7- Solved Problem Case 1- Mohr’s Circle of Inertia.

Last Updated on October 5, 2026 by Maged kamel

Solved problem-case-1-Mohr’s Circle of Inertia.

In this Post, we will introduce solved Problem case 1: Mohr’s Circle of Inertia, where Ix is greater than Iy, and the Product of Inertia Ixy is positive.

For a Triangle with a breadth of 150mm and a height of 200mm, Mohr’s Circle of Inertia must be used to find the maximum and minimum moments of Inertia about the two principal axes and their orientation relative to the Triangle’s left corner.

Solved problem-case-1-Mohr’s Circle of Inertia.

For a right-angle Triangle, we have already proved that at a corner Point, the Moment of Inertia about the x-axis is b*h^3/12. The Value of the Moment of Inertia about the y-axis, Iy, is equal to h*b^3/12, while the Product of Inertia is equal to h^2*b^2/24.

Using the given data for b = 150 mm and h = 200 mm, we can obtain the Moment of Inertia about the x-axis as 100 × 10^6 mm^4. The Moment of Inertia about the y-axis is 56.25*10^6 mm4.

The Product of Inertia Ixy is 37.50*10^6 mm4. Since Ix is greater 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 Moment of Inertia. The vertical axis represents the Product of Inertia, with positive values pointing upward.

We start by locating Point A, which has Ix and Ixy values of (100, 37.50) units and lies above the axis of Inertia by a positive Ixy and, apart from the vertical axis, by a distance of (100) units.

Similarly, we can draw Point B, which has coordinates (Iy, -Ixy). The coordinates will be (56.25, -37.50); for the negative Value, the Point will be below the axis of Inertia by 37.50 units and 56.25 units from the vertical axis. We will join the 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 finding the Midpoint O, which has coordinates 0.50*(Ix+Iy). From the given data, this Value is (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 × 10^6 mm4.Iy=56.25×10^6 mm4.Ixy=37.50×10^6mm4. Using the data, we can calculate the radius of Mohr’s Circle of Inertia; applying the known formula, we get a radius of 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 radius of the Circle to the center Point. The maximum Value of the Moment of Inertia is (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_minequa11 units, the Product of Inertia is equal, and the Point of minimum Value 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 between 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 locate Point A’, the mirror image of Point A. Point A’ has coordinates (100, -37.50) units. We will join Point O with Point A’ to 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’s Value will be equal to Imax. We plug in the Value of (2φp 1) as -59.743 and estimate the Value of Ix’.

We find that the Value matches the estimate from Mohr’s Circle for Imax. 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’s Value will be equal to Imin. If we plug in 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 Imin.

Orient the datums so the x-direction is horizontal.

In the normal view, the x-axis is horizontal, 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 make the x-axis horizontal in Mohr’s Circle of Inertia, represent the x-axis as the line OE. We adjust the datums of Ix and Ixy to define the direction of U we have already estimated; we will find that the directions of U and V are achieved.

Point B is at an angle of 180 from Line OE and still has the same coordinates of (56.25, -37.50). The complete 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.

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