An anisotropic micromechanics model for predicting the rafting direction in Ni-based single crystal superalloys

Abstract

An anisotropic micromechanics model based on the equivalent inclusion method is developed to investigate the rafting direction of Ni-based single crystal superalloys. The micromechanical model considers actual cubic structure and orthogonal anisotropy properties. The von Mises stress, elastic strain energy density, and hydrostatic pressure in different inclusions of micromechanical model are calculated when applying a tensile or compressive loading along the [001] direction. The calculated results can successfully predict the rafting direction for alloys exhibiting a positive or a negative mismatch, which are in agreement with pervious experimental and theoretical studies. Moreover, the elastic constant differences and mismatch degree of the matrix and precipitate phases and their influences on the rafting direction are carefully discussed.

Appendices

Appendix 1

For anisotropic materials, the elastic constants matrix is symmetric and can be expressed as

$$\begin{aligned} {\varvec{C}}=\left[ {{ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} {C_{11}} &{} {C_{12}} &{} {C_{13}} &{} {C_{14}} &{} {C_{15}} &{} {C_{16}} \\ &{} {C_{22}} &{} {C_{23}} &{} {C_{24}} &{} {C_{25}} &{} {C_{26}} \\ &{} &{} {C_{33}} &{} {C_{34}} &{} {C_{35}} &{} {C_{36}} \\ &{} &{} &{} {C_{44}} &{} {C_{45}} &{} {C_{46}} \\ &{} {\hbox {Sym.}} &{} &{} &{} {C_{55}} &{} {C_{56}} \\ &{} &{} &{} &{} &{} {C_{66}} \\ \end{array}}} \right] . \end{aligned}$$

(24)

According to Ting [26], the Stroh eigenvalues $p_{r}$ and the eigenvectors ${\varvec{a}}_{r}$ satisfy the following eigenrelation

$$\begin{aligned} \left[ {{\varvec{Q}}+p\left( {{\varvec{R}}+{\varvec{R}}^{\mathrm{T}}} \right) +p^{2}{\varvec{T}}} \right] {\varvec{a}}=0, \end{aligned}$$

(25)

where the superscript T denote the transpose of matrix, and ${\varvec{Q}}, {\varvec{R}}, {\varvec{T}}$ are given by

$$\begin{aligned} \begin{array}{l} {\varvec{Q}}=\left[ {{\begin{array}{ccc} {C_{11}} &{}\quad {C_{16}} &{}\quad {C_{15}} \\ {C_{16}} &{}\quad {C_{66}} &{}\quad {C_{56}} \\ {C_{15}} &{}\quad {C_{56}} &{}\quad {C_{55}} \\ \end{array}}} \right] ,\quad {\varvec{R}}=\left[ {{ \begin{array}{ccc} {C_{16}} &{}\quad {C_{12}} &{}\quad {C_{14}} \\ {C_{66}} &{}\quad {C_{26}} &{}\quad {C_{46}} \\ {C_{56}} &{}\quad {C_{25}} &{}\quad {C_{45}} \\ \end{array}}} \right] , \\ {\varvec{T}}=\left[ {{ \begin{array}{ccc} {C_{66}} &{}\quad {C_{26}} &{}\quad {C_{46}} \\ {C_{26}} &{}\quad {C_{22}} &{}\quad {C_{24}} \\ {C_{46}} &{}\quad {C_{24}} &{}\quad {C_{44}} \\ \end{array}}} \right] . \\ \end{array}. \end{aligned}$$

(26)

Due to ${\varvec{a}}\ne {\varvec{0}}$, p satisfies the sextic equation

$$\begin{aligned} \left| {{\varvec{Q}}+p\left( {{\varvec{R}}+{\varvec{R}}^{\mathrm{T}}} \right) +p^{2}{\varvec{T}}} \right| =0. \end{aligned}$$

(27)

According to Eq. (27), the solutions to the sextic equation are three pairs of complex conjugate roots for p. We take $\hbox {Im}p_{r}> 0$, $r = 1, 2, 3$.

Introduce the matrix ${\varvec{N}}$ and vector ${\varvec{b}}$, and the expression of ${\varvec{N}}$ is

$$\begin{aligned} {\varvec{N}}=\left[ {{\begin{array}{c@{\quad }c} {-{\varvec{T}}^{-1}{\varvec{R}}^{\mathrm{T}}}&{} {{\varvec{T}}^{-1}} \\ {-{\varvec{Q}}+{\varvec{RT}}^{-1}{\varvec{R}}^{\mathrm{T}}}&{} {-{\varvec{RT}}^{-1}} \\ \end{array}}} \right] . \end{aligned}$$

(28)

Meanwhile, ${\varvec{p}}, {\varvec{a}}, {\varvec{b}}$, and ${\varvec{N}}$ satisfy the following relation

$$\begin{aligned}&{\varvec{N}}\left[ {{\begin{array}{l} {{\varvec{a}}_{r}} \\ {{\varvec{b}}_{r}} \\ \end{array}}} \right] =p_{r}\left[ {{\begin{array}{l} {{\varvec{a}}_{r}} \\ {{\varvec{b}}_{r}} \\ \end{array}}} \right] \quad r=1,2,3, \end{aligned}$$

(29)

$$\begin{aligned}&{\varvec{b}}_{\alpha }^{\mathrm{T}} {\varvec{a}}_{\beta } +{\varvec{a}}_{\alpha }^{\mathrm{T}} {\varvec{b}}_{\beta } =\delta _{\alpha \beta } , \end{aligned}$$

(30)

where $\delta _{\alpha \beta }$ is Kronecker delta.

Appendix 2

For the integral of Eq. (20), if the inclusion is rectangular, the contribution of each straight-line segment along the boundary of the inclusion can be obtained. We suppose the start point of certain straight-line segments is $\left( {x_{1},y_{1}} \right)$, and the end point is $\left( {x_{2},y_{2}} \right)$. As in Pan [25], the contribution of this straight-line segment along the boundary of the inclusion is

$$\begin{aligned} \varepsilon _{\beta \alpha }= & {} 0.5n_{i}C_{ijlm} \varepsilon _{lm}^{**} \frac{l}{\uppi }\hbox {Im}\left\{ {\sum \limits _{r=1}^{3}{A{ }_{jr}A_{\beta r} h_{r,\alpha }}} \right\} \nonumber \\&+\,0.5n_{i}C_{ijlm} \varepsilon _{lm}^{**} \frac{l}{\uppi }\hbox {Im}\left\{ {\sum \limits _{r=1}^{3}{A{ }_{jr}A_{\alpha r} h_{r,\beta }}} \right\} , \nonumber \\ \varepsilon _{\alpha 3}= & {} 0.5n_{i}C_{ijlm} \varepsilon _{lm}^{**} \frac{l}{\uppi }\hbox {Im}\left\{ {\sum \limits _{r=1}^{3}{A{ }_{jr}A_{3r} h_{r,\alpha }}} \right\} , \end{aligned}$$

(31)

where $\alpha$, $\beta = 1, 2$; $r = 1, 2, 3$; $n_{i}$ is the outward normal component along the line segment: $n_{1}=\frac{y_{2}-y_{1}}{l}$, $n_{2}=-\frac{x_{2}-x_{1}}{l}$; $l=\sqrt{\left( {x_{2}-x_{1}} \right) ^{2}+\left( {y_{2}-y_{1}} \right) ^{2}}$ is the length of the line segment; and $h_{r,1}$ , $h_{r,2}$ are expressed as follows

$$\begin{aligned} h_{r,1}= & {} \frac{-1}{\left( {x_{2}-x_{1}} \right) +p_{r}\left( {y_{2}-y_{1}} \right) }\ln \left( {\frac{x_{2}+p_{r}y_{2}-s_{r}}{x_{1}+p_{r}y_{1}-s_{r}}} \right) , \nonumber \\ h_{r,2}= & {} \frac{-p_{r}}{\left( {x_{2}-x_{1}} \right) +p_{r}\left( {y_{2}-y_{1}} \right) }\ln \left( {\frac{x_{2}+p_{r}y_{2}-s_{r}}{x_{1}+p_{r}y_{1}-s_{r}}} \right) .\nonumber \\ \end{aligned}$$

(32)