# 7- A Solved problem-case-1-Mohr’s circle of inertia.

Last Updated on February 24, 2024 by Maged kamel

## Solved problem-case-1-Mohr’s circle of inertia.

In this post, we will introduce a solved problem case 1: Mohr’s circle of inertia, where Ix is bigger than Iy and The product of inertia Ixy is positive. For a given triangle of breadth equals 150mm and a height of 200mm, it is required to use Mohr’s circle of inertia to get the maximum and minimum moments of inertia for the two major principal axes and their orientation concerning the left corner point of the triangle.

### Solved problem-case-1-Mohr’s circle of inertia.

For the case of the right-angle triangle, we have already proved that for a corner point of a right-angle triangle, we have the value of the moment of inertia about the x-axis x equal to b*h^3/12. The value of the moment of inertia about the y-axis Iy is equal o h*b^3/12, while the product of inertia is equal to h^2*b^2/24.

Using the given data for b=150mm and h=200mm, we can get the value of the moment of inertia about the x-axis as equal to 100*10^6mm4. The moment of inertia about the Y-axis will be equal to 56.25*10^6 mm4.

The product of inertia Ixy will be equal to 37.50*10^6 mm4. Since Ix is bigger than Iy and Ixy is positive, this is case-1-Mohr’s circle of inertia.

### Solved problem-case-1-Mohr’s circle of inertia-Imax and Imin.

For the Solved problem case-1-Mohr’s circle of inertia, We start by drawing two intersecting axes. The horizontal axis represents the value of the moment of inertia. The vertical axis represents the value of the product of inertia, with a positive value pointing up.

We start by locating Point A, which has Ix, Ixy values as(100, 37.50) units and will be located above the axis of inertia by a value of positive Ixy, and apart from the vertical axis Ixy by a distance of (100) units.

Similarly, we can draw point B, which has a coordinate of (Iy,-Ixy). In units will be (56.25,-37.50), for the negative value the point will be below the axis of inertia by 37.50 units and apart from the vertical axis by the positive value of 56.25 units. We will join both two points. Line AB will intersect the horizontal axis at point O, which will be the center of the circle.

We can draw the circle by getting the middle point O, which has a coordinate of 0.50*(Ix+Iy), from the given data this value is equal to (100+56.25)*0.50=78.125 units.

The radius of the circle is estimated from the equation, R=sqrt ((Ix-Iy)^2+Ixy^2). we have Ix=100*10^6mm4.Iy=56.25×10^6 mm4.Ixy=37.50×10^6mm4. We can use the data to get the radius of Mohr’s circle of inertia, applying the known formula, we can get the radius value is equal to 43..4138*10^6 mm4. Please refer to the slide image for more details.

To get the maximum value of inertia Imax we can add the value of the radius of the circle to the value of the center point, the maximum value of the moment of inertia is equal to (34.414+78.125)=121.539 units, the product of inertia is zero, this point is point E.

For the minimum value of inertia, the value will be equal to (0.50*(Ix+Iy)-R, where R is the radius value.

I minimum is equal to 34.711 units, the product of inertia is equal to zero, and the point of minimum value of inertia is point C.

### Solved problem-case-1-Mohr’s circle of inertia-direction of principal axis U.

The direction of the principal axis U, can be obtained by estimating the tangent value of the angle from that principal axis and the horizontal axis X. The value will be(-Ixy)/(Ix-Iy)/2).
We have Ix=100 units, Iy=56.25 units, Ixy=37.50 units, the tan value is (2*(37.50)/(100-56.25)=-1.714. Since tan is negative, the value of the angle (2φp1 ) is minus (59.743).

To draw the direction of the major axis U in Mohr’s circle of inertia, we will locate point A’, which is the mirror point of Point A. Point A’ has a coordinate of (100,-37.50) units. we will join point O , with point A’ we will get the direction of the major Axis U.

The angle (- 59.743) is a central angle, by joining line CA’, we can get the direction of U but in the normal view, where x and Y have a 90-degree angle between them. Line CA’ is shown in Mohr’s circle of inertia, in the slide image.

### Solved problem-case-1-Mohr’s circle of inertia- Check the value of the sum of Iu &Iv.

It is important to check that the sum of Ix and Iy, will be the same value as the sum of Iu and Iv, where Iu is the maximum value of inertia, While Iv is the minimum value of inertia. we have already proved that the two sums have the same values.

### Solved problem case-1- Mohr’s circle of inertia- Check the value of Ix’ from the general equation =Imax.

We can use the general expression for ix’ to check the value of Imax, provided that when the angle 2θ = (2φp 1), which is (-59.743), Ix’ value will be equal to Imax. We plugin with the value of (2φp 1) as=-59.743, and estimate the value of Ix’, we will find that the value is the same as estimated from Mohr’s circle for I max. This means that we have the correct value of the principal angle.

### Solved problem case-1- Mohr’s circle of inertia- Check the value of Iy’ from the general equation =Imin.

We can use the general expression for ix’ to check the value of Imin, provided that when the angle 2θ = (2φp 1), which is (-59.743), Iy’ value will be equal to Imin. We plugin with the value of (2φp 1) as=-59.743, and estimate the value of Iy’, we will find that the value is the same as estimated from Mohr’s circle for I min.

### Oriented datums to let x-direction as a horizontal axis.

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

To let the x-axis as horizontal Mohr’s circle of inertia, the x-axis is represented by line OE. The datums of Ix&Ixy are adjusted to represent the direction of U we have already estimated, we will find that the directions of U and V are achieved.

Point B is at angle 180 from Line OE and still has the same coordinate of (56.25,-37.50). The full data that covers our Solved problem case-1-Mohr’s circle of inertia is included in the slide image.

In the next post, we will introduce case 2 in Mohr’s circle of inertia -the 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.

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