# 11-Moment of inertia Iy-Case-2 for a right-angle triangle.

## Inertia For Right-Angle Triangle-Iy-Case-2.

### Brief content of the first video.

In this segment, we are going to estimate the moment of inertia for the right angle Iy-Case -2, for which the opposition to the right side and the base at the bottom of the hypotenuse on the left side.

Our left corner is located at the intersection of both axes X and Y. These lines are intersecting at point A and our triangle is ABC with the base is =b  and the height =h and CG  located at X distance x̅ =2/3 b and Y distance y̅=(h/3). This is a part of the first 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.

### Moment of inertia Iy-case-2-for y-axis at the left side.

There are two conditions for the y-axis, the first condition is when the y-axis is passing by the intersection point between the base and the hypotenuse.
1-For Iy estimation,  for case -2, 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 AC equation y=h*x/b, for x value it will be=b*y/h.
3- Inertia due to that strip=dA*(x^2).

4- Integrate from x=0 to x=b, the final answer is Iy=(h*b^3)/4. this is the Inertia Iy-case-2 for a right-angle triangle.
The radius of gyration Ky for right-angle case-2.

5- from the expression that Ky=sqrt(Iy/A), we can get the radius of gyration about the y-axis, since A=(1/2)b*h. the value of the moment of inertia Iy-Case-2, Iy=h*b^3/4. Then Ky=sqrt(h*b^3/4)/(1/2*b*h)= h/sqrt(2).

### Moment of inertia Iy-Case-2 for right-angle.

Suppose we want to change the location of the y-axis to make it, for example, coincide with the opposite side of the triangle.
The x-axis points to the left and we have the y-axis coincide with the opposite side of the triangle.
Again we are going to use a vertical strip with width =  dx and height y.

1-For the estimation of Iy -case -2, 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 AC equation y=(h/b)*(b-x).
3- the moment of inertia due to that strip=dA*(x^2).

4- we will integrate from x=0 to x=b, the details of integration are shown in the next slide. The final answer is Iy=h*b^3/12.
When comparing this value by the previous value for iy when the y-axis is at the left, we find that Iy value becomes smaller. The radius of gyration Ky for right-angle case-2 when the y-axis is at the right side.

5- from the expression that Ky=sqrt(Iy/A), we can get the radius of gyration about the y-axis, since A=(1/2)b*h.
Iy=h*b^3/12. Then Ky=sqrt(h*b^3/12)/(1/2*b*h)= h/sqrt(6).

### Brief content of the second video.

We want to estimate the Iy at the Cg and the Polar moment of inertia, we have a right angle triangle ABC for which we have listed all the calculated values of the Iy, for the case of y is at the intersection of the base and the hypotenuse.

We assume that the axis of y, which we call y1, and the value of the moment of inertia about y1 will be Iy1=h*b^3/4, while if we take another y-axis, let us call y2.
Let us scroll its way to passing by BC, the opposite side. This is a part of the video, which has a closed caption in English.

### Inertia Iy-case-2 at the CG for a right-angle triangle.

1-For Iyg estimation, we will deduct the product of A*xbar^2, from the Iy-inertia-case-2-Iy value which is =h*b^3/4.
2-We get the value of Iyg=h*b^3/36, which was derived from the case where the y-axis is located on the left side, I call it Y1 as shown in the slide.

3- the moment of inertia about the CG-Iyg value for the second case, where the y-axis is located at the right side, is the same value after subtracting A*x bar^2 from Iy value =h*b^3/12.

### Polar Moment of inertia Jo for right-angle case-2 for y-axis at the left side.

The polar moment of inertia can be estimated by adding the sum of Ix and Iy. We have Ix value =bh^3/12, while Iy=hb^3/4 when the y-axis is located at the left. that position will be called Y1.
Jo=(bh^3/12)+(hb^3/4). The final value for J0G, which is the polar moment of inertia at the Cg can be written as:

\begin{aligned} J_{0} &=I_{x}+I_{y} \\ &=\frac{b h^{3}}{12}+\frac{h b^{3}}{4} \\ &=\frac{4 b h}{12}\left(h^{2}+b^{2}\right) \\ &=(1 / 3)* b h\left(h^{2}+b^{2}\right) \end{aligned}

The steps to get the polar moment of inertia at the CG.

The final value for J0G, which is the polar moment of inertia at the Cg can be written as:

$$J_{0}g \ =b h\left(h^{2}+b^{2}\right) / 36.$$

### Moment of inertia Jo for right-angle case-2 for y-axis at the right side.

The polar moment of inertia can be estimated by adding the sum of Ix and Iy. We have Ix value =bh^3/12, while Iy=hb^3/12 when the y-axis is located at the right that position will be called Y. Jo=(bh^3/12)+(hb^3/12).

\begin{aligned} J_{0} &=I_{x}+I_{y} \\ &=\frac{b h^{3}}{12}+\frac{h b^{3}}{12} \\ &=\frac{ b h}{12}\left(h^{2}+b^{2}\right) \\ &=(1 / 12)* b h\left(h^{2}+b^{2}\right) \end{aligned}

The final value for J0G, which is the polar moment of inertia at the Cg can be written as:

$$J_{0g}=\frac{b h^{3}}{12}+\frac{h b^{3}}{12}-\frac{b^{2} h^{2}}{18}$$

This is the final value of the polar moment of inertia for the right-angle triangle. at the CG.

This is the table for the plain shape areas and their values 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, Product of inertia Ixy– for the right-angle triangle-case-2.

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