In this article (and video above), we reveal the SECRET to overcoming the challenging topic of moment of inertia and help you achieve success on your engineering licensing exam.
We’ll start by revisiting the basics of inertia and then dive deeper into essential topics like centroidal area and the parallel axis theorem. These concepts are critical for understanding how the moment of inertia works and will prepare you to handle the different scenarios you’re likely to encounter on the exam.
Question:
In this question, we are presented with an I-beam, and we’re asked to calculate its moment of inertia about its centroidal x-axis. We’re also given a list of possible answers, all presented in mm – so straight off the bat, we know that we’ll be sticking to this unit in our calculations to keep things simple and prevent possible conversion mistakes. Now – before we jump into the solution to this problem, let’s review some key concepts so you have a better understanding of the theory behind this question.
Problem Context: Area Moment of Inertia
The area moment of inertia is a parameter that defines how much resistance a shape – like that of the cross-section of a beam – has to bending due to its geometry. For example – when we apply a concentrated load to the center of three simply supported beams like we have here – we get an intuitive feel for how the beams would react under this stress. In this case, the beams are bending about their own central axis referred to as their centroidal axis. The beam to the left would deform easily under this stress due to its cross section having a small inertia or ‘resistance to bending’ about its central axis. But when you look at the second or third beam, they have larger inertias making them more resistant to bending in this plane. This resistance to bending is influenced by the distribution of area about the bending axis, meaning when a shape has more of its area distributed further away from the bending line, like with our I-beam, it’s also less likely to bend.
However, it’s important to realize that this concept of inertia is relative – in other words, the inertia of a shape can only be taken with respect to a defined axis. So, here’s a summary of the different axial cases that you might expect to see on the FE Exam. First off, you may be asked to calculate a shape’s inertia about its centroidal axis and, as we’ve shown before, this axis is positioned through the geometric center of the shape. Here, we name our axes xc and yc with c referring to the ‘centroidal’ axis.
In the second case, you may be asked to calculate the cross-section’s inertia about your coordinate axes, or the x– or y-axis, as shown here.
Lastly, you may be asked to calculate the shape’s inertia about any axis positioned at a predefined point. It’s important to identify which of these cases you’re working with while tackling these problems since this will later help you distinguish which formulas to use.
If you’re working with a basic shape like the rectangle we have here, the FE handbook conveniently defines all the equations we need to calculate the shape’s inertia about the x– and y– coordinate axes, as well as about its own centroidal x– and y-axes. They list these formulas for a range of simpler shapes like circles and triangles, and even for some more complex shapes like parabolas.
We can utilize these formulas to calculate the different inertias we’re interested in. However, what’s very important here is that you’re familiar with the notation used by the handbook – in other words, be sure to know that Ixc and Iyc refer to the shape’s inertia about its own centroidal axes while Ix and Iy refer to its inertia about the global x– and y– axes.
Problem Context: Parallel Axis Theorem
In the case of the I-beam, we’re working with a relatively complex shape when compared to something like a rectangle, but luckily there’s a theorem that allows us to tackle these complex composite shapes, and it’s referred to as the parallel axis theorem.
To put it into simpler terms – we can refer to the third axial case we had earlier. The theorem states that we can calculate the inertia of the shape about this arbitrary axis, and we’ll refer to this inertia as Ix. This is done by adding the shape’s centroidal inertia to a second term – consisting of the distance from this arbitrary axis to the shape’s centroidal axis multiplied by the shape’s area. This theorem allows you to calculate the inertia of a shape about any axes, and consequently it also allows for the evaluation of more complex shapes like our I-beam. So, without further ado – let’s get started on our problem!
Step 1 – Find the centroid
First, we need to recognize that this seemingly complex shape can be split up into three smaller rectangles. In other words, if we can manage to calculate the moment of inertia for each individual shape about the centroidal x-axis, and we sum these values together, we will have the total inertia of the shape. Second, we need to figure out where the centroid of this asymmetric shape is located – and here we’re referring to the centroid of area. Remember, since this cross-section has a larger area closer to its base, its centroid will also be shifted towards the bottom.
We can define the centroid of the I-beam as y̅, a distance measured from the x-axis. To find the position of this centroid, we’ll need to calculate the area of each rectangle as well as the distance from the reference x-axis to each respective rectangle. The centroid can then be found as the sum of all areas multiplied by their distances and then divided by the sum of all areas.
We start off by calculating the area of our top rectangle, which would be its base multiplied by its height, or 120 × 15 mm. This rectangle’s height from the x-axis to the centroid can be found to be 137.5 mm. We can then go ahead and define these values for our other two shapes to find the centroid of the I-beam. The second rectangle has an area of 100 × 15 mm and its height above the x-axis is 80 mm. Finally, the third rectangle has an area of 180 × 30 mm while its height is equal to 15 mm. We substitute these values into our equation to find that the I-beam has a centroid of 51.55 mm from the x-axis.
Step 2 – Calculate respective inertias
Now that we know where the centroid of the shape is located, we can continue by calculating the respective inertias of each shape about this axis – and this is where the parallel axis theorem can come in quite handy. We use this theorem to calculate the individual inertias of each shape about the centroidal axis at 51.55 mm. Starting off with the inertia of the orange rectangle, the theorem requires that we first calculate the moment of inertia for this shape about its own centroidal axis. We then substitute the base as 120 mm and the height as 15 mm, to find that this rectangle has an inertia of 33,750 mm4 about its own centroidal axis.
Now we need to find a value for dy1 but first let’s recall that this parameter is defined as the distance between the center of the shape to the axis of interest, and in our case this is y̅. So, intuitively, we can calculate this value by subtracting y̅ or 51.55 mm from y1 or 137.5 mm to find that dy1 has a value of 85.95 mm. We can now substitute these values into our parallel axis theorem, along with the area of the shape to find its inertia as 13.33 × 106 mm4.
We now repeat this process for the other two shapes. We start by calculating the green rectangle’s inertia about its own centroidal axis using the same formula as before. By substituting the base as 15 mm and the height as 100 mm, we find that it has an inertia of 1.25 × 106 mm4. Next, we calculate this shape’s distance from the centroidal axis by subtracting its central height from that of the centroidal axis as before to find dy2 as 28.45 mm. When we then substitute these values back into the parallel axis theorem, we find that this section has an inertia of 2.46 × 106 mm4 about the I-beam’s centroidal axis.
And finally, we find the inertia of the blue rectangle. For this we’ll use the same formula as before to find that it has an inertia of 405,000 mm4 about its own centroidal x-axis. And when inspecting its distance from the centroidal x-axis, we find dy3 as 36.55 mm. When we substitute these values into our parallel axis theorem one last time, we find that the blue rectangle has an inertia of 7.62 × 106 mm4 about the centroidal axis.
Step 3 – Combine inertias
All that’s left to do now is to add these values to find the total inertia of the entire shape. In doing so we find the inertia to be 23.41 × 106 mm4.
The Answer is D!
And that wraps up our problem! The correct answer is D. Always take a moment to carefully review all the options before making your final selection—at first glance, you might have been tempted to choose A here.
Final Remarks:
To wrap up, let’s review the concepts covered today. We wanted to calculate the I-beam’s moment of inertia about its centroidal axis. To do this, we first had to find the centroid of the compound shape, using the areas and relative distances of each rectangle. This point defined the centroidal axis for the system. Next, we calculated the inertias of each shape about this centroidal axis by utilizing the parallel axis theorem three times – once for each shape. But in order to apply this formula we also had the compute each shape’s inertia about its own centroidal axis. And we did this by using the readily available formulas provided by the FE handbook.
I hope you found this week’s FE Exam article helpful. In upcoming articles, I will answer more FE Exam questions and run through more practice problems. We publish videos bi-weekly on our Pass the FE Exam YouTube Channel. Be sure to visit our page here and click the subscribe button as you’ll get expert tips and tricks – to ensure your best success – that you can’t get anywhere else. Believe me, you won’t want to miss a single video.
Lastly, I encourage you to ask questions in the comments of the videos or here on this page, and I’ll read and respond to them in future videos. So, if there’s a specific topic you want me to cover or answer, we have you covered.
I’ll see you next week… on Pass the FE Exam
Anthony Fasano, P.E., AEC PM, F. ASCE
Leave a Reply