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

## Moment of inertia-Iy for the right-angle triangle-Case-1.

### Brief content of the video.

In this lecture, we are going to talk about how to determine the moment of inertia-Iy in the y-direction for the right-angle triangle case-1. We have our triangle ABC with base =b and height =h and d.

We are going to use, this time, a vertical strip with width=dx and height = Y, and this strip is a distance=x from the y-axis, for the x-axis and y-axis intersection. We are intersecting at the left corner of that triangle, again our expression of the line BC is y= (-h/b)*x+h, for the vertical strip.

We have dA= the width which is dx*y, and we are going to integrate from x=0 to x = b, we are going to substitute y value by writing y=(-h*x/b)+h, our dA is y* dx, So we are going to multiply this value, y,* (-h*x/b)+h it by dx .

For our Iy, we are going again to multiply this area dA *the horizontal distance ^2, which is x^2 and we are going to integrate so that Iy= ∫ from 0 to b of dA *x^2. Our dA =(-h*x/b)+h, will be multiplied by x^2 and we will get h out, so we have ∫ from 0 to b of (h*)-(x^3*dx/b) +x^2*dx). This is a part of the video, which has a closed caption in English. If you wish to see the video on U-tube, here is the link.

### Step-by-step guide for the calculation of moment of inertia-Iy.

We are interested in the left corner of that triangle again our expression of the line BC is y= (-h/b)*x+h, for the vertical strip, we have dA= the width which is dx*y.. We will integrate from X=0 to X = b. We substitute for Y- value by writing y=(-h*x/b)+h, our dA *x^2 will give the value of dIy about the y-axis. Multiply dA*x^2, dIy=(h*)-(x^3*dx/b) +x^2*dx).

The integration from x=0 to x=b will be carried out.

Finally, Iy=h*b^3/12, for which, h, is the triangle height and b is the base length. the inertia is about the y-axis which is located 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 axes theorem and deduct the product of A*xbar^2.
A=(1/2)*b*h, for the triangle Cg it is located at a distance=b/3 from the left corner and 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^2y 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, at the Y-axis passing 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, then Ipg=b*h(b^2+h^2)/36, we call it Jog, the value at Jog at the CG is shown as per the next slide image. The final value=(h^2+b^2)/18

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

The Polar of inertia at the external axis x,y passing the external corner. Jo=Ix+Iy, adding them together, then Jo=b*h(b^2+h^2)/12. The value at Jog at the CG is shown as per the next slide image.

### Using a horizontal strip as one option to estimate the moment of inertia-Iy value.

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, while 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 rectangle section moment of inertia Iy at the edge, which is = the height of the width^3. We can substitute and use the relation between x and y.

We are integrating into 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 a right-angle triangle case-1 about the Y-axis. The result is matching with the previous value of Iy. But this method took a larger time compared with the vertical strip.