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Viscous stress tensor

  1. Jan 28, 2014 #1

    joshmccraney

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    hey pf!

    in reading a book on viscous stresses i found the following: [tex]\tau_{ij}=2\mu\Big(s_{ij}-\frac{1}{3}s_{kk}\delta_{ij}\Big)[/tex] where einstein summation is used. now we have [tex]s_{ij}=\frac{1}{2}\Big(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\Big)[/tex] and then the claim is incompressible flow implies [itex]s_{kk}=0[/itex]. can someone explain why this is so?

    im using standard notation, but if i need to clarify let me know.

    thaks!
     
  2. jcsd
  3. Jan 28, 2014 #2
    Well, s_ij is the elastic stress tensor, isn't it?
    The diagonal components (s_ii) describe the compression stress so if there is no compression...
    The off diagonal components describe shear stress.
     
  4. Jan 28, 2014 #3

    joshmccraney

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    for sure, but does incompressible flow imply zero compression? (i suppose here i am not thinking of my control volume as an infintesimal unit of fluid, but instead, perhaps a solid submerged). in this case, would the above work, or are we just talking about fluids as our control volume (in which case, i totally agree/understand what you're saying)?
     
  5. Jan 28, 2014 #4

    AlephZero

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    If the OP, ##s_{ij}## it is the elastic strain tensor.

    ##s_{kk}## is just the change in volume of a fluid element. Incompressible means the change in volume is zero, by definition.
     
  6. Jan 28, 2014 #5

    joshmccraney

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    are you sure? [itex]\frac{\partial u_i}{\partial x_i}[/itex] represents change in velocity, [itex]u_i[/itex], with respect to a spacial dimension, [itex]x_i[/itex]. could you explain how this partial derivative is a volume change (i don't think the units add up), or are you referring to the vector (adopting einstein's summation notation), for which we'd have some sort of gradient [itex]\nabla \vec{u}[/itex], which still seems incorrect as a volume change.

    let me know what you think

    thanks
     
  7. Jan 28, 2014 #6

    joshmccraney

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    oh, and yes on the strain, not stress (which you evidently already know)
     
  8. Jan 28, 2014 #7
    sij is neither the elastic stress tensor nor the elastic strain tensor. For a viscous fluid, sij as used here is the rate of deformation tensor. The rate of deformation tensor is equal to the sum of the velocity gradient tensor and its transpose, divided by 2.

    Now for skk using Einstein summation notation:
    [tex]s_{kk}=\frac{1}{2}\Big(\frac{\partial u_k}{\partial x_k}+\frac{\partial u_k}{\partial x_k}\Big)=\frac{\partial u_k}{\partial x_k}[/tex]
    This is just the divergence of the velocity vector, which is zero for an incompressible fluid.
     
  9. Jan 28, 2014 #8

    AlephZero

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    Ahhh.... I'm a solid mechanics guy more than fluids, I assumed u was the displacement vector not the velocity vector. Sorry for any confusion.

    But it comes to the same thing either way. Think about a rectangular block of material with dimensions ##X## by ##Y## by ##Z##.

    The component ##\epsilon_{11}## of the strain tensor corresponds to a change in volume of the block from ##XYZ## to ##X(1+\epsilon_{11})YZ## and similarly for ##\epsilon_{22}## and ##\epsilon_{33}##.

    Combining all three strains the volume changes from ##XYZ## to ##XYZ(1+\epsilon_{11})(1+\epsilon_{22})(1+\epsilon_{33})##. If the strains are small, we can ignore terms higher than first order, so the volume is approximately ##XYZ(1+\epsilon_{11} + \epsilon_{22} + \epsilon_{33})##.

    The shear strains do not change the volume of the block, they only change its shape.

    If the material is incompressible the volume doesn't change, so ##XYZ(1+\epsilon_{11} + \epsilon_{22} + \epsilon_{33}) = XYZ##, which means that ##\epsilon_{11} + \epsilon_{22} + \epsilon_{33} = 0##.

    The argument is basically the same for the rate-of-strain tensor, except ##u_{11}##, ##u_{22}##, ##u_{33}## give the rate of change of the volume of the fluid element instead of the volume. For an incompressible fluid, the rate of change of the volume is zero.
     
  10. Jan 28, 2014 #9

    boneh3ad

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    You can also look at it in terms of the continuity equation:
    [tex]\dfrac{D\rho}{Dt} + \rho(\nabla \cdot \vec{u}) = 0.[/tex]
    An incompressible fluid implies that [itex]D\rho/Dt = 0[/itex], so that leaves
    [tex]\nabla \cdot \vec{u} = 0.[/tex]
    This, of course, is equivalent to what was above termed [itex]s_{kk}[/itex], as
    [tex]\nabla \cdot \vec{u} = (\partial_1,\partial_2,\partial_3) \cdot (u_1,u_2,u_3) = \dfrac{\partial u_1}{\partial x_1} + \dfrac{\partial u_2}{\partial x_2} + \dfrac{\partial u_3}{\partial x_3} = \dfrac{\partial u_k}{\partial x_k} = s_{kk}.[/tex]
     
  11. Jan 29, 2014 #10

    joshmccraney

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    thanks! i didn't realize the divergence, but this makes perfect sense now! but, my book calls [itex]s_{ij}[/itex] the instantaneous strain rate tensor
     
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