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Imagine there is a spherical cavity made of an elastic material. The diameter of the cavity is $1 \,\text{cm}$ so that the capacity of the cavity is $4\pi/3 \approx4.19 \,\text{cm}^3$. Now we pump $5\,\text{cm}^3$ of water into the cavity. How to calculate the water pressure? Which parameter of the cavity material correlates to the water pressure, for example, the stiffness or shear modulus?

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  • $\begingroup$ Is the cavity in an infinite medium, or is it a sphere with a hole? $\endgroup$
    – nicoguaro
    Commented Jul 1, 2019 at 14:48

2 Answers 2

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The elastic modulus $\lambda$:

$\lambda=\frac{\sigma}{\epsilon}$, with $\sigma$ the stress and $\epsilon$ the strain, together with the thickness of the elastic material is what would be used to calculate the pressure the water would find itself under.

Since as $5\:\mathrm{cm^3}$ is larger than $4.19\:\mathrm{cm^3}$, the elastic the ball-shaped material must strain (elongate or stretch, if you prefer), because liquids are quasi-incompressible. The pressure will be proportionally higher for higher $\lambda$ and higher material thickness.

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  • $\begingroup$ Yeah, the cavity will stretch, and the water pressure will increase. If we know the $\lambda$ and the thickness of the cavity, how to write the expression of the water pressure? Thank you very much! $\endgroup$
    – bushiwo
    Commented Oct 13, 2015 at 15:02
  • $\begingroup$ @bushiwo: Hello. For 1D problems already this problem is not easy because the thickness of the material reduces with strain (the material is itself incompressible). For 2D and 3D (here!) this involves stress and strain tensors and it becomes mathematically quite challenging. Prior to answering I googled for specific examples of this problem and came up with nothing. Unfortunately the derivation is also above my pay grade, so I owe you that answer! $\endgroup$
    – Gert
    Commented Oct 13, 2015 at 15:50
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Due to the symmetry of the problem, the differential equation for elasticity simplifies to

$$\nabla \nabla \cdot \mathbf{u} = 0\, ,$$

and $\mathbf{u} = u_r \hat{\mathbf{e}}_r$, or

$$\frac{1}{r^2} \frac{d}{dr} (r^2 u) = 3c_1\, ,$$

with solution

$$u_r = c_1 r + \frac{c_2}{r^2}\, ,$$

and stresses:

\begin{align*} \sigma_{rr} &= \frac{E}{1 - 2\nu}c_1 - \frac{2E}{1 + \nu}\frac{c_2}{r^3}\\ \sigma_{\theta\theta} &= \sigma_{\phi\phi} = \frac{E}{1 - 2\nu}c_1 + \frac{E}{1 + \nu} \frac{c_2}{r^3}\, , \end{align*}

where $E$ is the Young's modulus and $\nu$ the Poisson coefficient.

The exact solution would depend on the configuration of your problem.

In the case of a thin shell, we have

$$p = \frac{2 E}{(1 - \nu)} \frac{h \Delta R}{R^2}\, .$$

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