Slip in Single Crystals

Concept Check 7.1 Which of the following is the slip system for the simple cubic crystal structure Why  

Resolved shear stress—dependence on applied stress and orientation of stress direction relative to slip plane normal and slip direction 

A metal single crystal has a number of different slip systems that are capable of operating. The resolved shear stress normally differs for each one because the orientation of 

Yield strength of a single crystal— dependence on the critical resolved shear stress and the orientation of the most favorably oriented shp system 

In response to an apphed tensile or compressive stress, slip in a single crystal commences on the most favorably oriented shp system when the resolved shear stress reaches some critical value, termed the critical resolved shear stress it represents the minimmn shear stress required to initiate slip and is a property of the material that determines when yielding occurs. The single crystal plastically deforms or yields when T (max) = t ss, and the magnitude of the apphed stress required to initiate yielding (i.e., the yield strength o-y) is 

---

Figure 5.9 Schematic illustration of plastic deformation in single crystals by (a) slip and (b) twinning. From Z. Jastrzebski, The Nature and Properties of Engineering Materials, 2nd ed. Copyright 1976 by John Wiley Sons, Inc. This material is used by permission of John Wiley Sons, Inc.

It is noted that one of the possible slip systems in single crystals shows an anomalous temperature dependence for the yield stress (Minonishi, 1991 Umakoshi etal., 1993b), i.e. the yield stress increases with rising temperature until a maximum is reached. Such an anomalous temperature dependence is characteristic of various in-termetallics and has been analyzed in much detail for the well-known case of Nij Al, which is discussed in Sec. 4.1.2. The findings for NijAl, however, do not apply... 

There are many mechanisms that can lead to plastic deformation in single crystals, but the most important is slip. The two things that we need to consider are the inherent resistance to the movement of dislocations provided by the periodicity of the lattice and the orientation of the crystal with respect to the applied stress. 

Plastic deformation is more difficult in polycrystals than in single crystals because now we have to consider what happens at the grain boundaries. Grain boundaries act as barriers to dislocation motion and if adjacent grains are not favorably oriented for slip to continue, dislocations will pile up at the boundary. 

Typically around indents in single-crystal faces series of slip lines are visible as shown in Figure 3.1 and Figure 6.23. These are sketched on Figures 2.8 and 3.4. Such glide bands contain systems of loops lying in several... 

In single crystals, there are preferred planes where dislocations may propagate, referred to as slip planes. For a particular crystal system, the planes with the greatest atomic density will exhibit the most pronounced slip. For example, slip planes for bcc and fee crystals are 110 and 111, respectively other planes, along with those present in hep crystals, are listed in Table 2.9. Metals with bcc or fee lattices have significantly larger numbers of slip systems (planes/directions) relative to hep. For example, fee metals have 12 slip systems four unique 111 planes, each... 

The typically (100) oriented cleavage planes in single crystals grown from Eu-rich melts show 10 to 10 cm"2 dislocation etch pits. They sometimes form linear (100) or (110) oriented arrays, probably resulting from slip bands. Cubic voids with dimensions of some irn are seen in material annealed at 1600°C for 50 h. The total amount is a few percent by volume. Electrical data indicate the existence of a remarkable amount of Se vacancies in the as-grown crystals, Kuivalainen et al. [20], cf., Stubb [21]. 

A usual metal crystal contains about 10 dislocation lines/cm. Moving all of these out of the crystal under the influence of an applied stress does not account for the observed shear strain. Suppose our crystal is a cube 1 cm on a side and it is oriented in such a way that a shear stress is applied on the slip plane in a possible slip direction. Suppose one-third of the 10 dislocation lines have this slip direction and are oriented favorably for slip (this is clearly an overestimate), then taking b equal to 3 x 10 cm gives a total shear strain of 1 %. Shear strains much greater than this are observed, sometimes exceeding 100% in single crystals. Clearly, dislocation multiplication processes... 

At room temperature, NiAl deforms almost exclusively by (100) dislocations [4, 9, 10] and the availability of only 3 independent slip systems is thought to be responsible for the limited ductility of polycrystalline NiAl. Only when single crystals are compressed along the (100) direction ( hard orientation), secondary (111) dislocations can be activated [3, 5]. Their mobility appears to be limited by the screw orientation [5] and yield stresses as high as 2 GPa are reported below 50K [5]. However, (110) dislocations are responsible for the increased plasticity in hard oriented crystals above 600K [3, 7]. The competition between (111) and (110) dislocations as secondary slip systems therefore appears to be one of the key issues to explain the observed deformation behaviour of NiAl. 

---