Last Updated on March 31, 2024 by Maged kamel
- Moment of inertia-Iy for the right-angle triangle-Case-1.
- Step-by-step guide for the calculation of moment of inertia-Iy right-angle-case-1.
- Moment of inertia-Iy at the CG of the right-angle triangle.
- Polar moment of inertia at the CG for the right-angle triangle for case-1.
- Polar moment of inertia at the left corner for the right-angle triangle for case-1.
- Using a horizontal strip as one option to estimate the moment of inertia-Iy value.
Moment of inertia-Iy for the right-angle triangle-Case-1.
Step-by-step guide for the calculation of moment of inertia-Iy right-angle-case-1.
We are interested in the left corner of that triangle again our expression of the line BC is y= (-h/b)*x+h, for the vertical strip, its area is dA, and we have dA= the width which is dx*y.
We will integrate the value of X from X=0 to X = b. We substitute for the Y- value by writing the equation of y=(-h*x/b)+h, our dA *x^2 will give the value of dIy about the y-axis. Multiply dA*x^2. The expression for dIy=(h*)-(x^3*dx/b) +x^2*dx).
![](https://magedkamel.com/wp-content/uploads/2022/01/page-1-post-8-inertia.jpg)
The integration from x=0 to x=b will be carried out.
Finally, the value of the moment of inertia about the Y axis, Iy=h*b^3/12, for which, h, is the triangle height and b is the base length. the inertia is about the y-axis on the left side of the triangle.
![page 2 -post 8-inertia Moment of inertia-Iy for right-angle triangle-2/2.](https://magedkamel.com/wp-content/uploads/2022/01/page-2-post-8-inertia.jpg)
Moment of inertia-Iy at the CG of the right-angle triangle.
For Iy at the CG, we will use the parallel axes theorem and deduct the product of A*xbar^2. x bar represents the horizontal distance between Cg and the vertical axis Y.
The area A=(1/2)*b*h, for the triangle Cg is located at a distance=b/3 from the left corner and y=h/3 from the base of the triangle. Finally, we get Iyg=h*b^3/36. The square of the radius of gyration k^2y can be estimated by dividing Iyg/A. The value of K^2yg equals b^2/18.
![page 3 -post 8-inertia The radius of gyration about y’ axis at the Cg.](https://magedkamel.com/wp-content/uploads/2022/01/page-3-post-8-inertia.jpg)
Polar moment of inertia at the CG for the right-angle triangle for case-1.
Polar moment of inertia for the right-angle triangle case-1, at the Y-axis passing the CG. We have estimated Ix at the CG, from post-7-Moment of inertia for right-angle-Ix- Case-1.
The polar moment of inertia Ip=Ixg+Iyg, adding them together, then Ipg=b*h(b^2+h^2)/36, we call it Jog, the value at Jog at the CG is shown as per the next slide image. The final value of J0 at Cg=(h^2+b^2)/18
![](https://magedkamel.com/wp-content/uploads/2022/01/page-4-post-8-inertia.jpg)
Polar moment of inertia at the left corner for the right-angle triangle for case-1.
The Polar of inertia at the external axis x,y passing the external corner. Jo=Ix+Iy, adding them together, then Jo=b*h(b^2+h^2)/12. The value at Jog at the CG is shown as per the next slide image.
![page 5 -post 8-inertia Polar moment of inertia Jo for a right angle.](https://magedkamel.com/wp-content/uploads/2022/01/page-5-post-8-inertia.jpg)
Using a horizontal strip as one option to estimate the moment of inertia-Iy value.
The new option is to use a horizontal strip to get the value of the moment of inertia-Iy for a right-angle triangle case No-1. The following steps are followed: 1- establish a relation between x and y, where x is the strip width, while y is the distance from the base to that strip, this can be done by examining the slope of line CB.
![page 6 -post 8-inertia Getting Iy by using a horizontal strip for a right angle-1/3.](https://magedkamel.com/wp-content/uploads/2022/01/page-6-post-8-inertia.jpg)
2-What is the inertia of that strip about the y-axis? the answer is that this is a typical case of a rectangle section moment of inertia Iy at the edge, which is equal to the height of the width^3. We can substitute and use the relation between x and y.
![page 7 -post 8-inertia Getting Iy by using a horizontal strip for a right angle-2/3.](https://magedkamel.com/wp-content/uploads/2022/01/page-7-post-8-inertia.jpg)
We are integrating into the y-direction, we want to get rid of x.
3-The integration will be done from y=0 to y=h. Finally, we can get the Iy value or the moment of inertia Iy for a right-angle triangle case-1 about the Y-axis. The result is matching with the previous value of Iy. However, this method took a longer time compared with the vertical strip.
![page 8 -post 8-inertia Getting Iy by using a horizontal strip for a right angle-3/3.](https://magedkamel.com/wp-content/uploads/2022/01/page-8-post-8-inertia.jpg)
As we can see the final result matches the values included in the table of inertia for plane shapes.
![page 0-post 8-inertia Part of the list of inertia for plain shapes.](https://magedkamel.com/wp-content/uploads/2022/01/page-0-post-8-inertia.jpg?ver=1709634716)
This is the PDF used for the illustration of this post.
This is a link Second Moment of Area for Standard shapes.
This is the next post, Product of inertia Ixy– for the right-angle triangle-case-1.