Last Updated on December 31, 2025 by Maged kamel
Moment of Inertia Ix for the rectangular section.
The lecture’s objective includes 5 points for this [post and following posts. The next image explains the first three objectives as follows.
1- Estimate the moment of inertia for the x-axis passing through the CG.
2- Estimate the moment of inertia for the y-axis passing through the CG by using the parallel axes theorem.
3- Estimate the radius of gyration for the x and y-axes.
The next image explains the other two objectives as follows.
4- Estimate the product of inertia for the x and y-axes.
5- Estimate the polar moment of inertia.
The procedure for obtaining Ix for a 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 and the base of the rectangular section r is y.
3-The expression of the Moment of inertia for that strip is dIx =b*dy*y, which is the area of the strip multiplied by the vertical distance to the x-axis.
4-The moment of inertia about the x-axis, 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 an external axis passing through 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.
As another alternative method for Ix for the rectangular section about the CG, we can obtain the expression for 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 y = h/2 and y =- h/2.
IThe moemnt of inertia about x- axis 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 expressions are identical whether using the parallel theorem or the direct integration about an internal axis x’.
The radii of gyration for the rectangular section.
The radius of gyration can be estimated by considering that the square of the radius of inertia about the x-axis, K^2x=Ix/A. At the same time, k^2y is the square of the radius of gyration about the Y-axis, so Ky=sqrt(Iy/A). For Ix=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 or about a horizontal axis passing by the Cg of the rectangular section.
For the moment of inertia for a rectangle about the x-axis, Ix = b*h^3/12, the area is (b*h), so we can get the expression for Kxg as follows: h/sqrt (3).
The list of values for inertia for plain shapes.
The table shown lists inertias for plain sections. The case of the rectangle’s inertia is shown as the fourth item.
You can download and review the content of this post through the following pdf file.
For an external resource, Engineering core courses, the moment of inertia.
Next post: the moment of inertia Iy for the rectangular section.