 # 16-Polar Moment of inertia j- for the triangle.

## Polar Moment Of Inertia J- For The Triangle.

Our aim is to get the J for the triangle at point a, where the two axes x and y intersect.
1-We will add both Ix+Iy as follows: Ix at point a=b*h^3/12.

For more details about the moment of inertia at the x-axis, for the triangle refer to Moment of inertia Ix– for a given triangle. While the value of Iy=(b*h)*(a^2+b^2+ab)/12. For more details about the moment of inertia at the y-axis, for the triangle refer to Moment of inertia Iy– for a given triangle.

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This is the expression for Ix for a triangle.

[latexpage]\begin{equation}
I_{x}=\frac{h^{3} b}{12}
\end{equation}

This is the final value for Ix for the triangle as estimated earlier at the edge point a by adding the inertias Ix values for the two right-angle triangles.

2-: Iy at point a  can be obtained by adding the inertia as Iy for the two right-angle triangles.

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

## Polar Moment Of Inertia J- For the Triangle at the left corner point.

Add the Ix +Iy to get the final value for the polar moment of inertia for the triangle at the left corner point a.

3-:J at the point, a, after adding Ix+Iy

\begin{equation} J_{0}=\frac{h b}{12}\left[h^{2}+a^{2}+a b+b^{2}\right] \end{equation}

This is the final value of the polar moment of inertia for the triangle at edge point a.

### Polar moment of inertia J for The triangle at the CG.

.There are two ways to get the expression for Jg, the first way is to deduct the product of A*(xbar^2+y bar^2) from the J value. where x bar is the horizontal distance from the CG of the triangle to the vertical axis y at point a. Y bar is the vertical distance from the CG of the triangle to the horizontal axis -x axis y passing by the point a. For more details on how to find the x-bar.

For more details on how to find the y-bar, please refer to post 13 Moment of inertia Ix- for a given triangle.

The calculation for the polar moment of inertia at the centroid. of the triangle.

\begin{equation} J_{g}=\frac{h b}{3 \ 6}\left[h^{2}+a^{2}-a b+b^{2}\right] \end{equation}

### The second method is to get the Polar moment of inertia J for The triangle at the CG.

The second method to get the polar of inertia is to add ix at cg value plus the value of Iy at the Cg The second way is by adding Ixg+Iyg for the triangle. Both ways will lead to the same result.

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 moment of inertia for an isosceles triangle-Ix, Iy.

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