# Statics

## 🟦  16.1  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. ## 🟦  16.2  What is 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 and  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.  ## 🟦  16.3 The integrals

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 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.  ## 🟦  16.4 Useful formulas

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.

Important: you can compute a moment of inertia about other axes. Our focus is on the centroidal moments of inertia, so in this class we will limit our scope to axes that coincide with the centroid. Here is the derivation for the moment of inertia for a rectangle: ## 🟦  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 distance terms are small).

We can engineer better cross-sectional areas so that they have higher moments of inertia. For this reason, 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 shapes? We use the parallel axis theorem.

***note to self: need flipbook of derivation of parallel axis theorem*** Here is an infographic of the parallel axis theorem

We will use this to compute the centroidal moments of inertia for a composite (built-up) shape. Here is an example problem (and solution) for the parallel axis theorem.  Please take a close look at that last answer. The final answer is left in engineering notation, as there is no way to further simplify it. Please do not use prefixes. For example, do not write "M mm^4". Leave it as E6 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. Solution:

## 🟦  16.6 Product of Inertia

Watch this short video that provides an overview of the Product of Inertia.

Then, work through the example problem below carefully, using the flipbook as a guide.

Some prefer to approach these calculations in a tabular format: ## 🟦  16.7 Principal Moments of Inertia

Mines in Amsterdam students: please go through this flipbook thoroughly.

I will explain what this is for and why it is important in class.

## ➜ 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.  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. 