A semi-infinite edge dislocation model for the proportionality limit stress of metals under high strain rate

Abstract

Micromechanics of strain rate dependent elastic response, within the proportionality limit in metals is investigated, on the basis of dislocation kinetics. It is postulated that, the strain rate dependence of proportionality limit stress is dominated by inertia of dislocations, over drag controlled mechanisms. Subsequently, kinetic energy of accelerating edge dislocation at its incipient motion, is expressed. The proposed, inertia-dominated model is non dissipative in nature when compared with that of Frank-Read dislocation nucleation-based model and dislocation-drag mechanism-based model at high strain rates. Using Hamiltonian formalism, a new rate dependent slip criterion with corresponding threshold shear stress is derived. Experimental data on FCC samples, Aluminium-1100-0 and Oxygen free Copper; and BCC samples, pure Iron and mild steel, within a benchmark strain rate of 10⁴ s⁻¹, are used to validate the model prediction. Reported theory on dislocation drag controlled model is compared with the proposed inertia-based theory, using published experimental data.

Appendices

Appendix

In order to provide all information required for the development of model, additional remarks are given subsequently.

Appendix A: train energy of initial-state disregistry

From the theory of elasticity, at-rest (static) shear stress generated at any point (X, Y), for an edge dislocation is as follows [12],

$$ {\upsigma }_{{{\text{XY}}}} = \frac{{{\text{Gb}}}}{{2{\uppi }\left( {1 - {\upnu }} \right)}}\frac{{{\text{X}}\left( {{\text{X}}^{2} - {\text{Y}}^{2} } \right)}}{{\left( {{\text{X}}^{2} + {\text{Y}}^{2} } \right)^{2} }} $$

(42)

where G is the shear modulus of bulk material and ν is the Poisson’s ratio. X and Y are the distances between the point at which stress is evaluated (x, y) and the point at which disregistry exists (${\text{x}}^{\prime } ,{\text{y}}^{\prime }$). Following the Eq. (42), an infinitesimal dislocation of Burgers vector ${\rm{d}}\delta \left( {{{\rm{x}}^\prime },{\rm{t}}} \right)$, located at ${\text{x}}^{\prime }$(${\text{X}} = {\text{x}} - {\text{x}}^{\prime }$) produces a shear stress at some other point x in glide plane (${\rm{Y}} = {\rm{y}} - {{\rm{y}}^\prime } = 0$), as follows. Note that, at the point of discontinuity X = 0, ${\upsigma }_{{{\text{xy}}}}$ can’t be defined.

$$ {\upsigma }_{{{\text{xy}}}} \left( {{\text{x}},0} \right) = \frac{{\frac{{\text{G}}}{{2{\uppi }\left( {1 - {\upnu }} \right)}}{\text{d}}\delta \left( {{\text{x}}^{\prime } {\text{,t}}} \right)}}{{{\text{x}} - {\text{x}}^{\prime } }}\left. \right|_{{{\text{t}} = 0}} $$

(43)

Consequently, the elastic strain energy stored in the upper half of the crystal is equal to the work done by surface forces in generating a disregistry δ(x, t). Hence the strain energy due to disregistry is given by the following equation,

$${{\updelta d}}{{\rm{W}}^{\rm{E}}} = \frac{{\rm{G}}}{{2{{\uppi }}\left( {1 - {{\upnu }}} \right)}}\frac{{{\rm{d}}\delta\left( {{{\rm{x}}^\prime },{\rm{t}}} \right)}}{{{\rm{x}} - {{\rm{x}}^\prime }}}\delta \left( {{\rm{x}},{\rm{t}}} \right){\left. \right|_{{\rm{t}} = 0}}$$

(44)

Appendix B: At-rest critical shear stress: Peierls-Nabarro (PN) equation for self-stress

In this section, we explain self-stress of the dislocation based on Peierls-Nabarro model [12]. Self-stress is formed by the dislocation line curvature and it is also known as line tension [141]. In our study, the line tension for a semi-infinite straight dislocation is included in the dislocation core energy. Within the glide plane, at each point a distance ${\text{x}}^{\prime }$ from the dislocation line, disregistry δ $\left( {\text{x}} \right)$ of the upper half of the crystal (y > 0) with respect to the lower half (y < 0) results from the continuous distribution of infinitesimal dislocations with disregistry ${\updelta }^{\prime } \left( {{\text{x}}^{\prime } } \right){\text{dx}}^{\prime }$. Strain energy at t ≤ 0 can be written as follows,

$$ {\delta W}^{{\text{E}}} = - \frac{{\text{G}}}{{2{\uppi }\left( {1 - {\upnu }} \right)}}\mathop \int \limits_{{ - {\text{L}}}}^{{\text{L}}} \mathop \int \limits_{{ - {\text{L}}}}^{{\text{L}}} {\updelta }^{\prime } \left( {\text{x,0}} \right){\updelta }^{\prime } \left( {{\text{x}}^{\prime } {,}0} \right)\ln \left( {{\text{x}} - {\text{x}}^{\prime } } \right){\text{dxdx}}^{\prime } $$

(45)

For a dislocation at rest, the total energy is contributed by two parts. The first one is, surface energy contribution from the misfit of atomic half planes across the glide plane. This theoretical measure of misfit surface energy can be obtained by cutting a perfect crystal (having no stacking fault) along the glide plane and then displacing one of the half crystals relative to the other in the tangent direction of glide plane and then re-joining to get new bonding between the adjacent atoms. This surface energy is named as generalized stacking fault energy (GSFE) [39]. Under immobile condition, restoring force is calculated as the gradient of generalized stacking fault energy, γ [39]. This restoring force has the same formal interpretations as that of PN stress [39]. Misfit energy is given by the following equation,

$$ {\text{U}}_{{{\text{SFE}}}} = \mathop \int \limits_{ - L}^{L} {\upgamma }\left( {{\updelta }\left( {\text{x,0}} \right)} \right){\text{dx}} $$

(46)

Then the restoring force is calculated as follows,

$$ {\text{F}}_{{\text{b}}} \left( {\updelta } \right) = - \nabla ({\text{U}}_{{{\text{SFE}}}} ) $$

(47)

The second part of energy contribution is from the elastic energy stored in the two elastic half-spaces as a function of disregistry. Strain energy equation for an edge dislocation at rest is given in Eq. (45). Hence, the total energy contribution is given by adding Eqs. (45) and (46) to get the following equation,

$$ \begin{aligned}{{\rm{U}}_{{\rm{Total}}}}\left( {\delta \left( {{\rm{x}},0} \right)} \right) &= \mathop \int \limits_{ - {\rm{L}}}^{\rm{L}} {{\upgamma }}\left( {\delta \left( {{\rm{x}},0} \right)} \right){\rm{dx}} \\ &\quad+ ~\frac{{\rm{G}}}{{2{{\uppi }}\left( {1 - {{\upnu }}} \right)}}\mathop \int \limits_{ - {\rm{L}}}^{\rm{L}} \mathop \int \limits_{ - {\rm{L}}}^{\rm{L}} {\delta ^\prime }\left( {{\rm{x}},0} \right){\rm{dx}}~\left\{ {\ln \left( {{\rm{x}} - {{\rm{x}}^\prime }} \right)\frac{{{\rm{d}}\delta \left( {{{\rm{x}}^\prime },0} \right)}}{{{\rm{d}}{x^\prime }}}{\rm{d}}{{\rm{x}}^\prime }} \right\}\end{aligned} $$

(48)

Upon minimization of Eq. (48), one can get the static resisting force as follows,

$${{\rm{F}}_{\rm{b}}}\left( \delta \right) = {\sigma _{{\rm{PN}}}}{\rm{b}} = \frac{{{\rm{Gb}}}}{{2{{\uppi }}\left( {1 - {{\upnu }}} \right)}}\mathop \int \limits_{ - {\rm{L}}}^{\rm{L}} \left\{ {\frac{1}{{{\rm{x}} - {{\rm{x}}^\prime }}}\frac{{{\rm{d}}\delta \left( {{{\rm{x}}^\prime },0} \right)}}{{{\rm{d}}{{\rm{x}}^\prime }}}{\rm{d}}{{\rm{x}}^\prime }} \right\}$$

(49)

where ${\upsigma }_{{{\text{PN}}}}$ represents the Peierls-Nabarro stress.

Appendix C: Derivation of inertial mass factor

Value of coefficient α, is found by equating the scaled measures (by a factor, ${\upkappa }$) of total mass given in Eq. (28) followed by (29) and using Eq. (4), as follows,

$$\mathop \int \limits_{ - {\rm{L}}}^{\rm{L}} \mathop \int \limits_{ - \frac{{\rm{d}}}{2}}^{\frac{{\rm{d}}}{{\rm{2}}}} {{\upalpha }}{{{\uprho }}_{\rm{m}}}{\rm{dy~}}{\delta ^\prime }{\rm{dx}} = {{\upkappa }}\left[ {1 + {{\left( {\frac{{{{\rm{c}}_2}}}{{{{\rm{c}}_1}}}} \right)}^4}} \right]\frac{{{{{\uprho }}_{\rm{m}}}{{\rm{b}}^2}}}{{4{{\uppi }}}}\ln \left( {\frac{{8{{\rm{c}}_0}{{\uptau }}}}{{\rm{d}}}} \right)$$

(50)

$${{\upalpha }}{{{\uprho }}_{\rm{m}}}\mathop \int \limits_{ - {\rm{L}}}^{\rm{L}} \mathop \int \limits_{ - \frac{{\rm{d}}}{2}}^{\frac{{\rm{d}}}{2}} {\rm{dy}}~{\delta ^\prime }{\rm{dx}} = {{\upkappa }}\left[ {1 + {{\left( {\frac{{{{\rm{c}}_2}}}{{{{\rm{c}}_1}}}} \right)}^4}} \right]\frac{{{{{\uprho }}_{\rm{m}}}{{\rm{b}}^2}}}{{4{{\uppi }}}}\ln \left( {\frac{{8{{\rm{c}}_0}{{\uptau }}}}{{\rm{d}}}} \right)$$

(51)

$${{\upalpha }}{{{\uprho }}_{\rm{m}}}{\rm{d}}\mathop \int \limits_{ - {\rm{L}}}^{\rm{L}} {\delta ^\prime } = {{\upkappa }}\left[ {1 + {{\left( {\frac{{{{\rm{c}}_2}}}{{{{\rm{c}}_1}}}} \right)}^4}} \right]\frac{{{{{\uprho }}_{\rm{m}}}{{\rm{b}}^2}}}{{4{{\uppi }}}}\ln \left( {\frac{{8{{\rm{c}}_0}{{\uptau }}}}{{\rm{d}}}} \right)$$

(52)

Using Eq. (4) in Eq. (52), one can get the following equation,

$${{\upalpha }}{{{\uprho }}_{\rm{m}}}{\rm{bd}} = {{\upkappa }}\left[ {1 + {{\left( {\frac{{{{\rm{c}}_2}}}{{{{\rm{c}}_1}}}} \right)}^4}} \right]\frac{{{{{\uprho }}_{\rm{m}}}{{\rm{b}}^2}}}{{4{{\uppi }}}}\ln \left( {\frac{{8{{\rm{c}}_0}{{\uptau }}}}{{\rm{d}}}} \right)$$

(53)