Brief data for post-14a-inertia

14a-How to get of Moment of inertia Iy-at Cg for a triangle?

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Moment of inertia Iy-at Cg for a triangle.

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

To estimate the moment of inertia Iy-at Cg for a triangle 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 can be equated to the sum of the products of each area by its horizontal distance to the CG and can be written as (A1*x̅1 +A2*x̅2).

The first area equals 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)+ the second the area, which equals (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.

This is the link for the video on YouTube.

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

1-We need to estimate the x-bar of the triangle or the distance from the Center of gravity to the Y-axis, so we can utilize this value to get the Iyg.

Iy at the Cg for the triangle.
Iy at the Cg for the triangle.

2-To get the x̅ distance for the triangles by using the first moment of area expression, the A*x̅ which is the area of the triangle will be equal to 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 Cgof the first triangle to the y-axis, Y bar is passing by the external left edge. The first moment of the area can be written as equal first area multiplied by its CG.

From the y-axis, our first area is (1/2*a*h)*(2/3)*a) then we add the product of the second area, by the (2a+b/3). The first moment of area for the second triangle equals (1/2)(b-a)*(h)*(2a+b/3).

Divide the sum of the first areas by the area of the triangle to get the final value of the x bar as follows: 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).

x bar value for the triangle.
x bar value for the triangle.

3-We can now use the parallel axes theory to get the value of the moment of inertia Iy-at Cg-Iyg. We know that Iy=Iyg +(A*xbar^2).
4-We will subtract the product of (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 Moment of inertia Iy-at Cg, Iyg=1/18*(b^2-a^2-ab), where a is the distance from edge point a to point d, while b is the distance between points a and b.

The value of the radius of gyration about the y-axis at the CG can be estimated as shown by dividing the Moment of inertia Iy-at Cg, Iyg by the area.

moment of inertia Iy-at Cg

The square of the radius of gyration at the y direction for the triangle 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.

list of inertia for a  triangle.

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.

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