Cartesian Tensors in Engineering Science by L. G. Jaeger and B. G. Neal (Auth.)

By L. G. Jaeger and B. G. Neal (Auth.)

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For any other isotropic tensor the components are T = kô where k is a scalar. Invariance of form then follows from vq TS£ =• ^sp^-tqTpq = PQi λspλtq(kSpq) — kXspXtp — kost- This is the equivalent in suffix notation to the matrix development which led up to eqn. 6). The reader may now see the reason for the reservation made above in parentheses concerning the moments of inertia of the cylindrical bar shown in Fig. 23. For any given diameter of bar, there is one particular length for which h = h = h \ in this one case the moment of inertia tensor is isotropic with respect to all three axes, and the xi direction is no longer uniquely defined.

Fn =fn cos 0 + / sin 0 - I 2 (fll - /22\ cos20 2 ) 1 + ( ^ 2 , 2 2 sin20. 4) is clearly of the form [x — a] + y = r . By taking a unit force in the x' direction the reader will readily 2 2 2 2 verify that / 2 = I and /21 = 2 2 )~ \ - ^ 2 U )° 2^ j 22 s i n 2 0 - It may be noted in passing that f[ = / 2 1 in accordance with the reciprocal theorem. The steps 1, 2 and 3 given above enable one to give a physical significance to the tensor transformation as follows. (a) Starting with a unit load in the x{ direction, this is repre" 1 sented as the column vector 0 referred to X19X2 axes.

Step 4. Step pqrstw Write down the symbol λ six times, with reasonable spaces in between, viz. λ λ λ λ λ λ Put the dashed subscripts, one with each λ V I Z . λα Afe Aç λα λβ 9 λ/ 5. Add the undashed subscripts one with each λ9 viz. λαρλ \)qλ οτλάβλ et λ/u Step 6. According to choice put Tabcdef on the left and Tp tu on the right or vice versa. Both transformations are valid, viz. qr8 Tabcdef ~ and Tpqrstu — λapλbqλcrλdsλetλfuTpqrstu λαρλ^λοτλάβλβίλ/ηΤαάοάβ/ Having now established the pattern of the transformation laws for tensors it emerges from inspection of eqns.

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