# Overview of Elasticity of Materials/Introducing Stress

## Introduction

We will begin by developing the constitutive equations that describe relationship between stress, ${\displaystyle \sigma }$ , and strain, ${\displaystyle \varepsilon }$ . This is the subset of continuum mechanics that focuses on the purely elastic regime, and in particular, will focus on linear elasticity where Hook'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: surface forces that are either point forces or distributed forces that are applied over a surface or body force that are 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, ${\displaystyle \mathbf {P} }$ , 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, ${\displaystyle A}$ .

As you recall the stress is defined as the force divided by the area over which it is applied. The force, ${\displaystyle \mathbf {P} }$ , 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 ${\displaystyle \theta }$  yielding a normal stress ${\displaystyle \sigma _{33}={\frac {P\cos \!\theta }{A}}}$ . The tangential component of the force, ${\displaystyle P\sin \!\theta }$ , can further be projected into the two orthogonal directions identified in Figure 1 and ${\displaystyle x_{1}}$  and ${\displaystyle x_{2}}$ , yielding two orthogonal shear stresses. This is performed according to the angle ${\displaystyle \phi }$  giving ${\displaystyle \sigma _{31}={\frac {P\sin \!\theta \cos \!\phi }{A}}}$  and ${\displaystyle \sigma _{32}={\frac {P\sin \!\theta \sin \!\phi }{A}}}$ .

Note here that we've defined the coordinate system such that the ${\displaystyle x_{3}}$  direction normal to the cut surface. It is convenient to use ${\displaystyle (x_{1},x_{2},x_{3})}$  instead of ${\displaystyle (x,y,z)}$  because it allows us to pass the indexes to the stress and strain quantities. In this example the normal stress is given ${\displaystyle \sigma _{33}}$  to specify that the normal stress is applied to the surface with a normal in the ${\displaystyle x_{3}}$  with a force projected in the ${\displaystyle x_{3}}$  direction. The tangential components ${\displaystyle \sigma _{31}}$  and ${\displaystyle \sigma _{32}}$  specify the surface having a normal ${\displaystyle x_{3}}$  with forces projected in the ${\displaystyle x_{1}}$  and ${\displaystyle x_{2}}$  directions. Cutting an infinitesimal cuboid the stresses are defined in all three directions as shown in Figure Y. For comparison, the notation used in some textbooks will write normal stresses ${\displaystyle \sigma _{x}}$  whereas here we'll use ${\displaystyle \sigma _{11}}$ . These textbooks also use ${\displaystyle \tau }$  to denote shear stress, such as ${\displaystyle \tau _{xy}}$  whereas here we'll use ${\displaystyle \sigma _{12}}$ . This allows the stress state to be succinctly written in matrix (tensor) form

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

Figure 2: An infinitesimal cuboid of material with the stresses defined according to the ${\displaystyle (x_{1},x_{2},x_{3})}$  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 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 thin 2D object, e.g., a piece of paper, which has no stress out of the plane. This allows us to write ${\displaystyle \sigma _{33}=0}$ . Further there is no shear in the ${\displaystyle x_{3}}$  direction such that ${\displaystyle \sigma _{13}=\sigma _{23}=0}$ . For an object in the plain stress condition our goal to determine the state of stress at some point for any orientation of the axis.

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

For this object, the direction with zero force is ${\displaystyle x_{3}}$  coming out of the page and the non-zero stress state in the ${\displaystyle x_{1}}$  and ${\displaystyle x_{2}}$  directions have components ${\displaystyle \sigma _{11}}$ , ${\displaystyle \sigma _{22}}$ , and ${\displaystyle \sigma _{12}=\sigma _{21}}$ .

Imagine a new area defined on a plane rotated about ${\displaystyle x_{3}}$  such that the normal, defined ${\displaystyle x_{1}'}$  is related to ${\displaystyle x_{1}}$  by ${\displaystyle \theta }$  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

{\displaystyle {\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}}}

where the elements ${\displaystyle A\cos \!\theta }$  and ${\displaystyle A\sin \!\theta }$  are the projection of the A in the original orientation, shown in Figure 3 (a), ${\displaystyle S_{1}}$  and ${\displaystyle S_{1}}$  are the total stresses in the ${\displaystyle x_{1}}$  and ${\displaystyle x_{2}}$  directions, and ${\displaystyle \sigma _{12}=\sigma _{21}}$ . Then dividing by A yields
{\displaystyle {\begin{aligned}S_{1}=\sigma _{11}\cos \!\theta +\sigma _{12}\sin \!\theta \\S_{2}=\sigma _{22}\sin \!\theta +\sigma _{12}\cos \!\theta .\end{aligned}}}

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

${\displaystyle \sigma _{11}'=S_{1}\!\cos \!\theta +S_{2}\!\sin \!\theta }$

. In a similar fashion we project tangential to the plane and yield
${\displaystyle \sigma _{12}'=S_{2}\!\cos \!\theta -S_{2}\!\sin \!\theta }$

. Resulting in
{\displaystyle {\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 \\\\\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}}}

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

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

{\displaystyle {\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 \\\\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}}}

Projecting the total stress in the normal direction yields

{\displaystyle {\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}}}

Substituting the above equations for ${\displaystyle S_{1}}$  and ${\displaystyle S_{2}}$  into the equation for ${\displaystyle \sigma _{22}'}$  yields

{\displaystyle {\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}}}

The well-known trigonometric identities

${\displaystyle {\begin{array}{lcl}\cos ^{2}\!\theta ={\cos 2\theta +1 \over 2}&\quad &\sin ^{2}\!\theta ={1-\cos 2\theta \over 2}\\2\sin \!\theta \cos \!\theta =\sin 2\theta &&\cos ^{2}\!\theta -\sin ^{2}\theta =\cos 2\theta \end{array}}}$

are applied to ${\displaystyle \sigma _{11}'}$ , ${\displaystyle \sigma _{12}'}$ , and ${\displaystyle \sigma _{22}'}$  resulting in
{\displaystyle {\begin{aligned}\sigma _{11}'&={\sigma _{11}+\sigma _{22} \over 2}+{\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }+\sigma _{12}\sin {2\theta }\\\sigma _{22}'&={\sigma _{11}+\sigma _{22} \over 2}-{\sigma _{11}-\sigma _{22} \over 2}\cos {2\theta }-\sigma _{12}\sin {2\theta }\\\sigma _{12}'&={\sigma _{22}-\sigma _{11} \over 2}\sin 2\theta +\sigma _{12}\cos 2\theta .\end{aligned}}}

### 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 ${\displaystyle \sigma '}$  is ${\displaystyle \sigma _{11}'+\sigma _{22}'=\sigma _{11}+\sigma _{22}}$ , for all ${\displaystyle \theta }$ . This means that the trace of the stress tensor${\displaystyle \sigma }$  is invariant.

A second result is that the maximum normal stresses and shear stresses vary as a sine wave with period ${\displaystyle \pi }$ . With in 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 ${\displaystyle {\frac {\pi }{4}}}$ , (3) the maximum and minimum normal stresses are shifted from each other by ${\displaystyle {\frac {\pi }{2}}}$ , and (4) the maximum and minimum shear stresses are shifted by ${\displaystyle {\frac {\pi }{4}}}$  from the minimum and maximum normal stresses.

Any stress state can be rotated to yield ${\displaystyle \sigma _{12}=\sigma _{21}=0}$ . 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 ${\displaystyle \sigma _{p1}}$  to be the largest and the sequentially smaller principal stresses ${\displaystyle \sigma _{p2}}$  and ${\displaystyle \sigma _{p3}}$ , although here we have limited ourselves to 2D plane stress and only enumerate ${\displaystyle \sigma _{p1}}$  and ${\displaystyle \sigma _{p2}}$ .

We know ${\displaystyle \sigma _{12}=0}$  in the principal orientation, which means we can use our equation for ${\displaystyle \sigma _{12}'}$  to determine the angle (${\displaystyle \theta }$ ) needed to rotate the tensor ${\displaystyle \sigma }$  into ${\displaystyle \sigma '}$  which is principal,

{\displaystyle {\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}}\\\tan 2\theta &={2\ \sigma _{12} \over \sigma _{11}-\sigma _{22}}.\end{aligned}}}

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

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

For a simple right triangle with hypotenuse ${\displaystyle c}$  and sides ${\displaystyle a}$  and ${\displaystyle b}$  we know

{\displaystyle {\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, ${\displaystyle a^{2}+b^{2}=c^{2}}$  and the above trigonometric derivation to find
{\displaystyle {\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

{\displaystyle {\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}}\\\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}}}

These equations tell us for a given stress state ${\displaystyle \sigma }$  what rotation is needed align ${\displaystyle \sigma }$  with the principal axis.

Substituting these equations into ${\displaystyle \sigma _{11}'}$ , determines the principal stresses

{\displaystyle {\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)\\&={\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}}}

To find the maximum shear stress we take the derivative with respect to theta of our simplified equation for ${\displaystyle \sigma _{12}'}$ .

{\displaystyle {\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 \\{\sigma _{22}-\sigma _{11} \over 2\ \sigma _{12}}&={\sin 2\theta \over \cos 2\theta }=\tan 2\theta \end{aligned}}}

Notice that this is the negative reciprocal of our earlier ${\textstyle \tan 2\theta }$  equation when the principal orientation was derived. This is indicative of

{\displaystyle {\begin{aligned}\tan \phi ={\frac {a}{b}}\\\tan \phi +{\frac {\pi }{2}}={\frac {-b}{a}}\end{aligned}}}

which implies from our ${\displaystyle \tan 2\theta }$  that the orientation with maximum shear stress is rotated ${\displaystyle {\frac {\pi }{2}}}$  from the principal orientation and consequently
${\displaystyle \sigma _{12MAX}=\pm \left[\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}}^{2}.\right]^{1 \over 2}}$

### Mohr's Circle

A convenient means of visualizing angular relationships is through Mohr's circle, which we derive here. Rearrange the expression for ${\displaystyle \sigma _{11}'}$ ,

{\displaystyle {\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}}}

and square the expression
{\displaystyle {\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}}}

Next add this to the square of the expression for ${\displaystyle \sigma _{12}'}$
{\displaystyle {\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}}}

to yield
{\displaystyle {\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: ${\displaystyle (x-h)^{2}+y^{2}=r^{2}}$
${\displaystyle \underbrace {\left(\sigma _{11}'-{\sigma _{11}+\sigma _{22} \over 2}\right)^{2}} _{(x-h)^{2}}+\underbrace {{\sigma _{12}'}^{2}} _{y^{2}}=\underbrace {\left({\sigma _{11}-\sigma _{22} \over 2}\right)^{2}+{\sigma _{12}}^{2}} _{r^{2}}}$

From this expression Mohr's circle is drawn in Figure 6. For a given stress state ${\displaystyle \sigma }$  the center of the circle is ${\displaystyle h={\frac {\sigma _{11}-\sigma _{22}}{2}}}$  and the radius ${\displaystyle r=\left(\left({\frac {\sigma _{11}-\sigma _{11}}{2}}\right)^{2}+\sigma _{12}^{2}\right)^{\frac {1}{2}}}$ . A bisecting line intercepts the circle such that that projection onto the x-axis identified ${\displaystyle \sigma _{11}}$  and ${\displaystyle \sigma _{22}}$ . The projection onto the y-axis identifies ${\displaystyle \sigma _{12}}$ . Rotating the bisection is equivalent to transforming the stress state by ${\displaystyle 2\pi }$ , i.e., a rotation of by ${\displaystyle \phi }$  on the diagram is equivalent to rotating by ${\displaystyle 2\pi }$  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 on the diagram by ${\displaystyle \pi }$  is equivalent to rotating the system by ${\displaystyle \theta ={\frac {\pi }{2}}}$ , which can be imagined as rotating the cuboid faces until the system is back in registry, i.e., it comes returns to the original stress state. Further, rotating on the diagram by ${\displaystyle {\frac {\pi }{2}}}$  is equivalent to rotating by ${\displaystyle \theta ={\frac {\pi }{4}}}$ , which is known to be the orientation with maximum shear stress.

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

Thus, from a given initial stress state, ${\displaystyle \sigma }$ , all stress states that can be achieved through rotation are visualized on 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 ${\displaystyle A}$  that passes through our ${\displaystyle x_{1}}$ , ${\displaystyle x_{2}}$ , and ${\displaystyle x_{3}}$  coordinate system, as shown in Figure 7. The plane intercept the axis at (${\displaystyle J}$ , ${\displaystyle K}$ , and ${\displaystyle L}$ ) 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.

<FIGURE> Figure 7: "Coordinate Plane"

Define ${\displaystyle \ell }$ , ${\displaystyle m}$ , and ${\displaystyle n}$  to be the direction cosine between ${\displaystyle x_{1}}$ , ${\displaystyle x_{2}}$ , and ${\displaystyle x_{3}}$  and the normal to stress. Using the unit vectors \hat{i}, \hat{j}, and \hat{k} parallel to ${\displaystyle x_{1}}$ , ${\displaystyle x_{2}}$ , and ${\displaystyle x_{3}}$  we have

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

The projection of stress along ${\displaystyle x_{1}}$ , ${\displaystyle x_{2}}$ , and ${\displaystyle x_{3}}$  direction give the total stresses ${\displaystyle S_{1}}$ , ${\displaystyle S_{2}}$ , and ${\displaystyle S_{3}}$

${\displaystyle S_{1}=\sigma _{p}\ell ;\quad S_{2}=\sigma _{p}m;\quad S_{3}=\sigma _{p}n}$

As in the biaxial derivation the area is projected into the three directions giving the triangles in Figure 7 ${\displaystyle LOK=A\ell }$ , ${\displaystyle JOL=Am}$  and ${\displaystyle JOK=An}$ . We can now equate the forces in the two reference frames

{\displaystyle {\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}\\\sigma _{p}\ell &=\sigma _{11}\ell +\sigma _{21}m+\sigma _{31}n\end{aligned}}}

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

{\displaystyle {\begin{aligned}\sigma _{p}m&=\sigma _{12}\ell +\sigma _{22}m+\sigma _{32}n\\\sigma _{p}n&=\sigma _{13}\ell +\sigma _{23}m+\sigma _{33}n\end{aligned}}}

These equations rearrange to

{\displaystyle {\begin{aligned}0&=(\sigma _{11}-\sigma _{p})\ell +\sigma _{12}m+\sigma _{13}n\\0&=\sigma _{12}\ell +(\sigma _{22}-\sigma _{p})m+\sigma _{23}n\\0&=\sigma _{13}\ell +\sigma _{23}m+(\sigma _{33}-\sigma _{p})n\end{aligned}}}

This set of equations can be solved for ${\displaystyle \left[\ell ,m,n\right]}$  for a particular value of ${\displaystyle \sigma _{p}}$ . This set of secular equations can be solved for eigenvalues ${\displaystyle \sigma _{p}}$  and eigenvectors ${\displaystyle \left[\ell ,m,n\right]}$ . The non-trivial solutions, when ${\displaystyle \ell ,m,}$  and ${\displaystyle n}$  are non-zero, involves setting the determinant

${\displaystyle det\ \left|{\begin{matrix}\sigma _{11}-\sigma _{p}&\sigma _{12}&\sigma _{13}\\\sigma _{12}&\sigma _{22}-\sigma _{p}&\sigma _{23}\\\sigma _{13}&\sigma _{23}&\sigma _{33}-\sigma _{p}\end{matrix}}\right|}$

to zero and solving for the eigenvalues and subsequent eigenvectors.

Upon rearranging we get

{\displaystyle {\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}}}

The three roots of this cubic equation give the principal stresses, \sigma_{p1}, \sigma_{p2}, and \sigma_{p3}. The eigenvalues, once determined are substituted back into the secular equations to determine the eigenvectors corresponding ${\displaystyle \left[\ell ,m,n\right]}$ , also recognizing that ${\displaystyle \ell ^{2}+m^{2}+n^{2}=1}$ .

Solving the cubic equation is not the focus of this text, but the equation is important because the coefficients in front of the principal stress must be invariant, i.e., the same principal coordinates must exists no matter the orientation of the coordinate system. From the cubic equation the three invariants are

{\displaystyle {\begin{aligned}I_{1}&=(\sigma _{11}+\sigma _{22}+\sigma _{33})\\I_{2}&=(\sigma _{11}\sigma _{22}+\sigma _{22}\sigma _{33}+\sigma _{33}\sigma _{11}-{\sigma _{12}}^{2}-{\sigma _{23}}^{2}-{\sigma _{31}}^{2})\\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}}}

Is this useful? Yes, yes, and yes! These invariant relations determine the relationship between stresses in different orientations.

-=-=-=-=-=-=-=-=-=-=-=-=-=

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:

{\displaystyle {\begin{aligned}S_{x_{1}}A=F_{x_{1}}&=\sigma _{11}\ell A+\sigma _{12}mA+\sigma _{31}n\\S_{x_{1}}&=\sigma _{11}\ell +\sigma _{12}m+\sigma _{33}n\\\therefore \qquad S_{x_{2}}&=\sigma _{12}\ell +\sigma _{22}m+\sigma _{23}n\\\therefore \qquad S_{x_{3}}&=\sigma _{13}\ell +\sigma _{23}m+\sigma _{33}n\end{aligned}}}

Which gives the total stress: ${\displaystyle S^{2}={S_{x_{1}}}^{2}+{S_{x_{2}}}^{2}+{S_{x_{3}}}^{2}}$

From this, the projection onto the normal component is: ${\displaystyle \sigma '=S_{x_{1}}\ell +S_{x_{2}}m+S_{x_{3}}n}$

Further substituting some earlier equations gives us:

{\displaystyle {\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\\\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}}}

The magnitude of the shear component can be determined utilizing ${\displaystyle S^{2}={\sigma '}^{2}+\tau ^{2}}$ , 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 planes of principal normal stresses. Starting by setting our known stress state as the principal axis such that ${\displaystyle \sigma _{11}=\sigma _{p_{1}}}$ , ${\displaystyle \sigma _{22}=\sigma _{p_{2}}}$ , and ${\displaystyle \sigma _{33}=\sigma _{p_{3}}}$ , our direction cosine is between the principal axis and the normal of the plane with the maximum shear stress. This means that our earlier projection equation is rewritten as:

${\displaystyle \sigma '=\sigma _{p_{1}}\ell ^{2}+\sigma _{p_{2}}m^{2}+\sigma _{p_{3}}n^{3}}$

Taking the square of this equation gives us:

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

We can then combine this with the principal components to get:

${\displaystyle \tau _{MAX}^{2}=(\sigma _{11}-\sigma _{22})^{2}\ell ^{2}m^{2}+(\sigma _{11}-\sigma _{33})^{2}\ell ^{2}n^{2}+(\sigma _{22}-\sigma _{33})^{2}m^{2}n^{2}}$

With this solution, we now have three possible planes. One plane bisects ${\displaystyle \sigma _{11}}$ , and ${\displaystyle \sigma _{22}}$ , another plane bisects ${\displaystyle \sigma _{11}}$ , and ${\displaystyle \sigma _{33}}$ , and the final plane bisects ${\displaystyle \sigma _{22}}$ , and ${\displaystyle \sigma _{33}}$ . (Bisecting means ${\textstyle \theta ={\pi \over 4}}$ , and ${\textstyle \cos {\pi \over 4}={{\sqrt {2}} \over 2}}$ )

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

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

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

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

Note that we know there are two planes of maximum shear stress, rotated ${\textstyle \pi \over 2}$  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.