Last Updated on September 8, 2026 by Maged kamel
Moment of Inertia for right-angle-Ix, Case 1.
What is the difference between case 1 and case 2?
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.
For Case No. 1, the Moment of Inertia for right-angle-Ix is the second case in the attached reference table from the FE Exam reference manual 3-5. Case 2 is the first case in the same table. This is the list of the first Moment of Area and Inertia for the common plane shapes.
The next slide image shows the difference between Case 1 and Case 2 of the right-angled triangle. For Case 1, the opposite side of the triangle is on the left side, and for Case 2, the opposite side of the triangle is on the right.
How can I determine the Inertia Ix for a right-angle strip using a horizontal strip?
For an x-axis passing through the base and the Y-axis passing through the left side, using the following steps, we will estimate the Inertia for right-angle-Ix using a horizontal strip:
For line CB, we will write the equation for that line as y = (-h*x/b) + h.
We can check the validity of this equation by substituting the x-coordinates of both C and B and getting the corresponding y-values.
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 equal to the base multiplied by the height, x*dy. To obtain dy, we will differentiate the equation for y, which we have already estimated.
4- Our Moment of Inertia due to the strip is dIx=dA*y^2; remember that dA=x*dy.
5- We can write dIx=b*(h-y)/h*(y^2*dy), after substitution by the value of dy.
6-the value of Ix=∫dIx=∫b(h-y)/h(y^2dy), from y=0 to y=h.6- after integration and substitutionwe get Ix=bh^3/12.
The value of Inertia for a right-angle-Ix=base*height^3/12. The following slide image shows the steps.
How can I get the Inertia for a right-angle triangle, Ix, using a vertical strip?
Using the following steps, we will estimate 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, is shown in the next slide picture.
6-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=bh^3/12.8- k^2x=Ix/A=bh^3/(12(0.50b*h))=h^2/6.
How to get the Inertia for a right-angle Ix at the CG?
To get the Inertia for right-angle-Ix at the CG, we can use the Expression of Ix and then find Ixg at the CG from the theorem of parallel axes, by deducting the Product of the Area by the square of the distance between the CG and the X-axis.
The Moment at the CG, Ixg=Ix-A*ybar^2. Substitute the value of Ix; write Ixg as (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(bh^3/36)/(0.50bh))=h/(3*sqrt(2)).
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This is a link to the next post, ‘Moment of Inertia – for the Right-Angle Triangle Case 1-Iy ‘.
This is a useful external link on the Moment of Inertia: “The Moment of Inertia of a Rectangle.”
This is a link to the video that accompanies this post.