5- Easy approach for Product of inertia Ixy for a rectangle

Last Updated on December 31, 2025 by Maged kamel

Product Of Inertia Ixy for a Rectangle at the CG.

The following post presents three methods for estimating Ixy at the CG. 

The first method for computing the product of inertia Ixy for a Rectangle.

The first method for computing the product of inertia Ixy for a Rectangle, by examining the intersection of the two axes x dash and y dash at the Cg.
1—The rectangle can be considered composed of 4 equal areas, if we consider the total area as equal to A, like quarters; each quarter =A/4; for the first quarter, the x distance is at the left of the y dash-axis, so the x distance is =-b/4, while the y1 is =h/4 with a positive sign. 

 2—We will check the signs of x and y for the remaining three quarters, based on the positive directions of the two axes X dash and Y dash. For the second area, x is +ve and y is +ve; for the third area, x is -ve, and y is also -ve. For the last quarter, the x-distance is -ve, and the y-distance is also-ve.

3- Summing all the products of A_i *xi*y_i, we will get Ixy at the Cg =0; since quarters A1 and A3 will give a positive value of Ixy, while quarters A1 and A4 will give a negative value for Ixy. The following slide provides a detailed overview of the first estimation method.

The second method for finding the product of inertia Ixy for a rectangle using a horizontal strip.

The second method to get the value of the product of inertia for the external edge and also at the Cg is as follows:
1-Introduce a horizontal strip of width dy and breadth=b.  
2- Estimate the product of inertia Ixy=∫(b*dx)*(x/2)*y from y=0 to y=h.    

3-The value of integration will be Ixy=b*b/2∫ydy=b^2/2*(y^2/2) from zero to h. The Ixy=1/2*b^2*(h^2/2)=(b*h)*(b*h/4)=A*b*h/4. 4. While the product of inertia at the cg will be Ixyg=Ixy-A(b/2)*(h/2)=0. The details are shown in the next slide image.       

The third method for finding the product of inertia Ixy for a rectangle using a rectangular strip.

The third method to obtain the value of the product of inertia for the external edge and at the Cg is as follows: 1. Introduce a strip of width dy and breadth dx.  

2-The product of inaerta about the Cg, Ixyg=∬dx*dy*x*y, x, and y are the distances to the X-CG and Y-CG, respectively; these axes intersect at the Cg. The integration will be performed from x =- b/2 to x = b/2 and from y =- h/2 to y = h/2.   
3- The value of integration will be Ixyg=0.           

Ixy at the external axes.

We continue using the third method to get the value of product of inertia Ixy at the external axes, which does not equal 0; we will integrate the area (dA*x*y) from x=0 to x=b and from y=0 to y=h to obtain the product of inertia at the two axes intersection at the external corner of the rectangular section.

This is a list of the data for the rectangle, from the first moment of area to the product of inertia.

Polar moment of inertia.

The polar moment of inertia is Ix + Iy. If we want the polar moment of inertia at the CG, we add the moments of inertia at the CG, and we get J0G = b*h *(b2 + h2)/12. At the external axes, the polar moment of inertia is J0 = (bh/3 )*(h2+b2).

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

For an external resource on Engineering core courses: the moment of inertia.

This is a link to the next post for the Moment of inertia.