Cross-Section Analysis – Mohr’s Circle (3)

„Przekroje kołem się toczą…”

Welcome to the third part of the article on analyzing cross-sectional parameters. This part will focus on the analytical calculation of principal central moments of inertia and the graphical interpretation of moments of inertia, known as Mohr’s circle.

In this part of the article, I will cover the following topics:

• Rotation of the coordinate system
• Mohr’s circle and principal moments of inertia

Rotation of the coordinate system – and the moments change…

In the previous part of the article, I discussed the theoretical basics and applications of principal central moments of inertia in stress and displacement calculations. In this part, we will discuss how to calculate them and even represent them graphically ????.

We’ll start with three formulas that allow us to determine moments of inertia and the moment of deviation for any rotation of the coordinate system—it might get a bit tricky:

I_ξ={I_x+I_y \over2} +{I_x-I_y\over2}  cos⁡2φ-D_{xy} sin⁡2φ

I_η={I_x+I_y \over2} -{I_x-I_y\over2}  cos⁡2φ+D_{xy}  sin⁡2φ

D_{ξη}={I_x-I_y\over2}  sin⁡2φ+D_{xy}  cos⁡2φ

Note: These formulas are correct only for this specific representation of the xy system, angle φ, and the rotated ξη system. In literature or class notes, these formulas may differ in signs, for example, because the vertical or horizontal axis might have opposite directions!!!

Using the presented formulas, we could even calculate the principal moments, by finding the function’s extremum… Calculating the derivatives with respect to angle φ… setting them equal to 0 and substituting the calculated angle into the formula… It would be a mathematical massacre, so I do not recommend it. Instead, you can derive the formulas from the graphical representation called Mohr’s circle.

Mohr’s circle and principal moments of inertia – nothing beats a nice picture…

Mohr’s circle is, let’s say, a representation of the current orientation of the coordinate system in the moments of inertia system. This sounds quite complicated, but I guarantee that constructing Mohr’s circle is as simple as building a hammer.

The procedure is as follows and is shown in the graphic below:

1. Calculate the values of the moments of inertia and the moment of deviation for the current location and orientation of the coordinate system.
2. Draw a preliminary coordinate system, where the horizontal axis will represent the moments of inertia, and the vertical axis will represent the moment of deviation. Pay attention to the calculated values of the moments of inertia and the moment of deviation—the system must be drawn in such a way that these values “fit” in a certain scale, and the vertical axis allows for the notation of both positive and negative values. It’s also important to ensure that the vertical and horizontal scales are the same—otherwise, we’ll get Mohr’s ellipse instead of a circle…
3. Mark the calculated moments of inertia on the horizontal axis and determine the center of Mohr’s circle, located between them at c = (I_x + I_y)/2.
4. Above (or below—it doesn’t matter) the points corresponding to  I_x i I_y mark the coordinate corresponding to the value of the moment of deviation.
5. Draw a circle centered at point c (see step 3) with a radius determined by the points marked in step 4.
6. Done—you’ve drawn Mohr’s circle! You can now proceed to analyze the results.

Using Mohr’s circle, we can quickly determine the values of the principal moments of inertia—all you need to do is read the coordinates where Mohr’s circle intersects the horizontal axis. This corresponds to an orientation of the coordinate system where the moment of deviation equals 0—everything checks out. The same values can be calculated analytically by determining the center of the circle (left part of the formula) and its radius using the Pythagorean theorem (right part of the formula). The first principal moment (maximum) is the sum of the center and the radius, and the second principal moment (minimum) is the difference between the center and the radius:

I_1={I_x+I_y \over2}+\sqrt{({I_x-I_y \over2})^2+D_{xy}^2}

I_2={I_x+I_y \over2}+\sqrt{({I_x-I_y \over2})^2+D_{xy}^2}

From Mohr’s circle, you can also read the inclination angle of the principal axes of inertia—it is half the angle between the current orientation of the coordinate system (shown in bold on the graphic) and the horizontal axis:

tg⁡2φ={-2D_{xy} \over{I_x-I_y }}

If you’re wondering why it’s half the angle, it’s because a 180° rotation in Mohr’s circle brings us to the same moment of inertia values (just swapped), which corresponds to a 90° rotation of the coordinate system associated with the cross-section.

One last note—even if we rotate the system associated with the cross-section by the calculated angle φ, we must know which axis corresponds to I₁, and which to I₂. This can be easily determined using two rules:

• If I_x > l_y then the axis resulting from the rotation of the x-axis by the angle φ corresponds to I₁, if l_x < l_y it corresponds to I₂ .
• If I_x = l_y, a D_xy < 0, the axis corresponding to I₁​ is rotated 45° relative to the x-axis, and if D_xy > 0, it is rotated -45°.

To summarize—using Mohr’s circle, we can calculate the principal moments of inertia, determine the orientation of the principal central axes of inertia, and calculate moments of inertia and deviation for any orientation of the coordinate system.

Now, I invite you to the EquiBeam application where you’ll find unlimited possibilities for calculating cross-section parameters—it’s definitely worth applying this knowledge in practice! In the video below, you’ll find a tutorial explaining how to input data into EquiBeam for cross-sectional property calculations.

It’s great that you made it to the end of this article! I am confident that the topic of calculating basic cross-sectional parameters is now clear, and completing tasks will not be a problem for you.