In a previous tutorial, Transformation of Stress, we learned about how to transform a stress field from one coordinate system to another. We did this by multiplying the transformation matrix, $ [A] $, by the stress field, $ [\sigma] $, and the transposed transformation matrix $ [A]^T $ to get
[ 1 ]
$$ [\sigma '] = [A] [\sigma] [A]^T $$ $$ \begin{bmatrix} \sigma_{xx}' & \tau_{xy}' \\ \tau_{xy}' & \sigma_{yy}' \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \sigma_{xx} & \tau_{xy} \\ \tau_{xy} & \sigma_{yy} \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} $$
Now this is specifically for a 2D transformation, but 3D transformations are used as well. Multiplying the matrices also gave 3 equations, one for the transformed $ \sigma_x $, $ \sigma_y $, and $ \tau_{xy} $.
[ 2 ]
$$ \sigma_{x}' = (\frac{\sigma_x + \sigma_y}{2}) + (\frac{\sigma_x - \sigma_y}{2}) \cos 2\theta + \tau_{xy} \sin 2\theta $$ $$ \sigma_{y}' = (\frac{\sigma_x + \sigma_y}{2}) - (\frac{\sigma_x - \sigma_y}{2}) \cos 2\theta - \tau_{xy} \sin 2\theta $$ $$ \tau_{xy}' = -(\frac{\sigma_x - \sigma_y}{2}) \sin 2\theta + \tau_{xy} \cos 2\theta $$
The reason a 2D transformation was covered was to lead into Mohr's Circle. Imagine you have a pencil and you are looking at it in one 2D frame. In this frame, it is being acted upon by normal and/or shear stresses. However, in another reference frame, say if the pencil is rotated, we would observe different normal and/or shear stresses than before.
Mohr's Circle is a way for us to analyze the current stresses to find a frame where the tensile and compressive stresses are maximized. Let's learn more about it!
With Mohr's Circle, imagine drawing a circle on a 2D axis, where the x-axis corresponds to normal stress and the y-axis corresponds to shear stress. One thing to note is the y-axis is flipped such that the positive direction is downwards. An example of Mohr's Circle can be found in Figure 1.
Fig. 1 - Example of Mohr's Circle
Now there is a lot to dissect in Figure 1, but we will step through each part; this is merely an example of what Mohr's Circle could look like. The circle will always be centered on the x-axis, but the radius and the center of the circle will change depending on the system and the stresses applied. The x-axis corresponds to the normal stresses where $ \sigma_x $ is on the right side of the center point and $ \sigma_y $ is on the left side of the center point. The y-axis corresponds to the shear stresses, where the positive $ \tau_{max} $ is located on the downward y-axis and the negative $ \tau_{max} $ is located on the upward y-axis. I know it is backwards, and it's accepted this way in order to conserve standard angle convention of counterclockwise being positive.
With a given point $ (\sigma_x, \tau_{xy}) $ and its corresponding point $ (\sigma_y, -\tau_{xy}) $, a rotation of $ 2\theta $ can be made to translate to the new frame points $ (\sigma_x', \tau_{xy}') $ and $ (\sigma_y', -\tau_{xy}') $. Now, there are two other thetas written in Figure 1, $ \theta_p $ and $ \theta_s $. $ \theta_s $ is the degree of rotation needed to transform a system to where its maximum shear stress occurs, and $ \theta_p $ is the degree of rotation needed to transform a system to where the principal stresses occur. The principal stresses are the maximum and minimum normal stresses acting on a material in a specific plane, but we will talk more about principal stresses in another tutorial!
Also notice $ \sigma_{I} $ and $ \sigma_{II} $. If $ \sigma_{I} > 0 $, then that value is the maximum tensile stress, and if $ \sigma_{II} < 0 $, then that is the maximum compressive stress. Now, from the figure we can see that $ 2 \theta_s = 2 \theta_p + \frac{\pi}{2} $, or
[ 3 ]
$$ \theta_s = \theta_p + \frac{\pi}{4} $$
Eq. (3) will allow us to transition between the maximum shear stress frame and the principal stress frame. We can draw a horizontal line going through $ (\sigma_y, -\tau_{xy}) $ and then we can form a relation between $ \sigma_x $, $ \sigma_y $, $ \tau_{xy} $, and $ \theta_p $.
Fig. 2 - Mohr's Circle derivation of principal stress degree.
Using Figure 2, we can find that
[ 4 ]
$$ \tan 2 \theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y} $$
Now that we have a way to find the angle of rotation needed to transform a frame to get principal stresses, we can see how to find the center and radius of the circle.
[ 5 ]
$$ C = \frac{\sigma_x + \sigma_y}{2} $$
[ 6 ]
$$ R = \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + \tau_{xy}^2} $$
The center of the circle is calculated by taking the average of the x and y normal stresses. As for the radius of Mohr's Circle, it is found by using Pythagorean Theorem.
What is nice about Mohr's Cirlce is it makes finding $ \sigma_{I} $ and $ \sigma_{II} $ in Figures 1 and 2 very easy!
[ 7 ]
$$ \sigma_{I} = C + R $$ $$ \sigma_{II} = C - R $$
Now we know how to use Mohr's Circle to find the maximum shear stresses for a given system and what angle is needed to rotate the current frame to that system.