## Mohr’s circle of inertia part-2.

In this post, Mohr’s circle of Inertia part-2, we will continue 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.

In this video, the content of this post starts from the time: 7.52 till the end. The expression for Iy prime and Ix’y’ is to be derived by the use of Mohr’s circle of inertia part-2 and through the study of trigonometry of the various triangle.

### Locate point D’ of (Iy, Ix’y’) on Mohr’s circle of inertia part-2.

We locate a new point, which is point D’, and the D’ coordinate is(Iy’, -Ix’y’). To locate this point that extends the DC line till intersects with the Mohr’s circle at point D’ as we can see in the next slide image.

The coordinate of point B is (Iy, -Ixy) measured from the Ixy vertical axis and from Ix, Iy horizontal axis.

### The expression of Iy prime by using Mohr’s circle of inertia part-2.

To get the expression for Iy prime by using Mohr’s circle-part 2. We will examine the sector CBD’ which is enclosed by an angle of 2θ

Setting the positive directions at point c pointing to the left and downwards, the distance from point d’ to the axis of Ixy is the distance from point c to that axis minus the horizontal distance between points C and D’.

Iy’=(Ix+Iy)/2-R cos( 2α +2 θ). From our study of trigonometry, we know that R cos( 2α +2 θ) =R cos 2α cos 2 θ -R sin 2α sin 2 θ

From the triangle of CBB’, we find that R* cos(2 θ) is the difference between(Ix+Iy)/2) and Iy, which will yield to (Ix-Iy/2).

From the triangle of CBB’, we find that R* sin(2 α ) is the Ixy distance.

Iy’ is the difference from the distance of point c to the vertical axis of Ixy minus the horizontal distance between points c and D’. We can use trigonometry to get an expression for Iy’ as follows. Iy’=(Ix+Iy)/2-R cos( 2α +2 θ) can be readjusted to Iy’=(Ix+Iy)/2- (Ix-Iy/2) cos 2 θ +Ixy sin2 θ .

The expression matches with the expression for Iy’ obtained from the moment of inertia estimation about inclined axes.

### The expression of Ix’y’ by using Mohr’s circle of inertia part-2.

To get the expression for Ix’y’, we return back to our sector DCA which is enclosed by angle 2 θ. Point D has a coordinate of (Ix’, Ix’y’).

Ix’y’ is= R*sin ( 2α +2 θ) .from our study of trigonometry, R*sin ( 2α +2 θ) =R*sin 2α cos2 θ + R*cos 2α sin2 θ . sin 2α =Ixy/R, while cos 2α =(Ix-Iy)/2/R. R will go with R and the expression will be adjusted.

Setting all these terms together, we can rewrite the expression for Ix’y’=(Ix-Iy)/2*sin 2 θ +Ixy cos 2 θ.

The expression matches with the expression for Ix’y’ obtained from the moment of inertia estimation about inclined axes.

We will continue the discussion of Mohr’s circle in the next post. The discussion will be focussing on the direction of the major axis, for our case of Ix which is bigger than Iy, and Ixy is positive.

This is the next post, which is titled: Easy Approach to Mohr’s circle of inertia-First case.

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

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