- Mohr's circle of inertia-second case.
- Post content.
- Mohr's circle of inertia- second case.
- What is the orientation of the principal axes for Mohr's circle of inertia-second case?
- The steps used to draw Moh's circle of inertia-second case.
- Need for two mirror points A' and B' in Mohr's circle of inertia-second case.
- The direction of U and V axes in Mohr's circle of inertia-second case.

## Mohr’s circle of inertia-second case.

### Video for case-2 Mohr’s circle of inertia.

A step-by-step guide to understanding case 2-Mohr’s circle of inertia. the video has a closed caption.

## Post content.

In this post, we will be talking about Mohr’s circle of inertia-second case. Mohr’s circle of the inertia-second case is the case where the moment of inertia about the x-axis is lesser than the moment of inertia about axis Y, and the product of inertia Ixy is positive.

Mohr’s circle of the inertia-second case can be found as an example in two cases, the first case is where we have a rectangular shape where the height is lesser than the width of the rectangle. The value of the moment of inertia about the x-axis and Y-axis together with the value of the product of inertia about the left corner is included in the next slide image.

### Mohr’s circle of inertia- second case.

Mohr’s circle of the inertia-second case is the case of inertia about the left corner of the unequal angle of dimension (b,h).

### What is the orientation of the principal axes for Mohr’s circle of inertia-second case?

We have the general expression for the moment of inertia for any oriented axis, we call it Ix’. For Mohr’s circle of the inertia-second case, the value can be a maximum value when the angle 2θ has a cosine of a negative value and the sine value is also negative, this angle is called 2φp1 or ( 2θp1). This angle is measured from the x-axis.

The other orthogonal direction has an angle of 2φp2 or ( 2θp2) is also measured from the x-axis, but this angle will have a positive cosine value and a positive sine value. The angle 2θp1 will be located in the third quarter.

This is the case where Ix’ is minimal when 2θp2 will be located in the first quarter.

How to get the U and V direction for the normal view? by rotating the x-axis by an angle of 2θp1in the clockwise direction?

### The steps used to draw Moh’s circle of inertia-second case.

The first two steps are to draw the two orthogonal axes. The first axis represents the Moment of inertia values, Ix, Iy, and Imax, and minimum values, while the second axis represents the values of the product of Inertia Ixy.

The coordinate of point A(Ix, Ixy), +Ixy is given from the date of the case. Point B has a coordinate (Iy,-Ixy), and the value of Ixy is the positive value of the Ixy of point A. Locating Points A and B with their respective values of Ix and Iy, As shown in the next slide image.

Join Points A and B and get the middle point of line AB, this point is the point of the Circle center.

The radius value is the sqrt( (Iy-Ix/2)^2+Ixy^2).

Start from point O and draw the circle with the radius value just estimates. The circle will interest the line in two points C and E.

Angle 2θp1 is the angle between the X-axis and Major axis U, which is the angle enclosed between OA and Line OE. Point E is the point of the maximum value of inertia.

Angle 2θp2 is the angle between the x-axis and Minor axis V, which is the angle enclosed between OA and Line OC. Point C is the point of the minimum value of inertia.

The distance from the vertical axis Ixy to point C will give the Minimum value of inertia, while the distance from the same axis to point E will give the maximum value of inertia.

### Need for two mirror points A’ and B’ in Mohr’s circle of inertia-second case.

In the normal view the x-axis is a horizontal axis, while in Mohr’s circle of the inertia-second case, the x-axis is oriented by an angle of 2φp1 from the U-direction. We need a mirror point A’ to let the x-direction in Mohr’s circle of the inertia-second case be a horizontal line.

Setting point A’, which is a mirror of point A and has a coordinate of (Ix,-Ixy) and gives the direction of the maximum moment axis U. This point A’ will enable us to rotate the x-axis or line OA a clockwise rotation by the value of 2θp1. The x-axis will be horizontal, the U line will also have a new direction represented by line OA’.

Point B’ is the mirror of point B and has a coordinate of (Iy,+Ixy) and is used to indicate the direction of the V-axis, which is the direction of the minimum moment of the inertia axis.

From the relation of tan 2φp, we have a positive value of tangent which means that the V axis will have an enclosed angle measured in the anti-clockwise direction.

### The direction of U and V axes in Mohr’s circle of inertia-second case.

Join points OA’ to get the U-direction. Join the two points O and B’ to get the V- direction. The directions of both the U and V axes are shown in the next slide image.

The V axis is rotated by an angle of (θp2) anti-clockwise from axis X, While the U axis is rotated by an angle of ( φθ1) in the clockwise direction from axis X in the normal view.

For the orientation of two axes Ixy and Ix, so the x-axis is horizontal, we can find that point E is the point at which Ix’ = Ix at 2θ = zero and Ix’y’ =+Ixy since the line OE is the horizontal line, while point A’ has Ix’=Iy and Ix’y’=-Ixy.

It is shown that this drawing fulfills the requirement of Case-2, that Ix is less than Iy and Ixy is positive. The line OA’ is the direction of U, This line has an angle equal to ( -2θp1) measured from the horizontal line OE. The line OB’ is the direction of V. This line has an angle equal to (+2θp2) measured from the horizontal line OE as an anti-clock direction.

In the next post, we will solve a problem that covers a solved problem for Mohr’s circle of inertia-second case.

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