Dislocation glide

Conservative motion within a plane containing both the dislocation line and its Burgers vector. For a nonscrew line their cross product determines the plane normal; a screw line can belong to more than one such plane.

A lattice-friction stress for dislocation glide, associated with Rudolf Peierls and Frank Nabarro. A simple model gives ${\tau }_P\sim 3G\exp (-2\pi w/b)$, where $w$ is core width and $b$ the Burgers vector magnitude. Spreading slip over many atomic rows reduces the periodic energy barrier; narrower cores are harder to move in this model.

A crystallographic plane together with an allowed dislocation glide direction in that plane. Geometry alone does not determine its activation stress.

Plastic flow involving more than one active slip system. Interactions between dislocations on intersecting systems can give substantially faster work hardening than easy single slip.

Dislocation glide in a hexagonal crystal on its basal $(0001)$ plane, ordinarily along a close-packed basal direction in the $⟨11\overline{2}0⟩$ family. Direction indices must be converted using the actual hexagonal metric, including the ratio $c/a$.

For uniaxial stress $\sigma$, projection onto a slip-plane normal and a slip direction gives the displayed shear stress. The angles $\varphi$ and $\lambda$ are measured from the tensile axis to that normal and direction respectively.

The ratio $M=\sigma /{\tau }_c$ connecting a macroscopic tensile flow stress to a reference critical resolved shear stress in a specified polycrystal model. Texture, slip families and compatibility assumptions matter. An order-three factor is a conventional estimate for randomly oriented cubic grains, distinct from the factor two for a perfectly oriented single slip system.

The ideal activation criterion compares resolved shear stress with a critical resolved shear stress. Non-glide stresses, notably in some body-centred cubic crystals, can modify this ideal criterion.

In an ideal single-slip tensile test neglecting elastic stretch, the specimen end-to-end vector expressed in crystal coordinates changes as $\mathbf{r}={\mathbf{r}}_0+\gamma (\mathbf{n}\cdot {\mathbf{r}}_0)\mathbf{s}$, with unit slip direction $\mathbf{s}$ and slip-plane normal $\mathbf{n}$. It moves towards the slip direction along a great circle. The perpendicular component is unchanged, so $L\sin \lambda =L_0\sin {\lambda }_0$. Secondary slip changes this kinematics; equal Schmid factors determine onset only if the systems have equal current critical stresses.

The orientation factor converting the magnitude of uniaxial stress to resolved shear stress. It lies between zero and $1/2$ because the slip-plane normal and slip direction are perpendicular.