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Axial normal stresses

The polystyrene simulation followed the experiments of Bell and Edie (12) with good agreement. Figure 14.8 shows the simulation results for fiber spinning nylon-6.6 with a draw ratio of 40. The figure demonstrates the wealth of information provided by the model. It shows the velocity, temperature, axial normal stress, and crystallinity fields along the threadline. We see the characteristic exponential-like drop in diameter with locally (radially) constant but accelerating velocity. However, results map out the temperature, stress, and crystallinity fields, which show marked variation radially and axially. [Pg.831]

Fig. 14.8 Simulation results for velocity, temperature, axial normal stress, and crystallinity fields for low-speed spinning of nylon-6.6. [Reprinted with permission from Joo et al., Two-dimensional Numerical Analysis of Nonisothermal Melt Spinning with and without Phase Transition, J. Non-Newt. Fluid Mech., 102, 37-70 (2002).]... Fig. 14.8 Simulation results for velocity, temperature, axial normal stress, and crystallinity fields for low-speed spinning of nylon-6.6. [Reprinted with permission from Joo et al., Two-dimensional Numerical Analysis of Nonisothermal Melt Spinning with and without Phase Transition, J. Non-Newt. Fluid Mech., 102, 37-70 (2002).]...
In the preceding equations, cross-section determined by the following equation ... [Pg.296]

Figure 23 Pressure effects on the profiles of (a) axial liquid velocity (b) Reynolds axial normal stress in a bubble column (Pc = 5.08 cm, nitrogen water). (From Lee et al., 2001.)... Figure 23 Pressure effects on the profiles of (a) axial liquid velocity (b) Reynolds axial normal stress in a bubble column (Pc = 5.08 cm, nitrogen water). (From Lee et al., 2001.)...
The computed turbulent axial and radial normal stresses and shear stresses increase with increasing superficial gas velocity. Axial normal stresses are considerably higher than their radial counterpart and both exceed Reynolds shear stresses. The maximum in Reynolds shear stress increases remarkably as the gas velocity is raised from 6 cm/s to 10 cm/s and its location is in the neighborhood of the inversion point for the axial velocity profile. [Pg.368]

Where a is the axial normal stress uniformly applied owz = L. X is shear-lag parameter, and Ef are elastic modules of CNT and matrix respectively. For the known stress and strain distribution under RVE we can calculate elastic effective properties quantifications [7]. The effective module > can be calculated as follow... [Pg.34]

To capture stress variation through the adhesive thickness and free edge stress in the adhesive ends, models based on the 2D elastic theory have been used. Allman (1977), and then Chen and Cheng (1983) assumed linear variation of peel stress and constant shear stress across the adhesive thickness. Ojalvo and Eidinoff (1978), and then Carpenter (1980) adopted linear variation of shear stress and constant peel stress through the thickness. Luo and Tong (2004) chose linear and higher order variations of shear, peel, and axial normal stresses across the adhesive thickness. [Pg.604]

Similar reasoning leads to the conclusion that in cone-and-plate or disk-plate viscometers (Sections 16.7 and 16.8) there will be forces tending to push the cone and plate or the disk and plate apart. Indeed, there are, and the measurement of these forces provides a means of determining the functions Ni j) and Also, for the case of torsional flow, Maxwell and Scalora showed that if a hole is drilled through the plate along the axis of rotation, a screwless extruder is obtained, as the axial normal stress pushes a polymer melt through the hole. Unlike a screw extruder, however, this device works only with viscoelastic fluids. [Pg.295]

In the context of elastic deformation two parameters, known as stress and strain respectively, are very relevant. Stress is an internal distributed force which is the resultant of all the interatomic forces that come into play during deformation. In the case of the solid bar loaded axially in tension, let the cross sectional area normal to the axial direction be A0. From a macroscopic point of view the stress may be considered to be uniformly distributed on any plane normal to the axis and to be given by o A0 where o is known as the normal stress. The stress has to balance the applied load, F, and one must, therefore have o Aq = F or o = F/Aq. The units of stress are those of force per unit area, i.e., newtons per square... [Pg.11]

When a viscoelastic liquid flows through a tube, the normal stress differences cause the liquid to be under an axial tension while a normal... [Pg.131]

Average axial stresses (over the pipe cross section) due to longitudinal forces caused by displacement strains are not normally considered in the determination of displacement stress range, since this stress is not significant in typical piping layouts. In special cases, however, consideration of average axial displacement stress... [Pg.109]

Fig. 8.13. (a) Distribution of interlaminar normal stress, <7 , and interlaminar shear stress, r-,. in (0°/90°]s laminate under axial tension, (b) Effect of stacking sequence on through-the-thickness distribution of... [Pg.343]

I/zp Superficial particle velocity in one-dimensional axial flow A Ratio of vertical normal stress at wall to averaged vertical normal... [Pg.367]

Let us consider an apparently simple situation of compaction of solids in a cylinder (Fig. 4.9). Assuming a uniform stress field, a normal force Fo applied to the top ram generates within the solids a certain normal stress %zz, as well as a radial stress xrr. The frictional shear force due to the latter acts in the opposite direction to the applied force. Hence, the transmitted force to the lower ram, FL, will be smaller than the applied force. By making a force balance similar to that made in deriving the Janssen equation, and assuming that the wall friction is fully mobilized, that the ratio of axial to radial stresses is a constant throughout, and that the coefficient of friction at the wall is constant, we obtain the following simple exponential relationship between the applied and transmitted force ... [Pg.155]

Hence, in dealing with steady motion of particulate solids, it is evident that the axial stress or pressure drops exponentially, whereas in the case of liquid flow, it drops linearly with distance. This difference stems, of course, from the fact that frictional forces on the wall are proportional to the absolute local value of normal stress or pressure. In liquids, only the pressure gradient and not the absolute value of the pressure affects the flow. Furthermore, Eq. 4.7.2 indicates that the pushing force increases exponentially with the coefficient of friction and with the geometric, dimensionless group CL/A, which for a tubular conduit becomes 4 L/D. [Pg.158]

Clearly, this is an elastic rather than a simple viscous type of response and simple fluids like water do not behave in this fashion. A qualitative understanding of this is relatively straightforward the chains in the melt are deformed as they are pushed into the capillary (or die, if we are talking about extrusion) and as they are sheared in the narrow channel. Normal stresses then develop in a direction perpendicular to the flo (i.e., axially in the capillary). These stresses are relieved and deformation is recovered when the polymer exits the die, so the polymer extrodate swells. (It s a bit more complicated than this, because some relaxations can also occur in the capillary, if it is long enough— but you get the idea.)... [Pg.444]

Therefore the normal stress distribution, r (r, 2), can be obtained from the solution to this integral equation if the axial load for the surfaces of the die is known. Thompson [75] has assumed that a parabolic radial distribution of axial pressure on the top and bottom punch faces of the cylindrical die, in accord with Unkel s experimental measurements [76]. In addition, Thompson has assumed that this parabolic distribution is valid for the length of the cylindrical die as well, giving... [Pg.662]

Fig. 5 Results for the axial stress along the crack plane and the interlaminar normal stress in axial direction starting from the crack tips... Fig. 5 Results for the axial stress along the crack plane and the interlaminar normal stress in axial direction starting from the crack tips...
This may be contrasted to, e.g., the isotropic pressnre developed in a fluid under pressure, with only nonnewtonian fluids able to develop and sustain a nonisotropic distribution of normal stress. In addition, the radial normal stress acting at the wall develops a wall shear stress that opposes gravity and helps support the weight of the powder. As originally developed by Janssen [Zeits. D. Vereins Deutsch Ing., 39(35), 1045 (1895)], from a balance of forces on a differential slice, the axial stress as a function of depth z is given by... [Pg.2271]

Boundary layer approximation. The Landau problem, which was described above, is an example of an exact solution of the Navier-Stokes equations. Schlichting [427] proposed another approach to the jet-source problem, which gives an approximate solution and is based on the boundary layer theory (see Section 1.7). The main idea of this method is to neglect the gradients of normal stresses in the equations of motion. In the cylindrical coordinates (71, ip, Z), with regard to the axial symmetry (Vv = 0) and in the absence of rotational motion in the flow (d/dip = 0), the system of boundary layer equations has the form... [Pg.22]

The axial load and axial displacement are logged in PC during triaxial deformation test. Using these logged data, it is possible to derive the shear stress and normal stress acting on the slip surface by using the following equations. [Pg.686]


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See also in sourсe #XX -- [ Pg.368 ]




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