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Continuum Aspects of Plastic Flow
Publisher Summary
Plastic flow occurs by shear. Dislocations move when the shear stress on the slip plane exceeds the dislocation yield strength τy of a single crystal. If this is averaged over all grain orientations and slip planes, it can be related to the tensile yield strength σy of a polycrystal by σy = 3τy. In solving problems of plasticity, however, it is more useful to define the shear yield strength k of a polycrystal. It is equal to σy/2, and differs from τy because it is an average shear resistance over all orientations of slip plane. When a structure is loaded, the planes on which shear may occur can often be identified or guessed, and the collapse load calculated approximately by requiring that the stress exceed k on these planes. This chapter shows that k = σy/2, and uses k to relate the hardness to the yield strength of a solid. The chapter examines the tensile instabilities that appear in the drawing of metals and polymers.
12.1 Introduction
Plastic flow occurs by shear. Dislocations move when the shear stress on the slip plane exceeds the dislocation yield strength τy of a single crystal. If this is averaged over all grain orientations and slip planes, it can be related to the tensile yield strength σy of a polycrystal by σy = 3τy (Chapter 11).
In solving problems of plasticity, however, it is more useful to define the shear yield strength k of a polycrystal. It is equal to σy/2, and differs from τy because it is an average shear resistance over all orientations of slip plane. When a structure is loaded, the planes on which shear will occur can often be identified or guessed, and the collapse load calculated approximately by requiring that the stress exceed k on these planes.
In this chapter, we show that k = σy/2, and use k to relate the hardness to the yield strength of a solid. We then examine tensile instabilities which appear in the drawing of metals and polymers.
12.2 Onset of yielding and shear yield strength, k
A tensile stress applied to a piece of material will create a shear stress at an angle to the tensile stress. Let us examine the stresses in more detail. Resolving forces in Figure 12.1 gives the shearing force as 
. The area over which this force acts in shear is 
Thus the shear stress, τ is
If we plot this against θ as in Figure 12.2 we find a maximum τ at θ = 45° to the tensile axis. This means that the highest value of the shear stress is found at 45° to the tensile axis, and has a value of σ/2.
Now, from what we have said in Chapters 10 and 11, if we are dealing with a single crystal, the crystal will not in fact slip on the 45° plane—it will slip on the nearest lattice plane to the 45° plane on which dislocations can glide (Figure 12.3). In a polycrystal, neighboring grains each yield on their nearest-to-45° slip planes. On a microscopic scale, slip occurs on a zigzag path; but the average slip path is at 45° to the tensile axis. The shear stress on this plane when yielding occurs is therefore τ = σy/2, and we define this as the shear yield strength k:
12.3 Analyzing the hardness test
The concept of shear yielding—where we ignore the details of the grains in the polycrystal and treat the material as a continuum—is useful in many respects. For example, we can use it to calculate the loads that would make the material yield for all sorts of quite complicated geometries.
A good example is the problem of the hardness indenter that we referred to in the hardness test in Chapter 9. Then, we stated that the hardness
We assume that the material does not work-harden so as the indenter is pushed into the material, the yield strength does not change. For simplicity, we consider a two-dimensional model. (A real indenter is three-dimensional, but the result is, for practical purposes, the same.)
As we press a flat indenter into the material, shear takes place on the 45° planes of maximum shear stress shown in Figure 12.4, at a value of shear stress equal to k. By equating the work done by the force F as the indenter sinks a distance u to the work done against k on the shear planes, we get:
This simplifies to
from which
But F/A = H so
(Strictly, shear occurs not just on the shear planes we have drawn, but on myriad 45° planes near the indenter. If our assumed geometry for slip is wrong it can be shown rigorously by a theorem called the upper-bound theorem that the value we get for F at yield—the so-called “limit” load—is always on the high side.)
