 # 4-Easy introduction to Mohr’s circle of inertia part-1.

## Mohr’s circle of inertia.

In this post, we will be starting our discussion about 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 distance for the same area but about horizontal and vertical axes x,y. The Inertia Ix is bigger than Iy and Ixy is positive.

### Mohr’s circle of inertia video.

This video is an introduction to Mohr’s circle of inertia. We will be talking about the Mohr circle of inertia and how can be used to estimate the maximum and minimum value of inertia. Mohr’s circle is the graphical representation of the moment of inertia about inclined axes.

Step by step how to derive the expressions for Ix’ and Iy’ will be presented. the content of this post is from time 0 to 8.00.

0:00– Introduction to the content of the video.

1:49– How to draw Mohr’s circle of inertia?

07:29- The point of maximum value of inertia.

## Introduction to Mohr’s circle of inertia.

### Brief content of the post.

Here is a brief description of the post items that will be explained.

We will start to learn how to draw Mohr’s circle of inertia from the given data for two points when Ix is bigger than the Iy value. We can see that the perpendicular distances from the small area dA to x’ and axes will be as follows, y’=y*cos θ -x sin θ. While for x’=x*cos θ +y sin θ.

We have two ideas, about the nature of Mohr’s circle of inertia whether it is a centric circle where the center point is the point of intersection of Ix, Iy, and Ixy or a shifted circle?

The third point is what are the principal angle 2θp value and its direction?

The fourth point is what are the values for Iu and Iv or the maximum point of inertia and the minimum point of inertia and its direction?

The fifth point is Where is the pole point in Mohr’s circle and in the normal view.

### How to draw Mohr’s circle of inertia?

First, in the case of Ix value being bigger than Iy value, how can we draw Mohr’s circle of inertia?

We start to locate point X with a coordinate of  Ix, Ixy as positive values and there is another point which is point Y  that has a coordinate of (Iy,-Ixy). Joining these two points will intersect with the vertically shifted axis at the center of a circle.

The radius of this circle equals the square root of (Ix-Iy/2)^2 and the square of Ixy. We can represent the X-axis as the axis joining point X with the center of the circle.

While the Y-axis is the axis joining point Y with the center of the circle. We have axis X’ which can be obtained by joining the center of Mohr’s circle of inertia by point x’ which has a coordinate of (Ix’, Ix’y’). The opposite point is point y’ which has a coordinate of (Iy’,- Ix’y’).

The enclosed angle between axis X and axis x’ is 2θ which is twice the value of theta from the normal view from which we obtained the three equations Ix’, Iy’, and Ix’y’.

Since Mohr’s circle of inertia has both points X and X’, and that concludes the radius of that circle can be also obtained by the square of the value of (Ix’-(Ix-Iy)/2) plus the square of Ix’y’ value.

### How to prove that Mohr’s circle of inertia is a shifted circle?

From the previous post, we have three equations for Ix’, Iy, and Ix’y’ as shown in the slide.

The three equations represent two points on Mohr’s circle of inertia, which is a shifted circle drawn with the x-axis representing the values of the moment of inertia  Ix, Iy.

To prove that the Mohr circle is shifted, we want to check the sum of the square of (Ix’-(Ix+Iy)/2 plus the square of Ix’y’ will be equal to the sum of the square of (Ix-Iy)/2 and the square of Ixy.

From equation 1, if we deduct from the two sides the value of (Ix+Iy)/2, we can use the term a, as equal to (Ix-Iy)/2, while b term equals Ixy. Ix’-1/2(Ix-Iy)=a*cos(2θ)-b*sin(2θ). For Ix’y its value will be equal to (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).

If we have a look at the fourth slide. Adding together will give (a^2+b^2), which we will rewrite as (Ix-Iy)^2/4+Ixy^2.

This value represents the square value of the radius the square of the distance of Mohr’s circle of inertia and point X.

We will move to the next slide, where we can check what the value of the square of ( Iy’ minus (Ix+Iy)/2 plus the square of Ixy will give, Using the same terms a and b, we will proceed to equation II and will deduct from both sides the value of (Ix+Iy)/2.

The new right-hand side can be written as equal to  (-a *cos(2 θ) +b sin(2θ).

Adding together the square (Iy’ 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 with radius value from point X and the center of the circle.

### How to find the point of maximum value for inertia?

The point of the maximum value of inertia is point Z where the Ixy value equals zero.

What about the direction of the principal axis?  The direction is obtained from Mohr’s circle of inertia by an enclosed angle of 2θp in the clockwise direction from the x-axis. while in the normal view, it has an angle of θp from the x-axis in the clockwise direction.

The direction is obtained from Mohr’s circle of inertia by an enclosed angle of 2θp in the clockwise direction from the x-axis. while in the normal view, it has an angle of θp from the x-axis in the clockwise direction.

### The expression of Imax using Mohr’s circle.

The value of Imax or Iu can be estimated as equal to the shift which is (Ix+Iy)/2 plus the radius value. Checking the tan 2θp expression we can draw an angle and get the radius value in terms of Ixy, Ix and Iy.

Thanks a lot and peace be upon you all.

The next post will be an Easy introduction to Mohr’s Circle of Inertia part 2.

This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial and polar Area Moment of Inertia, and Section Modulus. Scroll to Top
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