 # 10- Easy Approach to Mohr’s circle of inertia-third case.

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

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

We are expecting that the tangent value of 2θp is positive from the equation of tan 2θp since Ixy is negative.

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.

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

We have the general expression for the moment of inertia for any oriented axis, we call it Ix’. For Mohr’s circle of inertia-third case, the value can be a maximum value when the angle 2θ has a cosine of a positive value and the sine value is also positive, this angle is called 2φp1 or ( 2θp1). This angle is measured from the x-axis.
The other 180- direction has an angle of 2φp2 or ( 2θp2) is also measured from the x-axis, but this angle will have a negative cosine value and also a negative sine value.

The angle 2φp1 will be located in the first quarter, while angle 2φp2 will be located in the third quarter.

### The steps used to draw Moh’s circle of inertia-third 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 data of the case. Point B has a coordinate (Iy,+Ixy), the value of Ixy is the negative value is identical to 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( (Ix-Iy/2)^2+Ixy^2).
Start from point O and draw the circle with the radius value just estimates. The circle will interest with the line in two points C and E.

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

Setting point A’, which is a mirror of point A and has a coordinate of (Ix,+Ixy) and is giving 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 2φp1. 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.

Angle 2φp1 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 2φp2 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 value of Minimum value of inertia, while the distance from the same axis to point E will give the maximum value of inertia

In the normal view the x-axis is a horizontal axis, while in Mohr’s circle of the inertia-third case, the x-axis is oriented by an angle of 2φp1 from the U-direction.

From the relation of tan 2φp, we have a positive value of tangent which means that the U 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-third case.

Join points OA’ to get The U- direction. Join the two points O and B’ to get the V- direction for the normal view, the view from which x and y axes are orthogonal 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 B’ has Ix’=Iy and Ix’y’=+Ixy.

It is shown that this drawing fulfills the requirement of Case-3, that Ix is bigger than Iy and Ixy is negative.

The V axis is rotated by an angle of (φp2) in a clockwise direction from axis X, While the U axis is rotated by an angle of ( 2φp1) in the anticlockwise direction from axis X.

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

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

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