Limiting yield stress

If slip-line fields do not control plastic indentation, what does The answer is not the beginning of the plastic deformation, but the end of it. The end means after deformation hardening has occurred. That is, it is not the initial yield stress, Y0, that controls indentation, but the limiting yield stress, Y. This is... 

If the decrease in viscosity is very large at small shear rates, the system is sometimes called pseudoplastic (curves 4 and 5). Commonly, concentrated suspensions show a plastic behavior, that is, there is no response until a limiting yield stress oy has been exceeded. If the flow is linear above oy, the system is called Bingham plastic (curve 4) and can be expressed by the Bingham model (5) ... 

Gels are.essentially dispersions in which the attractive interactions between the elements of the disperse phase are so strong that the whole system develops a rigid network structure and, under small stresses, behaves elastically. In some instances the gel may flow plastically above a limiting yield stress [Chapter 8, Figure 8.3(e)] the gel then often exhibits thixotropic behaviour, the rigid gel state being re-formed when the stress is removed. 

Figure 17 11 lustration curve of shear stress vs. shear rate for defining the clastic-limit yield stress, ly, static yield stress, x, and the dynamic yield stress, T(j. Reproduced with permission from R.T. Bonnecaze and J.F. Brady, J. Rheol, 36(1X1992)73... 

A composite material used for rock-drilling bits consists of an assemblage of tungsten carbide cubes (each 2 fcm in size) stuck together with a thin layer of cobalt. The material is required to withstand compressive stresses of 4000 MNm in service. Use the above equation to estimate an upper limit for the thickness of the cobalt layer. You may assume that the compressive yield stress of tungsten carbide is well above 4000 MN m , and that the cobalt yields in shear at k = 175 MN m . What assumptions made in the analysis are likely to make your estimate inaccurate ... 

At room temperature, NiAl deforms almost exclusively by (100) dislocations [4, 9, 10] and the availability of only 3 independent slip systems is thought to be responsible for the limited ductility of polycrystalline NiAl. Only when single crystals are compressed along the (100) direction ( hard orientation), secondary (111) dislocations can be activated [3, 5]. Their mobility appears to be limited by the screw orientation [5] and yield stresses as high as 2 GPa are reported below 50K [5]. However, (110) dislocations are responsible for the increased plasticity in hard oriented crystals above 600K [3, 7]. The competition between (111) and (110) dislocations as secondary slip systems therefore appears to be one of the key issues to explain the observed deformation behaviour of NiAl. 

Though such data are ambiguous, and sometimes even contradictory [12], they can be rationally explained on the basis of qualitative considerations on intermolecular interactions of a polymer with a filler. Of practical importance is the fact that varying the nature of the dispersion medium and the filler and thus controlling the intensity of net-formation, we can vary the yield stress of filled polymers within wide limits and in different directions. 

Stress-strain curves at the conditions of product application. If applicable, this would usually indicate the toughness of material by sizing up the area under the curve (Chapter 2). It would also show the proportional limit, yield point, corresponding elongations, and other relevant data. 

Linear viscoelasticity Linear viscoelastic theory and its application to static stress analysis is now developed. According to this theory, material is linearly viscoelastic if, when it is stressed below some limiting stress (about half the short-time yield stress), small strains are at any time almost linearly proportional to the imposed stresses. Portions of the creep data typify such behavior and furnish the basis for fairly accurate predictions concerning the deformation of plastics when subjected to loads over long periods of time. It should be noted that linear behavior, as defined, does not always persist throughout the time span over which the data are acquired i.e., the theory is not valid in nonlinear regions and other prediction methods must be used in such cases. 

As an example eonsider a thin walled tube, free to pivot at both ends, where we want it to yield (exeeed its elastie limit) just as it buckles. Assume it is a high strength material (yield stress O.OIE ). We then obtain L/r = 31 or L/a = 22. 

Bingham number Nm N - T°° Moo / r0 = yield stress = limiting viscosity (Yield/viscous) stresses Flow of Bingham plastics... 

A film of paint, 3 mm thick, is applied to a flat surface that is inclined to the horizontal by an angle 9. If the paint is a Bingham plastic, with a yield stress of 150 dyn/cm2, a limiting viscosity of 65 cP, and an SG of 1.3, how large would the angle 9 have to be before the paint would start to run At this angle, what would the shear rate be if the paint follows the power law model instead, with a flow index of 0.6 and a consistency coefficient of 215 (in cgs units) ... 

The Bingham plastic model can describe acrylic latex paint, with a yield stress of 112 dyn/cm2, a limiting viscosity of 80 cP, and a density of 0.95 g/cm3. What is the maximum thickness of this paint that can be applied to a vertical wall without running ... 

You must determine the horsepower required to pump a coal slurry through an 18 in. diameter pipeline, 300 mi long, at a rate of 5 million tons/yr. The slurry can be described by the Bingham plastic model, with a yield stress of 75 dyn/cm2, a limiting viscosity of 40 cP, and a density of 1.4 g/cm3. For non-Newtonian fluids, the flow is not sensitive to the wall roughness. 

A pipeline has been proposed to transport a coal slurry 1200 mi from Wyoming to Texas, at a rate of 50 million tons/yr, through a 36 in. diameter pipeline. The coal slurry has the properties of a Bingham plastic, with a yield stress of 150dyn/cm2, a limiting viscosity of 40 cP, and an SG of 1.5. You must conduct a lab experiment in which the measured pressure gradient can be used to determine the total pressure drop in the pipeline. 

You want to predict how fast a glacier that is 200 ft thick will flow down a slope inclined 25° to the horizontal. Assume that the glacier ice can be described by the Bingham plastic model with a yield stress of 50 psi, a limiting viscosity of 840 poise, and an SG of 0.98. The following materials are available to you in the lab, which also may be described by the Bingham plastic model ... 

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