Overview of Elasticity of Materials/Introducing Stress

Introduction

We will begin by developing the constitutive equations that describe the relationship between stress, $\sigma$, and strain, $\epsilon$. This is the subset of continuum mechanics that focuses on the purely elastic regime, and in particular, will focus on linear elasticity where Hooke's Law holds true.

Figure 1: (a) The external forces, $𝐏,$ 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, $A$.

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 2: An infinitesimal cuboid of material with the stresses defined according to the $(x_1,x_2,x_3)$ coordinate system.

As you recall, the stress is defined as the force over the area which it is applied. The force, $𝐏$, is a vector quantity, allowing the components to be projected into the normal and tangential directions. As shown in Figure 1, the normal component is defined according to the angle $\theta$, yielding a normal stress ${\sigma }_{33}=\frac{P\cos \theta }{A}$. The tangential component of the force, $P\sin \theta$, can be further projected into the two orthogonal directions identified in Figure 1 as $x_1$ and $x_2$, yielding two orthogonal shear stresses. This is performed according to the angle $\varphi$, giving ${\sigma }_{31}=\frac{P\sin \theta \cos \varphi }{A}$ and ${\sigma }_{32}=\frac{P\sin \theta \sin \varphi }{A}$.

Note here that we've defined the coordinate system such that the $x_3$ direction is the direction normal to the cut surface. It is convenient to use $(x_1,x_2,x_3)$ instead of $(x,y,z)$ because it allows us to pass the indexes to the stress and strain quantities. In this example, the normal stress is given by ${\sigma }_{33}$ to specify that the normal stress is applied to the surface with a normal in the $x_3$ direction with a force projected in the $x_3$ direction. The tangential components ${\sigma }_{31}$ and ${\sigma }_{32}$ specify the surface having a normal $x_3$ with forces projected in the $x_1$ and $x_2$ directions, respectively. Cutting an infinitesimal cuboid, the stresses are defined in all three directions as shown in Figure 2. For comparison, the notation used in some textbooks will write normal stresses ${\sigma }_x$, whereas here we will use ${\sigma }_{11}$. These textbooks also use $\tau$ to denote shear stress, such as ${\tau }_{xy}$ whereas here we will use ${\sigma }_{12}$. This allows the stress state to be succinctly written in matrix (tensor) form

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

The imaginary slice taken through point in the body in Figure 1 could have been any plane, but the force would remain the same. This would result in a new definition of the surface normal, and potentially a new expression for the stress. The physical presence of the stress does not change, but the description does, i.e., the coordinate system is modified. The remainder of this section is devoted to expressing the coordinate transformation and analysis of the stresses.

Plane Stress

We will begin by simplifying the picture we are working with. The plane stress condition is observed for a thin 2D object, e.g., a piece of paper, which has no stress out of the plane. This allows us to write ${\sigma }_{33}=0$. Further, there is no shear in the $x_3$ direction such that ${\sigma }_{13}={\sigma }_{23}=0$. For an object in the plane stress condition, our goal is to determine the state of stress at some point for any orientation of the axis.

Figure 3: (a) An area, $A$, defined for the plane stress condition in which the normal of the area is ${x_1}^{\prime }$, 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 ${x_1}^{\prime }$, 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:

$\begin{array}{c}{F}_1={\sigma }_{11}A\cos \theta +{\sigma }_{12}A\sin \theta =S_{1}A\\ F_2={\sigma }_{22}A\sin \theta +{\sigma }_{12}A\cos \theta =S_{2}A\end{array}$ Template:Spaces[1 & 2]

where the elements $A\cos \theta$ and $A\sin \theta$ are the projection of the A in the original orientation, shown in Figure 3 (a), and $S_1$ and $S_2$ are the total stresses in the $x_1$ and $x_2$ directions, where ${\sigma }_{12}={\sigma }_{21}$. Then, dividing by A yields:

$\begin{array}{c}{S}_1={\sigma }_{11}\cos \theta +{\sigma }_{12}\sin \theta \\ S_2={\sigma }_{22}\sin \theta +{\sigma }_{12}\cos \theta \end{array}$ Template:Spaces[3 & 4]

Projecting the total stresses shown in Figure 3 (b) into the normal direction in the ${x_1}^{\prime }$ coordinate yields

${\sigma }_{1^{\prime }1}=S_1\cos \theta +S_2\sin \theta$ Template:Spaces[5]

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

${\sigma }_{1^{\prime }2}=S_2\cos \theta -S_1\sin \theta$ Template:Spaces[6]

Resulting in

$\begin{array}{cc}{\sigma }_{1^{\prime }1}& =({\sigma }_{11}\cos \theta +{\sigma }_{12}\sin \theta )\cos \theta +({\sigma }_{22}\sin \theta +{\sigma }_{12}\cos \theta )\sin \theta \\ & ={\sigma }_{11}{\cos }^2\theta +{\sigma }_{22}{\sin }^2\theta +2{\sigma }_{12}\sin \theta \cos \theta \end{array}$ Template:Spaces[7]

and

$\begin{array}{cc}{\sigma }_{1^{\prime }2}& =({\sigma }_{22}\sin \theta +{\sigma }_{12}\cos \theta )\cos \theta -({\sigma }_{11}\cos \theta +{\sigma }_{12}\sin \theta )\sin \theta \\ & ={\sigma }_{12}(co{s}^2\theta -{\sin }^2\theta )+({\sigma }_{22}-{\sigma }_{11})\sin \theta \cos \theta \end{array}$ Template:Spaces[8]

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

It is known that ${\sigma }_{1^{\prime }2}={\sigma }_{2^{\prime }1}$ and therefore only ${\sigma }_{2^{\prime }2}$ needs determining. To do so, we define a new area that is rotated by $\pi$/2 relative to our original plane as shown in Figure 4. In this new orientation,

$\begin{array}{cc}{S}_{1}A& ={\sigma }_{11}A\cos (\theta +\frac{\pi }{2})+{\sigma }_{12}A\sin (\theta +\frac{\pi }{2})\\ S_1& =-{\sigma }_{11}\sin \theta +{\sigma }_{12}\cos \theta \end{array}$ Template:Spaces[9]

and

$\begin{array}{cc}{S}_{2}A& ={\sigma }_{22}A\sin (\theta +\frac{\pi }{2})+{\sigma }_{12}A\cos (\theta +\frac{\pi }{2})\\ S_2& ={\sigma }_{22}\cos \theta -{\sigma }_{12}\sin \theta \end{array}$ Template:Spaces[10]

Projecting the total stress in the normal direction yields

$\begin{array}{cc}{\sigma }_{2^{\prime }2}& =S_1\cos (\theta +\frac{\pi }{2})+S_2\sin (\theta +\frac{\pi }{\theta })\\ & =-S_{x_1}\sin \theta +S_{x_2}\cos \theta \end{array}$ Template:Spaces[11]

Substituting Equations 9 and 10 for $S_1$ and $S_2$ into Equation 11 for ${\sigma }_{2^{\prime }2}$ yields

$\begin{array}{cc}{\sigma }_{2^{\prime }2}& =-(-{\sigma }_{11}\sin \theta +{\sigma }_{12}\cos \theta )\sin \theta +({\sigma }_{22}\cos \theta -{\sigma }_{12}\sin \theta )\cos \theta \\ & ={\sigma }_{11}{\sin }^2\theta +{\sigma }_{22}{\cos }^2\theta -2{\sigma }_{12}\sin \theta \cos \theta \end{array}$ Template:Spaces[12]

The well-known trigonometric identities

$\begin{array}{ccc}{\cos }^2\theta =\frac{\cos 2\theta +1}{2}& & {\sin }^2\theta =\frac{1-\cos 2\theta }{2}\\ 2\sin \theta \cos \theta =\sin 2\theta & & {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta \end{array}$

are applied to Equations 7, 8, and 12 for ${\sigma }_{1^{\prime }1}$, ${\sigma }_{1^{\prime }2}$, and ${\sigma }_{2^{\prime }2}$ respectively, resulting in

$\begin{array}{cc}{\sigma }_{1^{\prime }1}& =\frac{{\sigma }_{11}+{\sigma }_{22}}{2}+\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\cos 2\theta +{\sigma }_{12}\sin 2\theta \end{array}$ Template:Spaces[13]

$\begin{array}{cc}{\sigma }_{2^{\prime }2}& =\frac{{\sigma }_{11}+{\sigma }_{22}}{2}-\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\cos 2\theta -{\sigma }_{12}\sin 2\theta \end{array}$ Template:Spaces[14]

and

$\begin{array}{cc}{\sigma }_{1^{\prime }2}& =\frac{{\sigma }_{22}-{\sigma }_{11}}{2}\sin 2\theta +{\sigma }_{12}\cos 2\theta \end{array}$ Template:Spaces[15]

Principal Stress

There are numerous immediate results that come from this derivation, from which we can gain greater insights. One result that comes from the equations for ${\sigma }^{\prime }$ is ${\sigma }_{1^{\prime }1}+{\sigma }_{2^{\prime }2}={\sigma }_{11}+{\sigma }_{22}$, for all $\theta$. This means that the trace of the stress tensor $\mathit{\sigma }$ is invariant.

A second result is that the maximum normal stresses and shear stresses vary as a sine wave with period $\pi$. Within this oscillation, the normal and shear stresses are shifted by a phase factor that results in (1) the maximum and minimum normal stresses occur when the shear is zero, (2) the maximum and minimum shear stresses are shifted from each other by $\pi$/4, (3) the maximum and minimum normal stresses are shifted from each other by $\pi$/2, and (4) the maximum and minimum shear stresses are shifted by $\pi$/4 from the minimum and maximum normal stresses.

Any stress state can be rotated to yield ${\sigma }_{12}={\sigma }_{21}=0$. This diagonalizes the stress tensor and gives normal stresses that are extreme. In this orientation, the planes are called the principal planes and the normal stresses are called the principal stresses. The directions that give these principal stresses are called the principal axis. As a matter of convention, we define the first principal stress ${\sigma }_{p1}$ to be the largest and the sequentially smaller principal stresses to be ${\sigma }_{p2}$ and ${\sigma }_{p3}$, although here we have limited ourselves to 2D plane stress and only enumerate ${\sigma }_{p1}$ and ${\sigma }_{p2}$.

We know ${\sigma }_{12}=0$ in the principal orientation, which means we can use Equation 8 for ${\sigma }_{1^{\prime }2}$ to determine the angle ($\theta$) needed to rotate the tensor $\mathit{\sigma }$ into ${\sigma }^{\prime }$ which is principal,

$\begin{array}{cc}0& ={\sigma }_{1^{\prime }2}={\sigma }_{2^{\prime }1}\\ 0& ={\sigma }_{12}({\cos }^2\theta -{\sin }^2\theta )+({\sigma }_{22}-{\sigma }_{11})\sin \theta \cos \theta \\ -({\sigma }_{22}-{\sigma }_{11})\sin \theta \cos \theta & ={\sigma }_{12}({\cos }^2\theta -{\sin }^2\theta )\\ \frac{\sin \theta \cos \theta }{{\cos }^2\theta -{\sin }^2\theta }& =\frac{{\sigma }_{12}}{{\sigma }_{11}-{\sigma }_{22}}\end{array}$

Resulting in

$\begin{array}{cc}\tan 2\theta & =\frac{2{\sigma }_{12}}{{\sigma }_{11}-{\sigma }_{22}}\end{array}$ Template:Spaces[16]

It is observed graphically by plotting $\tan 2\theta$ in Figure 5 that adjacent roots are each separated by $\pi$/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 separated by $\frac{\pi }{2}$.

For a simple right triangle with hypotenuse $c$ and sides $a$ and $b$ we know

$\begin{array}{c}\sin \xi =\frac{b}{c}\\ \cos \xi =\frac{a}{c}\\ \tan \xi =\frac{b}{a}\end{array}$

which can be combined with the Pythagorean Theorem, $a^2+b^2=c^2$ and Equation 16,

$\begin{array}{cc}a& ={\sigma }_{12}\\ b& =\frac{1}{2}({\sigma }_{11}-{\sigma }_{22})\\ c& =\pm {(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}\end{array}$

These can be further combined which yields

$\begin{array}{cc}\sin 2\theta & =\pm \frac{{\sigma }_{12}}{{(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}}\end{array}$ Template:Spaces[17]

and

$\begin{array}{c}\\ \cos 2\theta & =\pm \frac{\frac{1}{2}({\sigma }_{11}-{\sigma }_{22})}{{(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}}\end{array}$ Template:Spaces[18]

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

Substituting these equations into Equation 13 for ${\sigma }_{1^{\prime }1}$, determines the principal stresses

$\begin{array}{cc}{\sigma }_p& =\frac{{\sigma }_{11}+{\sigma }_{22}}{2}+\frac{{\sigma }_{11}-{\sigma }_{22}}{2}(\pm \frac{\frac{1}{2}({\sigma }_{11}-{\sigma }_{22})}{{(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}})+{\sigma }_{12}\left(\frac{{\sigma }_{12}}{\pm {(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}}\right)\\ & =\frac{{\sigma }_{11}+{\sigma }_{22}}{2}+\frac{\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2}{\pm {(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}}+\frac{{{\sigma }_{12}}^2}{\pm {(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}}\\ & =\frac{{\sigma }_{11}+{\sigma }_{22}}{2}\pm \left(\frac{\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2}{{(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}}\right)\end{array}$

Resulting in

$\begin{array}{cc}& =\frac{{\sigma }_{11}+{\sigma }_{22}}{2}\pm {(\frac{1}{4}({\sigma }_{11}-{\sigma }_{22}{)}^2+{{\sigma }_{12}}^2)}^{\frac{1}{2}}\end{array}$ Template:Spaces[19]

Use Equation 19 in Equation 16 to find $\theta$ for $0<\theta <\frac{\pi }{4}$.

To find the maximum shear stress, we take the derivative with respect to theta of our simplified Equation 15 for ${\sigma }_{1^{\prime }2}$ and set it equal to $0$.

$\begin{array}{cc}0& =\frac{\mathrm{d}}{\mathrm{d}\theta }(\frac{{\sigma }_{22}-{\sigma }_{11}}{2}\sin 2\theta +{\sigma }_{12}\cos 2\theta )\\ & =2\frac{{\sigma }_{22}-{\sigma }_{11}}{2}\cos 2\theta -2{\sigma }_{12}\sin 2\theta \\ & =({\sigma }_{22}-{\sigma }_{11})\cos 2\theta -2{\sigma }_{12}\sin 2\theta \end{array}$

Resulting in an expression for $2\theta$:

$\begin{array}{c}\\ \frac{{\sigma }_{22}-{\sigma }_{11}}{2{\sigma }_{12}}& =\frac{\sin 2\theta }{\cos 2\theta }=\tan 2\theta \end{array}$ Template:Spaces[20]

Notice that Equation 20 and Equation 16 are negative reciprocals which means that $2{\theta }_p$ and $2{\theta }_{MAX}$ are shifted by $\pi$/2. This is indicative of

$\begin{array}{c}\tan \varphi =\frac{a}{b}\\ \tan \varphi +\frac{\pi }{2}=\frac{-b}{a}\end{array}$

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 ${\sigma }_{12MAX}$:

${\sigma }_{12MAX}=\pm {[{\left(\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\right)}^2+{{\sigma }_{12}}^2]}^{\frac{1}{2}}$ Template:Spaces[21]

Mohr's Circle

A convenient means of visualizing angular relationships is through Mohr's circle, which we derive here. Rearrange Equation 13 for ${\sigma }_{1^{\prime }1}$ and Equation 15 for ${\sigma }_{1^{\prime }2}$,

$\begin{array}{cc}{\sigma }_{1^{\prime }1}-\frac{{\sigma }_{11}+{\sigma }_{22}}{2}& =\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\cos 2\theta +{\sigma }_{12}\sin 2\theta \end{array}$ Template:Spaces[22]

$\begin{array}{c}{\sigma }_{1^{\prime }2}=\frac{{\sigma }_{22}-{\sigma }_{11}}{2}\sin 2\theta +{\sigma }_{12}\cos 2\theta \end{array}$ Template:Spaces[23]

Square both expressions,

$\begin{array}{cc}{({\sigma }_{1^{\prime }1}-\frac{{\sigma }_{11}+{\sigma }_{22}}{2})}^2& ={(\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\cos 2\theta +{\sigma }_{12}\sin 2\theta )}^2\end{array}$

$\begin{array}{cc}{\left({\sigma }_{1^{\prime }2}\right)}^2& ={(\frac{{\sigma }_{22}-{\sigma }_{11}}{2}\sin 2\theta +{\sigma }_{12}\cos 2\theta )}^2\end{array}$

Next, add them together to yield

$\begin{array}{cc}{({\sigma }_{1^{\prime }1}-\frac{{\sigma }_{11}+{\sigma }_{22}}{2})}^2+{{\sigma }_{1^{\prime }2}}^2& ={\left(\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\right)}^2({\cos }^{2}2\theta +{\sin }^{2}2\theta )+{{\sigma }_{12}}^2({\cos }^{2}2\theta +{\sin }^{2}2\theta )\\ {({\sigma }_{1^{\prime }1}-\frac{{\sigma }_{11}+{\sigma }_{22}}{2})}^2+{{\sigma }_{1^{\prime }2}}^2& ={\left(\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\right)}^2+{{\sigma }_{12}}^2\end{array}$

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 ${\sigma }^{\prime }$ corresponds to rotating by $2\theta$ on the diagram.

$\underset{(x-h{)}^2}{\underbrace{{({\sigma }_{1^{\prime }1}-\frac{{\sigma }_{11}+{\sigma }_{22}}{2})}^2}}+\underset{y^2}{\underbrace{{{\sigma }_{1^{\prime }2}}^2}}=\underset{r^2}{\underbrace{{\left(\frac{{\sigma }_{11}-{\sigma }_{22}}{2}\right)}^2+{{\sigma }_{12}}^2}}$ Template:Spaces[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 $r={({\left(\frac{{\sigma }_{11}-{\sigma }_{11}}{2}\right)}^2+{\sigma }_{12}^2)}^{\frac{1}{2}}$. 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 $\varphi$ on the diagram is equivalent to rotating by $2\pi$ in our equations. This allows the new stress state to be read from the diagram. When the bisector is horizontal, the principal orientation is identified. Rotating the bisection on the diagram by $\pi$ is equivalent to rotating the system by $\theta =\pi$/2, which can be imagined as rotating the cuboid faces until the system is back in registry, i.e., it returns to the original stress state. Further, rotating the bisection on the diagram by $\pi$/2 is equivalent to rotating by $\theta =\pi$/4, which is known to be the orientation with maximum shear stress. Thus, from a given initial stress state, $\sigma$, all stress states that can be achieved through rotation are visualized on the circle.

Generalizing from 2D to 3D

Generalizing from 2D to 3D, we move from a biaxial plane stress system to a triaxial system. Determining the principal axis and angular relations is similar to the case of 2D and will be shown below. Note as a matter of convention, when two of the three principal stresses are equal, we call the system "cylindrical", and if all three principal stresses are equal, we call the system "hydrostatic" or "spherical".

As in the case of the biaxial system, we begin by defining a plane with area $A$ that passes through our $x_1$, $x_2$, and $x_3$ coordinate system, as shown in Figure 7. The plane intercepts the axis at ($J$, $K$, and $L$) as demonstrated in the figure. To simplify the problem and allow us to make progress toward our derivation, we will say that the plane is one of the principal planes so that the shear stress components are zero. Thus, we only need to consider our principal stress that is normal to the plane.

Define $\ell$, $m$, and $n$ to be the direction cosine between $x_1$, $x_2$, and $x_3$ and the normal to the stress. Using the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ parallel to $x_1$, $x_2$, and $x_3$, we have

Figure 7: Coordinate plane JKL in 3D that passes through the x, y, z coordinate system with positive shear stresses acting on it where $O$ is the origin.

$\ell =\cos {\theta }_1=\frac{\hat{i}\cdot \mathit{\sigma }}{|\mathit{\sigma }\mathit{|}}m=\cos {\theta }_2=\frac{\hat{j}\cdot \mathit{\sigma }}{|\mathit{\sigma }\mathit{|}};n=\cos {\theta }_3=\frac{\hat{k}\cdot \mathit{\sigma }}{|\mathit{\sigma }\mathit{|}}$

The projection of stress along $x_1$, $x_2$, and $x_3$ direction give the total stresses $S_1$, $S_2$, and $S_3$:

$S_1={\sigma }_p\ell ;S_2={\sigma }_{p}m;S_3={\sigma }_{p}n$

In the biaxial derivation, the area is projected into three directions, producing the triangles in Figure 7 which have areas $LOK=A\ell$, $JOL=Am$ and $JOK=An$. We can now equate the forces in the two reference frames:

$\begin{array}{cc}{S}_{1}A={\sigma }_p\ell A=F_x& =LOK{\sigma }_{11}+JOL{\sigma }_{21}+JOK{\sigma }_{31}+\\ & =A\ell {\sigma }_{11}+Am{\sigma }_{21}+An{\sigma }_{31}\end{array}$

So,

$\begin{array}{cc}{\sigma }_p\ell & ={\sigma }_{11}\ell +{\sigma }_{21}m+{\sigma }_{31}n\end{array}$ Template:Spaces[25]

By a similar process, the $F_{x_2}$ and $F_{x_3}$ components yield

$\begin{array}{cc}{\sigma }_{p}m& ={\sigma }_{12}\ell +{\sigma }_{22}m+{\sigma }_{32}n\end{array}$ Template:Spaces[26]

$\begin{array}{cc}{\sigma }_{p}n& ={\sigma }_{13}\ell +{\sigma }_{23}m+{\sigma }_{33}n\end{array}$ Template:Spaces[27]

These equations rearrange to

$\begin{array}{cc}0& =({\sigma }_{11}-{\sigma }_p)\ell +{\sigma }_{12}m+{\sigma }_{13}n\end{array}$ Template:Spaces[28]

$\begin{array}{cc}0& ={\sigma }_{12}\ell +({\sigma }_{22}-{\sigma }_p)m+{\sigma }_{23}n\end{array}$ Template:Spaces[29]

$\begin{array}{cc}0& ={\sigma }_{13}\ell +{\sigma }_{23}m+({\sigma }_{33}-{\sigma }_p)n\end{array}$ Template:Spaces[30]

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

$det\left|\begin{array}{ccc}{\sigma }_{11}-{\sigma }_p& {\sigma }_{12}& {\sigma }_{13}\\ {\sigma }_{12}& {\sigma }_{22}-{\sigma }_p& {\sigma }_{23}\\ {\sigma }_{13}& {\sigma }_{23}& {\sigma }_{33}-{\sigma }_p\end{array}\right|$

to zero and solving for the eigenvalues and subsequent eigenvectors.

Upon rearranging, we get

$\begin{array}{cc}0={{\sigma }_p}^3-& ({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33}){{\sigma }_p}^2\\ & +({\sigma }_{11}{\sigma }_{22}+{\sigma }_{22}{\sigma }_{33}++{\sigma }_{33}{\sigma }_{11}-{{\sigma }_{12}}^2-{{\sigma }_{23}}^2-{{\sigma }_{31}}^2){\sigma }_p\\ & -({\sigma }_{11}{\sigma }_{22}{\sigma }_{33}+2{\sigma }_{12}{\sigma }_{23}{\sigma }_{31}-{\sigma }_{11}{{\sigma }_{23}}^2-{\sigma }_{22}{{\sigma }_{13}}^2-{\sigma }_{33}{{\sigma }_{12}}^2)\end{array}$ Template:Spaces[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 $[\ell ,m,n]$, 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

$\begin{array}{c}{I}_1=({\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33})\end{array}$ Template:Spaces[32]

$\begin{array}{c}{I}_2=({\sigma }_{11}{\sigma }_{22}+{\sigma }_{22}{\sigma }_{33}+{\sigma }_{33}{\sigma }_{11}-{{\sigma }_{12}}^2-{{\sigma }_{23}}^2-{{\sigma }_{31}}^2)\end{array}$ Template:Spaces[33]

$\begin{array}{c}{I}_3=({\sigma }_{11}{\sigma }_{22}{\sigma }_{33}+2{\sigma }_{12}{\sigma }_{23}{\sigma }_{31}-{\sigma }_{11}{{\sigma }_{23}}^2-{\sigma }_{22}{{\sigma }_{13}}^2-{\sigma }_{33}{{\sigma }_{12}}^2)\end{array}$ Template:Spaces[34]

This is useful because these invariant relations determine the relationship between stresses in different orientations, i.e. given $\sigma$, you can now directly determine ${\sigma }_1$, ${\sigma }_2$, and ${\sigma }_3$.

Now, we will generalize our solution to include not only the principal stresses. Just as we did earlier, we can write out the total forces:

$\begin{array}{c}{S}_{x_1}A=F_{x_1}={\sigma }_{11}\ell A+{\sigma }_{12}mA+{\sigma }_{31}nA\end{array}$

$\begin{array}{c}{S}_{x_1}={\sigma }_{11}\ell +{\sigma }_{12}m+{\sigma }_{33}n\end{array}$ Template:Spaces[35]

$\begin{array}{c}{S}_{x_2}={\sigma }_{12}\ell +{\sigma }_{22}m+{\sigma }_{23}n\end{array}$ Template:Spaces[36]

$\begin{array}{c}{S}_{x_3}={\sigma }_{13}\ell +{\sigma }_{23}m+{\sigma }_{33}n\end{array}$ Template:Spaces[37]

Which gives the total stress:

$S^2={S_{x_1}}^2+{S_{x_2}}^2+{S_{x_3}}^2$ Template:Spaces[38]

From this, the projection onto the normal component is:

${\sigma }^{\prime }=S_{x_1}\ell +S_{x_2}m+S_{x_3}n$ Template:Spaces[39]

Substituting Equations 34-36 into Equation 38 gives us:

$\begin{array}{cc}{\sigma }^{\prime }& =({\sigma }_{11}\ell +{\sigma }_{12}m+{\sigma }_{31}n)\ell +({\sigma }_{12}\ell +{\sigma }_{22}m+{\sigma }_{31}n)m+({\sigma }_{13}\ell +{\sigma }_{31}m+{\sigma }_{33}n)n\end{array}$

Which simplifies to,

${\sigma }^{\prime }={\sigma }_{11}{\ell }^2+{\sigma }_{22}{m}^2+{\sigma }_{33}{n}^2+2{\sigma }_{12}\ell m+2{\sigma }_{23}mn+2{\sigma }_{31}n\ell$ Template:Spaces[40]

The magnitude of the shear component can be determined utilizing $S^2={{\sigma }^{\prime }}^2+{\tau }^2$, 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:

${\sigma }^{\prime }={\sigma }_{p_1}{\ell }^2+{\sigma }_{p_2}{m}^2+{\sigma }_{p_3}{n}^2$ Template:Spaces[41]

Squaring this equation gives us:

${{\sigma }^{\prime }}^2={{\sigma }_{p_1}}^2{\ell }^4+{{\sigma }_{p_2}}^2{m}^4+{{\sigma }_{p_3}}^2{n}^4+2{\sigma }_{p_1}{\sigma }_{p_2}{\ell }^2{m}^2+2{\sigma }_{p_1}{\sigma }_{p_3}{\ell }^2{n}^2+2{\sigma }_{p_2}\sigma p_3{m}^2{n}_2^p$ Template:Spaces[42]

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

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

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

${\tau }_{MAX}^2=({\sigma }_{11}-{\sigma }_{22}{)}^2{\ell }^2{m}^2+({\sigma }_{11}-{\sigma }_{33}{)}^2{\ell }^2{n}^2+({\sigma }_{22}-{\sigma }_{33}{)}^2{m}^2{n}^2$ Template:Spaces[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 $\theta =\frac{\pi }{4}$, and $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$). Here are the values of $\ell ,m,n$, and $\tau$ for these three planes:

$\mathit{\ell }$	$𝒎$	$𝒏$	$\mathit{\tau }$
$$0$$	$$\frac{\sqrt{2}}{2}$$	$\frac{\sqrt{2}}{2}$	$$\frac{1}{2}({\sigma }_{22}-{\sigma }_{33})$$
$$\frac{\sqrt{2}}{2}$$	$$0$$	$\frac{\sqrt{2}}{2}$	$$\frac{1}{2}({\sigma }_{11}-{\sigma }_{33})$$
$$\frac{\sqrt{2}}{2}$$	$\frac{\sqrt{2}}{2}$	$$0$$	$$\frac{1}{2}({\sigma }_{11}-{\sigma }_{22})$$

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

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

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 $\pi$/2 from each other. Thus, the direction cosine above are actually $\pm \frac{\sqrt{2}}{2}$.

Because these axial rotations are decoupled, we can represent 3D stress states using Mohr's Circles as seen in Figure 8.

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