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Law of Laplace

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This example verifies\ \ the computed stress and the computed work using a simple spherical shell geometry\ \ and time-varying Neumann boundary condition applied on the inner boundary. To\ \ verify the computed stress, we use the law of Laplace, see<a href=\"#fn1\" class=\"\ footnote-ref\" id=\"fnref1\" role=\"doc-noteref\">¹, to estimate\ \ the wall stress and compare it against the mean computed stress. For the work\ \ varification, we compute the external work analytically and compare it against\ \ the computed internal work.

\n\n<h2 id=\"stress-verification\">Stress\ \ Verification\n

To verify the stress, we use the law of Laplace\ \ to estimate the wall stress (wall tension).The <a href=\"https://en.wikipedia.org/wiki/Young%E2%80%93Laplace_equation\"\

law of Laplace can be used to describe the wall stress of a spherical shell\ \ in terms of the surface pressure applied on the inner boundary, see Figure <code\ \ class=\"interpreted-text\" role=\"numref\">fig-lol-setup. To estimate the\ \ wall tension, we consider the geometry at time <span class=\"math inline\">\\ (t = T\) with pressure <span class=\"math inline\">\(p(t=T) = p\)\ \ and radii <span class=\"math inline\">\(r(t=T) = r, R(t=T) = R\).

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\n<img src=\"02_laplace/lawoflaplace_setup.png\"\ \ class=\"align-center\" style=\"width:6.94444in\" alt=\"\" />\n

\n\n<h2 id=\"derivation-of-laplaces-law\"\ Derivation Of Laplace's Law\n

To derive Laplace's law we want to\ \ use the balance of forces. We assume that the center of the spherical shell is\ \ the origin and we denote the inner radius of the deformed geometry by <span class=\"\ math inline\">\(r\) and the outer radius by <span class=\"math inline\"\ \(R\) with <span class=\"math inline\">\(R > r\), see figure\ \ <code class=\"interpreted-text\" role=\"numref\">fig-lol-geometry.

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\n<img src=\"02_laplace/lawoflaplace_geometry.png\"\ \ class=\"align-center\" style=\"width:3.47222in\" alt=\"\" />\n

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By <span class=\"math inline\">\\ (h\) we denote the wall thickness which is given by <span class=\"math inline\"\ \(h = R - r\). Next, we clip the geometry using the x-y-plane as a clipping\ \ plane and obtain two equally sized shells, see figure <code class=\"interpreted-text\"\ \ role=\"numref\">fig-lol-forces.

\n<div id=\"fig-lol-forces\">\n

\n\ <img src=\"02_laplace/lawoflaplace_force.png\" class=\"align-center\" style=\"width:3.47222in\"\ \ alt=\"\" />\n

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The\ \ force vector acting on the inner surface is given in <a href=\"https://en.wikipedia.org/wiki/Spherical_coordinate_system\"\ spherical coordinates by

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<span class=\"math display\">\[\begin{aligned}\n\ \mathbf{F}(\theta, \varphi) = p\n\left(\n\begin{array}{c}\n \sin(\theta)\ \ \, \cos(\varphi) \\\n \sin(\theta) \, \sin(\varphi) \\\n \cos(\\ theta)\n\end{array}\n\right)\n\end{aligned}\]

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with <span class=\"\ math inline\">\(\theta \in [0, \pi]\) and <span class=\"math inline\"\ \(\varphi \in [0, 2 \pi]\). The components of the total force vector\ \ <span class=\"math inline\">\(\mathbf{F}^{tot}\) are

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<span class=\"\ math display\">\[\begin{aligned}\nFx^{tot} & = \int \limits{\theta=0}^{\\ frac{\pi}{2}} \int \limits_{\varphi=0}^{2 \pi} \big( \mathbf{F} \cdot \\ mathbf{e}x \big) \, r^2 \, \sin(\theta) \mathrm{d} \varphi \, \mathrm{d}\ \ \theta\n= p \, r^2 \int \limits{\theta=0}^{\frac{\pi}{2}} \int \limits_{\\ varphi=0}^{2 \pi} \sin^2(\theta) \, \cos(\varphi) \mathrm{d} \varphi \\ , \mathrm{d} \theta = 0, \\\nFy^{tot} & = \int \limits{\theta=0}^{\\ frac{\pi}{2}} \int \limits_{\varphi=0}^{2 \pi} \big( \mathbf{F} \cdot \\ mathbf{e}y \big) \, r^2 \, \sin(\theta) \mathrm{d} \varphi \, \mathrm{d}\ \ \theta\n= p \, r^2 \int \limits{\theta=0}^{\frac{\pi}{2}} \int \limits_{\\ varphi=0}^{2 \pi} \sin^2(\theta) \, \sin(\varphi) \mathrm{d} \varphi \\ , \mathrm{d} \theta = 0, \\\nFz^{tot} & = \int \limits{\theta=0}^{\\ frac{\pi}{2}} \int \limits_{\varphi=0}^{2 \pi} \big( \mathbf{F} \cdot \\ mathbf{e}z \big) \, r^2 \, \sin(\theta) \mathrm{d} \varphi \, \mathrm{d}\ \ \theta\n= p \, r^2 \int \limits{\theta=0}^{\frac{\pi}{2}} \int \limits_{\\ varphi=0}^{2 \pi} \sin(\theta) \, \cos(\theta) \mathrm{d} \varphi \, \\ mathrm{d} \theta = p \, r^2 \, \pi \\\n\end{aligned}\]

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with\ \ <span class=\"math inline\">\(\theta \in (0, \frac{\pi}{2})\) since\ \ we only integrate over one half of the shell.

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For the derivation of Laplace's\ \ law we approximate the wall stress <span class=\"math inline\">\(\sigma\)\ \ by its mean value <span class=\"math inline\">\(\bar{\sigma}\), see\ \ figure <code class=\"interpreted-text\" role=\"numref\">fig-lol-forces.\ \ Since the geometry and the acting forces are symmetric, the tangential stress\ \ in any direction must be the same and there will be zero shear stress. Due to\ \ the assumption, the mean wall stress at the cut face just acts just in z-direction,\ \ thus we have <span class=\"math inline\">\(\bar{\sigma} \cdot \mathbf{e}_x\ \ = 0\) and <span class=\"math inline\">\(\bar{\sigma} \cdot \mathbf{e}_y\ \ = 0\). The components of the total wall stress vector <span class=\"math\ \ inline\">\(\bar{\sigma}^{tot}\) are

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<span class=\"math display\"\ \[\begin{aligned}\n\bar{\sigma}x^{tot} & = \int \limits{r=r}^{R} \\ int \limits_{\varphi=0}^{2 \pi} \big( \bar{\sigma} \cdot \mathbf{e}_x \\ big) \, r^2 \, \sin(\theta) \mathrm{d} \varphi \, \mathrm{d} r = 0, \\\ \n\bar{\sigma}y^{tot} & = \int \limits{r=r}^{R} \int \limits_{\varphi=0}^{2\ \ \pi} \big( \bar{\sigma} \cdot \mathbf{e}_y \big) \, r^2 \, \sin(\theta)\ \ \mathrm{d} \varphi \, \mathrm{d} r = 0, \\\n\bar{\sigma}z^{tot} &\ \ = \int \limits{r=r}^{R} \int \limits_{\varphi=0}^{2 \pi} \big( \bar{\\ sigma} \cdot \mathbf{e}_z \big) \, r^2 \, \sin(\theta) \mathrm{d} \varphi\ \ \, \mathrm{d} r = \pi \, \big( R^2 - r^2 \big) \, \sigma_W\n\end{aligned}\\ ]

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with <span class=\"math inline\">\(\sigma_W := \bar{\sigma}\ \ \cdot \mathbf{e}_z\). From equations <code class=\"interpreted-text\"\ \ role=\"eq\">equ-lol-totforcevec and <code class=\"interpreted-text\" role=\"\ eq\">equ-lol-totstensionvec we have

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<span class=\"math display\"\ \[\mathbf{F}^{tot} = \bar{\sigma}^{tot} \quad \Leftrightarrow \quad p \\ , r^2 = (R^2 - r^2) \, \sigma_W.\]

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From <code class=\"interpreted-text\"\ \ role=\"eq\">equ-lol-equalcond we conclude that

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<span class=\"math\ \ display\">\[\sigma_W = \frac{p \, r^2}{R^2 - r^2} = \frac{p \, r}{2 \,\ \ h \left( 1 + \frac{h}{ 2 \, r } \right) }\]

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and for <span\ \ class=\"math inline\">\(h \ll r\) we obtain the law of Laplace

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\\[\\sigma_W \\approx \\sigma_{L} = \\frac{p \\\ , r}{2 \\, h}\\]

01_boundary_conditions 03_unloading