# Calculate elasticity of rubber balloon

1. Aug 4, 2015

### AlexMentink

For a Biomedical Engineering project I have to develop a model for the respiratory system of a neonate. This will be done by using pressure/volume data of the lungs. To model these, I would like to use a rubber balloon. However the calculation are more difficult than I thought.. (Especially because rubber behaves non-linear and with a hysteresis effect, I will ignore these effects) As assumptions I used the fact that the balloon is perfectly spherical, behaves linearly and the strain is dependent on the radius relative to the original radius (R0).

Could anyone help me?

Thanks!

2. Aug 4, 2015

### Staff: Mentor

Well, for one thing, E = σ/ε. And, you are aware the p in your equation is the inside pressure minus the outside pressure, correct?

Chet

3. Aug 4, 2015

### AlexMentink

Ah of course, thanks for noticing..

Yes the p should be the difference in pressure. Should the formula be correct now?

4. Aug 4, 2015

### nasu

You don't have proportionality with V^(-1/3).

By the way, E and p have the same dimensions and can be measured in the same units (Pa or N/m^2).

What is t? The thickness of the membrane?

5. Aug 4, 2015

### AlexMentink

Hmm ok, the thing is I have a linear relationship between the lung volume and pressure, therefore I would like to use V/p or p/V in the equation to solve for E.

I don't have any idea how to extract p/V or V/p from this relation..

t is indeed the thickness of the membrane

6. Aug 4, 2015

### nasu

You could call that ratio "x" and express p as x*V in the equation.
Then solve for this x.

Or just solve for p and then divide the result by V. It will depend on volume as well as the parameters of the balloon.

7. Aug 4, 2015

### Staff: Mentor

The relationship may be linear, but it is not a direct proportion (at least not in the case of your lungs). When the pressure difference goes to zero, your lung volume does not go to zero. In the case of a balloon, when the pressure difference goes to zero, the volume of the balloon does go to zero. So maybe a balloon is not such a good quantitative model of a lung.

Chet

8. Aug 4, 2015

### Nidum

Volumetric expansion of a balloon with pressure can never be very linear over large ranges

Membrane thickness will reduce considerably for non trivial volumetric expansions .

Maybe something more like a cylinder and piston with a spring ?

Or a cylindrical bellows with a spring ?

9. Aug 4, 2015

### AlexMentink

The thing is that I'm building a box as a model for the respiratory system including diaphragm and abdomen stiffness. In most surgeries CO2 is inflated in either the thorax or abdomen to make room for using the instruments. The box should model the relationships between lungs, diaphragm and abdomen stiffness.

During all real procedures, a minimum pressure is inflated into the lungs, to make sure they don't stick together. Therefore in this model, the balloon will be kept at a minimal pressure and will have a minimal volume.

Finally, the lung pressure/volume relationship is not really linear, but in the form of y=ax^2-bx. The problem is that I don't know how to setup a nonlinear mathematical relationship using the material properties of a rubber balloon. Does anyone know what the general formula is for calculating the elasticity of rubber (non linear)?

10. Aug 4, 2015

### nasu

The elasticity of rubber is given, it is a material property.
It may depend on the strain in the material (this will make the material non-linear).

What do you actually try to calculate the elasticity? Starting from what?
What are these p/v data for the lungs mentioned in the OP?
P/v versus what parameter?

11. Aug 4, 2015

### Nidum

Last edited by a moderator: May 7, 2017
12. Aug 4, 2015

### sophiecentaur

Apart from looking a bit like a lung, how does a balloon provide a better model than a cylinder and piston? A piston could actually be driven with servos to give it all sorts of pv characteristics or a number of pistons in cylinders with springs could provide a mechanical model to any degree of agreement. the diaphragm is as much like a piston in structure as a balloon.

13. Aug 5, 2015

### AlexMentink

Thanks for all quick replies!

I will make it a bit more clear.
In most surgeries CO2 is inflated in either the thorax or abdomen to make room for using the instruments. In the case of a thorax surgery, a needle is placed between the lungs and ribcage inflating CO2 (causing a pneumothorax). Increase in pressure will squeeze the lungs together and the diaphragm down. Data about these pressure-volume relationships are known.

In the case of an abdominal surgery, a needle is placed between the intestines and CO2 is inflated. Increase in pressure will cause the diaphragm to go up and the abdominal wall expand. Data about these pressure-volume relationships are known as well.

The box should model the relationships between lungs, diaphragm and abdomen stiffness to research the effect of pressure in the thorax on the abdominal wall etc.

Professional inflators with adjustable flow/pressure/volume are available.

The box will be air-tight and easy to disassemble.

Last edited: Aug 5, 2015
14. Aug 5, 2015

### AlexMentink

The box will look something like this. Open sides will be covered with a rubber membrane and pressed together to make the box air-tight. The openings on the sides are for CO2 inflation and measurement of pressure differences. The opening on top is for the respiratory inflation (model for the lungs will be placed here).

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Last edited: Aug 5, 2015