# Statics

## đźź¦Â  16.1Â  The second moment of area

In this course, we have often computed the "moment of a force," which is a tendency to rotate.

Engineers have adopted the habit of saying "moment of" when referring to the idea of multiplying something by a perpendicular distance.

### moment of = times distance

Let's adopt that terminology: we will say "moment of an area" to mean area times distance, as seen in the image.

We used the first moment of area (area times distance) in order to compute centroids. Specifically, we set the first moment of area equal to zero, to solve for the location of the centroid.

In this lesson, we will use the second moment of area, which can be thought of as the product of area and distance squared. The second moment of area is more commonly called the moment of inertia. The units are length to the fourth power (in^4 or mm^4).

## đźź¦Â  16.2Â  Significance of the moment of inertia

The term moment of inertia is an interesting one. There are two common sub-categories:

In dynamics classes, the mass moment of inertia is used to measure rotational inertia (e.g. tying a weight to a string andÂ  swinging it in a circle). The mass moment of inertia has units of mass times distance squared.

In solid mechanics classes, the area moment of inertia is used to determine how a solid member performs under a given load. It correlates to the beam's resistance to bending. The area moment of inertia has units of length to the fourth.

In this class, we focus on learning how the area moment of inertia calculated. You will learn how to apply the moment of inertia in future studies.

## đźź¦Â  16.3 Integrals for moment of inertia

We will use the capital letter I as a symbol for moment of inertia. Here are the integrals that define the (area) moment of inertia about an axis. Note that the origin of the coordinate system is at the centroid of the cross-section. This detail is critical if you want to calculate the centroidal moments of inertia. Also, note that the distance we want to square is perpendicular to the axis of interest.

### âťŹ Formulas for Ix and Iy

Big picture, there are four variants on the moment of inertia, as depicted here.

In this course, we will learn how to compute three out of the four.

The fourth (J, the polar moment of inertia) is a second moment of area in polar coordinates.

Depending on your major, you may learn about J, the polar moment of inertia, in future studies. It's only used for torsion (twisting).

## đźź¦Â  16.4 Formulas for moment of inertia

Let's memorize the formulas for the moment of inertia of a circle (about a centroidal axis) and the moment of inertia of a rectangle about the two centroidal axes.

Note: you can compute a moment of inertia about other axes. In this class, our focus is on the centroidal moments of inertia. We will only solve the moments of inertia about axes that coincide with the centroid.

### âťŹ Formulas to memorize

Here is the derivation for the moment of inertia for a rectangle about a centroidal axis.

## đźź¦Â  16.5 Parallel axis theorem

Unfortunately, solid rectangular and circular cross-sections are not very efficient. This is because in rectangles and circles, the area is lumped close to the centroid. That means that the moment of inertia is low (because the perpendicular distance terms are small).

We can engineer better cross-sectional areas so that they have higher moments of inertia. This is why we tend to use built-up shapes, such as the ones in the image, so that we minimize the cost of the material while increasing the moment of inertia.

But how do we calculate the moment of inertia for these types of built-up shapes? We use the parallel axis theorem.

This infographic of the parallel axis theorem is our guide for compute the centroidal moments of inertia for a composite (built-up) shape (Ix and Iy).

### âťŹ Built-up shapes

***Author's note: I need to create a flipbook that has a derivation of the parallel axis theorem. Probably can't do this by June 18th 2024 though. It's not mission critical for this class - it's more important that you know how to use the P.A.T.***

### âťŹ Infographic: the parallel axis theorem (P.A.T.)

Here is an example problem (and solution) for the parallel axis theorem. You must calculate y bar as the first step (if you need to review Lesson 6 Centroids, please do so).

### âťŹ P.A.T. solution

In the answer, the solution is written as 5.18 E6 mm^4. That is the best way to express that answer. We would not add a prefix to it (e.g. we would not write "M mm^4").

Sometimes, it is advantageous to use solids and voids when using the parallel axis theorem. Here is an example that shows how that works. It's similar to the logic we used when we learned in the centroids lesson.

### âťŹ Prompt: P.A.T. with solids and voids

Determine the moment of inertia of this hollow square cross-section. First, use an additive method. Then, use a subtractive method.

## đźź¦Â  16.6 Product of inertia

The product of inertia, Ixy, is a calculation that is only useful for cross-sections that do not have a line of symmetry.

It's defined as:

### âťŹ Video: product of inertia of a square

This flipbook solves Ix, Iy, and Ixy for this anti-symmetric shape.

Neither the centroidal x-axis nor the centroidal y-axis is a line of symmetry.

This is why the product of inertia, Ixy, is non-zero.

Pay special attention to the signsÂ  when computing the product of inertia. Each term will be positive, negative, or zero.

### âťŹ Flipbook: product of inertia for a "Z" shape

Some engineers prefer to approach these calculations in a tabular format. This is an alternate solution format to the same problem as above. The results are exactly the same.

The product of inertia is always equal to zero when one or both centroidal axes are a line of symmetry. The significance of the product of inertia, Ixy, is explained in Section 16.9. First, you need to learn a little about rotating axes in Section 16.7 and principal moments of inertia in Section 16.8.

## đźź¦Â  16.7 Rotation of the centroidal axes

A cross-sectional area can have many different pairs of centroidal axes.

Sometimes the cross-sectional axes align to the geometry, such as this rectangle. We have already learned how to calculate Ix and Iy for these axes.

But in some applications, we may want to determine the moments of inertia about a different pair of axes.

We might call these the u and v axes, as shown.

We could calculate moments of inertia about these axes, but it would take a bit of effort to write the functions that would describe the rectangle in the uv coordinate system.

Note the Î¸z in the image. It measures the change in angle from one coordinate system (xy)Â  to another (uv). The z axis is perpendicular to your screen.

## đźź¦Â  16.8 Introduction to the principal moments of inertia

In engineering, when we use the word principal, we usually mean most important.

The principal moments of inertia refer to the maximum and minimum moments of inertia for a given cross-sectional area.

Moments of inertia are different for each angular rotation about the z-axis.

For instance, consider a (colorful and) oddly-shaped cross-sectional area.

We could compute the moment of inertia about any of these centroidal axes (a-a, b-b, c-c, d-d, etc.). We'd get a different value for each computation.Â

But, as we consider rotating these axes about z (the axis coming out of the screen) in all possible orientations, we'd ultimately discover an axis that yields the largest moment of inertia. We call that Imax.

The Imax axis has a pair at 90 degrees. The moment of inertia about that axis is the smallest one that you can calculate. We call it Imin.

Imin and Imax are called the principal moments of inertia.

For the (colorful and) oddly-shaped cross-sectional area depicted here, Imax is about axis g-g. Imin is about axis h-h. Axes g-g and h-h can be called the principal axes.

## đźź¦Â  16.9 Procedure for calculating the principal moments of inertia

You need to calculate the principal moments of inertia for a certain cross-section.

Here is a quick reference that shows you when you should and should not bother computing Ixy, the product of inertia.

In a nutshell, if your centroidal x-axis or y-axis is a line of symmetry, you're already looking at the principal axes! There is no need to consider any rotated axis pairs.

If neither your centroidal x-axis nor your centroidal y-axis are a line of symmetry, then you must do the following:

### âťŹ Mohr's Circle: a visual representation of the moment of inertia

I like to describe Mohr's Circle as a tool. It helps us transform moments of inertia from one pair of axes (x and y) to another pair of axes (u and v).

Mohr's Circle is one of those concepts in engineering that may not "click" for you the first time through. You'll need to spend a little time with Mohr's Circle before it starts making sense. With time and effort, it will make sense, I promise. Its application is not limited to Statics - you'll also see it used in other courses.

This flipbook shows how to compute the principal moments of inertia.

It's based on a cross-section with Ix = 5.440 in^4; Iy = 4.012 in^4; and Ixy = 1.286 in^4.

The Mohr's Circle procedure, in a nutshell:

### âťŹ Flipbook: Getting acquainted with Mohr's Circle

This flipbook provides an EXAMPLE for use of Mohr's Circle to compute the principal moments of inertia.

### âťŹ Flipbook: Example problem for Mohr's Circle

Did you notice that all angles in Mohrâ€™s Circle are twice the change in angle in real life?

We can use Mohr's Circle to transform the moments of inertia from any input pairs of axes to any output pairs of axes, but the most common use of Mohr's Circle is to determine the principal moments of inertia and sketch the principal axes on the cross-sectional geometry.

## âžś Practice Problems

Problem 1. Determine the centroid (x bar, y bar) of the rectangular cross-sectional area of the piece of lumber. Then, determine the centroidal moments of inertia of that cross-sectional area.

Fun fact: the dimensions of a 2 by 4 are actually 1.5 x 3.5 inches.Â

Problem 2. Determine the centroid (x bar, y bar) of the total rectangular cross-sectional area. Then, determine the centroidal moments of inertia of that cross-sectional area.

Problem 3. Determine the centroid (x bar, y bar) of the T-shaped cross-sectional area. Then, determine the centroidal moments of inertia.

By the way, please assume that this cross-section and others are all glued together, even when the glue isn't explicitly illustrated.

Problem 4. Determine the centroid (x bar, y bar) of the I-shaped cross-sectional area. Then, determine the centroidal moments of inertia.

Problem 5. Determine the centroid (x bar, y bar) of the C-shaped cross-sectional area. Then, determine the centroidal moments of inertia.

Problem 6. Determine the centroid (x bar, y bar) of the oddly-shaped cross-sectional area. Then, determine the centroidal moments of inertia.

Since this cross-section does not have a line of symmetry, also calculate the Product of Inertia.

Problem 7. You have computed the following properties for a cross-section:

Ix = 160 in4

Iy = 100 in4Â

Ixy = +40 in4Â

Your task is (1) to construct Mohrâ€™s Circle, (2) to report the principal stresses, and (3) to solve đťžˇp.

Solution (credit: YouTube user David Spears)

That's it for Summer 2024!