If ɛ 0 is constant, then from Eq. 62 the strain within the inhomogeneity
will also be constant provided that the moduli differ but by constant
values and the following tensor field is constant within the inhomogeneity”:
∫
P ijkl (x) = − G ijkl (x − x ′ )dx ′ (63)
So that:
But:
so that:
W
ɛ ij = ɛ 0 ij − P ijkl [C] lkmn ɛ mn (64)
I ijmn ɛ mn = ɛ ij (65)
(I ijmn + P ijkl [C] lkmn )ɛ mn = ɛ 0 ij (66)

Therefore, letting:
we have:
Or, with:
B ijmn = I ijmn + P ijkl [C] lkmn (67)
B ijmn ɛ mn = ɛ 0 ij (68)
B = I + P : [C] (69)
Markov [6] identifies a fourth-order tensor A given by:
A = (I + P : [C]) −1 = B −1 (70)
Finally, Markov [6] connects the constant strain ɛ 0 with the constant strain in
the ellipsoidal cavity:
ɛ = A : ɛ 0 (71)
In the next two sections, it will be shown that the tensor P is constant, which
will allow us to derive the tensor A.
6 Role of Eshelby’s tensor in integral equation
for strains
In order to make further progress from the integral equation above toward Kroner’s
cubic, we need to derive an important connection between the tensor P
and another tensor T related to Eshelby’s tensor, as described below.
For this purpose, we will follow Weinberger et al [4]. They introduce the
Eshelby tensor S (See [3] Eq. (3.3) p. 104) that relates the ”constrained strain
inside the inclusion to the (eigen)strain in the inclusion in response to zero
stress”. The Eshelby tensor can be written as:
Weinberger et al [4] Eq. (2.13) write:
where by letting:
S = T : C (72)
S ijmn = − 1 2 C lkmn(D iklj + D jkli ) (73)
T ijkl = − 1 2 (D iklj + D jkli ) (74)
we can relate T to P, and where, for an ellipsoidal inclusion, Weinberger et al
[4] Eq. (2.16) give:
∫
D ijkl (x) = G ij,kl (x − x ′ )dV (x ′ ) (75)
V 0
From Eqs. 61, 63, 73, 74, and 75, it is clear that:
T = P (76)

7 Symmetry properties and model of tensor D
In order to show that tensor T (and therefore that P) is a constant, we need
to establish some symmetry properties of tensor D. Weinberger et al [4] Eq.
(2.61) calculate the D ijkl tensor when x lies inside the ellipsoidal inclusion as:
D ijkl = − abc
4π
where, from [4] Eq. (1.70):
∫ π ∫ 2π
0
0
(zz) −1
ij z sinΦ
kz l dΘdΦ (77)
β3 (zz) −1
ij = 1 (
δ ij − λ )
1 + µ 1
z i z j
µ 1 λ 1 + 2µ 1
and from [4] Eqs. (2.17), (2.42), (2.43), (2.44):
For a unit sphere, from [4] Eq. (2.46):
D ijkl = − 1
4πµ 1
∫ π
0
(78)
z 1 = sinΦcosΘ (79)
z 2 = sinΦsinΘ (80)
∫ 2π
0
z 3 = cosΦ (81)
(δ ij + 2κ 1 z i z j ) z k z l sinΦdΘdΦ (82)
where, given Lame’s constant λ 1 , shear modulus µ 1 , Poisson’s ratio ν 1 , and bulk
modulus k 1 for the isotropic matrix:
κ 1 = − 1 1
= − 1 λ 1 + µ 1
(83)
4 1 − ν 1 2 λ 1 + 2µ 1
ν 1 = 3k 1 − 2µ 1
2(3k 1 + µ 1 )
κ 1 = (− 1 2 ) (3k 1 + µ 1 )
(3k 1 + 4µ 1 )
From Eq. 82, it is clear that D satisfies the following minor symmetries:
(84)
(85)
D ijkl = D jikl = D ijlk (86)
However, it is also possible to show that D satisfies the major symmetry:
This is shown in Section 14 so that we can write:
D ijkl = − 1 (1 + 2 3µ 1 5 κ 1)δ ij δ kl − 2
15
D ijkl = D klij (87)
or, in terms of the isotropic basis tensors:
D = − 1 µ 1
[
(1 + 2 3 κ 1)I ′ + 4
15 κ 1I ′′ ]
κ 1
µ 1
(δ ik δ jl + δ il δ jk ) (88)
(89)

We now return to the tensor T defined in Eq. 74:
D iklj = γδ ik δ lj + ω(δ il δ kj + δ ij δ kl ) (90)
D jkli = γδ jk δ li + ω(δ jl δ ki + δ ij δ kl ) (91)
T ijkl = − 1 2 (D iklj + D jkli ) (92)
T ijkl = 1 ( 2
3µ 1 5 κ 1δ ij δ kl + 1 2 (1 + 4 )
5 κ 1)(δ ik δ lj + δ il δ kj )
(93)
or, in terms of the isotropic basis tensors:
Using Eq. 85:
T = t ′ I ′ + t ′′ I ′′ (94)
t ′ = 1 + 2κ 1
3µ 1
(95)
t ′′ = 1 + 4 5 κ 1
3µ 1
(96)
t ′ =
1
3k 1 + 4µ 1
(97)
t ′′ = 3
5µ 1
k 1 + 2µ 1
3k 1 + 4µ 1
(98)
Thus, the tensor T and therefore tensor P is a constant.
8 Derivation of tensor A for isotropic matrix
and isotropic inhomogeneity
Since P = T , we have from the above:
P = p 1 I ′ + p 2 I ′′ (99)
p 1 = t ′ (100)
p 2 = t ′′ (101)
For the special case where both the matrix and the inhomogeneity are isotropic,
with subscript ”1” for the isotropic matrix and subscript ”2” for the isotropic
inhomogeneity, the corresponding elastic moduli are written as:
so that:
C 1 = 3k 1 I ′ + 2µ 1 I ′′ (102)
C 2 = 3k 2 I ′ + 2µ 2 I ′′ (103)
P : [C] = 3[k]p 1 I ′ + 2[µ]p 2 I ′′ (104)

where:
[k] = k 2 − k 1 (105)
[µ] = µ 2 − µ 1 (106)
We now proceed to derive the tensor A in Eq. 70. Putting Eq. 104 into Eq. 70
and using Eq. 11 we have:
I + P : [C] ≡ B = b ′ I ′ + b ′′ I ′′ (107)
b ′ = 1 + 3[k]p 1 (108)
b ′′ = 1 + 2[µ]p 2 (109)
Note, that because of the properties of fourth-order tensors described in Section
3:
which can be written as:
A = B −1 = 1 b ′ I′ + 1
b ′′ I′′ (110)
A = a ′ I ′ + a ′′ I ′′ (111)
a ′ = 1 b ′ = k 1
k 1 + 3[k]k 1 p 1
(112)
a ′′ = 1
b ′′ = µ 1
µ 1 + 2[µ]µ 1 p 2
(113)
Comparing Eqs. 112 and 113 with Markov’s [6] Eq. (4.58), we find that:
3k 1 p 1 = α 1 (114)
2µ 1 p 2 = β 1 (115)
Note that Markov has an error in his Eq. (4.58) for the basis tensor I ′′
ijkl which
he writes as:
which should be corrected to read:
1
2 (δ ijδ kl + δ il δ jk − 2 3 δ ijδ kl ) (116)
I ′′
ijkl = 1 2 (δ ikδ jl + δ il δ jk − 2 3 δ ijδ kl ) (117)
This concludes the derivation of tensor A for isotropic matrix and isotropic
inhomogeneity.
9 Self consistent scheme for cubic polycrystals
The goal is to derive an equation (Kroner’s cubic) that relates the shear modulus
of the aggregate of cubic crystals to the elastic coefficients of a cubic crystal.

Markov [6] assumes that ”...this aggregate (polycrystal) is an assembly of
monocrystals, homogeneous grains with one and the same elastic properties,
defined by the elastic tensor C. The crystallographic axes of each grain vary.
Hence the tensor C is rotated in a complicated manner when one moves across
the solid and exactly this is what makes the polycrystal heterogeneous”. To
keep things simple, Markov assumes that ”there exist no preferable orientations
of grains, i.e. of the crystallographic axes. Hence there is no texture presented
so that the polycrystal is macroscopically isotropic, with a tensor of effective
moduli”:
C ∗ = 3k ∗ I ′ + 2µ ∗ I ′′ (118)
where I ′ and I ′′ are given by Eqs. 11, 4, and 10. Markov continues. ”Our aim
is to develop a certain simple approximate scheme of self-consistent type for
evaluating C ∗ by means of the given elastic tensor C for a single grain. Imagine
each grain is a sphere, immersed into an unbounded matrix with the effective,
but yet unknown, [assumed isotropic] properties C ∗ . Fix one of the grains, W .
According to Eq. 71 the strain within such a grain is constant”:
ɛ gr = A(C, C ∗ ) : ¯ɛ (119)
where ¯ɛ is the ”prescribed macrostrain tensor, applied to the polycrytalline representative
volume element (RVE)”, interpreted here as the strain ɛ 0 resulting
when the moduli C 1 and C 2 match. Moreover, in virtue of Eqs. 70 and 99,
Markov claims that given:
so that:
A(C, C ∗ ) ≡ A ∗ (120)
A ∗ = B ∗(−1) (121)
B ∗ = [I + P ∗ : (C − C ∗ )] (122)
P ∗ = p ∗ 1I ′ + p ∗ 2I ′′ (123)
p ∗ 1
1 =
3k ∗ + 4µ ∗ (124)
p ∗ 2 = 3
5µ ∗ k ∗ + 2µ ∗
3k ∗ + 4µ ∗ (125)
It appears that Markov’s argument is that Eq. 99 applies to both the isotropic
matrix and the approximately isotropic aggregate of cubic (anisotropic) crystals
in perfectly random orientations. The difference between these two configurations
is that in the case of the isotropic aggregate, the elastic modulus C ∗ is
unknown.
Markov is expanding the role of tensor A to apply to the case where the
inhomogeneity is not isotropic. Note that the tensor P was derived from the
Green’s function for the isotropic matrix alone. The constitutive tensor C is now
allowed to be anisotropic. This is consistent with the previous development.

Markov now describes the critical condition to be satisfied for determining
this aggregate effective elastic modulus: ”Let us now average Eq. 119 with
respect to all possible crystallographic orientations of the axes of the grain.
This operation will be denoted by 〈·〉 Ω
”. Then:
”The key observation now is the identity”:
〈ɛ gr 〉 Ω
= 〈A ∗ 〉 Ω
: ¯ɛ (126)
〈ɛ gr 〉 Ω
= ¯ɛ (127)
”i.e. the strain averaged over all crystallographic orientations in a grain equals
the macrostrain. Inserting Eq. 126 into Eq. 127:
〈A ∗ 〉 Ω
= I (128)
through which the unknown effective tensor C ∗ can be found”.
10 Elastic moduli of a cubic polycrystal
For a cubic polycrystal the calculation of the unknown effective tensor leads to
Kroner’s cubic for the effective shear modulus.
The tensor of the elastic cubic crystal moduli C in this case is given by
Markov as:
C = 3kI ′ + 2µ 2 I ′′ + 2(µ 1 − µ 2 )O h (129)
O h = e 4 1 + e 4 2 + e 4 3 (130)
Markov does not define e 4 i , saying only that O h is the ”basic fourth-rank tensor
with cubic symmetry (whose axes are along the orthonormal crystallographic
basis e i , i = 1, 2, 3)”. Turning to [8], Bertram defines e 4 i as (no sum on i):
e 4 i = e i ⊗ e i ⊗ e i ⊗ e i (131)
To simplify the calculation of the shear modulus, Markov introduces ”three
basic fourth-rank tensors with cubic symmetry”:
Σ 1 = I ′ (132)
Σ 2 = I ′ + I ′′ − O h (133)
Σ 3 = O h − I ′ (134)
Markov claims that these three tensors are orthogonal, which is true after applying
the correction in Eq. 117; that is, for each i, j = 1, 2, 3, i ≠ j:
This correction also allows the relation:
Σ i : Σ j = 0 (135)
Σ 1 + Σ 2 + Σ 3 = I (136)

Finally, this correction also confirms that for each i = 1, 2, 3, with no implied
sum on i:
Σ i : Σ i = Σ i (137)
With respect to this basis, Markov’s elastic coefficients take the form:
where:
C = 3K ′ Σ 1 + 2µ 2 Σ 2 + 2µ 1 Σ 3 (138)
K ′ = k + 2 3 (µ 1 − µ 2 ) (139)
To interpret Markov’s formulation for the elastic coefficients of the cubic crystal
in Eq. 129, we introduce the elastic coefficients of the cubic crystal in the form
described by Lubarda [9] in Eq. (3.1) (Note Lubarda’s sign error: +2c 44 should
be −2c 44 ):
C ijkl = c 12 δ ij δ kl + 2c 44 I ijkl + (c 11 − c 12 − 2c 44 )Âijkl (140)
which is equivalent in Markov’s notation to:
or,
C = 3c 12 I ′ + 2c 44 (I ′ + I ′′ ) + (c 11 − c 12 − 2c 44 )O h (141)
C = (3c 12 + 2c 44 )I ′ + 2c 44 (I ′′ ) + (c 11 − c 12 − 2c 44 )O h (142)
but, in terms of Markov’s basis:
C = (c 11 + 2c 12 )Σ 1 + 2c 44 Σ 2 + (c 11 − c 12 )Σ 3 (143)
Comparing Eqs. 138 with 143 shows that:
In the sequel, the symbolic notation:
will be useful.
K ′ = 1 3 (c 11 + 2c 12 ) (144)
µ 1 = 1 2 (c 11 − c 12 ) (145)
µ 2 = c 44 (146)
C = (α, β, γ) ⇔ C = αΣ 1 + βΣ 2 + γΣ 3 (147)

11 Rotational averages
Before proceeding with the averaging process of the constitutive coefficients
and, in particular, of the tensor A, it will be useful to describe the details of
calculating rotational averages.
According to Eqs. (2) and (3) in Andrews [10], the rotational average of a
fourth order tensor C is given by:
〈C〉 i1i 2i 3i 4
= Λ i1i 2i 3i 4;λ 1λ 2λ 3λ 4
C λ1λ 2λ 3λ 4
(148)
and where Λ is defined by Eq. (19) in Andrews [10] as:
⎛
Λ i1i 2i 3i 4;λ 1λ 2λ 3λ 4
= 1 ( )
δi1i
30
2
δ i3i 4
δ i1i 3
δ i2i 4
δ i1i 4
δ i2i 3
⎝
4 −1 −1
−1 4 −1
−1 −1 4
⎞ ⎛
⎠ ⎝
Let’s begin with finding the rotational average of tensor I ′ .
In the following, the subscripts h and Ω are understood. Following Andrews,
we have:
where we must have the correspondence:
and where
〈I ′ 〉 ijkl
= Λ ijkl;αβγω I ′ αβγω (150)
(i 1 , i 2 , i 3 , i 4 ) ⇔ (i, j, k, l) (151)
(λ 1 , λ 2 , λ 3 , λ 4 ) ⇔ (α, β, γ, ω) (152)
Carrying out the above calculations results in:
Similarly, with the correspondence:
we find:
and, with the correspondence:
I ′ αβγω = 1 3 δ αβδ γω (153)
〈I ′ 〉 = I ′ (154)
(i 1 , i 2 , i 3 , i 4 ) ⇔ (i, k, j, l) (155)
(λ 1 , λ 2 , λ 3 , λ 4 ) ⇔ (α, γ, β, ω) (156)
〈δ ik δ jl 〉 = δ ik δ jl (157)
(i 1 , i 2 , i 3 , i 4 ) ⇔ (i, l, j, k) (158)
(λ 1 , λ 2 , λ 3 , λ 4 ) ⇔ (α, ω, β, γ) (159)
δ λ1λ 2
δ λ3λ 4
δ λ1λ 3
δ λ2λ 4
δ λ1λ 4
δ λ2λ 3
⎞
⎠(149)

we find:
〈δ il δ jk 〉 = δ il δ jk (160)
so that:
〈I〉 = I (161)
and, therefore:
〈I ′′ 〉 = I ′′ (162)
Let’s now find the rotational average of tensor O h . Following Andrews, we have:
From Eqs. 130 and 131:
So that:
〈O〉 ijkl
= Λ ijkl;αβγω O αβγω (163)
e 4 i = δ αi e α ⊗ δ βi e β ⊗ δ γi e γ ⊗ δ ωi e ω (164)
O αβγω =
3∑
(δ αi δ βi δ γi δ ωi ) (165)
i=1
and from Eqs. 149, 163, and 165 we obtain:
Now:
so that:
〈O ijkl 〉 = 1
30 (δ αβδ γω A ijkl + δ αγ δ βω B ijkl + δ αω δ βγ C ijkl )O αβγω (166)
A ijkl = 4δ ij δ kl − δ ik δ jl − δ il δ jk (167)
B ijkl = −δ ij δ kl + 4δ ik δ jl − δ il δ jk (168)
C ijkl = −δ ij δ kl − δ ik δ jl + 4δ il δ jk (169)
δ αβ δ γω O αβγω = 3 (170)
δ αγ δ βω O αβγω = 3 (171)
δ αω δ βγ O αβγω = 3 (172)
〈O ijkl 〉 = 1
10 (A ijkl + B ijkl + C ijkl ) = 1 5 (δ ijδ kl + δ ik δ jl + δ il δ jk ) (173)
We now define H as:
so that:
H ijkl = δ ij δ kl + δ ik δ jl + δ il δ jk (174)
〈O〉 = 1 5 H (175)

Note that from Eqs. 10, 11, and 174:
H = 5I ′ + 2I ′′ (176)
We now prove the following identities which we will need in the sequel:
From 133, 154, 162,173, and 176:
〈Σ 2 〉 = 3 5 I′′ (177)
〈Σ 3 〉 = 2 5 I′′ (178)
〈Σ 2 〉 = 〈I ′ 〉 + 〈I”〉 − 〈O〉 (179)
〈Σ 2 〉 = I ′ + I” − 1 5 H (180)
〈Σ 2 〉 = I ′ + I” − 1 5 (5I′ + 2I”) (181)
〈Σ 2 〉 = 3 I” (182)
5
Eq. 178 is shown is a similar fashion. From Eqs. 133 and 134:
so that:
Similarly, from Eq. 134:
Σ 2 + Σ 3 = I ′′ (183)
〈Σ 1 〉 = 〈I ′ 〉 = Σ 1 (184)
〈Σ 2 〉 = 3 5 I” = 3 5 (Σ 2 + Σ 3 ) (185)
〈Σ 3 〉 = 〈O〉 − 〈I ′ 〉 (186)
〈Σ 3 〉 = 1 5 H − I′ (187)
〈Σ 3 〉 = 1 5 (5I′ + 2I ′′ ) − I ′ (188)
〈Σ 3 〉 = 2 5 I′′ (189)
〈Σ 3 〉 = 2 5 (Σ 2 + Σ 3 ) (190)
12 Derivation of tensor A for anisotropic inhomogeneity
and Kroner’s cubic
For the tensor C of a single grain, see Eq. 138, we have:
C = (α, β, γ), α = 3K ′ , β = 2µ 2 , γ = 2µ 1 (191)

Due to the orthogonal properties of the Σ ′ is, inversion and multiplication take
the forms:
C −1 = ( 1 α , 1 β , 1 γ ) (192)
With C ′ = (α ′ , β ′ , γ ′ ) and C ′′ = (α ′′ , β ′′ , γ ′′ ):
C ′ : C ′′ = (α ′ α ′′ , β ′ β ′′ , γ ′ γ ′′ ) (193)
The most important formula, however, concerns the tensors of the form given
by Eq. 147, averaged over all orientations, using Eqs. 184, 185, 190:
From Eqs. 122 and 191 the quantity:
has the form:
〈C〉 Ω
= 〈(α, β, γ)〉 Ω
= (α, ¯β, ¯β) (194)
¯β = 1 (3β + 2γ) (195)
5
B ∗ = I + P ∗ : (C − C ∗ ) (196)
B ∗ = (1, 1, 1) + (p ∗ 1, p ∗ 2, p ∗ 2) · [(3K ′ , 2µ 2 , 2µ 1 ) − (3k ∗ , 2µ ∗ , 2µ ∗ )] (197)
which reduces to:
B ∗ = (1 + 3p ∗ 1(K ′ − k ∗ ), 1 + 2p ∗ 2(µ 2 − µ ∗ ), 1 + 2p ∗ 2(µ 1 − µ ∗ )) (198)
But Eq. 198 is the inverse of Eq. 120, so that:
A ∗ 1
= (
1 + 3p ∗ 1 (K′ − k ∗ ) , 1
1 + 2p ∗ 2 (µ 2 − µ ∗ ) , 1
1 + 2p ∗ 2 (µ 1 − µ ∗ ) ) (199)
whose average follows Eqs. 194 and 154:
〈A ∗ 〉 Ω
= (α ∗ , β ∗ , β ∗ ) (200)
α ∗ 1
=
1 + 3p ∗ 1 (K′ − k ∗ )
(201)
β ∗ = 3 1
5 1 + 2p ∗ 2 (µ 2 − µ ∗ ) + 2 1
5 1 + 2p ∗ 2 (µ 1 − µ ∗ )
(202)
Finally, the average of A ∗ must satisfy Eq. 128 subject to Eq. 136, so that:
α ∗ Σ 1 + β ∗ (Σ 2 + Σ 3 ) = I = I ′ + I ′′ (203)
Σ 2 + Σ 3 = I ′′ (204)
α ∗ I ′ + β ∗ I ′′ = I ′ + I ′′ (205)
(α ∗ − 1)I ′ + (β ∗ − 1)I ′′ = 0 (206)
(207)

Since I ′ and I ′′ are orthogonal:
α ∗ = 1 (208)
β ∗ = 1 (209)
from which:
k ∗ = K ′ (210)
and substituting Eq. 125 into Eq. 202 and applying Eq. 209 gives a cubic
equation for the aggregate shear modulus µ ∗ in the form:
(µ ∗ ) 3 + a(µ ∗ ) 2 − bµ ∗ − c = 0 (211)
a = 9k∗ + 4µ 1
8
b = 3µ 2(k ∗ + 4µ 1 )
8
c = 3k∗ µ 1 µ 2
4
Using Eq. 210, 144, 145, and 146, the above coefficients of the cubic become:
(212)
(213)
(214)
a = 1 8 (5c 11 + 4c 12 ) (215)
b = 1 8 c 44(7c 11 − 4c 12 ) (216)
c = 1 8 c 44(c 11 + 2c 12 )(c 11 − c 12 ) (217)
Kroner’s [2] Eq. (22) shows the cubic for the aggregate shear modulus G as:
Converting to Voigt notation:
G 3 + ᾱG 2 + ¯βG + ¯γ = 0 (218)
ᾱ = (3 ¯K + 4¯ν)
8
¯β = −( ¯K + 12¯ν) ¯µ 8
(219)
(220)
¯γ = − ¯K ¯µ ¯ν 4
(221)
¯K = c 1111 + 2c 1122 (222)
¯µ = c 1212 (223)
¯ν = (c 1111 − c 1122 )
2
(224)
¯K = c 11 + 2c 12 (225)
¯µ = c 66 (226)
¯ν = (c 11 − c 12 )
2
(227)

But because of cubic symmetry, c 66 = c 44 , so that Kroner’s coefficients ᾱ, ¯β, ¯γ
are consistent with the above coefficients a, b, c, namely:
ᾱ = 1 8 (5c 11 + 4c 12 ) = a (228)
¯β = − c 44
8 (7c 11 − 4c 12 ) = −b (229)
¯γ = − c 44
8 (c 11 + 2c 12 )(c 11 − c 12 ) = −c (230)
as required, thus completing the derivation of Kroner’s cubic.
13 Comparison with experimental data
Hashin and Shtrikman [11], [12] have derived bounds for the isotropic shear
modulus of aggregates of cubic crystals. Gschneider [13] provides experimental
data, and Simmons [14] provides cubic crystal elastic coefficients in Mbar. Sisodia
[15] compares some experimental data with calculated shear modulus and
reports the formulation of the Hashin-Shtrikman bounds, G ∗ 1 and G ∗ 2 in terms
of the single crystal cubic coefficients c ij as:
where
(
) −1
G ∗ 5
1 = G 1 + 3 − 4β 1 (231)
G 2 − G 1
(
G ∗ 2 = G 2 + 2
5
G 1 − G 2
− 6β 2
) −1
(232)
β 1 = − 3 K + 2G 1
5 G 1 (3K + 4G 1 )
β 2 = − 3 K + 2G 2
5 G 2 (3K + 4G 2 )
An anisotropy index à is defined by:
K = c 11 + 2c 12
3
(233)
(234)
(235)
G 1 = c 11 − c 12
2
(236)
G 2 = c 44 (237)
à = G 2
G 1
= 2c 44
c 11 − c 12
(238)
which reduces to unity for isotropic material. Table 1 shows the cubic coefficients
and the anisotropy index for several materials. Notice that Tungsten (W) and
Molybdenum (Mo) come closest to being isotropic.
To compare the shear modulus determined by Kroner’s cubic with experimental
values, we use a simple iterative Newton’s method, beginning with a trial

value at the arithmetic average of the Hashin-Shtrikman’s bounds, as described
in Section 15. Table 1 shows these bounds and compares this calculated value
with the experimental value. Notice that for elements which are not in Group
IV-A, the error is under 20%, while for Group IV-A, the error varies from 40%
to 74%.
Sisodia [15] mentions that if the anisotropic index à > 1, then G∗ 2 > G ∗ 1
and vice versa. Notice also from Table 1, for those (most) elements where the
anisotropy index exceeds unity, it is the case that G ∗ 2 > G ∗ 1. For Molybdenum
and Tungsten which are practically isotropic, the bounds coincide. Finally,
Niobium is the only element with à < 1, and there G∗ 2 < G ∗ 1.
Units:
kg f
cm 2
Gschneider’s experimental values are given in units of × 10 6 . This is
converted to GPa by multiplying by gravitational acceleration as 9.80665 m s
, so 2
that Gschneider’s units are converted as follows:
kg f
cm 2 × 106 × 10 4 cm2
m 2 × 9.80665 m s 2 × N
kg m s 2
× P a
N
m 2
× GP a
10 9 P a
(239)
and 1 kg f
cm
×10 6 then corresponds to 98.0665 GPa. For example, with Aluminum,
2
Schneider’s value of 0.271 kg f
cm
× 10 6 corresponds to 26.6 GPa.
2

Table 1: Calculated (Gk) vs. Experimental (Gx) Shear Modulus (GPa)
Element Group c11 c12 c44 G1* G2* Gk Gx Error Anisotropy
Ag11650 I-B 124 93.4 46.1 29 30.9 30.2 28.6 0.06 3.01
Ag11649 I-B 120 89.7 43.7 28 29.6 29 28.6 0.01 2.88
Ag11651 I-B 121 89.9 42.6 28 29.4 28.9 28.6 0.01 2.74
Ag11678 I-B 122 90.7 45.4 29.2 30.8 30.2 28.6 0.06 2.90
Au10555 I-B 187 155 42.9 28.6 29.9 29.4 27.6 0.07 2.68
Cu10380 I-B 170 115 61 44.1 45.2 44.8 45.1 -0.01 2.22
Cu10381 I-B 176 128 75.2 46 49.4 48.2 45.1 0.07 3.13
Cu10382 I-B 170 124 64.5 41.9 44.2 43.3 45.1 -0.04 2.80
Cu10383 I-B 168 121 75.4 46 49.4 48.2 45.1 0.07 3.21
Cu10384 I-B 170 123 75.3 46 49.5 48.2 45.1 0.07 3.20
Cu10385 I-B 171 124 75.5 46 49.5 48.2 45.1 0.07 3.21
Cu10423 I-B 168 121 75.7 46.2 49.6 48.4 45.1 0.07 3.22
Al10037 III-A 106 59 28.5 26.3 26.4 26.3 26.6 -0.01 1.21
Ge10531 IV-A 132 50.9 66.9 54.5 54.8 54.7 39.2 0.40 1.65
Ge10532 IV-A 129 47.9 67 54.7 55 54.9 39.2 0.40 1.65
Ge10538 IV-A 129 48.3 67.1 54.5 54.9 54.8 39.2 0.40 1.66
Pb10681 IV-A 47.7 40.3 14.4 8 9 8.66 5.39 0.61 3.89
Si11644 IV-A 168 66 84 68.6 69 68.9 39.7 0.74 1.65
Si11645 IV-A 166 63.9 79.6 66.5 66.7 66.6 39.7 0.68 1.56
Nb11140 V-B 246 139 29.3 37.7 37.2 37.4 37.5 0.00 0.55
Ta11928 V-B 260 154 82.6 69 69.3 69.2 68.6 0.01 1.56
V12078 V-B 196 133 67 49.3 50.3 49.9 46.5 0.07 2.13
W12020 VI-B 513 206 153 153 153 153 153 0.00 1.00
Mo10974 VI-B 470 168 107 123 123 123 116 0.06 0.71
Mo10995 VI-B 441 172 122 127 127 127 116 0.09 0.91
Fe10650 VIII 231 135 116 80.8 83.4 82.4 81.5 0.01 2.42
Ni11110 VIII 248 155 124 82.4 86.1 84.7 75 0.13 2.67
Ni11111 VIII 247 144 124 86.1 88.9 87.8 75 0.17 2.41
Pd11222 VIII 227 176 71.7 46.7 49.2 48.3 51.1 -0.05 2.81
Pt11224 VIII 347 251 76.5 63.5 63.8 63.7 61 0.04 1.59
Th12001 actinide 75.3 48.9 47.8 27.3 30 29 27.9 0.04 3.62

14 Major symmetries of tensor D
Write:
D ijkl = − 1
4πµ (M ijkl + 2κ 1 N ijkl ) (240)
M ijkl =
N ijkl =
∫ π ∫ 2π
0 0
∫ π ∫ 2π
0
0
(δ ij z k z l sinΦ) dΘdΦ (241)
(z i z j z k z l sinΦ) dΘdΦ (242)
It is clear that N satisfies both major and minor symmetries. It remains to
show that M ijkl satisfies major symmetry. Clearly, if i ≠ j, then M ijkl = 0. It
can also be shown that if k ≠ l, then M ijkl = 0. For this purpose, since M ijkl
satisfies minor symmetries, it suffices to show that:
M ij12 = M ij13 = M ij23 = 0 (243)
Putting Eqs. 79,80,81 into Eq. 241 as needed to form M ij12 , M ij13 , M ij23 , we
find that in each case the Θ-integral vanishes.
For M ij12 , the Θ-integral involves:
∫ 2π
0
(cosΘsinΘ)dΘ = 1 4
∫ 2π
For M ij13 , the Θ-integral involves:
∫ 2π
For M ij23 , the Θ-integral involves:
0
∫ 2π
0
0
(sin2Θ)d(2Θ) = (− 1 4 )[cos2Θ]2π 0 = 0 (244)
cosΘdΘ = [sinΘ] 2π
0 = 0 (245)
sinΘdΘ = −[cosΘ] 2π
0 = 0 (246)
From the above, it is clear that the only non-zero values of M ijkl are obtained
only if i = j and k = l. With i = j, M ijkl reduces to the second order quantity:
M kl =
∫ π ∫ 2π
0
0
(z k z l sinΦ) dΘdΦ (247)
With k = l = 1, 2, 3 and putting Eqs. 79,80,81 into Eq. 247 as needed, we find
that:
M 11 =
∫ π ∫ 2π
0
0
sin 2 Φcos 2 ΘsinΦdΘdΦ =
[ Θ
=
2 + 1 ] 2π
4 sin2Θ
0
∫ 2π
0
cos 2 ΘdΘ
∫ π
[
− 1 3 cosΦ(sin2 Φ + 2)
0
sin 3 ΦdΦ (248)
] π
0
= 4 3 π (249)

Similarly:
M 33 =
Thus, for:
∫ π ∫ 2π
M 22 =
∫ 2π
sin 2 ΘdΘ
∫ π
0
0
cos 2 ΦsinΦdΘdΦ = (−1)2π
0 0
0
sin 3 ΦdΦ = 4 3 π (250)
∫ π
cos 2 Φd(cosΦ) = 4 3 π (251)
ijkl = 1111, 1122, 1133, 2211, 2222, 2233, 3311, 3322, 3333 (252)
M ijkl = 4 3 π (253)
These results clearly show that, given all the non-zero terms, M ijkl does have
major symmetry as well as minor symmetry. Now that we have shown that
both N ijkl and M ijkl have both major and minor symmetry, we can construct
the following model for D ijkl which assumes both major and minor symmetry:
D ijkl = γδ ij δ kl + ω(δ ik δ jl + δ il δ jk ) (254)
where γ is given by any member of the list in Eq. 252, say 1122 for which:
=
∫ π ∫ 2π
0
0
N 1122 =
∫ π ∫ 2π
0
0
(
z
2
1 z 2 2sinΦ ) dΘdΦ (255)
(
sin 2 Φcos 2 Θsin 2 Φsin 2 ΘsinΦ ) dΘdΦ (256)
=
∫ 2π
0
cos 2 Θsin 2 ΘdΘ
∫ π
0
∫ π
( ∫ 1 2π
) ( )
4
= sin 2 (2Θ)d(2Θ) sin 3 ΦdΦ
8 0
5 0
( [ ] ) 2π (
1 1
=
8 2 (2Θ) ( 4 [
0
5 ) − 1 ] π )
3 (cosΦ)(2) = 4π
0
15
sin 5 ΦdΦ (257)
(258)
(259)
so that:
D 1122 = − 1 (M 1122 + 2κ 1 N 1122 )
4πµ 1
(260)
= − 1 ( 4 4πµ 1 3 π + 2κ 4π
1
15 ) (261)
= − 1
3µ 1
(1 + 2 5 κ 1) (262)
γ = D 1122 = − 1
3µ 1
(1 + 2 5 κ 1) (263)

Finally, ω represents any D ijkl term for which M ijkl = 0 but N ijkl ≠ 0, that is,
for the following list of (ijkl) indices for which (i ≠ j and k ≠ l) but for which
(i = k and j = l) or (i = l and j = k):
1212, 2121, 1313, 3131, 2323, 3232, 1221, 2112, 1331, 3113, 2332, 3223 (264)
For D 1313 , M 1313 = 0 and:
=
∫ 2π
0
=
=
∫ 2π
0
∫ π ∫ 2π
0
N 1313 =
0
∫ 2π
=
0
∫ π ∫ 2π
0
0
(
z
2
1 z 2 3sinΦ ) dΘdΦ (265)
(
sin 2 Φcos 2 Θcos 2 ΦsinΦ ) dΘdΦ (266)
cos 2 ΘdΘ
∫ π
0
sin 3 Φcos 2 ΦdΦ (267)
(∫ π
cos 2 ΘdΘ sin 3 Φ ( 1 − sin 2 Φ ) )
dΦ
(∫ π
cos 2 ΘdΘ sin 3 ΦdΦ −
0
0
∫ π
0
(268)
)
sin 5 ΦdΦ = 4
15 π (269)
so that:
ω = D 1313 = − 2κ 1
N 1313 = − 2κ 1 4π
4πµ 1 4πµ 1 15 = − 2 15
κ 1
µ 1
(270)
giving our model as:
D ijkl = − 1 (1 + 2 3µ 1 5 κ 1)δ ij δ kl − 2 κ 1
(δ ik δ jl + δ il δ jk ) (271)
15 µ 1

15 Iterative Newton’s method solution of Kroner’s
cubic
Rewrite Eq. 211 as:
x 3 + ax 2 − bx − c = 0 (272)
where x = µ ∗ . It turns out that it is easier to use Newton’s method rather than
the formulas for solving a cubic equation. To this end, we can write an exact
Taylor’s expansion in the form:
f(x + ∆x) = f(x) + f ′ (x)∆x + 1 2 f ′′ (x)(∆x) 2 + 1 6 f ′′′ (x)(∆x) 3 (273)
since all derivatives greater than third order vanish. Now apply Newton’s
method to first, second, or third order, as follows. Write:
f(x n+1 ) = f(x n ) + f ′ (x n )∆x + 1 2 f ′′ (x n )(∆x) 2 + 1 6 f ′′′ (x n )(∆x) 3 (274)
x n+1 = x n + ∆x (275)
Newton’s method is usually based only on a first order expansion, solving for
∆x assuming that the first derivative does not vanish:
f(x n+1 ) = 0 ≈ f(x n ) + f ′ (x n )∆x (276)
∆x = −f(x n)
f ′ (x n )
(277)
We consider the case where the first derivative is too small by going to a second
order approximate solution, and if both first and second derivatives are too
small, by going to a third order solution. Details follow. ”Too small” means that
the normalized value of the derivative is less than 10 −10 . If the first derivative
is too small, then we check the second derivative. If it is also too small, then
we use the third order approximation, and the Taylor’s expansion reduces to:
f n+1 = f(x n+1 ) = 0 = f n + 1 6 f ′′′ (x n )(∆x) 3 (278)
But from Eq.272, f ′′′ (x n ) = 6, so that we can solve for ∆x:
f n = f(x n ) (279)
∆x = (−f n ) 1 3 (280)
If the first derivative is too small, but the second derivative is not too small,
then from the second order approximation, we can solve for ∆x provided that
< 0, so that:
f n
f ′′ n
∆x =
√
−2 f n
f ′′ n
(281)

otherwise there is no solution (More below). If the first derivative is sufficiently
large, then we can consider again the second order approximation:
f(x n+1 ) = 0 ≈ f(x n ) + f ′ (x n )∆x + 1 2 f ′′ (x n )(∆x) 2 (282)
Checking the discriminant D:
D = (f ′ n) 2 − 2f ′′
nf n (283)
If this discriminant is negative, then we revert to the first order solution Eq.
277. If the discriminant is positive but the second derivative is too small, the
we also revert to the first order solution Eq. 277. Finally, if the discriminant is
positive and the second derivative is sufficiently large, then we get the second
order solution for ∆x:
∆x = −f ′ n + √ D
f ′′ n
(284)
We choose the root with the ”+” sign because as f n ⇒ 0, ∆x ⇒ 0. Note that
all cases reported here fell in this second order solution, where both first and
second derivatives were large and the discriminant was strictly positive.
The case above without solution is interesting. This is the case where f n ′ is
have the same sign. If they are both positive,
this means that locally (about x n ), f(x) opens upward but does not cross the
x-axis. Similarly, if they are both negative, f(x) opens downward but does not
cross the x-axis. In both of these cases, there is clearly no solution.
too small and both f n and f ′′
n

References
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crystals. Journal of Applied Mechanics, 21:236–240, September 1954.
[2] E. Kroner. Berechnung der elastischen konstanten des vielkristalls aus den
konstanten des einkristalls. Zeitschrift fur physik, 151:504–518, 1958.
[3] J. D. Eshelby. Elastic inclusions and inhomogeneities. In I.N. Sneddon
and R. Hill, editors, Progress in solid mechanics, volume II, pages 88–140.
North-Holland, 1961.
[4] C. Weinberger, W. Cai, and D.M. Barnett. Lecture notes - elasticity of
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