Chapter 5: Stress Transformations

We have introduced our critical constitutive equations for relating stress and strain for linear elastic isotropic cubic materials. This is great and it was a lot of work to reach out to these expressions; however, in all of the stress states that we worked with, the stresses were applied parallel to our coordinate system axes.

This was very fortuitous, but there are a myriad of applications where this will not be the case and in that instance it will be important that we are able to resolve or determine the stress state at arbitrary rotations or other coordinate systems. This will be particularly pertinent and relevant when we get into yield and plasticity, as certain materials will tend to fail when shear stresses are maximized or when normal stresses are maximized. So the toolkit that we will develop will be multipronged and we will take several approaches, many of which will require a little bit of linear algebra—so I hope that you are ready.

Resolving Stress on Plane of Interest

Now I know that all of you are statics experts, so one way that we can resolve the stresses in a new coordinate system is by first applying our static equilibrium conditions for the forces in our new coordinate system. First, we must define a new coordinate system that aligns with the plane of interest, i.e. we must rotate our old coordinate axes (x, y) to align with our new coordinate axes (x′, y′).

Then we balance out the forces assuming static equilibrium:

\[
\sum F_{x'} = 0 = \sigma_{x'x'}A' - (\tau_{xy}A' \sin \theta)\cos \theta - (\sigma_{yy} A' \sin \theta)\sin \theta - (\tau_{xy} A' \cos \theta)\sin \theta - (\sigma_{xx} A' \cos \theta)\cos \theta
\]

After some rearranging and trigonometric transformations:

\[
\sigma_{x'x'}(\theta) = \frac{\sigma_{xx} + \sigma_{yy}}{2} + \frac{\sigma_{xx} - \sigma_{yy}}{2}\cos 2\theta + \tau_{xy}\sin 2\theta
\]

\[
\sigma_{y'y'}(\theta) = \frac{\sigma_{xx} + \sigma_{yy}}{2} - \frac{\sigma_{xx} - \sigma_{yy}}{2}\cos 2\theta - \tau_{xy}\sin 2\theta
\]

\[
\tau_{x'y'}(\theta) = -\left(\frac{\sigma_{xx}-\sigma_{yy}}{2}\right)\sin 2\theta + \tau_{xy}\cos 2\theta
\]

We have successfully transformed the stress state to our new coordinate system!

Now, in addition to finding the stress state in a new coordinate system, it is also helpful to find the principal stress state, which is the stress state where there exist only normal stresses and all shear stresses are equal to zero. To do this we can simply solve for when the normal stresses are maximized or when the shear stress values are zero, and we will find that the principal stress angle will be simply:

\[
\tan 2\theta = \frac{\tau_{xy}}{\frac{\sigma_{xx} - \sigma_{yy}}{2}}
\]

We can then substitute this back into the above equations and get:

\[
\sigma_{1} = \frac{\sigma_{xx} + \sigma_{yy}}{2} + \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^{2} + \tau_{xy}^{2}}
\]

\[
\sigma_{2} = \frac{\sigma_{xx} + \sigma_{yy}}{2} - \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^{2} + \tau_{xy}^{2}}
\]

\( \sigma_{1} \) and \( \sigma_{2} \) are the principal normal stresses, and you can see by definition that \( \sigma_{1} > \sigma_{2} \). We can do the same for the maximum shear stress and find that:

\[
\tau_{xy,\text{max}} = \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^{2} + \tau_{xy}^{2}}
\]

Now, how many of you can do this upon request at a job interview or memorize these equations for 30+ years? My guess is probably not many of you can do so. Additionally, I like to refer to this as the plug-and-chug method as it does not leave much intuition or thought—students tend to just plug and chug into the equations. Plus I will tend to not ask questions that you can solve easily using these expressions, so I would not utilize this method. Additionally, it is not general for the 3D stress state, but let's hold off on that for a while.

Mohr’s Circle: A Graphical Approach

Another approach that is often the bane of many mechanics students is Mohr's Circle. But today we will demystify Mohr's Circle and, more importantly, learn when it is useful to utilize this technique and when it is better and perhaps more appropriate to use our linear algebra approach.

Let’s take the following example. I have the following stress tensor, all in MPa, as seen below:

\[
\sigma =
\begin{bmatrix}
15 & 4 & 0\\
4 & 5 & 0\\
0 & 0 & 0
\end{bmatrix}
\]

Now what is the principal stress state and the values of principal stress? What about the stress state for a 15° clockwise rotation around the 3-axis?

Now we can find this by plugging and chugging it into our equations, and we will find the values as seen below...

However, this is not physical and we don’t really gain any intuition and, most importantly, it is a pain to do, and when I ask harder questions this will be very difficult to do.

Now I would suggest instead to utilize another more graphical, visual, and helpful tool, which is Mohr’s Circle construction. Otto Mohr, a German engineer in the 1800s, recognized that we can represent the principal stress state graphically. We can re-write the equations above and sum the square of \( \text{Eq.}(\sigma_{x'x'}) \) and \( \text{Eq.}(\tau_{x'y'}) \) to give us:

\[
\left[\sigma_{1} - \left(\frac{\sigma_{xx} + \sigma_{yy}}{2}\right)\right]^{2} + \tau_{xy,\text{max}}^{2} = \left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^{2} + \tau_{xy}^{2}
\]

You should notice that this form has a similar structure to the equation of a circle:

\[
(x - c)^{2} + y^{2} = r^{2}
\]

where the center of the circle is \( \frac{\sigma_{xx} + \sigma_{yy}}{2} \) and the radius of the circle is \( \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^{2} + \tau_{xy}^{2}} \). We can plot Mohr’s Circle for an arbitrary initial stress state.

The way that we will practically utilize Mohr’s Circle is as follows (note that this works when considering plane stress states only):

1. Draw a plot with normal stress on the x-axis and shear stress \( \tau \) on the y-axis.
2. Treat the stress tensor as a coordinate system with x-axis values for normal stress and y-axis values for shear stress. Plot the coordinates on the graph.
3. The 4 coordinates that you may possibly have (or 2 coordinates) are points that will lie on your Mohr’s Circle. Draw a line connecting two of the points.
4. When drawing the line, pick the line such that the smaller normal stress has a positive shear stress value and the larger stress has the negative value—i.e. draw the top part of the line on the left and bottom on the right.
5. Determine the length of this line as this will determine the diameter of your circle and thus the radius as well.
6. Determine where the line crosses the x-axis; this is the center of your Mohr’s Circle.
7. Determine the angle between your line and the x-axis; this angle is \( 2\theta \). It is \( 2\theta \) due to the equations and transformations described above.
8. Once you have the center, radius, and \( 2\theta \) values, you can find the principal stress state, principal stress angle, maximum shear stress value and angle, and then you can use geometry to get other stress states.

This is best illustrated by doing an example below for the previous stress state.

Now if you are still unsure about Mohr’s Circle you are in luck—I have about 5 hours of videos of worked examples posted on my YouTube channel, so go nuts!

Plane Stress Linear Algebra Transformations

Now as we see above, Mohr’s Circle can be very useful for the geometrically gifted or inclined—but I am not one of those people. Also, I find that when trying to determine the stress state at an arbitrary rotation, Mohr’s Circle can become a bit cumbersome, and I always fear that I may possibly forget the \( 2\theta \) caveat.

So I would suggest instead—or better yet, to supplement Mohr’s Circle—we can use a linear algebra approach and a rotation matrix to convert stresses in one coordinate system to another coordinate system. Again, remember that we are still only working with plane stress states here, not 3D stress state transformations. So let's get started.

Let's start by first defining our transformation matrix \( T \) where:

\[
\begin{bmatrix}
T
\end{bmatrix}
=
\begin{bmatrix}
c^2 & s^2 & 2sc\\
s^2 & c^2 & -2sc\\
-sc & sc & c^2 - s^2
\end{bmatrix}
\]

where \( c = \cos \theta \) and \( s = \sin \theta \).

So we now write that:

\[
\sigma' = T \sigma
\]

or similarly:

\[
\begin{bmatrix}
\sigma_{11'} \\
\sigma_{22'} \\
\sigma_{12'}
\end{bmatrix}
=
\begin{bmatrix}
c^2 & s^2 & 2sc\\
s^2 & c^2 & -2sc\\
-sc & sc & c^2 - s^2
\end{bmatrix}
\begin{bmatrix}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{12}
\end{bmatrix}
\]

So we can utilize this linear algebra approach and solve the same question. As you can see, it is a very elegant way to solve these problems and requires minimal coding, but you still have a clear intuition for how to solve these problems. However, often I will still draw the circle to keep the rotations clear.

Mohr’s Circle and Linear Algebra for Strain

Now before we move on to dealing with some complex 3D stress states and rotations of our representative volume element (RVE), you often see Mohr’s Circle and transformations of stress only, but I was never personally taught how to deal with strain transformations. But no worries—let's learn how to do this together right now.

We can follow a very similar procedure for Mohr’s Circle. If we do some transformations, you will see the equations look very similar in form to those for our stress equations:

\[
\epsilon_{x'x'}(\theta) = \frac{\epsilon_{xx} + \epsilon_{yy}}{2} + \frac{\epsilon_{xx} - \epsilon_{yy}}{2}\cos 2\theta + \frac{\gamma_{xy}}{2}\sin 2\theta
\]

\[
\epsilon_{y'y'}(\theta) = \frac{\epsilon_{xx} + \epsilon_{yy}}{2} - \frac{\epsilon_{xx} - \epsilon_{yy}}{2}\cos 2\theta - \frac{\gamma_{xy}}{2}\sin 2\theta
\]

\[
\frac{\gamma_{xy}}{2}(\theta) = -\left(\frac{\epsilon_{xx} - \epsilon_{yy}}{2}\right)\sin 2\theta + \frac{\gamma_{xy}}{2}\cos 2\theta
\]

So for Mohr’s Circle we can simply follow very similar steps as we did for stress, but we just have to make the following adjustments:

1. Draw a plot with normal strain on the x-axis and shear strain \( \frac{\gamma}{2} \) on the y-axis.
2. Treat the strain tensor as a coordinate system with x-axis values for normal strain and y-axis values for shear strain. Plot the coordinates.
3. The 4 coordinates that you may possibly have (or 2 coordinates) are points that will lie on your Mohr’s Circle. Draw a line connecting two of the points.
4. Determine the length of this line; this will determine the diameter of your circle and thus the radius as well.
5. Determine where the line crosses the x-axis; this is the center of your Mohr’s Circle.
6. Determine the angle between your line and the x-axis; this angle is \( 2\theta \). IT IS \( 2\theta \) DUE TO THE EQUATIONS AND TRANSFORMATIONS DESCRIBED ABOVE.
7. Once you have the center, radius, and \( 2\theta \) values, you can find the principal strain state, principal strain state angle, maximum shear strain value and angle, and then you can use geometry to get other stress states.

One question you may be asking as we get started is: why do we plot \( \frac{\gamma}{2} \) and not \( \gamma \)? Well, that goes back to infinitesimal strain theory, and we said that the tensorial definition of shear strain and engineering shear strain had the following relationship \( \epsilon_{12} = \frac{\gamma_{12}}{2} \). Aren't you glad we covered that in lecture?

Let's get to work and use the following strain tensor, all in microstrain:

\[
\epsilon =
\begin{bmatrix}
340 & 90 & 0\\
90 & 110 & 0\\
0 & 0 & 0
\end{bmatrix}
\]

We can also do a similar linear algebra transformation for strain as well, but we have to account for the following complication: we have the definition that \( \epsilon_{xy} = \frac{1}{2}\gamma_{xy} \), so we have to account for this in our transformation specifically via the introduction of the Reuter’s Matrix.

\[
\begin{bmatrix}
R
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 2
\end{bmatrix}
\]

So to transform to new coordinate systems for strain we have to do the following:

\[
\epsilon' = R T R^{-1} \epsilon
\]

This is assuming that we are writing our strain tensor \( \epsilon \) in the following form, working with engineering shear strain values:

\[
\epsilon =
\begin{bmatrix}
\epsilon_{11} \\
\epsilon_{22} \\
\gamma_{12}
\end{bmatrix}
\]

If you have already converted to the tensorial definition of stress and strain you are good and can skip the Reuter’s matrix and use our usual transformation:

\[
\epsilon' = T \epsilon
\]

Isotropic Stress-Strain Transformations for Plane Stress-Strain States

Now if we want to relate or transform stress and strain tensors in one coordinate system to another, we can do the following transformation:

\[
\begin{bmatrix}
\epsilon_{11}\\
\epsilon_{22}\\
\epsilon_{33}\\
\gamma_{23}\\
\gamma_{13}\\
\gamma_{12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E} & -\frac{\nu}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & \frac{1}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & -\frac{\nu}{E} & \frac{1}{E} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{G} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{G} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{G}
\end{bmatrix}
\begin{bmatrix}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{33}\\
\sigma_{23}\\
\sigma_{13}\\
\sigma_{12}
\end{bmatrix}
\]

\[
\begin{bmatrix}
\epsilon_{11}\\
\epsilon_{22}\\
0\\
0\\
0\\
\gamma_{12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E} & -\frac{\nu}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & \frac{1}{E} & -\frac{\nu}{E} & 0 & 0 & 0 \\
-\frac{\nu}{E} & -\frac{\nu}{E} & \frac{1}{E} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{G} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{G} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{G}
\end{bmatrix}
\begin{bmatrix}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{33}\\
\sigma_{23}\\
\sigma_{13}\\
\sigma_{12}
\end{bmatrix}
\]

or more visually appealing:

\[
\begin{bmatrix}
\epsilon_{11}\\
\epsilon_{22}\\
\gamma_{12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{E} & -\frac{\nu}{E} & 0 \\
-\frac{\nu}{E} & \frac{1}{E} & 0 \\
0 & 0 & \frac{1}{G}
\end{bmatrix}
\begin{bmatrix}
\sigma_{11}\\
\sigma_{22}\\
\sigma_{12}
\end{bmatrix}
\]

Obviously, this can be written differently if you were to use the tensorial definition of shear strain, but often you will be given the engineering shear strain. However, we know previously how we can change coordinate systems for both stress and strain independently. We can now combine the two and change coordinate systems for both strain and stress as follows:

\[
\begin{bmatrix}
\epsilon_{11'}\\
\epsilon_{22'}\\
\gamma_{12'}
\end{bmatrix}
= R T R^{-1} S T^{-1}
\begin{bmatrix}
\sigma_{11'}\\
\sigma_{22'}\\
\sigma_{12'}
\end{bmatrix}
\]

A good exercise is to write out each transformation to see if this indeed makes sense, as you will get several problems where you will have to work out the correct transformation—i.e. no plug and chug in this class, so we will develop some linear algebra intuition.

3D Stress Transformations

Now we have up to this point only been working with stress transformations for plane stress states. But what happens if we have a complex 3D stress state? How do we rotate this and obtain our new stress tensor? What about finding principal stress states? How about the rotation to obtain principal strain or maximum shear stress? Wow, what great questions—let's get into it right now.

We have to introduce a new transformation matrix in order to deal with complex 3D stress state rotations, which we call \( Q \):

\[
Q =
\begin{bmatrix}
\cos (a) & \cos (b) & \cos (c)\\
\cos (d) & \cos (e) & \cos (f)\\
\cos (g) & \cos (h) & \cos (i)
\end{bmatrix}
\]

where \( a \) is the angle between 1 and \( 1' \), \( b \) is the angle between 1 and \( 2' \), \( c \) is the angle between 1 and \( 3' \), \( d \) is the angle between 2 and \( 1' \), \( e \) is the angle between 2 and \( 2' \), \( f \) is the angle between 2 and \( 3' \), \( g \) is the angle between 3 and \( 1' \), \( h \) is the angle between 3 and \( 2' \), and \( i \) is the angle between 3 and \( 3' \).

So in general we can transform to our new stress tensor in the new coordinate system as follows:

\[
\sigma' = Q^{T} \sigma Q
\]

where \( Q^{T} \) is the transpose of this transformation matrix.

Now you have all the tools to transform stress and strain for virtually every scenario, so I am excited to see what I write for your problem set...