5.1.1 Elastic Model

In this model the materials are treated as elastic solids parameterized by their Young Modulus E and Poisson ratio $\nu$. The stress tensor is calculated uniquely from the strain tensor which is solved from the Navier Stokes equations [Zie91] together with the displacement boundary conditions.

In theory of linear elasticity with small displacements the strain tensor can be defined as

$$\varepsilon \left(\vphantom{\begin{array}{c} \varepsilon _{xx}\\ \varepsilon... ...varepsilon _{xy}\\ \varepsilon _{yz}\\ \varepsilon _{zx} \end{array} }\right. \begin{array}{c} \varepsilon _{xx}\\ \varepsilon _{yy}\\ \varep... ...{zz}\\ \varepsilon _{xy}\\ \varepsilon _{yz}\\ \varepsilon _{zx} \end{array} \left.\vphantom{\begin{array}{c} \varepsilon _{xx}\\ \varepsilon... ...varepsilon _{xy}\\ \varepsilon _{yz}\\ \varepsilon _{zx} \end{array} }\right) \left(\vphantom{\begin{array}{ccc} \frac{\partial}{\partial x} & ... ...rac{\partial}{\partial z} & 0 &\frac{\partial}{\partial x} \end{array} }\right. \begin{array}{ccc} \frac{\partial}{\partial x} & 0 & 0\\ 0 & \fr... ... y}\\ \frac{\partial}{\partial z} & 0 &\frac{\partial}{\partial x} \end{array} \left.\vphantom{\begin{array}{ccc} \frac{\partial}{\partial x} & ... ...rac{\partial}{\partial z} & 0 &\frac{\partial}{\partial x} \end{array} }\right) \left(\vphantom{\begin{array}{c} u\\ v\\ w \end{array} }\right. \begin{array}{c} u\\ v\\ w \end{array} \left.\vphantom{\begin{array}{c} u\\ v\\ w \end{array} }\right) \theta$$ (5.1)

with the displacement field:

$$\theta \left(\vphantom{ \begin{array}{c} u(x,y,z)\\ v(x,y,z)\\ w(x,y,z) \end{array} }\right. \begin{array}{c} u(x,y,z)\\ v(x,y,z)\\ w(x,y,z) \end{array} \left.\vphantom{ \begin{array}{c} u(x,y,z)\\ v(x,y,z)\\ w(x,y,z) \end{array} }\right)$$ (5.2)

Assuming a linear material law, the stress tensor can now be calculated using the equation

$$\sigma \varepsilon \varepsilon_{0}^{} \sigma_{0}^{}$$ (5.3)

where $\varepsilon_{0}^{}$ are prestrains and $\sigma_{0}^{}$ are prestresses due to change of temperature, crystal growth or for, e.g., volumetric expansions as in case of oxidation. Assuming an isotropic material (5.3) can be written as:

$$\left(\vphantom{ \begin{array}{c} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{xy}\\ \sigma _{yz}\\ \sigma _{zx} \end{array} }\right. \begin{array}{c} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{xy}\\ \sigma _{yz}\\ \sigma _{zx} \end{array} \left.\vphantom{ \begin{array}{c} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{xy}\\ \sigma _{yz}\\ \sigma _{zx} \end{array} }\right) {\frac{E\cdot(1-\nu)}{(1+\nu)(1-2\nu)}} \left(\vphantom{ \begin{array}{cccccc} 1 & \frac{\nu}{1-\nu} & \f... ...-\nu)} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1-2\nu}{2(1-\nu)} \end{array} }\right. \begin{array}{cccccc} 1 & \frac{\nu}{1-\nu} & \frac{\nu}{1-\nu} &... ...2\nu}{2(1-\nu)} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1-2\nu}{2(1-\nu)} \end{array} \left.\vphantom{ \begin{array}{cccccc} 1 & \frac{\nu}{1-\nu} & \f... ...-\nu)} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1-2\nu}{2(1-\nu)} \end{array} }\right)$$ (5.4)

Introducing the distributed body forces

$$\left(\vphantom{ \begin{array}{c} f_x(x,y,z)\\ f_y(x,y,z)\\ f_z(x,y,z) \end{array} }\right. \begin{array}{c} f_x(x,y,z)\\ f_y(x,y,z)\\ f_z(x,y,z) \end{array} \left.\vphantom{ \begin{array}{c} f_x(x,y,z)\\ f_y(x,y,z)\\ f_z(x,y,z) \end{array} }\right)$$ (5.5)

the system of differential equations to be solved is finally found to:

$${\frac{\partial \sigma_{xx}}{\partial x}} {\frac{\partial \sigma_{xy}}{\partial y}} {\frac{\partial \sigma_{zx}}{\partial z}} {\frac{\partial f}{\partial x}}$$ (5.6)

$${\frac{\partial \sigma_{xy}}{\partial x}} {\frac{\partial \sigma_{yy}}{\partial y}} {\frac{\partial \sigma_{zy}}{\partial z}} {\frac{\partial f}{\partial y}}$$ (5.7)

$${\frac{\partial \sigma_{xz}}{\partial x}} {\frac{\partial \sigma_{yz}}{\partial y}} {\frac{\partial \sigma_{zz}}{\partial z}} {\frac{\partial f}{\partial z}}$$ (5.8)