Moment of inertia for right-angle-Ix- Case-1.
Brief content of the video.
We have an objective, which is how to determine the moment of inertia for right-angle-Ix about the X-axis, by using a horizontal strip parallel to the external X-axis, we have a right-angle triangle in two shapes. In the first shape, we have the triangle AB, the adjacent side at the base, and the opposite side at the left-hand side of the triangle, this is the hypotenuse side.
We are going to make the X-axis coincide with the adjacent side and the Y-axis will coincide with the opposite side. We are going to use a horizontal strip parallel to the x-axis at the adjacent base and with the same result, we can obtain it by using a strip vertical strip, which is perpendicular to the external x-axis. This is a part of the video, which has a closed caption in English.
You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.
Case no.1 is the one for which the x-axis coincides with the base and the y-axis coincides with the opposite side, while case no.2 is the one for which the x-axis coincides with the base and the y-axis is at the intersection point between the adjacent side and the hypotenuse side.
Case no.1, the moment of inertia for right-angle-Ix is the second case in the reference attached table from FE Exam reference manual 3-5.
This is the list of the first moment of area and inertia for the common plane shapes.
The next slide image represents the difference between Case -1 and Case-2 of the right-angle Triangle. for case -1 the opposite of the triangle is at the left side and for case-2, the opposite of the triangle is at the right.
How to get inertia for right-angle-Ix by using a horizontal strip?
Using the following steps, we will estimate the inertia for right-angle-Ix using a horizontal strip:
1-for line CB, we will write the equation as y=-h*x/b+h, we can check the validity of this equation by substituting the value of x coordinates of both C& b and get the corresponding value.
2- Our strip is a horizontal strip with width=dy and base=x. 3-dA, which is the area of the horizontal strip will be=x*dy, to get the dy, we will differentiate the y equation, which we have already estimated.
3- Our moment of inertia due to strip is dIx=dA*y^2, remember that dA=x*dy.
4- we can write dIx=b*(h-y)/h*(y^2*dy), after substitution by the value of dy.
5-Ix=∫dIx=∫b*(h-y)/h*(y^2*dy), from y=0 to y=h.6- after integration and substitution we get Ix=b*h^3/12.
The value of inertia for a right-angle-Ix=base*height^3/12.
How to get theinertia for right-angle-Ix by using a vertical strip?
Using the following steps, we will estimate the Ix the moment of inertia for the right-angle triangle by using a vertical strip: 1- we are going to move this strip horizontally, so we have to substitute the value of y=-h/b*x+h, as we will see later.
2- Our strip is a vertical strip with width=dx and height=y.
3- The strip with an area dA, will be=y*dx.
4-The moment of inertia for the small element -dIx from our study of a rectangular section can be estimated as dx*y^3.
5- The moment of inertia for the small element – dIx value is shown as per the next slide picture.
6-we will integrate from x=0 to x=b.
7- we will get the same value of Ix as estimated by using a horizontal strip, which is=b*h^3/12.8- k^2x=Ix/A=b*h^3/(12*(0.50*b*h))=h^2/6.
How to get inertia for right-angle-Ix but at the CG?
Ixg at the CG can thus be obtained from the theorem of parallel axes, Ixg=Ix-A*ybar^2=(b*h^3/12)-(0.50*b*h)*(h/3)^2=Ixg=b*h^3/36.
The radius of gyration at the Cg, Kxg=sqrt(IxG/Area)=sqrt(b*h^3/36)/(0.50*b*h))=h/(3*sqrt(2)).
This is the PDF used for the illustration of this post.
For an external resource, the definition of inertia and the second moment of area for common shapes.
This is a link for the next post, Moment of inertia – for the right-angle triangle-case-1-Iy.