# Slip via dislocation motion

It is possible to estimate the stress required for slip by movement of lattice planes past one another (*i.e.* in
a perfect crystal) and for slip by dislocation motion.

### Stress required for slip in a perfect crystal

One can estimate the shear stress required to cause slip in a perfect crystal as follows. Imagine two close-packed planes
with separation *h* and interatomic distance *b*. On application of a shear stress *τ*,
the top plane moves a distance *u* with respect to the bottom plane as shown in the diagram.

The shear stress *τ* needed for slip to proceed in this way is zero when *u*
= 0, *u* = ½ *b* and *u* =
*b* etc. As a first approximation we assume that *τ* varies sinusoidally with *u*, then a suitable
form for *τ* is:

$$\tau = {\tau _{crit}}\sin {{2\pi u} \over b}$$

where *τ** _{crit}* is a constant that equals the shear stress that must
be applied to move one close-packed plane continuously over the next, and hence corresponds to the maximum value of

*τ*. This maximum occurs when

*u*=

*b*/4.

*τ* is related to the shear strain *γ* by the shear modulus, *G*:

$$\tau = G\gamma $$

and the shear strain angle *γ* is given by

$$\gamma = {u \over h}$$

It is now possible to make the assumption that the ratio *u*/*b* is small since the macroscopic slip is the result of a small amount of slip on each plane, rather than a large amount on one plane. This simplifies the expression since sin *x* ~ *x*.

Combining equations [1] and [3],

$$\tau = {\tau _{crit}}\sin {{2\pi u} \over b} = 2\pi {\tau _{crit}}{{h\gamma } \over b}$$

Rearranging equation [2] and substituting,

$${\tau \over \gamma } = G = {{2\pi {\tau _{crit}}h} \over b}$$

hence

$${\tau _{crit}} = {{Gb} \over {2\pi h}}$$

For close-packed spheres, *b*/*h* = 2/3^{1/2}, which gives

$${\tau _{crit}} = {{Gb} \over {\pi \sqrt 3 }} \approx {G \over 5}$$

### Stress required for slip by means of dislocation motion

When a dislocation line moves through a crystal, the interatomic forces across the slip plane are overcome in a series of local movements, rather than breaking all of the atomic bonds across the slip plane at once as would occur in a perfect lattice. The stress needed to drive a dislocation through a crystal continuously is called the *Peierls-Nabarro stress*, τ_{p}. Calculation of τ_{p} is difficult, and it depends on the ratio of the dislocation
width *w* to the magnitude of the Burgers vector *b*. One simple model gives:

$${\tau _p} = 3G\exp \left( { - {{2\pi w} \over b}} \right)$$

Setting *w* = *b*, which is the minimum reasonable width for a dislocation, and
therefore will give the maximum τ_{p}, reduces equation [6] to:

$${\tau _p} = 3G\exp ( - 2\pi ) \approx {G \over {180}}$$

This approximate calculation shows that the Peierls-Nabarro stress to move a dislocation through a lattice (equation [7]) is orders of magnitude less than the stress required to cause slip in a perfect lattice (equation [5]). Therefore it follows that slip occurs by dislocation motion when dislocations are present in the lattice.