Block compression shear

When a mbber block of rectangular cross-section, bonded between two rigid parallel plates, is deformed by a displacement of one of the bonded plates in the length direction, the rubber is placed in a state of simple shear (Figure 1.1). To maintain such a deformation throughout the block, compressive and shear stresses would be needed on the end surfaces, as well as on the bonded plates [1,2]. However, the end surfaces are generally stress-free, and therefore the stress system necessary... 

In order to gain some knowledge about the degradation of the adhesive layer during the delamination test, the specimens which successfully passed the test were further tested in compression shear (Fig. 24). For these tests, blocks with a cross section of 50 by 50 mm were cut out of the tested specimens. As a control, specimens of a similar size were cut from the part of the original sample which had not been subjected to the delamination tests. The compression shear tests conhrmed that the five adhesive layers which passed the delamination tests did not suffer any degradation during the harsh treatment. Their shear resistance was hardly reduced after the harsh delamination tests. 

Another compressive shear test used for wood joints is ASTM D905-49. Test joints are prepared by bonding two blocks of wood, free from defects, with the grain of each block parallel to the length direction. Test specimens are cut from this bonded joint and tested in a special shearing apparatus (Figure 13). Loads are applied using a crosshead movement of 0.015 inch/min. The shear stress is calculated with the failure load and bonded shear area. 

D-4501. Test Method for Shear Strength of Adhesive Bonds Between Rigid Substrates by the Block-Shear Method. Fig. 9 shows the two specimen configurations and the test head for this new, general-purpose, compression shear test which should be suitable for testing adhesives for plastics, metals, glass, wood, and other substrates. 

FIGURE 1.1 Shear of a bonded block. The points A and B denote regions in tension and compression, respectively. 

Note that the normal (vertical) stress t22 is a second-order compressive stress in this case. However, as pointed out by Rivlin [1] and Ogden [2], stresses need to be apphed to the block end surfaces to maintain the shear deformation, consisting of a stress normal to the end surface in the deformed state, and a shear stress (Figure 1.1) ... 

This phenomenon is attributed to the absence on the end surfaces of the shear and compressive stresses that are needed in order to maintain a state of simple shear. As a result, the stresses throughout the block are affected (in contrast to a conventional end effect that would have a negligible effect far from the ends). One consequence is that high internal stresses can develop, sufficient in principle to cause failure. It is clear that the effect of the special conditions obtaining at the ends should be taken into consideration in the rational design of rubber springs. 

Most mechanical and civil engineering applications involving elastomers use the elastomer in compression and/or shear. In compression, a parameter known as shape factor (S—the ratio of one loaded area to the total force-free area) is required as well as the material modulus to predict the stress versus strain properties. In most cases, elastomer components are bonded to metal-constraining plates, so that the shape factor S remains essentially constant during and after compression. For example, the compression modulus E. for a squat block will be... 

In the extruder, not only shear flow is present, but also extensional flow occurs as well. This is illustrated in Fig. 3.20 for the deformation of a fluid element. Wherever cross-sections narrow, such as at the tips or between kneading blocks and the wall, the fluid elements are compressed and extended. This effect is particularly relevant for non-homogenous polymer melts, e.g., immiscible blends, in which the disperse phase can be split by extensional deformation. For more details, see Chapter 9. 

Figure 3.23 Effect of bonded steel sheets on the response of rubber blocks to compression and shear. (From Ref. 24.)...

Finally, Figure 14.5d shows a schematic of simple shear deformation. The specimen is clamped between steel blocks. The blocks must move parallel to each other in order to get a shear strain that is uniform along the waisted region. The shear stress is calculated as t = F/A, where F is the force applied to the plane of area A. In this test it is not necessary to distinguish between nominal and true stress because the shear strain does not affect A. The shear strain is defined as y = Ax/y, where Ax is the displacement of planes separated by a distance y. Ax being measured in the direction of the force applied, which is perpendicular to y. As in the case of the compression plane strain test, there is no change in the dimension of the sample along the z axis. 

Strength. A structure composed of "acid brick" or block in any other chemically-resistant masonry unit cannot be reinforced. Although it has good compressive strength, it is weak in tension and shear, and depends for its integrity on the bond of the mortar to the masonry face. 

The orientation of crystalline stems with respect to the interface of the microstructure in block copolymers depends on both morphology and the speed of chain diffusion, which is controlled by block copolymer molecular weight and the crystallization protocol (i.e. cooling rate). In contrast to homopolymers, where folding of chains occurs such that stems are always perpendicular to the lamellar interface, a parallel orientation was observed for block copolymers crystallized from a lamellar melt phase perpendicular folding was observed in a cylindrical microstructure. Both orientations are shown in Fig. 8. Chain orientation can be probed via combined SAXS and WAXS on specimens oriented by shear or compression. In PE, for example, the orientation of (110) and (200) WAXS reflections with respect to Bragg peaks from the microstructure in the SAXS pattern enables the unit cell orientation to be deduced. Since PE stems are known to be oriented along the c axis, the chain orientation with respect to the microstructure can be determined. 

The bulk modulus of rubber, which depends on the strength of the van der Waals forces between the molecules, is 2 GPa. Therefore, the compressive modulus of a rubber layer increases by a factor of a thousand as the shape factor increases from 0.2 (Fig. 4.3). The responses are not shown for S < 0.2 such tall, thin rubber blocks would buckle elastically (Appendix C, Section C. 1.4), rather than deforming uniformly. When laminated rubber springs are designed, Eqs (4.5) and (4.7) allow the independent manipulation of the shear and compressive stiffness. The physical size of the bearing will be determined by factors such as the load bearing ability of the abutting concrete material, or a limit on the allowable rubber shear strain to 7 < 0.5 and the compressive strain e < -0.1. 

An analysis of the shear band pattern in Fig. 8.1 predicted high tensile stresses parallel to the surface, at the point where the shear bands penetrate deepest into the block. The figure shows the crack that has formed at this location. Therefore, a compressive force, applied on the surface of a material that is brittle in tension, can cause fracture. 

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