Last Updated on August 18, 2026 by Maged kamel
Ix-the moment of inertia for a right-angle-Case-2.
The moment of inertia-case-2 for a right-angle by using a horizontal strip.
To get the area and Moment of inertia for regular shapes, we can refer to the NCEES reference handbook, 3.50. The Moment Of Inertia Ix-Case-2 is the first item in the table.
The right-angle triangle, which is on the right side of the next slide, represents case No.2. Please refer to the next slide image.
Step-by-step guide for the calculation of moment of inertia-case 2.
1-For Ix-Case-2 estimation, a horizontal strip will be used, the strip width is dy, and its height is x.
2- Since the strip height starts from the base and intersects with line AC, the y value is the same as the line AC equation, which is y=h*x/b.
3- The moment of inertia due to that strip=dA*y^2. Our strip’s left edge is at a distance =x.
4- the width of the strip=(b-x) and the depth is =dy.
5- We will substitute the value of dA and try to make all our items as a function of y, since we will integrate from y=0 to y=h for the Ix value.
6-After performing the integration, we get the value of inertia for a right-angle-Ix-case-2 as Ix=b*h^3/12 about the x-axis passing by the base.Ix=(base)*(height)^3/12.
7-For Ix at the CG, we will use the parallel axes theorem and deduce the product of A*x-bar*y-bar.
8- for x bar=b/3, while y bar=h/3. Finally, we get Ixg=b*h^3/36.
The moment of inertia-case-2 for a right-angle by using a vertical strip as an alternative method.
1-For the moment of inertia for a right-angle triangle in case 2, the second alternative method, which uses a vertical strip, will be used; the strip width is (dx ), and its height is y).
2- Since the strip height starts from the base and intersects with line AC, y- y-value is the same as the line Bc equation.
3-Inertia due to that strip=dx*y^3/3 as derived from the case of the rectangle.
4- Integrate the vertical strip of width dx from x=0 to x=b, the final answer is Ix=b*h^3/12, same as per the previous method by using a horizontal strip.
The radius of gyration can be calculated as the square root of (Ix/A). The inertia Ix at the base of the right-angle triangle- case 2 equals b*h^3/12. The area of the triangle equals =0.5*b*h. The square of the radius of gyration equals b*h^3/12/(0.5*b*h)=h^2/6. The kx value equals (h/sqrt (6).
For the estimation of Ixg at the Cg. We will subtract the product of the area and Y^2bar. We have the area=1/2bh, while ybar^2=(2/3b)^2. We will get Ixg = b*h^3/36, from which we can obtain K^2 g = h^2/18, as shown on the next slide. The kx value at the CG equals (h/sqrt (6).
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For an external resource, Engineering core courses, the moment of inertia.
This is a link to the playlist for all videos for inertia.
The following post is the moment of inertia for the right-angle triangle, Iy- Case 2.