# Overview of Elasticity of Materials/Introduction to Tensors

## Introduction to Tensors

We've been working with stress and in particular looking as 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 determination of the stress state in any orientation. This allows determination of the orientation and value of the maximum shear and normal forces, which are critical for engineering design. As you've see, computing this information requires either extensive use of equations or geometry/trigonometry. In this section tensors are going to be introduced, which will allow for a more elegant means of addressing coordination transformations.

Lets begin by thinking about vectors, such as

${\displaystyle \mathbf {S} =\left\langle S_{1},S_{2},S_{3}\right\rangle }$

This vector is represented using the ${\displaystyle x_{i}}$  coordinates, but it just as well could have been expressed relative to a different set, which we will call ${\displaystyle x'_{i}}$ . The direction cosines, the cosine between all the axis of the two coordinate systems, allows us to rewrite the vector
${\displaystyle S'_{1}=S_{1}cos\left(x_{1},x'_{1}\right)+S_{2}cos\left(x_{2},x'_{1}\right)+S_{3}cos\left(x_{3},x'_{1}\right)}$

where ${\textstyle cos\left(x_{i},x'_{j}\right)}$  is the cosine between ${\displaystyle x_{i}}$  and ${\displaystyle x'_{j}}$  and can be rewritten ${\textstyle a_{ji}=cos\left(x_{i},x'_{j}\right)}$  allowing forː
{\displaystyle {\begin{aligned}S'_{1}=a_{11}S_{1}+a_{12}S_{2}+a_{13}S_{3}\\S'_{2}=a_{21}S_{1}+a_{22}S_{2}+a_{23}S_{3}\\S'_{3}=a_{31}S_{1}+a_{32}S_{2}+a_{33}S_{3}\end{aligned}}}

recognizing that for the cosine of an angle ${\displaystyle a_{ji}=a_{ij}}$ . This set of three equations can be written in a compact for, known as Einstein notation
${\displaystyle S'_{i}=a_{ij}S_{j}}$

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 ${\displaystyle j}$  shows up twice on the right side of the equation, but the ${\displaystyle i}$  only once. This means that this equation becomes
${\displaystyle S'_{i}=a_{ij}S_{j}=\sum _{j=1}^{3}a_{ij}S_{j}=a_{i1}S_{1}+a_{i2}S_{2}+a_{i3}S_{3}.}$

In this equation ${\displaystyle i}$  is a dummy variable; substituting the value 1, 2, or 3 in for ${\displaystyle i}$  returns the equations above.

Here ${\displaystyle \mathbf {a} }$  is a rank 2 tensor that relates the two vectors ${\displaystyle \mathbf {S} }$  and ${\displaystyle \mathbf {S'} }$ . Tensors are geometric objects that describe the linear relationship between scalars, vectors, and other tensors. Are the vector's ${\displaystyle \mathbf {S} }$  and ${\displaystyle \mathbf {S'} }$  tensors? Although vectors can be tensors, in this case they are not because ${\displaystyle \mathbf {S} }$  and ${\displaystyle \mathbf {S'} }$  do not act to map linear spaces onto each other. The rank of a tensor is the number of indexes needed to describe it, therefore ${\displaystyle \mathbf {a} }$  is a rank two tensor because it requires ${\displaystyle i}$  and ${\displaystyle j}$ , ${\displaystyle a_{ij}}$ , to describe it.

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

${\displaystyle \mathbf {J} =\mathbf {\phi } \mathbf {E} }$

Both ${\displaystyle \mathbf {J} }$  and ${\displaystyle \mathbf {E} }$  are vectors since they have both a magnitude and direction. Interestingly, the off-axis terms in ${\displaystyle \mathbf {J} }$  implies cross interactions between the vectors, e.g., the current response in the ${\displaystyle x_{1}}$  direction is influenced by the electric field in the ${\displaystyle x_{2}}$  and ${\displaystyle x_{3}}$  directions, which is indeed true.

There are many other tensors that represent materials 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, ${\displaystyle \mathbf {n} }$ , to the stress vector at that point, ${\displaystyle \mathbf {S} }$ , as was discussed in the previous section.

### Tensor Transformations

The vectors that represent material properties also must be able to transform. This is useful to allow coordinate transformation, which essentially are 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 transformation themselves are enabled by transformation tensors.

In the section above we rotated vector ${\displaystyle \mathbf {S} }$  to ${\displaystyle \mathbf {S'} }$  by applying transformation tensor ${\displaystyle \mathbf {a} }$

${\displaystyle S'_{i}=a_{ij}S_{j}}$

What if we want to reverse this? We can simply reverse the equation
${\displaystyle S_{i}=a_{ji}S'_{i}}$

Note that there are implications here regarding the inversion of ${\displaystyle \mathbf {a} }$ . Since
${\displaystyle a_{ji}S'_{i}=a_{ji}a_{ij}S_{j}=S_{j}}$

we have that ${\displaystyle a_{ji}a_{ij}=I}$  which means that transposing ${\displaystyle \mathbf {a} }$  yields the inverse of ${\displaystyle \mathbf {a} }$ , written ${\displaystyle \mathbf {a^{-1}} =\mathbf {a^{T}} }$ .

Consider now that there is a second vector that we'll call ${\displaystyle \mathbf {Q} }$  is related to ${\displaystyle \mathbf {S} }$  by the rank two materials property tensor ${\displaystyle \mathbf {T} }$

${\displaystyle \mathbf {S} _{i}=\mathbf {T} _{ij}\mathbf {Q} _{j}}$

In a transformed coordinate system we'll express this as
${\displaystyle {\mathbf {S} _{i}}'={\mathbf {T} _{ij}}'{\mathbf {Q} _{j}}'}$

So we can now write

{\displaystyle {\begin{aligned}{\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{aligned}}}

This tells us that the transformation of ${\displaystyle x_{j}}$  to ${\displaystyle x_{j}'}$  causes the transformations from ${\displaystyle \mathbf {S} }$  to ${\displaystyle \mathbf {S} '}$ , ${\displaystyle \mathbf {Q} }$  to ${\displaystyle \mathbf {Q} '}$ , and ${\displaystyle \mathbf {T} }$  to ${\displaystyle \mathbf {T} '}$ , where the vector transformations are given by the above equations, and the tensor transformation is given by ${\displaystyle {\mathbf {T} _{i\ell }}'={\mathbf {T} _{ij}}{\mathbf {T} _{jk}}{\mathbf {a} _{\ell k}}}$  and ${\displaystyle {\mathbf {T} _{i\ell }}={\mathbf {a} _{ji}}{\mathbf {T} _{jk}}'{\mathbf {a} _{k\ell }}}$ . Note that these solutions are really double sums over ${\displaystyle j}$  and ${\displaystyle k}$ , due to Einstein Notation.

Because the order of the summation is not important, we can writeː ${\displaystyle {\mathbf {a} _{ij}}{\mathbf {T} _{jk}}{\mathbf {a} _{\ell k}}={\mathbf {a} _{ij}}{\mathbf {a} _{\ell k}}{\mathbf {T} _{jk}}}$

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

### Tensor Symmetry

The nature of a tensor is determined by it's application. There are subsets of tensors that we can classify according to their symmetry properties.

Symmetric tensors have a structure such asː ${\displaystyle \left[{\begin{matrix}\alpha _{1}&\beta _{1}&\beta _{2}\\\beta _{1}&\alpha _{2}&\beta _{3}\\\beta _{2}&\beta _{3}&\alpha _{3}\end{matrix}}\right]}$  where ${\displaystyle \mathbf {T} _{ij}=\mathbf {T} _{ij}}$

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

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

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

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

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

The properties of tensors are highly tied to the crystal symmetry of the material they represent. For example, let's say that we have two vector properties ${\displaystyle \mathbf {S} }$  and ${\displaystyle \mathbf {Q} }$  in a crystal which are related by a tensor ${\displaystyle \mathbf {T} }$ . If we rotate the reference farm according to a symmetry element of the crystal then ${\displaystyle {\mathbf {T} _{ij}}'=\mathbf {T} _{ij}}$ .

<FIGURE> "Rotating a Simple Cubic Crystal" (Due to symmetry, the crystal will periodically rotate such that it is practically at the same orientation it started at.)

Let's examine this by looking at a simple cubic crystal. 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.

Crystal Tensors
Crystal

Formation

Tensor Number of

Independent

Components

Cubic
${\displaystyle \left[{\begin{matrix}{\text{S}}&0&0\\0&{\text{S}}&0\\0&0&{\text{S}}\end{matrix}}\right]}$

1
Tetragonal

Hexagonal

Trigonal

${\displaystyle \left[{\begin{matrix}{\text{S}}_{1}&0&0\\0&{\text{S}}_{1}&0\\0&0&{\text{S}}_{3}\end{matrix}}\right]}$

2
Orthorhombic
${\displaystyle \left[{\begin{matrix}{\text{S}}_{1}&0&0\\0&{\text{S}}_{2}&0\\0&0&{\text{S}}_{3}\end{matrix}}\right]}$

3
Monoclinic
${\displaystyle \left[{\begin{matrix}{\text{S}}_{11}&0&{\text{S}}_{13}\\0&{\text{S}}_{22}&0\\{\text{S}}_{13}&0&{\text{S}}_{33}\end{matrix}}\right]}$

4
Triclinic
${\displaystyle \left[{\begin{matrix}{\text{S}}_{11}&{\text{S}}_{12}&{\text{S}}_{13}\\{\text{S}}_{21}&{\text{S}}_{22}&{\text{S}}_{23}\\{\text{S}}_{31}&{\text{S}}_{32}&{\text{T}}_{33}\end{matrix}}\right]}$

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. Let's now rederive the 3D stress relationships using tensors. The stresses normal to an oblique plane are writtenː

${\displaystyle \sigma _{ij}=a_{ni}\sigma _{ij}}$

Here the ${\displaystyle n}$  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 ${\displaystyle \sigma _{p}}$ , then we can write our equation asː

${\displaystyle \sigma _{nj}=a_{pj}\sigma _{p}}$

By combining these two equations, we getː

${\displaystyle \underbrace {a_{ni}\sigma _{ij}} _{sum\ over\ i}-a_{pj}\sigma _{p}=0}$

#### Kronecker Delta

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

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

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 won't be using this expression often, but in applying this to our previous equation ${\textstyle (a_{ni}\sigma _{ij}-a_{pj}\sigma _{p}=0)}$  we can replace the scalar ${\displaystyle a_{pj}}$  with the contraction of the second rank tensor such thatː

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

The rule for contraction here is to replace ${\displaystyle i}$  with ${\displaystyle j}$ , and remove the Kronecker Delta term.

Returning to our earlier equation, we replace the ${\displaystyle a_{pj}}$  with our Kronecker Delta expansion to getː

${\displaystyle a_{ni}\sigma _{ij}-\sigma _{p}a_{pi}\delta _{ji}=0}$

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

${\displaystyle (\sigma _{ij}-\sigma _{p}\delta _{ji})a_{pi}=0}$

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

${\displaystyle |\sigma _{ij}-\sigma _{p}\delta _{ji}|=\left[{\begin{matrix}\sigma _{x}-\sigma _{p}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{y}-\sigma _{p}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{z}-\sigma _{p}\end{matrix}}\right]}$

Which yields the same result as returned before.

#### 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 ${\displaystyle {\boldsymbol {\sigma }}}$ ː ${\displaystyle {\boldsymbol {\sigma }}_{ij}\delta _{ij}=\sigma _{ii}=\sigma _{11}+\sigma _{22}+\sigma _{33}={\text{I}}_{1}}$

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

${\displaystyle {\text{I}}_{2}=\left|{\begin{matrix}\sigma _{22}&\sigma _{23}\\\sigma _{32}&\sigma _{33}\end{matrix}}\right|+\left|{\begin{matrix}\sigma _{11}&\sigma _{13}\\\sigma _{31}&\sigma _{33}\end{matrix}}\right|+\left|{\begin{matrix}\sigma _{11}&\sigma _{12}\\\sigma _{21}&\sigma _{22}\end{matrix}}\right|}$

The third invariant is the determinant of ${\displaystyle {\boldsymbol {\sigma }}}$  where ${\displaystyle {\text{I}}_{3}=\det[{\boldsymbol {\sigma }}]}$ .