Plastically shearing

If the maximum resolved shear stress r and the plastic shear strain rate y are defined according to (it is assumed that the Xj and Xj directions are equivalent)... 

Steady-propagating plastic waves [20]-[22] also give some useful information on the micromechanics of high-rate plastic deformation. Of particular interest is the universality of the dependence of total strain rate on peak longitudinal stress [21]. This can also be expressed in terms of a relationship between maximum shear stress and average plastic shear strain rate in the plastic wave... 

Bai [48] presents a linear stability analysis of plastic shear deformation. This involves the relationship between competing effects of work hardening, thermal softening, and thermal conduction. If the flow stress is given by Tq, and work hardening and thermal softening in the initial state are represented... 

We first consider strain localization as discussed in Section 6.1. The material deformation action is assumed to be confined to planes that are thin in comparison to their spacing d. Let the thickness of the deformation region be given by h then the amount of local plastic shear strain in the deformation is approximately Ji djh)y, where y is the macroscale plastic shear strain in the shock process. In a planar shock wave in materials of low strength y e, where e = 1 — Po/P is the volumetric strain. On the micromechanical scale y, is accommodated by the motion of dislocations, or y, bN v(z). The average separation of mobile dislocations is simply L = Every time a disloca-... 

The plastic deformation patterns can be revealed by etch-pit and/or X-ray scattering studies of indentations in crystals. These show that the deformation around indentations (in crystals) consists of heterogeneous rosettes which are qualitatively different from the homogeneous deformation fields expected from the deformation of a continuum (Chaudhri, 2004). This is, of course, because plastic deformation itself is (a) an atomically heterogeneous process mediated by the motion of dislocations and (b) mesoscopically heterogeneous because dislocation motion occurs in bands of plastic shear (Figure 2.2). In other words, plastic deformation is discontinuous at not one, but two, levels of the states of aggregation in solids. It is by no means continuous. And, it is by no means time independent it is a flow process. 

In this book, elastic strain and plastic deformation will be differentiated by both words and symbols. Elastic strain is given the usual symbols e and y for extensional and shear elastic strains, respectively. For plastic shear deformation. 8 will be used, e and 8 are physically different entities, e and y are conservative quantities which store internal energy. 8 is not conservative. The work done to create it is dissipated as heat and structural defects. 

Neglecting the elastic forces, lumping the geometric factors into a constant, b, and assuming the plastic shear deformation is x/r, yields the plastic resistive force ... 

In the unstrained material far from the center of an indentation, dislocations can move freely at much lower stresses than in the material near the center where the stress (and the deformation) is much larger. Thus, local plastic shear bands can form at the edges of the indenter, and do (Chaudhri, 2004). The lengths of these shear bands are often several times the size of an indentation. The leading dislocations in these bands move in virgin (undeformed) material, so they can move at lower stresses than the dislocations in the strain-hardened material near the center of an indentation.. The patterns they form are called rosettes. ... 

The loops around the precipitates act as stress concentrators. They exert shearing stresses in addition to the applied stress on the particles. When enough of them have accumulated, the precipitates will be plastically sheared as the loops disappear one by one. This is the basis of a theory of precipitation hardening in an aluminum-copper alloy by Fisher, Hart, and Pry (1953). The precipitate in this case is CuA12, and the precipitates cause an increment of hardening added to the hardness of the solid-solution (Al-Cu) matrix. Quantitative agreement with experimental measurements is fair. 

Stresses can can be concentrated by various mechanisms. Perhaps the most simple of these is the one used by Zener (1946) to explain the grain size dependence of the yield stresses of polycrystals. This is the case of the shear crack which was studied by Inglis (1913). Consider a penny-shaped plane region in an elastic material of diameter, D, on which slip occurs freely and which has a radius of curvature, p at its edge. Then the shear stress concentration factor at its edge will be = (D/p)1/2.The shear stress needed to cause plastic shear is given by a proportionality constant, a times the elastic shear modulus,... 

The Ni octahedra derive their stability from the interactions of s, p, and d electron orbitals to form octahedral sp3d2 hybrids. When these are sheared by dislocation motion this strong bonding is destroyed, and the octahedral symmetry is lost. Therefore, the overall (0°K) energy barrier to dislocation motion is about COCi/47r where = octahedral shear stiffness = [3C44 (Cu - Ci2)]/ [4C44 + (Cu - C12)] = 50.8 GPa (Prikhodko et al., 1998), and the barrier = 4.04 GPa. The octahedral shear stiffness is small compared with the primary stiffnesses C44 = 118 GPa, and (Cn - C12)/2 = 79 GPa. Thus elastic as well as plastic shear is easier on this plane than on either the (100), or the (110) planes. 

Figure 8.6 Glide plane (111) of Ni3Al and unit glide-plane cells for levels (0—the plane below the plastic shear) and (1—the plane above the plastic shear).

Figure 10.7 illustrates the prototype hexaboride crystal structure, that of lanthanum hexaboride. It consists of a simple cubic array of boron octahedra surrounding a metal atom at the body center of each cube. The octahedra are linked by B-B bonds connecting their comers. This makes the overall structure relatively hard with approximately the hardness of boron itself since plastic shear must break B-B bonds. The open volumes surrounded by boron octahedra are occupied by the relatively large lanthanum atoms as the figure shows schematically. 

An alternative version of the lanthanum hexaboride crystal structure has the boron octahedra occupying the body centered positions of the cubic array of lanthanum atoms (Figure 10.8). This version makes it clear that in order to plastically shear the structure, the boron octahedra must be sheared. Note that the octahedra are linked together both internally and externally by B-B bonds. 

In glasses, dislocation lines are the boundaries between plastically sheared areas, and material in which plastic shear has not yet occurred (Gilman, 1968). 

Indeed, it has been observed that the onset of yielding of isotropic polymers is approximately constant, 0.02< [<0.025, which implies that 0.04shear yield strain, the plastic shear deformation of the domain satisfies a plastic shear law. For temperatures below the glass transition temperature, the continuous chain model enables the calculation of the tensile curve of a polymer fibre up to about 10% strain [6]. Figure 7 shows the observed stress-strain curves of PpPTA fibres with different moduli compared to the calculated curves. 

In order to simplify the discussion and keep the derivation of the formulae tractable, a fibre with a single orientation angle is considered. In a creep experiment the tensile deformation of the fibre is composed of an immediate elastic and a time-dependent elastic extension of the chain by the normal stress ocos20(f), represented by the first term in the equation, and of an immediate elastic, viscoelastic and plastic shear deformation of the domain by the shear stress, r =osin0(f)cos0(f), represented by the second term in Eq. 106. 

In Fig. 63 the occupation of state 1 is equal to Nly the occupation of state 2 is equal to N2 and the total occupation is equal to N1+N2=N. The viscoelastic and plastic shear strain is proportional to the decrease of the occupation of state 1 or proportional to the increase of the occupation of state 2. Without external stress the probability for transition from state 1 to state 2 (v ) is proportional to the Boltzmann factor expI-Uf T)-1], and for the inverse transition 2—>1 (v) the probability is proportional to N2exp[-U(kBT) 1]. 

As for the derivation of Eqs. 122,123 and 124 only the transitions 1—>2 have been counted, these equations do not describe recovery processes, where the transitions 2 —>1 are important as well. These approximations have been made for convenience s sake, but neither imply a limitation for the model, nor are they essential to the results of the calculations. Equation 124 is the well-known formula for the relaxation time of an Eyring process. In Fig. 65 the relaxation time for this plastic shear transition has been plotted versus the stress for two temperature values. It can be observed from this figure that in the limit of low temperatures, the relaxation time changes very abruptly at the shear yield stress Ty = U0/Q.. Below this stress the relaxation time is very long, which corresponds with an approximation of elastic behaviour. 

It is proposed that the viscoelastic and plastic shear strain of a domain, 5f3(f)=tan(0(f)-0o), is proportional to (NI2-NJ. Then it follows from Eqs. 125 and 126 that... 

The magnitude of the activated transition is denoted by I, where I=cN and c is an arbitrary constant. A transition density function is introduced to describe the viscoelastic and plastic shear deformation of the domain. Hence, following Eq. 107 the total shear strain of a domain in terms of the ERT model is given by... 

The descriptions presented in the foregoing sections are concerned mainly with composites containing brittle fibers and brittle matrices. If the composite contains ductile fibers or matrix material, the work of plastic deformation of the composite constituents must also be taken into account in the total fracture toughness equation. If a composite contains a brittle matrix reinforced with ductile libers, such as steel wire-cement matrix systems, the fracture toughness of the composite is derived significantly from the work done in plastically shearing the fiber as it is extracted from the cracked matrix. The work done due to the plastic flow of fiber over a distance on either side of the matrix fracture plane, which is of the order of the fiber diameter d, is given by (Tetelman, 1969)... 

Plastic deformation. If normal or tangential stresses become too high, plastic deformation of the contact area will occur. Under certain circumstances, this can lead to plastic shearing, parallel to the surface, without plastic penetration normal to the surface. 

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