## Solved problems for the moment of inertia.

### Brief content of the first video.

While for the expression for the radius of gyration for x, we have K^2 x= Ix/area, our Ix already obtained =0.0288 / area so 2x=3/25, we can get the kx, by taking the sqrt of 3 /25 which is = 0.3464m. 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.

### The First solved problem for a given rectangle.

We are going to solve our first problem, problem # 1

We have x and y axes, but we have a rectangular section. with width=0.40m. And the height is 0.60 m. This rectangle is apart from the y-axis by a distance of 0.20 m.

It is required to estimate the following for the given rectangle. IX, Kx, the radius of gyration at the x-axis, and b.

We want to estimate the moment of inertia at the y-axis and the radius of gyration for the y-axis.

We have our X and Y external axes. As we know that I x = the Ixg+ the area *(y^2).

We have, for the moment of inertia about the CG, for a rectangle section, we have b* h^3/12. Then we have this x-axis. We have to add the product of area multiplied by y^2, first our Ix g=b*h^3/12= (0.40*0.60^3/12, which will give us 0.0072 m4.**Our area** is (0.40*0.60), which is 0.24 m2. For item A* y^2, we have A=b*h, and y̅ =h/2 All raised to the power of 2. So we have b*h, which is (0.40*0.60)*(h/2)^2,h/2=0.30 divided by 4, which will give us 0.0216 m4.

Our required moment of inertia about the x-axis will be the summation of these two items, the first item is Ixg=0.0072+ A*y̅^2, which is =0.0216 adding both will give us 0.0288 m4.

For k^2x=Ix/A=0.0288/0.24=3/25, then the Kx=sqrt(3/25)=sqrt(3)/5.

For part b of the first solved problem, Iy=Iyg+A*xbar^2. Iyg=h*b^3/12=0.60*?(0.40)^3/12=0.0032 m4.

A*x^2g=(0.40)*(0.60)*(0.20+0.5*0.40)^2=0.0384m4. The final Iy=(0.0032+0.0384)=0.0416m4.

k^2y=Iy/A=0.0416/(0.40*0.60)=13/75. ky=sqrt(13/75)=0.4163 m2.

### Brief content of the video.

We are going to solve another solved problem, we call it problem #2. In this problem, It is required to estimate the following. **moment of inertia **Ix and to estimate the * product of inertia *of the given unequal angle.

In an application of the rectangle, we can constitute an equal angle, consist

of two rectangles, The first one with width=1.0 m and the height = 4 m

*This is a part of the video*, which has a

*closed caption*in English.

### The second solved problem for a given section.

We can consider the L shape as consisting of two rectangles. The first one of width=1.0 m and the height = 4 m and the other rectangle, the width of 2 m and the height is 1 meter, if we join them together, they will constitute an unequal angle with the same width, as we can see Unequal angle 3 by 4 and a width of 1.0m, can be evaluated as composed of two rectangles.

The first one A1 of width 1.0m and height=4.0 m, that’s why its area would become(1*4)=4.0 m^2 And its X1 will be=1/2 m. y1 from the x-axis will be. 4 / 2, which is 2 meters. For the second area, we have the A2=(3-1)*.this is the breadth, the height will be 1.0 m, so we have 2=(1*2)=2.0 m^2 Our X2, measured to the external y-axis, will be (1m+1/2 of the difference between (3-1.0) Which will give us = 2 meters, our y2= 0.50 m.

The Ix for the unequal angle is the summation of Ix for the first rectangle+ the Ix for the second rectangle.

The moment of inertia Ix2 value of the solved problem.

The final Ix=22.00 m4 is part a) of the solved problem.

For part c) of the solved problem, for each rectangle we estimate the product of inertia about the external axes, and then add them together.

The final Ixy=6.00 m4 is part b) of the solved problem.

This is the PDF used for the illustration of this post.

For the next post, Moment of inertia for the right-angle triangle-Ix- Case-1.

This is a useful external link for the moment of inertia- Moment Of Inertia Of Rectangle