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

Mohr’s circle of inertia-third case.

Video for Mohr’r circle of inertia-Third case.

The video describes how to draw Mohr’s circle of inertia- the third case where Ix>Iy and Ixy is negative. How do you find the expressions of Iu and iv by Mohr’s circle and by general expressions?

Content of the post.

In this post, we will be talking about Mohr’s circle of inertia-third case. Mohr’s circle of 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, and the product of inertia Ixy is negative. We expect that the tangent value of 2θp is positive from the equation of tan 2θp since Ixy is negative. Two examples of shapes that satisfy the requirements of Ix, Iy, and Ixy which are the unequal angle with b>a, we consider the Ix, Iy, and Ixy about the Cg, the second example is the right angle triangle case 1, where ix >Iy and ixy is negative at the Cg.

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 the 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 2θp1. This angle is measured from the X-axis to axis u in the anticlockwise direction. The next slide image shows how to derive the expression.

For Mohr’s circle of the inertia-third case, the value of Ix’ can be a minimal value when the angle 2θp2 has a cosine of a negative value and the sine value is also negative. This angle is measured from the X-axis to axis V, in the clockwise direction.

When Iy’ value can be maximum or minimum?

For Mohr’s circle of the inertia-third case, the value for Iy’ can be a maximum value when the angle 2θp2 has a cosine of a negative value and the sine value is also negative. This angle is measured from the Y-axis to U axis. The next slide image shows how to derive the expression.

For Mohr’s circle of the inertia-third case, the value for Iy’ can be minimal when the angle 2θp1 has a cosine and also sine of a positive value. This angle is measured from the Y-axis to V axis. The next slide image shows how to derive the expression.

For Mohr’r circle of inertia-third case, in the next slide, a discussion of Mohr’s circle of inertia reveals that the Circle gives the U and V directions in the horizontal direction, while the X-axis is titled. to get the x-axis to be in the horizontal position we need to rotate the x-axis by an angle of 2θp1 in the anticlockwise direction.

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), and the value of Ixy is the negative value 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 which is point O.

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 the line in two points C and E.

Point E is the maximum value of Ix and it equals (Ix+Iy)/2)+R. Point C is the minimum value of Ix and it equals (Ix+Iy)/2)-R.

We can find that line OA represents the x-axis in Mohr’s circle, while axis OB is the Y-axis in Mohr’s circle.

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 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.

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 direction of U and V axes in normal view using 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 the x and y axes are orthogonal in the normal view. The directions of both the U and V axes are shown in the next slide image.

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 and Polar Area Moment of Inertia, and Section Modulus.