Leibniz intuition on F1 carbon frames

F1 cars, modern racing bicycles, wind turbines, and the latest generation of planes, all have major portions of their bodies made of composites. Those materials are realized by soaking “stronger” fibers (e.g. carbon fibers) in a “weaker” matrix (e.g. epoxy resin), letting then the entire assembly dry and solidify. Years ago, while designing a mechanical component made of carbon fiber, I was struggling with core concepts about such mechanical structures. I had a fair understanding of how a playable cloth made of fibers (figure 1 a/b) could become something rigid and resistant (figure 1 c). However, I was still struggling with mechanical fundamentals. Here was my problem: fibers act as wires, they can therefore withstand only a stretch in the direction of their main axis – we can only stretch a wire. Then, how can an F1-car’s chassis return exceptional performance all-around even though it has fibers usually aligned on only a few main directions (figure 2)? The resin holding the fibers together cannot be doing much of the resisting job. As we will see in a few lines, the resin is usually much weaker than the fibers, weaker also than common mechanical materials like aluminum and steel. The superior performances of composites against virtually any load would then suggest that the resin is continuously serving the fibers rather than offering direct resistance. In particular, it seems the resin is “adjusting” external loads to the fiber’s only axial capability. The intuitive discussion below about how the resin exerts its role offers effective tools to anyone interested in knowing more about composite materials and anybody who is even wondering about practicing with them.

Figure 1

Figure 2: bi-directional carbon twill offering maximum performances along two directions perpendicular to each other

Important note: we will reference resins of the epoxy type and reinforcing fibers made of carbon; common choices in modern applications. Here are also a few technical references we will leverage below for our technical intuition. The Hooke’s Law:

Tension  = Elasticity Modulus x Stretch

Meaning: when we stretch materials we obtain a resisting tension determined by the Elasticity Modulus of the material, therefore specific to it (i.e. stretching a 1 m steel bar of 1 mm takes a bigger effort than stretching a 1 m soft-rubber bar of the same amount). With that in mind, we can point out that epoxy resin usually has mechanical properties about 2 orders of magnitude below carbon fiber. We give here some numbers just for reference (MPa below means N/mm^2 or, newton over squared millimeter):

• Resin’s maximum possible tension (strength) = 30 MPa in traction / 100 Mpa in compression
• Resin’s elastic module (rigidity) = 2.5 GPa
• Carbon fibers’ maximum possible tension (strength) = 3,500-4,150 MPa in traction / almost null in compression
• Carbon fibers’ axial elastic module (rigidity) = 250-350 GPa
• Steel’s strength is about one order of magnitude below carbon fibers
• Steel’s rigidity comparable to carbon fibers

The exact way the resin “adjusts” any load for the fibers within a composite assembly can be mathematically analyzed. However, that usually requires complex mathematical matrices originally built by some mathematical genius with a name from Eastern Europe. This article wants to provide an intuitive explanation possibly functioning as a rapid mental reference any time we need to make design decisions to be then refined through more exact math. Common answers trying to provide such an intuition about composite structures sound like the following: “the role of the resin is to spread the load across the fibers”. That sounds good, though, say a load is transversal to the fibers, the role of the resin cannot be a mere dilution of the load across multiple fibers, it must be something a bit more complex.

In my quest for a better understanding of the topic, I was finally put in the direction of what would become my intuitive answer by something unrelated to composite materials. That topic was the Leibniz catenary, the mathematical formulation of the shape adopted by a hanging wire fixed at its extremities (figure 3). A catenary has deep fundamentals, for example, it is related to the shape that a water-drop adopts to minimize its surface energy – if wondering, it is different from the segment of a circle. For sake of this article, rather than discussing further the catenary, we will directly give the intuition related to composites. It was the importance of the boundary conditions in solving Leibniz’s differential equation of a catenary that helped me gain my mental framework about composite assemblies. Here follows the result.

Figure 3: example of a catenary shape

How resin “spreads” the load across the fibers

Referencing figure 4, we can model a part of a bigger carbon-fiber frame (e.g. F1 or bicycle frame) through a sandwich with one layer of resin, a layer of carbon fibers (aligned horizontally in the picture), and another layer of resin.

Figure 4: composite sandwich with carbon fibers aligned horizontally in the picture

Considering the left end of the sandwich as fixed because clamped to the rest of the bigger frame, a transversal load at the right end would cause the sandwich to bend and stretch (figure 5). We should interpret figure 5 within the realm of small deformations: the sandwich is stretched while remaining virtually horizontal; both a longitudinal tension because of the stretch and vertical shear stress pushing down the assembly would arise. For our simplified discussion, we will deal only with the former (i.e. tension due to stretching), but the reasoning would apply to the shear stress too and in a different way to some compression.

Figure 5

Let us simplify the model further: we can here neglect the different elongation that figure 5 would suggest for the three layers because of the different distances from the central neutral axis – indeed, for the discussion below the load could be even thought of as applied normally to the plane of figure 3. If wondering, yes, we are neglecting also the interaction stretch/compression. Therefore, referencing the Hooke’s law (reported again immediately below) the resisting tension of each layer would differ only through the modulus of elasticity:

Tension = Elasticity Modulus x Elongation

As already mentioned in the initial note, the resin has an elasticity modulus of about two orders of magnitude below the fibers’ one. Its resisting tension would be similarly lower. Therefore, as soon as the sandwich is stretched a given amount, similarly to a stronger teammate in the game of rope pulling, the fibers are the ones providing most of the resisting tension. That high resisting action is then passed back to the external load by the matrix. Critical to the dynamic is that the resin offers at any given elongation a lower resistance as if it was willing to stretch more than the fibers under the same external load. Here is then our first critical takeaway on the intuition of composites:

Takeaway 1: The resin “transfers” external loads to the stronger fibers by being always a step behind in terms of resisting tension (figure 6).

Figure 6: the resin yields more than the fibers under the external load, therefore it tends to pull the fibers transferring them the external load

While the discussion may seem coherent so far, a problem now arises. Agreed, we need weaker matrices to let the fibers perform most of the resistance. Moreover, figures 5 and 6 give an intuition on how a transversal load, impossible to handle for naked fibers through a small deformation, can be resisted when the fibers are embedded within the solid resin. The resin is to some extent reorienting the load along the fibers allowing them to function axially. However, takeaway 1 implies that any load from the external environment and any reaction from the fibers pass back and forth through the matrix. Having the latter mechanical capabilities about two orders of magnitude below the fibers’ ones, that should constitute a limit to the entire assembly? In other words: the performance of a chain is determined by the weaker link, composites should then have overall performances comparable to the resin’s ones. Real applications show something different: composite parts have higher performances comparable to the fibers’ ones. We can solve the mystery through our second critical takeaway:

Takeaway 2: The external load, once transferred to the fibers from the resin as an axial stretch, is resisted across a relatively small section, the cross-section of fibers. That generates a high tension (i.e. force/area) which can be resisted thanks to the fibers’ max allowed tension as high as 4150Mpa (stated initially). Conversely, once the same external load and resistance pass back and forth through the resin, they are distributed across a wider surface constituted by the entire lateral surface of contact between the fibers and the resin. The interaction generates a lower tension, being the same load or resistance spread (i.e. divided) over a bigger surface.

While not rigorous, we can give an idea of how that dilution of the load happens while passing from the fibers to the resin; we will also then comment on the limitation of the calculation.

Having the resin a maximum allowed tension in the order of 30 Mpa (stated initially), we would need a resin's surface about x100 times bigger than the cross-section of the fibers for a coherent transfer of the tensions. That is because the fibers’ maximum allowed tension is about 4150 MPa, 100 times bigger. To have the two values work together, we need to divide the same load over a surface 100 times bigger to have the resulting stress 100 times lower. That sounds feasible considering that usually, composites are a few millimeters thick, while their length can be measured in centimeters or meters. Here follows an example.

Fibers’ cross-section, related to the thickness of the part of a few millimeters:

• Cross-section area of a cylindrical fiber: A = phi x r^2 -> ex. 1.5 mm^2 = 2.25 mm^2

Surface area between the resin and the fibers, related to the centimeter or meter length:

• Lateral surface area of a fiber (contact with the resin): fibers’ perimeter x length = 2 x phi x r x L -> [mm x cm or mm x m] -> ex. 1.5 mm x 100 mm (i.e. 10 cm) = 150 mm^2 or 1.5 mm x 1,000 mm (i.e. 1 m) = 1,500 mm^2

We can see above that we can easily reach values of x100 or x1,000 between the two areas. As we anticipated above, while what we have just quantified could serve well our intuition, it has its limitation. The stress which is axial within the fibers is then transferred from and to the resin as a combination of traction (or compression) and shear stress, respectively normal and parallel to the surface of contact with the fibers. Those two components should be treated separately based on the particular configuration assumed by the assembly. The needed surface could result in being scaled up or down. A detailed discussion on continuum mechanics is beyond the objective of this article. Our main point is to show that the ideas we are discussing are coherent among them; we are trying to build intuition based on sound engineering principles, not on personal “guesses”.

Important note: usually, frames have hollow cross-sections (e.g. round or oval tubes). Sometimes they could be considered even big hollow tabs (e.g. F1 car’s chassis). That geometry allows carbon assemblies to leverage additional principles of mechanics. However, the discussion presented here stays and constitutes the core idea for the interaction between the resin and the fibers.

Conclusion

The dynamics above do not imply that the orientations of the fibers are not important because the resin can then just transform any load in an axial one to the fibers. The Hooke’s law we leveraged tells us something critical about materials: no resisting force is possible if there is no deformation at all. The bones within our legs would not be able to sustain us if they were not experiencing a little shortening under our weight – that deformation is negative because it has to resist a compressive load. However, we also want to usually minimize the needed overall deformation of the assembly. That is because deformation in excess generates dissipation – think about a sprinting cyclist wasting part of its energy to bend the bicycle frame rather than to move forward. Therefore, the optimal orientation of the fibers is in general along the direction of the main stresses. A bending beam would indeed have major axial stretch and compression as per our model, while a transmission shaft would experience them at 45 degrees compared to the longitudinal axis – torque is usually transmitted through tensions spiraling around the longitudinal axis at 45 degrees. That correct positioning of the fibers would allow the deformation of the material to immediately load the resistance of the material through the Hooke’s law (the fibers in our case), and no additional deformation would be needed.

While nowadays engineering disposes of incredible tools like software able to figure the detailed profile of stresses within a composite structure, we should still always strive to reach an intuitive understanding of the fundamental dynamics. That intuition is what can guide us during the R&D phase towards solutions that can be then refined through those modern tools. Moreover, correct intuition is critical in questioning the results from the software, which sometimes interprets the model not as intended by the engineer. Being able to spot incoherence requires a correct understanding of the underlying dynamics.

To conclude by going back to the Leibniz catenary, the resin could be thought of as a set of new boundary conditions for the fibers. Those new conditions allow the fibers to do the resisting job entirely, not only in the axial portion they could perform as naked fibers.

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If interested in sharing additional or different ideas or in discussing some project related to these concepts, please feel free to reach out at riccardo[at]m-odi.com or to connect on LinkedIn.