This post about bicycle frame stiffness is the first in a series of articles that aim to debunk and clarify some of the myths surrounding modern bicycle design. Taking a popular-science approach, we'll go into the nitty-gritty details of mechanical engineering and materials science. Yes, that means that there will be equations!

While I think that hard facts and numbers are needed to tackle some of these myths, I understand that equations aren't for everyone. This is why you can choose how you would like to read this article: press the button below, and all equations will disappear. You will still get all relevant information, but you may need to trust me on the numbers. If you want to go through the calculations though, simply keep reading!

## It's stiffer... yet again!

Writing this article in 2018, it feels almost retrospective to pick up the topic of stiffness in bicycle frame design. You may remember how *stiff* became the new *lightweight *a couple of years ago, but then *aero* became the new *stiff *somewhere around 2016, and now it almost looks as if *lightweight* is going to go full circle and become the new *aero*.

By the laws of bicycle marketing, this means that we're still some time away from *stiff* becoming the new *lightweight* (yet again), but trust me, it's going to happen. And this time, you want to be prepared!

## What's stiffness?

Practically speaking, stiffness is a measure of how much an object is going to deform if you compress, stretch, bend, shear or twist it. Therefore, stiffness is measured in units of load per deformation.

Risking to bring back bad memories of boring physics classes, let's have a look at the example of a simple spring. If you take both ends of the spring in your hands and start to pull with a force *F*, the spring will lengthen. The more you pull, the longer it will get; the longer it gets, the more you'll have to pull to make it even longer. And once you let go of it, it will go back to its original length (unless you pull way too hard, but we'll talk about that next time).

To calculate the stiffness $k$ of the spring you would devide the pulling force $F$ by the change in length $\Delta L$:

$$ k = {F \over \Delta L} $$

Thus, in this case, the unit of stiffness is Newton per meter.

Now consider a round, thin-walled steel tube as you would typically have it in a bicycle. If you take both ends of the tube in your hands and start to pull, the tube will lengthen ever so slightly. You won't be able to see it by eye, but the more you pull, the longer it will get; the longer it gets, the more you'll have to pull to make it even longer. And once you let go of it, it will go back to its original length.

Thus, both the spring and tube behave in similar ways; however, their stiffness is very different.

This brings us to our first definition of stiffness: To say that two objects have the same stiffness is to say that, when you press or pull on the objects with a given force, the change in length will be the same.

## Let's make it stiffer!

Now imagine we are cyclists in the summer of 2012. Wiggo just won the Tour de France on his Pinarello Dogma, and we're all on the verge of becoming stiffness enthusiasts. What can we do to stiffen our tube?

We could either increase the wall thickness of the tube (essentially have a tube with more material), or we could change the material to a material that's even more resistant to elastic deformation than steel.

The fact that we have two options here is very important since it highlights that stiffness is not a property of shape *or* material; it's a property of shape *and* material.

Thus, saying things like "carbon fiber is stiffer than steel" makes about as much sense as saying that pop music is louder than reggae. It's not just the type of music – it's also how you play it!

## Stiffness and materials

That being said, I hope it's clear that there's no such thing as a stiff material. However, different materials have different *resistances to elastic deformation*.

Simply speaking, we would say that a deformation is *elastic* if it's fully reversible once we take away the load (think of a spring). In contrast, we would say that a deformation is *plastic* if we apply so much load that the shape of the object changes permanently (think of stepping on a soda can).

In engineering, a material's resistance to elastic deformation can be quantified by its ** Young's modulus**, also referred to as

*modulus of elasticity*. It's a physical property that tells you how much force per unit surface area you need to apply in order to (elastically) stretch a solid rod of material by a given length.

You can determine the Young's modulus of a material if you perform the above pulling-experiment with a number of tubes of different material, but identical shape: a steel tube will be more resistant to stretching than a rubber tube, and a glass tube will be more resistant to stretching than a wooden tube. Thus, steel has a higher Young's modulus than rubber, and glass has a higher Young's modulus than wood.

The Young's modulus $E$ of a material is defined as:

$$ E = {\sigma(\epsilon) \over \epsilon} = {F/A \over \Delta L / L_0} = {F L_0 \over A \Delta L},$$

where $\sigma$ is the uniaxial force $F$ per unit area $A$ (= the normal stress), $\epsilon$ is the dimensionless change in length in response to the stress, $L_0$ is the initial length, and $\Delta L$ is the change in length.

The unit of $E$ is Newton per meter square, also referred to as *Pascal* (Pa). Since $E$ is rather large for most engineering materials, it is often more convenient to use units of megapascal (MPa, which is the same as Newton per millimeter square) or gigapascal (GPa) instead.

## On alloys and superalloys

In the following, we will focus on the three most common metal alloys used for bicycle design: steel alloys, titanium alloys, and aluminium alloys.

Firstly, note that all of these materials are **alloys** (we'll talk more about that next time). Secondly, please also tell that to the salesperson in your local bike shop who's always trying to sell you an "alloy" frame.

Now here's where it becomes interesting: Contrary to popular belief, steel has by far the highest Young's modulus out of these three alloys. Titanium alloys, despite their high-performance ("superalloy") image, come in second place, with a Young's modulus about 50 % lower than that of steel. Finally, aluminium alloys rank third, with about one-third of the Young's modulus of steel.

This means that if you compare bicycle tubes with the same diameter, wall thickness and length, the tensile stiffness of a steel tube will be about two times that of a tube made from a titanium alloy and about three times that of the aluminium alloy version.

If that isn't fun-fact-trivia gold for your next club ride... !

To provide some numbers, the Young's moduli of these materials are:

- steel ~200 GPa
- titanium alloys ~110 GPa
- aluminium alloys ~69 GPa

There are many different types of steel alloys, titanium alloys, and aluminium alloys, and adding small amounts of alloying elements to these materials can have a large effect on some of their physical properties. However, it has (practically) no effect on their Young's modulus.

For example, the Young's modulus of a simple hardware-store-grade mild steel alloy is essentially the same as that of a high-performance steel alloy used for gas turbine blades.

"Then what's so great about high-performance alloys anyway?", you may ask. We'll get to that next time!

## Stiff and lightweight

As we said above, looking at the material side of stiffness only gives us half of the story. For a fair comparison, we also need to look at the shape of the tube.

Assuming that we want to end up with three tubes of equal stiffness, we could now either reduce the wall thickness of the steel tube (to make it less stiff), or we could increase the wall thickness of the tubes made from titanium and aluminium alloys (to make them stiffer).

As cyclists, we know that there's really only one option here: let's make it stiffer!

As we said before, two objects have the same stiffness if the elastic displacement is the same for a given load. Using the above equations, we can write the change in length as follows:

$$ \Delta L = {F L_0 \over A E}$$

If we want to get the same change in length for a steel tube (st) as for a tube made from a titanium alloy (ti), we can say that:

$$ {F_{st} L_{0,st} \over A_{st} E_{st}} \stackrel{!}{=} {F_{ti} L_{0,ti} \over A_{ti} E_{ti}} $$

Given that the force and initial length are the same in both cases, this can be simplified to:

$$ {A_{ti}} = A_{st} {E_{st} \over E_{ti}} $$

This gives us:

$$ {A_{ti}} = A_{st} {200 \ \text{GPa} \over 110 \ \text{GPa}} \approx 1.8 \cdot A_{st} $$

Similarly, in the case of the aluminium alloy tube (al) we find that:

$$ {A_{al}} = A_{st} {200 \ \text{GPa} \over 69 \ \text{GPa}} \approx 2.9 \cdot A_{st} $$

To do so, we need to compensate for the lower Young's modulus of the titanium and aluminium alloys by increasing the tube wall thickness. In fact, we would need to (almost) double the cross-section area of the titanium alloy tube, and we would need to (almost) triple the cross-section area of the aluminium alloy tube. For equal stiffness, the figure below shows a comparison of the cross-sections of our three tubes.

Now here comes the interesting part: Obviously, increasing the wall thickness will increase the volume of the tube; however, the densities of the three alloys are almost perfectly correlated with their Young's moduli: the lower the density, the lower the Young's modulus.

More precisely, the density of titanium alloys is about half that of steel, and the density of aluminium alloys is about one third that of steel. Sounds familiar?

This means that, after increasing the wall thickness of our titanium and aluminium alloy tubes, not only do we now have three tubes with equal stiffness, but we also have three tubes with (essentially) the same mass. You can't cheat the universe!

To calculate the mass $m$ of a tube we multiply the cross-sectional area, the length and the density $\rho$ of the material:

$$m = A \cdot L_0 \cdot \rho,$$

Given that the length of all tubes is the same, we can say that:

$$m \sim A \cdot \rho$$

The density of steel is around 7.8 g/cm^{3}; most titanium and aluminium alloys have densities of 4.5 and 2.7 g/cm^{3} respectively. Together with our results from above, we then find that:

$${m_{ti} \over m_{st}} = {A_{ti} \cdot \rho_{ti} \over A_{st} \cdot \rho_{st}} = {1.8 \cdot A_{st} \cdot 4.5 \ \text{g/cm}^3 \over A_{st} \cdot 7.8 \ \text{g/cm}^3} = {1.8 \cdot 4.5 \over 7.8} = 1.04$$

Thus, for the same stiffness, the titanium alloy tube is 4 % *heavier* than the steel tube. Following the same approach, it can be shown that the weight difference between the aluminim alloy tube and the steel tube is smaller than 1 %.

Take that, popular belief!

To sum up: In our attempt to follow in the footsteps of Bradley Wiggins, we set out to build a lightweight and stiff bike. Thus, we exchanged our low-quality, hardware-store-grade steel tubing for aerospace-approved titanium and aluminium alloy tubes, spent hours typing equations (literally!), and tried to remember high school physics and undergrad engineering. Yet, in the end, we're left with a tube set that's just as heavy and stiff as what we started out with!

"What's the point?", I hear you ask. And rightfully so!

## How to cheat the universe

To answer this questions, we need to go back to our humble steel tube. In fact, we need to have a look at its history.

Traditionally, bicycles are made from steel tubing because steel tubes are easily available and cheap to manufacture. In addition, they provide plenty of stiffness and structural integrity to the bike.

To save weight, it makes sense to reduce the wall thickness of the tube as much as possible; however, there are some constraints: tubes with very thin walls are more difficult to manufacture. Also, tubes become increasingly difficult to weld and more prone to denting as wall thickness decreases.

Just to give you an example: a modern steel bicycle may have a downtube wall thickness somewhere around 1 mm, and a tube diameter of 35 mm. This allows for easy manufacturing and handling while ensuring structural integrity.

As we said before, if we want to switch from steel to a titanium alloy, we need to add material to the tube. So far, we chose to increase the wall thickness and keep the outer diameter constant. For a tube of 35 mm diameter, we would now have a wall thickness of 1.85 mm in case of the titanium alloy.

However, what if instead we keep the wall thickness constant and use the extra material to increase the tube diameter? In this case, we could have a titanium alloy tube with 1 mm wall thickness, and a tube outer diameter of 62.2 mm. For the aluminium alloy tube, we could increase the diameter to 99.6 mm and still have a wall thickness of 1 mm.

To calculate the wall outer diameter of a titanium alloy tube with the same wall thickness and tensile stiffness as a steel tube with 35 mm diameter and 1 mm wall thickness, we know that we need to increase the surface area of the original tube by a factor of 1.8. Thus, we can write that:

$$ A_{ti} \stackrel{!}{=} 1.8 \cdot A_{st} $$

Using the equation for the area of a ring with outer radius $r_o$ and inner radius $r_i$, we can say that:

$$ A = ({r_o^2 - r_i^2}) \cdot \pi $$

Inserting this in the above equation, we find that:

$$ ({r_{o,ti}^2 - r_{i,ti}^2}) \cdot \pi \stackrel{!}{=} 1.8 \cdot ({r_{o,st}^2 - r_{i,st}^2}) \cdot \pi $$

The target wall thickness $t$ can be expressed as:

$$ t_{st} = r_{o,st} - r_{i,st} ,$$

Now we want to find the outer diameter $r_o$ of the titanium tube for which the wall thickness and overall stiffness is the same as for the steel tube. Expressing the inner diameter of the titanium tube using the target wall thickness, we can write:

$$ ({r_{o,ti}^2 - (r_{o,ti} - t_{st})^2}) \cdot \pi \stackrel{!}{=} 1.8 \cdot ({r_{o,st}^2 - r_{i,st}^2}) \cdot \pi $$

Finally, we solve the equation for $r_{o,ti}$ and get rid of $\pi$ on both sides

$$r_{o,ti} = {1.8 \cdot (r_{o,st}^2 - r_{i,st}^2) + t^2 \over 2t} $$

Plugging in our numbers, we find that:

$$r_{o,ti} = {1.8 \cdot ((17.5 \ \text{mm})^2 - (16.5\ \text{mm})^2) + (1 \ \text{mm})^2 \over 2\cdot 1 \ \text{mm}} = 31.1 \ \text{mm} $$

Thus, the outer diameter of the tube is 62.2 mm. Following the same approach for an aluminium alloy tube, we find that:

$$r_{o,al} = {2.9 \cdot ((17.5 \ \text{mm})^2 - (16.5\ \text{mm})^2) + (1 \ \text{mm})^2 \over 2\cdot 1 \ \text{mm}} = 49.8 \ \text{mm}$$

Therefore, the outer diameter of the aluminium alloy tube is 99.6 mm.

Obviously, increasing the diameters of the tubes has no influence on their stiffness in tension or compression: the tubes still have the same cross-section area and they are still made from the same material. However, as we increased the diameter, we increased the bending stiffness and torsional stiffness!

## Bending and torsional stiffness

Let's start with bending stiffness. When we talk about bending here, it's important to note that we're *not* referring to the sort of bending that will have a *lasting* effect on the shape of the tube. As for our spring, we're interested in the kind of bending that is fully reversible, i.e., we're talking about *elastic* bending.

As we said earlier, stiffness is measured in units of load per deformation. While stretching (tension and compression) is a deformation caused by a force $F$ (see above), bending occurs due to what is referred to as a *bending moment* $M$. Thus, the unit of bending stiffness is Newton-meter per meter.

Intuitively, we would expect that a tube with a larger diameter is harder to bend than a tube with a smaller diameter. But by how much exactly?

To calculate the bending stiffness of a tube with a given length, we need to know two things: the Young's modulus of the tube material (hello again), and something that is called the ** second moment of area** of the cross-section. The bending stiffness is then directly proportional to the product of these two parameters.

In cartesian coordinates ($x-y$), the second moment of area $I_x$ with respect to the $x$-axis can be calculated for an arbitrary area shape $R$ as:

$$ I_x = {\iint\limits_{R}} y^2 \ \text{d}x \ \text{d}y $$

Now this looks awfully complicated, but for our case of a circular tube with inner diameter $r_1$ and outer diameter $r_2$, integrating the above equation will yield that:

$$I_x = {(r_2^4 - r_1^4) \pi \over 4} $$

Without derivation, it can be shown that the bending stiffness of a tube is directly proportional to its Young's modulus and its second moment of area:

$$ k_{bend} \sim E \cdot I_x $$

(Image by Wikipedia user IngenieroLoco. Licensed under CC BY-SA 4.0, from Wikimedia Commons.)

While the Young's modulus of the material is obviously independent of the tube shape, the second moment of area is not: if we double the diameter of our steel tube and keep the wall thickness constant, its second moment of area will increase by a factor of eight – and so will the bending stiffness.

As a rule of thumb, the further the tube material is away from the dot-dashed line in the illustration above (also referred to as the *neutral axis* of the tube), the higher the bending stiffness. That's also the reason why beams with $I$-shaped cross-sections are used to build skyscrapers.

Going back to our optimized titanium alloy tube with diameter 62.2 mm and wall thickness 1 mm, we can now see that its bending stiffness is about three times larger than that of the original steel tube; for the aluminium alloy tube with 99.6 mm diameter it's a factor of eight!

Similar behavior can be observed for a tube that is subjected to a torsional load. Just as the bending stiffness, the torsional stiffness is directly proportional to the second moment of area. However, instead of the Young's modulus, we would now need to consider something that is called the ** shear modulus**. For metals, this physical property is in good approximation proportional to the Young's modulus.

Thus, for our purposes, we can say that if you increase the tube diameter, both the bending stiffness and torsional stiffness will increase disproportionately.

We said that the bending stiffness is directly proportional to $E$ and $I$:

$$ k_{bend} \sim E \cdot I $$

Thus, to compare the bending stiffness of the steel tube and the titanium tube, we can say that:

$$ {k_{bend,ti} \over k_{bend,st}} = {{E_{ti} \cdot I_{ti}} \over {E_{st} \cdot I_{st}}} = {E_{ti} \over E_{st}} \cdot {{(r_{o,ti}^4 - r_{i,ti}^4) \pi \over 4} \over {(r_{o,st}^4 - r_{i,st}^4) \pi \over 4} } = {E_{ti} \over E_{st}} \cdot {{r_{o,ti}^4 - r_{i,ti}^4} \over {r_{o,st}^4 - r_{i,st}^4 }} $$

Plugging in some numbers, we find that:

$$ {k_{bend,ti} \over k_{bend,st}} = {110 \ \text{GPa} \over 200 \ \text{GPa} } \cdot {{(31.1 \ \text{mm} )^4 - (30.1 \ \text{mm} )^4} \over {(17.5 \ \text{mm} )^4 - (16.5 \ \text{mm} )^4}} = 3.2$$

Similarly, for the aluminium tube we find that:

$$ {k_{bend,al} \over k_{bend,st}} = {69 \ \text{GPa} \over 200 \ \text{GPa} } \cdot {{(49.8 \ \text{mm} )^4 - (48.8 \ \text{mm} )^4} \over {(17.5 \ \text{mm} )^4 - (16.5 \ \text{mm} )^4}} = 8.4$$

## Stiffness isn't everything

Many road cyclists will intuitively relate to the above findings since cycling wisdom has it that aluminium alloy bikes are way too stiff and harsh, steel bikes aren't stiff and harsh enough, and titanium bikes are just right. Thus, good frame design is about more than just maximizing tube diameters.

That being said, the theoretical tube diameters for our aluminium and titanium alloy tubes are not very practical anyways: You can't have a downtube diameter of 99.6 mm if your bottom bracket is only 68 mm wide. (Well, I suppose you could, but it's going to look odd.)

Also, our titanium and aluminium alloy tubes don't offer any weight savings yet – remember that we added material to compensate for the lower Young's modulus –, so probably we would rather take away some material than try to maximize the bending stiffness. (We'll talk about how much material we can take away next time!)

## From stiff tubes to stiff frames

Of course, when we look at a frame rather than a single tube, things become a little more complicated: the overall stiffness of a frame results from its stiffness in tension, compression, bending, torsion and shear.

If we look at a frame from the side, it's easy to see that it is essentially comprised of two large triangles (thanks to UCI regulations, but mostly for structural reasons).

Now imagine yourself riding along a deserted mountain pass as you hit a bump in the road with your front wheel. This will create an upward force in the contact point between the tire and the road. The force then travels through your spokes, front hub, and fork all the way up to the headtube.

Here, the force will be countered by the lower headset bearing. As a result, the lower end of the headtube will try to move up and rotate in a counterclockwise direction. To counteract this motion, the downtube will need to step in and hold the lower end of the headtube in place. To do so, it needs to sustain a tensile force.

Similarly, the toptube tries to hold the upper end of the headtube in place; thus, it needs to sustain a compressive force. These tensile and compressive forces then travel further through the frame, where they are again countered by tensile and compressive forces in the seatstays and chainstays.

Yes, this is a crude simplification of what will actually happen. But for the sake of argument, let's keep it simple!

While it's a little more complicated in reality, the takeaway message is that bicycles are designed so that most in-plane loads are sustained in tension and compression, rather than bending and torsion. And as we saw earlier, the tensile and compressive stiffness of a steel tube – for the same tube weight – is more or less the same as for a titanium tube.

However, things change once we're looking at out-of-plane loading: when you get out of the saddle and grip your handlebars tightly, rocking heavily from side to side as you stomp on the pedals, you're essentially trying to bend and twist the frame around an axis that runs somewhere from the center of the seattube to the center of the headtube. For this style of riding, bending and torsional stiffness matters!

## What's next?

So far, we saw that no matter what tube shape or material we choose, the tensile and compressive stiffness won't change much. However, we found that the bending and torsional stiffness of a bicycle frame can be increased if we find a way to increase the tube diameters.

Yet, as of now, our new set of shiny, large-diameter tubes provides no weight saving compared to our initial steel design (remember the correlation between the density and the Young's modulus?). So how can it be that aluminium and titanium alloy bikes are usually lighter than steel frames?

Also, if all steel alloys have the same Young's modulus, shouldn't those ultra-thin-tubed, Italian-made steel frames from the early 90s be less stiff than regular steel frames? (Yes) Then why do people always say that those frames are stiffer?

And finally, addressing a matter of life and death: if you're aiming for a bullet-proof design, should you follow the advice of **Sia and David Guetta** and go for titanium?

We'll find out next time!

Frans BonnevierDid not see the cliffhanger coming! I’m looking forward to the next post.

MikaelThx for the information. Can you please compare some facts with carbon frames? How much would a rider feel that difference?

Moritz PlossI contemplated whether I should include carbon fiber composite materials, but ultimately decided not to. the reason is that this article was never meant to be a comparison of different frame materials, but rather an overview of the fundamental engineering principles that apply to any frame design and material. in a way, it’s not even complete yet since the above is mostly about tubes, and not complete frames, so there’s a lot more to be said.

to answer your question, all of the above is valid for carbon fiber composite materials as well; the only difference is that the Young’s modulus for composite materials can be dependent on the direction of loading. whether or not you will feel a difference will very much depend on your style of riding and the overall frame geometry. the easiest (and only) way to answer that question is for you to go for a test ride on a bunch of different bikes (feel free to get in touch!).

I wouldn’t want to make a general statement here, and I certainly wouldn’t want to say that one type of frame is always stiffer than another (for example, from the above equations it follows that the length of the tubes can have a far greater influence on stiffness than the material). it depends, and that’s sort of the point of this article: there’s no short answer. thus, I wouldn’t recommend to use this article to inform a buying decision for a new bike; rather, it can help you to understand why two bikes that are very similar in design may feel slightly different.