Overview of Elasticity of Materials/Introduction to Tensors

Introduction to Tensors
We have been working with stress and in particular looking at stress at a point and the impact of rotating the reference frame. Using the tools in the previous sections, it is possible to identify the principal stresses and the orientation of the reference frame relative to the principal axis, which allows for the determination of the stress state in any orientation. This also allows for the determination of the orientation and value of the maximum shear and normal forces, which are critical for engineering design. As you have seen, computing this information requires either extensive use of equations or geometry/trigonometry. In this section, tensors are introduced, which allows for a more elegant means of addressing coordination transformations.

We will begin by thinking about vectors, such as

\mathbf{S} = \left \langle S_1, S_2, S_3 \right \rangle $$ This vector is represented using the $$x_i$$ coordinates, but it just as well could have been expressed relative to a different set, which we will call $$x'_i$$. The direction cosines, the cosine between all the axis of the two coordinate systems, allows us to rewrite the vector

S'_1 = S_1 \cos\left ( x_1, x'_1 \right ) + S_2 \cos\left ( x_2, x'_1 \right ) + S_3 \cos\left ( x_3, x'_1 \right ) $$ where $\cos\left ( x_i, x'_j \right )$ is the cosine between $$x_i$$ and $$x'_j$$ and can be rewritten as $a_{ji} = \cos\left ( x_i, x'_j \right )$  which allows forː
 * $$\begin{align}

S'_1 = a_{11} S_1 + a_{12}S_2 + a_{13} S_3 \end{align} $$     [1]


 * $$\begin{align}

S'_2 = a_{21} S_1 + a_{22}S_2 + a_{23} S_3 \end{align} $$     [2]


 * $$\begin{align}

S'_3 = a_{31} S_1 + a_{32}S_2 + a_{33} S_3 \end{align}$$     [3] recognizing that for the cosine of an angle, $$a_{ji} = a_{ij}$$. Equations 1-3 can be written in a compact form, known as Einstein notation:



S'_i=a_{ij}S_j $$    [4] In Einstein notation, if a subscript is seen two or more times on a side of an equation, a summation is performed. In the example above, the $$j$$ shows up twice on the right side of the equation, but the $$i$$ only once. This means that this equation becomes:

S'_i = a_{ij}S_j = \sum_{j=1}^{3} a_{ij}S_j = a_{i1} S_1 + a_{i2}S_2 + a_{i3} S_3 $$ In this equation, $$i$$ is a dummy variable; substituting the value 1, 2, or 3 in for $$i$$ returns the equations above.

Here, $$ \mathbf{a} $$ is a rank two tensor that relates the two vectors $$ \mathbf{S} $$ and $$ \mathbf{S'} $$. Tensors are geometric objects that describe the linear relationship between scalars, vectors, and other tensors. The rank of a tensor is the number of indexes, or directions needed to describe it which directly translates to the dimensionality of the respective array. Therefore, $$ \mathbf{a} $$ is a rank two tensor because it requires $$i$$ and $$j$$ ($$a_{ij}$$) to describe it, and as such, is defined by a two dimensional array. Other texts may refer to rank as dimensionality or order as the terms can be used interchangeably. The table below may be helpful in understanding the concept of rank:
 * {| class="wikitable"

!Name !Rank/dimensionality/order of tensor !Example (11) $$ \left ( \begin{matrix} {25} & {32} & {11} \end{matrix} \right ) $$ or $$ \left ( \begin{matrix} {25} \\ {32} \\ {11} \end{matrix} \right ) $$ \left ( \begin{matrix} {25} & {32} & {11} \\ {14} & {56} & {4} \\ {79} & {42} & {51} \end{matrix} \right ) $$ Are the vectors $$ \mathbf{S} $$ and $$ \mathbf{S'} $$ tensors? Although vectors can be tensors, in this case they are not because $$ \mathbf{S} $$ and $$ \mathbf{S'} $$ do not act to map linear spaces onto each other.
 * Scalar
 * Zero
 * Zero
 * Vector
 * One
 * One
 * Matrix
 * Two
 * Two
 * }

Tensors are used frequently to represent the intrinsic physical properties of materials. A good example is electrical conductivity, $$\boldsymbol{\phi}$$, which is a rank two tensor that expresses the current density in a material, $$\mathbf{J}$$, induced by the application of an electric field, $$\mathbf{E}$$.

\mathbf{J}=\mathbf{\phi}\mathbf{E} $$ Both $$\mathbf{J}$$ and $$\mathbf{E}$$ are vectors since they have both a magnitude and direction. Interestingly, the off-axis terms in $$\mathbf{J}$$ implies cross interactions between the vectors, e.g., the current response in the $$x_1$$ direction is influenced by the electric field in the $$x_2$$ and $$x_3$$ directions, which is indeed true.

There are many other tensors that represent material properties including the thermal conductivity, diffusivity, permittivity, dielectric susceptibility, permeability, and magnetic susceptibility to name a few. We will see that stress and strain also are tensors. Stress relates the surface normal to an arbitrary imaginary surface, $$\mathbf{n}$$, to the stress vector at that point, $$\mathbf{S}$$, as was discussed in the previous section.

Tensor Transformations
The vectors that represent material properties also must be able to transform. This is useful for coordinate transformation, which are essentially rotations. It also allows the tensors that represent material responses to transform according to crystallographic symmetry. These can involve rotations, mirror operations, and inversions. Because these transformations involve linear one-to-one mapping, the transformations themselves are enabled by transformation tensors.

In Equation 4 we rotated vector $$\mathbf{S}$$ to $$\mathbf{S'}$$ by applying transformation tensor $$\mathbf{a}$$

S'_i=a_{ij}S_j $$ What if we want to reverse this? We can simply reverse the equation:



S_i=a_{ji}S'_i $$    [5] Note that there are implications here regarding the inversion of $$\mathbf{a}$$. Since

a_{ji}S'_i=a_{ji}a_{ij}S_j=S_j $$ we have $$a_{ji}a_{ij}=I$$, which means that transposing $$\mathbf{a}$$ yields the inverse of $$\mathbf{a}$$, written as $$\mathbf{a^{-1}}=\mathbf{a^{T}}$$.

Consider now that there is a second vector, that we will call $$\mathbf{Q}$$, which is related to $$\mathbf{S}$$ by the rank two material property tensor $$\mathbf{T}$$:



\mathbf{S}_{i} = \mathbf{T}_{ij} \mathbf{Q}_j $$     [6] In a transformed coordinate system, we can express this as



{\mathbf{S}_{i}}' = {\mathbf{T}_{ij}}' {\mathbf{Q}_{j}}' $$    [7] So we can now write



\begin{align} {\mathbf{S}_{i}}' &= {\mathbf{a}_{ij}} {\mathbf{S}_{j}} \\ &= {\mathbf{a}_{ij}} {\mathbf{T}_{jk}} {\mathbf{Q}_{k}} \\ &= \underbrace{{\mathbf{a}_{ij}} {\mathbf{T}_{jk}} {\mathbf{a}_{lk}}}_{\mathbf{T}_{i\ell}} {\mathbf{Q}_{\ell}}' \end{align} $$

This tells us that the transformation of $$ x_j $$ to $$ x_j' $$ causes the transformations from $$ \mathbf{S} $$ to $$ \mathbf{S}' $$, $$ \mathbf{Q} $$ to $$ \mathbf{Q}' $$, and $$ \mathbf{T} $$ to $$ \mathbf{T}' $$, where the vector transformations are given by Equations 4 & 5, and the tensor transformation is given by

{\mathbf{T}_{i\ell}}' = {\mathbf{T}_{ij}} {\mathbf{T}_{jk}} {\mathbf{a}_{\ell k}} $$    [8] and



{\mathbf{T}_{i\ell}} = {\mathbf{a}_{ji}} {\mathbf{T}_{jk}}' {\mathbf{a}_{k\ell}} $$    [9] Note that these solutions are really double sums over $$ j $$ and $$ k $$, due to Einstein Notation.

Because the order of the summation is not important, we can writeː



{\mathbf{a}_{ij}} {\mathbf{T}_{jk}} {\mathbf{a}_{\ell k}} = {\mathbf{a}_{ij}} {\mathbf{a}_{\ell k}} {\mathbf{T}_{jk}} $$    [10]

This is a Tensor that relates $$ \mathbf{T} $$ and $$ \mathbf{T}' $$. Since it is a double sum, each term in $$ \mathbf{T} $$ has nine elements and the total tensor mapping the relationship between $$ \mathbf{T} $$ and $$ \mathbf{T}' $$ must have a total of $$ 81 $$ terms as $$ (9 \times 9 = 81) $$.

Tensor Symmetry
The nature of a tensor is determined by its application. There are subsets of tensors that we can classify according to their symmetry properties.

Symmetric tensors have a structure such asː $$ \left [ \begin{matrix} \alpha_1 & \beta_1 & \beta_2 \\ \beta_1 & \alpha_2 & \beta_3 \\ \beta_2 & \beta_3 & \alpha_3 \end{matrix} \right ] $$ where $$ \mathbf{T}_{ij} = \mathbf{T}_{ij} $$

Antisymmetric Tensors have a structure such asː $$ \left [ \begin{matrix} 0 & -\alpha & -\gamma \\ \alpha & 0 & \beta \\ \gamma & \beta & 0 \end{matrix} \right ] $$ where $$ \mathbf{T}_{ij} = -\mathbf{T}_{ij} $$

Note that the main diagonal of an antisymmetric tensor must be zero and the overall symmetry, or antisymmetry, depends on the reference frame selected. Any second rank tensor can be expressed as a sum of a symmetric and antisymmetric tensor asː



\mathbf{T}_{ij} = \underbrace{{1 \over 2} (\mathbf{T}_{ij} + \mathbf{T}_{ji})}_{Symmetric} + \underbrace{{1 \over 2} (\mathbf{T}_{ij} - \mathbf{T}_{ji})}_{Antisymmetric} $$    [11]

We will find this useful in the next section dealing with strain. Meanwhile, any symmetric tensor can be transformed by rotations to be aligned along its principal axis, such thatː



\left [ \begin{matrix} \text{T}_{11} & 0 & 0 \\ 0 & \text{T}_{22} & 0 \\ 0 & 0 & \text{T}_{33} \end{matrix} \right ] $$

The properties of tensors are highly related to the crystal symmetry of the material they represent. For example, say that we have two vector properties $$ \mathbf{S} $$ and $$ \mathbf{Q} $$ in a crystal which are related by a tensor $$ \mathbf{T} $$. If we rotate the reference frame according to a symmetry element of the crystal, then

{\mathbf{T}_{ij}}' = \mathbf{T}_{ij} $$ We will examine this by looking at a simple cubic crystal such as the one in Figure 1. When rotated, this crystal will periodically rotate back on itself, and the properties of the relevant tensors should do the same. By applying this theory to each possible crystal formations, we can develop simplified tensors for each, which represent this symmetry.
 * {| class="wikitable"

!Crystal Formation !Tensor !Number of Independent
 * +Crystal Tensors

Components \left [ \begin{matrix} \text{S} & 0 & 0 \\ 0 & \text{S} & 0 \\ 0 & 0 & \text{S} \end{matrix} \right ] $$ Hexagonal
 * Cubic
 * Cubic
 * 1
 * Tetragonal
 * Tetragonal

Trigonal \left [ \begin{matrix} \text{S}_{1} & 0 & 0 \\ 0 & \text{S}_{1} & 0 \\ 0 & 0 & \text{S}_{3} \end{matrix} \right ] $$ \left [ \begin{matrix} \text{S}_{1} & 0 & 0 \\ 0 & \text{S}_{2} & 0 \\ 0 & 0 & \text{S}_{3} \end{matrix} \right ] $$ \left [ \begin{matrix} \text{S}_{11} & 0 & \text{S}_{13} \\ 0 & \text{S}_{22} & 0 \\ \text{S}_{13} & 0 & \text{S}_{33} \end{matrix} \right ] $$ \left [ \begin{matrix} \text{S}_{11} & \text{S}_{12} & \text{S}_{13} \\ \text{S}_{21} & \text{S}_{22} & \text{S}_{23} \\ \text{S}_{31} & \text{S}_{32} & \text{T}_{33} \end{matrix} \right ] $$
 * 2
 * Orthorhombic
 * Orthorhombic
 * 3
 * Monoclinic
 * Monoclinic
 * 4
 * Triclinic
 * Triclinic
 * 6
 * }

Tensor Contractions and Invariant Relations in Stress
Much of this discussion has been about property relations, but here our interest is in the stress tensor; a symmetric tensor that can therefore be arranged to be aligned in the principal axis. We will now rederive the 3D stress relationships using tensors. The stresses normal to an oblique plane are writtenː



\sigma_{nj} = a_{ni} \sigma_{ij} $$    [12]

Here, $$ n $$ is the direction of the normal to the plane and is the original stress state. If the oblique plane is a principal direction, with a normal stress of $$ \sigma_p $$, then we can write our equation asː



\sigma_{nj} = a_{pj} \sigma_p $$    [13]

By combining Equations 12 & 13, we getː



\underbrace{a_{ni} \sigma_{ij}}_{sum \ over \ i} - a_{pj} \sigma_p = 0 $$    [14]

Kronecker Delta
Additionally, there is a handy expression called the Kronecker Delta $$ (\delta_{ij}) $$ that has the propertiesː

\delta_{ij} = \left [ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right ] = \begin{cases} 1 \ \text{if} \ i=j \\ 0 \ \text{if} \ i \neq j \end{cases} $$

When applied to a tensor, the Kronecker Delta is said to "contract" the tensor's rank by two. This turns a 4th rank tensor into a 2nd rank, a 3rd rank tensor into a 1st rank, etc... For the purpose of this text, we will not be using this expression often, but in applying this to Equation 14 $ (a_{ni} \sigma_{ij} - a_{pj} \sigma_p = 0) $, we can replace the scalar $$ a_{pj} $$ with the contraction of the second rank tensor such thatː

a_{pj} = \mathbf{a}_{pi} \delta_{ji} $$

The rule for contraction here is to replace $$ i $$ with $$ j $$ and remove the Kronecker Delta term.

Returning to Equation 14, we replace the $$ a_{pj} $$ with our Kronecker Delta expansion to getː



a_{ni} \sigma_{ij} - \sigma_p a_{pi} \delta_{ji} =0 $$    [15]

This equation can be entirely summed over $$ i $$, and because $$ p $$ is normal to the plane, this makes $$ a_{ni} $$ equal to $$ a_{pi} $$ and our equation evolves toː



(\sigma_{ij} - \sigma_p \delta_{ji}) a_{pi} = 0 $$    [16]

This gives us a set of three equations where $$ j = 1, 2, 3 $$. By substituting the direction cosines into the left term and using $$ a_{p1} = \ell $$, $$ a_{p2} = m $$, $$ a_{p3} = n $$ and $$ \delta_{ji} = 0 $$ when $$ j \neq i $$, we can solve for the non-trivial (non-zero) solution by taking the determinant ofː

\left| \begin{matrix} \sigma_x - \sigma_p & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y - \sigma_p & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z - \sigma_p \end{matrix} \right | =0 $$ Which yields the same result as returned before.
 * \sigma_{ij} - \sigma_p \delta_{ji}| =

The Three Invariants
We also identify the invariant relations. It should be noted that these also can come from the stress tensor. First, let's apply a contraction to $$\boldsymbol{\sigma}$$ː
 * $$\boldsymbol{\sigma}_{ij} \delta_{ij} = \sigma_{ii} = \sigma_{11} + \sigma_{22} + \sigma_{33} = \text{I}_1$$

This is our first invariant. The second invariant comes from the minors of $$\boldsymbol{\sigma}$$, which can be used to expand the determinant.


 * $$\text{I}_2 =

\left | \begin{matrix} \sigma_{22} & \sigma_{23} \\ \sigma_{32} & \sigma_{33} \end{matrix} \right | + \left | \begin{matrix} \sigma_{11} & \sigma_{13} \\ \sigma_{31} & \sigma_{33} \end{matrix} \right | + \left | \begin{matrix} \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} \end{matrix} \right |$$

The third invariant is the determinant of $$\boldsymbol{\sigma}$$ where $$\text{I}_3 = \det[\boldsymbol{\sigma}]$$.