How to calculate the clamp force of an elastic band

[HellomynameisTroy]

Hello! I am trying to calculate the clamp force of an elastic band around a handle bar. I want to prove it can hold the weight of a phone mount. I know I need to be able to resist the moment force and gravity at the worse load cases, but am having trouble figuring out how to do that. Do I need to estimate how much the band will stretch, then I can figure out how much pressure it getting put on the handle bar which can relate to a normal force? Any tips would be highly appreciative. I attached a photo of what the design would look like.

 

[LittleInch]

Nothing attached.

The key unknown will be the friction factor....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[3DDave]

Worst case is the phone acts as a mass on a spring-loaded arm and the natural frequency is in the range of operating the bike. The loads can get to be high; but I'd guess counting on 5x gravity would be safe.

If your elastic is a steel band clamp you may get enough tension with a high-friction rubber pad.

If the image was grey when you tried to upload or is from a protected source, like an on-line seller, it won't upload.

Don't use the attach function for images, use the "Image" button. If you tried "Upload" you didn't select the embed in message button or copy/paste the URL yourself.

 

[HellomynameisTroy]

 

[HellomynameisTroy]

Oh yeah, I should definitely take into account modal analysis. Thanks!
Thanks for the tip on the image upload, I think I got it.
I guess I'm wondering how do I show the free body diagram of the latch can withstand the weight of the phone?

 

[MintJulep]

Since you already seem to have a prototype, just do some tests.

Given the many already existing phone holders for bicycles available there doesn't seem to be much risk here.

 

[HellomynameisTroy]

The picture is just an example of how the design would work. I am interested in the math to prove it.

 

[MintJulep]

I think that you'll find all the necessary math in this paper:

https://repository.lib.ncsu.edu/bit...SMiRT-23_Paper_552.pdf?sequence=1&isAllowed=y

 

[3DDave]

That is a tiny light.

https://www.amazon.com/Beetle-Light-1-pc/dp/B00DTILRLG

housing molded of silicone.

Phone mounts are typically more rigid metal or plastic.

Here is an example calculation.

https://repository.lib.ncsu.edu/bit...SMiRT-23_Paper_552.pdf?sequence=1&isAllowed=y

Most make a prototype and fix what fails.

 

[dvd]

Which math? - thermal expansion (plastic is much larger CTE than steel), friction with mounting surface, tolerance stackup, moisture absorption, fatigue, etc. Agree with Mint Julep and test empirically against multiple combinations of the above.

 

[LittleInch]

Yes you will need to somehow estimate the force applied to the rubber band / strap.

It's the friction factor though for static friction which will be your enemy here and is highly variable depending on the rubber material, the smoothness of the handle bar and how much force you can actually reasonably pull on a band when securing it. Don't forget bikes get wet and water is a good lubricant

You need to avoid or reduce the moment applied as much as you can.

everything I have on a handle bar which can rotate, does so really quite easily, even with the band pulled as tight as it goes. Only those with extra rubber pads on underneath work well.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Compositepro]

The contact pressure applied by a strap or band is:

(T)/((pi)(r))

where T is the tension in pounds per inch of width of the strap, and r is the radius of curvature of the surface.

 

[Denial]

CompositePro. I might be misunderstanding the situation your equation is describing, but I suspect it is out by a factor of about 3.14.

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[Compositepro]

You are right. I derived the equation in my head, and in prior cases I must have needed to convert circumference to radius. That makes the equation about as simple as it can get:

T/r

 

[rb1957]

surely the pressure of the strap is P/A ... A = length * width ... width is taken as 1, but length is circumference of the circle = 2pi*r ...

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Compositepro]

The length of the cylinder wrapped by the strap is removed as a factor because the tension is in force per inch of strap width.

The projected area that this pressure acts upon is the diameter line (2r). The counter acting force keeping the two halves of the cylinder together is 2T.

 

[Scuka]

You've got two options:

You can spend hours trying to come up with a calculation that will be questionably accurate (while also using questionably accurate "worst load cases"), and still end up having to test it because you can't rely on calculations alone

or

you can just build a simplified test rig and test it.

 

[Denial]

As Compositepro points out in his 22Nov23@21:20 post, if a circular strap is placed around a cylinder of radius r and is tensioned to a uniform value T[sub]0[/sub], the radially-directed pressure between the strap and the cylinder is p = T[sub]0[/sub]/r (units of force/length).[ ] It follows from this that the strap will slip around the cylinder under the action of a torque of magnitude 2πµrT[sub]0[/sub] (where π is pi and µ is the coefficient of friction).[ ] If this torque is generated by a uniform, circumferentially-directed force applied around the strap's entire circumference the force's magnitude would be 2πµT[sub]0[/sub].

But this is not much use to the OP, whose "elastic strap" is not subjected to such a tractible uniform loading.[ ] As an alternative I investigated the situation where the band carries a single tangentially-directed concentrated force.[ ] This force will tend to increase the band's tension immediately behind it and stretch it a bit more, and to do the opposite immediately in front of it.

After a few false starts, I came up with the result that the strap will begin to slip when the single tangential force is
2πµT[sub]0[/sub]/(1-e[sup]-2πµ[/sup])

My workings are attached.
[Note to any potential user.[ ] As the Ancient Romans used to say about their calculus, "caveat emptor":[ ] a dictum that should be assumed to apply all the more when the price is zero.]


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[sup]Engineering mathematician / analyst.[ ] See my profile for more details.[/sup]