Institute of Metals Division - Theoretical Determination of the Slip System with Highest Resolved Shear Stress in a Fcc Crystal for any Orientation of the Tensile Axis (TN)

By computing the values of the resolved shear stress for a great many orientations of the tensile axis on all 12 (111) <110> slip systems, Taylor and Elam&apos; were able to map out a stereogram of slip systems, given here, Fig. 1, in Diehl&apos;s notation (see, for example Hauser and Jackson&apos;). The purpose of the present note is to derive such a stereogram by mathematical reasoning and to give a simple method of defining the system with highest resolved shear stress for any direction of the tensile axis, thereby eliminating tedious computation of the Schmid factor. Let the tensile axis, of direction cosines [xyz], make the angles 0 and relative to the slip plane normal [HKL] and the slip direction [hkl] respectively. Fig. 2 shows a simple cubic cell whose edges define a Cartesian coordinate system. Let A be a unit vector of origin 0 supported by the tensile axis. Its end point [xyz] then lies on a sphere of unit radius, on the positive hemisphere of which the planes z = 0, x = y, y = z limit the so called primitive triangle of a standard [loo] stereographic projection. For any point of this triangle, the following hold: 0«2 «y *sx « 1 [l] Now, for given [xpz], the most highly stressed system will be the one for which the following expression of the Schmid factor has the largest possible value: (HKL) belonging to the form (111) and [hkl] to the form <110>, the denominator on the right hand side of Eq. [2] is equal to 1.fi.a. The unknowns H, ... h, . . . can now be determined by correctly distributing +1 and -1 to H, K, L, and +1, -1,O to h,k, 1 giving the numerator of Eq. [2] its greatest positive value (the angles 0 and cpare chosen acute) subject to the condition: Hh+Kk+Ll = 0 [3] since the slip direction must lie in the slip plane. Condition [3] can be satisfied only by having one of its terms vanish and the other two equal to +1 and -1. Taking the case of the tensile axis lying in the primitive triangle, x2y, z and therefore Hh will be chosen equal to +1 with H = h = +1 in order to retain a maximum of + signs. There remain two alternatives: either 1 = 0 or k = 0. 1) take 2 = 0 Then Kk = -1 and the numerator of [2] becomes: Retaining a maximum of + signs gives L = k = +1 and becomes: 2) take k = 0 Then L1 = -1and, likewise, the numerator of [2] becomes: xz - z2 + xy + yz [5] But the difference [5] - [4] can be written: (y - z)(x+y +z) [6] positive or zero by virtue of [I]. If [6] is positive (y > z), then the largest possible value for the numerator of [2] for any xyz inside the primitive triangle is given by [5], i.e., for: That is to say, the active slip system corresponding to an orientat:~on of the tensile axis inside the primitive triangle is: