![]() ![]() the next slide image shows in detail the first method of estimation. To get the Ixy at the CG for a rectangle, we consider the moment of inertia about the two axes passing by the CG.ġ- 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 y 1 is =h/4 with a positive sign.Ģ- 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 are also-ve.ģ- Summing all the products of A i *xi*y i we will get Ixy at the Cg =0. Figure 10.6.1: (a) A barbell with an axis of rotation through its center (b) a. From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center. The first method used to get the product of inertia Ixy for a Rectangle. In the case with the axis at the end of the barbellpassing through one of the massesthe moment of inertia is. 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. But this time, our strip will be small with width =dx and height =dy. 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. This is a link to the first video via u-tube.Ī third method is presented in the following video to estimate the Ixy g as a third proof. This is a part of the video with a closed caption in English. for example, A1 is the area the width = (b/2) the height is (h/2). We divide this rectangle into 4 areas, namely A1, A2, A3, A4, and for each area, we are going to locate the CG of every area. To find the second moment of the area when the origin of the coordinate system does not coincide with the centroid, use the parallel axes theorem. Then we come to the CG and we draw another two axes, which are X’ and y’. ![]() We draw a rectangle for which the Base is b and the height is h, and we introduce the external two axes X and Y. We want to prove that the product moment of inertia for a rectangular section =0. In Unit number 5, we’re going to talk about the product moment of inertia for a rectangular section at the CG. Product Of Inertia Ixy for a Rectangle at the CG video. Product Of Inertia Ixy for a Rectangle at the CG.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |