Moment Of Inertia Ix for the rectangular section.
Moment Of Inertia Ix for the rectangular section-video.
We are going to talk about Ix for the rectangular section. The objective of this lecture is to estimate the moment of inertia Ix for the x-axis passing by CGby using the parallel axes theorem, which means that we are going to take internal X & Y axes.
We are going to estimate the Ix for the rectangular section about these two axes. Then we are going to deduct the multiplication of the area. by y̅^2. In the case of Ix. The second step is to estimate the moment of inertia about the y-axis Passing By the CG by using the parallel axis theory.
In the third step, once we have estimated the moment of inertia about x. and the moment of inertia about y, we can get easily the radius of gyration for Ix and Iy and the relative radius of gyrations kx and Ky.
The fourth step is to estimate the product moment of inertia for the rectangular section.
Number 5 is to estimate the polar moment of inertia. J0, Either at the CG or at external parallel axes. as we are going to see, in the first step, we are going to get the moment of inertia Ix.
We have two external axes namely x and y. This is a part of the video, which has a closed caption in English.
The objective of the lecture includes 5 points. The next image explains the first three objectives as follows.
The next image explains the other two objectives as follows.
The procedure to get Ix for the rectangular section.
1-Consider a horizontal strip of breadth=b, and thickness of dy.
2- The vertical distance of this strip from the external axis X passing by the left edge corner is y.
3-The expression of the Moment of inertia for that strip is dIx =bdyy.
4-Ix value, by integrating from y=0 to y=h, Ix=∫ dIx=∫b*y*dy=b*y^2 from y=h to y=0, the integration will be Ix=b*h^3/3, this is the moment of inertia about ann external axis passing by the left bottom corner of the rectangle.
Moment of inertia Ix for the rectangular section at the CG-Ixg.
5- For the Ixg , the moment of inertia about the Cg, we will use the parallel axes theorem and deduct the (area*(h/2)^2), so we can get the values as shown attached.
6- We have already estimated ix about the external axis as Ix=b*h^3/3, we will deduct(b*h)*(h/2)^2=b*h^3*(1/2-h^2)/(4)=b*h^3*(2-1)/4=b*h^3/12 which is the value of Ixg, which is the moment of inertia at the CG of the rectangular section of dimension (b*h).
Another alternative method for Ix for the rectangular section about the Cg.
We can get the expression of the moment of inertia about the CG directly by integrating the strip of area (b*dx) from y=-h/2 to y=+h/2, Ix=∫ b*dy*y^2=(b*y^3/3), we will substitute for y=h/2 and y=-h/2.
Ix=(b*(h/2)^3/3-b*(-h/2)^3e =(b/3)*(1/8)*(h^3+h^3)=2*b*h^3/24=b*h^3/12.
As we can see in the next slide, both two expressions are identical whether using the parallel theorem or the direct integration about an internal axis x’.
The radii of gyration for the rectangle section.
The radius of gyration can be estimated by considering that K^2x=Ix/A, where k^2x is the square of the radius of gyration about the x-axis, So Kx=sqrt(Ix/A). for tx=b*h^3/3, the area is(b*h) then we can get the expression for Kx as follows:
The radius of gyration at the CG can be estimated by considering that K^2xg=Ixg/A, where k^2xg is the square of the radius of gyration about the X’-axis.
For Ix=b*h^3/12, the area is(b*h) then we can get the expression for Kxg as follows:
The list of the values for inertia for plain shapes.
The shown table is a list of inertia for plain sections. The case of the inertia of the rectangle is shown as the fourth item.
This is the PDF used for the illustration of this post.
This is a link for a good external resource, to serve as a calculator, Calcresource.
Next post: the moment of inertia Iy for the rectangular section.
