## Full detailed illustration of the Moment of inertia Iy for a triangle.

### Brief content of the video.

We are going to estimate Iy or the moment of inertia Iy for the triangle.

Again we have the triangle ABC, with the base=b and the height=h, we have two perpendicular axes X and y, these axes are intersecting at point B of the triangle.

We can consider that this triangle is composed of two right-angle triangles one is Case -1, the other one is called case-2 and for triangle Number one we will find that the x and y axes are intersecting at point a, while for the other one the y-axis and x-axis intersection are apart from the line cd by a distance =a.

The moment of inertia for the triangle will be the summation of the Iy inertia for each individual right-angle triangles. We have estimated that for the right angle, we have the moment of inertia for the y-direction is Iy=(h*a^3/4) and this is about the y-axis. While our calculation for the Moment of Inertia at the y-direction based on That, y-dc, coincides with the opposite side of the triangle. This is a part of the first video, which has a *closed caption* in English

You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.

### Full detailed illustration of the Moment of inertia Iy for the triangle.

Our aim is to get the moment of inertia Iy for the triangle at the y-axis, which is passing by point a.

1-Since the triangle ABC consists of two triangles ACD and DBC, for the inertia for the triangle we are going to add the sum of two inertias for those two right-angles about the same axis y.

2-The inertia for an angle adc about axis y passing by point a=Iy1=h*a^3/12, where a is the base of the triangle ADC and h is the height. recall this value was estimated for the right-angle case no-2.

3-The inertia for an angle dbc about axis y-dc passing by point d=Iy-dc=(b-a)^3*h/12/12, where b is the base of the triangle abc.

We have to estimate the inertia at the Cg for right angle triangle case 1 and from Cg we can proceed to estimate the inertia at any place.

4-For the second triangle dbc , we need to estimate Iy-CG , then we add (area*x bar^2), the area is the area of the triangle dbc, x bar is the distance from its Cg t axis Y.

5- x bar distance as shown in the next slide, x bar=a+(b-a)*1/3=(2/3)*a+(b/3).

6- x^2-bar can be wriien asxbar^2=(1/9)*(2a+b)^2.

Iy final=Iy case -2 triangle+Iy-cg for triangle dbc+A*x^2 bar.

In the following images, there is more details of the summation process for Iy.

7-We substitute all the known values obtained earlier, the expression can be written as

[latexpage]\begin{equation}

I_{y}=\frac{b h}{36}(3)\left[b^{2}+a b+a^{2}\right]

\end{equation}

This is the final value of Iy for the triangle.

### The radius of gyration is about axis y for the triangle.

From the value of Iy by the area of the triangle, then take the square root of the product, then the value of ky can be found.

This is the final value of k^2y for the triangle.

We have completed the estimation of inertia Iy for a triangle and we can check the result by comparing with the inertia value with the NCEEs table.

This is the list of inertia for triangular shapes quoted from the NCEES tables of inertia.

This is the pdf file used in the illustration of this post.

For an external resource, the definition of the moment of inertia with solved problems, 2nd moment of inertia.

This is the next post, How to get of Moment of inertia Iy-at Cg for a triangle?