Elasticity tensor

The most general linear relation between two second-rank tensors ${\displaystyle \mathbf {T} ,\mathbf {E} }$ is

${\displaystyle T^{ij}=C^{ijkl}E_{kl}}$

where ${\displaystyle C^{ijkl}}$ are the components of a fourth-rank tensor ${\displaystyle \mathbf {C} }$.^{[1][note 1]} The elasticity tensor is defined as ${\displaystyle \mathbf {C} }$ for the case where ${\displaystyle \mathbf {T} }$ and ${\displaystyle \mathbf {E} }$ are the stress and strain tensors, respectively.

The compliance tensor ${\displaystyle \mathbf {K} }$ is defined from the inverse stress-strain relation:

${\displaystyle E^{ij}=K^{ijkl}T_{kl}}$

The two are related by

${\displaystyle K_{ijpq}C^{pqkl}={\frac {1}{2}}\left(\delta _{i}^{k}\delta _{j}^{l}+\delta _{i}^{l}\delta _{j}^{k}\right)}$

where ${\displaystyle \delta _{n}^{m}}$ is the Kronecker delta.^{[4][5][note 2]}

Unless otherwise noted, this article assumes ${\displaystyle \mathbf {C} }$ is defined from the stress-strain relation of a linear elastic material, in the limit of small strain.

For an isotropic material, ${\displaystyle \mathbf {C} }$ simplifies to

${\displaystyle C^{ijkl}=\lambda \!\left(X\right)g^{ij}g^{kl}+\mu \!\left(X\right)\left(g^{ik}g^{jl}+g^{il}g^{kj}\right)}$

where ${\displaystyle \lambda }$ and ${\displaystyle \mu }$ are scalar functions of the material coordinates ${\displaystyle X}$, and ${\displaystyle \mathbf {g} }$ is the metric tensor in the reference frame of the material.^[6][7] 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:

${\displaystyle C_{ijkl}=\lambda \!\left(X\right)\delta _{ij}\delta _{kl}+\mu \!\left(X\right)\left(\delta _{ik}\delta _{jl}+\delta _{il}\delta _{kj}\right)\quad {\text{[Cartesian coordinates]}}}$

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

${\displaystyle T^{ij}=\lambda \!\left(X\right)\cdot \left(\mathrm {Tr} \,\mathbf {E} \right)g^{ij}+2\mu \!\left(X\right)E^{ij}}$

where ${\displaystyle \mathrm {Tr} \,\mathbf {E} \equiv E_{\,i}^{i}}$ is the trace of ${\displaystyle \mathbf {E} }$. In this form, ${\displaystyle \mu }$ and ${\displaystyle \lambda }$ can be identified with the first and second Lamé parameters. An equivalent expression is

${\displaystyle T^{ij}=K\!\left(X\right)\cdot \left(\mathrm {Tr} \,\mathbf {E} \right)g^{ij}+2\mu \!\left(X\right)\Sigma ^{ij}}$

where ${\displaystyle K=\lambda +(2/3)\mu }$ is the bulk modulus, and

${\displaystyle \Sigma ^{ij}\equiv E^{ij}-(1/3)\left(\mathrm {Tr} \,\mathbf {E} \right)g^{ij}}$

are the components of the shear tensor ${\displaystyle \mathbf {\Sigma } }$.

The elasticity tensor of a cubic crystal has components

${\displaystyle {\begin{aligned}C^{ijkl}&=\lambda g^{ij}g^{kl}+\mu \left(g^{ik}g^{jl}+g^{il}g^{kj}\right)\\&+\alpha \left(a^{i}a^{j}a^{k}a^{l}+b^{i}b^{j}b^{k}b^{l}+c^{i}c^{j}c^{k}c^{l}\right)\end{aligned}}}$

where ${\displaystyle \mathbf {a} }$, ${\displaystyle \mathbf {b} }$, and ${\displaystyle \mathbf {c} }$ are unit vectors corresponding to the three mutually perpendicular axes of the crystal unit cell.^[8] The coefficients ${\displaystyle \lambda }$, ${\displaystyle \mu }$, and ${\displaystyle \alpha }$ 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.^[9]

In an orthonormal Cartesian coordinate basis, there is no distinction between upper and lower indices, and ${\displaystyle g^{ij}}$ is the Kronecker delta, so the expression simplifies to

${\displaystyle {\begin{aligned}C_{ijkl}&=\lambda \delta _{ij}\delta _{kl}+\mu \left(\delta _{ik}\delta _{jl}+\delta _{il}\delta _{kj}\right)\\&+\alpha \left(a_{i}a_{j}a_{k}a_{l}+b_{i}b_{j}b_{k}b_{l}+c_{i}c_{j}c_{k}c_{l}\right)\end{aligned}}}$

There are similar expressions for the components of ${\displaystyle \mathbf {C} }$ in other crystal symmetry classes.^[10] The number of independent elastic constants for several of these is given in table 1.^[9]

The elasticity tensor has several symmetries that follow directly from its defining equation ${\displaystyle T^{ij}=C^{ijkl}E_{kl}}$.^[11][2] The symmetry of the stress and strain tensors implies that

${\displaystyle C_{ijkl}=C_{jikl}\qquad {\text{and}}\qquad C_{ijkl}=C_{ijlk},}$

Usually, one also assumes that the stress derives from an elastic energy potential ${\displaystyle U}$:

${\displaystyle T^{ij}={\frac {\partial U}{\partial E_{ij}}}}$

which implies

${\displaystyle C_{ijkl}={\frac {\partial ^{2}U}{\partial E_{ij}\partial E_{kl}}}}$

Hence, ${\displaystyle \mathbf {C} }$ must be symmetric under interchange of the first and second pairs of indices:

${\displaystyle C_{ijkl}=C_{klij}}$

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.^[9]

Under rotation, the components ${\displaystyle C^{ijkl}}$ transform as

${\displaystyle C'_{ijkl}=R_{ip}R_{jq}R_{kr}R_{ls}C^{pqrs}}$

where ${\displaystyle C'_{ijkl}}$ are the covariant components in the rotated basis, and ${\displaystyle R_{ij}}$ are the elements of the corresponding rotation matrix. A similar transformation rule holds for other linear transformations.

The components of ${\displaystyle \mathbf {C} }$ 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.

${\displaystyle \mathbf {C} }$ possesses two linear invariants and seven quadratic invariants with respect to SO(3).^[12] The linear invariants are

${\displaystyle {\begin{aligned}L_{1}&=C_{\,\,\,ij}^{ij}\\L_{2}&=C_{\,\,\,jj}^{ii}\end{aligned}}}$

and the quadratic invariants are

${\displaystyle \left\{L_{1}^{2},\,L_{2}^{2},\,L_{1}L_{2},\,C_{ijkl}C^{ijkl},\,C_{iikl}C^{jjkl},\,C_{iikl}C^{jkjl},\,C_{kiil}C^{kjjl}\right\}}$

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.^[12]

A common strategy in tensor analysis is to decompose a tensor into simpler components that can be analyzed separately. For example, the displacement gradient tensor ${\displaystyle \mathbf {W} =\mathbf {\nabla } \mathbf {\xi } }$ can be decomposed as

${\displaystyle \mathbf {W} ={\frac {1}{3}}\Theta \mathbf {g} +\mathbf {\Sigma } +\mathbf {R} }$

where ${\displaystyle \Theta }$ is a rank-0 tensor (a scalar), equal to the trace of ${\displaystyle \mathbf {W} }$; ${\displaystyle \mathbf {\Sigma } }$ is symmetric and trace-free; and ${\displaystyle \mathbf {R} }$ is antisymmetric.^[13] Component-wise,

${\displaystyle {\begin{aligned}\Sigma ^{ij}\equiv W^{(ij)}&={\frac {1}{2}}\left(W^{ij}+W^{ji}\right)-{\frac {1}{3}}\left(\mathrm {Tr} \,\mathbf {W} \right)g^{ij}\\R^{ij}\equiv W^{[ij]}&={\frac {1}{2}}\left(W^{ij}-W^{ji}\right)\end{aligned}}}$

Here and later, symmetrization and antisymmetrization are denoted by ${\displaystyle (ij)}$ and ${\displaystyle [ij]}$, 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.^[11]

The elasticity tensor has rank 4, and its decompositions are more complex and varied than those of a rank-2 tensor.^[14] A few examples are described below.

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

${\displaystyle C^{ijkl}=M^{ijkl}+N^{ijkl}}$

where

${\displaystyle {\begin{aligned}M^{ijkl}\equiv C^{i(jk)l}={\frac {1}{2}}\left(C^{ijkl}+C^{ikjl}\right)\\N^{ijkl}\equiv C^{i[jk]l}={\frac {1}{2}}\left(C^{ijkl}-C^{ikjl}\right)\end{aligned}}}$

A disadvantage of this decomposition is that ${\displaystyle M^{ijkl}}$ and ${\displaystyle N^{ijkl}}$ do not obey all original symmetries of ${\displaystyle C^{ijkl}}$, 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.^[2]

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 ${\displaystyle \mathbf {S} }$ can be constructed from ${\displaystyle \mathbf {C} }$ by summing over all ${\displaystyle 4!=24}$ permutations of the indices

${\displaystyle {\begin{aligned}S^{ijkl}&={\frac {1}{4!}}\sum _{(i,j,k,l)\in S_{4}}C^{ijkl}\\&={\frac {1}{4!}}\left(C^{ijkl}+C^{jikl}+C^{ikjl}+\ldots \right)\end{aligned}}}$

where ${\displaystyle \mathbb {S} _{4}}$ is the set of all permutations of the four indices.^[2] Owing to the symmetries of ${\displaystyle C^{ijkl}}$, this sum reduces to

${\displaystyle S^{ijkl}={\frac {1}{3}}\left(C^{ijkl}+C^{iklj}+C^{iljk}\right)}$

The difference

${\displaystyle A^{ijkl}\equiv C^{ijkl}-S^{ijkl}={\frac {1}{3}}\left(2C^{ijkl}-C^{ilkj}-C^{iklj}\right)}$

is an asymmetric tensor (not antisymmetric). The decomposition ${\displaystyle C^{ijkl}=S^{ijkl}+A^{ijkl}}$ can be shown to be unique and irreducible with respect to ${\displaystyle \mathbb {S} _{4}}$. In other words, any additional symmetrization operations on ${\displaystyle \mathbf {S} }$ or ${\displaystyle \mathbf {A} }$ 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 ${\displaystyle G(3,\mathbb {R} )}$.^[2][15]

However, this decomposition is not irreducible with respect to the group of rotations SO(3). Instead, ${\displaystyle \mathbf {S} }$ decomposes into three irreducible parts, and ${\displaystyle \mathbf {A} }$ into two:

${\displaystyle {\begin{aligned}C^{ijkl}&=S^{ijkl}+A^{ijkl}\\&=\left(^{(1)}\!S^{ijkl}+\,^{(2)}\!S^{ijkl}+\,^{(3)}\!S^{ijkl}\right)+\,\left(^{(1)}\!A^{ijkl}+^{(2)}\!A^{ijkl}\right)\end{aligned}}}$

See Itin (2020)^[15] for explicit expressions in terms of the components of ${\displaystyle \mathbf {C} }$.

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

${\displaystyle {\mathcal {C}}=\left(^{(1)}\!{\mathcal {C}}\oplus \,^{(2)}\!{\mathcal {C}}\oplus \,^{(3)}\!{\mathcal {C}}\right)\oplus \,\left(^{(4)}\!{\mathcal {C}}\oplus \,^{(5)}\!{\mathcal {C}}\right)}$

with dimensions

${\displaystyle 21=(1\oplus 5\oplus 9)\oplus (1\oplus 5)}$

These subspaces are each isomorphic to a harmonic tensor space ${\displaystyle \mathbb {H} _{n}(\mathbb {R} ^{3})}$.^[15][16] Here, ${\displaystyle \mathbb {H} _{n}(\mathbb {R} ^{3})}$ is the space of 3D, totally symmetric, traceless tensors of rank ${\displaystyle n}$. In particular, ${\displaystyle ^{(1)}\!{\mathcal {C}}}$ and ${\displaystyle ^{(4)}\!{\mathcal {C}}}$ correspond to ${\displaystyle \mathbb {H} _{0}}$, ${\displaystyle ^{(2)}\!{\mathcal {C}}}$ and ${\displaystyle ^{(5)}\!{\mathcal {C}}}$ correspond to ${\displaystyle \mathbb {H} _{2}}$, and ${\displaystyle ^{(3)}\!{\mathcal {C}}}$ corresponds to ${\displaystyle \mathbb {H} _{4}}$.

• Continuum mechanics
• Solid mechanics
• Constitutive equation
• Strength of materials
• List of materials properties § Mechanical properties
• Representation theory of finite groups
• Voigt notation