Viscous Flow: Stress Strain Relationship
Objective:
Discuss assumptions which lead to the stress-strain relationship for
a Newtonian, linear viscous fluid:
 ∂ui ∂u j 
∂u
+
+δ λ k
 ∂x j ∂xi  ij ∂xk


∂uk ∂u ∂v ∂w
=
+
+
= ∇iV
∂xk ∂x ∂y ∂z
τ ij = µ 
where µ = dynamic viscosity coefficient
λ = bulk viscosity coefficient
0, i ≠ j
Note: δ ij = 
1, i = j
1  ∂ui ∂u j 
+
 ≡ shear strain rate in xi , x j plane
2  ∂x j ∂xi 
ε ij = 

Thus, written in terms of the strain rates, the stress tensor is:
τ ij =
2 µε ij
viscous stress
due to shearing
of a fluid element
+ δ ij λ (ε xx + ε yy + ε zz )
using indicial notation
this is ε kk
viscous stress due to an
overall compression or
expansion of the fluid
element's volume
This stress-strain relationship can be derived by the following two assumptions:
1. The shear stress is independent of a rotation of the coordinate system
2. The shear stress is at most a linear function of the strain rate tensor.
So, for example, τ xy :
τ xy = a11ε xx + a12ε xy + a13ε xz + a 22 ε yy + a 23ε yz + a33ε zz
Clearly, assumption #2 gives 6 unknowns per shear, a11 , a12 , etc. Note: why do
ε yx , ε zx & ε zy not appear in this expression for τ xy ? The total number of unknowns
for all stresses are: 36. But, this can be eventually reduced by applying #1 to the
most general linear form to the two unknowns µ & λ.
Viscous Flow Stress-Strain Relationship
Stokes’ hypothesis
Stokes hypothesized that the total normal viscous stress on a fluid element
surface,
τ xx + τ yy + τ zz = (2µ + 3λ )(ε xx + ε yy + ε zz )
should be zero, i.e. τ xx + τ yy + τ zz = 0 so that the normal force (stress) on a
surface is only that due to pressure. This requires that
2
2 µ + 3λ = 0 ⇒ λ = − µ
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Comments
•
•
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Experimental studies have indicated that λ ≠ − µ and in general is not
3
negative!
For incompressible flow, ε xx + ε yy + ε zz = ∇ ⋅ V = 0 so τ xx + τ yy + τ zz = 0
regardless of λ.
•
For most compressible flows ∇ ⋅ V is small compared to shearing strains (i.e.
ε xy , ε yz , etc.) so again, Stokes’ hypothesis has little impact. So, as a result,
2
common practice is to assume that λ = − µ .
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16.100 2002
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