MAT_EXPLOSIVE_HVRB
FE
Material properties
Beta command
This command is in the beta stage and the format may change over time.
"Optional title"
mid, $\rho_0$, $E$, $\nu$
$\sigma_0$, $Q$, $C$, $S$, $\Gamma$
$A$, $B$, $R_1$, $R_2$, $\omega$, $e_0$
$p_I$, $p_R$, $Z$, $M$, $X$, $\tau_R$, equilibrium
Parameter definition
Description
This is the History Variable Reactive Burn (HVRB) model, for predicting initiation of explosive materials. It is currently only implemented for Finite Elements.
The material pressure $p$ is defined as:
$\displaystyle{ p = (1 - F) \cdot p_s + F \cdot p_g }$
where $0 \leq F \leq 1$ is burn fraction, $p_s$ is the pressure of the solid material and $p_g$ is the pressure of generated gas products.
$\displaystyle{ p_s = \frac{K \eta}{(1-S \eta)^2} \cdot \left( 1 - \frac{\Gamma \eta}{2} \right) + \Gamma e_s }$
where:
$\displaystyle{ \eta = 1 - \rho_0 / \rho_s }$
$K$ is the bulk modulus, $\rho_0$ is the initial density and $\rho_s$ is the current density of the solid phase. $e_s$ is the sepcific internal energy per unit volume in the solid phase.
$\displaystyle{ p_g = A \left( 1 - \frac{\omega}{R_1 V} \right) \mathrm{e}^{-R_1 V} + B \left( 1 - \frac{\omega}{R_2 V} \right) \mathrm{e}^{-R_2 V} + \omega e_g}$
where:
$\displaystyle{ V = \rho_0 / \rho_g}$
$e_g$ and $\rho_g$ are the specific internal energy and density in the gas phase. The burn fraction $F$ is defined according to:
$\displaystyle{ F = 1 - \left(1 - \mathrm{min} \left( 1, \frac{\phi^M}{X} \right) \right)^X}$
$\displaystyle{ \phi(t) = \frac{1}{\tau_R} \int_0^t \left( \frac{\mathrm{max}(0, p - p_I)}{p_R} \right)^Z \mathrm{d}\tau }$
The combustion process generates thermal energy, and the energy release rate per unit volume is:
$\displaystyle{ \frac{\partial e}{\partial F} = \frac{\rho}{\rho_0} e_0 }$
Note that if equilibrium=0 then $\rho_s \equiv \rho_g$ and $p_s \not\equiv p_g$. equilibrium=1 activates an iterative scheme where the densities are balanced $(\rho_s \not\equiv \rho_g)$ such that $p_s \equiv p_g$. Temperature is not balanced. The thermal process is assumed to be slower and heat generated in the deflagration process is assumed to stay in the gas phase. Hence, $e_s \not\equiv e_g$.
The solid material is elasto-plastic with (J2) flow stress:
$\displaystyle{\sigma_y = \sigma_0 + Q \left( 1 - \mathrm{exp} (-C \varepsilon_p)\right)}$