External shear stresses

The core structure of the 1/2 [112] dislocation is shown in Fig. 4. This core is spread into two adjacent (111) plames amd the superlattice extrinsic stacking fault (SESF) is formed within the core. Such faults have, indeed, been observed earlier by electron microscopy (Hug, et al. 1986) and the recent HREM observation by Inkson amd Humphreys (1995) can be interpreted as the dissociation shown in Fig. 4. This fault represents a microtwin, two atomic layers wide, amd it may serve as a nucleus for twinning. Application of the corresponding external shear stress, indeed, led at high enough stresses to the growth of the twin in the [111] direction. 

Isolated rat hepatocytes were immobilized in cellulose multiporous microcarriers by Kino et al. [24]. The microcarriers had a pore size of 100 pm and protected the cells from external shear stress. A newly developed stirred tank reactor contained the microcarrier-immobilized hepatocytes. The 02-supply was improved by using an oxygenator. The performance of microcarrier-im-mobihzed hepatocytes in the reactor was as good as that in floating culture and they demonstrated good ammonia metaboUsm. 

According to the assumptions about structure with the superposing of thermal vibrations and the elastic energy A caused by external forces we obtain the well known Eyring differential equations4 for the difference z between the molecular displacements in the direction of the external shearing stress and that of the opposite direction. 

Complex flnids are viscoelastic when the fluid stiU maintains internal stress after the external shear stress has ceased. The internal stress decays with time the time reqnired for the fluid to recover to the initial state is called the relaxation time. For this case the shear modulus (G ) is a complex number ... 

Pure torsion tests were performed on ice single crystals at a constant imposed external shear stress". Softening was evidenced as the creep curves revealed a strain-rate increase, up to a cumulated plastic strain of 7%, see figure 1. Note that such a behaviour was also observed during compression and tension tests. ... 

Fibrous reinforced plates in practice are often made up of several layers, and the individual layers may be of different construction, such as mat, fabric, or roving. Furthermore, the various layers may be oriented at different angles with respect to each other in order to provide the best combination to resist some particular loading condition. Outside loads or stresses applied to a RP plate of this type result in internal stresses that are different in the individual layers. External direct stresses may result not only in internal direct stresses but in internal shear stresses and external shear stresses may result in internal direct stresses as well as internal shear stresses. 

If a sufficiently large shear stress acts on a dislocation, the dislocation moves through the crystal. How this happens is shown in figure 6.8 for an edge dislocation Near the dislocation line, the atoms are displaced from their equilibrium positions, stretching and compressing the atomic bonds. If an external shear stress is applied, trying to shift the upper crystal plane relative to the lower. 

Only in special cases will the external shear stress be exactly parallel to the slip system. Usually, it is thus necessary to calculate the resolved shear stress i. e., the stress component acting as shear stress on the considered slip system in the slip direction. If we restrict ourselves to the case of uniaxial loading as in a tensile test, the calculation of this component is not too difficult. We... 

Forces acting on dislocations Force by external shear stress... 

We first want to calculate the force exerted on a dislocation by an external shear stress. Consider a straight dislocation line of length l that is moved over a distance I2 by the external stress t (figure 6.22). For simplicity, we assume that the shear stress is perpendicular to the dislocation line and parallel to the slip plane. The external force Fext is Fext = TI1I2 because the stress r acts... 

Although the external shear stress on this so-called secondary slip plane or cross slip plane is smaller than on the primary one, moving along this path can be easier than trying to overcome the obstacle by cutting or the Orowan mechanism. This is the case if the effective shear stress t (see section 6.2.9) on the secondary slip plane is larger than on the primary one due to the absence of the obstacle force. Because screw dislocations can use this additional mechanism, they are frequently able to overcome obstacles more easily than edge dislocations. 

Rheological properties of foams (elasticity, plasticity, and viscosity) play an important role in foam production, transportation, and applications. In the absence of external stress, the bubbles in foams are symmetrical and the tensions of the formed foam films are balanced inside the foam and close to the walls of the vessel [929], At low external shear stresses, the bubbles deform and the deformations of the thin liquid films between them create elastic shear stresses. At a sufficiently large applied shear stress, the foam begins to flow. This stress is called the yield stress, Tq- Then, Equation 4.326 has to be replaced with the Bingham plastic model [930] ... 

In Cases 1 and 2 the orientation of the fibers with respect to the 1-2 directions chosen resulted in zero shear stresses associated with those directions, whereas in Case 3 the shear stresses were not zero. In all three cases, symmetry of the fiber orientations with respect to the stress directions resulted in internal stresses equal to the external stresses. These are special cases. In the more general case the internal direct stresses in the individual layers are not necessarily equal to the external direct stresses, nor are they the same in the various layers. Furthermore, even symmetrical Case 3 leads to internal shear stresses when external shear stresses are absent. In the more general case it is still more true... 

Figure 10.4. Illustration of how an edge dislocation can be moved by one lattice vector to the right by the slight displacement of few atomic columns near the core (left and middle panels), as indicated by the small arrows. Repeating this step eventually leads to a deformation of the solid, where the upper half and the lower half differ by one half plane (right panel), and the dislocation has been expelled from the crystal, r is the external shear stress that forces the dislocation to move in the fashion indicated over a width w, and b is the Burgers vector of the dislocation the length of the dislocation is I in the direction perpendicular to the plane of the figure.

Within the idealized situation of a single dislocation in an infinite crystal, it is possible to obtain the force per unit length of the dislocation due to the presence of external stress. If the width of the crystal is w and the length of the dislocation is I, then the work AW done by an external shear stress r to deform the crystal by moving an edge dislocation through it, in the configuration of Fig. 10.4, is... 

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