# Overview of Elasticity of Materials/Printable version

The current, editable version of this book is available in Wikibooks, the open-content textbooks collection, at
https://en.wikibooks.org/wiki/Overview_of_Elasticity_of_Materials

# Introduction

Currently, there is no suitable open source text for materials science that not only includes the basics of the topic, as seen in W.D. Callister's Materials Science and Engineering: An Introduction, but also covers additional concepts such as the derivation of Mohr's circle and an introduction to tensors. Never the less, these concepts are key for student to gain an understanding of more advanced topics in materials science and engineering. This text is built as an open-source companion for the currently available texts. It attempts to address these additional topics and give further detail on their application.
While this text will go in depth on the linear elastic behavior of materials, there are currently no plans to include boundary conditions, which are essential for advanced analysis. The non-linear response often observed in polymeric and biological materials also is neglected here. Thus, this content should only be considered as an introductory overview of the topic.
A basic level of understanding on the subject of materials science, including stress and strain, is required to understand the real-world applications of this text. While the resulting equations and concepts given here can be immediately applied by the reader, a basic understanding of the mathematical foundations presented here is recommended for advanced applications.
SPB -- April 2022

# Introducing Stress
## Introduction

We will begin by developing the constitutive equations that describe the relationship between stress, $\sigma$, and strain, $\varepsilon$. This is the subset of continuum mechanics that focuses on the purely elastic regime, and in particular, will focus on linear elasticity where Hooke's Law holds true.
The concepts of stress and strain originate by considering the forces applied to a body and its displacement. Beginning with forces, there are two types of forces that can be applied. First, there is surface force which can either be point forces or distributed forces that are applied over a surface. Second, there is body force which is applied to every element of a body, not just a surface (i.e., gravity, electric fields, etc.).
The body of interest has numerous forces acting on it and these are transmitted through the material. At any point inside the body, you can imagine slicing it to observe the forces present on the imagined cut surface, as pictured in Figure 1. These forces are the interactions between the material on either side of the imagined cut. We define the stress at a point in the body as the forces acting on the surface of such an imagined cut.
As you recall, the stress is defined as the force over the area which it is applied. The force, $\mathbf {P}$, is a vector quantity, allowing the components to be projected into the normal and tangential directions. As shown in Figure 1, the normal component is defined according to the angle $\theta$, yielding a normal stress $\sigma _{33}={\frac {P\cos \!\theta }{A}}$. The tangential component of the force, $P\sin \!\theta$, can be further projected into the two orthogonal directions identified in Figure 1 as $x_{1}$ and $x_{2}$, yielding two orthogonal shear stresses. This is performed according to the angle $\phi$, giving $\sigma _{31}={\frac {P\sin \!\theta \cos \!\phi }{A}}$ and $\sigma _{32}={\frac {P\sin \!\theta \sin \!\phi }{A}}$.
Note here that we've defined the coordinate system such that the $x_{3}$ direction is the direction normal to the cut surface. It is convenient to use $(x_{1},x_{2},x_{3})$ instead of $(x,y,z)$ because it allows us to pass the indexes to the stress and strain quantities. In this example, the normal stress is given by $\sigma _{33}$ to specify that the normal stress is applied to the surface with a normal in the $x_{3}$ direction with a force projected in the $x_{3}$ direction. The tangential components $\sigma _{31}$ and $\sigma _{32}$ specify the surface having a normal $x_{3}$ with forces projected in the $x_{1}$ and $x_{2}$ directions, respectively. Cutting an infinitesimal cuboid, the stresses are defined in all three directions as shown in Figure 2. For comparison, the notation used in some textbooks will write normal stresses $\sigma _{x}$, whereas here we will use $\sigma _{11}$. These textbooks also use $\tau$ to denote shear stress, such as $\tau _{xy}$ whereas here we will use $\sigma _{12}$. This allows the stress state to be succinctly written in matrix (tensor) form
       $\sigma =\left({\begin{aligned}\sigma _{11}\quad \sigma _{12}\quad \sigma _{13}\\\sigma _{21}\quad \sigma _{22}\quad \sigma _{23}\\\sigma _{31}\quad \sigma _{32}\quad \sigma _{33}\end{aligned}}\right)$
The imaginary slice taken through point in the body in Figure 1 could have been any plane, but the force would remain the same. This would result in a new definition of the surface normal, and potentially a new expression for the stress. The physical presence of the stress does not change, but the description does, i.e., the coordinate system is modified. The remainder of this section is devoted to expressing the coordinate transformation and analysis of the stresses.

## Plane Stress

We will begin by simplifying the picture we are working with. The plane stress condition is observed for a thin 2D object, e.g., a piece of paper, which has no stress out of the plane. This allows us to write $\sigma _{33}=0$. Further, there is no shear in the $x_{3}$ direction such that $\sigma _{13}=\sigma _{23}=0$. For an object in the plane stress condition, our goal is to determine the state of stress at some point for any orientation of the axis.
For this object, the direction with zero force is $x_{3}$ coming out of the page and the non-zero stress state in the $x_{1}$ and $x_{2}$ directions have components $\sigma _{11}$, $\sigma _{22}$, and $\sigma _{12}=\sigma _{21}$.
Imagine a new area defined on a plane rotated about $x_{3}$ such that the normal, defined $x_{1}'$, is related to $x_{1}$ by $\theta$ as shown in Figure 3.
The components of force on the area is determined by the application of the original stresses to the projection of the new area:
       ${\begin{aligned}F_{1}=\sigma _{11}A\cos \!\theta +\sigma _{12}A\sin \!\theta =S_{1}A\\F_{2}=\sigma _{22}A\sin \!\theta +\sigma _{12}A\cos \!\theta =S_{2}A\end{aligned}}$ [1 & 2]
where the elements $A\cos \!\theta$ and $A\sin \!\theta$ are the projection of the A in the original orientation, shown in Figure 3 (a), and $S_{1}$ and $S_{2}$ are the total stresses in the $x_{1}$ and $x_{2}$ directions, where $\sigma _{12}=\sigma _{21}$. Then, dividing by A yields:
       ${\begin{aligned}S_{1}=\sigma _{11}\cos \!\theta +\sigma _{12}\sin \!\theta \\S_{2}=\sigma _{22}\sin \!\theta +\sigma _{12}\cos \!\theta \end{aligned}}$ [3 & 4]
Projecting the total stresses shown in Figure 3 (b) into the normal direction in the $x_{1}'$ coordinate yields
       $\sigma _{11}'=S_{1}\!\cos \!\theta +S_{2}\!\sin \!\theta$ [5]
In a similar fashion, we project tangential to the plane and yield
       $\sigma _{12}'=S_{2}\!\cos \!\theta -S_{1}\!\sin \!\theta$ [6]
Resulting in
       ${\begin{aligned}\sigma _{11}'&=(\sigma _{11}\cos \!\theta +\sigma _{12}\sin \!\theta )\cos \!\theta +(\sigma _{22}\sin \!\theta +\sigma _{12}\cos \!\theta )\sin \!\theta \\&=\sigma _{11}\cos ^{2}\!\theta +\sigma _{22}\sin ^{2}\!\theta +2\ \sigma _{12}\sin \!\theta \cos \!\theta \end{aligned}}$ [7]
and
       ${\begin{aligned}\sigma _{12}'&=(\sigma _{22}\sin \!\theta +\sigma _{12}\cos \!\theta )\cos \!\theta -(\sigma _{11}\cos \!\theta +\sigma _{12}\sin \!\theta )\sin \!\theta \\&=\sigma _{12}(cos^{2}\!\theta -\sin ^{2}\!\theta )+(\sigma _{22}-\sigma _{11})\sin \!\theta \cos \!\theta \end{aligned}}$ [8]
It is known that $\sigma _{12}'=\sigma _{21}'$ and therefore only $\sigma _{22}'$ needs determining. To do so, we define a new area that is rotated by π{\textstyle \pi }/2 relative to our original plane as shown in Figure 4. In this new orientation,
       ${\begin{aligned}S_{1}A&=\sigma _{11}A\cos(\theta +{\pi \over 2})+\sigma _{12}A\sin(\theta +{\pi \over 2})\\S_{1}&=-\sigma _{11}\sin \!\theta +\sigma _{12}\cos \!\theta \end{aligned}}$ [9]
and
       ${\begin{aligned}S_{2}A&=\sigma _{22}A\sin(\theta +{\pi \over 2})+\sigma _{12}A\cos(\theta +{\pi \over 2})\\S_{2}&=\sigma _{22}\cos \!\theta -\sigma _{12}\sin \!\theta \end{aligned}}$ [10]
Projecting the total stress in the normal direction yields
       ${\begin{aligned}\sigma _{22}'&=S_{1}\cos(\theta +{\pi \over 2})+S_{2}\sin(\theta +{\pi \over \theta })\\&=-S_{x_{1}}\sin \!\theta +S_{x_{2}}\cos \!\theta \end{aligned}}$ [11]
Substituting Equations 9 and 10 for $S_{1}$ and $S_{2}$ into Equation 11 for $\sigma _{22}'$ yields
       ${\begin{aligned}\sigma _{22}'&=-(-\sigma _{11}\sin \!\theta +\sigma _{12}\cos \!\theta )\sin \!\theta +(\sigma _{22}\cos \!\theta -\sigma _{12}\sin \!\theta )\cos \!\theta \\&=\sigma _{11}\sin ^{2}\!\theta +\sigma _{22}\cos ^{2}\!\theta -2\ \sigma _{12}\sin \!\theta \cos \!\theta \end{aligned}}$ [12]
The well-known trigonometric identities
       ${\begin{array}{lcl}\cos ^{2}\!\theta ={\cos 2\theta +1 \over 2}&\quad &\sin ^{2}\!\theta ={1-\cos 2\theta \over 2}\\2\sin \!\theta \cos \!\theta =\sin 2\theta &&\cos ^{2}\!\theta -\sin ^{2}\theta =\cos 2\theta \end{array}}$
are applied to Equations 7, 8, and 12 for $\sigma _{11}'$, $\sigma _{12}'$, and $\sigma _{22}'$ respectively, resulting in
       ${\begin{aligned}\sigma _{11}'&={\sigma _{11}+\sigma _{22} \over 2}+{\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }+\sigma _{12}\sin {2\theta }\end{aligned}}$ [13]
       ${\begin{aligned}\sigma _{22}'&={\sigma _{11}+\sigma _{22} \over 2}-{\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }-\sigma _{12}\sin {2\theta }\end{aligned}}$ [14]
and
       ${\begin{aligned}\sigma _{12}'&={\sigma _{22}-\sigma _{11} \over 2}\sin 2\theta +\sigma _{12}\cos 2\theta \end{aligned}}$ [15]

### Principal Stress

There are numerous immediate results that come from this derivation, from which we can gain greater insights. One result that comes from the equations for $\sigma '$ is $\sigma _{11}'+\sigma _{22}'=\sigma _{11}+\sigma _{22}$, for all $\theta$. This means that the trace of the stress tensor ${\boldsymbol {\sigma }}$ is invariant.
A second result is that the maximum normal stresses and shear stresses vary as a sine wave with period $\pi$. Within this oscillation, the normal and shear stresses are shifted by a phase factor that results in (1) the maximum and minimum normal stresses occur when the shear is zero, (2) the maximum and minimum shear stresses are shifted from each other by $\pi$/4, (3) the maximum and minimum normal stresses are shifted from each other by $\pi$/2, and (4) the maximum and minimum shear stresses are shifted by $\pi$/4 from the minimum and maximum normal stresses.
Any stress state can be rotated to yield $\sigma _{12}=\sigma _{21}=0$. This diagonalizes the stress tensor and gives normal stresses that are extreme. In this orientation, the planes are called the principal planes and the normal stresses are called the principal stresses. The directions that give these principal stresses are called the principal axis. As a matter of convention, we define the first principal stress $\sigma _{p1}$ to be the largest and the sequentially smaller principal stresses to be $\sigma _{p2}$ and $\sigma _{p3}$, although here we have limited ourselves to 2D plane stress and only enumerate $\sigma _{p1}$ and $\sigma _{p2}$.
We know $\sigma _{12}=0$ in the principal orientation, which means we can use Equation 8 for $\sigma _{12}'$ to determine the angle ($\theta$) needed to rotate the tensor ${\boldsymbol {\sigma }}$ into $\sigma '$ which is principal,
       ${\begin{aligned}0&=\sigma _{12}'=\sigma _{21}'\\0&=\sigma _{12}(\cos ^{2}\!\theta -\sin ^{2}\!\theta )+(\sigma _{22}-\sigma _{11})\sin \!\theta \cos \!\theta \\-(\sigma _{22}-\sigma _{11})\sin \!\theta \cos \!\theta &=\sigma _{12}(\cos ^{2}\!\theta -\sin ^{2}\!\theta )\\{\sin \!\theta \cos \!\theta \over \cos ^{2}\!\theta -\sin ^{2}\!\theta }&={\sigma _{12} \over \sigma _{11}-\sigma _{22}}\end{aligned}}$
Resulting in
       ${\begin{aligned}\tan 2\theta &={2\ \sigma _{12} \over \sigma _{11}-\sigma _{22}}\end{aligned}}$ [16]
It is observed graphically by plotting $\tan 2\theta$ in Figure 5 that adjacent roots are each separated by π{\textstyle \pi }/2. Furthermore, we can now utilize the Pythagorean Theorem to solve for our principal stresses.
For a simple right triangle with hypotenuse $c$ and sides $a$ and $b$ we know
       ${\begin{aligned}\sin \xi ={b \over c}\\\cos \xi ={a \over c}\\\tan \xi ={b \over a}\end{aligned}}$
which can be combined with the Pythagorean Theorem, $a^{2}+b^{2}=c^{2}$ and Equation 16,
       ${\begin{aligned}a&=\sigma _{12}\\b&={1 \over 2}(\sigma _{11}-\sigma _{22})\\c&=\pm \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}\end{aligned}}$
These can be further combined which yields
       ${\begin{aligned}\sin 2\theta &=\pm {\sigma _{12} \over \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\end{aligned}}$ [17]
and
       ${\begin{aligned}\\\cos 2\theta &=\pm {{1 \over 2}(\sigma _{11}-\sigma _{22}) \over \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\end{aligned}}$ [18]
These equations tell us for a given stress state, $\sigma$, what rotation is needed to align $\sigma$ with the principal axis.
Substituting these equations into Equation 13 for $\sigma _{11}'$, determines the principal stresses
       ${\begin{aligned}\sigma _{p}&={\sigma _{11}+\sigma _{22} \over 2}+{\sigma _{11}-\sigma _{22} \over 2}\left(\pm {{1 \over 2}(\sigma _{11}-\sigma _{22}) \over \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\right)+\sigma _{12}\left({\sigma _{12} \over \pm \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\right)\\&={\sigma _{11}+\sigma _{22} \over 2}+{{1 \over 4}(\sigma _{11}-\sigma _{22})^{2} \over \pm \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}+{{\sigma _{12}}^{2} \over \pm \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\\&={\sigma _{11}+\sigma _{22} \over 2}\pm \left({{1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2} \over \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\right)\end{aligned}}$
Resulting in
          ${\begin{aligned}&={\sigma _{11}+\sigma _{22} \over 2}\pm \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}\end{aligned}}$ [19]
Use Equation 19 in Equation 16 to find $\theta$ for $0<\theta <{\pi \over 4}$.
To find the maximum shear stress, we take the derivative with respect to theta of our simplified Equation 15 for $\sigma _{12}'$ and set it equal to $0$.
       ${\begin{aligned}0&={\operatorname {d} \over \operatorname {d} \!\theta }\left({\sigma _{22}-\sigma _{11} \over 2}\sin 2\theta +\sigma _{12}\cos 2\theta \right)\\&=2{\sigma _{22}-\sigma _{11} \over 2}\cos 2\theta -2\sigma _{12}\sin 2\theta \\&=(\sigma _{22}-\sigma _{11})\cos 2\theta -2\sigma _{12}\sin 2\theta \end{aligned}}$
Resulting in an expression for $2\theta$:
       ${\begin{aligned}\\{\sigma _{22}-\sigma _{11} \over 2\ \sigma _{12}}&={\sin 2\theta \over \cos 2\theta }=\tan 2\theta \end{aligned}}$ [20]
Notice that Equation 20 and Equation 16 are negative reciprocals which means that $2\theta _{p}$ and $2\theta _{MAX}$ are shifted by $\pi$/2. This is indicative of
       ${\begin{aligned}\tan \phi ={\frac {a}{b}}\\\tan \phi +{\frac {\pi }{2}}={\frac {-b}{a}}\end{aligned}}$
which implies that $2\theta _{p}$ and $2\theta _{MAX}$ are separated by $\pi$/4. Through substitution of Equation 20 into Equation 15, we arrive at an expression for $\sigma _{12MAX}$:
       $\sigma _{12MAX}=\pm \left[\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}}^{2}\right]^{1 \over 2}$ [21]

### Mohr's Circle

A convenient means of visualizing angular relationships is through Mohr's circle, which we derive here. Rearrange Equation 13 for $\sigma _{11}'$ and Equation 15 for $\sigma _{12}'$,
       ${\begin{aligned}\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}&={\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }+\sigma _{12}\sin {2\theta }\end{aligned}}$ [22]
       ${\begin{aligned}\sigma _{12}'={\sigma _{22}-\sigma _{11} \over 2}\sin {2\theta }+\sigma _{12}\cos {2\theta }\end{aligned}}$ [23]
Square both expressions,
       ${\begin{aligned}\left(\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}\right)^{2}&=\left({\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }+\sigma _{12}\sin {2\theta }\right)^{2}\end{aligned}}$
       ${\begin{aligned}\left(\sigma _{12}'\right)^{2}&=\left({\sigma _{22}-\sigma _{11} \over 2}\sin 2\theta +\sigma _{12}\cos 2\theta \right)^{2}\end{aligned}}$
Next, add them together to yield
       ${\begin{aligned}\left(\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}'}^{2}&=\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}(\cos ^{2}\!2\theta +\sin ^{2}\!2\theta )+{\sigma _{12}}^{2}(\cos ^{2}\!2\theta +\sin ^{2}\!2\theta )\\{\left(\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}\right)^{2}}+{\sigma _{12}'}^{2}&=\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}}^{2}\end{aligned}}$
The resulting expression is the equation for a circle: $(x-h)^{2}+y^{2}=r^{2}$
       $\underbrace {\left(\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}\right)^{2}} _{(x-h)^{2}}+\underbrace {{\sigma _{12}'}^{2}} _{y^{2}}=\underbrace {\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}}^{2}} _{r^{2}}$ [24]
From this expression, Mohr's circle is drawn in Figure 6. For a given stress state, $\sigma$, the center of the circle is $h={\frac {\sigma _{11}-\sigma _{22}}{2}}$ and the radius $r=\left(\left({\frac {\sigma _{11}-\sigma _{11}}{2}}\right)^{2}+\sigma _{12}^{2}\right)^{\frac {1}{2}}$. A bisecting line intercepts the circle such that the projection onto the x-axis identifies $\sigma _{11}$ and $\sigma _{22}$. The projection onto the y-axis identifies $\sigma _{12}$. Rotating the bisection is equivalent to transforming the stress state by $2\pi$, i.e., a rotation by $\phi$ on the diagram is equivalent to rotating by $2\pi$ in our equations. This allows the new stress state to be read from the diagram. When the bisector is horizontal, the principal orientation is identified. Rotating the bisection on the diagram by $\pi$ is equivalent to rotating the system by $\theta =\pi$/2, which can be imagined as rotating the cuboid faces until the system is back in registry, i.e., it returns to the original stress state. Further, rotating the bisection on the diagram by $\pi$/2 is equivalent to rotating by $\theta =\pi$/4, which is known to be the orientation with maximum shear stress. Thus, from a given initial stress state, $\sigma$, all stress states that can be achieved through rotation are visualized on the circle.

## Generalizing from 2D to 3D

Generalizing from 2D to 3D, we move from a biaxial plane stress system to a triaxial system. Determining the principal axis and angular relations is similar to the case of 2D and will be shown below. Note as a matter of convention, when two of the three principal stresses are equal, we call the system "cylindrical", and if all three principal stresses are equal, we call the system "hydrostatic" or "spherical".
As in the case of the biaxial system, we begin by defining a plane with area $A$ that passes through our $x_{1}$, $x_{2}$, and $x_{3}$ coordinate system, as shown in Figure 7. The plane intercepts the axis at ($J$, $K$, and $L$) as demonstrated in the figure. To simplify the problem and allow us to make progress toward our derivation, we will say that the plane is one of the principal planes so that the shear stress components are zero. Thus, we only need to consider our principal stress that is normal to the plane.
Define $\ell$, $m$, and $n$ to be the direction cosine between $x_{1}$, $x_{2}$, and $x_{3}$ and the normal to the stress. Using the unit vectors ${\hat {i}}$, ${\hat {j}}$, and ${\hat {k}}$ parallel to $x_{1}$, $x_{2}$, and $x_{3}$, we have
       Figure 7: Coordinate plane JKL in 3D that passes through the x, y, z coordinate system with positive shear stresses acting on it where $O$ is the origin.$\ell =\cos \theta _{1}={{\hat {i}}\cdot {\boldsymbol {\sigma }} \over |{\boldsymbol {\sigma |}}}\quad m=\cos \theta _{2}={{\hat {j}}\cdot {\boldsymbol {\sigma }} \over |{\boldsymbol {\sigma |}}};\quad n=\cos \theta _{3}={{\hat {k}}\cdot {\boldsymbol {\sigma }} \over |{\boldsymbol {\sigma |}}}$
The projection of stress along $x_{1}$, $x_{2}$, and $x_{3}$ direction give the total stresses $S_{1}$, $S_{2}$, and $S_{3}$:
       $S_{1}=\sigma _{p}\ell ;\quad S_{2}=\sigma _{p}m;\quad S_{3}=\sigma _{p}n$
In the biaxial derivation, the area is projected into three directions, producing the triangles in Figure 7 which have areas $LOK=A\ell$, $JOL=Am$ and $JOK=An$. We can now equate the forces in the two reference frames:
       ${\begin{aligned}S_{1}A=\sigma _{p}\ell A=F_{x}&=LOK\sigma _{11}+JOL\sigma _{21}+JOK\sigma _{31}+\\&=A\ell \sigma _{11}+Am\sigma _{21}+An\sigma _{31}\end{aligned}}$
So,
       ${\begin{aligned}\sigma _{p}\ell &=\sigma _{11}\ell +\sigma _{21}m+\sigma _{31}n\end{aligned}}$ [25]
By a similar process, the $F_{x_{2}}$ and $F_{x_{3}}$ components yield
       ${\begin{aligned}\sigma _{p}m&=\sigma _{12}\ell +\sigma _{22}m+\sigma _{32}n\end{aligned}}$ [26]
       ${\begin{aligned}\sigma _{p}n&=\sigma _{13}\ell +\sigma _{23}m+\sigma _{33}n\end{aligned}}$ [27]
These equations rearrange to
       ${\begin{aligned}0&=(\sigma _{11}-\sigma _{p})\ell +\sigma _{12}m+\sigma _{13}n\end{aligned}}$ [28]
       ${\begin{aligned}0&=\sigma _{12}\ell +(\sigma _{22}-\sigma _{p})m+\sigma _{23}n\end{aligned}}$ [29]
       ${\begin{aligned}0&=\sigma _{13}\ell +\sigma _{23}m+(\sigma _{33}-\sigma _{p})n\end{aligned}}$ [30]
This set of equations can be solved for $\left[\ell ,m,n\right]$ for a particular value of $\sigma _{p}$. This set of secular equations can be solved for eigenvalues $\sigma _{p}$ and eigenvectors $\left[\ell ,m,n\right]$. The non-trivial solutions, when $\ell ,m,$ and $n$ are non-zero, involves setting the determinant
       $det\ \left|{\begin{matrix}\sigma _{11}-\sigma _{p}&\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}-\sigma _{p}&\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}-\sigma _{p}\end{matrix}}\right|$
to zero and solving for the eigenvalues and subsequent eigenvectors.
Upon rearranging, we get
       ${\begin{aligned}0={\sigma _{p}}^{3}-&(\sigma _{11}+\sigma _{22}+\sigma _{33}){\sigma _{p}}^{2}\\&\quad +(\sigma _{11}\sigma _{22}+\sigma _{22}\sigma _{33}++\sigma _{33}\sigma _{11}-{\sigma _{12}}^{2}-{\sigma _{23}}^{2}-{\sigma _{31}}^{2})\sigma _{p}\\&\qquad \qquad -(\sigma _{11}\sigma _{22}\sigma _{33}+2\sigma _{12}\sigma _{23}\sigma _{31}-\sigma _{11}{\sigma _{23}}^{2}-\sigma _{22}{\sigma _{13}}^{2}-\sigma _{33}{\sigma _{12}}^{2})\end{aligned}}$ [31]
The three roots of this cubic equation give the principal stresses, $\sigma _{p1}$, $\sigma _{p2}$, and $\sigma _{p3}$. The principle stresses, once determined, are substituted back into the secular Equations 28-30 to determine the eigenvectors corresponding to $\left[\ell ,m,n\right]$, also recognizing that $\ell ^{2}+m^{2}+n^{2}=1$.
Solving the cubic equation is not the focus of this text, but Equation 31 is important because the coefficients in front of the principal stress must be invariant, i.e., the same principal coordinates must exist no matter the orientation of the coordinate system. From the cubic equation, the three invariants are
       ${\begin{aligned}I_{1}=(\sigma _{11}+\sigma _{22}+\sigma _{33})\end{aligned}}$ [32]
       ${\begin{aligned}I_{2}=(\sigma _{11}\sigma _{22}+\sigma _{22}\sigma _{33}+\sigma _{33}\sigma _{11}-{\sigma _{12}}^{2}-{\sigma _{23}}^{2}-{\sigma _{31}}^{2})\end{aligned}}$ [33]
       ${\begin{aligned}I_{3}=(\sigma _{11}\sigma _{22}\sigma _{33}+2\sigma _{12}\sigma _{23}\sigma _{31}-\sigma _{11}{\sigma _{23}}^{2}-\sigma _{22}{\sigma _{13}}^{2}-\sigma _{33}{\sigma _{12}}^{2})\end{aligned}}$ [34]
This is useful because these invariant relations determine the relationship between stresses in different orientations, i.e. given $\sigma$, you can now directly determine $\sigma _{1}$, $\sigma _{2}$, and $\sigma _{3}$.
Now, we will generalize our solution to include not only the principal stresses. Just as we did earlier, we can write out the total forces:
       ${\begin{aligned}S_{x_{1}}A=F_{x_{1}}=\sigma _{11}\ell A+\sigma _{12}mA+\sigma _{31}nA\end{aligned}}$
       ${\begin{aligned}\qquad S_{x_{1}}=\sigma _{11}\ell +\sigma _{12}m+\sigma _{33}n\end{aligned}}$ [35]
       ${\begin{aligned}\qquad S_{x_{2}}=\sigma _{12}\ell +\sigma _{22}m+\sigma _{23}n\end{aligned}}$ [36]
       ${\begin{aligned}\qquad S_{x_{3}}=\sigma _{13}\ell +\sigma _{23}m+\sigma _{33}n\end{aligned}}$ [37]
Which gives the total stress:
       $S^{2}={S_{x_{1}}}^{2}+{S_{x_{2}}}^{2}+{S_{x_{3}}}^{2}$ [38]
From this, the projection onto the normal component is:
       $\sigma '=S_{x_{1}}\ell +S_{x_{2}}m+S_{x_{3}}n$ [39]
Substituting Equations 34-36 into Equation 38 gives us:
       ${\begin{aligned}\sigma '&=(\sigma _{11}\ell +\sigma _{12}m+\sigma _{31}n)\ \ell +(\sigma _{12}\ell +\sigma _{22}m+\sigma _{31}n)\ m+(\sigma _{13}\ell +\sigma _{31}m+\sigma _{33}n)\ n\end{aligned}}$
Which simplifies to,
       $\sigma '=\sigma _{11}\ell ^{2}+\sigma _{22}m^{2}+\sigma _{33}n^{2}+2\sigma _{12}\ell m+2\sigma _{23}mn+2\sigma _{31}n\ell$ [40]
The magnitude of the shear component can be determined utilizing $S^{2}={\sigma '}^{2}+\tau ^{2}$, but we cannot easily decompose our shear stress into its constituent elements. Fortunately, we are primarily interested in the maximum shear stress. We know that the plane containing the maximum shear stress is located midway between the planes of principal normal stresses. Starting by setting our known stress state as the principal axis such that $\sigma _{11}=\sigma _{p_{1}}$, $\sigma _{22}=\sigma _{p_{2}}$, and $\sigma _{33}=\sigma _{p_{3}}$, our direction cosine is between the principal axis and the normal of the plane with the maximum shear stress. This means that Equation 39 for projection is rewritten as:
       $\sigma '=\sigma _{p_{1}}\ell ^{2}+\sigma _{p_{2}}m^{2}+\sigma _{p_{3}}n^{2}$ [41]
Squaring this equation gives us:
       ${\sigma '}^{2}={\sigma _{p_{1}}}^{2}\ell ^{4}+{\sigma _{p_{2}}}^{2}m^{4}+{\sigma _{p_{3}}}^{2}n^{4}+2\sigma _{p_{1}}\sigma _{p_{2}}\ell ^{2}m^{2}+2\sigma _{p_{1}}\sigma _{p_{3}}\ell ^{2}n^{2}+2\sigma _{p_{2}}\sigma {p_{3}}m^{2}n_{2}^{p}$ [42]
We can then use the principal components and substitute Equations 34-36 into Equation 37 to get:
       $S^{2}={\sigma _{11}}^{2}\ell ^{2}+{\sigma _{22}}^{2}m^{2}+{\sigma _{33}}^{2}n^{2}$ [43]
After much algebra and putting Equations 41 & 42 into Equation 40, we get:
       $\tau _{MAX}^{2}=(\sigma _{11}-\sigma _{22})^{2}\ell ^{2}m^{2}+(\sigma _{11}-\sigma _{33})^{2}\ell ^{2}n^{2}+(\sigma _{22}-\sigma _{33})^{2}m^{2}n^{2}$ [44]
With this solution, we now have three possible planes. One plane bisects $\sigma _{11}$ and $\sigma _{22}$, another plane bisects $\sigma _{11}$ and $\sigma _{33}$, and the final plane bisects $\sigma _{22}$ and $\sigma _{33}$. (Bisecting means θ=π/4 {\textstyle \theta ={\pi \over 4}}, and cos⁡ π/4=√2/2 {\textstyle \cos {\pi \over 4}={{\sqrt {2}} \over 2}}). Here are the values of $\ell ,m,n$, and $\tau$ for these three planes:
       ${\boldsymbol {\ell }}$ ${\boldsymbol {m}}$ ${\boldsymbol {n}}$ ${\boldsymbol {\tau }}$ $0$ ${\frac {\sqrt {2}}{2}}$ ${\frac {\sqrt {2}}{2}}$ ${\frac {1}{2}}(\sigma _{22}-\sigma _{33})$ ${\frac {\sqrt {2}}{2}}$ $0$ ${\frac {\sqrt {2}}{2}}$ ${\frac {1}{2}}(\sigma _{11}-\sigma _{33})$ ${\frac {\sqrt {2}}{2}}$ ${\frac {\sqrt {2}}{2}}$ $0$ ${\frac {1}{2}}(\sigma _{11}-\sigma _{22})$
By convention, $\sigma _{11}>\sigma _{22}>\sigma _{33}$, and therefore our maximum shear stress is:
       $\tau _{MAX}={\sigma _{11}-\sigma _{33} \over 2}$
Note that we know there are two planes of maximum shear stress, rotated π{\textstyle \pi }/2 from each other. Thus, the direction cosine above are actually ± √2/2 {\textstyle \pm {{\sqrt {2}} \over 2}}.
Because these axial rotations are decoupled, we can represent 3D stress states using Mohr's Circles as seen in Figure 8.

# 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⁡(x$_i$, x$_j$$^′$){\textstyle \cos \left(x_{i}, x'_{j}\right)} is the cosine between $x_{i}$ and $x'_{j}$ and can be rewritten as a$_{ji}$=cos⁡(x$_i$, x$_j$$^′$){\textstyle a_{ji}=\cos \left(x_{i}, x'_{j}\right)} which allows forː
       ${\begin{aligned}S'_{1}=a_{11}S_{1}+a_{12}S_{2}+a_{13}S_{3}\end{aligned}}$ [1]
       ${\begin{aligned}S'_{2}=a_{21}S_{1}+a_{22}S_{2}+a_{23}S_{3}\end{aligned}}$ [2]
       ${\begin{aligned}S'_{3}=a_{31}S_{1}+a_{32}S_{2}+a_{33}S_{3}\end{aligned}}$ [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:
       Name Rank/dimensionality/order of tensor Example Scalar Zero $(11)$ Vector One $\left({\begin{matrix}{25}&{32}&{11}\end{matrix}}\right)$ or $\left({\begin{matrix}{25}\\{32}\\{11}\end{matrix}}\right)$ Matrix Two $\left({\begin{matrix}{25}&{32}&{11}\\{14}&{56}&{4}\\{79}&{42}&{51}\end{matrix}}\right)$
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.
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
       ${\begin{aligned}{\mathbf {S} _{i}}'&={\mathbf {a} _{ij}}{\mathbf {S} _{j}}\\&={\mathbf {a} _{ij}}{\mathbf {T} _{jk}}{\mathbf {Q} _{k}}\\&=\underbrace {{\mathbf {a} _{ij}}{\mathbf {T} _{jk}}{\mathbf {a} _{lk}}} _{\mathbf {T} _{i\ell }}{\mathbf {Q} _{\ell }}'\end{aligned}}$
This tells us that the transformation of $x_{j}$ 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.
       Crystal Tensors Crystal Formation Tensor Number of Independent Components Cubic $\left[{\begin{matrix}{\text{S}}&0&0\\0&{\text{S}}&0\\0&0&{\text{S}}\end{matrix}}\right]$ 1 Tetragonal Hexagonal Trigonal $\left[{\begin{matrix}{\text{S}}_{1}&0&0\\0&{\text{S}}_{1}&0\\0&0&{\text{S}}_{3}\end{matrix}}\right]$ 2 Orthorhombic $\left[{\begin{matrix}{\text{S}}_{1}&0&0\\0&{\text{S}}_{2}&0\\0&0&{\text{S}}_{3}\end{matrix}}\right]$ 3 Monoclinic $\left[{\begin{matrix}{\text{S}}_{11}&0&{\text{S}}_{13}\\0&{\text{S}}_{22}&0\\{\text{S}}_{13}&0&{\text{S}}_{33}\end{matrix}}\right]$ 4 Triclinic $\left[{\begin{matrix}{\text{S}}_{11}&{\text{S}}_{12}&{\text{S}}_{13}\\{\text{S}}_{21}&{\text{S}}_{22}&{\text{S}}_{23}\\{\text{S}}_{31}&{\text{S}}_{32}&{\text{T}}_{33}\end{matrix}}\right]$ 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. 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$_{ni}$σ$_{ij}$−a$_{pj}$σ$_p$=0){\textstyle (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
       $|\sigma _{ij}-\sigma _{p}\delta _{ji}|=\left|{\begin{matrix}\sigma _{x}-\sigma _{p}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{y}-\sigma _{p}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{z}-\sigma _{p}\end{matrix}}\right|=0$
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 ${\boldsymbol {\sigma }}$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 ${\text{I}}_{3}=\det[{\boldsymbol {\sigma }}]$.

# Introducing Strain

This section introduces strain and show tensor symmetry of strain tensor. We also will discuss special subsets of stress and strain including dilatation and deviatoric stresses and strains.

# Average Strain

In this section, we are going to revisit strain to consider it in the infinitesimal limit and to investigate its relationship to displacement using a tensor notation. When we began our discussion, we examined average strain, engineering strain in terms of linear elongation (Figure 1a) and shear deformations (Figure 1b).

# Infinitesimal Strain

While average strain generally looks at the strain of a volume, we will now consider how a point on an elastic body moves and how points near it do also.
We will begin in 2D. Say there is a point ($P$) on an elastic body that is located at coordinate $\{x_{1},\ x_{2}\}$ such as in Figure 2 (this notation will be more convenient when we want to work with tensors). If we deform the body, then $P$ is displaced to $P'$ which has the coordinates $\{x_{1}+u_{1},\ x_{2}+u_{2}\}$. We call $\mathbf {u}$ the displacement vector.
Looking at Figure 2, there is a point infinitesimally close to $P$, called $Q$, with coordinates $\{x_{1}+dx_{1},\ x_{2}+dx_{2}\}$. When $P$ is displaced to $P'$ by the deformation, $Q$ is similarly displaced to $Q'$ with coordinates $\{x_{1}+u_{1}+dx_{1},\ x_{2}+u_{2}+dx_{2}\}$. Thinking critically, the displacement experienced on a body depends on the position on the body. Therefore $\mathbf {u} =u(x_{1},\ x_{2})$. This allows us to use the chain rule to express infinitesimal displacements. Now define the following terms:
       ${\begin{aligned}\operatorname {d} \!u_{1}&={\partial u_{1} \over \partial x_{1}}\operatorname {d} \!x_{1}+{\partial u_{1} \over \partial x_{2}}\operatorname {d} \!x_{2}\end{aligned}}$ [3]
       ${\begin{aligned}\\\operatorname {d} \!u_{2}&={\partial u_{2} \over \partial x_{1}}\operatorname {d} \!x_{1}+{\partial u_{2} \over \partial x_{2}}\operatorname {d} \!x_{2}\end{aligned}}$ [4]
       ${\begin{matrix}e_{11}&={\partial u_{1} \over \partial x_{1}}\qquad e_{12}&={\partial u_{1} \over \partial x_{2}}\\e_{21}&={\partial u_{2} \over \partial x_{1}}\qquad e_{22}&={\partial u_{2} \over \partial x_{2}}\end{matrix}}$ [5-8]
This allows us to write our infinitesimal displacements using Einstein Notation:
       ${\begin{aligned}du_{i}=e_{ij}\ dx_{j}\end{aligned}}$ [9]

### Displacement Tensors

What is the physical significance of this? This is easier to see looking in special directions. Considering the points $P=\{x_{1},\ x_{2}\}$, $Q_{1}=\{x_{1}+dx_{1},\ x_{2}\}$ where $dx_{2}=0$ and $Q_{2}=\{x_{1},\ x_{2}+dx\}$ where $dx_{1}=0$ as seen in Figure 3. Then after the deformation:
       ${\begin{aligned}P'&=\{x_{1}+u_{1},\ x_{2}+u_{2}\}\\{Q_{1}}'&=\{x_{1}+u_{1}+dx_{1}+du_{1},\ x_{2}+u_{2}+du_{2}\}\\{Q_{2}}'&=\{x_{1}+u_{1}+du_{1},\ x_{2}+u_{2}+dx_{2}+du_{2}\}\end{aligned}}$
How do we interpret this? In the case of $Q_{1}$, we have $\operatorname {d} \!x_{2}=0$. Thus, based on Equations 3 & 5 and Equations 4 & 7 expressing the infinitesimal displacements, we can infer that $\operatorname {d} \!u_{1}=e_{11}\operatorname {d} \!x_{1}$, and $\operatorname {d} \!u_{2}=e_{21}\operatorname {d} \!x_{1}$.
This tells us that $e_{11}$ is an expression of uniaxial extension in the $x_{1}$ direction and $e_{21}$ is a rotation of $Q_{1}$ around the point $P$. Similarly, in the case of $Q_{2}$, we can again combine Equation 3 with Equation 6 and Equation 4 with Equation 8 which yields $\operatorname {d} \!u_{1}=e_{12}\ \operatorname {d} \!x_{2}$, and $\operatorname {d} \!u_{2}=e_{22}\ \operatorname {d} \!x_{2}$. Thus, $e_{22}$ is a uniaxial extension in the $x_{2}$ direction, and $e_{12}$ is a rotation of $Q_{2}$ around point $P$. These $\mathbf {e}$ are our displacement tensors.

### The Strain Tensor

Let's return to $P$, displacing to $P'$. What is the relationship between $x_{i}$, and ${x_{i}}'$?
       ${x_{1}}'=x_{1}+u_{1}$ [10]
From Equations 3, 5-8:
       ${\begin{aligned}\operatorname {d} \!u_{1}&=e_{11}\operatorname {d} \!x_{1}+e_{12}\operatorname {d} \!x_{2}\end{aligned}}$ [11]
Integrating, we have
       ${\begin{aligned}\int _{0}^{u_{1}}\operatorname {d} \!u_{1}&=e_{11}\int _{0}^{x_{1}}\operatorname {d} \!x_{1}+e_{12}\int _{0}^{x_{2}}\operatorname {d} \!x_{2}\\u_{1}&=e_{11}x_{1}+e_{12}x_{2}\\{x_{1}}'&=x_{1}+e_{11}x_{1}+e_{12}x_{2}\end{aligned}}$
Which can be rewritten as
       ${\begin{aligned}{x_{1}}'&=\left(1+e_{11}\right)x_{1}+e_{12}x_{2}\end{aligned}}$ [12]
In a similar fashion, we can also prove that
       ${\begin{aligned}{x_{2}}'=e_{21}x_{1}+(1+e_{22})x_{2}\end{aligned}}$ [13]
Looking at our tensor, we can see that our deformations also have translations and rotations. We are not interested in these because they do not tell us about material response such as dilatation (change in volume) or distortion (change in shape). Translations and rotations are a part of the field of mechanics called dynamics. Here we are interested in small scale elastic deformations. Our $e_{12}\neq e_{21}$, but we know that our stress tensor is symmetric as $\sigma _{12}=\sigma _{21}$. We can therefore rewrite our displacement tensor as a combination of a symmetric and antisymmetric tensor.
       $\underbrace {\left({\begin{matrix}e_{11}&e_{12}\\e_{21}&e_{22}\end{matrix}}\right)} _{\mathbf {e} }=\underbrace {\left({\begin{matrix}e_{11}&{1 \over 2}(e_{12}+e_{21})\\{1 \over 2}(e_{21}+e_{12})&e_{22}\end{matrix}}\right)} _{\boldsymbol {\varepsilon }}+\underbrace {\left({\begin{matrix}0&{1 \over 2}(e_{12}+e_{21})\\{1 \over 2}(e_{21}+e_{12})&0\end{matrix}}\right)} _{\boldsymbol {\omega }}$ [14]
Here, ${\boldsymbol {\varepsilon }}$ is the strain tensor and ${\boldsymbol {\omega }}$ is the rotation tensor. This can be seen schematically in Figure 4. In the scope of this text, we are only interested in ${\boldsymbol {\varepsilon }}$, but it is generally still worth remembering that displacement includes both shear and rotation components:
       ${\begin{aligned}u_{i}=\varepsilon _{ij}x_{j}+\omega _{ij}x_{j}\end{aligned}}$ [15]
If a deformation is irrotational, or in other words, the directions of the principal axes of strain do not change as a result of displacement, then $w_{ij}=0$ and
       $u_{i}=\varepsilon _{ij}x_{j}$ [16]
The strain tensor maps the irrotational displacement at a point to an imaginary plane, with normal in any direction that cuts through the point. Because the strain tensor $({\boldsymbol {\varepsilon }})$ is a tensor, it must transform in the same manner as the stress tensor did in earlier sections of this text. As a reminder:
       ${\begin{aligned}{\sigma _{kl}}'&=a_{ki}\ a_{\ell j}\ \sigma _{ij}\end{aligned}}$ [17]
       ${\begin{aligned}{\varepsilon _{k\ell }}'&=a_{ki}\ a_{\ell j}\ \varepsilon _{ij}\end{aligned}}$ [18]

#### Average Engineering Strain

Note that when we first started to look at this subject we defined shear strain as $\gamma =\tan \theta$, which is asymmetric. In terms of our strain tensor, this would be $\gamma _{ij}=2\varepsilon _{ij}$. (It must be rotated back so each side had an angle of 1/2 θ{\textstyle {1 \over 2}\theta }.) You may frequently see a matrix written as:
       $\left({\begin{matrix}\varepsilon _{x}&\gamma _{xy}\\\gamma _{yx}&\varepsilon _{y}\end{matrix}}\right)$
It is sometimes useful to write it this way. However, it is not a tensor, because it does not transform the same as Equations 17 & 18 do, due to its asymmetry. Textbooks generally like using this "average engineering strain" (γ$_{ij}$){\textstyle (\gamma _{ij})}, but we will not be using this here unless absolutely necessary.

### Generalizing 2D to 3D

The results we found for our 2D strain tensor can easily be generalized to 3D by writing them in Einstein Notation and using "3" in the place of "2" in the implicit sums. Equations 3 & 4 are
       $\operatorname {d} \!u_{j}={\partial u_{j} \over \partial x_{i}}\operatorname {d} \!x_{i}$ [19]
Which in 3D expresses 3 equations $j=1,\ 2,\ 3$, and expands to:
       $\operatorname {d} \!u_{j}={\partial u_{j} \over \partial x_{1}}\operatorname {d} \!{x_{1}}+{\partial u_{j} \over \partial x_{2}}\operatorname {d} \!{x_{2}}+{\partial u_{j} \over \partial x_{3}}\operatorname {d} \!{x_{3}}$
The displacement tensor is:
       $e_{ij}={\partial u_{i} \over \partial x_{j}}$ [20]
The strain tensor is:
       $\varepsilon _{ij}={1 \over 2}\left(e_{ij}+e_{ji}\right)$ [21]
The rotation tensor is:
       $\omega _{ij}={1 \over 2}\left(e_{ij}-e_{ji}\right)$ [22]
Which gives us the displacement:
       $u_{i}=\varepsilon _{ij}x_{j}+\omega _{ij}x_{j}=e_{ij}x_{j}$ [23]
And the new displaced coordinates are:
       $x_{j}'=x_{j}+u_{j}$ [24]
Now that we have a symmetric strain tensor with properties analogous to stress, we can examine the other properties using similar methods as when we analyzed stress. For small strains where $\Delta \approx \varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=I_{1}$, we define the mean strain as:
       ${\begin{aligned}{\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33} \over 3}={\varepsilon _{kk} \over 3}=\varepsilon _{m}\end{aligned}}$
Thus,
       $\varepsilon _{m}\approx {\Delta \over 3}$ [25]
The total strain tensor can be broken into dilatation and deviatoric components:
       $\varepsilon _{ij}={\varepsilon _{ij}}'+\varepsilon _{m}=\left(\varepsilon _{ij}-{\Delta \over 3}\delta _{ij}\right)+{\Delta \over 3}\delta _{ij}$ [26]
In a similar fashion, we also have deviatoric stresses and hydrostatic stresses which are analogous to the deviatoric and dilatation strains. The hydrostatic, or mean, stress is:
       $\sigma _{m}={\sigma _{kk} \over 3}={\sigma _{11}+\sigma _{22}+\sigma _{33} \over 3}$ [27]
Therefore, the deviatoric stress can be deduced because:
       ${\begin{aligned}\sigma _{ij}&={\sigma _{ij}}'+{1 \over 3}\delta _{ij}\sigma _{kk}\\{\sigma _{ij}}'&={\sigma _{ij}}+\sigma _{m}\delta _{ij}\\\\{\sigma _{ij}}'&=\left({\begin{matrix}\sigma _{11}-\sigma _{m}&\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}-\sigma _{m}&\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}-\sigma _{m}\end{matrix}}\right)\end{aligned}}$ [28-31]
The principal components of ${\sigma _{ij}}'$ break down to:
       ${\sigma _{1}}'={2 \over 3}\left({\sigma _{1}-\sigma _{2} \over 2}+{\sigma _{1}-\sigma _{3} \over 2}\right)$ [32]
And we know that these are just the maximum shear stresses:
       ${\sigma _{1}}'={2 \over 3}\left({\sigma _{12}}^{MAX}+{\sigma _{13}}^{MAX}\right)$ [33]
Keep in mind that we took this from:
       $0=\left(\sigma '\right)^{3}-J_{1}\left(\sigma '\right)^{2}-J_{2}\sigma '-J_{3}$ [34]
Where:
       ${\begin{aligned}J_{1}&=(\sigma _{11}-\sigma _{m})+(\sigma _{22}-\sigma _{m})+(\sigma _{33}-\sigma _{m})\\J_{2}&={1 \over 6}\left[\left(\sigma _{11}-\sigma _{22}\right)^{2}+\left(\sigma _{22}-\sigma _{33}\right)^{2}+\left(\sigma _{33}-\sigma _{11}\right)^{2}+6\left(\sigma _{12}^{2}+\sigma _{23}^{2}+\sigma _{31}^{2}\right)\right]\\J_{3}&=\det |{\boldsymbol {\sigma }}'|\end{aligned}}$ [35-38]

# Isotropic Response

This section covers linear isotropic response.
We now understand tensor relations and have established two tensors that we need for elasticity; ${\boldsymbol {\sigma }}$ and ${\boldsymbol {\varepsilon }}$. In this chapter we will learn how to relate these to each other. In this particular section we will begin studying elasticity in the case of isotropic elasticity where we assume that all directions have the same material response.

#### Modulus of Elasticity

In earlier sections we learned that Hooke's Law can relate uniaxial loading to uniaxial strain:
       $\sigma _{11}=E\ \varepsilon _{11}$ [1]
Here, $E$ is the modulus of elasticity. Note that we picked the $x_{1}$ direction arbitrarily. It could have easily been $x_{2}$ or $x_{3}$, or any other direction in this crystal, and found the same material response, $E$. This is what comes from working with an isotropic solid.

#### Poisson's Ratio

When we load stress in the $x_{1}$ direction, there is also a natural contraction in the transverse directions, $x_{2}$ and $x_{3}$. This is expressed as Poisson's Ratio, $\nu$.
       $\varepsilon _{22}=\varepsilon _{33}=-\nu \varepsilon _{11}=-{\nu \ \sigma _{11} \over E}$ [2]
For a perfectly incompressible material, which means it conserves volume, we will have a Poisson's Ratio of 0.50. However, most real materials such as common metals have a Poisson's Ratio of around 0.33.
For linear elastic, isotropic materials, normal stresses will not induce shear strain and shear stress will not cause normal stress. In linear elasticity, the various contributions to strain can be superimposed. We can use the two above equations for the Modulus of Elasticity and Poisson's Ratio to determine that if a triaxial normal stress is applied, the strain will be:
       ${\begin{aligned}\varepsilon _{11}&={\sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\quad {\begin{cases}\varepsilon _{11}&={1 \over E}\left[\sigma _{11}-\nu \left(\sigma _{22}\sigma _{33}\right)\right]\\\varepsilon _{22}&={1 \over E}\left[\sigma _{22}-\nu \left(\sigma _{33}\sigma _{11}\right)\right]\\\varepsilon _{33}&={1 \over E}\left[\sigma _{33}-\nu \left(\sigma _{11}\sigma _{22}\right)\right]\end{cases}}\end{aligned}}$ [3-5]
Moving to shear relations, the shear stress and strain are related by the shear modulus, $G$.
       $\sigma _{12}=G\gamma _{12}=G2\varepsilon _{12}$ [6]
Again, due to the isotropic nature, this relation holds in all the other directions as well. These three constants of proportionality are sufficient to describe the isotropic linear response. They take typical values (for common engineering metals):
       ${\begin{aligned}50<\ &E<200\quad GPa\\25<\ &G<75\quad GPa\\0.25<\ &\nu <0.35\end{aligned}}$
There are many other useful relations that we will summarize here. The Block Modulus, $K$, is the ratio of the hydrostatic pressure to the dilatation it produces:
       K=−p/Δ=σ$_m$/Δ=1/β {\textstyle K={-p \over \Delta }={\sigma _{m} \over \Delta }={1 \over \beta }} [7]
Where −p{\textstyle -p} is the hydrostatic pressure and $\beta$ is the compressibility.
Adding together Equations 3-5 for the triaxial normal strain, we get:
       ${\begin{aligned}\underbrace {\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}} _{\Delta }&={1-2\nu \over E}(\underbrace {\sigma _{11}+\sigma _{22}+\sigma _{33}} _{3\sigma _{m}})\\\Delta &={1-2\nu \over E}3\sigma _{m}\end{aligned}}$ [8]
Applying this to Equations 7 & 8 for the Block Modulus results in:
       $K={\sigma _{m} \over \Delta }={E \over 3(1-2\nu )}$ [9]
These next solutions are given without proof, because several advanced topics are used to show the relationships.
       $G={E \over 2(1+\nu )}$ [10]
       $E={9K \over 1+3K/G}$ [11]
       $\nu ={1-2G/3K \over 2+2G/3K}$ [12]
       $G={3(1-2\nu )K \over 2(1+\nu )}$ [13]
       $K={E \over 9-3E/G}$ [14]
The stress-strain relationships can be expressed in compact tensor notation:
       $\varepsilon _{ij}={1+\nu \over E}\sigma _{ij}-{\nu \over E}\sigma _{kk}\delta _{ij}$ [15]
Looking specifically at a normal strain, $\varepsilon _{22}$ for example:
       $\varepsilon _{22}={1+\nu \over E}\sigma _{22}-{\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\delta _{22}$
Similarly, for a shear strain, $\varepsilon _{12}$, gets us:
       $\varepsilon _{12}={1+\nu \over E}\sigma _{12}-{\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\delta _{12}$
From G=E/(2(1+ν)) {\textstyle G={E \over 2(1+\nu )}}, we find that:
       ${\begin{aligned}{1+\nu \over E}&={1 \over 2G}\\\varepsilon _{12}&={1 \over 2G}\ \sigma _{12}\\\sigma _{12}&=2G\ \varepsilon _{12}\end{aligned}}$
For a given stress state:
       ${\boldsymbol {\sigma }}=\left({\begin{matrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\end{matrix}}\right)$
We can only write the strain tensor as:
       ${\boldsymbol {\varepsilon }}=\left({\begin{matrix}{1 \over E}(\sigma _{11}-\nu (\sigma _{22}+\sigma _{33}))&{\sigma _{12} \over 2G}&{\sigma _{13} \over 2G}\\{\sigma _{12} \over 2G}&{1 \over E}(\sigma _{22}-\nu (\sigma _{33}+\sigma _{11}))&{\sigma _{23} \over 2G}\\{\sigma _{13} \over 2G}&{\sigma _{23} \over 2G}&{1 \over E}(\sigma _{33}-\nu (\sigma _{11}+\sigma _{22}))\end{matrix}}\right)$ [16]
So we have an expression for strain in terms of the applied load, the stress. This can certainly be useful for predicting the deformation to anticipate when loading in application. There is another application we are interested in. Given we deform a part, what stress does it feel? This answers questions such as, "how much can I bend this part before I reach the yield strength?" Basically, we want to take our existing solution and invert it.
To do this, we can take our first triaxial strain equation, Equation 3, and rearrange it:
       ${\begin{aligned}\varepsilon _{11}&={\sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\\&\quad +0={\nu \sigma _{11} \over E}-{\nu \sigma _{11} \over E}\\&={\sigma _{11} \over E}+{\nu \sigma _{11} \over E}-{\nu \sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\end{aligned}}$
And we arrive at:
       ${\begin{aligned}\varepsilon _{11}&={1+\nu \over E}\sigma _{11}-{\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\end{aligned}}$ [17]
We can additionally take our earlier equation for $\Delta$, Equation 8, and rearrange it to get:
       ${\begin{aligned}\Delta &={1-2\nu \over E}3\sigma _{m}\\\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}&={1-2\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\end{aligned}}$
And
       $\sigma _{11}+\sigma _{22}+\sigma _{33}={E \over 1-2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})$ [18]
Then by substituting Equation 18 into Equation 17, we get:
       ${\begin{aligned}\varepsilon _{11}&={1+\nu \over E}\sigma _{11}-{\nu \over E}\left({E \over 1+2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\right)\\\sigma _{11}&=\varepsilon _{11}\left({E \over 1+\nu }\right)+{E \over 1+\nu }{\nu \over E}{E \over 1-2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\end{aligned}}$
Which becomes
       ${\begin{aligned}\sigma _{11}&=\left({E \over 1+\nu }\right)\varepsilon _{11}+{\nu E \over (1+\nu )(1-2\nu )}(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\end{aligned}}$ [19]
Thinking about shear stress is simpler:
       ${\begin{aligned}\sigma _{12}&=2G\ \varepsilon _{12}\end{aligned}}$
Substituting in Equation 10 results in
       ${\begin{aligned}\sigma _{12}&=2\ {E \over 2(1+\nu )}\ \varepsilon _{12}\end{aligned}}$
Which becomes
       ${\begin{aligned}\sigma _{12}&=\left({E \over 1+\nu }\right)\varepsilon _{12}\end{aligned}}$ [20]
Taking Equations 19 & 20 gives us the tensor expression for stress in terms of strain, which is:
       $\sigma _{ij}={E \over 1+\nu }\varepsilon _{ij}+\underbrace {\nu E \over (1+\nu )(1-2\nu )} _{\lambda }\varepsilon _{kk}\delta _{ij}$ [21]
Here, $\lambda$ is the Lamé Constant. We can analyze these results and extract useful equations. We will begin by extracting the deviatoric and hydrostatic contributions to stress.
       $\sigma _{ij}'=2G\ \varepsilon _{ij}'$ [22]
We can show this is true by considering shear and normal components.
Shear:
       $\sigma _{12}'=\sigma _{12}={E \over (1+\nu )}\varepsilon _{12}=2G\ \varepsilon _{12}=2G\ \varepsilon _{12}'$
Normal:
       ${\begin{aligned}\sigma _{11}'&={2\sigma _{22}-\sigma _{22}-\sigma _{33} \over 3}\\&={1 \over 3}\left[2\ (2G\varepsilon _{11}+\lambda \varepsilon _{11}+\lambda \varepsilon _{22}+\lambda \varepsilon _{33})-(2G\varepsilon _{22}+\lambda \varepsilon _{11}+\lambda \varepsilon _{22}+\lambda \varepsilon _{33})-(2G\varepsilon _{33}+\lambda \varepsilon _{11}+\lambda \varepsilon _{22}+\lambda \varepsilon _{33})\right]\\&={1 \over 3}(4G\varepsilon _{11}-2G\varepsilon _{22}-2G\varepsilon _{33})\\&=2G{2\varepsilon _{11}-\varepsilon _{22}-\varepsilon _{33} \over 3}\\&=2G\varepsilon _{11}'\end{aligned}}$
The relationship between the hydrostatic stress and the mean strain is:
       $\sigma _{ii}={E \over 1-2\nu }\varepsilon _{kk}=3K\varepsilon _{kk}$ [23]
Note that we've seen this expression several times before!

## Simplified Cases

Plane stress $\sigma _{33}=0$.
Once again looking at our earlier triaxial strain equations, Equations 3 & 4, we will start by adding the first two together:
       ${{\begin{aligned}\varepsilon _{11}&={1 \over E}[\sigma _{11}-\nu \sigma _{22}]={1 \over E}\sigma _{11}-{\nu \over E}\sigma _{22}\\+\nu (\varepsilon _{22}&={1 \over E}[\sigma _{22}-\nu \sigma _{11}]=-{\nu \over E}\sigma _{11}+{1 \over E}\sigma _{22})\end{aligned}} \over \varepsilon _{11}+\nu \varepsilon _{22}=\left({1 \over E}-{\nu ^{2} \over E}\right)\sigma _{11}}{\begin{aligned}\\\\\rightarrow \sigma _{11}&={E \over 1-\nu ^{2}}(\varepsilon _{11}+\nu \varepsilon _{22})\\\sigma _{22}&={E \over 1-\nu ^{2}}(\varepsilon _{22}+\nu \varepsilon _{11})\end{aligned}}$
Plane stress encountered either as loaded sheet, or more likely a pressure vessel.
Another simplification is Plane Strain, where $\varepsilon _{33}=0$, typically when one dimension is much greater than the other two so $\varepsilon _{11}$ and $\varepsilon _{22}$ are much greater than $\varepsilon _{33}$, such as a long rod where strain along the length of the rod is constrained. Here, Equation 5 rearranges to:
       ${\begin{aligned}\varepsilon _{33}&={1 \over E}\left[\sigma _{33}-\nu (\sigma _{11}+\sigma _{22})\right]=0\\\sigma _{33}&=\nu (\sigma _{11}+\sigma _{22})\end{aligned}}$
Note that $\sigma _{33}$ does not equal zero just because $\varepsilon _{33}$ equals zero. Substituting this $\sigma _{33}$ equation into Equations 3-5 yields us:
       ${\begin{aligned}\varepsilon _{11}&={1 \over E}\left[(1-\nu ^{2})\sigma _{22}-\nu (1+\nu )\sigma _{11}\right]\\\varepsilon _{22}&={1 \over E}\left[(1-\nu ^{2})\sigma _{22}-\nu (1+\nu )\sigma _{11}\right]\end{aligned}}$

# Anisotropic Response

This section covers linear anisotropic response.
At this point, we have developed our notation for describing stress and strain, and have written expressions relating the two in terms of material parameters such as $E$, $G$, and $\nu$. This is all within the assumption of a homogeneous, isotropic solid. This is often times a fairly good approximation, and you are likely to recognize the expressions derived here. Hopefully, this section will provide an insight to mechanics of solids as well as other materials-focused topics.
We will begin by looking at an example. Consider a brittle material that is hot at first and then suddenly quenched to low temperature. This can cause thermal shock and breakage. So the part is quenched from $T_{max}$ to $T_{min}$, where $\Delta T=T_{max}-T_{min}$.
It is known that, without proof, the thermal stress is
       $\sigma _{TH}=-{\frac {\alpha \Delta TE}{(1-2\nu )}}$ [1]
Where $\alpha$ is the coefficient of thermal expansion.
Where is this from? We know that this is the hydrostatic stress:
       $\sigma _{ii}={\frac {E}{1-2\nu }}\varepsilon _{kk}$
Which can be rewritten as
       $\sigma _{11}+\sigma _{22}+\sigma _{33}={\frac {E}{1-2\nu }}(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})$
Where
       $\sigma _{m}={\frac {1}{3}}\sigma _{ii}$
Then it must be approximated that
       ${\frac {1}{3}}\Delta =\alpha \Delta T$
For spherical, circular flaws, or cracks with radius $r$, the energy is
       $U_{TOT}=U_{o}-U_{STRAIN}+U_{SURF}$ [2]
Where $U_{TOT}$ is the total energy of the system and $U_{o}$ is the energy of the stress-free and crack-free system with volume $V_{o}$.
The strain energy is
       $U_{STRAIN}={\frac {V_{o}\sigma _{TH}^{2}}{2E}}-{\frac {N\sigma _{TH}^{2}}{2E}}{\frac {4}{3}}\pi r^{3}$ [3]
Where $N$ is the number of cracks.
Where is this from? Our expression for strain energy when normal stress is applied is
       $U_{o}^{NORMAL}={\frac {1}{2}}{\frac {\sigma _{11}^{2}}{E}}$
Where $U_{o}^{NORMAL}$ is the energy per unit volume.
Therefore, the first term in Equation 3 is simply the energy due to the thermal strain. The second term is the strain relieved when $N$ number of cracks of volume ${\frac {4}{3}}\pi r^{3}$ open (we are subtracting this term because the strain is being relieved):
       $N\left({\frac {4}{3}}\pi r^{3}\right){\frac {\sigma _{TH}^{2}}{2E}}$
Is this reasonable? This is probably slightly underestimated. Finally, the last term is
       $U_{SURF}=NG_{c}\pi r^{2}$ [4]
Where $G_{c}$ is the toughness energy required to create new surface (units of $J/m^{2}$). Note for ideal brittle materials, $G_{c}=2\gamma$.
In this example, isotropic elasticity is used, and in all likelihood is a reasonable approximation for bulk polycrystalline or amorphous solids. Sometimes we cannot assume an isotropic solid. This is most common when dealing with systems where single crystals are studied such as the one in Figure 1. Looking at Figure 1, even if $|P_{1}|=|P_{3}|$, pulling in direction $|P_{1}|$ will encounter different resistance (constant of proportionality in Hooke's Law) than pulling in direction $|P_{2}|$.
The anisotropic expression of Hooke's Law is a tensor relation:
       $\sigma _{ij}=C_{ijkl}\varepsilon _{kl}$ [5]
       $\varepsilon _{ij}=S_{ijkl}\sigma _{kl}$ [6]
Where $C_{ijkl}$ is the stiffness or elastic constant and $S_{ijkl}$ is the elastic compliance.
Here, the primary focus will be on $C_{ijkl}$, but the two behave similarly. The elastic constant tensor is a fourth rank tensor that connects two second rank tensors. In Equation 5, the right hand side is a double sum over $k$ and $l$, resulting in 9 terms in the sum. Since there are 9 expressions for $\sigma _{ij}$, there is a total of 81 elements in ${\boldsymbol {C}}$. This seems like a large number, but are they unique? We know ${\boldsymbol {\sigma }}$ and ${\boldsymbol {\varepsilon }}$ are symmetric, therefore
       $\sigma _{ij}=\sigma _{ji}\rightarrow C_{ijkl}=C_{jikl}$
And
       $\varepsilon _{kl}=\varepsilon _{lk}\rightarrow C_{ijkl}=C_{ijlk}$
This reduces the number of elastic constants from 81 to 36 unique values. To further simplify, we must consider the elastic energy. We know that the energy of the system is
       $U_{o}={\frac {1}{2}}\sigma _{ij}\varepsilon _{ij}$ [7]
As depicted in Figure 2.
For homogeneous elastic loading, we can write
       $dU_{o}=\sigma _{ij}d\varepsilon _{ij}$ [8]
Which is the superposition of
       ${\begin{aligned}dU_{o}&=\sigma _{11}d\varepsilon _{11}+\sigma _{12}d\varepsilon _{12}+\sigma _{13}d\varepsilon _{13}\\&+\sigma _{21}d\varepsilon _{21}+\sigma _{22}d\varepsilon _{22}+\sigma _{23}d\varepsilon _{23}\\&+\sigma _{31}d\varepsilon _{31}+\sigma _{32}d\varepsilon _{32}+\sigma _{33}d\varepsilon _{33}\end{aligned}}$ [9]
Now we will consider starting in an initial state $\sigma =\varepsilon =0$ and straining $\varepsilon _{11}$. This increases the internal energy, the energy stored in bonds, by
       $dw_{1}=\sigma _{11}d\varepsilon _{11}=C_{1111}\varepsilon _{11}d\varepsilon _{11}$ [10]
Where $\sigma _{11}=C_{1111}\varepsilon _{11}$ from Equation 5 (all strain except $\varepsilon _{11}$ are zero).
Integrating this
       $\int _{0}^{w_{1}}dw_{1}=\int _{0}^{\varepsilon _{11}}C_{1111}\varepsilon _{11}d\varepsilon _{11}$
Becomes
       $w_{1}={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}$ [11]
Now apply a second strain deformation $\varepsilon _{22}$ and integrate:
       $\int _{0}^{w_{2}}dw_{2}=\int _{0}^{\varepsilon _{22}}\sigma _{22}d\varepsilon _{22}=\int _{0}^{\varepsilon _{22}}(C_{2222}\varepsilon _{22}+C_{2211}\varepsilon _{11})d\varepsilon _{22}$
Using Equation 5 and the fact that all $\varepsilon =0$ except $\varepsilon _{11}$ and $\varepsilon _{22}$. This becomes
       $w_{2}={\frac {1}{2}}C_{2222}\varepsilon _{22}^{2}+C_{2211}\varepsilon _{11}\varepsilon _{22}$ [12]
Then the total work is
       $w_{TOT}=w_{1}+w_{2}={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}+{\frac {1}{2}}C_{2222}\varepsilon _{22}^{2}+C_{2211}\varepsilon _{11}\varepsilon _{22}$ [13]
Imagine now we reverse the order; first applying $\varepsilon _{22}$ then $\varepsilon _{11}$.
       $w_{1}'={\frac {1}{2}}C_{2222}\varepsilon _{22}$
       $w_{2}'={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}+C_{1122}\varepsilon _{22}\varepsilon _{11}$
       $w_{TOT}'={\frac {1}{2}}C_{1111}\varepsilon _{11}^{2}+{\frac {1}{2}}C_{2222}\varepsilon _{22}^{2}+C_{1122}\varepsilon _{11}\varepsilon _{22}$ [14]
In the linear regime superposition holds, therefore the order is not important and
       ${\begin{aligned}w_{TOT}=w_{TOT}'\\\therefore C_{2211}=C_{1122}\end{aligned}}$
This is generalized to
       $C_{ijkl}=C_{klij}$
Which reduces the number of unique values to 21. This is the fewest number of elastic constants that must be specified for an arbitrary crystal.

## Crystal Symmetry Simplifications

Fortunately, tensor relations are tied to the symmetry of a crystal. For high symmetry crystals, such as cubic, only 3 unique values are needed and further many zeroes exist. The following rules can be utilized to simplify the notation:
* All $C_{iiii}=C_{1111}$
* All $C_{iijj}=C_{1122}$
* All $C_{ijij}=C_{2323}$
* All other $C_{ijkl}=0$ (not in Equations 1-3).
This makes the situation much simpler. Next, we will simplify the notation using Voigt notation as described in the table below.
       ${\boldsymbol {ij}}$ pair 11 22 33 23 31 12 m 1 2 3 4 5 6
This is often (not always) found in textbooks and manuscripts. In this notation, work is
       ${\begin{aligned}dw&=\sigma _{1}d\varepsilon _{1}+\sigma _{6}d\varepsilon _{6}+\sigma _{5}d\varepsilon _{5}\\&+\sigma _{6}d\varepsilon _{6}+\sigma _{2}d\varepsilon _{2}+\sigma _{4}d\varepsilon _{4}\\&+\sigma _{5}d\varepsilon _{5}+\sigma _{4}d\varepsilon _{4}+\sigma _{3}d\varepsilon _{3}\\dw&=\sigma _{1}d\varepsilon _{1}+\sigma _{2}d\varepsilon _{2}+\sigma _{3}d\varepsilon _{3}+2\sigma _{4}d\varepsilon _{4}+2\sigma _{5}d\varepsilon _{5}+2\sigma _{6}d\varepsilon _{6}\end{aligned}}$
Using Voigt notation, we can now write the stress-strain relation
       ${\boldsymbol {\sigma }}={\boldsymbol {C}}{\boldsymbol {\varepsilon }}$
In 2D instead of the original 4D format.
       $\left[{\begin{matrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{matrix}}\right]=\left[{\begin{matrix}{\text{C}}_{11}&{\text{C}}_{12}&{\text{C}}_{13}&{\text{C}}_{14}&{\text{C}}_{15}&{\text{C}}_{16}&{\text{C}}_{14}&{\text{C}}_{15}&{\text{C}}_{16}\\{\text{C}}_{12}&{\text{C}}_{22}&{\text{C}}_{23}&{\text{C}}_{24}&{\text{C}}_{25}&{\text{C}}_{26}&{\text{C}}_{24}&{\text{C}}_{25}&{\text{C}}_{26}\\{\text{C}}_{13}&{\text{C}}_{23}&{\text{C}}_{33}&{\text{C}}_{34}&{\text{C}}_{35}&{\text{C}}_{36}&{\text{C}}_{34}&{\text{C}}_{35}&{\text{C}}_{36}\\{\text{C}}_{14}&{\text{C}}_{24}&{\text{C}}_{34}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}\\{\text{C}}_{15}&{\text{C}}_{25}&{\text{C}}_{35}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}\\{\text{C}}_{16}&{\text{C}}_{26}&{\text{C}}_{36}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}\\{\text{C}}_{14}&{\text{C}}_{24}&{\text{C}}_{34}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}\\{\text{C}}_{15}&{\text{C}}_{25}&{\text{C}}_{35}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}\\{\text{C}}_{16}&{\text{C}}_{26}&{\text{C}}_{36}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}\end{matrix}}\right]\left[{\begin{matrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\\\varepsilon _{4}\\\varepsilon _{5}\\\varepsilon _{6}\end{matrix}}\right]$ [15]
In this format, it is still possible to transform the vectors and tensor ${\boldsymbol {C}}$ together. This can further be shortened to
       $\left[{\begin{matrix}\sigma _{1}\\\sigma _{2}\\\sigma _{3}\\\sigma _{4}\\\sigma _{5}\\\sigma _{6}\end{matrix}}\right]=\left[{\begin{matrix}{\text{C}}_{11}&{\text{C}}_{12}&{\text{C}}_{13}&{\text{C}}_{14}&{\text{C}}_{15}&{\text{C}}_{16}\\{\text{C}}_{12}&{\text{C}}_{22}&{\text{C}}_{23}&{\text{C}}_{24}&{\text{C}}_{25}&{\text{C}}_{26}\\{\text{C}}_{13}&{\text{C}}_{23}&{\text{C}}_{33}&{\text{C}}_{34}&{\text{C}}_{35}&{\text{C}}_{36}\\{\text{C}}_{14}&{\text{C}}_{24}&{\text{C}}_{34}&{\text{C}}_{44}&{\text{C}}_{45}&{\text{C}}_{46}\\{\text{C}}_{15}&{\text{C}}_{25}&{\text{C}}_{35}&{\text{C}}_{45}&{\text{C}}_{55}&{\text{C}}_{56}\\{\text{C}}_{16}&{\text{C}}_{26}&{\text{C}}_{36}&{\text{C}}_{46}&{\text{C}}_{56}&{\text{C}}_{66}\end{matrix}}\right]\left[{\begin{matrix}\varepsilon _{1}\\\varepsilon _{2}\\\varepsilon _{3}\\\gamma _{4}\\\gamma _{5}\\\gamma _{6}\end{matrix}}\right]$ [16]
However, this has shear strain components ($\gamma$), which means that in this representation, $C$ is not a tensor but a matrix. This is because the rotational properties of tensors are not satisfied. We can see how these two are related by writing out one of the stresses from Equation 15:
       $\sigma _{1}=C_{11}\varepsilon _{1}+C_{12}\varepsilon _{2}+C_{13}\varepsilon _{3}+C_{14}\varepsilon _{4}+C_{15}\varepsilon _{5}+C_{16}\varepsilon _{6}+C_{14}\varepsilon _{4}+C_{15}\varepsilon _{5}+C_{16}\varepsilon _{6}$
       $\sigma _{1}=C_{11}\varepsilon _{1}+C_{12}\varepsilon _{2}+C_{13}\varepsilon _{3}+C_{14}2\varepsilon _{4}+C_{15}2\varepsilon _{5}+C_{16}2\varepsilon _{6}$
Where $\gamma _{4}=2\varepsilon _{4}$, $\gamma _{5}=2\varepsilon _{5}$, and $\gamma _{6}=2\varepsilon _{6}$.
Applying our simplifications due to crystalline symmetry we can write for a cubic crystal:
       $C_{ij}^{CUBIC}=\left[{\begin{matrix}{\text{C}}_{11}&{\text{C}}_{12}&{\text{C}}_{12}&0&0&0\\{\text{C}}_{12}&{\text{C}}_{11}&{\text{C}}_{12}&0&0&0\\{\text{C}}_{12}&{\text{C}}_{12}&{\text{C}}_{11}&0&0&0\\0&0&0&{\text{C}}_{44}&0&0\\0&0&0&0&{\text{C}}_{44}&0\\0&0&0&0&0&{\text{C}}_{44}\end{matrix}}\right]$ [17]
Returning to our definition of strain energy (Equation 7) we can write this in Voigt notation:
       $U_{o}={\frac {1}{2}}\sigma _{i}\varepsilon _{i}$ [18]
And we can write the stress-strain relation (Equation 5) as
       $\sigma _{j}=C_{jk}\varepsilon _{k}$ [19]
Which allows us to write
       $U_{o}={\frac {1}{2}}C_{ij}\varepsilon _{i}\varepsilon _{j}$ [20]
Which is a double sum over $i$ and $j$. Putting $C_{ij}^{CUBIC}$ from Equation 17 into Equation 20 yields
       $U_{o}^{CUBIC}={\frac {1}{2}}C_{11}(\varepsilon _{1}^{2}+\varepsilon _{2}^{2}+\varepsilon _{3}^{2})+2C_{44}(\varepsilon _{4}^{2}+\varepsilon _{5}^{2}+\varepsilon _{6}^{2})+C_{12}(\varepsilon _{1}\varepsilon _{2}+\varepsilon _{2}\varepsilon _{3}+\varepsilon _{3}\varepsilon _{1})$ [21]
Here are a few useful relationships for cubic crystals which are given without proof:
       $C_{11}={\frac {S_{11}+S_{12}}{(S_{11}-S_{12})(S_{11}+2S_{12})}}$ [22]
       $C_{12}={\frac {-S_{12}}{(S_{11}-S_{12})(S_{11}+2S_{12})}}$ [23]
       $C_{44}={\frac {1}{S_{44}}}$ [24]
       ${\frac {1}{E}}=S_{11}-2\left[(S_{11}-S_{12})-{\frac {1}{2}}S_{44}\right](l^{2}m^{2}+m^{2}n^{2}+l^{2}n^{2})$ [25]
Here $l$, $m$, and $n$ are direction cosine from $x_{1}$, $x_{2}$, and $x_{3}$ and are the direction of applied uniaxial loading. Mapping the anisotropic elastic properties to the isotropic limit, we have
       $S_{11}={\frac {1}{E}}$ [26]
       $S_{12}={\frac {-\nu }{E}}$ [27]
       $S_{44}={\frac {1}{G}}$ [28]
Yet for a truly isotropic medium, only 2 material parameters are needed. We find
       $S_{44}=2(S_{11}-S_{12})$ [29]
And
       $C_{12}=\lambda$ [30]
       $C_{11}=2G+\lambda$ [31]
       $C_{44}={\frac {1}{2}}(C_{11}-C_{12})$ [32]
       $A={\frac {2C_{44}}{C_{11}-C_{12}}}$ [33]
Where $A$ is the Zener anisotropy ratio ($A=1$ for isotropic material).

# Appendix A: Coordinate Notation

This is a discussion of the notational choices used in the book including the coordinate choices.

# Appendix B: List of Variables

This section lists the variables used throughout the text.