## 🟦 6.1 A conceptual introduction

Two books are stacked on a table, but not in a symmetric pile.

We can use a moment equilibrium equation to determine the (x, y) location of the force of the table on the lower surface of the bottom book.

This procedure is similar to the way we will calculate centroids (center of area), and therefore serves as a useful introduction to the topic.

Work through the flipbook, paying close attention to how each moment calculation is set up.

### ❏ Flipbook: an asymmetric stack of books

## 🟦 6.2 Moment of area

Up until now, we have the used the word moment for a tendency to rotate (a force multiplied by a perpendicular distance to an axis). We might explain this idea as:

The moment of a force = force times perpendicular distance.

To calculate the location of the centroid of an area, we will use the moment of an area. This idea is extremely abstract, but can be expressed like this:

The moment of an area = area times perpendicular distance.

In other words, we can use the phrase "moment of" as a mathematical operation: it means to multiply something (a force, an area, a mass, etc.) by a perpendicular distance.

### ❏ Force times distance

### ❏ Area times distance

## 🟦 6.3 Two analogies for centroids

First, say you cut out a random shape out of cardboard, such as the blue body below. You punch a hole in it somewhere (not necessarily at the centroid), and tie a string through the hole as shown.

When you let go, the body swings and oscillates until it reaches static equilibrium. The line of the cable defines one of the body's centroidal axes -- an axis coincident with its centroid or center of mass. For this example, since the material is homogenous, the center of mass is also the center of area.

Another way to think about the center of area is to imagine having to balance a planar area on a pyramid-shaped support. It will only balance if the pyramid (a 3D pin connection) is precisely located at the centroid. If the pyramid were to be moved to any other point, the planar area will not balance -- it will rotate.

## 🟦 6.4 Why we calculate centroids

While there are many applications for calculating the centroid of an area (or a weight, or a mass, etc.), the most important application for this course is to calculate the centroid of the cross-sectional area of a prismatic member.

Prismatic members are formed by extruding a cross-sectional shape in the longitudinal direction (the z-axis depicted below). Non-prismatic members have variations in the cross-sectional geometry along the length of the member (e.g. a tapered beam).

## 🟦 6.5 Centroids you already know

You should already know the location of the centroid for three common shapes: rectangles, squares, and triangles.

Please note the terminology in the drawings below:

Centroid

Centroidal axes

Extreme bottom left fiber

x bar and y bar

When you need to find the centroid of a triangle, find the third-points of each leg. The centroid lies at the intersection of lines that are perpendicular to each side, located at the third-point, on the "heavy" or long side of the triangle.

## 🟦 6.6 Integration method (for integrable functions)

The integration method for determining the location of the centroid of area is nearly identical to the procedure for calculating the resultant force of a line load (in the last chapter).

You probably did this type of problem in your Calculus I course.

Note that in the practical world of engineering, very few cross-sectional areas are integrable functions. We tend to use built-up sections. The component area method (in the next section) is more useful and practical than the integration method. It's based on the same mathematical principle.

## 🟦 6.7 Component area method (for built-up cross-sections)

Most engineering cross-sections can be split into rectangles.

This is true for the commonly used I-shaped member, T-shaped member, C-shaped member, and L-shaped member. Each can be created by combining multiple rectangles.

Our strategy is to break up the composite area into component areas with known centroids (rectangles, triangles, and circles).

To use the component area method:

break the composite shape into components

compute the area of each component

sum them to get the total composite area

measure the perpendicular distance from the centroid of each component area to the extreme bottom left fiber

the sum of the first moment of area of the composite shape is equal to the sum of the first moment of area of the component shapes

Work your way through the interactive visual for an example.

The centroid calculation can be thought of as a weighted average.

Here is the same problem as above in written format.

Sometimes it's advantageous to use a combination of solids plus voids in centroid calculations. Here is the same problem, but worked with a 6"x10" solid component and a 8"x4" void component. All you have to do is assign a negative sign to the terms associated with the voids.

## 🟦 6.8 Single-symmetry, double-symmetry, and anti-symmetry

Before spending time calculating a centroid, determine whether your area happens to be symmetric.

If you find two perpendicular lines of symmetry, then the centroid lies at the intersection of those lines.

These types of areas are sometimes called doubly-symmetric:

Some shapes have one line of symmetry. These shapes can be called singly-symmetric. The centroid lies somewhere on the line of symmetry.

Some shapes are anti-symmetric.

Anti-symmetry does not mean asymmetry.

The diagram below shows how an anti-symmetric shape can be folded over two lines of anti-symmetry.

The centroid will lie on lines of anti-symmetry.

## 🟦 6.9 Other applications for centroidal calculations

(We won't do this Summer I 2024.)

We can also calculate the center of mass and the center of weight.

## ➜ Practice Problems

Here are scrambled answers to the practice problems. Use them to check your work.

0.75" 1.00" 1.75" 2.00" 2.00" 2.60" 2.75" 3.50" 5.00" 6.27"

12 mm 12 mm 33.3 mm 57.7 mm 79.4 mm 90.9 mm 107 mm 120 mm 161 mm 196 mm

Problem 1.

Problem 3.

Problem 4.

Problem 5.

Problem 6.

Problem 7.

Problem 8.

Problem 9.

Problem 10.

THE END :)