Why a fuller doesn't make a sword stiffer

"Fullers make swords stiffer"

You've all probably heard it at least once, and possibly even said it. If you said it with me around, I probably made sure you never said it again.

For the impatient: No, they don't. Stop saying it. But kudos for not calling it a 'blood-groove'.
For the impatient who call it a blood-grove: Just stop. Please.

For the patient, read on.
For the patient wondering about hammered fullers and other things, read on... and I'll get to that.



If you're following the Mechanics 101 posts, I glossed over the idea of the area moment of inertia, aside from a basic definition, because I wanted to make this post to deal with it in gory detail.

To summarize the mechanics of beam bending, Assume we have a slender beam of a linear elastic material subjected to a moderate transverse load P, such as that shown below:

Cantilever Beam

 Which looks an awful lot like a sword that has been bent:

Convincing, no?

The displacement at the end of the beam can be written in terms of the applied load as follows:
\[\Delta = \frac{Pl^3}{3EI}\]
Where E is the Young's modulus (slope of the linear elastic portion of the stress-strain curve), and I is the area moment of inertia. The product of the two, \(EI\), is referred to as the bending stiffness.

From Mechanics 101, part 4, we know that the Young's modulus is just the stiffness of the material the beam is made from. What exactly is the area moment then?

As I defined earlier, the area moment of inertia is a measure of the distribution of area within a cross-section, relative to some rotational axis. In other words, it's the geometric contribution to the bending stiffness. If we know the exact geometry of a cross-section, as well as the rotational axis of interest, we can calculate the area moment. For example, if we had two cross-sections like the following:

A) Rectangular cross-section. B) I-beam cross-section. The dotted line is at the middle of the sections.

For the rectangular section, the area moment about the dotted line is as follows:
\[I_{rect,x} = \frac{1}{12} \frac{a}{2} a^3 = \frac{1}{24} a^4\]

For the I-beam section above, we can use the formula for the area moment of a rectangle and the parallel axis theorem. The result is as follows:
\[I_{I-beam,x} = \left(\frac{a^4}{384}\right) + 2\left(\frac{7a^4}{384}\right) = \frac{15a^4}{384}\]

If we take \(a=1.0\) cm, then we end up with the following values:
\[I_{rect,x} \approx 0.0417~\mathrm{cm}^4\]
\[I_{I-beam,x} \approx 0.0390~\mathrm{cm}^4\]
In other words, the I-beam is about 6% less stiff than the rectangular section of the same outer dimensions. Obviously, the amount of difference will vary depending on the exact dimensions of the portions of the I-beam, but it will always be less than the solid rectangular section of the same outer dimensions.

"But Sparky", you say, "That bending axis is wrong for a sword!" And I say: you're right, though the point will still stand, but even more dramatically. A more reasonable axis would look like this:

Sections with a vertical bending axis

For this choice of axis, our area moment for the rectangular section (A) is:
\[I_{rect,y} = \frac{1}{12} a \left(\frac{a}{2}\right)^3 = \frac{1}{96} a^4\]
For the I-beam we get (this time it doesn't require the parallel axis theorem. Yay!):
\[I_{I-beam,y} =\frac{1}{12}\left(\frac{a}{2}\right)\left(\frac{a}{4}\right)^3 + \frac{2}{12}\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)^3 = \frac{9}{1536}a^4\]
Using \(a = 1.0\) cm we get:
\[I_{rect,y} \approx 0.0104~\mathrm{cm}^4\]
\[I_{I-beam,y} \approx 0.0059~\mathrm{cm}^4\]
So about this bending axis, our I-beam is actually about half the stiffness of the solid section of the same outer dimensions! This is why I-beams in structural applications are usually not oriented such that they bend about this axis. Like before, however, the magnitude of difference will depend on the exact dimensions of the beam sections.

So why do we use I-beams at all if they don't make things stiffer? Well, let's look at the total cross-sectional area: The rectangular section has an area of \(a^2/2\), while the I-beam has an area of \(3a^2/8\): 1/4 less area than the rectangular section. That means for the same length, the I-beam will require less material. This translates to weight and material cost savings. And when designed and installed properly, the minor change in bending stiffness generally isn't as big of a deal as the savings in weight and cost in most applications. And for purposes where you want some flexibility without losing too much thickness, an I-beam may help.

Hopefully I've convinced you: Fullers do not make swords stiffer, in fact they do the opposite. Their greater purpose is to make swords lighter.

With me? Good. Now let's make it a bit more complicated.

About those hammered fullers

So the I-beam profile created by a fuller isn't responsible for any increase in stiffness over an equivalently sized solid section. But what if the fuller wasn't created by removing material, but rather by moving material around?

This is actually pretty simple: In that case, because of conservation of mass, the outer dimensions of the piece will no longer be the same. So it is possible for a hammered fuller to increase the stiffness, because you are not removing material... assuming that is that you don't grind that excess off to make an edge.

About those risers/ridges on blades

Risers, the opposite of a fuller, can make swords stiffer, depending on your reference point. If you start with some cross-section, and you add some amount of material to make the riser, then you stiffen the blade compared to the initial cross-section. 

However, if you instead remove material from a larger section, you end up with a less stiff section. But it would still be stiffer than if you'd made a uniform thickness reduction. To put this graphically:

Graphical description of riser possibilities

How about work hardening and heat treatment?

So, we've dealt with the geometric contribution to the bending stiffness thanks to the area moment of inertia. So what about the material part? Say work hardening due to hammering the fuller or heat treatment?

Well, one way to deal with it is to simply say 'that's not the fuller, that's material processing'. But that's kind of cheesy, at least for the work hardening. For the heat treatment, it's not: a fuller has nothing to do with heat treatment. A maker could apply a given heat treatment to a blade regardless of if it had a fuller, so it becomes meaningless to talk about as part of the fullering operation. I may write a post specifically on heat treatment of metals because it is a really interesting topic, though.

Let's talk work hardening instead:

Work hardening, as briefly mentioned in Mechanics 101, part 4, is the process that occurs in materials (particularly steels) after yielding. For a certain range of strain, you can actually start to support an increasing stress. Also when you unload, regardless of which region you're in, you return along a line parallel to the linear elastic region.

Typical loading and unloading stress-strain curve for a ductile metal

If you were to then re-load that piece, assuming minimal hysteresis, you'd follow the unloading line back up to \(\sigma\) at a strain of \(\epsilon_{max}\) (relative to the initial configuration). In other words, the Young's modulus would not have changed. What does change, however, is the effective yield stress: the stress that the piece can carry before behaving plastically has increased. This is called strengthening or hardening and does not affect the elastic stiffness of the material, which is the slope.

So remember: hardening =/= stiffening. Different concepts.

Still with me? You rock.