Ideal gas for quasi-static calculations

A special case of a linear elastic isotropic material is an ideal gas for small pressure deviations. From the ideal gas equation one finds that the pressure deviation $ dp$ is related to a density change $ d \rho$ by

$\displaystyle dp=\frac{d \rho}{\rho_0} \rho_0 r T,$ (303)

where $ \rho_0$ is the density at rest, $ r$ is the specific gas constant and $ T$ is the temperature in Kelvin. From this one can derive the equations

$\displaystyle t_{11}=t_{22}=t_{33}=(\epsilon_{11}+\epsilon_{22}+\epsilon_{33}) \rho_0 r T$ (304)


$\displaystyle t_{12}=t_{13}=t_{23}=0,$ (305)

where $ \boldsymbol{t}$ denotes the stress and $ \boldsymbol{\epsilon}$ the linear strain. This means that an ideal gas can be modeled as an isotropic elastic material with Lamé constants $ \lambda=\rho_0 r T$ and $ \mu=0$. This corresponds to a Young's modulus $ E=0$ and a Poisson coefficient $ \nu=0.5$. Since the latter values lead to numerical difficulties it is advantageous to define the ideal gas as an orthotropic material with $ D_{1111}=D_{2222}=D_{3333}=D_{1122}=D_{1133}=D_{2233}=\lambda$ and $ D_{1212}=D_{1313}=D_{2323}=0$.