SYMMETRY OF THE STRESS TENSOR

1. SYMMETRY OF THE STRESS TENSOR The stress tensor ij satisfies the symmetry condition This condition is a consequence of the conservation of moment of momentum. Consider a volume of moving fluid (rather than a fixed volume through which fluid flows in and out) containing mass m, accelerating at rate and subjected to surface force and body (gravitational) force . Newton’s second law requires that Conservation of momentum requires the following. Where denotes an arbitrarily chosen moment arm,

2. SYMMETRY OF THE STRESS TENSOR A complete proof that the stress tensor ij is symmetric is rather tedious. Here we simplify the problem by a) considering only the surface force and b) demonstrating that 12 = 21. At the end of the lecture we show how the result generalizes to the other shear stresses (23 = 32 and 13 = 31) and the case for which the body force (gravity) is included. We demonstrate the desired result (12 = 21) by taking moments about the x3 axis of the illustrated control volume, which is moving with the fluid. Since we have dropped the body force, the relevant balance equation is

3. x3 x1 x3 x2 x1 x2 SYMMETRY OF THE STRESS TENSOR The control volume has dimensions x1, x2 and x3. The acceleration vectors a1, a2 and a3 are located at the center of the control volume. We wish to compute the moment of the acceleration vector about the x3 axis, as a prelude to computing . The acceleration a3 contributes nothing to this moment, as it is parallel to x3. (1/2)x2 a3 The arm for computing the moment of a1 about x3 is (1/2)x2. a2 a1 The contribution to the moment from a1 is thus a1 (1/2)x2

4. x3 x1 x3 x2 x1 x2 SYMMETRY OF THE STRESS TENSOR The contribution to the moment about the x3 axis from a2 is computed as follows. The arm for computing the moment of a2 about x3 is (1/2)x1. The contribution to the moment from a2 is thus a3 (1/2)x1 The x3 component of is thus given as a2 a1 a2 (1/2)x1

5. SYMMETRY OF THE STRESS TENSOR We now wish to compute where We do this by considering the contributions from each of the six faces of the control volume.

6. x3 x3 x1 x2 x1 B A x2 SYMMETRY OF THE STRESS TENSOR Face A (left) No contribution from 21 because 21 is parallel to arm (1/2)x1. No contribution from 23 because 23 is parallel to x3 21 23 22 23 Only contribution is: 22 21 Face B (right) No contribution from 23 because 23 is parallel to x3 Contribution is: Since the total contribution from Faces A & B is

7. 11 13 12 13 12 11 x3 x3 x1 x2 x1 C D x2 SYMMETRY OF THE STRESS TENSOR Face C (back) No contribution from 12 because 12 is parallel to arm (1/2)x2. No contribution from 13 because 13 is parallel to x3 Only contribution is: Face D (front) No contribution from 13 because 13 is parallel to x3 Contribution is: Since the total contribution from Faces C & D is

8. 31 33 32 33 32 31 x3 x3 x1 x2 x1 E F x2 SYMMETRY OF THE STRESS TENSOR Face E (bottom) No contribution from 33 because 33 is parallel to x3. Contribution is: Face F (top) No contribution from 33 because 33 is parallel to x3 Contribution is: Manipulating as before, the total contribution from Faces E & F is

9. SYMMETRY OF THE STRESS TENSOR The sum of moments of the surface force about x3 axis is equal to so that the x3 component of the equation reduces to

10. SYMMETRY OF THE STRESS TENSOR Now divide both sides of the equation by x1x2x3 to get Taking the limit as x1, x2 and x3 all  yields the desired result:

11. Including the body force will not change this result, because in computing , takes the form SYMMETRY OF THE STRESS TENSOR Repetition of the same analysis for the x1 axis and the x2 axis yields the results or in general and in addition the relevant arm must be (1/2)x1, (1/2)x2 or (1/2)x3 so the body force will scale as (x)4 and will thus  0 as x1, x2 and x3  0 when the body force is included in the equation at the top of the previous slide.