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Torsional strength

Hardness, Shore Tensile strength Elongation at break Tensile modulus Tear strength Torsional modulus (Clash-Berg)... [Pg.52]

One of the major difficulties with molecular mechanics procedures (MMh- or otherwise) is that they almost always fail. That is, you find that force constants are not available for the molecule of interest. This is both the strength and weakness of molecular mechanics it uses atom types to introduce specific chemical environments for the atoms within a molecule (to obtain accuracy in the calculations) but then requires knowledge of force constants specific to that chemical environment (as specific as stating that an atom is in a five-member ring containing one oxygen and one carbon, for example). As the number, N, of atom types rises the number of force constants needed to describe all possible occurrences of these atom type becomes very large. For torsions, for... [Pg.204]

A significant aspect of hip joint biomechanics is that the stmctural components are not normally subjected to constant loads. Rather, this joint is subject to unique compressive, torsion, tensile, and shear stress, sometimes simultaneously. Maximum loading occurs when the heel strikes down and the toe pushes off in walking. When an implant is in place its abiUty to withstand this repetitive loading is called its fatigue strength. If an implant is placed properly, its load is shared in an anatomically correct fashion with the bone. [Pg.189]

Figure 4.41 shows the Stress-Strength Interference (SSI) diagrams for the two assembly operation failure modes. The instantaneous stress on the relief section on first assembly is composed of two parts first the applied tensile stress,. v, due to the pre-load, F, and secondly, the torsional stress, t, due to the torque on assembly, M, and this is shown in Figure 4.41(a) (Edwards and McKee, 1991). This stress is at a maximum during the assembly operation. If the component survives this stress, it will not fail by stress rupture later in life. [Pg.204]

It is required to find the torque without slippage that can be transmitted by a hub that is assembled by an interference fit to a powered shaft. The hub outside diameter D = 070 mm, and the shaft diameter d = 050 mm, as shown in Figure 4.55. The length of the hub is 100 mm. Both hub and shaft are machined from hot rolled steel SAE 1035 with a yield strength Sy A(342,26) MPa (see Table 4.6). Given that the hub is stopped suddenly in service due to a malfunction, and considering only the torsional stresses, what is the probability that the shaft will yield ... [Pg.223]

The shear yield strength for duetile metals is a linear funetion of the uniaxial yield strength. Therefore, for pure torsion from equation 4.56 ... [Pg.227]

Most materials scientists at an early stage in their university courses learn some elementary aspects of what is still miscalled strength of materials . This field incorporates elementary treatments of problems such as the elastic response of beams to continuous or localised loading, the distribution of torque across a shaft under torsion, or the elastic stresses in the components of a simple girder. Materials come into it only insofar as the specific elastic properties of a particular metal or timber determine the numerical values for some of the symbols in the algebraic treatment. This kind of simple theory is an example of continuum mechanics, and its derivation does not require any knowledge of the crystal structure or crystal properties of simple materials or of the microstructure of more complex materials. The specific aim is to design simple structures that will not exceed their elastic limit under load. [Pg.47]

Torsions-faden, m. torsion filament or wire, -festigkeit, /, torsional strength, -kraft, /, torsional force, -spaonung, /, torsional stress, twisting stress, -wa(a)ge,/. torsion balance, winkeh m. angle of torsion. [Pg.449]

The section modulus. Z, of the box should be 2 / times greater than the section modulus, z, of the pin in a drill collar connection. On the right side of the connection are the spots at which the critical area of both the pin (Ap) and box (A ) should be measured for calculating torsional strength. [Pg.722]

Torsional Yield Strength of the Drill Pipe is Baaed on a Shear Strength of S7.7%of the Minimum Yield Strength. [Pg.753]

Torsional Yield Strength of the Tool Joint Baaed on Tensile Yield Strength of the Pin and Compressive Yield Strength of the Box - Lowest Value Prevailing. [Pg.753]

Tensile Yield Strength of the Tool Joint Pin is based on 120.000 psi Yield and the Cross Sectional Area at the Root of the Thread 5/8 inch from the Shoulder. Torsional Yield Strength of the Drill Pipe is Based on a Shear Strength of 57.7% of the Minimum Yield Strength. [Pg.754]

OTool Joint with dimensions shown has a lower torsional yield strength than the drill pipe to which it is attached. [Pg.759]

Normally, if the drill pipe is sufficiently strong for tension, it will have satisfactory strength in torsion, collapse and burst however, if there is any doubt, additional checkup calculations must be performed. [Pg.767]

There is a wide variety of commercially available sprockets. While they may vary in design, methods of manufacture, and materials of constmction, they all have some common features. They will all have hardened teeth designed to mate with a specific type of chain, sufficient web strength to effectively transmit their rated horsepower or torsional forces, and a boss or hub that can be bored to the mating shaft s diameter. [Pg.988]

The shear modulus of a material can be determined by a static torsion test or by a dynamic test employing a torsional pendulum or an oscillatory rheometer. The maximum short-term shear stress (strength) of a material can also be determined from a punch shear test. [Pg.60]

Because strain measurements are difficult if not impossible to measure, few values of yield strength can be determined by testing. It is interesting to note that tests of bolts and rivets have shown that their strength in double shear can at times be as much as 20% below that for single shear. The values for the shear yield point (kPa or psi) are generally not available however, the values that are listed are usually obtained by the torsional testing of round test specimens. [Pg.60]

A shaft subject to torque is generally considered to have failed when the strength of the material in shear is exceeded. For a torsional load the shear strength used in design should be the published value or one half the tensile strength, whichever is less. The maximum shear stress on a shaft in torsion is given by the following equation ... [Pg.147]


See other pages where Torsional strength is mentioned: [Pg.886]    [Pg.361]    [Pg.517]    [Pg.367]    [Pg.886]    [Pg.361]    [Pg.517]    [Pg.367]    [Pg.204]    [Pg.290]    [Pg.78]    [Pg.79]    [Pg.88]    [Pg.229]    [Pg.65]    [Pg.451]    [Pg.508]    [Pg.7]    [Pg.154]    [Pg.191]    [Pg.193]    [Pg.123]    [Pg.413]    [Pg.432]    [Pg.759]    [Pg.1326]    [Pg.997]    [Pg.1381]    [Pg.1297]    [Pg.60]    [Pg.318]    [Pg.31]    [Pg.670]    [Pg.271]   
See also in sourсe #XX -- [ Pg.20 ]




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