1. 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. 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 x 3 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. 3 SYMMETRY OF THE STRESS TENSOR x1x1 x2x2 x3x3 x1x1 x3x3 x2x2 a1a1 a2a2 a3a3 The control volume has dimensions  x 1,  x 2 and  x 3. The acceleration vectors a 1, a 2 and a 3 are located at the center of the control volume. We wish to compute the moment of the acceleration vector about the x 3 axis, as a prelude to computing. The acceleration a 3 contributes nothing to this moment, as it is parallel to x 3. The arm for computing the moment of a 1 about x 3 is (1/2)  x 2. (1/2)  x 2 a1a1 The contribution to the moment from a 1 is thus

4. 4 SYMMETRY OF THE STRESS TENSOR x1x1 x2x2 x3x3 x1x1 x3x3 x2x2 a1a1 a2a2 a3a3 The contribution to the moment about the x 3 axis from a 2 is computed as follows. The arm for computing the moment of a 2 about x 3 is (1/2)  x 1. (1/2)  x 1 a2a2 The contribution to the moment from a 2 is thus The x 3 component of is thus given as

5. 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. 6 x1x1 x2x2 x3x3 x1x1 x3x3 x2x2  22  21  23  21  22  23 AB Face A (left) No contribution from  21 because  21 is parallel to arm (1/2)  x 1. No contribution from  23 because  23 is parallel to x 3 Only contribution is: Face B (right) No contribution from  23 because  23 is parallel to x 3 Contribution is: Since the total contribution from Faces A & B is SYMMETRY OF THE STRESS TENSOR

7. 7 x1x1 x2x2 x3x3 x1x1 x3x3 x2x2  11  12  13  12  11  13 DC Face C (back) No contribution from  12 because  12 is parallel to arm (1/2)  x 2. No contribution from  13 because  13 is parallel to x 3 Only contribution is: Face D (front) No contribution from  13 because  13 is parallel to x 3 Contribution is: Since the total contribution from Faces C & D is SYMMETRY OF THE STRESS TENSOR

8. 8 x1x1 x2x2 x3x3 x1x1 x3x3 x2x2  31  32  33  32  31  33 EF Face E (bottom) No contribution from  33 because  33 is parallel to x 3. Face F (top) No contribution from  33 because  33 is parallel to x 3 Contribution is: Manipulating as before, the total contribution from Faces E & F is Contribution is: SYMMETRY OF THE STRESS TENSOR

9. 9 The sum of moments of the surface force about x 3 axis is equal to so that the x 3 component of the equation reduces to SYMMETRY OF THE STRESS TENSOR

10. 10 Now divide both sides of the equation by  x 1  x 2  x 3 to get SYMMETRY OF THE STRESS TENSOR Taking the limit as  x 1,  x 2 and  x 3 all  yields the desired result:

11. 11 SYMMETRY OF THE STRESS TENSOR Repetition of the same analysis forthe x 1 axisand the x 2 axis yields the results or in general Including the body force will not change this result, because in computing, takes the form and in addition the relevant arm must be (1/2)  x 1, (1/2)  x 2 or (1/2)  x 3 so the body force will scale as (  x) 4 and will thus  0 as  x 1,  x 2 and  x 3  0 when the body force is included in the equation at the top of the previous slide.