Elasticity tensor explained

The elasticity tensor is a fourth-rank tensor describing the stress-strain relation ina linear elastic material. Other names are elastic modulus tensor and stiffness tensor. Common symbols include

and

The defining equation can be written as

T^ij=C^ijklE_kl

where

T^ij

and

E_kl

are the components of the Cauchy stress tensor and infinitesimal strain tensor, and

C^ijkl

are the components of the elasticity tensor. Summation over repeated indices is implied.^[1] This relationship can be interpreted as a generalization of Hooke's law to a 3D continuum.

A general fourth-rank tensor

in 3D has 3⁴ = 81 independent components

F_ijkl

, but the elasticity tensor has at most 21 independent components. This fact follows from the symmetry of the stress and strain tensors, together with the requirement that the stress derives from an elastic energy potential. For isotropic materials, the elasticity tensor has just two independent components, which can be chosen to be the bulk modulus and shear modulus.

Definition

The most general linear relation between two second-rank tensors

T,E

T^ij=C^ijklE_kl

where

C^ijkl

are the components of a fourth-rank tensor

.^[1] The elasticity tensor is defined as

for the case where

and

are the stress and strain tensors, respectively.

The compliance tensor

is defined from the inverse stress-strain relation:

E^ij=K^ijklT_kl

The two are related by

K_ijpqC^pqkl=

	1
	2

	k
\left(\delta
	i

	l
\delta
	j

	l
\delta
	i

	k
\delta
	j

\right)

where

	m
\delta
	n

is the Kronecker delta.^[2]

Unless otherwise noted, this article assumes

is defined from the stress-strain relation of a linear elastic material, in the limit of small strain.

Special cases

Isotropic

For an isotropic material,

simplifies to

C^ijkl=λ\left(X\right)g^ijg^kl+\mu\left(X\right)\left(g^ikg^jl+g^ilg^kj\right)

where

and

\mu

are scalar functions of the material coordinates

, and

is the metric tensor in the reference frame of the material. In an orthonormal Cartesian coordinate basis, there is no distinction between upper and lower indices, and the metric tensor can be replaced with the Kronecker delta:

C_ijkl=λ\left(X\right)\delta_ij\delta_kl+\mu\left(X\right)\left(\delta_ik\delta_jl+\delta_il\delta_kj\right) [Cartesiancoordinates]

Substituting the first equation into the stress-strain relation and summing over repeated indices gives

T^ij=λ\left(X\right) ⋅ \left(TrE\right)g^ij+2\mu\left(X\right)E^ij

where

TrE\equiv

	i
E
	i

is the trace of

.In this form,

\mu

and

can be identified with the first and second Lamé parameters.An equivalent expression is

T^ij=K\left(X\right) ⋅ \left(TrE\right)g^ij+2\mu\left(X\right)\Sigma^ij

where

K=λ+(2/3)\mu

is the bulk modulus, and

\Sigma^ij\equivE^ij-(1/3)\left(TrE\right)g^ij

are the components of the shear tensor

\Sigma

Cubic crystals

The elasticity tensor of a cubic crystal has components

\begin{align} C^ijkl&=λg^ijg^kl+\mu\left(g^ikg^jl+g^ilg^kj\right)\ &+\alpha\left(aⁱa^ja^ka^l+bⁱb^jb^kb^l+cⁱc^jc^kc^{l\right)
\end{align}}

where

, and

are unit vectors corresponding to the three mutually perpendicular axes of the crystal unit cell. The coefficients

\mu

, and

\alpha

are scalars; because they are coordinate-independent, they are intrinsic material constants. Thus, a crystal with cubic symmetry is described by three independent elastic constants.

In an orthonormal Cartesian coordinate basis, there is no distinction between upper and lower indices, and

g^ij

is the Kronecker delta, so the expression simplifies to

\begin{align} C_ijkl&=λ\delta_ij\delta_kl+\mu\left(\delta_ik\delta_jl+\delta_il\delta_kj\right)\ &+\alpha\left(a_ia_ja_ka_l+b_ib_jb_kb_l+c_ic_jc_kc_{l\right)
\end{align}}

Other crystal classes

There are similar expressions for the components of

in other crystal symmetry classes. The number of independent elastic constants for several of these is given in table 1.

Table 1: Number of independent elastic constants for various crystal symmetry classes.! Crystal family! Point group! Independent components
Triclinic		21
Monoclinic		13
Orthorhombic		9
Tetragonal	C₄, S₄, C_4h	7
Tetragonal	C_4v, D_2d, D₄, D_4h	6
Rhombohedral	C₃, S₆	7
Rhombohedral	C_3v, D₆, D_3d	6
Hexagonal		5
Cubic		3

Properties

Symmetries

The elasticity tensor has several symmetries that follow directly from its defining equation

T^ij=C^ijklE_kl

. The symmetry of the stress and strain tensors implies that

C_ijkl=C_jikl and C_ijkl=C_ijlk,

Usually, one also assumes that the stress derives from an elastic energy potential

T^ij=

	\partialU
	\partialE_ij

which implies

C_ijkl=

	\partial²U
	\partialE_ij\partialE_kl

Hence,

must be symmetric under interchange of the first and second pairs of indices:

C_ijkl=C_klij

The symmetries listed above reduce the number of independent components from 81 to 21. If a material has additional symmetries, then this number is further reduced.

Transformations

Under rotation, the components

C^ijkl

transform as

C'_ijkl=R_ipR_jqR_krR_lsC^pqrs

where

C'_ijkl

are the covariant components in the rotated basis, and

R_ij

are the elements of the corresponding rotation matrix. A similar transformation rule holds for other linear transformations.

Invariants

The components of

generally acquire different values under a change of basis. Nevertheless, for certain types of transformations,there are specific combinations of components, called invariants, that remain unchanged. Invariants are defined with respect to a given set of transformations, formally known as a group operation. For example, an invariant with respect to the group of proper orthogonal transformations, called SO(3), is a quantity that remains constant under arbitrary 3D rotations.

possesses two linear invariants and seven quadratic invariants with respect to SO(3). The linear invariants are

\begin{align} L₁&=

	ij
C
	ij

\\ L₂&=

	ii
C
	jj

\end{align}

and the quadratic invariants are

\left\{

	2,
L
	1

	2,
L
	2

L₁L_2,C_ijklC^ijkl,C_iiklC^jjkl,C_iiklC^jkjl,C_kiilC^kjjl\right\}

These quantities are linearly independent, that is, none can be expressed as a linear combination of the others. They are also complete, in the sense that there are no additional independent linear or quadratic invariants.

Decompositions

See also: Tensor decomposition. A common strategy in tensor analysis is to decompose a tensor into simpler components that can be analyzed separately. For example, thedisplacement gradient tensor

W=\nabla\xi

can be decomposed as

	1
	3

\Thetag+\Sigma+R

where

\Theta

is a rank-0 tensor (a scalar), equal to the trace of

;

\Sigma

is symmetric and trace-free; and

is antisymmetric. Component-wise,

\begin{align} \Sigma^ij\equivW^(ij)&=

	1
	2

\left(W^ij+W^ji\right)-

	1
	3

\left(TrW\right)g^ij\\ R^ij\equivW^[ij]&=

	1
	2

\left(W^ij-W^ji\right) \end{align}

Here and later, symmeterization and antisymmeterization are denoted by

(ij)

and

[ij]

, respectively. This decomposition is irreducible, in the sense of being invariant under rotations, and is an important tool in the conceptual development of continuum mechanics.

The elasticity tensor has rank 4, and its decompositions are more complex and varied than those of a rank-2 tensor. A few examples are described below.

M and N tensors

This decomposition is obtained by symmeterization and antisymmeterization of the middle two indices:

C^ijkl=M^ijkl+N^ijkl

where

\begin{align} M^ijkl\equivC^i(jk)l=

	1
	2

\left(C^ijkl+C^ikjl\right)\\ N^ijkl\equivC^i[jk]l=

	1
	2

\left(C^ijkl-C^ikjl\right) \end{align}

A disadvantage of this decomposition is that

M^ijkl

and

N^ijkl

do not obey all original symmetries of

C^ijkl

, as they are not symmetric under interchange of the first two indices. In addition, it is not irreducible, so it is not invariant under linear transformations such as rotations.

Irreducible representations

See also: Young tableau and Irreducible representation. An irreducible representation can be built by considering the notion of a totally symmetric tensor, which is invariant under the interchange of any two indices. A totally symmetric tensor

can be constructed from

by summing over all

4!=24

permutations of the indices

\begin{align} S^ijkl&=

	1
	4!

\sum
	(i,j,k,l)\inS₄

C^ijkl\ &=

	1
	4!

\left(C^ijkl+C^jikl+C^ikjl+\ldots\right) \end{align}

where

S₄

is the set of all permutations of the four indices. Owing to the symmetries of

C^ijkl

, this sum reduces to

S^ijkl=

	1
	3

\left(C^ijkl+C^iklj+C^iljk\right)

The difference

A^ijkl\equivC^ijkl-S^ijkl=

	1
	3

\left(2C^ijkl-C^ilkj-C^iklj\right)

is an asymmetric tensor (not antisymmetric). The decomposition

C^ijkl=S^ijkl+A^ijkl

can be shown to be unique and irreducible with respect to

S₄

. In other words, any additional symmetrization operations on

will either leave it unchanged or evaluate to zero. It is also irreducible with respect to arbitrary linear transformations, that is, the general linear group

G(3,R)

However, this decomposition is not irreducible with respect to the group of rotations SO(3). Instead,

decomposes into three irreducible parts, and

into two:

\begin{align} C^ijkl&=S^ijkl+A^ijkl\\ &=\left(⁽¹⁾S^ijkl+⁽²⁾S^ijkl+⁽³⁾S^ijkl\right)+\left(⁽¹⁾A^ijkl+⁽²⁾A^ijkl\right) \end{align}

See Itin (2020) for explicit expressions in terms of the components of

This representation decomposes the space of elasticity tensors into a direct sum of subspaces:

l{C}=\left(⁽¹⁾l{C} ⊕ ⁽²⁾l{C} ⊕ ⁽³⁾l{C}\right) ⊕ \left(⁽⁴⁾l{C} ⊕ ⁽⁵⁾l{C}\right)

with dimensions

21=(1 ⊕ 5 ⊕ 9) ⊕ (1 ⊕ 5)

These subspaces are each isomorphic to a harmonic tensor space

	3)
H
	n(R

. Here,

	3)
H
	n(R

is the space of 3D, totally symmetric, traceless tensors of rank

. In particular,

⁽¹⁾l{C}

and

⁽⁴⁾l{C}

correspond to

H₁

⁽²⁾l{C}

and

⁽⁵⁾l{C}

correspond to

H₂

, and

⁽³⁾l{C}

corresponds to

H₄

Bibliography

The Feynman Lectures on Physics - The tensor of elasticity
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Notes and References

Here, upper and lower indices denote contravariant and covariant components, respectively, though the distinction can be ignored for Cartesian coordinates. As a result, some references represent components using only lower indices.
Combining the forward and inverse stress-strain relations gives .Due to the minor symmetries and, this equation does not uniquely determine . In fact, is a solution for any . However, only preserves the minor symmetries of K, so this is the correct solution from a physical standpoint.