## Moment of inertia Iy-at Cg for a triangle.

### Brief content of the video.

In order to estimate the moment of inertia Iy-at Cg for a triangle which we call it Iyg.

First, we have to find out the x̅ distance, for the triangles by using the first moment of area expression, that the A*x̅, which is the area of the triangle =A1*x̅1 +A2*x̅2.

We have one 1/2(b*h)* x̅, the distance of Cg to the y-axis = the first area multiplied by its CG, From the y-axis.

Our first area is (1/2*a*h)*(2/3a)+ second the area, whichis= (1/2)(b-a)*(h)*Again we are going to multiply by the Cg distance from our y-axis, we have (1/3(b-a)), + a distance which is between the y-axis blessing by cd and the original at y-axis for which we are evaluating the moment of inertia. This is a part of the video, which has a closed caption in English.

### Moment of inertia Iy-at the Cg.

1-We need to estimate the x bar of the triangle, so we can utilize this value to get the Iyg.

2-To get the x̅ distance for the triangles by using the first moment of area expression, that the A*x̅ which is the area of the triangle will be equal tyo A1*x̅1 +A2*x̅2.

The value of the first moment of area for the triangle can be estimated as 1/2(b*h) multiplied by x̅, which is the distance of Cg to the y-axis = the first area multiplied by its CG.From the y-axis, our first area is (1/2*a*h)*(2/3)*a) plus the second the area, which is= (1/2)(b-a)*(h)*(2a+b/3).

The final value of x bar is as follows:

[latexpage]\begin{equation}

x_ b=\frac{b}{3 b}(b+a)=\frac{1}{3}(b+a)

\end{equation}

Please refer to the slide image for more details. the final x bar distance for the triangle is found to be equal to 1/3*(a+b).

3-We can now use the parallel axes theory to get the value of the moment of inertia Iy-at Cg-Iyg.

4-We will subtract the (A*xbar^2) from the value of Iy, which is equal to (b*h/12)*(b^2+a^2+a*b) to get the value of Iycg-moment of inertia Iy-at Cg

The value of the radius of gyration about the y axis at the CG can be estimated as shown.

[latexpage]\begin{equation}

I_{y c g}=\frac{b h}{36}\left[b^{2}+a^{2}-a b\right]

\end{equation}

[latexpage]\begin{equation}

k^{2} y c_{9}=\frac{I_{y c \cdot 9}}{A}=\frac{1}{18}\left[b^{2}+a^{2}-a b\right]

\end{equation}

The square of radius of gyration at the y direction at the Cg is equal to 1/18*(b^2+a^2-ab).

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

This is a link for 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 a link for the next post, Product of inertia Ixy– for the triangle.