Last Updated on March 1, 2024 by Maged kamel
Product Of Inertia Ixy for a Rectangle at the CG.
A third method is presented in the following video to estimate the Ixy g as a third proof. Again, we have our rectangle with base b and height h, and we have external two axes x and y at the left edge left corner of the rectangle, and we have x CG and y CG. But this time, our strip will be small with width =dx and height =dy.
We will estimate the horizontal distance as X to y CG and y to the CG. This is a part of the video with a closed caption in English.
The first method used to get the product of inertia Ixy for a Rectangle.
To get the Ixy at the CG for a rectangle, we consider the moment of inertia about the two axes passing by the CG.
1- The rectangle can be considered as composed of 4 equal areas, like quarters, each quarter =A/4; for the first quarter, the x distance is at the left of y’ axis, so the x distance is =-b/4, while the y1 is =h/4 with a positive sign.
2- For the remaining three quarters, we will check the signs of x&y; for the second area, the xis+ve & y is +ve; for the third area, the xis-ve and y are also-ve, for the last quarter, the x-distance is -ve, and y distance is also-ve.
3- Summing all the products of Ai *xi*yi we will get Ixy at the Cg =0. The next slide image shows in detail the first method of estimation.
![Pict 1- post 5- inertia The product of inertia for a rectangle](https://magedkamel.com/wp-content/uploads/2021/11/Pict-1-post-5-inertia.jpg)
The second method to find the product of inertia Ixy for a Rectangle.
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 strip of width dy and breadth=b.
2- estimate the Ixy=∫h*dy*x/2*y from y=0 to y=h.
3-the value of integration will be Ixy=Abh/4. 4 the value of Ixyg=Ixy-A(b/2)*(h/2)=0. the details are shown in the next slide image.
![Pict 2- post 5- inertia The second method is to get the value of the product of inertia.](https://magedkamel.com/wp-content/uploads/2021/11/Pict-2-post-5-inertia.jpg)
The third 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 strip of width dy and breadth=dx.
2-Ixyg=∬dx*dy*x*y, x, and y are the distances to the X-CG and Y-CG respectively, the integration will be done 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.
![Pict 3- post 5- inertia The third method is to get the value of the product of inertia for a rectangle.](https://magedkamel.com/wp-content/uploads/2021/11/Pict-3-post-5-inertia.jpg)
Ixy at the external axes.
Ixy at the external axes does not=0, but we will integrate the area (dAxy) from x=0 to x=b and from y=0 to y=h, to get the product of inertia at the two axes at the external corner of the rectangle section.
![Pict 3- post 5- inertia The third method is to get the value of the product of inertia for a rectangle.](https://magedkamel.com/wp-content/uploads/2021/11/Pict-3-post-5-inertia.jpg)
This is a list of the data for the rectangle starting from the first moment of area to the value of the product of inertia.
![Pict 5- post 5- inertia The product of inertia from a table.](https://magedkamel.com/wp-content/uploads/2021/11/Pict-5-post-5-inertia.jpg)
Polar moment of inertia.
The polar moment of inertia=Ix+Iy, if we want the polar moment of inertia value at the CG, we add both the moments of inertia at the CG, and we will get J0G=b*h(b^2+h^2)/12. While at the external axes, The polar moment of inertia j0=bh/3(h^2+b^2).
![Pict 6- post 5- inertia The polar moment of inertia for the rectangular section](https://magedkamel.com/wp-content/uploads/2021/11/Pict-6-post-5-inertia.jpg)
This is the PDF used for the illustration of this post.
For an external resource, engineering core courses, the moment of inertia.
This is a link to the next post, for the Moment of inertia.