R3iCe#
CTI formulation#
How to build the CTI law is describe in the page : CTI demonstration
Description of the Continuous Transverse Isotropic (CTI) formulation#
A transverse isotropic material is a material with a different behavior in one main direction \(e_3\) and perpendicular to this direction. But the properties perpendicular to \(e_3\) are identical. Therefore the properties transverse to \(e_3\) are isotropic.
Such formulation can be suitable model for modelling single crystal behavior for hexagonal material such as ice, quartz, magnesium alloy.
Single crystal linear behavior#
\(\eta_1\) is the viscosity for shear parallel to the basal plane
\(\beta\) is the viscosity ratio between shear parallel a basal plane and within the basal plane
\(\gamma\) is the viscosity ratio between compression (or traction) along the \(e_3\) axis (\(\eta'\)) and in one direction perpendicular \(e_r\)
Therefore for a same solicitation in different direction \(S=S_{33}=S_{11}\); it gives :
CTI equation#
Note
To make the equation dimensionless we use :
typical viscosity \(\eta_n \sim Pa.s^{\frac{1}{n}}\)
If stress boundary condition is applied :
typical stress \(\Sigma\sim Pa\).
Therefore a characteristic time \(T\) is defined as :
\(T=\left(\frac{\eta_n}{\Sigma}\right)^{n}\)
If strain rate condition is applied :
typical strain rate : \(\dot{\varepsilon}_m\sim s^{-1}\)
\(T=\frac{1}{\dot{\varepsilon}_m}\sim s\)
\(\Sigma=\eta_n \dot{\varepsilon}_m^{\frac{1}{n}}\sim Pa\)
This is the general form for the CTI equation. The linear formulation is obtained with \(n=1\).
with :
\(\alpha_1= \frac{1}{2\beta}\)
\(\alpha_2=\frac{\gamma}{\beta}-1\)
\(\alpha_3=1-\frac{1}{\beta}\)
with \(\eta^\star\) the apparent viscosity that depend of the strain rate \(D\).
with :
\(\alpha_1= \frac{1}{2\beta}\)
\(\alpha_2=\frac{\gamma}{\beta}-1\)
\(\alpha_3=1-\frac{1}{\beta}\)
with :
\(\alpha_1= \frac{1}{\beta}\)
\(\alpha_2=2\left(\frac{\gamma}{\beta}-1\right)\)
\(\alpha_3=2\left(1-\frac{1}{\beta}\right)\)
Description of the orientation evolution equation#
with
\(\mathcal{Mo}=\left(\frac{\eta_n}{\Sigma}\right)^{n}\frac{1}{\Gamma_{RX}}\)