symmetry of the stress tensor：应力张量的对称性

The acceleration a3 contributes nothing to this
moment, as it is parallel to
x3. The arm for computing the
(1/2)x2 a3
moment of a1 about x3 is
(1/2)x2.
x3
x3
force, the relevant balance equation is
x1 x1 x2

m rxarxFS
x2
2
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 rxaabout the x3 axis, as a prelude to computing mrxa.
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.
mass m, and body
(gravitational) force Fg . Newton’s second law requires that
maFSFg
Conservation of momentum requires the following. Where rdenotes an arbitrarily chosen moment arm,
The arm for computing the
moment of a2 about x3 is (1/2)x1.
The contribution to the
moment from a2 is thus
a2
1 2
x1

The x3 component of mrxa
is thus given as
x3
x1x2x3a22 1x1a12 1x2
a2
a3
a1 a2 x3 x1 x1 x2
(1/2)x1
(1/2)x1
4
x2
SYMMETRY OF THE STRESS TENSOR
We now wish to compute
rxபைடு நூலகம்S
where
FSi jinjdA S
x3
Face B (right)
No contribution from 23 because 23 is parallel to x3
SYMMETRY OF THE STRESS TENSOR
The stress tensor ij satisfies the symmetry condition
ij ji
This condition is a consequence of the conservation of moment of
m rxarx(F SF g)
1
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.
The contribution to the
a1 a2 x3
moment from a1 is thus

a1
1 2
x
2
a1
(1/2)x2
x1 x1 x2
3
x2
SYMMETRY OF THE STRESS TENSOR
The contribution to the moment about the x3 axis from a2 is computed as follows.
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.
x3
Since we have dropped the body
momentum. Consider a volume of moving fluid (rather than a fixed
volume through which fluid flows in and accelerating at rate a and subjected to
out) containing surface force FS
We do this by considering the contributions from each of the six faces of the control volume.
x3
x3 x1 x1 x2
x2
5
SYMMETRY OF THE STRESS TENSOR
Face A (left)
A
No contribution from 21
B
because 21 is parallel to
arm (1/2)x1.
No contribution from 23 because 23 is parallel to x3
Only contribution is: 22x2x1x321x1