Understanding the Area Moment of Inertia

Let’s say we have a plank of wood which we would like to use to cross a canal. It has a rectangular cross-section and so we could either use it like we have done here, or we could rotate it onto its side, like this. Intuitively we can tell that the plank will be stiffer if the load is applied to the shorter side of the cross-section. Some cross-sections are much more efficient at resisting bending than others. The further the material is spread from the bending axis, the stiffer a cross-section tends to be.

 

The cross-section on the right has more material located far from the bending axis and so is better at resisting bending, even though both cross-sections have the same area. This concept of resistance to bending can be quantified by calculating the area moment of inertia, which is also sometimes called the second moment of area. The area moment of inertia reflects how the area of a cross-section is distributed relative to a particular axis, and so is a measure of how much resistance the cross-section has to bending.

 

The I-beam locates the majority of the material as far as possible from the bending axis, and so is a very efficient cross-section. This is why it is so commonly used in construction. In this video we’re going to take a detailed look at the area moment of inertia. Let’s start by seeing how it can be calculated for an arbitrary cross-section like the one shown here. The first thing to note is that the area moment of inertia is not a unique property of a cross-section. It quantifies the resistance to bending about a particular axis, and so its value changes depending on where we place this reference axis.

 

We can approximate the area moment of inertia of a cross-section by splitting it into small elements. Each element contributes to the total area moment of inertia by a quantity equal to its area dA multiplied by Y^2, where Y is the distance to the reference axis, which is the X axis in this case. We can sum up the values for all of the small elements to obtain the area moment of inertia for the entire cross-section. It is denoted by the letter I, and because the X axis is our reference axis, we will give it the subscript X. We can define the area moment of inertia more precisely using integration, like this. And if the Y axis is our reference axis, we can calculate I-Y in the same way.

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The area moment of inertia has the unit of length to the fourth power, and because of the squared term it is always a positive quantity. Let’s work through an example where we calculate the I-X and I-Y values for a rectangular cross-section, using those equations. We can consider that the rectangle is made up of multiple thin strips which each have a height equal to dy.

 

Each strip has an area equal to b * dy, which gives us the following integral. The limits of the integral are from the bottom to the top of the rectangle, so from negative h over 2, to h over 2. If we solve this definite integral we end up with I-X being equal to b multiplied by h to the power 3, divided by 12. We can obtain the equation for I-Y by switching the height and the width terms. So there you go, we just calculated the area moment of inertia for a rectangular cross-section using integration.

 

But calculating integrals can be difficult, so to help us out area moment of inertia equations are often provided in reference texts for a range of common shapes. A few examples are shown here. Usually the equations are provided for centroidal axes, which are axes that pass through the centroid of the cross section. Remember that the centroid is the geometric centre of a cross-section. X-C and Y-C are called the centroidal axes. So what do we do if we need to obtain an area moment of inertia equation for an axis that is not a centroidal axis, but it is too difficult to calculate it using integration? Fortunately there is a method we can use to calculate an adjusted area moment of inertia for any axis that is parallel to a centroidal axis, like this one.

 

The adjusted area moment of inertia I-X can be calculated by summing the moment of inertia of the centroidal axis, and the product of the cross-sectional area A and the square of distance d between both axes. This is called the parallel axis theorem. It’s useful because we can take the equations found in reference texts for centroidal axes and adjust them to obtain the moments of inertia for any parallel axes.

 

Let’s look at an example. Earlier we determined an equation for the area moment of inertia of a rectangular cross-section, about an axis passing through the centroid of the rectangle. We can use the parallel axis theorem to calculate I for an axis shifted to the bottom of the cross-section. All we have to do is take the equation for I we derived earlier, and add the area of the rectangle b*h multiplied by the distance h over 2, squared. One useful property of area moments of inertia is that they can be added to and subtracted from each other. This means we can calculate the value of I for a shape like this by taking I for section A and subtracting I for section B.

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If we are calculating I based on equations, we need to make sure that the equations we are using correspond to the correct reference axis. In the case of this T section for example, we can’t just add the I values for two rectangles given in a textbook because these equations are for centroidal axes only, and the reference axis doesn’t pass through the centroid of the top rectangle. Fortunately we can use the parallel axis theorem to overcome this. If h and b are the height and the width of the rectangles, the distance between the reference axis and the centroid of rectangle 2 is (h1+h2)/2.

 

We can calculate I for the composite cross-section using the parallel axis theorem, like this. This is another reason the parallel axis theorem is really important. It gives us an easy way to calculate area moments of inertia for all sorts of different composite shapes. It’s worth clarifying at this point that the area moment of inertia should not be confused with the mass moment of inertia, which is a parameter used to describe the resistance of a body to changes in rotational velocity. There are similarities in the way the two parameters are calculated, but they have different units and completely different uses. So, when might you actually need to use the area moment of inertia? Well it is a particularly important parameter for the analysis of beams and columns.

 

This equation, for example, defines the deflection of a beam for an applied bending moment M. You will notice that the term E-I appears in many equations. It is called “flexural rigidity”. It quantifies the resistance of a beam to bending. As we have just seen, I is the resistance due to the geometry of the beam cross-section. But the stiffness of the beam material also contributes to its total resistance to bending, and is captured by Young’s modulus E. Flexural rigidity also appears in the analysis of columns, where it can be used to calculate the critical buckling load.

 

Another important parameter related to the area moment of inertia that appears in the analysis of columns is the radius of gyration. It represents the theoretical distance at which we could condense the entire area of a cross-section into a narrow strip, to get the same moment of inertia as the original cross-section. It can be calculated using this equation. So far we have only calculated the area moments of inertia relative to the X and Y axes. We can calculate a third area moment of inertia for a reference axis that is perpendicular to the plane of the cross-section. This quantity is called the polar moment of inertia, and it is usually denoted using the letter J. It represents the resistance of the cross-section to twisting about the reference axis. It is calculated in the same way as I-X and I-Y, but using the distance Rho to the axis, rather than the perpendicular distance to the X or Y axes. We can calculate J using integration, like this. Rho squared is equal to X squared plus Y squared, so we can expand the equation. And by doing a little more work we can figure out that J is equal to I-X plus I-Y. This is known as the perpendicular axis theorem. The polar moment of inertia is mainly relevant for situations which involve torsion. Check out my video on torsion if you want to learn more about this. To fully master the area moment of inertia there is one last thing we need to cover, which is the rotation of the reference axes. We can use the transformation equations shown here to calculate moments of inertia for rotated axes. The I-X-Y term in these equations is the product of inertia, and it is calculated using this equation. Rotating the reference axes works in a very similar way to transformation of stresses, which I covered in a video on stress transformation. This is because like stress, the area moment of inertia is a tensor quantity. In fact you can even use Mohr’s circle to determine moments of inertia for rotated axes, like you can with stresses. It has I on the horizontal axis, and the product of inertia I-XY on the vertical axis. You can use Mohr’s circle to find the principal moments of inertia, which are the maximum and minimum values for any angle of rotation. That’s it for what ended up being quite a detailed review of the area moment of inertia. Thanks for watching, and don’t forget to subscribe if you haven’t already!

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