## 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.**

### Brief description of the 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. This is a part of the video. The content of the video which is covered in this post is from time 0.00 to 7.51.

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

## Introduction to Mohr’s circle of inertia.

Before drawing Mohr’s circle of inertia, In the first slide, we have two axes x and y, and an area, for which we want to estimate the moment of inertias for two inclined axes, namely x’ and y’.

We will start calculating the moment of inertia for an area of dA about x’ and y’, then integrate for the whole area to get the general expression for Ix’ & Iy’ and Ix’y’.

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.

The content of this slide is quoted from Engineering mechanics statics -5th edition. Ix’ will be the integration of the product of dA*y’^2, while Iy’

will be the integration of the product of dA*x’^2.

We can see that we have two expressions. The first expression is Ix prime, which is=1/2*(Ix+Iy)+1/2*(Ix-Iy)*cos 2θ -Ixy sin 2θ. The second expression is for Iy prime, which is=1/2*(Ix+Iy)-1/2*(Ix-Iy)*cos 2θ +Ixy sin 2θ.

Adding Ix’+Iy’ will give Ix+Iy.

In the fourth slide, there is an expression for Ix’y’ value, it will be =1/2*(Ix-Iy)*sin 2θ +Ixy cos 2θ.

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

We will have a look at the graphical representation of the moment of inertias about inclined axes. This is Called Mohr’s circle for inertia.

How to check the two equations for Ix prime and Iy prime by graphing? There are the following steps to implement.

The first step is to draw two axes X-axis that represents the value of Ix, Iy, I max, and I min. The second axis is the Y-axis which represents the product of inertia Ixy value.

For a given section, we have three known values, Ix &Iy and Ixy. Our case is the Case For Ix is > Iy and Ixy value is positive.

We start to locate point A which has a coordinate value of (Ix, Ixy). We draw another point, point B with a coordinate of (Iy,-Ixy).

We will join both points and locate a new point at the intersection of the line AB with the X-axis, this point is point c, its coordinate is ((1/2*(Ix+Iy),0). To match with the expression of Ix’ and Iy prime.

### Locate point D of (Ix’, Ix’y’) on Mohr’s circle of inertia.

We locate a new point, is point D, the enclosed angle is 2θ, in the counter-clockwise direction from line CA. This is Twice the value of θ used to derive the expressions for Ix’ and Iy’. The length CA=CD. This length will be called R. Consider that line CA is making an angle of 2α with the x-axis.

We proceed to the next slide. The coordinate of point D is (Ix’, Ix’y’). We will estimate the length AC, from the right angle triangle ACA1, we find that R=sqrt(1/2*(Ix+Iy)^2+Ixy^2)).

We are interested in getting the value of Ix’, the horizontal distance from the Ixy axis to point D. Point D is at the left side of the point and the enclosed angle is 2 θ represented by angle ACD. The x-axis is reprinted by CA, while axis- x’ is represented by line CD.

The Mohr Circle doubles the angles as compared with the normal view.

It can be seen, that this distance, Ix’, is= OC+R cos( 2α +2 θ). The OC value=1/2*(Ix+Iy).

The overall length of Ix’=1/2(Ix+Iy)+R cos( 2α +2 θ).

We will move to a new slide, as we know from our study of in math, that R cos( 2α +2 θ)=cos(2α) cos(2 θ)-sin(2α) sin(2 θ).

From the shown upper left triangle, we can see that Cos 2α =1/2(Ix-Iy)/R. While 2α =(Ixy)/R.

The item of R cos( 2α +2θ) can be further simplified.

### The expression of Ix prime by using Mohr’s circle.

This is the final expression for Ix prime which matches with the expression derived by using the moment of inertia, please refer to image no 3.

This is the pdf file used in the illustration of this post and the next post.

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 & Polar Area Moment of Inertia and Section Modulus.