Last Updated on October 5, 2026 by Maged kamel
Mohr’s Circle of Inertia.
In this Post, we will begin discussing Mohr’s circle of Inertia. Still, we will use the relation of x’ and y’ for an Area about an inclined axis, with the x and y distances for the same location but about the horizontal and vertical axes x and y. The Inertia Ix is bigger than Iy, and Ixy is positive.
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Introduction to Mohr’s circle of Inertia.
Brief content of the Post.
The Post summarizes the items that will be explained.
When Ix is greater than Iy, we will begin learning to draw Mohr’s circle of Inertia using the provided data for two points. As we can see, y’=ycos θ -x sin θ represents the perpendicular distances from the little Area dA to x’ and the axes. In contrast, x’ = x cos θ + y sin θ.
We have two theories regarding the composition of Mohr’s circle of Inertia: is it a shifted circle, or a centric circle with the Ix, Iy, and Ixy values as its center?
What is the Value and direction of the primary angle 2θp at the third Point?
What are the values of Iu and Iv, or the maximum and minimum moments of Inertia, along with their directions, that constitute the fourth Point?
Where is the pole Point in Mohr’s circle, and what is the normal view at the fifth Point?
How do you draw Mohr’s circle of Inertia?
First, how do you draw Mohr’s circle of Inertia when Ix exceeds Iy?
We begin by locating Point X, which has positive Ix and Ixy values. Point Y, on the other hand, has coordinates (Iy, -Ixy). The circle’s center is the intersection of these two locations with the vertically shifted axis.
This circle’s radius equals the square root of Ixy^2 plus the difference between (Ix-Iy/2)^2. The line connecting Point X to the circle’s center is called the X-axis.
The Y-axis connects Point Y to the circle’s center. Our axis, X’, is produced by connecting the Point x’, which has the coordinates (Ix’, Ix’y’), to the center of Mohr’s circle of Inertia. Point y’, with coordinates (Iy’,- Ix’y’), is the opposite Point.
The enclosed angle of 2θ, twice the Value of theta from the normal view—from which we derived the three equations Ix’, Iy’, and Ix’y’—is between axis X and axis x’.
Given that Mohr’s circle of Inertia has both points X and X’, it follows that the square of the values of (Ix’-(Ix-Iy)/2) and the square of the Value of Ix’y’ can likewise be used to determine the radius of that circle.
How to prove that Mohr’s circle of Inertia is shifted?
As seen in the slide, we have three equations for Ix’, Iy, and Ix’y’ from the previous Post.
Two points on Mohr’s circle of Inertia—a shifted circle with the values of the moment of Inertia, Ix, and Iy—are represented by the three equations.
The sum of the squares of (Ix’-(Ix+Iy)/2 and Ix’y ‘) should equal the sum of the squares of (Ix-Iy)/2 and the Ixy Value to demonstrate that the Mohr circle has been relocated.
The term a in equation 1 can be used to equal (Ix-Iy)/2, while the term b can be used to equal the Ixy Value
if we subtract the Value of (Ix+Iy)/2 from both sides. Acos(2θ)-bsin(2θ) equals Ix’-1/2(Ix-Iy). Its Value for Ix’y will be (a* sin(2θ)+b*sin(2θ).
Square (Ix’-1/2*(Ix+Iy) will give a^2*cos^2(2θ)+b^2 sin^2((2θ)-2a*b*sin((2θ)*cos((2θ).
While squaring Ix’y’ will give a^2*sin^2(2θ)+b^2 cos^2((2θ)+2a*b*sin(2θ)*cos(2θ).
Suppose we have a look at the fourth slide. Adding them together gives (a^2+b^2), which we rewrite as (Ix-Iy)^2/4 + Ixy^2.
This Value represents the square of the radius, the square of the distance from Mohr’s circle of Inertia to Point X.
We’ll proceed to the following slide to see what the square of (Iy’ minus (Ix+Iy)/2 plus the square of Ixy Value will equal. We will move on to equation II and subtract (Ix + Iy)/2 from both sides, using the same terms, a and b.
The formula for the new right-hand side is (-a os(22θ+ + b sin (2θ).
Adding together the square (I of y’ minus (Ix+Iy)/2 plus the square of Ix’y’ will give (a^2+b^2), which we will rewrite as (Ix-Iy)^2/4+Ixy^2. The result matches the distance from Point X to the circle’s center.
How do we find the Point of maximum Inertia?
The Point Z, where the Ixy Value equals zero, is the Point of maximum Inertia.
What about the primary axis’s direction? By using an enclosed angle of 2θp measured clockwise from the x-axis, one can determine the direction from Mohr’s circle of Inertia. In contrast, it has an angle θp relative to the x-axis, measured clockwise when viewed from the normal.
By using an enclosed angle of 2θp measured clockwise from the x-axis, one can determine the direction from Mohr’s circle of Inertia. In contrast, it has an angle θp from the x-axis, measured clockwise when viewed from the common orientation.
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The following Post will be Post 5– an Easy introduction to Mohr’s Circle of Inertia, part 2.
This is a Link to a useful external resource—a calculator for cross-section, Mass, Axial and Polar Area Moment of Inertia, and Section Modulus.