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Resolved shear stress equation

Equations (5.20) and (5.21) are valid for an applied tensile or compressive stress and can be used in the case of twinning as well. However, Ter for twinning is usually greater than Xcr for shear. Some values of the critical resolved shear stress for slip in some common metals, and their temperature dependence, are shown in Figure 5.13. Note that the critical resolved shear stress for HCP and FCC (close-packed) structures rises only modestly at low temperatures, whereas that for the BCC and rock salt structures increases significantly as temperature decreases. [Pg.395]

This equation is nothing more than the statement that plastic flow will begin on the a slip system when the resolved shear stress (i.e. a n ) on that slip... [Pg.59]

Equation (4.6) gives the resolved shear stress . The product in the equation is known as the Schmid factor and determines whether the orientation is favorable for shp. The conditions for slip are given by Schmid s Law and the value of Eq. (4.6), often represented in the literature by Tj, indicating the onset of plastic deformation and called critical resolved shear stress . CRSS is a structure-sensitive property, since it is very dependent on impurities and the way the crystal was grown and handled. [Pg.295]

Putting this into equation (6.6) results in the resolved shear stress (or Schmid stress)... [Pg.180]

This equation determines the shear stress in a slip system resulting from the external stress and the orientation of the system. The factor cos A cos 0 is known as the Schmid factor. If the resolved shear stress reaches the critical value Tcrit, the material yields. The yield criterion for uniaxial loading is thus... [Pg.180]

Figure 3.20. Reciprocal of the cosine and rotational constraint terms of equation (3.9)—that is, the effective resolved shear stress—on (110) plane of MgO using either compressive or tensile forces with the slip system 110 (110>. Figure 3.20. Reciprocal of the cosine and rotational constraint terms of equation (3.9)—that is, the effective resolved shear stress—on (110) plane of MgO using either compressive or tensile forces with the slip system 110 (110>.
The Schmid-Boas equation will give the distribution of the resolved shear stresses as a function of sliding direction. [Pg.60]

Edge cracks, 168 Edge dislocation, 67 Effective resolved shear stress correlation with Knoop hardness, 99 equation, 99, 100, 107, 110 mean, 99... [Pg.163]

The effect this has on developing predictive equations is discussed in Section 3.6.1 where resolved shear stress models are considered. [Pg.221]

According to equation (3.10), when the axis of rotation of the slip system is parallel to the axis drawn at 90° to the sliding direction that lies in the plane of sliding, i.e., XY, then sin to = 0. This produces a decreased penetration and therefore an increased hardness. At < = 90° to XY, the slip system rotation axis will be normal to the surface axis XY, the effective resolved shear stress is maximized, leading to a maximum penetration, and this direction will be the softest. [Pg.227]

We will learn in the subsequent chapters that ductile metals yield by shearing, so now we will attempt to estimate their theoretical peld strength, or what will be called their critical resolved shear stress (erss). Shearing requires planes of atoms to slide over each other, rather like shearing a deck of cards as illustrated in Figure 7.7a. For cubic crystals. Equation 7.5 gives Sz = 446 = GQ y = Gxjd where G is the shear coefficient and x < d. [Pg.153]

Resolved shear stress is the shear stress resulting from an applied tensile stress that is resolved onto a plane that is neither parallel nor perpendicular to the stress direction. Its value is dependent on the applied stress and orientations of plane and direction according to Equation 7.2. [Pg.242]

Critical resolved shear stress is the minimum resolved shear stress required to initiate dislocation motion (or slip) and depends on yield strength and orientation of slip components per Equation 7.4. [Pg.242]

Consider a cylindrical single crystal with cross-sectional area A. It yields at a tensile load P. The angles between the tensile axis and the normal to the slip plane and the slip direction are (p and X, respectively, as illustrated in Figure 3.8. The area of the slip plane inclined at the angle (p is y4/cos< ), and the stress component in the slip direction is PcosX. The critical resolved shear stress, can therefore be calculated by Equation 3.12. The single crystal starts to yield when the shear stress reaches the critical resolved shear stress (Figure 3.8) ... [Pg.80]

When rotated into the principal coordinates, all the shear stresses vanish, and it is reasonable to think of the average normal compressive stress as a pressure. It is apparent from Eq. 2.183, however, that the average compressive stress is not equal to the thermodynamic pressure p as evaluated from an equation of state. Stokes made this interesting observation and recognized its concomitant dilemma in his famous 1845 paper. He hypothesized that the dilemma could be resolved by assuming that... [Pg.57]

Details of the method of resolution have been described in [9] and [11]. Considering the Poiseuille-Couette flow, turbulence equations are resolved by a half implicit method, consisting into impliciting the terms which will increase the diagonal dominance of the matrix. Then I and 3 are evaluated and velocity profiles are obtained from equations (6). These steps are iterated until a convergence criterion is satisfied. Together with velocity profiles, the coefficients Gy and and the shear stress at the wall are... [Pg.419]

The creation terms embody the changes in momentum arising from external forces in accordance with Newton s second law (F = ma). The body forces arise from gravitational, electrostatic, and magnetic fields. The surface forces are the shear and normal forces acting on the fluid diffusion of momentum, as manifested in viscosity, is included in these terms. In practice the vector equation is usually resolved into its Cartesian components and the normal stresses are set equal to the pressures over those surfaces through which fluid is flowing. [Pg.108]


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