Breif description of Post 15-inertia.

15-Product of inertia Ixy-for the triangle.

Product Of Inertia Ixy- For The Triangle.

Brief content of the video.

We will estimate, the product-moment of inertia Ixy for the triangle ABC, the base =b, and the height equal to h, the two intersecting axes X-axis and Y-axis are intersecting at point A.

If we drop a perpendicular line from point c to the base ab, it will be intersecting at point d, with distance to the left =a, for the Ixy. We can consider this triangle ABC as composed of two right angles triangles, the first one which is triangle ADC.

The second task is to estimate The IXY at the Cg of that triangle. This is a part of the 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.

Product of inertia Ixy for The triangle.

Our aim to get the product of inertia Ixy for the triangle at the intersection between axis x and axis y at point a.

We will use the previous data for the product of inertia that was obtained from the previous posts for both case 1 and case 2.

1-Since the triangle ABC consists of two triangles ACD and dbc, then our Ixy for the triangle =Ixy1, exactly as estimated from case-2, but for case 1, the value for Ixy to be adjusted since it is estimated at the ydc.

The adjustment is done by taking Ixy value at the CG of the triangle cdb and add a new value for Ixy from the CG to the new y-axis at Point a. The value is estimated as the triangular Area of the case-1*x1^2 to be added. x1 is the Cg distance to Y-axis. xcg =(1/3*(a+b)), the y- cg =h/3.

Product of Inertia Ixy- for the Triangle.

The matching items should be cleared, the product of inertia Ixy estimation in detail is shown in the next slides.

Product of inertia at the left corner.

Omit(+ 3a^2) with the (-3a^2), proceed to the calculation for the product of inertia Ixy for the triangle.

The final value of Ixy is

\begin{equation} \ I x y=b^{*} h^{2 *}(1 / 24)^{*}(2 a+b) \end{equation}

The value of Ixy in terms of the total area.

If we wish to express the Ixy for the triangle in terms of the area of the triangle, we call it At, we can substitute for At as=0.50*b*h or half the base by the height.
Rewrite the Ixy as

\begin{equation} \mid x y=A t^{*}(1 / 12)^{*}(2 a+b)(h) \end{equation}

The final expression for Ixy for a triangle

The final value of the product of inertia Ixy at the edge point- a where the x-axis and y-axis intersects.

The value of Ixy for the triangle at the Cg.

The Cg of the triangle is located at x distance=1/3*(a+b), while y distance=h/3. These values are from Cg to the vertical axis Y at a and from the x-axis at the base of the triangle.

The final value of Ixy at the Cg is to be obtained by using the parallel axes theorem, and subtracting the product of the triangle area*xcg*ycg)

The final value of Ixy at the CG after subtracting (A*x-bar*y-bar)  is   

\begin{equation} \ I x y \operatorname{cg}=b^{*} h^{2 *}(1 / 72)^{*}(2 a-b) \end{equation}
Product of Inertia Ixy at the CG- for the triangle.

This is the final value of the product of inertia at the Cg

list of inertia

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.

For the next post, 16-Polar Moment of inertia– for the triangle.

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