Technical Notes - Stress Analysis of a Single Crystal in Pure Torsion

IT has been observed that a hexagonal-close-packed crystal will undergo the same macroscopic displacements as an isotropic material if the basal plane is perpendicular to the axis of twist.&apos; Other orientations may produce both bending and twisting, but this analysis will be confined to pure torsion of a right circular cylinder. The z axis, the axis of twist, and the specimen axis are coincident. The displacements are given by Uq = rz, Ur = Uz = 0 [1] where 4 is the angle of twist per unit length. This requires rotational slip, as discussed by Frank&apos; and Wilman, which may be observed readily. The relative rotation of adjoining slip planes is simply produced by a movement of a planar grid of screw dislocations."." The screw dislocations are of the same sign and they move radially inward. For hexagonal-close-packed metals, the grid is hexagonal and the screw dislocations lie in the < 2110 > directions. The displacement of a point is the resultant of two displacements produced by intersecting screws sweeping over the point. The dislocations must pass over the point (r, o) in such a manner that its resultant displacement corresponds to the observed macroscopic displacement. If 6&apos; is measured from the [1010] direction, then for 0 < ? < ~/3, only the [1210] and 121101 type screw dislocations are needed to produce the required displacement. If n1 screws of type [1210] and n2 of type [2110] pass over a point, then, in order that Ur = 0 n, sin ? - n2 sin (/3 — 6&apos;) = 0. [2] If the tangential displacement is to be independent of 0, then n1 cos ? + n2cos (/3 —?) = N [3] where N is independent of ? and thus depends only on r. Nb is the tangential displacement where b is the Burgers vector. Since U? is a linear function of r, then N must be a linear function of r. If a2 is the outer radius and a, is the inner radius to which the plastic deformation has extended, then N = Nm2 r-a1 [4] This function meets the conditions that at