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Resolved shear strain

After the dislocation nucleates, the strain distribution changes drastically. As Fig. 14(c) shows, the passage of the dislocation adds a negative strain, which makes nucleation of a second dislocation of this type unlikely until the indenter has produced more strain. On the intersecting glide plane, however, the resolved shear strain remains high [Fig. 14(d)], which can lead to the nucleation of a different dislocation loop, such as the one in Fig. 13(h). [Pg.256]

Fig. 14. Strain distribution during dislocation nucleation under an indenter. The grayscale indicates the level of the local resolved shear strain, y = 2a,j, the sign of which is indicated by the arrows. In (a)-(c), the resolved shear stress is on the (1 1) plane along the [11 2] direction in (d) it is on the (1 11) plane in the [11 2] direction, (a) and (b) correspond to a subcritical fluctuation at t = 91 min. (a) x-z plane (vertical), centered below the indenter tip at y = 29 pm. (b) y-z plane (horizontal), located 10 pm below the needle at z = 10 pm. (c) and (d) correspond to a just-nucleated dislocation loop at t = 119 min both are y-z planes (vertical) at x = 33 pm, which intersect the loop diametrically, (e) Plot of the resolved shear strain along the dashed line at z = 23 pm in (c), for all particles within 1.6 pm of the line. From Schall et al. [31]. Fig. 14. Strain distribution during dislocation nucleation under an indenter. The grayscale indicates the level of the local resolved shear strain, y = 2a,j, the sign of which is indicated by the arrows. In (a)-(c), the resolved shear stress is on the (1 1) plane along the [11 2] direction in (d) it is on the (1 11) plane in the [11 2] direction, (a) and (b) correspond to a subcritical fluctuation at t = 91 min. (a) x-z plane (vertical), centered below the indenter tip at y = 29 pm. (b) y-z plane (horizontal), located 10 pm below the needle at z = 10 pm. (c) and (d) correspond to a just-nucleated dislocation loop at t = 119 min both are y-z planes (vertical) at x = 33 pm, which intersect the loop diametrically, (e) Plot of the resolved shear strain along the dashed line at z = 23 pm in (c), for all particles within 1.6 pm of the line. From Schall et al. [31].
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)... [Pg.223]

The continuous chain model includes a description of the yielding phenomenon that occurs in the tensile curve of polymer fibres between a strain of 0.005 and 0.025 [ 1 ]. Up to the yield point the fibre extension is practically elastic. For larger strains, the extension is composed of an elastic, viscoelastic and plastic contribution. The yield of the tensile curve is explained by a simple yield mechanism based on Schmid s law for shear deformation of the domains. This law states that, for an anisotropic material, plastic deformation starts at a critical value of the resolved shear stress, ry =/g, along a slip plane. It has been... [Pg.20]

The shear component of the applied stress appears to be the major factor in causing yielding. The uniaxial tensile stress in a conventional stress-strain experiment can be resolved into a shear stress and a dilational (negative compressive) stress normal to the parallel sides of test specimens ofthe type shown in Fig. 11-20. Yielding occurs when the shear strain energy reaches a critical value that depends on the material, according to the von Mises yield criterion, which applies fairly well to polymers. [Pg.421]

Figure 13.3 Generalized response to oscillating shear for a fluid showing linear viscoelasticity. The shear stress r results from the imposed shear strain y, and is offset by S X can be resolved into the in-phase contribudon r and the out-of-pha contribution r"... Figure 13.3 Generalized response to oscillating shear for a fluid showing linear viscoelasticity. The shear stress r results from the imposed shear strain y, and is offset by S X can be resolved into the in-phase contribudon r and the out-of-pha contribution r"...
Figure 6.22 Schematic illustration of the dependence of normalized critical resolved shear stress on temperature for two strain rates,... Figure 6.22 Schematic illustration of the dependence of normalized critical resolved shear stress on temperature for two strain rates,...
FIGURE 17.7 (a) Stress-strain curve for a crystal suitably oriented for plastic flow, (b) Temperature dependence of the normalized critical resolved shear stress for two strain rates, where y, >... [Pg.314]

These tests were performed using an Instron Model 8561 (single screw) machine in air and the furnace was adapted to perform four-point bend tests. The rates indicated in Fig. 2.3 relate to crosshead displacement. Figure 2.4 shows the resolved shear stress at yield for the specimens tested ate = 4.2 x 10 s above Tc at the indicated orientations. The mechanism for slip is dislocation glide, which explains the orientation dependence of yield, as seen in Fig. 2.4. Thus, the BDT temperature, Tc, of the sapphire (AI2O3) varies not only with the strain rate, but also with the crystallographic orientation of the fracture plane. [Pg.116]

Resolved shear stress is a concept related to plastic deformation and associated with shear stress. Similarly, it is reasonable to talk about ceramics exhibiting ductility as a consequence of acting shear stress. To do so, one must consider stress and strain tensors. The stress tensor in Sects. 1.22 and 1.23 (Eqs. 1.13-1.13b) is rewritten here as ... [Pg.293]

The authors [6] suggest the term shearability , Sm, for the maximum shear strain that a homogeneous crystal can withstand. It is defined by Sm = argmax o-(s), where a(s), is the resolved shear stress and s is the engineering shear strain in a specified slip system. The relaxed shear stress, (7 in Table 4.2 is normalized by Gr. In this table, experimental and calculated values of the relaxed shear vales of Gr are given. For details on these calculations, refer to the work of Ogata et al. [6]. [Pg.298]

In contrast with the Takayanagi model, which considers only extensional strains, a major deformation process involves shear in the amorphous regions. Rigid lamellae move relative to each other by a shear process in a deformable matrix. The process is activated by the resolved shear stress a sin y cosy on the lamellar surfaces, where y is the angle between the applied tensile stress o and the lamellar plane normals, which reaches a maximum value for y = 45° (see Chapter 11 for discussion of resolved shear stress in plastic deformation processes). [Pg.179]

From the above example it can be seen that a complex system needing careful analysis is present in each case, but the underlying fact is that the type of dislocation and their interactions are intimately concerned with the stress-strain field imposed by the geometry of the indenter. The implication of this is that hardness anisotropy is an obvious manifestation of dislocation interactions and indenter facet geometry. Simplified interpretations of this have been sought, of which the Brookes Resolved Shear Stress model, given in Section 3.6.1, is an important development. [Pg.211]

For a single slip the slip direction in the crystal rotates always towards the direction of maximum extension in uniaxial tension it rotates towards the tensile axis, while in uniaxial compression it rotates away from the compression axis. The angle through which the crystal rotates is a simple function of the applied strain [107]. It must be mentioned that the slip takes place when the resolved shear stress on the slip plane reaches a critical value known as critical resolved shear stress. Currently the critical resolved shear stresses for slips are well known for only few polymers. [Pg.35]


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