# 13- Easy approach to Moment of inertia Ix-for a triangle.

## Moment of inertia Ix-for a triangle.

### Brief content of the video.

We are going to talk about the moment of inertia Ix for a given triangle. The Triangle has base b and height =h and with X and Y axis intersecting on the left-hand corner. The triangle ABC can be considered as two right- angles triangles added together.

When we are dropping a perpendicular from c to line AB, the point of intersection will be considered as d, the length ad=a, while the remaining distance from length db will be=(b-a).

If we consider this triangle ABC has composed of two triangles, the first right-angle triangle is adc and the x-axis and y-axis are intersecting at point a.

This triangle, we call it triangle no.1. has a base =a to which it can be considered as case no.2, while for the other triangle, triangle number two which is dbc just 90 degrees the base =(b-a), and the height = h.

But regarding the y axis, we will find dc side having distance =a, we can call this the right angle as case no.#1, in order to estimate the moment of inertia in The x-direction, we are going to write down all the expression which we have obtained previously.

That for Triangle Number one, we have Ix=the base * the (height)^3/12. 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.

### List of the area moment of inertia for triangle.

The value of the area and Moment of inertia for the triangle is the third item in the shown table from the NCEES reference handbook-3.50. the shown table lists the values of Ix and Iy for the triangle.

### Step-by-step guide for the calculation of Ix-for a triangle.

1-For the moment of inertia  Ix estimation for a triangle, we will consider that triangle, as composed of two right-angle triangles, for the first right-angle triangle, it will be considered as case No.2.For case number #2 y-axis is passing by the left corner a.

The first right-angle triangle has a base =a and a height equals h as shown in the next slide image.

While another right-angle triangle, will be considered as case no.1, we will list the values of inertia as Ix1 and Ix2.

2-For Ix estimation adding the two values of inertia (a)*h^3/12+(b-a)*h^3/12 will give us

$$I_{x}=\frac{h^{3} b}{12}$$

3-For the k^2x value, we will divide the Ix value /area of the triangle, so we get

$$\ {k}_{x}^{2}=\frac{\ I x_{F}}{A}=\frac{h^{3} b}{12} \times \frac{1}{(1 / 2)(h b)}=\frac{h^{2}}{6}$$

y bar value for the triangle will be found as equal to h/3.

4- For the moment of inertia Ixg at the CG, we are going to estimate the y-bar for the triangles shown in the next slide from the first-moment area.

5- Ixg=Ix-A*y bar^2, after substitution we will get the moment of inertia Ix for a triangle at the Cg as :

$$I_{x c_{g}}=b^{*} \frac{h^{3}}{6}$$

6-The radius of the gyration of the triangle can be estimated from the following relation.

$$k_{c g}=\sqrt{\frac{h^{2}}{18}}=\frac{h}{3 \sqrt{2}}$$

The square value of the radius of gyration for a triangle at the Cg can be found to be equal to h^2/18.

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 Iy– for the triangle.

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