Modeling of De-cohesion and the Initiation of Hot Tearing in Coherent Mushy Zones of Metallic Alloys

Abstract

The initiation of a hot tear in the coherent mushy zone of metallic alloys is associated commonly with the opening up of the solid skeleton caused by thermally induced deformation. A previously established constitutive model for the continuum modeling of coherent mushy zones has been further developed in the current study to address the opening up, or decohesion, of the solid skeleton associated with volumetric tensile deformation. Whereas the original model accounts for the cohesion of the solid skeleton caused by the deformation by means of an internal variable, an additional internal variable accommodating the decohesion has been introduced in the new model. The modeled decohesion is interpreted as the initiation of a hot tear.

Appendix Constraint on C

The general tensorial stress in the main directions is given by

$$ {\varvec{\sigma}} = \left(\begin{array}{lll} \sigma _{11} & 0 & 0\\ 0 & \sigma _{22} & 0 \\ 0 & 0 & \sigma _{33} \\ \end{array} \right) $$

(A1)

The deviatoric part of stress is then

$$ {\varvec{\tau}} = \left(\begin{array}{lll} {\frac{1}{3}}(2\sigma _{11}-\sigma _{22}-\sigma _{33}) & 0 & 0 \\ 0 & {\frac{1}{3}}(-\sigma _{11}+2\sigma _{22}-\sigma _{33}) & 0 \\ 0 & 0 & {\frac{1}{3}}(-\sigma _{11}-\sigma _{22}+2\sigma _{33})\\ \end{array}\right) $$

(A2)

According to Eq. [24], the viscoplastic strain rate is given by

$$ \begin{aligned} {\dot{\varvec{\epsilon}}}_{\rm s}^{p}=&\left( \begin{array}{lll}\dot\epsilon _{11} & 0 & 0 \\ 0 & \dot\epsilon _{22}& 0 \\ 0 & 0 & \dot\epsilon _{33}\\ \end{array} \right)= {\frac{\dot{\epsilon}_{0}\exp{\left({\frac{-Q}{RT}}\right)}} {\sigma_{0}^{n}}}\left({\frac{1}{9}}A_2(g_{\rm s})J_1^2+3A_3(g_{\rm s})J_2 \right)^{{\frac{n-1}{2}}}\\ & \left\{ {\frac{-A_{2}}{9D^n}}\left(\begin{array}{lll} -\sigma _{11} & -\sigma _{22} & -\sigma _{33} \\ \end{array}\right) \left( \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{array} \right) \right.\\ & + \left. {\frac{A_{3}}{2C^n}} \left( \begin{array}{lll} (2\sigma _{11}-\sigma _{22}-\sigma _{33}) & 0 & 0 \\ 0 & (-\sigma _{11}+2\sigma _{22}-\sigma _{33}) & 0 \\ 0 & 0 & (-\sigma _{11}-\sigma _{22}+2\sigma _{33})\\ \end{array} \right)\right\} \end{aligned} $$

(A3)

Now, if the loading direction is in one direction only, the 11-direction, say ($i.e., \sigma _{11}\neq 0, \sigma _{22}=\sigma _{33}=0$), the viscoplastic strain rates in the two other main directions are

$$ \dot{\varepsilon}^{p}_{22}=\dot{\varepsilon}^{p}_{33}= \left({\frac{A_{2}}{9D^n}} -{\frac{A_{3}}{2C^n}}\right) \sigma _{11} $$

(A4)

Obviously, the quantity cannot be larger (smaller) than zero if σ₁₁ is larger (smaller) than zero. This gives the constraint

$$ \left({\frac{A_{2}}{9D^n}}-{\frac{A_{3}}{2C^n}}\right)\leq 0 \Rightarrow C\leq D/k\hbox{ where } k=\left[{\frac{2}{9}}{\frac{A_{2}(g_{\rm s})}{A_{3}(g_{\rm s})}}\right]^{{\frac{1}{n}}} $$