Brief description for post 21-inertia.

1-Introduction to the moment of inertia.

Introduction to the moment of inertia.

Content of the video.

We are going to talk about the moment of inertia, the moment of inertia falls into two categories the first category is the area moment of inertia or sometimes called the second moment of area.


This term is used for determining the stresses due to shear and moment and torsion acting on a given section and dealing with the static Problems and how to solve these Static Problems by use of the Moment of inertia. 

The second category is the mass moment of inertia, this represents the resistance of a body to angular acceleration due to torque and this topic is very important in the study of Dynamics.
In this course, we are going to talk about the area moment of inertia, And the mass moment of inertia, it will be later on discussed. This is a part of the video, which has a closed caption in English.

A summary of the different types for the moment of inertia.

There is the area moment of inertia and mass moment of inertia, the difference in between lies in the fact that the second moment of inertia is used for the static areas, and used to determine the various types of stress, like bending moment and shear stress, while the torsional stress depends on the polar moment of inertia.

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.

What is a Moment of inertia?

The second category is used to express the resistance of bodies to angular rotations. the next slide shows the main differences.

The difference between the second moment of area about x and y axes.

For the second moment of area, it can b estimated about x or y axes, We will introduce a strip and we call that strip area dA.

If we are going to evaluate the inertia of the X-axis, we need to know y dimension of the normal dimension to the x-axis.

The difference between Ix and Iy.

The moment of inertia about the x-axis is equal to the integration of d A * y^2, while, the moment of inertia about the y-axis, as we can see, is equal to the integration of d A *x^ 2

X is the normal distance to the y -axis. In this regard, we are going to use a vertical strip, this strip, the area we call it dA.

The difference between  Ix and Iy.

The integration symbol is the same as a summation of individual areas. In that regard, each area will be multiplied by its X component, but a symbol integration means that we are going.

Move this is strip multiplication by x^2. Starting from one point Ending at another point.


What is the product of inertia Ixy?

The next item will be the product of inertia, suppose we have a big area like this, where is the area we call it A.
We will introduce only a small strip of area =dA from that strip, we are going to estimate the CG distance for both the x, y coordinates, for x is a normal horizontal distance perpendicular to Y- axis, while y, is the normal distance perpendicular to the x-axis.

What is the product of inertia?

The expression for Ixy = the integration of dA* (x*y), for that the strip, this = the summation, for example, A*x1*y1+A2*x2 y2, etc, where i=1 till i=n. In the end, instead of performing this task


We are going to introduce only one small strip and we start integration from our start point until our end.


What are the steps to estimate the moment of inertia?


Number#1: Draw the element, to know what is the shape of that element, either a rectangle triangle, Trapezoidal, etc.

Number #2: Derive nan expression for dA, by introducing the strip we will find out this is a horizontal or vertical strip.

Number #3 -Write the expression for the normal distances, for both X and Y, and make an interrelation between them. We need y if we are going to estimate the inertia at the x-axis, and the x distance if you are going to calculate the Iy, and of course, we need both if we are going to calculate the product moment of inertia.

Number# 4 Set the limits moving the strip, what point of start? until what point it will end, and in some cases especially in the triangle.
We are taking the integration in two steps. As we are going to see.

How to determine the moment of inertia?

Number # 5-Perform the integration, after evaluating, we will get the moment of area value.

Parallel axes theorem proof for Ix.


There is a very important, theory, which is called parallel axis theory, we are going to prove for Ix, suppose we have two intersecting axes, namely x and y this is the area we are going to estimate. The moment of inertia by introducing a small area dA.

We are going to discuss the first point, the first question to ask where are you going to estimate? The moment of inertia, you are going to estimate regarding the x-axis or to the y- axis. From the sketch shown, let x, y are two external axes, while x’ and y’ are two axes passing by the CG.

Deriving an expression for Ix.

We are going to estimate the moment of inertia, for example about the x-axis, we need the vertical distance from the CG of the dA small infinitesimal area dA.

This y is can be estimated as, the sum of two components, the first component is called y1,y1 is the vertical distance between the CG of the dA to the x’s, which is passing by the c.g.+ y bar, which we know, that the vertical CG distance from the external chosen x y.

If you’re going to write Ix= to the integration of dA*big distance Y^2, it can be equated to (y1+y bar)^2.

And if we’re going to estimate for the I y for the moment of inertia about the y axis, we are going to integrate d A x *x^2, as we are going to see. We will put it inside the bracket and will be multiplied by dA. Then Ix= the integration of dA*y^2+the integration of dA*2* y1*y bar+the integration of d A*y bar ^2.

Part a, the proof of parallel axes theorem for Ix.

This first item, the multiplication of area *y1^2 will resemble the moment of inertia about the x ‘, which is the axis passing by the CG.


The moment of inertia about the x’ will resemble the moment of inertia about the x ‘, which is the axis passing by the CG, will resemble the moment of inertia about the x ‘, which is the axis passing by the CG, which is an axis X’ passing by the CG.

We are going to write as x ‘+ will resemble the moment of inertia about the x ‘, which is the axis passing by the CG which is an X’passing by the CG, we are going to write as x ‘+ 2 *( y bar, being constant will come outside the bracket*, we left with dA*y1, the dA*y1 is the first moment of area, and since this is the axis about the CG.

As we know the first moment of area =0, this term will become 0 +A*Y bar^2.

Part b, the proof of parallel axes theorem for Ix.

So, finally the moment of inertia = Ix’, about the axis passing by the CG, +A*y bar^2, in this regard, sometimes it is easy to estimate the moment of inertia about. external axis subtracting the multiplication of the area by y^2.

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

For an external resource, Engineering core courses, the moment of inertia.

The next post: Moment of inertia Ix– for the rectangular section.

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