Overview of Elasticity of Materials/Printable version

Overview of Elasticity of Materials

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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 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 hold's 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.

Figure 1: (a) The external forces, $\mathbf {P} ,$ administered to the body will be transmitted internally. A point on an imaginary slice taken through the body will have force on the surface. (b) The force on this slice can be projected into components acting normal or tangential to the area, $A$.

As you recall the stress is defined as the force divided by the area over 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 further be projected into the two orthogonal directions identified in Figure 1 and $x_{1}$ 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'll use $\sigma _{11}$. These textbooks also use $\tau$ to denote shear stress, such as $\tau _{xy}$ whereas here we'll use $\sigma _{12}$. This allows the stress state to be succinctly written in matrix (tensor) form

$\sigma =\left({\begin{aligned}\sigma _{11}\quad \sigma _{12}\quad \sigma _{13}\\\sigma _{21}\quad \sigma _{22}\quad \sigma _{23}\\\sigma _{31}\quad \sigma _{32}\quad \sigma _{33}\end{aligned}}\right)$

Figure 2: An infinitesimal cuboid of material with the stresses defined according to the $(x_{1},x_{2},x_{3})$ coordinate system.

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

Let's begin by simplifying the picture we're 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.

Figure 3: (a) An area, $A$, defined for the plane stress condition in which the normal of the area is $x_{1}'$, rotated from $x_{1}$ by $\theta$. The projection of A into the $x_{1}$ and $x_{1}$ directions are shown. (b) The components of the total stress on the area are shown.

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:

${\begin{aligned}F_{1}=\sigma _{11}A\cos \!\theta +\sigma _{12}A\sin \!\theta =S_{1}A\\F_{2}=\sigma _{22}A\sin \!\theta +\sigma _{12}A\cos \!\theta =S_{2}A\end{aligned}}$           [eq 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), $S_{1}$ and $S_{2}$ are the total stresses in the $x_{1}$ and $x_{2}$ directions, and $\sigma _{12}=\sigma _{21}$. Then dividing by A yields:

${\begin{aligned}S_{1}=\sigma _{11}\cos \!\theta +\sigma _{12}\sin \!\theta \\S_{2}=\sigma _{22}\sin \!\theta +\sigma _{12}\cos \!\theta \end{aligned}}$           [eq 3 & 4]

Projecting the total stresses shown in Figure 3 (b) into the normal direction in the $x_{1}'$ coordinate we get

$\sigma _{11}'=S_{1}\!\cos \!\theta +S_{2}\!\sin \!\theta$           [eq 5]

In a similar fashion we project tangential to the plane and yield

$\sigma _{12}'=S_{2}\!\cos \!\theta -S_{1}\!\sin \!\theta$           [eq 6]

Resulting in

${\begin{aligned}\sigma _{11}'&=(\sigma _{11}\cos \!\theta +\sigma _{12}\sin \!\theta )\cos \!\theta +(\sigma _{22}\sin \!\theta +\sigma _{12}\cos \!\theta )\sin \!\theta \\&=\sigma _{11}\cos ^{2}\!\theta +\sigma _{22}\sin ^{2}\!\theta +2\ \sigma _{12}\sin \!\theta \cos \!\theta \end{aligned}}$           [eq 7]

and

${\begin{aligned}\sigma _{12}'&=(\sigma _{22}\sin \!\theta +\sigma _{12}\cos \!\theta )\cos \!\theta -(\sigma _{11}\cos \!\theta +\sigma _{12}\sin \!\theta )\sin \!\theta \\&=\sigma _{12}(cos^{2}\!\theta -\sin ^{2}\!\theta )+(\sigma _{22}-\sigma _{11})\sin \!\theta \cos \!\theta \end{aligned}}$           [eq 8]

Figure 4: A new area that is rotated by ${\frac {\pi }{2}}$ from the original shown in Figure 3.

It is known that $\sigma _{12}'=\sigma _{21}'$ therefore only $\sigma _{22}'$ needs determining. To do so we define a new area that is rotated by ${\textstyle \pi }$/2 relative to our original plane as shown in Figure 4. In this new orientation,

${\begin{aligned}S_{1}A&=\sigma _{11}A\cos(\theta +{\pi \over 2})+\sigma _{12}A\sin(\theta +{\pi \over 2})\\S_{1}&=-\sigma _{11}\sin \!\theta +\sigma _{12}\cos \!\theta \end{aligned}}$           [eq 9]

and

${\begin{aligned}S_{2}A&=\sigma _{22}A\sin(\theta +{\pi \over 2})+\sigma _{12}A\cos(\theta +{\pi \over 2})\\S_{2}&=\sigma _{22}\cos \!\theta -\sigma _{12}\sin \!\theta \end{aligned}}$           [eq 10]

Projecting the total stress in the normal direction yields

${\begin{aligned}\sigma _{22}'&=S_{1}\cos(\theta +{\pi \over 2})+S_{2}\sin(\theta +{\pi \over \theta })\\&=-S_{x_{1}}\sin \!\theta +S_{x_{2}}\cos \!\theta \end{aligned}}$           [eq 11]

Substituting Equations 9 and 10 for $S_{1}$ and $S_{2}$ into Equation 11 for $\sigma _{22}'$ yields

${\begin{aligned}\sigma _{22}'&=-(-\sigma _{11}\sin \!\theta +\sigma _{12}\cos \!\theta )\sin \!\theta +(\sigma _{22}\cos \!\theta -\sigma _{12}\sin \!\theta )\cos \!\theta \\&=\sigma _{11}\sin ^{2}\!\theta +\sigma _{22}\cos ^{2}\!\theta -2\ \sigma _{12}\sin \!\theta \cos \!\theta \end{aligned}}$           [eq 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

${\begin{aligned}\sigma _{11}'&={\sigma _{11}+\sigma _{22} \over 2}+{\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }+\sigma _{12}\sin {2\theta }\end{aligned}}$           [eq 13]

${\begin{aligned}\sigma _{22}'&={\sigma _{11}+\sigma _{22} \over 2}-{\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }-\sigma _{12}\sin {2\theta }\end{aligned}}$           [eq 14]

and

${\begin{aligned}\sigma _{12}'&={\sigma _{22}-\sigma _{11} \over 2}\sin 2\theta +\sigma _{12}\cos 2\theta \end{aligned}}$           [eq 15]

Principal Stress

There are numerous immediate results that come from this derivation, from which we can gain greater insights. One results 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 $\sigma$ 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 $\sigma _{p2}$ 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 $\sigma$ 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

${\begin{aligned}\tan 2\theta &={2\ \sigma _{12} \over \sigma _{11}-\sigma _{22}}\end{aligned}}$           [eq 16]

It is observed graphically by plotting $\tan 2\theta$ in Figure 5 that adjacent roots are each separated by ${\textstyle \pi }$/2. Furthermore, we can now utilize the Pythagorean Theorem to solve for our principal stresses.

Figure 5: Graphical demonstration that the roots of $tan2\theta$ are separate by ${\frac {\pi }{2}}$.

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 by yield

${\begin{aligned}\sin 2\theta &=\pm {\sigma _{12} \over \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\end{aligned}}$           [eq 17]

and

${\begin{aligned}\\\cos 2\theta &=\pm {{1 \over 2}(\sigma _{11}-\sigma _{22}) \over \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}}\end{aligned}}$           [eq 18]

These equations tell us for a given stress state $\sigma$ what rotation is needed 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

${\begin{aligned}&={\sigma _{11}+\sigma _{22} \over 2}\pm \left({1 \over 4}(\sigma _{11}-\sigma _{22})^{2}+{\sigma _{12}}^{2}\right)^{1 \over 2}\end{aligned}}$           [eq 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 $2\theta$:

${\begin{aligned}\\{\sigma _{22}-\sigma _{11} \over 2\ \sigma _{12}}&={\sin 2\theta \over \cos 2\theta }=\tan 2\theta \end{aligned}}$           [eq 20]

Notice that Equations 20 and 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 $\sigma _{12MAX}$:

$\sigma _{12MAX}=\pm \left[\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}}^{2}\right]^{1 \over 2}$           [eq 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 $\sigma _{12}'$,

${\begin{aligned}\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}&={\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }+\sigma _{12}\sin {2\theta }\end{aligned}}$           [eq 22]

${\begin{aligned}\sigma _{12}'={\sigma _{22}-\sigma _{11} \over 2}\sin {2\theta }+\sigma _{12}\cos {2\theta }\end{aligned}}$           [eq 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: $(x-h)^{2}+y^{2}=r^{2}$

Figure 6: Mohr's circle for the plane stress condition. The initial stress state is $\sigma$ and rotation of the system by $\theta$ to $\sigma '$ corresponds to rotating by $2\theta$ on the diagram.

$\underbrace {\left(\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}\right)^{2}} _{(x-h)^{2}}+\underbrace {{\sigma _{12}'}^{2}} _{y^{2}}=\underbrace {\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}}^{2}} _{r^{2}}$           [eq 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 intercept 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.

$\ell =\cos \theta _{1}={{\hat {i}}\cdot \sigma \over |\sigma |}\quad m=\cos \theta _{2}={{\hat {j}}\cdot \sigma \over |\sigma |};\quad n=\cos \theta _{3}={{\hat {k}}\cdot \sigma \over |\sigma |}$

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 the 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,

${\begin{aligned}\sigma _{p}\ell &=\sigma _{11}\ell +\sigma _{21}m+\sigma _{31}n\end{aligned}}$           [eq 25]

By a similar process, the $F_{x_{2}}$ and $F_{x_{3}}$ components yield

${\begin{aligned}\sigma _{p}m&=\sigma _{12}\ell +\sigma _{22}m+\sigma _{32}n\end{aligned}}$           [eq 26]

${\begin{aligned}\sigma _{p}n&=\sigma _{13}\ell +\sigma _{23}m+\sigma _{33}n\end{aligned}}$           [eq 27]

These equations rearrange to

${\begin{aligned}0&=(\sigma _{11}-\sigma _{p})\ell +\sigma _{12}m+\sigma _{13}n\end{aligned}}$           [eq 28]

${\begin{aligned}0&=\sigma _{12}\ell +(\sigma _{22}-\sigma _{p})m+\sigma _{23}n\end{aligned}}$           [eq 29]

${\begin{aligned}0&=\sigma _{13}\ell +\sigma _{23}m+(\sigma _{33}-\sigma _{p})n\end{aligned}}$           [eq 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

${\begin{aligned}0={\sigma _{p}}^{3}-&(\sigma _{11}+\sigma _{22}+\sigma _{33}){\sigma _{p}}^{2}\\&\quad +(\sigma _{11}\sigma _{22}+\sigma _{22}\sigma _{33}++\sigma _{33}\sigma _{11}-{\sigma _{12}}^{2}-{\sigma _{23}}^{2}-{\sigma _{31}}^{2})\sigma _{p}\\&\qquad \qquad -(\sigma _{11}\sigma _{22}\sigma _{33}+2\sigma _{12}\sigma _{23}\sigma _{31}-\sigma _{11}{\sigma _{23}}^{2}-\sigma _{22}{\sigma _{13}}^{2}-\sigma _{33}{\sigma _{12}}^{2})\end{aligned}}$           [eq 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

${\begin{aligned}I_{1}&=(\sigma _{11}+\sigma _{22}+\sigma _{33})\end{aligned}}$           [eq 32]

${\begin{aligned}I_{2}&=(\sigma _{11}\sigma _{22}+\sigma _{22}\sigma _{33}+\sigma _{33}\sigma _{11}-{\sigma _{12}}^{2}-{\sigma _{23}}^{2}-{\sigma _{31}}^{2})\end{aligned}}$           [eq 33]

${\begin{aligned}I_{3}&=(\sigma _{11}\sigma _{22}\sigma _{33}+2\sigma _{12}\sigma _{23}\sigma _{31}-\sigma _{11}{\sigma _{23}}^{2}-\sigma _{22}{\sigma _{13}}^{2}-\sigma _{33}{\sigma _{12}}^{2})\end{aligned}}$           [eq 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, let's generalize our solution to include not only the principal stresses. Just as we did earlier we can write out the total forces:

${\begin{aligned}S_{x_{1}}A=F_{x_{1}}&=\sigma _{11}\ell A+\sigma _{12}mA+\sigma _{31}n\end{aligned}}$

${\begin{aligned}S_{x_{1}}&=\sigma _{11}\ell +\sigma _{12}m+\sigma _{33}n\end{aligned}}$           [eq 35]

${\begin{aligned}\qquad S_{x_{2}}&=\sigma _{12}\ell +\sigma _{22}m+\sigma _{23}n\end{aligned}}$           [eq 36]

${\begin{aligned}\qquad S_{x_{3}}&=\sigma _{13}\ell +\sigma _{23}m+\sigma _{33}n\end{aligned}}$           [eq 37]

Which gives the total stress:

$S^{2}={S_{x_{1}}}^{2}+{S_{x_{2}}}^{2}+{S_{x_{3}}}^{2}$           [eq 38]

From this, the projection onto the normal component is:

$\sigma '=S_{x_{1}}\ell +S_{x_{2}}m+S_{x_{3}}n$           [eq 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,

${\begin{aligned}\\\sigma '&=\sigma _{11}\ell ^{2}+\sigma _{22}m^{2}+\sigma _{33}n^{2}+2\sigma _{12}\ell m+2\sigma _{23}mn+2\sigma _{31}n\ell \end{aligned}}$           [eq 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:

$\sigma '=\sigma _{p_{1}}\ell ^{2}+\sigma _{p_{2}}m^{2}+\sigma _{p_{3}}n^{2}$           [eq 41]

Squaring this equation gives us:

${\sigma '}^{2}={\sigma _{p_{1}}}^{2}\ell ^{4}+{\sigma _{p_{2}}}^{2}m^{4}+{\sigma _{p_{3}}}^{2}n^{4}+2\sigma _{p_{1}}\sigma _{p_{2}}\ell ^{2}m^{2}+2\sigma _{p_{1}}\sigma _{p_{3}}\ell ^{2}n^{2}+2\sigma _{p_{2}}\sigma {p_{3}}m^{2}n_{2}^{p}$           [eq 42]

We can then use the principal components and substitute Equations 34-36 into Equation 37 to get:

$S^{2}={\sigma _{11}}^{2}\ell ^{2}+{\sigma _{22}}^{2}m^{2}+{\sigma _{33}}^{2}n^{2}$           [eq 43]

After much algebra and putting Equations 41-42 into Equation 40, we get:

$\tau _{MAX}^{2}=(\sigma _{11}-\sigma _{22})^{2}\ell ^{2}m^{2}+(\sigma _{11}-\sigma _{33})^{2}\ell ^{2}n^{2}+(\sigma _{22}-\sigma _{33})^{2}m^{2}n^{2}$           [eq 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 ${\textstyle \theta ={\pi \over 4}}$, and ${\textstyle \cos {\pi \over 4}={{\sqrt {2}} \over 2}}$).

${\begin{matrix}{\boldsymbol {\ell }}&\mathbf {m} &\mathbf {n} &{\boldsymbol {\tau }}\\0&{{\sqrt {2}} \over 2}&{{\sqrt {2}} \over 2}&{1 \over 2}(\sigma _{22}-\sigma _{33})\\{{\sqrt {2}} \over 2}&0&{{\sqrt {2}} \over 2}&{1 \over 2}(\sigma _{11}-\sigma _{33})\\{{\sqrt {2}} \over 2}&{{\sqrt {2}} \over 2}&0&{1 \over 2}(\sigma _{11}-\sigma _{22})\end{matrix}}$

By convention, $\sigma _{11}>\sigma _{22}>\sigma _{33}$, and therefore our maximum shear stress is:

$\sigma _{MAX}={\sigma _{11}-\sigma _{33} \over 2}$

Figure 8: A 3D Mohr's Circle includes three circles, one for each axis, and follows the $\sigma _{11}>\sigma _{22}>\sigma _{33}$ convention.

Note that we know there are two planes of maximum shear stress, rotated ${\textstyle \pi }$/2 from each other. Thus, the direction cosine above are actually ${\textstyle \pm {{\sqrt {2}} \over 2}}$.

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'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

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

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 ${\textstyle cos\left(x_{i},x'_{j}\right)}$ is the cosine between $x_{i}$ and $x'_{j}$ and can be rewritten ${\textstyle a_{ji}=cos\left(x_{i},x'_{j}\right)}$ allowing forː

${\begin{aligned}S'_{1}=a_{11}S_{1}+a_{12}S_{2}+a_{13}S_{3}\end{aligned}}$           [eq 1]

${\begin{aligned}S'_{2}=a_{21}S_{1}+a_{22}S_{2}+a_{23}S_{3}\end{aligned}}$           [eq 2]

${\begin{aligned}S'_{3}=a_{31}S_{1}+a_{32}S_{2}+a_{33}S_{3}\end{aligned}}$           [eq 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:

$S'_{i}=a_{ij}S_{j}$           [eq 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 2 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. Are the vector's $\mathbf {S}$ 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. The rank of a tensor is the number of indexes needed to describe it, therefore $\mathbf {a}$ is a rank two tensor because it requires $i$ and $j$, $a_{ij}$, to describe it.

Tensors are used frequently, to represent the intrinsic physical properties of materials. A good example is the electrical conductivity, ${\boldsymbol {\phi }}$, 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 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, $\mathbf {n}$, 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 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 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

$S_{i}=a_{ji}S'_{i}$           [eq 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 that $a_{ji}a_{ij}=I$ which means that transposing $\mathbf {a}$ yields the inverse of $\mathbf {a}$, written $\mathbf {a^{-1}} =\mathbf {a^{T}}$.

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

$\mathbf {S} _{i}=\mathbf {T} _{ij}\mathbf {Q} _{j}$            [eq 6]

In a transformed coordinate system we'll express this as

${\mathbf {S} _{i}}'={\mathbf {T} _{ij}}'{\mathbf {Q} _{j}}'$           [eq 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

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

and

${\mathbf {T} _{i\ell }}={\mathbf {a} _{ji}}{\mathbf {T} _{jk}}'{\mathbf {a} _{k\ell }}$           [eq 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

${\mathbf {a} _{ij}}{\mathbf {T} _{jk}}{\mathbf {a} _{\ell k}}={\mathbf {a} _{ij}}{\mathbf {a} _{\ell k}}{\mathbf {T} _{jk}}$           [eq 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 it's 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

$\mathbf {T} _{ij}=\underbrace {{1 \over 2}(\mathbf {T} _{ij}+\mathbf {T} _{ji})} _{Symmetric}+\underbrace {{1 \over 2}(\mathbf {T} _{ij}-\mathbf {T} _{ji})} _{Antisymmetric}$           [eq 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 tied to the crystal symmetry of the material they represent. For example, let's 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 farm according to a symmetry element of the crystal then

${\mathbf {T} _{ij}}'=\mathbf {T} _{ij}$

Figure 1: 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 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. Let's now rederive the 3D stress relationships using tensors. The stresses normal to an oblique plane are writtenː

$\sigma _{ij}=a_{ni}\sigma _{ij}$           [eq 12]

Here the $n$ 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

$\sigma _{nj}=a_{pj}\sigma _{p}$           [eq 13]

By combining equations 12 & 13, we getː

$\underbrace {a_{ni}\sigma _{ij}} _{sum\ over\ i}-a_{pj}\sigma _{p}=0$           [eq 14]

Kronecker Delta

Additionally, there is a handy expression called a 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 won't be using this expression often, but in applying this to equation 14 ${\textstyle (a_{ni}\sigma _{ij}-a_{pj}\sigma _{p}=0)}$ 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

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

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

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

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

This is where we talk about average strain

Infinitesimal Strain

While average strain generally looks at the strain of a volume, we will not consider how a point on an elastic body moves and how points near it do also.

<FIGURE> "2D Strain at a Point" (Description)

Let's begin in 2D. Say there is a point ($P$) on an elastic body that is located at coordinate $\{x_{1},\ x_{2}\}$. (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 $\{x_{1}+u_{1},\ x_{2}+u_{2}\}$. <FIGURE> We call $\mathbf {u}$ the displacement vector.

Now let's say that 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.

${\begin{aligned}\operatorname {d} \!u_{1}&={\partial u_{1} \over \partial x_{1}}\operatorname {d} \!x_{1}+{\partial u_{1} \over \partial x_{2}}\operatorname {d} \!x_{2}\\\\\operatorname {d} \!u_{2}&={\partial u_{2} \over \partial x_{1}}\operatorname {d} \!x_{1}+{\partial u_{2} \over \partial x_{2}}\operatorname {d} \!x_{2}\end{aligned}}$

Now define the following terms:

${\begin{matrix}e_{11}&={\partial u_{1} \over \partial x_{1}}\qquad e_{12}&={\partial u_{1} \over \partial x_{2}}\\e_{21}&={\partial u_{2} \over \partial x_{1}}\qquad e_{22}&={\partial u_{2} \over \partial x_{2}}\end{matrix}}$

This allows us to write our infinitesimal displacements using Einstein Notation: $du_{i}=e_{ij}\ dx_{j}$

Displacement Tensors

<FIGURE> "Title" (Description)

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 $dx_{1}=0$. <FIGURE> 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 our initial equations 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 equations 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 ${x_{i}}'$?

${x_{1}}'=x_{1}+u_{1}$

${\begin{aligned}\operatorname {d} \!u_{1}&=e_{11}\operatorname {d} \!x_{1}+e_{12}\operatorname {d} \!x_{2}\\\int _{0}^{u_{1}}\operatorname {d} \!u_{1}&=e_{11}\int _{0}^{x_{1}}\operatorname {d} \!x_{1}+e_{12}\int _{0}^{x_{2}}\operatorname {d} \!x_{2}\\u_{1}&=e_{11}x_{1}+e_{12}x_{2}\\{x_{1}}'&=x_{1}+e_{11}x_{1}+e_{12}x_{2}\\&=\left(1+e_{11}\right)x_{1}+e_{12}x_{2}\end{aligned}}$

In a similar fashion we can also prove that ${x_{2}}'=e_{21}x_{1}+(1+e_{22})x_{2}$.

Looking at our picture <FIGURE>, we can see that our deformations also have translations and rotations. We're 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.

$\underbrace {\left({\begin{matrix}e_{11}&e_{12}\\e_{21}&e_{22}\end{matrix}}\right)} _{\mathbf {e} }=\underbrace {\left({\begin{matrix}e_{11}&{1 \over 2}(e_{12}+e_{21})\\{1 \over 2}(e_{21}+e_{12})&e_{22}\end{matrix}}\right)} _{\boldsymbol {\varepsilon }}+\underbrace {\left({\begin{matrix}0&{1 \over 2}(e_{12}+e_{21})\\{1 \over 2}(e_{21}+e_{12})&0\end{matrix}}\right)} _{\boldsymbol {\omega }}$

<FIGURE> "Visual representation of the displacement tensor $(\mathbf {e} )$." (Here the displacement tensor $(\mathbf {e} )$ has been broken up into the strain tensor $({\boldsymbol {\varepsilon }})$, and the rotation tensor $({\boldsymbol {\omega }})$.)

Here ${\boldsymbol {\varepsilon }}$ is the strain tensor and ${\boldsymbol {\omega }}$ is the rotation tensor. Schematically this looks like <FIGURE>. 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: $u_{i}=\varepsilon _{ij}x_{j}+\omega _{ij}x_{j}$

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:

${\begin{aligned}{\sigma _{kl}}'&=a_{ki}\ a_{\ell j}\ \sigma _{ij}\\{\varepsilon _{k\ell }}'&=a_{ki}\ a_{\ell j}\ \varepsilon _{ij}\end{aligned}}$

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 ${\textstyle {1 \over 2}\theta }$.) 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 the above equations do, due to its asymmetry. Textbooks generally like this "average engineering strain" ${\textstyle (\gamma _{ij})}$, 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.

$\operatorname {d} \!u_{j}={\partial u_{j} \over \partial x_{i}}\operatorname {d} \!x_{i}$

Which in 3D express 3 equation. $j=1,\ 2,\ 3$, 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: $e_{ij}={\partial u_{i} \over \partial x_{j}}$

The strain tensor is: $\varepsilon _{ij}={1 \over 2}\left(e_{ij}+e_{ji}\right)$

The rotation tensor is: $\omega _{ij}={1 \over 2}\left(e_{ij}-e_{ji}\right)$

Which gives us the displacement: $u_{i}=\varepsilon _{ij}x_{j}+\omega _{ij}x_{j}=e_{ij}x_{j}$

The new displaced coordinates: $x_{j}'=x_{j}+u_{j}$

Now that we have a symmetric strain tensor with properties analogous to stress, we can examine the other properties these other properties using similar methods as when we analyzed stress. For small strains where $\Delta \approx \varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}$, we define the mean stress as:

${\begin{aligned}{\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33} \over 3}={\varepsilon _{kk} \over 3}=\varepsilon _{m}\\\varepsilon _{m}\approx {\Delta \over 3}\qquad \qquad \end{aligned}}$

The total strain tensor can be broken into dilatation and deviatoric components.

$\varepsilon _{ij}={\varepsilon _{ij}}'+\varepsilon _{m}=\left(\varepsilon _{ij}-{\Delta \over 3}\delta _{ij}\right)+{\Delta \over 3}\delta _{ij}$

In a similar fashion we also have deviatoric stresses and hydrostatic stresses which are analogous to the deviatoric and hydrostatic stresses which are analogous to the deviatoric and dilatation strains. The hydrostatic, or mean, stress is:

$\sigma _{m}={\sigma _{kk} \over 3}={\sigma _{11}+\sigma _{22}+\sigma _{33} \over 3}$

Therefore, the deviatoric stress can be reduced because:

${\begin{aligned}\sigma _{ij}&={\sigma _{ij}}'+{1 \over 3}\delta _{ij}\sigma _{kk}\\{\sigma _{ij}}'&={\sigma _{ij}}+\sigma _{m}\delta _{ij}\\\\{\sigma _{ij}}'&=\left({\begin{matrix}\sigma _{11}-\sigma _{m}&\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}-\sigma _{m}&\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}-\sigma _{m}\end{matrix}}\right)\end{aligned}}$

The principal components of ${\sigma _{ij}}'$ break down to:

${\sigma _{1}}'={2 \over 3}\left({\sigma _{1}-\sigma _{2} \over 2}+{\sigma _{1}-\sigma _{3} \over 2}\right)$

And we know that these are just the maximum shear stresses:

${\sigma _{1}}'={2 \over 3}\left({\sigma _{12}}^{MAX}+{\sigma _{13}}^{MAX}\right)$

Keep in mind that we took this from:

$0=\left(\sigma '\right)^{3}-J_{1}\left(\sigma '\right)^{2}-J_{2}\sigma '-J_{3}$

Where:

${\begin{aligned}J_{1}&=(\sigma _{11}-\sigma _{m})+(\sigma _{22}-\sigma _{m})+(\sigma _{33}-\sigma _{m})\\J_{2}&={1 \over 6}\left[\left(\sigma _{11}-\sigma _{22}\right)^{2}+\left(\sigma _{22}-\sigma _{33}\right)^{2}+\left(\sigma _{33}-\sigma _{11}\right)^{2}2+6\left(\sigma _{12}^{2}+\sigma _{23}^{2}+\sigma _{31}^{2}\right)\right]\\J_{3}&=\det |{\boldsymbol {\sigma }}'|\end{aligned}}$

<FIGURE> "Title" (Strain Gauges only measures uniaxial strain.)

Strain Gauges only measures uniaxial strain, no stress and no shear.

Isotropic Response

This section covers linear isotropic response.

We now understand tensor relations and have established two tensors that we need to elasticity; ${\boldsymbol {\sigma }}$ and ${\boldsymbol {\varepsilon }}$. In this chapter we will learn how to relate these to each other. In this particular lecture 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 chapters we learned that Hooke's Law can relate uniaxial loading to uniaxial strain:

$\sigma _{11}=E\ \varepsilon _{11}$

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, $\nu$.

$\varepsilon _{22}=\varepsilon _{33}=-\nu \varepsilon _{11}=-{\nu \ \sigma _{11} \over E}$

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:

${\begin{aligned}\varepsilon _{11}&={\sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\quad {\begin{cases}\varepsilon _{11}&={1 \over E}\left[\sigma _{11}-\nu \left(\sigma _{22}\sigma _{33}\right)\right]\\\varepsilon _{22}&={1 \over E}\left[\sigma _{22}-\nu \left(\sigma _{33}\sigma _{11}\right)\right]\\\varepsilon _{33}&={1 \over E}\left[\sigma _{33}-\nu \left(\sigma _{11}\sigma _{22}\right)\right]\end{cases}}\end{aligned}}$

Moving to shear relations, the shear stress and strain are related by the shear modulus, $G$.

$\sigma _{12}=G\gamma _{12}=G2\varepsilon _{12}$

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. ${\textstyle K={-p \over \Delta }={\sigma _{m} \over \Delta }={1 \over \beta }}$, where ${\textstyle -p}$ is the hydrostatic pressure and $\beta$ is the compressibility.

Adding together our earlier equations for the triaxial normal strain, we get:

${\begin{aligned}\underbrace {\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}} _{\Delta }&={1-2\nu \over E}(\underbrace {\sigma _{11}+\sigma _{22}+\sigma _{33}} _{3\sigma _{m}})\\\Delta &={1-2\nu \over E}3\sigma _{m}\end{aligned}}$

Applying this to our equations for the Block Modulus results in:

$K={\sigma _{m} \over \Delta }={E \over 3(1-2\nu )}$

These next solutions are given without proof, because several advanced topics are used to show the relationships.

<TABLE> Moduli Relationships

The stress-strain relationships can be expressed in compact tensor notation.

$\varepsilon _{ij}={1+\nu \over E}\sigma _{ij}-{\nu \over E}\sigma _{kk}\delta _{ij}$

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 ${\textstyle G={E \over 2(1+\nu )}}$, 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:

${\boldsymbol {\varepsilon }}=\left({\begin{matrix}{1 \over E}(\sigma _{11}-\nu (\sigma _{22}+\sigma _{33}))&{\sigma _{12} \over 2G}&{\sigma _{13} \over 2G}\\{\sigma _{12} \over 2G}&{1 \over E}(\sigma _{22}-\nu (\sigma _{33}+\sigma _{11}))&{\sigma _{23} \over 2G}\\{\sigma _{13} \over 2G}&{\sigma _{23} \over 2G}&{1 \over E}(\sigma _{33}-\nu (\sigma _{11}+\sigma _{22}))\end{matrix}}\right)$

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 and rearrange it.

${\begin{aligned}\varepsilon _{11}&={\sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\\&\quad +0={\nu \sigma _{11} \over E}-{\nu \sigma _{11} \over E}\\&={\sigma _{11} \over E}+{\nu \sigma _{11} \over E}-{\nu \sigma _{11} \over E}-{\nu \sigma _{22} \over E}-{\nu \sigma _{33} \over E}\\\varepsilon _{11}&={1+\nu \over E}\sigma _{11}-{\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\end{aligned}}$

We can additionally take our earlier equation for $\Delta$, and rearrange it to get:

${\begin{aligned}\Delta &={1-2\nu \over E}3\sigma _{m}\\\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}&={1-2\nu \over E}(\sigma _{11}+\sigma _{22}+\sigma _{33})\\\sigma _{11}+\sigma _{22}+\sigma _{33}&={E \over 1-2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\end{aligned}}$

Then combining the two above equations we get:

${\begin{aligned}\varepsilon _{11}&={1+\nu \over E}\sigma _{11}-{\nu \over E}\left({E \over 1+2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\right)\\\sigma _{11}&=\varepsilon _{11}\left({E \over 1+\nu }\right)+{E \over 1+\nu }{\nu \over E}{E \over 1-2\nu }(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\\\sigma _{11}&=\left({E \over 1+\nu }\right)\varepsilon _{11}+{\nu E \over (1+\nu )(1-2\nu )}(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33})\end{aligned}}$

Thinking about shear stress is simpler:

${\begin{aligned}\sigma _{12}&=2G\ \varepsilon _{12}\\\sigma _{12}&=2\ {E \over 2(1+\nu )}\ \varepsilon _{12}\\\sigma _{12}&=\left({E \over 1+\nu }\right)\varepsilon _{12}\end{aligned}}$

Taking the above two equations gives us the tensor expression for stress in terms of strain, which is:

$\sigma _{ij}={E \over 1+\nu }\varepsilon _{ij}+\underbrace {\nu E \over (1+\nu )(1-2\nu )} _{\lambda }\varepsilon _{kk}\delta _{ij}$

Here, $\lambda$ is the Lamé Constant. We can analyze these results and extract useful equations. Let's begin by extracting the deviatoric and hydrostatic contributions to stress.

$\sigma _{ij}'=2G\ \varepsilon _{ij}'$

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:

$\sigma _{ii}={E \over 1-2\nu }\varepsilon _{kk}=3K\varepsilon _{kk}$

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, let's 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, our third triaxial strain equation 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 the triaxial strain equations 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.

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.