Vectors are denoted by small, bold letters, e.g. v. Their components will be denoted by  $v_{i}$ , where, unless otherwise stated, Latin under scripts such as $i$ or $I$ will range from 1 to 3. In matrix form a vector will be represented as a column vector and its transpose as a row vector: $$\begin{equation} \begin{array}{cc} \mathbf{v}=\left(\begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array}\right),\quad & \mathbf{v}^{T}=\left(\begin{array}{ccc} v_{1}, & v_{2}, & v_{3}\end{array}\right)\end{array}.\label{eq1} \end{equation}$$ The following products are defined between vectors. Assume u, v are vectors. Also note that the Einstein summation convention is used throughout this manual [46].

The dot or scalar product: $$\begin{equation} \mathbf{u}\cdot\mathbf{v}=u_{i}v_{i}\,.\label{eq2} \end{equation}$$ The cross product: $$\begin{equation} \mathbf{u}\times\mathbf{v}=\left[\begin{array}{c} u_{2}v_{3}-u_{3}v_{2}\\ u_{3}v_{1}-u_{1}v_{3}\\ u_{1}v_{2}-u_{2}v_{1} \end{array}\right]\,.\label{eq3} \end{equation}$$ The vector outer product: $$\begin{equation} \left(\mathbf{u}\otimes\mathbf{v}\right)_{ij}=u_{i}v_{j}\,.\label{eq4} \end{equation}$$ Note that vectors are also known as first order tensors. Scalars are known as zero order tensors. The outer product, defined by equation (2.1-4), is a second order tensor.

Second order tensors are denoted by bold, capital letters, e.g. $\mathbf{A}$ . Some exceptions will be made to remain consistent with the literature. For instance, the Cauchy stress tensor is denoted by $\mathbf{\boldsymbol{\sigma}}$ . However, the nature of the objects will always be clear from the context. The following operations on tensors are defined. Assume $\mathbf{A}$ and $\mathbf{B}$ are second-order tensors.

The double contraction or tensor inner product is defined as: $$\begin{equation} \mathbf{A}:\mathbf{B}=A_{ij}B_{ij}\,.\label{eq5} \end{equation}$$ The trace is defined as: $$\begin{equation} \tr\mathbf{A}=\mathbf{I}:\mathbf{A}=A_{ii}\,.\label{eq6} \end{equation}$$ Here $\mathbf{I}$ is the second order identity tensor with components $\delta_{ij}$ .

In general the components of tensors will change under a change of coordinate system. Nevertheless, certain intrinsic quantities associated with them will remain invariant under such a transformation. The scalar product between two vectors is such an example. The double contraction between two second-order tensors is another example. The following set of invariants for second-order tensors is commonly used: $$\begin{equation} \begin{aligned}I_{1} & =\tr\mathbf{A},\\ I_{2} & =\frac{1}{2}\left(\left(\tr\mathbf{A}\right)^{2}-\tr\mathbf{A}^{2}\right)\\ I_{3} & =\det\mathbf{A}. \end{aligned} \,.\label{eq7} \end{equation}$$ A tensor $\mathbf{S}$ is called symmetric if it is equal to its transpose: $$\begin{equation} \mathbf{S}=\mathbf{S}^{T}\,.\label{eq8} \end{equation}$$ A tensor $\mathbf{W}$ is called anti-symmetric if it is equal to the negative of its transpose: $$\begin{equation} \mathbf{W}=-\mathbf{W}^{T}\,.\label{eq9} \end{equation}$$ Any second order tensor $\mathbf{A}$ can be written as the sum of a symmetric tensor $\mathbf{S}$ and an anti-symmetric tensor $\mathbf{W}$ : $$\begin{equation} \mathbf{A}=\mathbf{S}+\mathbf{W},\label{eq10} \end{equation}$$ where $$\begin{equation} \mathbf{S}=\frac{1}{2}\left(\mathbf{A}+\mathbf{A}^{T}\right),\mbox{ and}\mathbf{W}=\frac{1}{2}\left(\mathbf{A}-\mathbf{A}^{T}\right).\label{eq11} \end{equation}$$ Also note that for any tensor $\mathbf{B}$ the following holds: $$\begin{equation} \mathbf{B}:\mathbf{A}=\mathbf{B}:\mathbf{S}\,,\quad\mathbf{B}:\mathbf{W}=\mathbf{0}\,.\label{eq12} \end{equation}$$ With any anti-symmetric tensor a dual vector $\mathbf{w}$ can be associated such that, $$\begin{equation} \mathbf{\hat{w}}\cdot\mathbf{u}=\mathbf{w}\times\mathbf{u}\,,\label{eq13} \end{equation}$$ where the second order tensor $\mathbf{\hat{w}}$ is defined as, $$\begin{equation} \mathbf{\hat{w}}=\left[\begin{array}{ccc} 0 & -w_{3} & w_{2}\\ w_{3} & 0 & -w_{1}\\ -w_{2} & w_{1} & 0 \end{array}\right]\,.\label{eq14} \end{equation}$$ A second order $\mathbf{Q}$ tensor is called orthogonal if $\mathbf{Q}^{-1}=\mathbf{Q}^{T}$ .

In the implementation of the FE method it is often convenient to write symmetric second-order tensors using Voigt notation. In this notation the components of a 2$^{\mathrm{nd}}$ order symmetric tensor $\mathbf{A}$ are arranged as a column vector: $$\begin{equation} \left[\mathbf{A}\right]=\left[\begin{array}{c} A_{11}\\ A_{22}\\ A_{33}\\ A_{12}\\ A_{23}\\ A_{13} \end{array}\right]\,.\label{eq15} \end{equation}$$ Higher order tensors will be denoted by bold, capital, script symbols, e.g. $\boldsymbol{\mathcal{A}}$ . An example of a third-order tensor is the permutation tensor $\mathcal{E}_{ijk}$ , whose components are 1 for an even permutation of $\left(1,2,3\right)$ , -1 for an odd permutation of $\left(1,2,3\right)$ and zero otherwise. The permutation symbol is useful for expressing the cross-product of two vectors in index notation: $$\begin{equation} \left(\mathbf{u}\times\mathbf{v}\right)_{i}=\mathcal{E}_{ijk}u_{j}v_{k}\,.\label{eq16} \end{equation}$$ An example of a fourth-order tensor is the elasticity tensor $\boldsymbol{\mathcal{C}}$ which, in linear elasticity theory, relates the small strain tensor $\mathbf{\boldsymbol{\varepsilon}}$ and the Cauchy stress tensor $\boldsymbol{\sigma}=\boldsymbol{\mathcal{C}}:\boldsymbol{\varepsilon}$ .

Higher order tensors can be constructed from second order tensors in a similar way as second order tensors can be constructed from vectors. If $\mathbf{A}$ and $\mathbf{B}$ are second order tensors, then the following fourth order tensors can be defind by requiring that the following must hold for any second order tensor $\mathbf{X}$ : $$\begin{equation} \left(\mathbf{A}\otimes\mathbf{B}\right):\mathbf{X}=\left(\mathbf{B}:\mathbf{X}\right)\mathbf{A}\,,\label{eq17} \end{equation}$$ $$\begin{equation} \left(\mathbf{A}\,\underline{\otimes}\,\mathbf{B}\right):\mathbf{X}=\mathbf{A}\cdot\mathbf{X}\cdot\mathbf{B}^{T}\,,\label{eq18} \end{equation}$$ $$\begin{equation} \left(\mathbf{A}\,\overline{\otimes}\,\mathbf{B}\right):\mathbf{X}=\mathbf{A}\cdot\mathbf{X}^{T}\cdot\mathbf{B}^{T}\,,\label{eq19} \end{equation}$$ $$\begin{equation} \left(\mathbf{A}\,\underline{\overline{\otimes}}\,\mathbf{B}\right):\mathbf{X}=\frac{1}{2}\left(\mathbf{A}\cdot\mathbf{X}\cdot\mathbf{B}^{T}+\mathbf{A}\cdot\mathbf{X}^{T}\cdot\mathbf{B}^{T}\right)\,.\label{eq20} \end{equation}$$ The Cartesian component forms of the operators $\otimes$ , $\underline{\otimes}$ , $\overline{\otimes}$ and $\underline{\overline{\otimes}}$ are defined as follows: $$\begin{equation} \left(\mathbf{A}\otimes\mathbf{B}\right)_{ijkl}=\mathbf{A}_{ij}\mathbf{B}_{kl}\,,\label{eq21} \end{equation}$$ $$\begin{equation} \left(\mathbf{A}\,\underline{\otimes}\,\mathbf{B}\right)_{ijkl}=\mathbf{A}_{ik}\mathbf{B}_{jl}\,,\label{eq22} \end{equation}$$ $$\begin{equation} \left(\mathbf{A}\,\overline{\otimes}\,\mathbf{B}\right)_{ijkl}=A_{il}B_{jk}\,,\label{eq23} \end{equation}$$ $$\begin{equation} \left(\mathbf{A}\,\underline{\overline{\otimes}}\,\mathbf{B}\right)_{ijkl}=\frac{1}{2}\left(A_{ik}B_{jl}+A_{il}B_{jk}\right)\,.\label{eq24} \end{equation}$$ The fourth order identity tensors are defined as: $$\begin{equation} \begin{aligned}\mathbf{A} & =\mathcal{I}:\mathbf{A}\,,\\ \mathbf{A}^{T} & =\overline{\mathcal{I}}:\mathbf{A}\,, \end{aligned} \label{eq25} \end{equation}$$ where $\boldsymbol{\mathcal{I}}=\mathbf{I}\,\underline{\otimes}\,\mathbf{I}$ and $\overline{\boldsymbol{\mathcal{I}}}=\mathbf{I}\,\overline{\otimes}\,\mathbf{I}$ . The components are given by: $$\begin{equation} \begin{aligned}\mathcal{I}_{ijkl} & =\delta_{ik}\delta_{jl}\,,\\ \overline{\mathcal{I}}_{ijkl} & =\delta_{il}\delta_{jk}\,. \end{aligned} \label{eq26} \end{equation}$$