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

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.

2019 ◽  
Vol 87 (3) ◽  
Author(s):  
Brian Nyvang Legarth ◽  
Viggo Tvergaard

Abstract Full three-dimensional cell models containing a small cavity are used to study the effect of plastic anisotropy on cavitation instabilities. Predictions for the Barlat-91 model (Barlat et al., 1991, “A Six-Component Yield Function for Anisotropic Materials,” Int. J. Plast. 7, 693–712), with a non-quadratic anisotropic yield function, are compared with previous results for the classical anisotropic Hill-48 quadratic yield function (Hill, 1948, “A Theory of the Yielding and Plastic Flow of a Anisotropic Metals,” Proc. R. Soc. Lond. A193, 281–297). The critical stress, at which the stored elastic energy will drive the cavity growth, is strongly affected by the anisotropy as compared with isotropic plasticity, but does not show much difference between the two models of anisotropy. While a cavity tends to remain nearly spherical during a cavitation instability in isotropic plasticity, the cavity shapes in an anisotropic material develop toward near-spheroidal elongated shapes, which differ for different values of the coefficients defining the anisotropy. The shapes found for the Barlat-91 model, with a non-quadratic anisotropic yield function, differ noticeably from the shapes found for the quadratic Hill-48 yield function. Computations are included for a high value of the exponent in the Barlat-91 model, where this model represents a Tresca-like yield surface with rounded corners.


1980 ◽  
Vol 53 (5) ◽  
pp. 1133-1144 ◽  
Author(s):  
L. S. Porter ◽  
E. A. Meinecke

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.


1991 ◽  
Vol 28 (2) ◽  
pp. 276-281 ◽  
Author(s):  
Gianni Rossato ◽  
Paolo Simonini

The behaviour of a natural sand in triaxial compression and direct simple shear tests was compared by means of dimensionless analysis of parameters controlling the evolution of stresses and strains. The secant triaxial compression and direct simple shear moduli were interpreted in a dimensionless form. A criterion based on the equivalence between major principal strain in the two tests was considered to compare the results. Key words: sand, stress–strain behaviour, triaxial test, direct simple shear test, shear modulus, triaxial compression modulus.


2010 ◽  
Author(s):  
J. J. Gracio ◽  
H. J. Kim ◽  
G. Vincze ◽  
B. B. Panigrahi ◽  
F. Barlat ◽  
...  

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