Influence of Compression upon the Shear Properties of Bonded Rubber Blocks

1980 ◽  
Vol 53 (5) ◽  
pp. 1133-1144 ◽  
Author(s):  
L. S. Porter ◽  
E. A. Meinecke
Keyword(s):  
Shear Stress ◽  
Shear Modulus ◽  
Simple Shear ◽  
Compressive Force ◽  
Strain Curve ◽  
Shear Loading ◽  
Total Force ◽  
Strain Response ◽  
Stress Strain ◽  
Spring Rate

Abstract Rubber has a stress-strain response to compression-shear loadings that is the same as its stress-strain response to simple shear loadings. However, its load-deflection response to the compression-shear loading is not the same as its simple shear response. In determining the stress-strain relationship of the compression-shear loading from the load-deflection responses, three factors must be considered. First, the compression of the sample gives a lower rubber thickness. After calculating the strain, the lower thickness will give a higher strain than the original thickness at an equal deflection. Second, the compression gives a larger surface area due to bulging of the rubber. The higher area would result in a lower stress than the original area at an equal load. Third, the force that is necessary to compress the rubber block is stored in the rubber. When the rubber is sheared, the shear vector of the compressive force aides in deflecting the rubber. Therefore, the shear force vector would be added to the recorded load to determine the total force needed to shear the rubber. The resulting shear stress would be higher than the shear stress calculated by using the recorded load in calculating the shear stress. With all three factors accounted for, the shear stress-strain of the rubber is the same for the compressed part as it is for the uncompressed part. Therefore, the rubber's shear modulus, the slope of the shear stress-strain curve, has not been affected by the superimposed compression and remains an inherent property of the rubber. When designing a part to be used in a compression-shear application, one can use the shear and compression moduli normally obtained for shear and compression applications. The compression modulus would be used for determining the compressive spring rate and the amount of force used in lowering the shear spring rate. The shear modulus would be used to determine the shear rate by taking into account the geometry changes and the force due to compression.

Shear modulus of cohesionless soil: variation with stress and strain level

1992 ◽  
Vol 29 (1) ◽  
pp. 157-161 ◽  
Author(s):  
Martin Fahey
Keyword(s):  
Shear Stress ◽  
Shear Modulus ◽  
Shear Strain ◽  
Stress Level ◽  
Strain Curve ◽  
Strain Level ◽  
Cohesionless Soil ◽  
Backbone Curve ◽  
Stress Strain ◽  
Shear Stress Level

The value of the secant shear modulus (G) of sand measured in cyclic tests reduces as the amplitude of cycling increases. As a first approximation, it is assumed that the curve joining the extreme points of stress–strain (τ–γ) loops of different amplitudes (a so-called "backbone curve") is hyperbolic. The shear strength (τmax) of sand is directly proportional to the mean effective confining pressure (p′), whereas the maximum shear modulus (G0) is proportional to (p′)n, with n being between 0.4 and 0.5. Based on these assumptions, it is shown that at the same shear strain level, different G/G0 values should be expected at different p′ values. One of the features of a hyperbolic τ–γ curve is that there is a unique linear relationship between G/G0 and normalized shear stress level (defined as τ/τmax), independent of p′. Therefore, considering the normalized shear stress level rather than the shear strain level may be a more logical and unifying way of examining the variation in G/G0. Key words : shear modulus, hyperbolic stress–strain curve, pressuremeter test.

scholarly journals An Experimental Study on Mechanical Behavior of Parallel Joint Specimens under Compression Shear

2018 ◽  
Vol 2018 ◽  
pp. 1-12
Author(s):  
Fei Wang ◽  
Ping Cao ◽  
Yu Chen ◽  
Qing-peng Gao ◽  
Zhu Wang
Keyword(s):  
Shear Stress ◽  
Shear Strength ◽  
Failure Mode ◽  
Shear Failure ◽  
Strain Curve ◽  
Shear Loading ◽  
Plane Shear ◽  
Joint Spacing ◽  
Dip Angle ◽  
Joint Inclination

In order to investigate the influence of the joint on the failure mode, peak shear strength, and shear stress-strain curve of rock mass, the compression shear test loading on the parallel jointed specimens was carried out, and the acoustic emission system was used to monitor the loading process. The joint spacing and joint overlap were varied to alter the relative positions of parallel joints in geometry. Under compression-shear loading, the failure mode of the joint specimen can be classified into four types: coplanar shear failure, shear failure along the joint plane, shear failure along the shear stress plane, and similar integrity shear failure. The joint dip angle has a decisive effect on the failure mode of the specimen. The joint overlap affects the crack development of the specimen but does not change the failure mode of the specimen. The joint spacing can change the failure mode of the specimen. The shear strength of the specimen firstly increases and then decreases with the increase of the dip angle and reaches the maximum at 45°. The shear strength decreases with the increase of the joint overlap and increases with the increase of the joint spacing. The shear stress-displacement curves of different joint inclination samples have differences which mainly reflect in the postrupture stage. From monitoring results of the AE system, the variation regular of the AE count corresponds to the failure mode, and the peak value of the AE count decreases with the increase of joint overlap and increases with the increase of joint spacing.

Effect of Material Frame Rotation on the Hardening of an Anisotropic Material in Simple Shear Tests

2018 ◽  
Vol 85 (12) ◽  
Author(s):  
Kelin Chen ◽  
Stelios Kyriakides ◽  
Martin Scales
Keyword(s):  
Simple Shear ◽  
Plastic Anisotropy ◽  
Yield Function ◽  
Strain Response ◽  
Shear Tests ◽  
Stress Strain ◽  
Anisotropic Yield Function ◽  
Torsion Experiment ◽  
Thin Walled Tube ◽  
Material Frame

The shear stress–strain response of an aluminum alloy is measured to a shear strain of the order of one using a pure torsion experiment on a thin-walled tube. The material exhibits plastic anisotropy that is established through a separate set of biaxial experiments on the same tube stock. The results are used to calibrate Hill's quadratic anisotropic yield function. It is shown that because in simple shear the material axes rotate during deformation, this anisotropy progressively reduces the material tangent modulus. A parametric study demonstrates that the stress–strain response extracted from a simple shear test can be influenced significantly by the anisotropy parameters. It is thus concluded that the material axes rotation inherent to simple shear tests must be included in the analysis of such experiments when the material exhibits anisotropy.

Compression Stress Strain of Rubber

1933 ◽  
Vol 6 (1) ◽  
pp. 126-150 ◽  
Author(s):  
J. R. Sheppard ◽  
W. J. Clapson
Keyword(s):  
Hollow Sphere ◽  
Compressive Force ◽  
Strain Curve ◽  
The Other ◽  
One Dimension ◽  
Compression Stress ◽  
Stress Strain ◽  
Strain Data ◽  
Tensile Forces ◽  
Strain Axis

Abstract 1. A relation of simple form between compressive force and equivalent two-way tensile forces is developed. 2. Based on this relation, a new method for determining the compression stress strain of rubber is outlined, which avoids difficulties and errors inherent in direct compression. It consists in applying tensile forces simultaneously in two directions, and, from these and the strained dimensions, in computing the compressive force that would have produced the same deformation. 3. The mode of applying the two-way tensiles is to inflate a hollow sphere of rubber; the experimental data required to determine the compression stress strain are pressure of gas in, and dimensions of, the inflating hollow sphere. 4. The method has been applied to cold-cured pure-gum rubber in the form of toy balloons which, in its ordinary elongation stress strain, shows a breaking elongation of about 650 to 700 per cent and a tensile of 30 to 40 kg. per square centimeter. While the numerical values obtained on this stock have no special significance, as they will vary from stock to stock, the following are examples: breaking compression, about 97.3 per cent; breaking compressive force, 6000 to 9000 kg. per square centimeter (on original cross section); hysteresis, 29 to 35 per cent of work of compression to near rupture. 5. As a common measuring stick by which to gage degree of strain in deformations of different types—e. g., increasing one dimension (and diminishing the other two) as against diminishing one dimension (and increasing the other two)—energy seems the best. Energy at break for ordinary elongation stress strain was 50 to 70 kg. cm. per cubic centimeter, and for compression stress strain was 89 to 103 kg. cm. per cubic centimeter. 6. The compression stress-strain data may, if desired, be expressed in terms of two-way tensiles vs. two-way elongations. Energy of compression may be computed either as twice the area subtended between such a curve and the strain axis, or as the area between the compression stress strain and the strain axis. 7. It is strongly indicated that the compression stress strain of rubber is continuous with the ordinary elongation stress strain when both are plotted in the same units, and that the complete stress strain should accordingly be considered as a single continuous curve having an elongation branch and a compression branch with the origin as dividing point. 8. The analytic features of the complete stress strain are described. 9. Granting the observed concavity of the upper part of the elongation stress strain, and the thesis of continuity between elongation and compression, a point of inflection is bound to exist theoretically. 10. Implications of the thesis of continuity are: (1) An equation for the stress-strain curve must fit the complete curve; it is not sufficient that it fit the elongation branch only. (2) It is impossible to compute the compression stress strain from the ordinary (one-way) elongation stress-strain data. The two sets of data are related empirically. 11. When compressive force and equivalent two-way tensiles are based on actual cross sections, stress conditions at a point are expressed and we have the simple rule: Pressure at a point is numerically equal to the transverse tensions which, substituted therefor, will maintain the same strain.

scholarly journals Non-coaxiality of sand under bi-directional shear loading

2018 ◽  
Vol 5 (5) ◽  
pp. 172076 ◽  
Author(s):  
Yao Li ◽  
Yunming Yang
Keyword(s):  
Shear Stress ◽  
Strain Rate ◽  
Simple Shear ◽  
Shear Test ◽  
Shear Loading ◽  
Shear Behaviour ◽  
Shear Stresses ◽  
Simple Shear Test ◽  
Principal Axes ◽  
Angle Difference

This study aims to investigate the effect of consolidation shear stress magnitude on the shear behaviour and non-coaxiality of soils. In previous drained bi-directional simple shear test on Leighton Buzzard sand, it is showed that the level of non-coaxiality, which is indicated by the angle difference between the principal axes of stresses and the corresponding principal axes of strain rate tensors, is increased by increasing angle difference between the direction of consolidation shear stress and secondary shearing. This paper further investigated the relation and includes results with higher consolidation shear stresses. Results agree with the previous relation, and further showed that increasing consolidation shear stresses decreased the level of non-coaxiality in tests with angle difference between 0° and 90°, and increased the level of non-coaxiality in tests with angle difference between 90° and 180°.

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