Institute of Metals Division - Calculation of Martensite Nucleus Energy Using the Reaction-Path Model

ACCORDING to the "reaction-path" modell,2 of martensite nucleation, the shear angle of the embryonic martensite plate must be treated as a variable, and included in any calculation of nucleus critical size. Also, as can be deduced from this model, the interfacial free energy between austenite and martensite does not reach its final value until the shear is completed. It is zero for zero shear angle. However, in order to account for the kinetics of the martensite transformation, some sort of interfacial energy barrier appears to be necessary even with the reaction-path model, for otherwise the volume and the energy of formation of the critical size nucleus both collapse to zero.3 Cohen independently suggested that surface energy could be incorporated into the reaction-path model, with the overall free energy of a martensite embryo being a function of its volume and shear angle.' It is possible to estimate the energy associated with the formation of a critical-size martensite nucleus starting with the reaction-path model and including a surface free-energy barrier. As the dependence of interfacial free energy upon shear angle is unknown, a simple type of dependence will be assumed, with the belief that the true dependence would not lead to appreciably different results. Consider the work required to form a lenticular martensite plate with radius r, thickness t, and shear angle 8. There are three contributions; one being the interfacial free energy, one being the free energy change in the martensite plate, and one being the free energy increase in the surrounding austenite. The interfacial free energy u is assumed to depend upon the shear angle 0 according to the relationship s=s0(?/?0)n [1] where 8, is the equilibrium shear angle and n is an exponent that may lie in the range 0 n 2. The work required to form the interfaces of a martensite plate then is W. = 2pr² s0(?/?0)n [2] The free energy change per unit volume of martensite is composed of two parts, one the ordinary volume free energy ?f1. which is negative, and the other the elastic strain energy G?m²/2, where G is the shear modulus and 7, the shear strain relative to the martensite structure. This expression for the strain energy is valid only when the shear strain ym, is sufficiently small that the martensite is within its linear elastic range. There is no doubt that ym, lies beyond the linear elastic range for embryos that are considerably subcritical. However, for critical nuclei it will be shown that ym, is 1.5 pct or less, within the linear elastic range of martensite. For embryos of nearly critical size, then, the strain energy of the martensite is correctly given by G?m²/2. The shear strain in the martensite is ym, = 8, — 8, and the work required to form the strained martensite is Wm --= (pr²t/2) [?fv + G(?O - ?)²/2] [3] The free energy change in the austenite is entirely that due to elastic distortion. The elastic strain is not uniformly distributed in the austenite, being large near the martensite plate and small elsewhere. Approximately, however, the energy corresponds to a uniform shear strain ya= (?t/2)/r [4] throughout the volume 4pr³/3 surrounding the plate. The work required to strain the surrounding austenite then is Wa = (4pr³/3) (G?a²/2) = (G?²/6) prt² [51 For simplicity, the same shear modulus G is assumed for each structure. The total free energy for forming a plate then is W = W3 + Wm + Wa. = 2pr² s0 (?/p?0)n + (pr²t/2) [?fr+G(?0-?)²/2] + (G?²6) prt2 [6] This expression is correct for nuclei and for embryos of nearly critical size, where, as will be shown, the strain energy in the martensite is correctly given by the expression G (? — ?)². Having W as a function of r, t, and 8, as in Eq. 6, there is a saddle-point where W has a stationary value, W subsequently decreasing indefinitely as the nucleus volume increases along the reaction path. The stationary value of W is the energy of the critical nucleus. The critical nucleus has radius, thickness, and shear angle such that ?W/?r - awlat: = ?W/?p? = 0. Performing these differentiations and calculating the critical nucleus energy, W* = [8192p(G?/6)²;s/27 ?fv4] [7] where a= (?/?0)3n+1[l +G(8"-8)'/2af.]' [7a] and where 8 is to be determined from the equation (1 + 3n/4) + G8(6O - (9)/[Af. +G(6>o-6>)72] = 0 [8] For ?f, near —200 cal per mol or —10" ergs per cc, and 8, near 1/6, as for iron-base alloys, Eq. 8 gives ?0 - ? ~ - (4 + 3n) ?f1./4G0O [9] as the difference between the equilibrium shear angle and the actual shear angle for a critical nu-