A multi-field coupled FEM model for one-step-forming process of spark plasma sintering considering local densification of powder material

Abstract

A mechanical constitutive model of powder material is introduced to a fully coupled thermal–electric–mechanical finite element model to simulate the one-step-forming spark plasma sintering (SPS) process of metal powders. The effects of displacement field and local density distribution on sintering are considered in this article, which are generally neglected in the existing SPS models. The mechanical, thermal, and electrical parameters of powders are assumed as functions of local relative density and temperature. The simulated varying displacement field remodels the distributions of temperature and electric potential by changing the contact thermoelectric resistances. For the 20, 40, and 60 MPa external pressures, the simulation indicates that the sintering temperature and the temperature gradient within powders are decreased by enhancing the external pressure, and the comprehensive effect of stress promotes the densification of the colder regions. Thus, the interrelationship between the temperature gradient and the intrinsic stress distribution plays an important role in the densification mechanism of SPS powders.

Appendix

Iron properties

$$ \begin{aligned} \rho \left( {{\text{kg m}}^{ - 3} } \right) \,& = { 788}0 \\ E_{\text{s}} \left( T \right)\left( {\text{GPa}} \right) & = 1.96 \times 10^{2} (1 - 2.1 \times 10^{ - 4} (T - 298) - 1.7 \times 10^{ - 7} (T - 298)^{2} - 3.2 \times 10^{ - 10} (T - 298)^{3} ) \\ \alpha \left( T \right)\left( {1/{\text{K}}} \right) & = \left( {4.994 + 0.030T - 2.696 \times 10^{ - 5} T^{2} + 8.390 \times 10^{ - 9} T^{3} } \right) \times 10^{ - 6} \\ K_{\text{s}} (T)\left( {{\text{W/}}\left( {\text{m K}} \right)} \right) & = 73.014 - 5.0 \times 10^{{^{ - 2} }} \left( {T - 273} \right) - 7.096 \times 10^{{^{ - 5} }} \left( {T - 273} \right)^{2} + 1.366 \times 10^{{^{ - 7} }} \left( {T - 273} \right)^{3} - \\ \, & \quad 5.323 \times 10^{{^{ - 11} }} \left( {T - 273} \right)^{4} \\ R_{\text{s}} (T)\left( {1/\left( {\Upomega {\text{ m}}} \right)} \right) & = \left( {0.096 + 2.139 \times 10^{ - 4} \left( {T - 273} \right) + 2.240 \times 10^{ - 6} \left( {T - 273} \right)^{2} - 1.803 \times 10^{ - 9} \left( {T - 273} \right)^{3}} \right. \\ \, & \left. {\quad + 3.571 \times 10^{ - 13} \left( {T - 273} \right)^{4} } \right)^{ - 1} \times 10^{6} \\ c_{\user2{p}} \left( T \right)\left( {{\text{J/}}\left( {\text{kg K}} \right)} \right) & = 317.202 + 0.433 \times T \\ \end{aligned} $$

Graphite properties

$$ \begin{aligned} \rho \left( {{\text{kg m}}^{ - 3} } \right) \, & = { 185}0 \\ E\left( {\text{GPa}} \right) & = 200 \\ \alpha \left( T \right)\left( {1/{\text{K}}} \right) & = 8 \times 10^{ - 6} \\ K(T)\left( {{\text{W/}}\left( {\text{m K}} \right)} \right) & = 65 - 1.7 \times 10^{ - 2} T \\ R(T)\left( {1/\left( {\Upomega {\text{ m}}} \right)} \right) & = \left( {26 - 3 \times 10^{ - 2} T + 2 \times 10^{ - 5} T^{2} - 6.4 \times 10^{ - 9} T^{3} + 7.8 \times 10^{ - 13} T^{4} } \right)^{ - 1} \times 10^{6}\,[23] \\ c_{\user2{p}} \left( T \right)\left( {{\text{J/}}\left( {\text{kg K}} \right)} \right) & = 310.5 + 1.7 \times T \\ \end{aligned} $$

Silicon–nitride properties

$$ \begin{aligned} \rho \left( {{\text{kg m}}^{ - 3} } \right) \, & = { 3184} \\ E\left( {\text{GPa}} \right) & = 330 \\ \alpha \left( T \right)\left( {1/{\text{K}}} \right) & = 2.7 \times 10^{ - 6} \left( {0.856 + 4.919 \times 10^{ - 4} T} \right) \\ K\left( T \right)\left( {{\text{W/}}\left( {\text{m K}} \right)} \right) & = 34.271 - 0.0141 \times T \\ c_{\user2{p}}\left( {{\text{J/}}\left( {\text{kg K}} \right)} \right) & = 740 \\ \end{aligned} $$

Copper properties [9]

$$ \begin{aligned} \rho \left( {{\text{kg m}}^{ - 3} } \right) \, & = { 785}0 \\ K\left( T \right)\left( {{\text{W/}}\left( {\text{m K}} \right)} \right) & = 420.66 + 0.07 \times T \\ R\left( T \right)\left( { 1 /\left( {\Upomega {\text{ m}}} \right)} \right) & = \left( {5.5 + 0.038 \times T} \right)^{ - 1} \times 10^{9} \\ c_{\user2{p}} \left( T \right)\left( {{\text{J/}}\left( {\text{kg K}} \right)} \right) & = 355.3 + 0.1 \times T \\ \end{aligned} $$