Last Updated on June 15, 2025 by Maged kamel
Ix-the moment of inertia-right-angle triangle-Case-2.
The moment of inertia-case-2 for right-angle by using a horizontal strip.
To get the value of 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 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 deduct the product of the 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 right-angle by using a vertical strip as an alternative method.
1-For The moment of inertia for a right-angle-triangle-case-2, the second alternative method is by using 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 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 sqrt of (Ix/A).
For the estimation of Ixg at the Cg. We will subtract the product of the area*Y^2bar. We have the area=1/2bh, while ybar^2=(2/3b)^2. We will get Ixg=bh^3/36, from which we can get K^2g=h^2/18, as shown in the next slide.
For an external resource, Engineering core courses, the moment of inertia.
This is a link to the playlist for all videos for inertia.
The next post is the moment of inertia for the right-angle triangle-Iy- Case-2.