Basics of Fluid Mechanics, 2014a

248 CHAPTER 8. DIFFERENTIAL ANALYSIS
all the shear stresses contribute equally. For the assumption of a linear fluid 15 .
τ xy = μ Dγ (
xy dUy
= μ
Dt dx + dU )
x
dy
(8.69)
where, μ is the “normal” or “ordinary” viscosity coefficient which relates
the linear coefficient of proportionality and shear stress. This deformation
angle coefficient is assumed to be a property of the fluid.
In a similar fashion it can be written to other directions for xz as
τ xz = μ Dγ xz
Dt
= μ
( dUz
dx + dU x
dz
and for the directions of yz as
τ yz = μ Dγ yz
Dt
= μ
)
( dUz
dy + dU )
y
dz
(8.70)
(8.71)
Note that the viscosity coefficient (the linear coefficient
16 ) is assumed to be the same regardless
of the direction. This assumption is referred as
isotropic viscosity. It can be noticed at this stage,
the relationship for the two of stress tensor parts
was established. The only missing thing, at this
stage, is the diagonal component which to be
dealt below.
)
Advance material can be skipped
In general equation (8.69) can be written as
where i ≠ j and i = x or y or z.
τ ij = μ Dγ ij
Dt
y
τ xx
τ xy
( dUj
= μ + dU )
i
di dj
End Advance material
B
A
y’
τ yx
τ x’ y ’
τ yy
x’
D
τ x’ x ’
C
x
45 ◦
Fig. -8.10. Shear stress at two coordinates
in 45 ◦ orientations.
(8.72)
Normal Stress
The normal stress, τ ii (where i is either ,x, y, z) appears in the shear matrix
diagonal. To find the main (or the diagonal) stress the coordinates are rotate by 45 ◦ .
The diagonal lines (line BC and line AD in Figure 8.9) in the control volume move
to the new locations. In addition, the sides AB and AC rotate in unequal amount
which make one diagonal line longer and one diagonal line shorter. The normal shear
stress relates to the change in the diagonal line length change. This relationship can be
obtained by changing the coordinates orientation as depicted by Figure 8.10. The dx is
15 While not marked as important equation this equation is the source of the derivation.
16 The first assumption was mentioned above.