In classical elasticity, it is assumed that the stress at each point depends only on the elastic strain components at that same point.¹ In other words, the theory is local — the mechanical response at a point is determined solely by the state of deformation there: \[ \sigma_{ij} = F_{ij}(\epsilon_{11}, \epsilon_{22}, \dots, \epsilon_{12}) \quad (i,j=1,2,3) \] Additionally, we assume that there is no stress if all components of strain are zero (unless there are residual stresses).

Stiffness and Compliance Tensors

Expanding \(F_{ij}\) in power series: \[ \begin{aligned} \sigma_{11}&=C_{1111}\epsilon_{11}+C_{1112}\epsilon_{12}+C_{1113}\epsilon_{13}+\cdots+C_{1133}\epsilon_{33}\\ \sigma_{12}&=C_{1211}\epsilon_{12}+C_{1112}\epsilon_{12}+C_{1213}\epsilon_{13}+\cdots+C_{1233}\epsilon_{33}\\ \vdots\\ \sigma_{33}&=C_{3311}\epsilon_{11}+C_{3312}\epsilon_{12}+C_{3313}\epsilon_{13}+\cdots+C_{3333}\epsilon_{33} \end{aligned}\tag{*} \] or more concisely: \[ \sigma_{ij}=\sum_{k=1}^3\sum_{l=1}^3 C_{ijkl} \epsilon_{kl}. \] where \(C_{ijkl}\) is called the stiffness tensor or the tensor of elastic constants or elastic moduli of the material.²

In general, \(C_{ijkl}=C_{ijkl}(\mathbf{x})\); that is, the elastic coefficients vary from point to point. If \(C_{ijkl}\) is independent of the position, we say that the material is homogeneous.

The system of linear equations in (*) can be solved for \(\epsilon_{11}\), \(\epsilon_{12}\), …, \(\epsilon_{33}\), and we can express them in terms of \(\sigma_{11}\), \(\sigma_{12}\), …, \(\sigma_{33}\). The result will be another system of linear equations: \[ \begin{aligned} \epsilon_{11}&=S_{1111}\sigma_{11}+S_{1112}\sigma_{12}+\cdots+S_{1133}\sigma_{33}\\ \vdots\\ \epsilon_{33}&=S_{3311}\sigma_{11}+S_{3312}\sigma_{12}+\cdots+S_{3333}\sigma_{33} \end{aligned} \] or more concisely: \[ \epsilon_{ij}=\sum_{k=1}^3\sum_{l=1}^3S_{ijkl}\sigma_{kl}. \] Here we are dealing with another fourth rank tensor S_ijkl, which is called compliance tensor.

Since \(i, j, k\), and \(l\), each vary between 1, 2, and 3, \(C_{ijkl}\) (or \(S_{ijkl}\)) has 81 components. However, not all of them are independent. Using some symmetry, we can reduce the number of independent components. In this section, we will see that, if the material is isotropic, meaning that its properties are the same in all directions, only two elastic moduli are enough to specify all stiffness components.

The first thing that we can use to reduce the number of independent components is the symmetry of stress and strain. Symmetry of stress \(\sigma_{ij}=\sigma_{ji}\) implies that is symmetric with respect to the first two indices: \[C_{{\color{red}{ij}}kl}=C_{{\color{red}{ji}}kl}.\]Similarly, symmetry of stain \(\epsilon_{kl}=\epsilon_{lk}\) implies that \(C_{ijkl}\) is symmetric with respect to the last two indices \[ C_{ij{\color{red}{kl}}}=C_{ij{\color{red}{lk}}}. \] Since each pair of \((ij)\) and \((kl)\) can take on six different value (1,1), (2,2), (3,3), (1, 2) = (2,1), (1,3) = (3,1), (2, 3)=(3,2), there are only 36 different elastic constants, at most.

Voigt Notation

Using double notation (e.g., σᵢⱼ) and dealing with a fourth-rank tensor C_ijkl is cumbersome. Therefore, we introduce Voigt notation to simplify the representation of stress and strain.

The stress tensor components are arranged into a 6x1 column vector: \[ \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{13} \\ \sigma_{12} \end{bmatrix} = \begin{bmatrix} \sigma_1 \\ \sigma_2 \\ \sigma_3 \\ \sigma_4 \\ \sigma_5 \\ \sigma_6 \end{bmatrix} \] That is, we number them according to the following: \[ \begin{bmatrix} ① & {\color{red}{⑥}} & {\color{red}{⑤}}\\ & ② & {\color{red}{④}}\\ & & ③ \end{bmatrix} \] Similarly, the corresponding strain components are arranged into a 6x1 vector. Note the factors of 2 for the shear strains, which are introduced to ensure work conjugacy between the stress and strain vectors.

\[ \begin{bmatrix} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ 2\epsilon_{23} \\ 2\epsilon_{13} \\ 2\epsilon_{12} \end{bmatrix} = \begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \epsilon_3 \\ \epsilon_4 \\ \epsilon_5 \\ \epsilon_6 \end{bmatrix} = \begin{bmatrix} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \gamma_{23} \\ \gamma_{13} \\ \gamma_{12} \end{bmatrix} \]

Note: Do not confuse σ₁, σ₂, σ₃ and ε₁, ε₂, ε₃ with the principal values of stresses and strains. Here, we are using them with a totally different meaning.

Elastic Coefficient Matrix

Using Voigt notation, the linear relationship between stress and strain can be written using a matrix [c_mn]:

\[ \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{13} \\ \sigma_{12} \end{bmatrix} = \begin{bmatrix} c_{11} & c_{12} & c_{13} & c_{14} & c_{15} & c_{16} \\ c_{21} & c_{22} & c_{23} & c_{24}& c_{25}& c_{26}\\ \vdots & & \ddots & & & \vdots \\ & & & & & \\ & & & & & \\ c_{61}& & \dots & & & c_{66} \end{bmatrix} \begin{bmatrix} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \gamma_{23} \\ \gamma_{13} \\ \gamma_{12} \end{bmatrix} \]

From the above, it is clear that to express stress in terms of strain, we need 36 independent constants in the stiffness matrix [c].

Note: While Cᵢⱼₖₗ is a fourth-order tensor, the 6x6 matrix cₘₙ in Voigt notation is not a tensor. Therefore, we cannot use tensor transformation rules to find its components in a new coordinate system. We must transform the stress and strain components first and then deduce the new stiffness matrix.

Strain Energy

Previously we showed that the strain energy density, U₀,is given by \[ \begin{aligned} dU_0&=\sigma_{11} d\epsilon_{11}+\sigma_{12} d\epsilon_{12}+\cdots+\sigma_{33} d\epsilon_{33}\\ &=\sum_{i=1}^3\sum_{j=1}^3 \sigma_{ij}\ d\epsilon_{ij} \end{aligned} \] If there is a strain energy density function we can express stress as: \[ \sigma_{ij} = \frac{\partial U_0}{\partial \epsilon_{ij}} \] Using Voigt notation, we can write the stress energy density as \[ dU_0 = \sigma_1 d\epsilon_1 + \sigma_2 d\epsilon_2 + \sigma_3 d\epsilon_3 + \sigma_4 d\epsilon_4 + \dots + \sigma_6 d\epsilon_6 \] and \[ \sigma_{i} = \frac{\partial U_0}{\partial \epsilon_{i}} \] If the material is linear elastic, we have \(\sigma_i = \sum_{j=1}^{6} c_{ij} \epsilon_j\). Substituting this into the expression for dU₀: \[ dU_0 = \sum_{i=1}^{6} \sigma_i d\epsilon_i = \sum_{i=1}^{6} \sum_{k=1}^{6} c_{ik} \epsilon_k\, d\epsilon_i \] Integrating this expression gives the strain energy density: \[ U_0 = \sum_{i=1}^{6} \sum_{k=1}^{6} \frac{1}{2} c_{ik} \epsilon_k \epsilon_i \] By taking partial derivatives twice, we find: \[ \frac{\partial^2 U_0}{\partial \epsilon_i \partial \epsilon_j} = \frac{\partial}{\partial \epsilon_i}\left(\frac{\partial U_0}{\partial \epsilon_j}\right)=\frac{\partial \sigma_j}{\partial \epsilon_i} = c_{ji} \] \[ \frac{\partial^2 U_0}{\partial \epsilon_j \partial \epsilon_i} = \frac{\partial \sigma_i}{\partial \epsilon_j} = c_{ij} \] Since the order of differentiation does not matter for a function³: \[ \frac{\partial^2 U_0}{\partial \epsilon_i \partial \epsilon_j} = \frac{\partial^2 U_0}{\partial \epsilon_j \partial \epsilon_i} \] we must have:⁴ \[ c_{ji} = c_{ij} \] This means the stiffness matrix [c] must be symmetric. This symmetry reduces the number of different elastic constants to 21 (6 diagonal elements + 15 above diagonal elements).

The Effect of Material Symmetry

The number of independent elastic constants (21 for the most general anisotropic case) can be reduced if the material’s structure has some form of symmetry.

Material with One Plane of Symmetry

Let’s assume the material has a plane of material symmetry at z=0 (the xy-plane). This means the material’s response is identical if we reflect the coordinate system across this plane. Let’s define a new coordinate system (x’, y’, z’) such that: \[ x' = x, \quad y' = y, \quad z' = -z \] In the new coordinate system, due to the symmetry of the material, elastic coefficients remain invariant, but under this transformation, the stress and strain components change as follows: \[ \begin{aligned} \sigma_{x'x'} &= \sigma_{xx} & \sigma_{x'y'} &= \sigma_{xy} & \sigma_{x'z'} &= -\sigma_{xz} \\ \sigma_{y'y'} &= \sigma_{yy} & \sigma_{y'z'} &= -\sigma_{yz} & \sigma_{z'z'} &= \sigma_{zz} \end{aligned} \] \[ \begin{aligned} \epsilon_{x'x'} &= \epsilon_{xx} & \epsilon_{x'y'} &= \epsilon_{xy} & \epsilon_{x'z'} &= -\epsilon_{xz} \\ \epsilon_{y'y'} &= \epsilon_{yy} & \epsilon_{y'z'} &= -\epsilon_{yz} & \epsilon_{z'z'} &= \epsilon_{zz} \end{aligned} \] The constitutive relation for σₓₓ must be the same in both coordinate systems. In the original system: \[ \sigma_{xx} = c_{11}\epsilon_{xx} + c_{12}\epsilon_{yy} + c_{13}\epsilon_{zz} + c_{14}\gamma_{yz} + c_{15}\gamma_{xz} + c_{16}\gamma_{xy} \] In the new system: \[ \sigma_{x'x'} = c_{11}\epsilon_{x'x'} + c_{12}\epsilon_{y'y'} + c_{13}\epsilon_{z'z'} + c_{14}\gamma_{y'z'} + c_{15}\gamma_{x'z'} + c_{16}\gamma_{x'y'} \] Substituting the transformed components: \[ \sigma_{xx} = c_{11}\epsilon_{xx} + c_{12}\epsilon_{yy} + c_{13}\epsilon_{zz} - c_{14}\gamma_{yz} - c_{15}\gamma_{xz} + c_{16}\gamma_{xy} \] For the two equations to be identical for all possible strains, we must have: \[ c_{14} = -c_{14} \implies c_{14} = 0 \] \[ c_{15} = -c_{15} \implies c_{15} = 0 \] By applying the same logic to other stress components, we find that c₂₄ = c₂₅ = c₃₄ = c₃₅ = c₄₆ = c₅₆ = 0. The elastic matrix simplifies to: \[ [c_{mn}] = \begin{bmatrix} c_{11} & c_{12} & c_{13} & 0 & 0 & c_{16} \\ c_{12} & c_{22} & c_{23} & 0 & 0 & c_{26} \\ c_{13} & c_{23} & c_{33} & 0 & 0 & c_{36} \\ 0 & 0 & 0 & c_{44} & c_{45} & 0 \\ 0 & 0 & 0 & c_{45} & c_{55} & 0 \\ c_{16} & c_{26} & c_{36} & 0 & 0 & c_{66} \end{bmatrix} \] This reduces the number of elastic constants required to specify the stiffness tensor to 13.

Orthotropic Material

An orthotropic material has three mutually perpendicular planes of symmetry and three corresponding orthogonal axes. Many materials, such as wood, rolled metal sheet, and fiber-reinforced laminates, reinforced concrete can be treated as orthotropic.

If we align our coordinate axes with the normals of these planes, the matrix simplifies further. Applying a second symmetry plane (e.g., the xz-plane) will zero out additional coefficients. The resulting matrix for an orthotropic material is: \[ [c_{mn}] = \begin{bmatrix} c_{11} & c_{12} & c_{13} & 0 & 0 & 0 \\ c_{12} & c_{22} & c_{23} & 0 & 0 & 0 \\ c_{13} & c_{23} & c_{33} & 0 & 0 & 0 \\ 0 & 0 & 0 & c_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & c_{55} & 0 \\ 0 & 0 & 0 & 0 & 0 & c_{66} \end{bmatrix} \] The number of independent constants is now 9.

Cubic Material

A cubic material has the symmetry of a cube. In addition to being orthotropic, its properties are identical if we swap the coordinate axes (e.g., x → y, y → z, z → x). This imposes further constraints: \[ \begin{aligned} c_{11} = c_{22} = c_{33} \\ c_{12} = c_{13} = c_{23} \\ c_{44} = c_{55} = c_{66} \end{aligned} \] The matrix of elastic constants for a cubic material has only 3 independent constants: \(c_{11}, c_{12}\) ,and \(c_{44}\): \[ [c_{mn}] = \begin{bmatrix} c_{11} & c_{12} & c_{12} & 0 & 0 & 0 \\ c_{12} & c_{11} & c_{12} & 0 & 0 & 0 \\ c_{12} & c_{12} & c_{11} & 0 & 0 & 0 \\ 0 & 0 & 0 & c_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & c_{44} & 0 \\ 0 & 0 & 0 & 0 & 0 & c_{44} \end{bmatrix} \] Many materials including those with FCC, BCC, and diamond cubic structures have cubic symmetry. Examples of such materials include aluminum, copper, nickel, silver, gold, iron, silicon, germanium.

Isotropic Material

An isotropic material has the same properties in every direction. This is the highest level of material symmetry. For a material to be isotropic, its constitutive relations (and thus its elastic constants) must remain unchanged after any arbitrary rotation of the coordinate system.

We begin with the stiffness matrix for a cubic material, which has three independent constants (\(c_{11}\), \(c_{12}\), and \(c_{44}\)). An isotropic material is a special case of a cubic material, so we can find the additional constraint required for isotropy by enforcing rotational invariance.

This leaves only 2 independent constants for an isotropic material. These are typically expressed as the Lamé parameters, λ and μ (where μ is the shear modulus, G). \[ \begin{aligned} c_{44} &= \mu = G \\ c_{12} &= \lambda \\ \end{aligned} \] \[ \Rightarrow c_{11} = \lambda + 2\mu \] Substituting these into the constitutive equations, for example σ₁₁: \[ \begin{aligned} \sigma_{11} &= c_{11}\epsilon_{11} + c_{12}\epsilon_{22} + c_{12}\epsilon_{33} \\ &= (\lambda+2\mu)\epsilon_{11} + \lambda\epsilon_{22} + \lambda\epsilon_{33} \\ &= \lambda(\epsilon_{11}+\epsilon_{22}+\epsilon_{33}) + 2\mu\epsilon_{11} \end{aligned} \] Similarly, for all components: \[ \begin{aligned} \sigma_{22} &= \lambda(\epsilon_{11}+\epsilon_{22}+\epsilon_{33}) + 2\mu\epsilon_{22} \\ \sigma_{33} &= \lambda(\epsilon_{11}+\epsilon_{22}+\epsilon_{33}) + 2\mu\epsilon_{33} \\ \sigma_{23} &= 2\mu\epsilon_{23} \\ \sigma_{13} &= 2\mu\epsilon_{13} \\ \sigma_{12} &= 2\mu\epsilon_{12} \end{aligned} \] These relationships can be written compactly using tensor notation: \[ \boxed{\sigma_{ij} = \lambda \left(\sum_{k=1}^{3} \epsilon_{kk}\right) \delta_{ij} + 2\mu\epsilon_{ij}} \] where δᵢⱼ is the Kronecker delta: \[ \delta_{ij} = \begin{cases} 1 & \text{if } i=j \\ 0 & \text{if } i \neq j \end{cases} \] Using the Einstein summation convention (where summation is implied over a repeated index), the equation becomes even more compact: \[ \bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{\sigma_{ij} = \lambda \epsilon_{kk} \delta_{ij} + 2\mu\epsilon_{ij}} \]

Anisotropy Factor

In the section on isotropic materials, we derived the specific condition that the elastic constants of a cubic material must satisfy for it to be isotropic: \[ c_{44} = \frac{c_{11} - c_{12}}{2} \] This relationship provides a useful way to quantify the degree of elastic anisotropy in a cubic crystal. We can rearrange this expression into a ratio. This gives rise to the Zener anisotropy ratio (or anisotropy factor), denoted by A, which is defined as: \[ A = \frac{2c_{44}}{c_{11} - c_{12}} \] This dimensionless factor provides a measure of how anisotropic a material is:

• If A = 1, the condition for isotropy is perfectly met. The material’s elastic properties are the same in all directions.
• If A deviates from 1, the material is anisotropic. The magnitude of the deviation indicates the degree of anisotropy. For many cubic metals, this value can be significantly different from 1. For example, the value for iron is approximately 2.41, while for niobium, it is 0.49.

Elastic Constants of Various Materials

Reference

References

1. Nye, J. F. (1985). Physical properties of crystals: Their representation by tensors and matrices. Oxford University Press.
2. Shames, I. H., & Cozzarelli, F. A. (1992). Elastic and inelastic stress analysis (2nd ed.). Prentice Hall.
3. Sokolnikoff, I. S. (1956). Mathematical theory of elasticity (2nd ed.). McGraw-Hill.