# Calculate elasticity of rubber balloon

I am developing a model for the respiratory system of a neonate (newborn child) for a BME project by using pressure/volume data of the lungs. To model these, I plan to use a rubber balloon. However. the calculations are more difficult than I thought (especially because rubber behaves non-linearly and with a hysteresis effect, but I will ignore these effects).

I will assume the balloon is perfectly spherical, that it behaves linearly and that the strain is dependent on the radius relative to the original radius ($R_0$).

Here are the equations for the basis of my model:

$$V = \frac{4}{3} \pi R^3_1 \\ R_1 = \left( \frac{3V}{4\pi} \right)^\frac{1}{3} \\ \sigma = \frac{pR_1}{2t} \\ \epsilon = \frac{R_1 - R_0}{R_0} \\ E = \frac{\sigma}{\epsilon} = \frac{pR_0R_1}{2t(R_1-R_0)} = \frac{pR_0}{2t\left(1-\dfrac{R_0}{\left(\frac{3V}{4\pi}\right)^\frac{1}{3}}\right)} \\$$

My conclusion is: $$E \propto \frac{p}{V^\frac{1}{3}}$$

Is this correct?

• How will the sphere deform? From a uniform pressure? Is this a very thin ballon? like the outer radius and very similar to the inner radius? – willpower2727 Aug 5 '15 at 0:36
• The sphere will expand with pressure (of course), only steady state pressure-volume relationships are known. The balloon's wall is far smaller than the radius : R >>>> t. – Alexm Aug 5 '15 at 15:10
• It would seem that you have done a linear strain, as if the balloon is expanding along the radial direction. But it isn't, it is expanding in a normal direction and the strain would be proportional to the change in circumference. Would the strain being in two dimensions affect the modulus too? – user5367 Mar 8 '16 at 10:10

You need to be careful with the subscripts, e.g in the equation for $\sigma$, which $R$ do you want? However, what you seem to be saying is: $$E \propto \frac{p}{R}$$ and, if you take $p$ and $R$ as increments from initial values, that seems reasonable. Getting the constant of proportionality might be problematic.