13- A solved Problem for Mohr’s Circle of Inertia, case 4.

Last Updated on October 5, 2026 by Maged kamel

A solved Problem for Mohr’s Circle of Inertia, case 4.

This Post introduces a solved Problem for Mohr’s Circle of Inertia, case 4. The Moment of Inertia about the X-axis, Ix, is less than the Moment of Inertia about the Y-axis, Iy, and the Product of Inertia, Ixy, is negative.

The given solved Problem is for an unequal angle measuring 3 inches by 2 inches, and the thickness of the steel angle is (1/4) inch.

We need to determine the principal Moment of Inertia. We will use Mohr’s Circle of Inertia to get the directions of the principal axes and later verify them with the general equation to check the estimated Values. The long leg 3″ is parallel to the x-axis, while the short leg is parallel to the Y-axis.

We will use Table 1-7 for angles to get more information about Area and other dimensions.

Area, Cg, and inertias for the angle in the solved Problem for Mohr’s Circle of Inertia case 4.

The Area of the angle-3″x2″x1/4″ equals 1.19 in2, and the horizontal distance from the right edge of the unequal angle to the Cg is 0.98″. The vertical distance from the Cg of the unequal angle to the upper edge equals 0.487 inches.

We can get the Moment of Inertia Ix for the given angle from Table 1-7. We will reverse the values and use the Ix Value from the Table as Iy in the solved example since the angle is rotated.

The long leg is parallel to the x-axis, unlike the data shown in Table 1.7, where the long portion is parallel to the y-direction. The Area and Cg distances are shown in the Table on the next slide.

The data for the solved Problem is shown on the next slide.

The Product of Inertia for the angle in the solved Problem for Mohr’s Circle of Inertia case 4.

The x-axis and Y-axis are placed at the CG of the unequal angle. We want to estimate the Product of Inertia, Ixy. We divide the upper leg into two areas, while treating the short leg as one Area.

For each Area, we estimate the Product of Inertia and then add each Area multiplied by the Product of x*y to the CG axes.

The detailed estimations of the product-of-inertia values are shown in the next slide image.

The final Value of the Product of Inertia equals (-0.3798) inch4. The values of Ix, Iy, and Ixy are collected, and it is found that Ix is less than Iy and Ixy is negative. This case is case 4.

Solved Problem for Mohr’s Circle of Inertia, Case 4: Drawing the Circle.

For the solved Problem on 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 pointing up. We could make 1 unit equal to 0.10 inch4.

We start by locating Point A, which has Ix and Ixy values of (0.39, -0.3798) inch4 and will be located below the axis of Inertia by a Value of negative Ixy equals 0.3978 inch4, and apart from the vertical axis Ixy by a distance of +0.39 inch4.

Similarly, we can draw Point B, which has coordinates (Iy, +Ixy). In inch4 will be (1.39, +0.3798). We will join the two points. Line AB will intersect the horizontal axis at Point O, which will be the Circle’s center.

We can draw the Circle by getting the middle Point O, which has a coordinate of (0.50*(Ix+Iy),0) from the given data; this Value is equal to (0.39+1.09)*0.50=0.74 inch^4.

The radius of the Circle is estimated from the equation R=sqrt ((Ix-Iy)^2+Ixy^2); this has been applied since. We have Ix=0.39 inch4.Iy=1.09 inch4.Ixy=-0.3798 inch4.

The next slide image shows the calculations for the center Point and the Circle’s radius.

A solved Problem for Mohr’s Circle of Inertia, case 4: The Value of Imax and Imin.

The values of Imax and Imin are estimated as follows: Imax = (Ix + Iy) * 0.50 + R = (0.39 + 1.09) * 0.50 + 0.5165 = 1.2565 in^4.^ The estimation of Imin is equal to =(Ix+Iy)*0.50-R=(0.39+1.09)*0.50-0.5165=0.2235 inch^4.

We need two mirror points: the first Point is A’, with coordinates (0.39, +0.3789) in inches^4, and the second Point is B’, with coordinates (1.09, -0.3798).

To get the U direction, join Point O with Point A’. For the V direction, join Point O with Point B’.

Value of Angle 2φp2 for the Mohr’s Circle of Inertia case 4.

The principal axis V makes a clockwise angle of 2φp2. This angle can be estimated from tan (2φp2) = (2*Ixy)/(Iy-Ix), or (+*(-0.3798)/1.09-0.39)=-1.0851.

The angle is 47.338 degrees.

The Value of 2φp1 equals (180-47.338)=+132.662 degrees anticlockwise. The angle φp1 is 66.831 degrees, while the angle φp2 is -23.669 degrees.

To get the U and V directions in the normal view, join Point C with A’.For the V direction in the normal view, join Point C with Point B’.

The solved Problem for Mohr’s Circle- case 4- using the general Expression.

We can use the general Expression for Ix’ to check the Value of Imax, provided that (2φp1) is 132.662 degrees; the Ix’ Value will be equal to Imax. We plug in with Ix value= 0.39 inch4,Iy value=1.09 inch4, and Ixy=-0.3798 inch4.

The Value of Ix = 1.2565 in^4 matches Imax estimated using Mohr’s Circle—the complete details of Ix’ estimation using the general equation.

We can use the general Expression for Ix’ to check the Value of Imin, provided that (2φp1) is 132.662 degrees; the Iy’ Value will be equal to Imin. We plug in with Ix value= 0.39 inch4,Iy value=1.09 inch4, and Ixy=-0.3798 inch4.

The Value of Iy ‘ = 0.2235 in^4 is the same as Imin estimated using Mohr’s Circle—the complete details of Iy’s estimation by the general equation.

This compares the estimated Value of Imin with the Value from the steel angle Table. Imin is the Product of r^2*A = 0.22106 in^4. There is a slight difference between iv and the estimated Imin from both Mohr’s Circle of Inertia and the general equation.

You can download and review this Post’s content in the following PDF file.

The next Post, No. 14, is a solved Problem: Inertia Value for a given direction.

This is a Link to a useful external resource: a calculator for Cross-Section, Mass, Axial, and polar Area moments of Inertia, and Section Modulus.