# Friction between a pre-tensed steel hoop and a cylinder

1. Jan 22, 2012

### meldraft

Hey all,

I am designing a cup holder. It is supposed to be made out of a few steel rings. Its geometry is such that the hoops can open (so the radius increases), and you place the cup inside. Then, as the hoops spring back, they press upon the cup and the friction holds it in place as you lift the entire thing.

Making some rough assumptions (good accuracy is really not important at this point, I just want an order of magnitude), I thought I could use Laplace's Law for a cylindrical container, since my problem is basically the inverse (pressure is pointing inwards instead of outwards):

$$σ=\frac{PR}{t}$$

where σ is the hoop tension, P the pressure of the hoop to the cup, r the radius of the ring, and t the ring's thickness. To simplify the problem a little, I can assume that my rings are infinitely thin (not in the radial direction where I do have thickness. in the z-direction.). Therefore, if N is the total normal force (and this is tricky, because the vector sum is 0):

$$P=\frac{N}{L}$$

so P is a distributed force (Newton/m), much like a beam problem.

Now, since this is an elastic phenomenon, I used Hooke's Law, saying that:

$$εE=\frac{PR}{t}=\frac{NR}{Lt} => N=\frac{εELt}{R}$$

Let traction be fmax:

$$fmax=μN$$

By combining the last two equations:

$$fmax=\frac{μεELt}{R}$$

I would have been perfectly fine with this result, but for strains above ε=10^-6, this formula yields outlandishly high values of friction. For instance, for ε=0.7, I would get something like 0.4 GigaNewtons. I probably went wrong somewhere, but I really can't locate the mistake! I would be grateful for any advice you can give me!

P.S. All calculations were done in SI, so this is after the units check :P

Last edited: Jan 22, 2012
2. Jan 22, 2012

### AlephZero

You said "its geometry is such that the hoops can open so the radius increases".

If that means the hoop is split rather than continuous, your whole calculation is wrong. You should be treating the hoop as a curved beam, which will be orders of magnitude more flexible than a continuous ring.

If you try enlarging the diameter of a metal ring by forcing a cone shaped object into the hole, you will soon discover it does need a very large force!

Even if the ring is continuous, you are ignoring the flexibility of the cup. If it's a plastic cup, it will will be compressed (and possibly buckle) rather than the ring expanding.

3. Jan 22, 2012

### meldraft

You are right, the hoop is not continuous, and of course this explains the immense stress needed to open it (duh!)

My geometry is actually a helix, and I thought I could approximate it with circles, but apparently there is a deficiency in that plan.

I have to sleep now, so I'll post an update in the morning :tongue:

4. Jan 24, 2012

### meldraft

After many hours of thinking , I modelled the hoop as two semi-circles with a hinge. I got the friction as a function of the new radius, which yielded about 16 Newton per hoop, which is pretty reasonable, considering the assumptions :tongue:

I might try a more accurate model in the future, but for now I am happy

Thanks again for offering your insight!