kamtsa

Do clincher rims really compress when the tire is inflated?

I would think that the force that the inner part of the tube applies on the rim is balanced with the force of the outer part of the tube which pulls the rim out via the beads.

This is different from the compression applied on a submarine since there is no pulling force. (compression in a submarine is very visible, if you tie a string from side to side and dive, you notice how the string sags).

Kam

DannoXYZ 

Fluid-pressure in a closed vessel is distributed evenly in all directions. Air and liquids all have fluid properties.

In the submarine example, you can replace the string with a long spoke and tighten it. Under compression forces, the submarine will still compress and the spoke will go slack.

In the wheel/tyre example, the air-pressure pushes on all the inside surfaces evenly in all directions. You can calculate the inward pressure on the rim by measuring the surface-area in contact with tube and multiplying by the pressure.

This is also why a larger tyre at the same pressure exerts more outward pushing force on the rim's edges; it has more surface-area that's being pushed outward by the air-pressure.


Q: When an Irish car-bomb goes off, does the ground feel any pressure?

kamtsa

You ignored the argument about the outer pull. Since the same pressure is applied on the outer surface of the tube, the tire beads pull the rim out, balancing the compression force from the inner part of the tube.

Any ME or physics type cares to comment about this analysis?

Thanks,

Kam

Mike Mills

I would have thought the inward component was the lateral cross sectional area times the tire pressure, not thet "surface area in contact with the tube". Maybe this is a nit.

Mike Mills

I'm not sure the outward pressure negates the inward pressure. I'm not sure but will think about it some more.

Here's a counter argument - the pressure ensures there is adequate frictional force to hold the tire to the rim. The internal pressure is reacted by a hoop stress in the cords of the tire, that gets reacted at the beads by a frictional force.

We need a free body diagram and a real world test to calibrate the model. Anyone have a tension gage? Measure tension defalted and inflated. Does it change?

Kimmo 

The tyre does something to contain the pressure, and translates it to what I bet is mostly lateral force from the beads, so the rim only experiences spreading of the braking surfaces and radial compression from tyre pressure.

The beads prevent radial expansion, and if they didn't, the tyre would blow off.

DannoXYZ 

The tyre bead pushes the rim-edges out laterally sideways. It does not pull it away from the hub.

kamtsa

Here is a figure that demonstrates the pulling theory. The pressure pushes the tire out ("OUT" arrow) and this creates a pulling vector on the beads in the same direction (red arrows).

Kam

wheel_001.png

Kimmo 

What it doesn't show is the laterally tensile resistance of the tyre, leaving a net force between the beads. The beads may stretch very slightly under pressure, but they're meant to maintain an exact diameter.

noglider 

kamtsa, I believe DannoXYZ is right. The red arrows should point out to the sides, because air pressure inside the tire goes out from the center along radii. I don't think the tire construction can change the direction of the pressure. I'm saying this based on what DannoXYZ wrote and my intuition, so I could be wrong, as I don't have any ME background.

dit

I am no engineer or physicist (sic ) but since gas acts as a liquid I would think the the pressure in the tube would push in all directions with exactly the same pressure.

Interesting discussion nontheless.

Kimmo 

Consider what happens if you inflate the tyre off the rim: the tube bulges inwards as the tyre contains the pressure radially, but the beads are also forced apart.

DannoXYZ 

Yes, let's first review our Bernoulli and Pascal's principles in relation to gases. Then apply these basic principles to the bicycle-wheel example by examining each component individually and build up to a complete wheel. At the core of the wheel is the tube and the enclosed gas. Let's say we are magically able to straighten out a tube into a log and suspend it in mid-air by its geometric centre. Then we pump in air:



Outside atmospheric pressure P1 is roughly 14.5psi absolute. The pressure you see on the pump's gauge is guage-pressure, gpsi or pressure above atmospheric. When you make the 1st pump into the tube, you increase the number of gas molecules inside the same volume and increase its pressure to P2. Since P2 > P1, there is a net force inside the tube pushing outwards in all directions evenly. This force expands the tube evenly in all directions around its centre (assuming rubber-thickness is even).

At some point, the expansion will stop where internal-pressure P3 is the same as outside-pressure; the sum of atmopsheric pressure plus contraction force from the rubber. Since the tube is so thin and so weak, the first couple of pumps will expand the tube to easily 50% larger in diameter with minimal increase in internal pressure. In this illustration and in the vast majority of cases with a bare inner-tube being inflated, P3 is roughly equal to P1. And the change in volume is roughly equal to the extra volume of air pumped in.

Next up, enclosing the tube... partially...

Mike Mills

I think hoop stress is a better way to explain it. Hoop stress is a tensile stress which is tangent to the pressure vessel surface at all locations. In this case, the the pressure vessel is the tire casing. In the diagram, I have resolved the hoop stress (red arrow pointing upwards at an angle away from rim) into two components:

1. a force acting radially outward on the bead (red arrow pointing up)
2. A force acting laterally on the rim (red arrow pointing left/right)

I stole kamtsa's diagram and modified it. At the rim-tire interface, the hoop stress (red arrow pointing upwards at an angle away from rim) is trying to pull the bead off the rim. In order to do so, it has to stretch the bead enough to get it over the rim's edge. This radial outward force is resisted by the internal stress in the bead (the green arrow pointing down). Luckily, it has to do this at all locations simultaneously (you have to stretch the diameter of the bead to the full outer diameter of the rim, all at the same time. Luckily, the tire designer makes the bead stiff enough to prevent this much elongation under normal internal pressures.

The hoop stress also acts on the rim, trying to bow the sidewalls outward. This force is reacted by the internal stresses developed within the rim sidewalls (purple arrow pointing inwards).

There is also some force exerted on the bead which counter-acts some of the hoop stress in the tire casing. This is the internal pressure times the area of the casing below the hook on the rim. I believe this is small but I could be wrong. I did not show this on my free body diagram. Still, the main effect is the bead cannot be stretched in OD enough for it to leave the rim.

When all these forces are in equilibrium, the tire stays mounted to the rim.

This kind of analysis/perspective also explains why it is oh-so-important to be sure your bead is fully and properly seated on the rim (basicaly a circular bead) before you go to high pressure. You need for the internal pressure to act uniformly on the bead. If it is not uniform, if it is not circular, you get into an unstable configuration (elliptical shape) an the edge can lift locally.

Below the bead the rim will also see hoop (tangential) stress. Similar arguments apply.

It is interesting the picture does not use a modern double wall box section rim.

I am not a mechanical engineer and no mechanical engineers were harmed in the making of this post.

jr-14

Now, just where in the the world do you find the time to post 25,000+ times with all the wheel and bike building you do?

Panthers007

The same place most of us offering mechanical advice do - out of the goodness of our hearts to spend time helping strangers. Think about it.

imi

Yeah! This is what blows me away! Thanks all ye splendid Wrenches! (almost forgot the "r" in that last word... what would Dr.Freud have said?!)

Panthers007

"That's a lovely slip you're wearing. Where can I get one?"

DArthurBrown

I have heard the "inflate a tube on the wheel" while truing argument. You're right,in theory, the wheel shouldn't be affected by that kind of thing, but sometimes the wheels I've built or relaced can be, in fact, affected by inflating a tube, not by much, but a little, similar to the way a wheel my seem true until you ride it around the block, hear the spokes pinging, and recheck the it. I think the issue is that the spokes do not always seat properly when initially tightened. Alternatively, I just really grasp spoke pairs hard to check for this sort of thing, and it seems to work, but I've seen professional wheelbuilders true wheels and stick them right in the box without ever double-checking.

Mike Mills

I was thinking more about this last night. I came up with about 5ksi tensile (hoop) stress in the aluminum of the rim. I made a lot of assumptions to get to that number but I believe it is 5ksi not 500 psi and not 50ksi, so I'm close. :-)

I looked up the tensile strength of aluminum alloys and feel that 30ksi is a conservative, realistic number to use. I also found the elastic modulus which is in the 10Msi range at room temperature. Using this, I get a strain of 0.0005 (=5ksi/10Msi).

For a rim that is 1/2" wide, it would stretch 0.00025". Good luck seeing that with your eyes, even if the wheel is in a truing jig.

I would think the radially inward force on the spoke head would be of greater consequence. This force would reduce the tensile stress induced by tensioning the spokes. I'll need to do a few more calculations to get to that number. Stay tuned.

DannoXYZ 

What pressure did you use? I'm assuming 130psi. Also wouldn't you want to use the length of the rim's inner surface rather than just overall width? A flat single-section 1/2" wide rim would face less deflection than a one with a longer curved inner section. We should also take into account the wall thickness of the rim as well and calculate both its inner and outer radius changes. But this is jumping ahead...

Mike Mills

I did say I made a LOT of assumptions. :-)

IIRC, I was thinking of 100 psi. I don't think the calculations are accurate enough to worry about the difference between 100 psi and 130 psi. For example, I came up with 0.00025". Would it really make any difference if the strain/deflection were 30% higher, 0.00033" instead of 0.00025"? I think not. It is not significant to the margin of safety calculations, either.

DannoXYZ 

Ok, let's use 0.00025-0.00033" rim deflection. Let's compare this to the amount of spoke-stretch at 110kgf on a 2.0mm and a 2.0/1.8 DB spoke...

Mike Mills

I wish you luck with that one. :-)

I say that because it is a lot more complicated. For example, there are three series elements all of which will strain when you torque the spoke - the hub flange, the spoke and the rim. It is simpler to do just the spoke, as an element, isolated from everything else. What I do not know, and my engineering "gut" tells me is not correct, is whether spoke strain is the only important element.

I suspect the hub flange deflection is irrelevant.

I suspect the double wall construction of moderne hubs was developed, at least in part, to isolate tire pressure from spoke stress, as well as stiffen the rim. I suspect rim inner wall deflection may be important. However, as this question is the main topic of this thread, I would not want to make assumptions. I'd want to analyze that inner wall. Agree?

I would want to do some more serious calculations before attempting that one.

DannoXYZ 

Given the short length of the nipple-head and hub-flange (material on-axis with spoke), can we make the assumption that most of the stretch is in the spoke?