Update elasticity_scaling.md

chapter2/elasticity_scaling.md

@@ -18,6 +18,6 @@ where $\beta = 1+\frac{\lambda}{\mu}$ is a dimensionless elasticity parameter an
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 is a dimensionless variable reflecting the ratio of the load $\rho g$ and the shear stress term $\mu \nabla^2u \sim \mu \frac{U}{L^2}$ in the PDE.
 
-One option for the scaling is to chose $U$ such that $\gamma$ is of unit size ($U=\frac{\rho g L^2}{\mu}$). However, in elasticity, this leads to displacements of the size of the geometry. This can be achieved by choosing $U$ equal to the maximum deflection of a clamped beam, for which there actually exists a formula: $U=\frac{3}{2} \rho g L^2\frac{\delta^2}{E}$ where $\delta=\frac{L}{W}$ is a parameter reflecting how slender the beam is, and $E$ is the modulus of elasticity. Thus the dimensionless parameter $\delta$ is very important in the problem (as expected $\delta\gg 1$ is what gives beam theory!). Taking $E$ to be of the same order as $\mu$, which in this case and for many materials, we realize that $\gamma \sim \delta^{-2}$ is an appropriate choice. Experimenting with the code to find a displacement that "looks right" in the plots of the deformed geometry, points to $\gamma=0.4\delta^{-2}$ as our final choice of $\gamma$.
+One option for the scaling is to choose $U$ such that $\gamma$ is of unit size ($U=\frac{\rho g L^2}{\mu}$). However, in elasticity, this leads to displacements of the size of the geometry. This can be achieved by choosing $U$ equal to the maximum deflection of a clamped beam, for which there actually exists a formula: $U=\frac{3}{2} \rho g L^2\frac{\delta^2}{E}$ where $\delta=\frac{L}{W}$ is a parameter reflecting how slender the beam is, and $E$ is the modulus of elasticity. Thus the dimensionless parameter $\delta$ is very important in the problem (as expected $\delta\gg 1$ is what gives beam theory!). Taking $E$ to be of the same order as $\mu$, which in this case and for many materials, we realize that $\gamma \sim \delta^{-2}$ is an appropriate choice. Experimenting with the code to find a displacement that "looks right" in the plots of the deformed geometry, points to $\gamma=0.4\delta^{-2}$ as our final choice of $\gamma$.
 
-The simulation code implements the problem with dimensions and physical parameters $\lambda, \mu, \rho, g, L$ and $W$. However, we can easily reuse this code for a scaled problem: Just set $\mu=\rho=L=1$, $W$ as $W/L(\delta^{-1})$, $g=\gamma$ and $\lambda=\beta$.
+The simulation code implements the problem with dimensions and physical parameters $\lambda, \mu, \rho, g, L$ and $W$. However, we can easily reuse this code for a scaled problem: Just set $\mu=\rho=L=1$, $W$ as $W/L(\delta^{-1})$, $g=\gamma$ and $\lambda=\beta$.