Tensor Transformations and the Maximum Shear Stress (Draft 1, 1/28/07) Introduction The order of a tensor is the number of subscripts it has. For each subscript it is multiplied by a direction cosine array to rotate it to a new (primed) coordinate system. For example, the rotational transformations of a scalar, vector, and stress tensor between x k and x k axes are: K = K, V j = a jk V k, σ mp = a mj a pk σ jk, K = K, V k = a jk V j, σ jk = a mj a pk σ mp. A direction cosine array is an orthogonal array, that is, its transpose is its inverse. The number of range (non-repeated) indices denotes the order of the tensor and must be the same in every term (and with the same names). The summation indices (appearing twice) are dummies and can be replaced with any other pair of unused indices. The summation is taken to be from 1 to 3, unless otherwise stated. The index notation also is used to represent partial derivatives. That is denoted by placing a comma before the index. Therefore, ( ), k = ( ) / x k, and ( ), km = 2 ( ) / x k x m. To help remember the above tensor rotation definition rule, follow this procedure: 1. Write down the valid equation without subscripts, but with as many direction cosine array products (to the right of the equals) as there are range indices. 2. Insert range indices on the primed term. 3. Use that sequence of range indices as the first subscript for each of the direction cosine arrays. 4. Assign an equal number of new range indices to the quantity in the original system. 5. Use that sequence of range indices as the second subscript of each of the direction cosine arrays. Note that this last step creates the pairs of indices to be used in the required summations. Maximum shear stress calculation Historically, the transformation law for second order tensors (stress, strain, inertia, etc.) was computed graphically by Mohr s circle. While not as accurate and fast as computer calculations, Mohr s circle is still recognized as a useful visualization tool, especially for 3D transformations. In 2D applications Mohr s circle (and the above equations) are utilized to find the principal normal stresses and maximum shear stress in the 2D plane. However, it is not uncommon to find the actual maximum shear stress occurs in a plane perpendicular to the one studied and has a value higher than the 2D transformation yields. Fortunately, a quantity called the Octahedral shear stress can be shown to provide both lower and upper bounds on the true maximum shear stress. From the solution of the stress cubic equation, it can be shown that the following inequality is true in 2D and 3D: Page 1 of 7. Copyright J.E. Akin. All rights reserved.
1 [τ Oct (3/2) 1/2 ] / τ Max 2 / (3) 1/2 1.155. This information means that when solving planar stress problems you must either consider the Octahedral shear stress or solve the full cubic stress equation for the true maximum shear stress. Sample 2D stress tensor transformation To illustrate the previous point, a TK Solver implementation of the 2D tensor transformation is given in Figure 1. Note that the variable Bound has been added to establish the upper bound of the true 3D maximum shear. If you are using a failure criteria based on maximum shear then the Bound value is more conservative. It also hints when a 3D study is needed. Figure 1 Page 2 of 7. Copyright J.E. Akin. All rights reserved.
A typical 2D set of stress input values (σ xx = 40, σ yy = 20, σ xy = 10 psi) are shown in Figure 2. The Angle can have any desired value. First Angle was left blank and its principal value (- 67.5 deg.) was found. That value was cut and pasted as an input value to Angle, and 45 degrees was added to it to find the direction normal to a maximum shear plane. Then a second solve gave the results tabulated in Figure 2. Note that the (maximum) shear stress, ST12, on the plane normal to Angle has a value of 14.1, but the upper bound value for the maximum shear stress is 22.4 psi. You should use the upper bound, or solve a full 3D principal stress calculation with the same input data. Figure 2 The algorithm for solving the stress cubic for the principal stresses is well known, and is given in the top portion of Figure 3. Near the bottom of that figure some auxiliary calculations are shown for some common failure criterion: The Von Mise s effective stress (a measure of the stored distortional energy), the Octahedral shear stress, and the upper bound on the maximum shear stress. Page 3 of 7. Copyright J.E. Akin. All rights reserved.
The same stress tensor components given above (padded with zeros) were input as a 3D stress tensor. With that as the only input a direct solve with TK yields the results in Figure 4. In that figure you will notice that the 2D estimated max shear value of 14.1 psi does appear, but as the intermediate of the three maximum shear stresses. The true maximum value is 22.1 psi which is much closer to the upper bound value of 22.4 psi. The corresponding 3D Mohr s circles are given in Figure 5, plus the max shear bound and Von Mise s values. Figure 3 The cubic equation rules in Figure 3 are valid for the principal values of any second order tensor. To use them for a strain tensor or mass moment of inertia tensor you would just have to change the comments, and the units. Since the main items of interest all have the same Page 4 of 7. Copyright J.E. Akin. All rights reserved.
units you are really interested in just the numerical values. However, safety suggests that you should also change the calculation and display units. Figure 4 Deviatoric stress tensor Another commonly used form of the stress tensor splits it into hydrostatic and pure shear components. The hydrostatic pressure is the average of the diagonal of the stress tensor, and the deviatoric stress tensor is p = σ kk / 3 = (σ xx + σ yy + σ zz ) / 3, φ ab = σ ab δ ab p where δ ab is the Kronecker Delta tensor. The deviatoric stress tensor components are also computed in the above TK Solver rule and variable sheets. Page 5 of 7. Copyright J.E. Akin. All rights reserved.
Figure 5 Engineering shear strain While mathematicians and engineers working in the theory of elasticity and the theory of plasticity like to use the tensor definition of the strain-displacement relations (ε jk = [u j,k + u k,j ] / 2) most engineering applications use a shear strain measure that is twice as large, namely γ jk = [u j,k + u k,j ]. In the 2D case it is γ xy = u / y + v / x. By adding this simple rule the strain tensor transformation (within TK Solver) will work with either shear stress definition as an input. A 2D strain tensor transformation (and Mohr s transformation for engineering strains) is shown in Figure 6. It is basically the rules in Figure 1 with new comments and units (strain units are m/m and were left blank). In addition it has a rule stating that engineering shear is twice the tensor shear. The data illustrated in Figure 7, were input in the engineering form. The tensor strain ε xy was computed from γ xy automatically by TK. Then TK evaluated the transformation rules for all the components, including ε xy. Finally, ε xy was converted to γ xy. Page 6 of 7. Copyright J.E. Akin. All rights reserved.
Figure 6 Figure 7 Page 7 of 7. Copyright J.E. Akin. All rights reserved.