Taylor factor

In the previous section, we saw how the resolved shear stress required to activate a slip system in a single crystal can be calculated from the external stress. If Tcrit is the critical stress needed to activate slip, the external stress is connected to Tcrit by the Schmid factor cos A cos 0 for a single crystal. 

Most engineering alloys are poly crystalline. To calculate the yield strength from the critical resolved shear stress in an isotropic, polycrystalline material, we have to take into account that the grains are oriented in an arbitrary manner. We thus have to take the average of all possible crystal orientations. 

That the number of slip systems to be activated is five can be explained as follows An arbitrary deformation has six independent components of the strain tensor (see section 2.4.2). Because plastic deformation does not change the volume, each one of these components is dependent on the others, and five independent components remain, corresponding to the required five slip systems. 

This relation between the components of the strain tensor at constant volume can be derived by considering the deformation of a cuboid with edges h, I2, h that is deformed until the edge lengths are l + AZi, 

For small strains, products of strains can be neglected. This yields the equation 

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But we want the tensile yield strength, A tensile stress a creates a shear stress in the material that has a maximum value of t = a/2. (We show this in Chapter 11 where we resolve the tensile stress onto planes within the material.) To calculate cr from t,, we combine the Taylor factor with the resolution factor to give... 

This equation is similar to the HP relation Eq. (4.18a) and, in order to make it equivalent in terms of the yield stress, ffy, under tension, Ty is multiplied by the Taylor factor, M, to obtain Eq. (4.18a) as ... 

If we take these effects into account, the Schmid factor has to be replaced in a polycrystalline material by another number, the Taylor factor M. For a face-centred cubic material, M takes a value of 3.1 [34]. The relation between the critical resolved shear stress Tcrit and the yield strength measured in uniaxial tension ap thus is... 

This value of the Taylor factor has also been confirmed experimentally. Throughout section 6.4, we will use the Taylor factor to calculate the influence of strengthening mechanisms, which affect the critical shear stress, on the uniaxially measured yield strength. 

The derivation of the Taylor factor is rather involved. Here, we only want to sketch the main ideas for the example of a face-centred cubic lattice. A detailed discussion can be found in Cottrell [34. ... 

Then, the average of the resulting probability distribution has to be taken. If this is done, an incorrect value of 2.2 results for the Taylor factor. 

For a correct calculation, it has to be taken into account that five different slip systems must be activated in each grain to enable an arbitrary deformation. Some of these are oriented less favourably, thus increasing the value of the Taylor factor. To precisely determine its value, the five best-oriented slip systems have to be determined for each crystal orientation. As there are ( g ) = 792 possibilities to choose five systems out of twelve, this calculation is involved. In addition, normal stresses on the grain boundaries have to be continuous, and this has also to be accounted for. Finally, the average over all possible grain orientations is taken to arrive at the Taylor factor. 

Use the von Mises yield criterion to decide whether the material yields Can you decide which of the two results is correct Justify your answer In experiments on single crystals, the yield strength of the slip systems was determined as Tc- i = 60 MPa. Use the von Mises yield criterion to check whether a significant amount of slip systems in the polycrystal is activated at the stress value given The Taylor factor is M = 3.1. Calculate the stress deviator a for the given stress state ... 

A sheet of aluminium is rolled from an initial thickness of 10 mm to a final thickness of 5 mm. Using a transmission electron microscope, it was found that the dislocation density increases from = 10 m to q = 10 m . The prefactor in equation (6.20), needed to calculate the hardening contribution, is ky = 0.1. The length of the Burgers vector in a face-centred cubic lattice is h = y/2/2 a. The lattice constant of pure aluminium is a = 4.049 x 10 mm, the shear modulus is G = 26 200 MPa, and the Taylor factor is M = 3.1. Calculate the increase in strength due to this deformation ... 

For large systems comprising 36,000 atoms FAMUSAMM performs four times faster than SAMM and as fast as a cut-off scheme with a 10 A cut-off distance while completely avoiding truncation artifacts. Here, the speed-up with respect to SAMM is essentially achieved by the multiple-time-step extrapolation of local Taylor expansions in the outer distance classes. For this system FAMUSAMM executes by a factor of 60 faster than explicit evaluation of the Coulomb sum. The subsequent Section describes, as a sample application of FAMUSAMM, the study of a ligand-receptor unbinding process. 

The described method can generate a first-order backward or a first-order forward difference scheme depending whether 0 = 0 or 0 = 1 is used. For 9 = 0.5, the method yields a second order accurate central difference scheme, however, other considerations such as the stability of numerical calculations should be taken into account. Stability analysis for this class of time stepping methods can only be carried out for simple cases where the coefficient matrix in Equation (2.106) is symmetric and positive-definite (i.e. self-adjoint problems Zienkiewicz and Taylor, 1994). Obviously, this will not be the case in most types of engineering flow problems. In practice, therefore, selection of appropriate values of 6 and time increment At is usually based on trial and error. Factors such as the nature of non-linearity of physical parameters and the type of elements used in the spatial discretization usually influence the selection of the values of 0 and At in a problem. 

Next we consider how to evaluate the factor 6p. We recognize that there is a local variation in the Gibbs free energy associated with a fluctuation in density, and examine how this value of G can be related to the value at equilibrium, Gq. We shall use the subscript 0 to indicate the equilibrium value of free energy and other thermodynamic quantities. For small deviations from the equilibrium value, G can be expanded about Gq in terms of a Taylor series ... 

Although Eq. (10.50) is still plagued by remnants of the Taylor series expansion about the equilibrium point in the form of the factor (dn/dc2)o, we are now in a position to evaluate the latter quantity explicitly. Equation (8.87) gives an expression for the equilibrium osmotic pressure as a function of concentration n = RT(c2/M + Bc2 + ) Therefore... 

To correlate to the acentric factor a quadratic Taylor series in terms of the compressibility factor was formulated. This equation is represented as... 

The temperature dependence of Z for CO is very similar to the above two systems (Figure 3). Experimental data of Millikan and White and of Gaydon and Hurle (10) at temperatures above 1000 °K. are again larger than our values by a factor of 5. Windsor, Davidson, and Taylor (34) obtained a larger Z value than the ones above. 

The Earth is a highly unusual planet because life did evolve on it and it thrived to the extent that the surface and atmosphere of the planet were greatly modified. The Earth is unique in this respect relative to all known astronomical bodies (Taylor, 1999). The Earth s location, composition, and evolutionary history are all significant factors in the planet s success in nurturing life. Critical factors include its temperature, its atmosphere, its oceans, its long-term stability and its "just right" abundance of water and other light element compounds. 

Without loss of generality y = y can be assumed. If the dipole moment can be assumed to be a linear function of coordinate within the spread of the frozen Gaussian wave packet, the matrix element (gy,q,p, Pjt(r) Y,q, p ) can be evaluated analytically. Since the integrand in Eq. (201) has distinct maxima usually, we can introduce the linearization approximation around these maxima. Namely, the Taylor expansion with respect to bqp = Qq — Qo and 8po = Po — Po is made, where qj, and pj, represent the maximum positions. The classical action >5qj, p , ( is expanded up to the second order, the final phase-space point (q, p,) to the first order, and the Herman-Kluk preexponential factor Cy pj to the zeroth order. This approximation is the same as the ceUularization procedure used in Ref. [18]. Under the above assumptions, various integrations in U/i(y, q, p ) can be carried out analytically and we have... 

Oxygen isotopic fractionation factors used for the calculation were taken from Taylor (1997). Initial 8 0 value of hydrothermal solution (0%o) was estimated from 8 0 values of K-feldspar and quartz in the veins and homogenization temperatures (Shikazono and Nagayama, 1993), and that of groundwater (—7%c) was estimated from meteoric water value of the south Kyushu district (—7%c) (Matsubaya et al., 1975). 

Pauwels (1999) argues that the certified values of CRMs should be presented in the form of an expanded combined uncertainty according to the ISO Guide on the expression of uncertainty in measurement, so that coverage factor should always be clearly mentioned in order to allow an easy recalculation of the combined standard uncertainty. This is needed for uncertainty propagation when the CRM is used for calibration and the ISO Guide should be revised accordingly. The use of the expanded uncertainty has been pohcy in certification by NIST since 1993 (Taylor and Kuyatt 1994). 

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