## 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 reference attached table from** FE Exam reference manual 3-5.** While 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 represents the difference between Case -1 and Case-2 of the right-angle Triangle. For case -1 the opposite of the triangle is at the left side and for case-2, the opposite of the triangle is at the right.

### How to get inertia for right-angle-Ix by using a horizontal strip?

For an x-axis passing by the base and The Y-axis passing by the left side, using the following steps, we will estimate the inertia for right-angle-Ix using a horizontal strip:

1-for line CB, we will write the equation as y=-h*x/b+h, we can check the validity of this equation by substituting the value of x coordinates of both C& b and get the corresponding coordinate y value.

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 get the dy, we will differentiate the y equation, 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^2*dy), from y=0 to y=h.6- *after integration and substitution* we get Ix=b*h^3/12.

The value of inertia for a right-angle-Ix=base*height^3/12. the next slide image shows the steps.

### How to get the inertia for right-angle-Ix by using a vertical strip?

Using the following steps, we will estimate the 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 value is shown as per 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=b*h^3/12.8- k^2x=Ix/A=b*h^3/(12*(0.50*b*h))=h^2/6.

### How to get inertia for right-angle-Ix but at the CG?

To get the moment of inertia 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 area by the square of the distance between Cg and the X-axis.

The moment at the Cg, Ixg=Ix-A*ybar^2. Substitute by the value of Ix, write Ixg=(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(b*h^3/36)/(0.50*b*h))=h/(3**sqrt(2)).

This is a link for the next post, Moment of inertia – for the right-angle triangle-case-1-Iy.

This is a useful external link for the moment of inertia- The Moment Of Inertia Of a Rectangle.