We already know that shear stress causes slip in crystals. The resolved shear com­ponent based on the magnitude of the external load, geometry of the crystal

structure, and orientation of the active slip systems needs to reach or exceed a criti­cal value in order for the slip to occur. This critical shear stress is called critical resolved shear stress. The CRSS is the single crystal equivalent of yield stress of a polycrystalline material as obtained in a standard stress-strain curve. The CRSS value for a slip system in a crystal depends on purity and temperature. The following derivation of resolved shear stress makes clear how a single crystal can undergo slip.

Erich Schmid [15] was the first to analyze the problem, one of his contributions to understanding crystal plastic deformation. Let us take a cylindrical single crystal under tensile force F with a cross-sectional area A, as illustrated in Figure 4.4. We need to find out the resolved component of stress.

The normal stress (o) is given by F/A. Now we can define the slip system with respect to the tensile axis and normal to the slip plane. Slip plane contains the slip direction along which the resolved shear stress (tR) would act. As shown in Figure 4.4, the included angle between the tensile axis (along F) and the slip plane normal (along N) is 0, and the angle between the tensile axis and the slip direction is ф. The slip plane is inclined at 0 to the tensile axis, and thus its area (A0) is given by A/cos 0. The normal force F can be resolved along the slip direction as F-cos ф. Hence,

Equation (4.1) is also known as Schmid’s law, while the term cos 0 cos ф is called Schmid factor. Schmid factor gives a measure of the slip system orientation. The resolved shear stress is the maximum when 0 = ф = 45° so that tR = o — (1Д/2) — (1Д/2) = (1/2)o. In the extreme cases, where either 0 or ф is 90°, there will be no resolved shear stress in the slip plane, and the material tend to fracture rather than

deform. A numerical example on this topic is worked out in Example 4.2. Note that even though the derivation of resolved shear stress has been made in a uniaxial tensile situation, it is equally applicable to plastic deformation under compression.

■ Example 4.2

Calculate the critical resolved shear stress for a metallic single crystal (BCC) that starts deforming in the slip system (110)[111] when the tensile stress along [010] is 100 MPa.

Solution

To solve this problem, let us first refer to Eq. (4.1).

Given the tensile stress (o) = 100 MPa; we need to calculate cos 0 and cos ф. Refer to Figure 4.4 to locate the slip plane, slip direction, and tensile axis with their Miller indices. Note that the orientation of slip plane and slip direction can be taken as arbitrary with respect to the tensile axis. It is just a representation of the problem statement so that the steps below can be followed easily.

Before we start calculation of the cosines, let us turn our attention to Section 2.1. We know that in a cubic system the normal direction to a crys­tallographic plane bears the same digits of the Miller indices. So, the normal direction to the slip plane, (110), is [110].

Following Eq. (2.4), the cosine of the angle between the tensile axis [010] and slip plane normal [110] is given by the dot product of the unit vectors along these two directions:

u1u2 + v1v2 + w1w2
sj u2 + vj + w2^/ u2 + v2 + w2

(0)(-1) + (1) ( 1 ) + (0) (0)
p02 + 12 + 0^/ (-1)2 + 12 + 02

Furthermore, the cosine of the angle between the tensile axis [010] and the slip direction [111] is

cos ф = 1 .

V02 + 12 + 0V12 + 12 + 12

Hence, using Eq. (4.1), we obtain the resolved shear stress: tR = (100MPa)(1/P2)(1/p3) =~ 40.8MPa.

The problem statement states that the single crystal starts deforming at the tensile stress of 100 MPa on the given slip system. So, the resolved shear stress must reach the corresponding critical value to accomplish that. Hence, it is clear that the resolved shear stress of 40.8 MPa calculated above is indeed the CRSS.

CRSS value in a single crystal is determined by the interactions between various dislocations and their interactions with other types of defects present. However, this value is quite smaller than the theoretical shear strength of the crystal. Table 4.1 lists comparative values of the CRSS for four different metals. The theoretical shear strength values for respective metals are calculated from G/10 (where G is the shear modulus of the metal), and CRSS values are all experimentally determined [1].

4.1.2