Statics

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).

❏ First moment vs. second moment

🟦 18.2 Significance of the moment of inertia (MoI)

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

⇢ the area moment of inertia

⇢ the mass moment of inertia

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 is calculated. You will learn how to apply the area moment of inertia in future studies.

❏ Applications in dynamics problems

If you're designing a wrecking ball, the longer the cable, the more damage you do.

❏ Applications in solid mechanics problems

If you're designing a beam, it needs to be appropriately stiff to resist applied loads within deformation tolerances.

🟦 18.3 Integrals for moment of inertia (MoI)

We use the capital letter I as the 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.

❏ Explanation of the integral

❏ Formulas for Ix and Iy

The moment of inertia about x is equal to the integral of y^2 over the area. The moment of inertia about y is equal to the integral of x^2 over the area. In each case, we take the cumulative effect of area times distance squared.

There are four variants on the moment of inertia.

In Statics, we learn to compute Ix, Iy, and Ixy (the product of inertia).

The polar moment of inertia (J) is typically studied in subsequent courses. It uses polar coordinates and is used for torsion (twisting).

❏ Four variants on the (area) Moment of Inertia

Did you notice that each one is a different type of area times distance times distance?

🟦 18.4 Formulas for moment of inertia (MoI)

Please 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.

❏ Formulas to memorize

The centroidal axes are the black dashed line. Refer back to Lesson 6 if you need to review how we calculate the location of the centroidal axes.

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

❏ Derivation for moment of inertia for a rectangle

Remember: in this formula, you cube the term that is perpendicular to the axis of interest (it's called the axis of bending in future courses).

🟦 18.5 Parallel axis theorem (P.A.T.)

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 to maximize the moment of inertia.

This is why we tend to use built-up shapes. 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 (P.A.T.).

❏ 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 Fall 2024. It's not mission critical for Statics - it's more important that you know how to use the P.A.T.***

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

❏ 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).

❏ P.A.T. prompt 1 of 2

❏ P.A.T. solution 1 of 2

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.

❏ P.A.T. prompt 2 of 2

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

❏ P.A.T. solution 2 of 2

🟦 18.6 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:

❏ Video: product of inertia of a square

This flipbook solves Ix, Iy, and Ixy for an 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 signs when computing the product of inertia. Each term will be positive, negative, or zero.

❏ 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.

❏ Full solution: product of inertia for a "Z" shape

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.

Before tackling that, you need to learn a little about rotating axes in Section 16.7 and principal moments of inertia in Section 16.8.

🟦 18.7 Rotation of the centroidal axes

A cross-sectional area can have an infinite number of centroidal axis pairs.

Sometimes the cross-sectional axes align to the overall geometry, such as the x- and y- axes for this L-shaped (angle) beam.

If we loaded this beam in the y-direction, it would bend about x, so we'd use the x- moment of inertia shown here (8.68 in^4).

❏ M.o.I. about x and y

But what if we take the same angle shape and rotate it about z so that it can sit on support ledges?

In this orientation, the load is in the the v- direction. It would bend about u, meaning that we'd need to use the u-axis moment of inertia.

We could calculate moments of inertia about these axes, but it would take some effort. We need to figure out an easy way to solve for the moments of inertia about axes u-u and v-v.

Note the θz in the image. It measures the change in angle from one coordinate system (xy) to another (uv). We are rotating about z. This is called a coordinate transformation.

❏ M.o.I. about u and v

🟦 18.8 Introduction to the principal moments of inertia

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

The principal moments of inertia are the maximum and minimum moments of inertia for a given cross-sectional area. We call them Imax and Imin.

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

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

The Imax axis has a partner axis at a rotation of 90 degrees. The moment of inertia about that axis is the smallest one. We call it Imin.

For the cross-sectional area depicted here, Imax is about axis g-g. Imin is about axis h-h. Axes g-g and h-h are therefore called the principal axes.

❏ So many axes!

🟦 18.9 How to calculate the principal moments of inertia

Let's say that you are asked to calculate the principal moments of inertia for a cross-section.

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. There is no need to compute the product of inertia.

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

Compute Ix
Compute Iy
Compute Ixy
Construct a Mohr's Circle

❏ Flowchart for calculating Imin and Imax

❏ Mohr's Circle: it's 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 in other courses.

This flipbook is an introduction to the procedure for computing the principal moments of inertia using Mohr's Circle.

The flipbook example is based on a cross-section with:

- Ix = 5.440 in^4
- Iy = 4.012 in^4
- Ixy = 1.286 in^4

❏ Flipbook: Getting acquainted with Mohr's Circle

Here is the procedure to create a Mohr's Circle:

Plot Ix and Iy on the horizontal axis; plot Ixy on the vertical axis
Create a circle by plotting two ordered pairs:
- - - (Ix, Ixy)
    - (Iy, -Ixy)
Use fundamental geometry to solve for the center of the circle and the radius of the circle.
The principal moments of inertia lie at the left and right quadrants of the circle (where Ixy = 0)
The angle from the input xy axes to the output principal axes is:
- - equal to 2θp in the circle
  - equal to θp in real life

Here is another EXAMPLE that shows how we use Mohr's Circle to compute the principal moments of inertia.

❏ Flipbook: Example problem for Mohr's Circle

Did you notice that all angles in Mohr’s Circle are twice the change in angle in real life?

If you rotate a coordinate system 30 degrees clockwise in real life, then you need to travel 60 degrees clockwise in Mohr’s Circle.
The x and y axes are separated by 90 degrees in real life, and 180 degrees apart in Mohr’s Circle.

We can use Mohr's Circle to transform the moments of inertia from any input pairs of axes to any output pairs of axes (such as the angle in section 16.7).

The most practical use of Mohr's Circle is to determine the principal moments of inertia and to sketch the principal axes on the cross-sectional geometry.

➜ Practice Problems

This video solves the first 6 problems for you. Please use this to check your work, or as a learning tool. This is not for you to copy. The only way to learn how to solve these types of problems is to practice, practice, practice.

Scrambled answers to problems 1, 2, and 3:

0.75 0.984 1.50 1.75 1.75 1.75 3.00 5.36 6.34 7.88 10.7 22.8

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.

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.

Problem 3.

Determine the centroid (x bar, y bar) of the T-shaped cross-sectional area. Then, use the Parallel Axis Theorem to 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.

Scrambled answers to 4, 5, and 6:

1.21 1.29 1.42 1.64 1.75 3.25 3.25 4.01 5.44 11.7 15.2 73.0 73.0

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, use the Parallel Axis Theorem to determine the centroidal moments of inertia.