Brief description of post 10-inertia.

10-Ix-the moment of inertia for right-angle-Ix-Case-2.

Ix-the moment of inertia-right-angle triangle-Case-2.

Brief content of the video.

Hello, In this lecture, we will talk about the Ix-the moment of inertia-right-angle triangle-Case-2. We have the opposite side to the right.

The base is still the same and the hypotenuse is on the left side.



Still, we have our end triangle ABC with base is =b and height=h. This time we’re going to use a horizontal strip to determine the Ix about the x-axis,  X-axis coincides with the base, and perpendicular to it is the Y-axis.

These axes are intersecting at the left quadrant of the triangle, ABC at point A. Our horizontal strip will be apart from the y-axis by a distance = x.

The remaining width of that strip will be =b, which is the base of the triangle minus x, (b-x). This is the width of the Strip and the height of the strip is the = dy, and this strip is located at y from the x-axis. This is a part of the first video, which has a closed caption in English.

Ix-the moment of inertia -case-2 for right-angle by using a horizontal strip.

The value of the area and Moment of inertia for regular shapes from FNCEES reference handbook-3.50. Moment Of Inertia Ix-Case-2 is the first item in the table.

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.

The Two methods used to get Ix for right-angle-Ix-case-2.

The right-angle triangle which is at the right side of the next slide represents case No.2.

Step by step guide for the calculation of moment of inertia Ix-case-2.

Ix-the moment of inertia

1-For Ix-Case-2. estimation, a horizontal strip will be used, the strip width is dy and its height is x.   
2- since the strip height starts from the base and intersects with line  AC, y value is the same as the line AC equation, which is y=h*x/b.

3- the moment of inertia due to that strip=dA*y^2. Our strip’s left edge is at distance =x.                                                     
   4- the width of the strip=(b-x) and the depth is =dy.

Moment of inertia for the right-angle triangle Ix-case-2- at the CG.

5- we will substitute the value of dA, and try to make all our items as a function of y, since we will integrate from y=0 to y=h, for the Ix value.    

6-After performing the integration, we get the value of inertia for a right-angle-Ix-case-2 as Ix=b*h^3/12 about the x-axis passing by the base.Ix=(base)*(height)^3/12.

7-For Ix at the CG, we will use the parallel axes theorem and deduct the product of the A*x-bar*y bar.    

8- for x bar=b/3, while y bar=h/3.  Finally, we get Ixg=b*h^3/36.                        

Brief content of the second video.

We are going to estimate the moment of inertia Ix. For the right-angle triangle. Case #2, but by using a vertical strip the width of the strip is dx and the =dy. Case #2 is the opposite to the right of the triangle, and the base at the bottom coincides with X-axis at Y-axis at the intersection of hypotenuse through the base on the left side portion

Now regarding the equation of the line AC which is y = (h/b) *x.
Since we are integrating into the x-direction.*y^3/3.(h^3*b/12). x=b*h^3/12.This is a part of the video, which has a closed caption in English.

Ix-the moment of inertia-case-2 for right-angle by using a vertical strip as an alternative method.

1-For Ix-the moment of inertia for a right-angle-triangle-case-2, the second alternative method is by using a vertical strip will be used, the strip width is (dx )and its height is( y).    
    2- Since the strip height starts from the base and intersects with line  AC, y- value is the same as the line Bc equation.
  3-Inertia due to that strip=dx*y^3/3 as derived from the case of the rectangle.   
 

Inertia by using a vertical strip for right-angle triangle Ix-case- 2.

4- Integrate from x=0 to x=b, the final answer is Ix=b*h^3/12 same as per the previous method by using a horizontal strip.

The radius of gyration for right-angle triangle Ix-case-2.

The radius of gyration can be calculated as the sqrt of(Ix/A).

How to get the radius of gyration at Cg, KxG?

For the estimation of Ixg at the Cg. We will subtract the product of the area*Y^2bar. We have the area=1/2*b*h, while ybar^2=(2/3b)^2. We will get Ixg=b*h^3/36, from which we can get K^2g=h^2/18, as shown in the next slide.

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.

The next post is the Moment of inertia for the right-angle triangle-Iy- Case-2.

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