I am to know what is the difference of using tensor strain (particularly its rate) and engineering strain-rate for two dimensional fluid flow analysis.
As you may know, the Navier-Stokes momentum equations require incorporating stress terms as:
τ(x,y)= 2 µ*tensor_strain=
2 µ×1/2(∂u/ ∂y+∂v/ ∂x)
Hence: τ (x,y)= µ (∂u/ ∂y+∂v/ ∂x)
Let me know what happens to the solution (Navier-Stokes equations) if I incorporate:
τ (x,y)= 2mu×engineering_strain=2µ (∂u/ ∂y+∂v/ ∂x)
Notice: first remind me if such a discussion is valid for fluids strain-rate (formulated in shear stress), as generalized from the knowledge on solids infinitesimal strain theory. Is there any tensor assumption for fluid flow, similar to solids. Please elaborate your idea.
What do you mean exactly for engineering strain? In general, the Newton law has a more complex relation of the type
T = (-p+lambda*div v)*I + 2*mu*D
being lambda the second viscosity coefficient and D the symmetric gradient velocity. If you extract the trace from D and include in the isotropic part, if you assume valid the Stokes hypothesis, you get
T = -p'*I + 2*mu*D0
Further simplification can be done for incompressible flows. For example, Div(2*mu*D0)=mu Lap (v).
Now, what about your question?
as I have seen in solids mechanics and also fluid mechanics texts, engineering strain-rate has NOT any 1/2 multipliers in its off-diagonal elements, while strain-rate tensor has it in off-diagonal elements. To be mathematically expressed, it means:
1. off-diagonal terms for engineering strain-rate in fluid are: τij = (∂ui/∂xj+∂uj/∂xi)
2. off-diagonal terms for strain-rate tensor in fluids are: τij = 1/2*(∂ui/ ∂xj+∂uj/∂xi)
notice the 1/2 multiplier.
What would be the difference if the case 1 or the case 2 is incorporated in shear-stress places of the Navier-Stokes momentum equations.
Have again a look to the fluid dynamics textbooks. The symmetry of the stress tensor is assumed as a consequence of the angular momentum.
A symmetric tensor is always denoted by (A+AT)/2, this is the reason of the 1/2 coefficient.
Then, the symmetric tensor is decomposed in its isotropic part plus the deviatoric part. The isotropic part adds a term to the pressure.
I've had no graduate-level fluid dynamics course in my academic curriculum. Through online sources I saw isotropic tensor is invariant to rotation. So the decomposition seems to probably disintegrate the strain-rate into translational (isotropic) and deviatoric (rotational). Is it correct?
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