Symmetric strain matrix

As with stresses, transformations of strain elements from one coordinate-axis system into another obtained by rotations of axes obey the same transformation laws as those of stresses, utilizing the same direction cosines and making the symmetrical strain matrix also a tensor (McClintock and Argon 1966). 

Deformation is measured by a quantity known as strain (strain is a relative extension or contraction of dimension). Strain is similarly a tensor of the second rank having nine components (3x3 matrix). The relation between stress and strain in the elastic regime is given by the classical Hooke s law. It is therefore obvious that the Hooke s proportionality constant, known as the elastic modulus, is a tensor of 4 rank and is represented by a (9 x 9) matrix. Before further discussion we note the following. The stress tensor consists of 9 elements of which stability conditions require cjxy=(jyx, stress components in the symmetric stress matrix are only six. Similarly there are only six independent strain components. Therefore there can only be six stress and six strain components for an elastic body which has unequal elastic responses in x, y and z directions as in a completely anisotropic solid. The representation of elastic properties become simple by following the well known Einstein convention. The subscript xx, yy, zz, yz, zx and xy are respectively represented by 1, 2, 3, 4, 5 and 6. Therefore Hooke s law may now be written in a generalized form as. 

Clearly, the strain matrix is also symmetrical across the diagonal, resulting in only six independent strain elements. However, since the six strain elements are derived from only three displacements at a point, their gradients and the... 

They bring in the stress tensor [S], which is a symmetrical 3x3 matrix. The mechanical behavior of a material is determined by the mathematical expression of the six terms in the stress tensor. The purpose of rheology is to estabhsh these relations (called constitutive equations) between the stresses apphed inside the material and the strains they cause. As such, rheology appears as a disciphne situated upstream of mechanics. The equations for the fundamental law of dynamics can only be solved after determining the rheological behavior of the medium. 

Again, it is possible to show that the strain matrix is symmetric or that,... 

This is a tensor of fourth order, and in the general case it should be described by a matrix of 81 members (9x9). Since the stress and strain tensors are symmetrical and each has six independent components, the tensor of fourth order derived from them has 6x6 components. 

Arbitrary three-dimensional straining shear flow. Such flow is characterized by the boundary condition (2.5.1) remote from the drop with symmetric matrix of shear coefficients, Gkj = Gjk The solution of the problem on an arbitrary three-dimensional straining shear flow past a drop leads to the following expressions for the velocities outside and inside the drop [26,475] ... 

There are thus six different munbers present in the strain tensor. The symmetric form of the strain tensor prevents rotation of the unit cell with respect to the Cartesian axis system. It is possible to use this matrix to relate how a vector r in the unstrained matrix is related to one r in the strained structure as follows ... 

By using the Voigt notation, we represent equation (4.89) by a 10x10 symmetric matrix. Figure 4.17 shows this matrix and its inverse. The energy change per unit volume of a crystal subjected to o, E and AT in a reversible manner is equal to the sum of the strain energy (4.62), the electric polarization... 

From Hooke s law, the six independent components of the stress tensor can be expressed as a function of the six components of the strain tensor in a symmetrical matrix of order 6 with 21 modulus components for a general anisotropic sample of material. For an isotropic body, there are only two independent components. The mode of deformation will determine which modulus will be measured. 

Socrate et al. (2000) considered an axially symmetric problem, with a rubber sphere in the centre of a short cylinder of matrix the spheres are in a row, aligned with the tensile stress axis. The potential positions of crazes were predetermined, initially running radially from the material interface, then becoming normal to the tensile stress along the cylinder. The initial stress concentration is greatest in the polymer near the equator of the sphere (Fig. 4.11a). The model, for a 20% volume fraction of rubber, predicts a yield point in the tensile stress-strain curve at an average strain of 1%, and 24 MPa stress, when the first craze propagates across the section. However, this relieves the stress in the polystyrene, and a tensile stress concentration... 

Since the elastic strain energy is a unique function of state, which is independent of how that state was reached, it is possible to demonstrate that the elastic-compliance and elastic-constant matrixes, as defined above, must be symmetrical. This follows directly from the observation, for example, that... 

Here is a generalized (6N dimensional) force-torque vector, U -u (6N dimensional) is the particle translational-angular velocity relative to the bulk fluid flow evaluated at the particle centre, (3x3 matrix) is the traceless symmetric rate of the strain tensor (supposed to be constant in space). The resistance matrices Rfu (6N x6N) and Rfe (6N x 3 x 3) which depend only on the instantaneous relative particle configurations (position and orientation) relate the force-torque exerted by the suspending fluid on the particles to their motion relative to the fluid and to the imposed shear flow, respectively. Note that in ER (MR) fluids torques can be neglected. 

Since the stress and strain tensors are symmetric. the piezoelectric coefficients can be converted from tensor to matrix notation. Table 13 provides the piezoelectric matrices for a-quartz together with several values rfyk and Cjjk. including the corresponding temperature coefficients [259], [261]. 

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