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I am doing a quick calculation on how to calculate the pressure needed to inflate a perfectly spherical balloon to a certain volume, however I have difficulties with the fact that the balloon (rubber) has resistance to stretching and how this affects the pressure needed. It has to do with the E-modulus of the material I think, but I can not think of a proper way to calculate it?

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The complete stress tensor, while accurate, is largely unnecessary for solving this problem, as it is a thin walled pressure vessel

Assuming the balloon is spherical, the strain can just be calculated from the current and initial radii.

$$\epsilon=\frac{r}{r_0}-1$$

The stress can be found using the modulus of elasticity:

$$\sigma=E\,\epsilon$$

The thin wall pressure equation can get you to pressure, if you know the thickness, by balancing outward pressure inside with the inward tension along a great circle of the sphere:

$$\pi\,r^2\,P=2\,\pi\,r\,\sigma\,t$$ $$P=\frac{2\,\sigma\,t}r$$

Because balloons get thinner as they stretch, the thickness will actually vary. Rubber typically has a poisson's ratio of 0.5 meaning it keeps a constant volume while being deformed. We can then calculate the thickness in terms of the radius: $$t\,r^2=t_0\,{r_0}^2$$ $$t=t_0\,\left(\frac{r_0}{r}\right)^2$$

Putting them all together:

$$P=\frac{2\,E\,\left(r-r_0\right)\,t_0\,r_0}{r^3}$$

To see what this looks like, we can make a generic plot:

Plot of pressure vs radius of balloon

As you can see, there is a maximum pressure after which it becomes easier and easier to inflate the balloon. We can solve for this maximum pressure by equating the derivative with zero, solving for r, and plugging back in:

$$0=\frac{dP}{dr}=2\,E\,t_0\,r_0\left(\frac1{r^3}-3\frac{r-r_0}{r^4}\right)$$

$$r=\frac32\,r_0$$

$$P_{max}=\frac{8\,E\,t_0}{27\,r_0}$$

Of course this assumes a constant modulus of elasticity, which never holds true for a large enough deformation.

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    $\begingroup$ This answer is very informative, but I was wondering if you could explain the axis and what you used to plot them? I'm tempted to repeat this but with a non-linear E and see what I get $\endgroup$ Feb 17 at 16:27
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    $\begingroup$ @AncientSwordRage the x-axis is the normalized balloon radius. So it's just the current radius divided by the initial radius. The y-axis is a number that should be proportional to the pressure if E were constant. I used a program called Mathcad, but you could use excel, or google sheets. I'd just make a column for radius, E, and then a pressure column that uses the formula to calculate the pressure. $\endgroup$
    – Rick
    Feb 19 at 15:07
  • $\begingroup$ The 'proportional to pressure' bit seems obvious when I look at the units. Thanks again! $\endgroup$ Feb 19 at 15:48
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The generic plot above has some experimental verification at https://www.youtube.com/watch?v=fwh-i0WB_bQ. screenshot of results at end of video A pity they didn't measure radius of balloon, but if we assume constant rate of expansion, the shape of the graph plotted against time should be similar. Indeed, it clearly shows the initial maximum pressure, which is shown above to be caused by the small radius of curvature of the balloon surface, followed by a decrease to roughly constant as the radius of curvature increases. Departure from the assumption of constant modulus of elasticity then causes the pressure to start to increase again.

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What you are missing is the fact that $p=\int \tau \cdot \hat n dA$, where $\tau$ is the stress tensor. I suggest you to check out Landau's book: Theory of Elasticity. He solves the problem of a hollow sphere with different pressures.

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