8-Moment of inertia-Iy for right-angle triangle case-1.

Last Updated on September 8, 2026 by Maged kamel

Moment of inertia-Iy- for right-angle triangle case 1.

Step-by-step guide for the calculation of Moment of inertia-Iy- for the right-angle case 1.

We are again interested in the left corner of that triangle. For the line BC, y = (-h/b) x + h. For the vertical strip, the area is dA = width dx* y.

We will integrate the value of x from x = 0 to x = b. We substitute for y using y = (-h*x/b) + h; our dA*x^2 term 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).

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, where 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.

Moment of inertia, Iy, at the CG of the right-angle triangle.

For Iy at the CG, we will use the parallel-axis theorem to derive the product I_A x bar^2. x bar represents the horizontal distance between the CG and the vertical axis Y.

The Area of triangle A equals (1/2) bh. The triangle CG is located at a distance b/3 from the left corner and at 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^2,y, can be estimated by dividing Iyg/A. The value of K^2yg equals b^2/18.

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, about the Y-axis passing through 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, we get pg = (b^2 + h^2)/36. We call this J0g. The value of J0 at the CG is shown in the next slide. The final value of J0 at Cg=(h^2+b^2)/18

Polar moment of inertia at the left corner for the right-angle triangle for case 1.

The polar Moment of inertia at the external axis x,y passing through the external corner. Jo = Ix + Iy; adding them together, then Jo = b*h (b^2 + h^2)/12. Jo’s value at the CG is shown on the next slide.

Using a horizontal strip as one option to estimate the Moment of inertia, Iy, for the right-angle triangle case 1

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. At the same Time, y is the distance from the base to that strip. This can be done by examining the slope of line CB.

2-What is the inertia of that strip about the y-axis? The answer is that this is a typical case of a rectangular section: the Moment of inertia Iy at the edge, which is equal to the height times the width^3. We can substitute and use the relation between x and y.

We are integrating in 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 case 1: right-angle triangle. The result matches the previous value of Iy. However, this method took longer than the vertical strip method.

As we can see, the final result matches the values included in the table of inertia for plane shapes.

This is the PDF file used for the illustration of this post. You can download it by clicking the following button.

This is the link to the video covering the data in this post.

To determine how to find the Moment of inertia Ix for a right-angle case, consider case 1. Refer to the previous post link.

This is a link to the Second Moment of Area for Standard shapes.

This is the next post: Product of inertia Ixy for the right-angle triangle case 1.