11- A Solved problem-case-3-Mohr’s circle of inertia.

Last Updated on January 4, 2026 by Maged kamel

A Solved problem-case-3-Mohr’s circle of inertia.

In this post, we will introduce a solved problem-case-3-Mohr’s circle of inertia, where Ix is bigger than Iy, and the product of inertia Ixy is negative.

The given solved problem concerns an unequal angle measuring 8 inches by 6 inches, with a thickness of 1 inch.

It is required to determine the principal moments of inertia. We will use Mohr’s circle of inertia to obtain the directions of the principal axes, and later verify the estimated values using the general equation.

We will check the X-bar for the unequal angle by dividing the area of the unequal angle into two rectangles, from which the difference will give the area. We set two axes x and Y at the Cg.

The first rectangle is eight by 6 inches, and the second is five by 7 inches. The x bar or the distance from the Cg can be found as equal to 1.65 inches measured from the Y-axis.

A Solved problem-case-3-Mohr’s circle of inertia-estimate Ix, Iy, and Ixy at the Cg.

We can proceed to estimate y-bar for the given unequal angles. The Y bar or the distance from the Cg can be found as equal to 1.65 inches measured from the axis Y”.

We will estimate the moment of inertia Ix about the Cg for the Unequal angle by considering two rectangles: the first is 1 inch wide by 8 inches high, and the second is 5 inches long by 1 inch thick.

We will add the product of the area for each rectangle by the square of the vertical distance between each rectangle Cg to the final Cg distance to get the Value of the moment of inertia about the x-axis passing through the Cg.

We will estimate the moment of inertia Iy about the Cg for the Unequal angle by considering two rectangles: the first is 1 inch wide by 8 inches high, and the second is 5 inches long by 1 inch thick. We will add the product of the area of each rectangle and the square of the horizontal distance between each rectangle, Cg, to the final Cg distance.

The last step is to estimate the product of inertia for the unequal angle by evaluating it for each rectangle, which is zero. We will add the product of each area by its horizontal and vertical distance to Cg.

Based on the previous estimation, we have case 3: Mohr’s circle of inertia, where Ix is larger than Iy and Ixy is negative.

Solved problem-case-3-Mohr’s circle of inertia-drawing the circle.

For the solved problem-case-3-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 inch4.

We start by locating Point A, which has Ix, Ixy values of (80.8, -32.31) units, is located below the axis of inertia by a value of negative Ixy = 32.31 units, and is apart from the vertical axis by a distance of (80.80) units.

Similarly, we can draw point B, which has a coordinate of (Iy,+Ixy). In units will be (38.8, +32.31). 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 by using the midpoint O, which has a coordinate of 0.50*(Ix + Iy). From the given data, this value is equal to (80.81 + 38.8) × 0.50 = 59.80 units.

The radius of the circle is estimated from the equation, R=sqrt ((Ix-Iy)^2+Ixy^2), which is applied since. We have Ix=80.8 inch4.Iy=38.80 inch4.Ixy=-32.31 inch4.

For the minimum value of inertia, the value will be equal to (0.50*(Ix+Iy)-R, where R is the radius value. 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 38.534 inch4. Please refer to the slide image for more details.

The Value of the product of inertia is found to be equal to (-32.31)inch4 as shown in the next slide image.

The minimum value of the moment of inertia is equal to 21.266 inch4; point C represents the point of minimum moment of inertia on Mohr’s circle of inertia.

To get the maximum value of inertia, Imax, we can add the radius of the circle to the distance from the center. The maximum value of the moment of inertia is equal to (59.80+38.534)=32 units, the product of inertia is zero, and this point is point E.

The directions for both axes X and Y are represented by lines OA and OB.

Solved problem-case-3-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 A’, which is the mirror point of Point A. Point A’ has a coordinate of (80.8,+32.31) units. We will join point O with point A’, and we will get the direction of the major Axis U.

Value of Angle 2φp1 for the Solved problem-case-3-Mohr’s circle of inertia

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)

The equation is written as (-Ixy)/(Ix-Iy)/2). We have Ix=80.80 units, Iy=38.80 units, Ixy=-32.31 units, the tan value is (-2*(-32.31)/(80.80-38.80)=+1.53857.

The plus sign indicates that this angle is anticlockwise. The angle’s value is 56.978; this is 2φp1. The value of 2φp2 is equal to 180-56.978=123.021 degrees in the clockwise direction. Please refer to the next slide image for more information.

The directions of U and V for the normal view.

There is a separate slide to show the orientation of both U and V axes in the normal view, construct a line between points C and A’ will give the U direction in the normal view while drawing a line between points C and B’ will give the direction of V axis in the normal view. The value of φp1 equals 28.489 degrees, while the value of φp2=-61.511degrees in the clockwise direction.

Solved problem-case-3-Mohr’s circle of inertia using the general expression.

We can use the general expression for Ix’ to check the value of Imax, provided that when the angle 2θ = (2φp1), which is (56.978 degrees), Ix’ value will be equal to Imax. We plug in with Ix value= 80.80 inch4,Iy value=38.80 inch4, and Ixy=-32.31 inch4.

The value of Ix’=98.334 inch4 is the same as Imax estimated by the use of Mohr’s circle. The details of the estimation of Ix’ when 2θ = (2φp1).

We can use the general expression for Iy’ to check the value of Imin, We can use the general expression for Ix’ to check the value of Imax, provided that when the angle 2θ = (2φp1), which is (56.978 degrees), Ix’ value will be equal to Imax. We plug in with Ix value= 80.80 inch4,Iy value=38.80 inch4, and Ixy=-32.31 inch4.

The details of the estimation of Iy’ when 2θ = (2φp1).

The value of Iy’=21.266 inch4 is the same as Imin estimated by the use of Mohr’s circle.

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. Ix+Iy=80.80+38.8=119.60 inch4. imin+Imax=98.334+21.266=119.60 inch4. Check that IP is the same value when adding Ix+Iy as compared to the sum of Iu plus Iv.

Oriented datums to let the x-direction be a horizontal axis.

To let the x-axis be the horizontal Mohr’s circle of inertia, the x-axis is represented by the line OE. The datums of Ix&Ixy are adjusted to define the direction of U.

Line OE represents the x-direction concerning two new datums, which are shown in the next slide image. Line OC represents the Y direction relative to two new datums.

The oriented two axes fulfill the condition of Mohr’s circle of inertia, case 3. Ix is bigger than Iy, and Ixy is negative. Please refer to point E in the next slide image and check the normal view angles.

This is an extract from the table of Unequal angles. For our unequal angle 6x8x1 inches, the VV axis is the axis of minimum moment of inertia, and its value is given by A*rmin^2. The product offers Imin as 21.299 inch4, which is close to our estimated value of 21.266 inch4. Also, the direction of the alpha angle is close to our angle φp1.

You can download and review the content of this post through the following pdf file.

In the next post, we will solve a problem that covers Mohr’s circle of inertia, fourth case.

There is another new solved problem in post no. 14 is a solved problem-inertia value for a given direction.

This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial and Polar Area Moment of Inertia, and Section Modulus.