9- A solved problem-case-2-Mohr’s Circle of Inertia.

Last Updated on October 5, 2026 by Maged kamel

Solved problem-case-2: Mohr’s Circle of Inertia.

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In this Post, we will introduce a solved problem- Case 2: Mohr’s Circle of Inertia, where Ix is less than Iy, and the Product of Inertia Ixy is positive.

For a given Area, the following data are presented: the Moment of Inertia Ix is 14 inches4—the Moment of Inertia about the y-axis,, IyyIIys 2in2 in^4he Product of Inertia, Ixy, is 12 in^4.

We need to determine the principal moments of Inertia and the angle alpha that the x-axis makes with the maximum Moment of Inertia.

In the first slide, I evaluated the coordinates of points A and B, as well as the radius R, which we will use to draw Mohr’s Circle of Inertia.

Solvedproblem-case-2-Mohr’s Circle of inertia-drawing the Circle.

For the solved Problem, case 2, 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 a positive Value indicating a positive Product of Inertia. One unit represents 1 inch.

We start by locating Point A, which has Ix and Ixy values of (14,12) units. It will be located above the axis of Inertia by axy = 12 units and a14 units from the vertical axis.

Similarly, we can draw Point B, which has coordinates (Iy, -Ixy). In units, it will be (24, -12); for the negative Value, the Point will be below the axis of inertia by 12 units and 24 units from the vertical axis. We will join both points. The line AB intersects the horizontal axis at point O, which is the center of the Circle.

We can draw the Circle using the Midpoint O, whose coordinates are 0.50*(Ix + Iy). From the given data, this Value equals (14 + 24) × 0.50 = 19 units.

The radius of the Circle is estimated from the equation R = sqrt ((Iy-Ix)^2+Ixy^2), which is applied since Iy is bigger than Ix. We have Ix = 14 in^4. Iy=24 inch4.Ixy=12 inch4. We can use the data to determine Mohr’s Circle Inertia radius.

Using the known formula, we get a radius of 13 in^4. Please refer to the slide image for more details.

This shows how to find 2θp1 for the solved Problem, A Solvedproblem-Case-2-Mohr’s Circle of Inertia.

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 (19 + 13) = 32 units, and 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.

The minimum Moment of Inertia I min is equal to 6 units, the Product of Inertia is equal to zero, and the Point of minimum Value of Inertia is Point C.

Solvedproblem-case-2-Mohr’s Circle of inertia-direction of principal axes U&V.

To draw the direction of the major axis U in Mohr’s Circle of Inertia, we will locate Point ”, which is the mirror Point of Point A. Point ” has coordinates (10, -12) units. We will join Point O and Point ” to get the direction of the major Axis U.

To draw the direction of the major axis V in Mohr’s Circle of Inertia, locate Point B”, which is the mirror Point of Point B. Point B ” has coordinates (24, +12) units. Join Point O and Point B”, and with Point ” we will get the direction of the major Axis V.

The directions of U and V for the normal view.

Draw a line from Point C to ”, this is the direction of the U-axis in the normal view. Draw a line from C to Point B”. This is the direction of the V-axis in the normal view.

Solved problem-case-2: Mohr’s Circle of Inertia using the general Expression.

We can use the general Expression for I” to check the Value of Imax, provided that when the angle 2θ = (2φp1), which is (-112.62 degrees ), the Ix” Value will be equal to Imax. We plug in Ix Value = 14 in^4, Iy Value = 24 in^4, and Ixy = 12 in^4.

The Value of Ix’=32 in^4 is the same as Imax estimated using Mohr’s Circle.

We can use the general Expression for I” to check the Value of Imin, provided that when the angle 2θ = (2φp1), which is (-112.62 degrees, the Iy” Value equals. We plug in Ix = 14 inch4, Iy = 24 inch4, and Ixy = 12 inch4.

The Value of Iy’ = 6.0 in^4 is the same as Imin estimated using Mohr’s Circle.

Check that Ix + Iy equals Iu + Iv, where Iu is the maximum Value of Inertia and Iv is the minimum Value of Inertia. Ix+Iy=14+24=38 inch4. imin+Imax=6+32=38 inch4.

Orient the datums so the x-direction is horizontal.

To make this horizontal, rotate the Inertia so the x-axis is represented by line OE. Adjust the Ix and Ixy datums to define the direction of U.

Line OE represents the x-direction with respect to the two new datums shown in the next slide image. Line OC represents the y-direction with respect to the two new datums. The complete data that covers our solved Problem case 1- Mohr’s Circle of Inertia is included in the slide image.

The following Post discusses Mohr’s Circle of Inertia, the third case. In this case, the Moment of Inertia about the x-axis is bigger than that about the y-axis, and the Product of Inertia Ixy is negative.

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This video illustrates the content of this Post: an introduction to Mohr’s Circle of Inertia.

This links to a valuable external resource: a calculator for Cross-Section, Mass, Axial, and Polar Area Moment of Inertia, and Section Modulus.