Many tables and charts exist to help us find the moment of inertia of a shape about its own centroid, usually in both x- & y-axes, but only for simple shapes. How can we use those simple shapes and tables to develop the section properties we need for complex shapes?
Definition
The moment of inertia of an area with respect to any axis in the plane of the area is equal to the moment of inertia of the area with respect to a parallel centroidal axis plus the product of the area and the square of the distance between the two axes. This proposition is called the parallel axis theorem.
Fred Seely & Newton Ensign – “Analytical Mechanics for Engineers”
The Parallel Axis Theorem tells us that the moment of inertia of a component of the section about an axis some distance “d” from its own neutral axis is equal to its own moment of inertia about a parallel axis through its centroid, plus an additional contribution equal to the area of the component times the distance squared. Mathematically, that looks like this:
Intuitively, this makes sense, as the moment of inertia is all about integrating a distance squared term over the area of a shape. Moving the shape away from the axis in question means we need to add another term to pick up that offset distance in our moment of inertia.
Proof
If we take a “structural potato” shape, as above, with a known moment of inertia I.y about the centroidal axis Y, and wish to find the moment of inertia about the parallel axis Y’, then we can take our ordinary equation for the second moment of area (moment of inertia), which as a reminder looks like this:
Defining our coordinate shift as:
Then we can substitute eq (3) into eq (2) and expand the squared term:
Now we note a few facts:
- The first integral term in Equation (6) is exactly the definition of the integral we started with back in Equation (2)
- The second term in (6) is exactly 0
- This was a bit confusing to me for a while, but it has to do with the rules of our initial coordinate system. The centroidal axes we used for the initial moment of inertia of the shape feature the y-axis centered left/right on the shape, so integrating x over the area must result in an equal positive and negative portion, which cancel each other out.
- The final term is just d² integrated over the area, which is a constant integrated over the area. The d² term passes through the integral, which just becomes the area of the shape itself.
This leads us to the final equation, equation (8) below, which is the same as the definition equation (1) at the beginning of this article.
Example from “Analytical Mechanics for Engineers”
Thumbing through an old edition of a Mechanics of Materials textbook, I found the below great demonstration of how the Parallel Axis Theorem can be used to come up with section properties for a common steel channel shape.
Here we’re looking at finding the weak-axis moment of inertia for a C12x20.7.
Problem Statement
Find the moment of inertia of the channel section shown in Fig. 229 with respect to line XX. Find also the moment of inertia with respect to the parallel centroidal axis.
Solution
We divide up the area into triangles and rectangles, as shown in the figure. This is one strategy, another good one would be to use trapezoids for both flanges instead of triangles and rectangles.
Using a table to keep track of variables, we can quickly come up with the following:
| Part | a | y.0 | a*y.0 | I.0 | a*y.0² | I’.x=I.0+a*y.0² |
| a.1 | .745 | 1.61 | 1.20 | 0.44 | 1.93 | 2.37 |
| a.2 | .745 | 1.61 | 1.20 | 0.44 | 1.93 | 2.37 |
| a.3 | .585 | 1.17 | 0.68 | 0.23 | 0.80 | 1.03 |
| a.4 | .585 | 1.17 | 0.68 | 0.23 | 0.80 | 1.03 |
| a.5 | 3.360 | 0.14 | 0.47 | 0.02 | 0.07 | 0.09 |
| Sum | 6.02 in² | 4.23 in² | 6.89 in⁴ |
So, our total moment of inertia with respect to line XX is 6.89 in⁴.
Along the way, we had to calculate the areas of each part, and we took the opportunity to compute the products of the areas and centroidal distances, which can quickly give us that:
And using our new favorite tool, the parallel axis theorem, we can quickly compute the centroidal moment of inertia as:
So we finally get a centroidal weak-axis moment of inertial of 3.94 in⁴, which is a little different from the stated 3.86 in⁴ in the steel manual, but this slight difference is easily attributable to the radii and fillets on the real shape, which cause our triangles and rectangles to be not quite an accurate reflection of the shape.
We could get around this limitation by splitting the overall section into more pieces, and using portions of circles in the mix to pick up the actual shape.
Modern Use of the Parallel Axis Theorem
As investigated more fully in my article “How to Numerically Compute the Moment of Inertia”, most modern engineers use numerical computer tools like Risa Section or IES Shapebuilder to more quickly and accurately come up with section properties for complex shapes.
The Parallel Axis Theorem plays a crucial role in all of these programs, which effectively break up the shapes into a fine mesh of triangles to approximate the shape, then run a huge number of equations to create a much larger version of the tabular solution we ran above.
For shapes that don’t demand quite so much rigor, tabular tools like Microsoft Excel or Google Sheets can be used to quickly compute the areas, centroidal moments of inertia, and centroidal distances needed to drive an approximate solution, just like in our example.
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