cations, thus mutually influencing their slip behavior. In essence, the strain fields hinder the dislocation motion during plastic de-formation, resulting in a need for higher applied stresses if con-tinued deformation is wanted. In the immediate vicinity of a dis-location line, that is, within a few atomic distances of it, the just-mentioned inverse distance law is no longer valid. This area is the already mentioned dislocation core in which the lattice is considerably distorted, as shown in Figure 3.21. Its size is about 3b, where b is the magnitude of the Burgers vector.
This complex behavior of dislocations makes it nearly impos-sible at this time to calculate, for example, the stress–strain curve by a simple physical model or to predict the plasticity of mate-rials when certain experimental conditions, such as temperature, deformation speed, etc., are changed. Thus, empirical laws are still predominantly used when assessing plastic deformation.
Nevertheless, the experimentally observed three regions in stress–strain curves, as depicted in Figure 3.26 for appropriately oriented single crystalline FCC, HCP, as well as for BCC metals above Tc, can be qualitatively explained. In Stage I, slip occurs in a single slip system only; it is called the easy glide range. The strain hardening is small since essentially no interference from other slip planes takes place. The size of the easy glide region de-pends on the purity of a metal and the geometry of the sample.
In Stage II (linear hardening region) multiple glide in intersect-ing slip systems occurs and the dislocation density grows rapidly with increasing strain. The crystal hardens continuously, and the slope in the stress–strain curve steepens and is nearly constant.
The rate of strain hardening in Stage III eventually decreases due to a diminished dislocation multiplication and increased cross-slipping. It is called the dynamic recovery stage.
I
II
III
5 10
(%) FIGURE3.26.Schematic
rep-resentation of a generalized stress–strain diagram for single crystals of FCC, HCP as well as for BCC metals above Tc(Figure 3.23).
Compare with Figure 2.3 which depicts a
stress–strain diagram of a polycrystalline material.
So far, we have implied that the lattice is continuous through-out an entire piece of material, that is, we considered mainly single crystals. If a shear stress is applied to polycrystals (Figure 3.15), slip preferably commences in those grains in which the slip planes are tilted 45° to the load axis, as we shall see mo-mentarily. However, slipping is somewhat impeded by the sur-rounding grains, which may have less favorable orientations.
This results in an increase in the yield stress, y, which is larger the smaller the grains are. The following relationship is often found:
y0 兹
k d苶
(3.5)
(Hall–Petch relation), where 0is the yield stress for very large grains, k is a constant, and d is the average grain size. Upon increasing the stress further, slip successively takes place in other grains that are less favorably oriented. In other words, yielding in polycrystals does not occur simultaneously in all grains at a given stress. This results in a constantly changing yield stress, y, (rather than in a fixed one) as depicted in Fig-ure 2.6(b).
One further point needs to be clarified. We said above that slip is caused by shear stress. In general, however, materials are tested in tension and not in shear for simplicity, and for better defined experimental conditions (see Chapter 2). It is thus de-sirable to relate the applied tensile stress to the shear stress re-solved in the slip plane and in the slip direction. As a rule, the applied tensile force and the slip direction are not parallel or perpendicular to each other, but instead form an angle as de-picted in Figure 3.27. Thus, only part of the applied tensile force becomes effective for shear. This partial force is called the re-solved shear force:
Fr F cos. (3.6)
On the other hand, the area of the slip plane is A co
A s
0
(3.7)
(see Figure 3.27), where A0is the cross section of the single crys-tal rod and is the angle between the applied force and the nor-mal to the slip plane. Knowing from Eq. (2.1) that
A F
,0 (3.8)
Polycrystals
and defining:
r F A
,r (3.9)
and combining Eqs. (3.6)–(3.9) yields the shear stress, r, resolved on the slip plane and in the slip direction, that is, the resolved shear stress:
r cos cos. (3.10)
Equation (3.10) teaches us that the largest resolved shear stress is obtained if both angles and are 45°, which results in r
/2. This means that the slip plane normal, the slip direction, and the load axis are all situated in the same plane. On the other hand,ris zero for 90° or 90° (see Figure 3.27). The term cos cos in Eq. (3.10) is sometimes called the Schmid factor.
The critical resolved shear stress,0, which was introduced at the beginning of this section, is that stress which eventually causes an appreciable movement of dislocations and thus a measurable deformation of the crystal. It is a constant for a given slip sys-tem in a given crystal and varies over several orders of magni-tude, as seen in Figure 3.23. 0depends on the purity of a mate-rial, the crystal orientation (in the case of single crystals), and for BCC metals also heavily on temperature, as explained above Slip direction
Slip plane
A A0
or F Normal
to slip plane
FIGURE3.27.A slip plane normal is shown under an angle to an applied force (or stress). Note: is usually not equal to 90°.
The two angles are not nec-essarily in the same plane.