Statics

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 (which is a kind of 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

❏ Centroids of common shapes

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. We use the differential world to understand something about the discrete world:

dA = differential (microscale) area
A = discrete (macroscale) total area

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 exact same mathematical principle, but instead of using differential elements, we will use discrete elements.

❏ Solving for the centroid location through integration

🟦 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).

❏ Common 'built-up' cross-sections in engineering

In this course, we will assume that shapes are made with right angles. In reality, the interface between two rectangular shapes is generally made with a fillet radius (curves). This is partially due to manufacturing reasons, and partially for engineering reasons that are beyond the scope of Statics.

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.

❏ Interactive: centroid of a composite shape

Here is the same problem as above in written format.

It can be called an additive method when you're adding together the effects of different shapes.

❏ Solve for x bar (additive method)

❏ Solve for y bar (additive method)

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. This can be called a subtractive method.

❏ Solve for x bar (subtractive method)

❏ Solve for y bar (subtractive method)

🟦 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:

❏ Doubly-symmetric cross-sections

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

❏ Singly-symmetric cross-sections

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 at the intersection of two lines of anti-symmetry.

❏ An anti-symmetric cross-section

🟦 6.9 Other centroidal calculations

Although our focus in this course is calculating the centroid of an area, we can also apply the basic procedure to other centroidal calculations:

center of volume (volumetric centroid)
center of mass (mass centroid)
center of weight (weight centroid)
center of length (length centroid)

The units will be different, but the principles are identical. We will be calculating:

the moment of volume (volume times distance)
the moment of mass (mass times distance)
the moment of weight (weight times distance)
the moment of length (length times distance)

❏ Generic equation for solving for a centroidal axis

❏ Example 1: Centroid of volume

Two volumes sit on a surface: a cylinder and a cube.

This flipbook shows how you would calculate the center of volume. It includes calculations for x bar, y bar, and z bar.

By the way, I don't know of a practical application for the center of volume calculation. The FE (Fundamentals of Engineering) handbook lists centroid of volume as a topic to cover in a Statics course, and that's the only reason I decided to include it here. If anyone knows a practical reason why you would want to do this calculation, I'd love to learn more!

❏ Flipbook: Two volumes sit on a surface

❏ Example 2: Centroid of mass

Let's do a related problem. We'll take the same scenario as above (an aluminum cylinder and a steel cube), but this time we will find the center of mass.

Steel is more dense than aluminum, so in this calculation, we expect the centroid of mass to nudge a little closer to the steel cube than in the prior problem.

For this problem, we will use the following densities (unit weights) for the materials:

aluminum: ρ = 2.71 E-3 g/mm^3
steel: ρ = 7.85 E-3 g/mm^3

Different alloys of these two metals have slightly different densities.

Pronunciation tip: pronounce ρ as "rho"

❏ Flipbook: Two masses sit on a surface

❏ Example 3: Centroid of weight (or force)

We could do another variant of this problem by doing the centroid of weight calculation. However, since W = mg, and since g is a constant, we would get the same results as in the previous example. No need to work this one!