Physics of Stretch: What pressure does a band apply on a cylinder?

[Broken glass arch]

TL;DR Summary

I’m working with a continuous elastic loop and trying to relate measured tensile force in a test rig to the surface pressure it would apply when fitted around a cylinder.

Scenario 1 (figure 1)
A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material.
The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm).
At this level of stretch, the load cell reads 45N tensile force.

Key numbers
Elastic material width: 250mm
Distance between bars: 1190mm
Bar thickness: 5mm
Elastic material internal loop length: 1200mm
Elastic material thickness: 2mm
Reactionary force: 45N

Scenario 2 (figure 2)
The same continuous loop of elastic material is now stretched around a cylinder that has a circumference of 1200mm. This is the same level of stretch as in scenario 1.

The question!
What is the pressure exerted by the elastic material against the surface of the cylinder in figure 2, and why?
(Assume no effects of friction and assume the material is not permanently deformed.)

 

[Baluncore]

Welcome to PF.

The key search term you need is "hoop stress".
https://en.wikipedia.org/wiki/Cylinder_stress#Hoop_stress

 

[Broken glass arch]

Thanks, happy to be here. Found PF looking for help, but I'll definately be hanging around.

So here we're I'm up to:
My first instinct is that if you have a force and an area, you have pressure? We’ve got 22.5N (You need to divide the 45N by two as we’re measuring two “legs”) and 0.3m2 which gives us 83.3Pa.
Fairly quickly realised that's probably not right.

Looking at the law of Laplace (which is sort of a re-arranged hoop stress equation without the thickness) we get something quite different.

Law of Laplace for a cylinder is T=PR (T=wall tension, P = pressure, R = radius).
I re-arranged this to P=T / R.
T seems to be in N/m which would be 22.5 / 0.25=90.
So P=90 / 0.191
P=471 Pa.

What do you think?

 

[Chestermiller]

Is the allowed to contract in width when scenario 1 is applied? Is it allowed to contract in with what scenario 2 is applied?

 

[Broken glass arch]

Is the allowed to contract in width when scenario 1 is applied? Is it allowed to contract in with what scenario 2 is applied?

The assumption would be no widthwise contraction in either scenario.

 

[Broken glass arch]

I have noticed a realy dumb mistake in the orginial example. It should say the following (updated numbers in bold italic)

"The lower bar is moved downwards until the two bars are 595mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (595mm + 595mm + 5mm + 5mm).
At this level of stretch, the load cell reads 45N tensile force.

Key numbers
Elastic material width: 250mm
Distance between bars: 595mm
Bar thickness: 5mm
Elastic material internal loop length: 1200mm
Elastic material thickness: 2mm
Reactionary force: 45N"

Update fig 1

Can't edit the orignial post, but happy for mods to do so if they can?

 

[Chestermiller]

The assumption would be no widthwise contraction in either scenario.

In the first scenario, there is definitely going to be significant widthwise contraction (and this would reduce the required tensile force) if, as indicated, the original width is much smaller than the original length. To avoid widthwise contraction in the first scenario, you would have to make the width much larger than the length. This would result in contraction only near the edges of the sample, while the middle region would essentially be constrained.

 

[Chestermiller]

In scenario 2, the radius of the cylinder is found by $2\pi R=1.2$or $$R=\frac{1.2}{2\pi}=\frac{0.6}{\pi}$$The pressure is given by $$P=\frac{T}{RW}=\frac{22.5\pi }{(0.6)(0.25)}$$