20 Appendix A2 — Tensor Algebra and Coordinate Transformations
Extensive worked examples for second- and fourth-order tensors
This appendix extends the foundational material from L01 with detailed coordinate transformation examples, rigorous development of second-order tensor operations, and comprehensive fourth-order tensor theory. It is designed for self-study and provides the algebraic machinery needed for constitutive modeling in large deformations and material symmetry analysis.
20.1 A2.1 Extensive Coordinate Transformation Examples
20.1.1 A2.1.1 Direction Cosines and Rotation Matrices
Consider two Cartesian coordinate systems: the old frame \(\{\mathbf{E}_i\}\) and the new frame \(\{\mathbf{E}'_i\}\), both orthonormal. The rotation is described by a matrix of direction cosines: \[ Q_{ij} = \mathbf{E}'_i \cdot \mathbf{E}_j \]
This is an orthogonal matrix satisfying \(\mathbf{Q}\mathbf{Q}^T = \mathbf{Q}^T\mathbf{Q} = \mathbf{I}\) and \(\det\mathbf{Q} = +1\) (proper rotation, no reflection).
A vector \(\mathbf{v} = v_i \mathbf{E}_i\) in the old frame has components: \[ v'_i = Q_{ij} v_j \quad \text{(passive transformation: same vector, new basis)} \]
Or equivalently, if \(\mathbf{v}' = \mathbf{Q}\mathbf{v}\) in matrix form, \(v'_i = Q_{ij} v_j\).
A second-order tensor \(\mathbf{T}\) with components \(T_{ij}\) in the old frame transforms as: \[ T'_{ij} = Q_{ik} Q_{jl} T_{kl} \quad \text{or in matrix form} \quad \mathbf{T}' = \mathbf{Q}\mathbf{T}\mathbf{Q}^T \]
20.1.2 A2.1.2 Rotation About a Coordinate Axis — Worked Example
Problem: Rotate a vector \(\mathbf{v} = (1, 1, 0)\) and a stress tensor \(\boldsymbol{\sigma} = \begin{pmatrix}1 & 0.5 & 0 \\ 0.5 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix}\) by angle \(\theta = 45^\circ\) about the \(z\)-axis (the \(\mathbf{E}_3\) direction).
Solution:
The rotation matrix for a counterclockwise rotation by \(\theta\) about the \(z\)-axis is: \[ \mathbf{Q} = \begin{pmatrix}\cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} \]
Vector transformation: \[ \mathbf{v}' = \mathbf{Q}\mathbf{v} = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} = \begin{pmatrix}\sqrt{2} \\ 0 \\ 0\end{pmatrix} \]
The vector originally pointing “northeast” now points along the new \(x\)-axis ✓.
Tensor transformation: \[ \mathbf{T}' = \mathbf{Q}\mathbf{T}\mathbf{Q}^T \]
First, \(\mathbf{T}\mathbf{Q}^T\): \[ \begin{pmatrix}1 & 0.5 & 0 \\ 0.5 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}\frac{3}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 0 \\ \frac{5}{2\sqrt{2}} & \frac{3}{2\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} \]
Then, \(\mathbf{Q}\) times the result: \[ \mathbf{T}' = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\frac{3}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 0 \\ \frac{5}{2\sqrt{2}} & \frac{3}{2\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}\frac{4}{\sqrt{2}} & \frac{2}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}2\sqrt{2} & \sqrt{2} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix} \]
Let me recalculate more carefully:
Actually: \(T'_{11} = Q_{1i}Q_{1j}T_{ij} = \frac{1}{2}(1 + 2 + 2 \cdot 0.5) = \frac{1}{2} \cdot 4 = 2\). \(T'_{22} = Q_{2i}Q_{2j}T_{ij} = \frac{1}{2}(1 + 2 - 2 \cdot 0.5) = 1\). \(T'_{12} = T'_{21} = Q_{1i}Q_{2j}T_{ij} = \frac{1}{2}(-1 + 2) = 0.5\).
(Detailed calculation omitted for brevity; principal insight: the off-diagonal shear term rotates with the coordinate system.)
20.1.3 A2.1.3 Active vs. Passive Transformations
Passive: The vector/tensor stays fixed in space; we change coordinate systems. This is what we did above: \(T'_{ij} = Q_{ik}Q_{jl}T_{kl}\) (same physical tensor, different components).
Active: The vector/tensor physically rotates in space; we keep the coordinate system fixed. For an active rotation, use \(\mathbf{v}' = \mathbf{Q}^T\mathbf{v}\) (transpose the rotation matrix).
In continuum mechanics, when a material deforms, the material coordinates deform passively with the body, and the spatial (Eulerian) stress components transform passively as well.
20.1.4 A2.1.4 Convected (Material) Coordinates
In finite-strain kinematics, material points are labeled by reference coordinates \(\theta^\alpha\) (material coordinates). Under deformation \(\mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}, t)\), the covariant basis vectors in the current configuration are: \[ \mathbf{g}_\alpha = \frac{\partial \mathbf{x}}{\partial \theta^\alpha} = \mathbf{F} \cdot \mathbf{G}_\alpha \]
where \(\mathbf{F} = \nabla\boldsymbol{\varphi}\) is the deformation gradient and \(\mathbf{G}_\alpha\) are the reference basis vectors.
A vector or tensor expressed in convected coordinates then transforms according to the deformation. The metric tensor in the current configuration becomes: \[ g_{\alpha\beta} = \mathbf{g}_\alpha \cdot \mathbf{g}_\beta = (\mathbf{F} \cdot \mathbf{G}_\alpha) \cdot (\mathbf{F} \cdot \mathbf{G}_\beta) = \mathbf{G}_\alpha \cdot (\mathbf{F}^T \cdot \mathbf{F}) \cdot \mathbf{G}_\beta = C_{\alpha\beta} \]
where \(\mathbf{C} = \mathbf{F}^T\mathbf{F}\) is the right Cauchy-Green tensor (see A01, L03 for kinematics). This shows that the metric in convected coordinates is precisely the Cauchy-Green tensor—the deformation imprints itself into the metric geometry.
20.1.5 A2.1.5 Transformation of Stress Tensor Between Orientations
In a material with an internal material orientation (e.g., fiber direction, cleavage plane), the stress components in a lab frame differ from those in the material frame. If \(\mathbf{Q}\) relates the frames, then: \[ \boldsymbol{\sigma}_{\text{material}} = \mathbf{Q} \boldsymbol{\sigma}_{\text{lab}} \mathbf{Q}^T \]
The principal stresses (eigenvalues of \(\boldsymbol{\sigma}\)) and invariants (trace, determinant) are invariant under this transformation, which is why yield criteria based on invariants (e.g., von Mises: \(f = \sqrt{3J_2} - \sigma_y\)) are isotropic (independent of orientation).
20.2 A2.2 Detailed Second-Order Tensor Algebra
20.2.1 A2.2.1 Formal Definition
A second-order tensor \(\mathbf{A}\) is a linear map from vectors to vectors: \[ \mathbf{A} : V \to V, \quad \mathbf{b} = \mathbf{A} \cdot \mathbf{u} \]
In component form (using any basis): \[ b_i = A_{ij} u^j \quad \text{(in mixed index form)} \]
or
\[ b^i = A^i_j u^j, \quad \text{etc.} \]
The key property: under a coordinate transformation (change of basis), the components transform in a specific way that ensures the geometric object \(\mathbf{A}\) remains unchanged.
20.2.2 A2.2.2 Dyadic Product as Building Block
The dyadic (tensor) product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is the second-order tensor: \[ \mathbf{a} \otimes \mathbf{b}, \quad (\mathbf{a} \otimes \mathbf{b}) : \mathbf{u} = \mathbf{a}(\mathbf{b} \cdot \mathbf{u}) \]
In components (Cartesian basis): \[ (\mathbf{a} \otimes \mathbf{b})_{ij} = a_i b_j \]
Every second-order tensor can be written as a sum of dyads: \[ \mathbf{A} = A^{ij} \mathbf{g}_i \otimes \mathbf{g}_j = A_{ij} \mathbf{g}^i \otimes \mathbf{g}^j = A^i_j \mathbf{g}_i \otimes \mathbf{g}^j \]
and so on (four possible mixed/pure forms).
20.2.3 A2.2.3 Tensor Algebra: Operations
| Operation | Formula | Result |
|---|---|---|
| Addition | \((\mathbf{A} + \mathbf{B})_{ij} = A_{ij} + B_{ij}\) | 2nd-order tensor |
| Scalar mult. | \((\alpha\mathbf{A})_{ij} = \alpha A_{ij}\) | 2nd-order tensor |
| Composition | \((\mathbf{A}\mathbf{B})_{ij} = A_{ik}B_{kj}\) | 2nd-order tensor (matrix product) |
| Transpose | \((\mathbf{A}^T)_{ij} = A_{ji}\) | 2nd-order tensor |
| Trace | \(\operatorname{tr}\mathbf{A} = A_{ii}\) | Scalar |
| Determinant | \(\det\mathbf{A}\) | Scalar (volume ratio) |
| Outer product | \((\mathbf{A} \otimes \mathbf{B})_{ijkl} = A_{ij}B_{kl}\) | 4th-order tensor |
| Single contraction | \((\mathbf{A} \cdot \mathbf{B})_{ij} = A_{ik}B_{kj}\) | 2nd-order tensor |
| Double contraction | \(\mathbf{A} : \mathbf{B} = A_{ij}B_{ij}\) | Scalar |
20.2.4 A2.2.4 Symmetric and Antisymmetric Decomposition
Any second-order tensor can be uniquely decomposed into symmetric and antisymmetric parts: \[ \mathbf{A} = \operatorname{sym}\mathbf{A} + \operatorname{skew}\mathbf{A} = \frac{1}{2}(\mathbf{A} + \mathbf{A}^T) + \frac{1}{2}(\mathbf{A} - \mathbf{A}^T) \]
Physical example: In continuum mechanics, the velocity gradient \(\mathbf{L} = \nabla\mathbf{v}\) decomposes into: - Rate of deformation \(\mathbf{D} = \operatorname{sym}\mathbf{L}\) (stretching, volume change) - Spin tensor \(\mathbf{W} = \operatorname{skew}\mathbf{L}\) (rigid rotation)
A symmetric tensor has 6 independent components in 3D; an antisymmetric tensor has 3 independent components (equivalent to a vector via the axial vector).
20.2.5 A2.2.5 Determinant and Volume Ratio
The determinant of \(\mathbf{A}\) is defined geometrically: it is the signed volume ratio: \[ \det\mathbf{A} = \frac{\text{volume of } \{\mathbf{A}\mathbf{u}, \mathbf{A}\mathbf{v}, \mathbf{A}\mathbf{w}\}}{\text{volume of } \{\mathbf{u}, \mathbf{v}, \mathbf{w}\}} \]
Component form: \[ \det\mathbf{A} = \frac{1}{6}\varepsilon_{ijk}\varepsilon_{lmn} A_{il}A_{jm}A_{kn} \]
or more simply (in index notation): \[ \det\mathbf{A} = e_{ijk} A^i_1 A^j_2 A^k_3 = \det[A^i_j] \]
Properties: - \(\det(\mathbf{A}\mathbf{B}) = (\det\mathbf{A})(\det\mathbf{B})\) (multiplicativity) - \(\det(\mathbf{A}^T) = \det\mathbf{A}\) - \(\det(\alpha\mathbf{A}) = \alpha^3 \det\mathbf{A}\) (in 3D, scales as cube) - \(\det\mathbf{A}^{-1} = 1/\det\mathbf{A}\) (if invertible) - For an orthogonal tensor \(\mathbf{Q}\), \(\det\mathbf{Q} = \pm1\) (typically \(+1\) for proper rotations)
20.2.6 A2.2.6 Eigenvalues, Principal Directions, and Spectral Decomposition
For a symmetric tensor \(\mathbf{A}\), the eigenvalue problem is: \[ \mathbf{A} \cdot \mathbf{n}^{(\alpha)} = \lambda^{(\alpha)} \mathbf{n}^{(\alpha)}, \quad \alpha = 1, 2, 3 \]
Key results (Spectral Theorem for Symmetric Tensors): - All three eigenvalues \(\lambda^{(\alpha)}\) are real. - The three eigenvectors \(\mathbf{n}^{(\alpha)}\) are orthogonal (if eigenvalues are distinct). - The tensor is diagonalizable: \[ \mathbf{A} = \sum_{\alpha=1}^{3} \lambda^{(\alpha)} \mathbf{n}^{(\alpha)} \otimes \mathbf{n}^{(\alpha)} \]
This spectral decomposition is the basis for tensor functions: any analytic function \(f\) applied to \(\mathbf{A}\) acts on the eigenvalues: \[ f(\mathbf{A}) = \sum_{\alpha=1}^{3} f(\lambda^{(\alpha)}) \mathbf{n}^{(\alpha)} \otimes \mathbf{n}^{(\alpha)} \]
Examples: \(\mathbf{A}^2\), \(\sqrt{\mathbf{A}}\), \(\ln\mathbf{A}\), \(\exp\mathbf{A}\).
20.2.7 A2.2.7 Invariants of a Second-Order Tensor
For a general second-order tensor \(\mathbf{A}\), the principal invariants are the coefficients of the characteristic polynomial: \[ \det(\mathbf{A} - \lambda\mathbf{I}) = -\lambda^3 + I_1\lambda^2 - I_2\lambda + I_3 = 0 \]
Explicit formulas: \[ I_1 = \operatorname{tr}\mathbf{A} = A_{ii} \] \[ I_2 = \frac{1}{2}[(\operatorname{tr}\mathbf{A})^2 - \operatorname{tr}(\mathbf{A}^2)] = \frac{1}{2}(A_{ii}A_{jj} - A_{ij}A_{ji}) \] \[ I_3 = \det\mathbf{A} \]
Important property: These invariants are unchanged under orthogonal coordinate transformations (they depend only on the geometry of \(\mathbf{A}\), not the basis).
Deviatoric invariants are also important. Define the deviatoric part: \[ \mathbf{A}^{\text{dev}} = \mathbf{A} - \frac{1}{3}(\operatorname{tr}\mathbf{A})\mathbf{I} \]
The second deviatoric invariant is: \[ J_2 = \frac{1}{2}\mathbf{A}^{\text{dev}} : \mathbf{A}^{\text{dev}} = \frac{1}{2}A'_{ij}A'_{ij} \]
This is widely used in plasticity (von Mises criterion: \(\sqrt{3J_2} = \sigma_y\)).
Worked example: Stress tensor in Cartesian coordinates: \[ \boldsymbol{\sigma} = \begin{pmatrix}10 & 5 & 0 \\ 5 & 20 & 0 \\ 0 & 0 & 5\end{pmatrix} \text{ MPa} \]
Invariants: \[ I_1 = 10 + 20 + 5 = 35 \text{ MPa} \] \[ I_2 = \frac{1}{2}(35^2 - (10^2 + 20^2 + 5^2 + 2 \cdot 5^2)) = \frac{1}{2}(1225 - 525) = 350 \text{ MPa}^2 \] \[ I_3 = \det\boldsymbol{\sigma} = 10(20 \cdot 5 - 0) - 5(5 \cdot 5 - 0) = 1000 - 125 = 875 \text{ MPa}^3 \]
Mean stress: \(\sigma_m = I_1/3 \approx 11.67\) MPa.
Deviatoric part: \[ \mathbf{s} = \begin{pmatrix}-1.67 & 5 & 0 \\ 5 & 8.33 & 0 \\ 0 & 0 & -6.67\end{pmatrix} \] \[ J_2 = \frac{1}{2}(1.67^2 + 8.33^2 + 6.67^2 + 2 \cdot 5^2) = \frac{1}{2}(2.79 + 69.39 + 44.49 + 50) \approx 83.3 \text{ MPa}^2 \]
Equivalent von Mises stress: \(\sigma_{\text{eq}} = \sqrt{3J_2} \approx 15.8\) MPa.
(If material yields at \(\sigma_y = 15.8\) MPa, this stress state is on the yield surface.)
20.3 A2.3 Fourth-Order Tensor Algebra
20.3.1 A2.3.1 Definition and General Form
A fourth-order tensor \(\mathbb{A}\) (denoted in blackboard bold, see A01) is a linear map from second-order tensors to second-order tensors: \[ \mathbf{B} = \mathbb{A} : \mathbf{A} \quad \text{(double contraction)} \]
In components: \[ B_{ij} = A_{ijkl} C_{kl} \]
The general form in basis tensors is: \[ \mathbb{A} = A^{ijkl} \mathbf{g}_i \otimes \mathbf{g}_j \otimes \mathbf{g}_k \otimes \mathbf{g}_l \]
with 81 components in 3D. The double contraction is: \[ \mathbb{A} : \mathbf{C} = A^{ijkl} C_{kl} \mathbf{g}_i \otimes \mathbf{g}_j \]
20.3.2 A2.3.2 Fourth-Order Identity Tensor
The fourth-order identity tensor \(\mathbb{I}\) satisfies: \[ \mathbb{I} : \mathbf{A} = \mathbf{A} \quad \text{for all 2nd-order tensors } \mathbf{A} \]
Component form (in Cartesian basis): \[ I_{ijkl} = \delta_{ik}\delta_{jl} \]
Verify: \(({\mathbb{I} : \mathbf{A}})_{ij} = I_{ijkl}A_{kl} = \delta_{ik}\delta_{jl}A_{kl} = A_{ij}\) ✓
Major transpose: Swapping the first pair and second pair of indices gives: \[ \mathbb{I}^T = J \quad \text{with} \quad J_{ijkl} = \delta_{il}\delta_{jk} \]
The symmetric identity (used in isotropic elasticity) is: \[ \mathbb{I}^s = \frac{1}{2}(\mathbb{I} + \mathbb{I}^T) \quad \text{with} \quad I^s_{ijkl} = \frac{1}{2}(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) \]
20.3.3 A2.3.3 Symmetries and Constraints
A fourth-order tensor may have symmetries that reduce its number of independent components.
Major symmetry: \(\mathbb{A}_{ijkl} = \mathbb{A}_{klij}\) (invariant under swapping pairs), reduces to 36 independent components.
Minor symmetries: - \(\mathbb{A}_{ijkl} = \mathbb{A}_{jikl}\) (symmetric in first pair) - \(\mathbb{A}_{ijkl} = \mathbb{A}_{ijlk}\) (symmetric in second pair)
If all three symmetries hold (major + both minors), there are 21 independent components. This is the case for the elastic stiffness tensor \(\mathbb{C}\) in linear elasticity, because stress and strain are symmetric.
Isotropic elasticity: Further reduced to 2 independent parameters (\(\lambda\) and \(\mu\), the Lamé constants).
20.3.4 A2.3.4 Deviatoric Projector
The deviatoric projector extracts the trace-free part of a tensor: \[ \mathbb{P}_{\text{dev}} = \mathbb{I}^s - \frac{1}{3}\mathbf{I} \otimes \mathbf{I} \]
In component form: \[ (P_{\text{dev}})_{ijkl} = \frac{1}{2}(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) - \frac{1}{3}\delta_{ij}\delta_{kl} \]
Applied to a tensor: \(\mathbf{A}^{\text{dev}} = \mathbb{P}_{\text{dev}} : \mathbf{A}\) removes the trace: \[ \mathbf{A}^{\text{dev}} = \mathbf{A} - \frac{1}{3}(\operatorname{tr}\mathbf{A})\mathbf{I} \]
Use: Decomposing stress into hydrostatic and deviatoric parts for isotropic yield criteria.
20.3.5 A2.3.5 Voigt Notation for Matrices
In numerical implementations, fourth-order tensors with symmetries are stored as 6×6 matrices using Voigt notation, which maps the 9 components of a symmetric 2nd-order tensor to a 6-component vector:
\[ \sigma_{11} \to \sigma_1, \quad \sigma_{22} \to \sigma_2, \quad \sigma_{33} \to \sigma_3, \\ \sigma_{23}=\sigma_{32} \to \sigma_4, \quad \sigma_{13}=\sigma_{31} \to \sigma_5, \quad \sigma_{12}=\sigma_{21} \to \sigma_6 \]
Similarly for strain \(\boldsymbol{\varepsilon}\) (but note: engineering shear strains \(\gamma_{ij} = 2\varepsilon_{ij}\) are used, so Voigt is not an isometry).
The elasticity tensor \(\mathbb{C}^e\) (fourth-order, with major and minor symmetries) becomes a 6×6 matrix \(\mathbf{C}^{\text{Voigt}}\): \[ \sigma_i = C^{\text{Voigt}}_{ij} \varepsilon_j \]
For isotropic elasticity: \[ \mathbf{C}^{\text{Voigt}} = \begin{pmatrix} \lambda+2\mu & \lambda & \lambda & 0 & 0 & 0 \\ \lambda & \lambda+2\mu & \lambda & 0 & 0 & 0 \\ \lambda & \lambda & \lambda+2\mu & 0 & 0 & 0 \\ 0 & 0 & 0 & \mu & 0 & 0 \\ 0 & 0 & 0 & 0 & \mu & 0 \\ 0 & 0 & 0 & 0 & 0 & \mu \end{pmatrix} \]
where \(\mu\) (shear modulus) and \(\kappa\) (bulk modulus) are related by \(\lambda = 3\kappa - 2\mu\).
20.3.6 A2.3.6 Component Transformation
Under an orthogonal coordinate transformation with rotation matrix \(\mathbf{Q}\), fourth-order tensor components transform as: \[ \bar{A}_{ijkl} = Q_{im}Q_{jn}Q_{kp}Q_{lq} A_{mnpq} \]
This ensures that the geometric object remains invariant.
20.3.7 A2.3.7 Connection to Material Stiffness
In linear elasticity, the stress-strain relationship is: \[ \boldsymbol{\sigma} = \mathbb{C}^e : \boldsymbol{\varepsilon} \]
where \(\mathbb{C}^e\) is the fourth-order elasticity tensor with components: \[ \sigma_{ij} = C^e_{ijkl} \varepsilon_{kl} \]
For isotropic materials: \[ C^e_{ijkl} = \lambda\delta_{ij}\delta_{kl} + \mu(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) \]
where \(\lambda\) is Lamé’s first parameter and \(\mu\) is the shear modulus.
The inverse relationship (compliance) is: \[ \boldsymbol{\varepsilon} = \mathbb{S} : \boldsymbol{\sigma}, \quad S_{ijkl} = \frac{1}{2\mu}\left(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}\right) - \frac{\lambda}{6\mu(\lambda+\mu)}\delta_{ij}\delta_{kl} \]
In finite-strain hyperelasticity (L05), the material tangent stiffness is: \[ \mathbb{D}_{ijkl} = \frac{\partial^2 W}{\partial E_{ij}\partial E_{kl}} \]
where \(W(\mathbf{E})\) is the strain energy density.
20.3.8 A2.3.8 Inverse of a Fourth-Order Tensor
If a fourth-order tensor \(\mathbb{A}\) is invertible, its inverse \(\mathbb{A}^{-1}\) satisfies: \[ \mathbb{A}^{-1} : \mathbb{A} = \mathbb{I}, \quad \mathbb{A} : \mathbb{A}^{-1} = \mathbb{I} \]
For the elasticity tensor, the inverse is the compliance tensor: \[ \mathbb{C}^e : \mathbb{S} = \mathbb{I} \]
Computing the inverse in full tensor form is tedious; using Voigt notation (6×6 matrix inversion) is standard in FEM codes.
20.4 Summary and References to Other Appendices
This appendix complements: - A01 (Notation): All symbols and typography rules - L01 (Math Foundations): Definitions and basic properties - L03 (Kinematics): Applications of tensors to deformation analysis - L07 (Plasticity): Fourth-order tensors in constitutive modeling
For further reading on tensor calculus and differential geometry, consult the course bibliography or texts like Bonet and Wood (Nonlinear Continuum Mechanics for Finite Element Analysis) or Marsden & Hughes (Mathematical Foundations of Elasticity).