Torsional shearing stress distribution

In order to find a beam s torsional stiffness, the shear stress distribution in the cross-sectional plane (the c, y plane) must be determined. This requires the solution of Poisson s equation for the cross section... 

Fig. 64. Shear-stress distributions for solid butt joints in torsion (aspect ratio, 40) (from Adams et al, 1978b).

Equations (P17.6.11) and (PI7.6.12) indicate that the shear stress at any point within the cross section of an elliptical cylinder under torsion is tangent to an ellipse passing through this point with the same axis ratio, a/b, as that of the boundary ellipse. In other words, the lines of shear stress are concentric ellipses. Consequently, the inner boundary also coincides with a line of shear stress. Therefore the shear stress acting normally on the internal surface parallel to the z axis in null. Moreover, if a concentric cylinder is removed from the rod, the stress distribution in the remaining portion will be the same as in the solid cylinder. For this reason, the stress function will be given by... 

A critical issue when designing adhesively bonded joints mating pipes is when the joint undergoes torsion, which could often be experienced by pipes and risers in their service lives. Again, without elaborate mathematical derivations, the distribution of the shear stress in the adhesive with reference to the geometry shown in Fig. 18.8 can be evaluated by the following equation ... 

Fig. 3.48 a Stress distribution on a circular section subjected to a pure torsional moment T. b elemental free body taken at any point in the bar which shows the biaxial stress state generated by pure shearing-stress... 

Standards ASTM D 4562 and ISO 10123 describe a shear test in which the specimen is a pin bonded inside a coUar. The test uses a press to force the pin through the collar, which rests on a support cylinder. The test results are the load required to initiate failure divided by the bonded area between the pin and the collar. This type of test is particularly suited to test anaerobic adhesives. The shear strength determined with this test is only an average value because the stress distribution is not uniform along the overlap (Neme et al. 2006 Martinez et al. 2008). ASTM E 229 also uses a pin-and-coUar type of specimen except that here torsional loadings cause failure. The adhesive stress distribution in this case is more uniform and may be used to determine the adhesive shear modulus and strength. However, the standard was withdrawn in 2003. 

When the imposed deformation consists of an inhomogeneous shear, as in torsion, the normal forces generated (corresponding to the stresses t2i in Figures 1.15 and 1.16) vary from point to point over the cross-section (Figure 1.17). The exact way in which they are distributed depends on the particular form of strain energy function obeyed by the rubber that is, on the values of Wi and W2, which obtain under the imposed deformation state (Rivlin, 1947). 

The chain molecule solution will be characterized by a set of rotational isomeric states. After the initial thermalization, this set will still be in a nonequilibrium state due to the overall shear deformation. For the internal torsional angle distribution to relax to equilibrium, it is necessary for the solution liquid structure to relax and the solvent molecule distribution to reach equilibrium. In a concentrated polymer solution, these processes are highly coupled. Rotational isomeric state changes depend on both internal potentials and the local viscosity and are often in the nanosecond range, well above the glass transition for the solution. Total stress relaxation cannot occur any faster than the chains can change their local rotational isomeric states. 

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