Replacing the bent weapon was a safe call for many reasons, and in my opinion the right call at the time. But how about that breaking part? Let's talk a bit about that and a mode of failure called

*fatigue*.

For the impatient: yes, bending the weapon back does indeed increase the likelihood of failure, especially if it is loaded the same way repeatedly (it likely will be) or has been bent back multiple times. How much it increases the likelihood depends on many factors, but it is a safe play to remove the weapon from use until it can be repaired properly (if it can be at all, which also depends on many factors). Flexible trainers of all types will tend to have a lifetime due to the cyclic loading and and a failure process called fatigue, but that lifetime will generally be fairly long (like years) if they are properly made, used and maintained. However, notches and other surface damage, and plastic deformation of the blade can dramatically lower this lifetime. And the main danger of fatigue failure is that it can occur with little to no warning during the use of the item, unlike most acute failure modes like fracture.

For the patient: read on.

### Fatigue

It's pretty easy to understand that materials will fail if we apply a stress greater than their ultimate strength, for example pulling a bar with an ever increasing force until it breaks. But it's also possible for materials to fail at much lower loads through variable or cyclic loading. For example, bending the tab on a soda (pop, if you prefer) can back and forth until it finally breaks off. This mode of failure through variable loading is called fatigue, and is a very important mode of failure in many engineering systems: vehicle frames, bridge structures, buildings, aircraft wings, and yes, even flexible sword trainers.The reason fatigue is so important is fairly straightforward: over the life of many systems, they are subjected to moderate but varying loads many, many times. These loads may be well below the ultimate strength of the materials (which is fairly easy to design around), and possibly even the yield strength yet still lead to failure. Thus it takes additional consideration to account for fatigue, because the failure may be even more sudden and dramatic than an acutely over-loaded system, which often give warning signs of impending failure.

If we go back to our soda can analogy, the first time we bend the tab (say to open the can), it's unlikely to fail (wouldn't be very useful if we couldn't even open the can, right?). But as you idly push it back and forth, say while waiting for a meeting to end, the material will yield and undergo small crack propagation events. Eventually the thing just comes off, like in this handy dandy video I made:

So how does fatigue work? It's a fairly complex failure mode and manifests differently in particular materials, but in metals it tends to be related to the opening and connecting of microcracks at stress concentrations. For example: a notch left by a lathe tool, a sharp corner, or one of the many minuscule cracks that naturally exist at the interface between phases or grains. The intensification of stresses in the vicinity of the stress concentration can lead to dramatically increased local stresses compared to the loading on the overall body. Thus, in the vicinity of the stress concentration you may have small localized plastic deformation or crack propagation while the bulk of the body is responding elastically. Because we are propagating a crack with each loading, relieving the load doesn't mean the crack actually closes (which would require bonds to be re-formed and is thermodynamically unlikely in real systems), and so cyclic damage tends to have a cumulative effect. Repeated a number of times, this loading can lead to significant propagation of cracks, and finally failure of the part.

Now that we have some basic understanding of what fatigue

*is*, how can we design around it or understand how it may occur? Engineering mechanics can provide us some insight. For example, one question we may ask is how long can we expect a component to survive a given cyclic loading pattern - say a rod being loaded in cyclic tension:

Simple schematic of a cyclic tension loading, \(\sigma\) denoting nominal applied stress. |

*fatigue strength*, the applied stress which a system can sustain without failing due to fatigue, versus the number of cycles to failure at that stress level. This plot is typically called an S-N plot, and an example is given below for a steel (note that the cycle axis is a logarithmic scale):

Schematic of S-N curves, showing typical steel and aluminum behavior (not to scale). |

There's one important way that the S-N curve is interesting for steels compared to other metals: typically, non-ferrous alloys do not exhibit an obvious infinite cycle life region. Instead, their S-N curves tend to be continuously decreasing, as seen in the above diagram.

There's one remaining hitch, though: while these S-N curves can in principle be determined for components as manufactured, 'material level' tests are often done in the laboratory using specially prepared polished samples to establish a basic understanding of the material's behavior. In order to extrapolate from the fatigue limit determined in idealized material level tests to component-level behavior, an empirical relationship is often used. One method is the calculation of a

*Marin factor*, a factor of safety that is applied to the material-level fatigue limit that accounts for factors such as surface conditions, size effects, loading type, temperature, and environmental aspects.

In other words: the fatigue limit isn't really an intrinsic material property, it depends on a great number of specific factors.

### Applications!

So now that we've covered a bit of what mechanics can tell us about fatigue, let's apply it to some cases of interest.The first is: can we estimate the cycle life of a flexible steel trainer under 'normal' use?

Let's make a few assumptions:

- No blade degradation (this is probably the biggest one we'll make) - so no notches, etc.
- The blade is subjected to a varying bending moment, \(M\), rather than an axial force, with pinned end constraints. The maximum applied moment is \(M_{max}\) and the transverse blade deflection reaches a known displacement, \(d_{max}\). Once buckling begins in the pinned-pinned mode (similar to what we saw in my thrust measurements) the blade is essentially undergoing bending so this isn't too crazy of an assumption.
- We're still operating in the linear elastic regime, and deformation locally is small enough that Euler-Bernoulli beam theory is applicable. As described in the buckling post, this also isn't a terrible assumption.

\[\sigma = \frac{My}{I}\]

where \(y\) is the distance from the neutral axis (signed, so above the neutral axis is positive). If we call the maximum distance from the neutral axis \(c\), which also corresponds to the highest axial stress, then the maximum tensile stress can be written as:

\[\sigma_{max} = \frac{Mc}{I}\]

If we put this in terms of the peak bending moment we get:

\[\sigma_{max} = \frac{M_{max}c}{I}\]

To make this a bit more interesting, we can go back to the thrust measurements and make use of some deflection and force measurements.

At the maximum deflection of about 8 cm, this corresponded to the plateau region of about 180 N with my Albion Lichtenauer. This results in a maximum bending moment of about 14.4 N*m. I earlier determined the effective bending stiffness, \(EI\) to be approximately 20 N*m\(^2\). Also recall that for steels, the typical elastic modulus is about 200 GPa. With all of this information, we can derive an effective second moment of area, \(I\) of about 1\(\mathrm{x}10^{-10}\) m\(^4\). If we assume the effective cross-section is a rectangle 12 mm wide, we end up with a height of about 4.5 mm: in the right ballpark for the thickness for the middle of the blade just outside the fuller (the thickest portion of the blade). If we toss these numbers into the above equation we get:

\[\sigma_{max} = \frac{M_{max}c}{I} = \frac{(14.4 \mathrm{N*m})(0.00225 \mathrm{m})}{10^{-10} \mathrm{m}^4} \approx 346 MPa\]

So the maximum tensile stress (also the maximum compressive stress, since we've assumed a section that is symmetric about the neutral axis) is about 346 MPa.

If we assume the Lichtenauer is made of AISI 6150 polished steel (fine tempered martensitic structure. oil quench then temper at ~370 C), the fatigue limit is about 750 MPa [1]: a bit over twice our estimate above (for reference, the ultimate tensile strength of this material is about 1500 MPa, and the yield strength is about 1400 MPa.). So it looks like we may have an essentially infinite lifetime for our trainer!

But if we assume that the surface is machined rather than mirror polished, the corresponding Marin factor for surface finish is about 0.64, meaning the fatigue limit drops by nearly half! So the estimated fatigue limit for our machined 6150 is now only about 480 MPa: significantly closer to the maximum stress value we estimated above, but still above it. However, if we take into account blade damage (stress concentrations), we could end up with local areas of higher stresses (possibly on the order of twice as large or more) than our calculation above. So it's entirely reasonable to suspect that our trainers can end up with a finite lifetime due to material fatigue, while being loaded well below the yield strength. How long, however, will depend on how often it is used, to what level of loading and how significant an effect the stress concentrations have during the cycling.

#### Plastic deformation and fatigue

Let's look at another example: a sword bent plastically. To keep the math easy, let's continue to assume it's the same sword as my last example but now we're loading it significantly more. The yield strength of our AISI 6150 steel is taken as 1400 MPa, so we'll be loading it somewhere above this in order to get plastic deformation. To get an idea of the bending moment associated with that, we can re-arrange the expression we used earlier:\[\sigma_{max} = \frac{Mc}{I}\]

\[M = \frac{I\sigma_{max}}{c}\]

So, plugging in the numbers (assuming a 1400 MPa stress), we end up with a bending moment of about 62 N*m - so at this moment and above, we'd expect plastic deformation to occur. When you start talking about cyclic loads high enough to cause plastic deformation, you generally are looking at the short cycle life regime. As it ends up, using stress here isn't as straightforward as if you consider strain:

\[\frac{\epsilon_p}{2} = \epsilon'_f (N)^c\]

The above is called the

*Manson-Coffin*relation, and relates the plastic strain amplitude (\(\epsilon_p\)) in a component to the number of cycles to failure (N, for our loading case). The remaining parameters, \(\epsilon'_f\) and \(c\) are empirically determined parameters known as the

*fatigue ductility coefficient*and the

*fatigue ductility exponent*respectively. For most common metals, the fatigue ductility coefficient is approximately equal to the

*true strain*at fracture. For metals undergoing time-independent fatigue (load rate independent) the fatigue ductility coefficient is between about -0.5 and -0.7. The true strain at fracture can be determined from the percent elongation (%EL, a tabulated value for materials) as follows:

\[\epsilon_{frac} = \ln \left[1+\frac{\%EL}{100}\right]\]

For our AISI 6150 steel, we'll assume the percent elongation is 9% (a total engineering strain of 0.09), so \(\epsilon_{frac}\) is about 0.086. For simplicity, we'll further assume the fatigue ductility exponent is -0.6.

Well, just as it's easier to talk in terms of strain, it's easier to think of the permanent deformation of a blade in terms of the permanent deflection of the point from its original shape. If we assume the deformation was similar to that of a cantilever beam, then the displacement of the neutral axis can be represented by a quadratic equation of the following form:

\[w(x) = -\delta \left(\frac{x}{l_p}\right)^2\]

Where \(l_p\) is the length projected onto the original neutral axis. If we continue to assume that our trainer is 92.4 cm long, then for a plastic displacement of 3 cm, the projected length is 92.335 (calculated using the arc length of the above parabola). Notice that this is only 0.07 % different: so we're probably OK assuming that the projected length is still 92.4 cm and the cross-section uniform. So our permanent deflection as a function of distance looks like:

\[w(x) = -3.0 \mathrm{cm} \left(\frac{x}{92.4 \mathrm{cm}}\right)^2\]

If we know the displacement, then we can determine the axial strain. Since we are in bending, the strain is linearly distributed along the cross-section, with the strain tensile above the neutral axis and compressive below:

Schematic of deformation due to bending about the neutral axis. |

\[\epsilon = \frac{y}{\rho}\]

Where \(y\) is the distance from the neutral axis (signed) and \(\rho\) is the radius of curvature of the bent section. Formally, the radius of curvature is defined as follows:

\[\rho = \left| \frac{(1+f'(x)^2)^(3/2)}{f''(x)}\right| \]

Since the slope of the beam at any given point is given by \(\frac{dw}{dx}\), and in our range of interest this value is relatively small (much less than 1), we can approximate the radius of curvature as \(\frac{d^2w}{dx^2}\), so our strain at any point on the above section becomes:

\[\epsilon(y) = \frac{2\delta}{l_p^2}y\]

Assuming the projected length is the same as the original length, the symmetry of the problem results in the maximum strain at the maximum distance from the neutral axis, \(c\):

\[\epsilon_{max} = \frac{2\delta}{l_p^2}c\]

Where if we continue to assume that our cross section is 12 mm by 4.5 mm, and recall that we're looking at the plastic deformation only at this point, the maximum plastic tensile strain is about 1.16x\(10^{-4}\). Re-arranging the Manson-Coffin relation to get cycle life, and plugging in numbers we get:

\[N \approx \left(\frac{\epsilon_p}{2\epsilon_{frac}}\right)^\frac{1}{c} = \left(\frac{0.000116}{2*0.086}\right)^{\frac{-1}{0.6}} \approx 1.15\mathrm{x}10^5 \]

So, for this off-the-cuff example, we end up with a cycle life of over a hundred thousand cycles: a surprisingly long number of cycles. But note that the plastic strain is significantly smaller than the elastic strain component which would have a maximum around 0.002 (the typical strain at which ductile materials begin to experience plastic deformation) that would be recovered upon unloading. As a result, this example isn't really in the plastic strain dominated regime.

So what sort of permanent deformation would be required for us to see a plastic strain on the order of the nominal 0.002 elastic strain? Here, we'll have to not assume as much regarding the projected length of the blade and the radius of curvature, so things will get a bit more messy. But to start, let's assume we bend the blade sufficiently that it forms a quarter circle (this would be an extreme of the bending beam case) after it's been unloaded.

In this case, we can assume the arc length of the quarter circle is the length of the blade, \(l\), which is just a quarter of the circumference of the circle. Thus, the radius of curvature is equal to \(4l/2\pi\): the radius of the circle created. Plugging the values into the maximum strain definition, we arrive at:

\[\epsilon_max = \frac{c}{\rho} = \frac{0.00225 m}{.558 m} \approx .0038\]

So our maximum plastic strain would be almost double the elastic strain for this dramatic permanent deformation. Plugging this into the rearranged Manson-Coffin relation, we can obtain a cycle life estimate:

\[N \approx \left(\frac{\epsilon_p}{2\epsilon_{frac}}\right)^\frac{1}{c} = \left(\frac{0.0038}{2*0.086}\right)^{\frac{-1}{0.6}} \approx 575\]

So now we're at 575 cycles under ideal circumstances: a significantly shorter life now that we're well into the plastic deformation dominated regime.

But remember: these calculations are based on phenomenological models for idealized situations (read: they roughly match observation under specific circumstances, and get you in the ball-park for design). This is one of those places, much like equipment design, where both analytical and experimental work are important to ensure safety.

### Some perspective: FIE requirements for weapons

The FIE requires blades be subjected to significant elastic bending for 7000 cycles for epee (18,000 for foil) without breaking or permanently deforming. The test requires that the bend or buckling (either mode is acceptable) result in an*axial*displacement of the point of about 22 cm (about a 25% displacement relative to the length of the weapon). Since it took some serious mental gymnastics to figure out what that meant, and still I couldn't quite get it... I finally just tried it on a foil, here's what that would look like:

22 cm of axial displacement of the point of a non-FIE foil is some serious bend. Also, Tandy doesn't pay me for product placements. |

Let's put these cycle life numbers into a perspective that may make more sense: Let's assume you do 100 identical thrusts per training day, each just like the above. For 1 hour of actual training time that number would correspond to about 2 thrusts every minute on average, probably a fairly liberal estimate for weapons that utilize other attacks. So If you train twice a week, 50 weeks a year (so 100 days of training total per year), a blade with a cycle life of 100,000 cycles means about a 10 year life. Therefore a 10,000 cycle life would mean only 1 year. In other words: a foil or epee fencer training at our assumed level may be looking to replace weapons every year or two to avoid them breaking due to fatigue (based on a quick survey of my modern fencing friends this isn't a terribly outrageous estimate for hard use, but lifetimes vary widely). Interestingly, but as a slide aside, I wasn't able to find an equivalent fatigue specification for sabre blades, though the same modern fencing friends said they tended to fail much more often.

I admit that I'd love to do some fatigue testing of new HES trainers, but that would be even more expensive and time consuming than the drop-testing I've been working on, since it would involve the destructive testing of a new trainer (or at least the full blade). Anyone want to donate new trainers to the cause?

### References

- Baxa, M.S, Chang, Y.A. & Burck, L.H.
*Effects of sodium chloride and shot peening on corrosion fatigue of AISI 6150 Steel.*Metallurgical Transactions A, vol. 9A, 1978.

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