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Shear in fluids: What is measured and called - strain?

  1. Feb 24, 2015 #1
    It's about shear stress and strain. We are talking about Newtonian fluids, a situation where we are lubricating a rotating shaft with oil.

    I cannot understand what we are measuring that we use as the strain.

    SOLIDS. As far as I know we can say: Stress (F/A) = G x (Δ l / l)

    Where G is shear modulous, Δl is the displacement, parallel with the force, due to the stress (or force) - probably in fractions of a millimeter - and l is the height of the zone where displacement is taking place, again probably in fractions of a millmeter. You get the strain by measuring two distances - Δl and l - and obtaining the ratio of them, which is a unitless number.

    FLUIDS: Okay, can we say?: Stress (F/A) = η x (Δl / Δt)

    Where η is the coefficient of dynamic viscosity, Δl is the distance travelled by the oil at the rotating shaft side, Δt is unit time of 1 second.

    With fluids strain is actually the rate of strain. Of course, the layers of oil are circling around, there no movement on one side of the boundary, and where the shaft is, the oil there is moving at the speed of the shaft. So, there is a velocity gradient in the oil. The thin layers within the oil are sliding alonside each other, which must be if there is a gradient in the velocity of the oil across the height or length of the film of oil.

    But, in the above formula, there is no l. Only Δl.

    Thinking about shear in solids, I want to see things in these terms: Stress (F/A) = η x (Δl /t ) / l).

    The above formula is measuring the distance, at the shaft side, the oil traveles in 1 second (rate of shear?) and l is the thickness of the oil film.

    There are probably errors of some description of the above. Anyway, I'm not really sure what is rate of strain. Practically, what would you be measuring to get rate of strain? Obtaining strain is easy with solids. It's simple distances, with no reference to time. Thanks.
    Last edited: Feb 24, 2015
  2. jcsd
  3. Feb 24, 2015 #2
    For fluids, the rate of strain (aka, shear rate), is the velocity gradient within the gap. It is the velocity of the moving wall divided by the gap spacing. This has units of 1/s.

  4. Feb 25, 2015 #3
    I see that strain is deformation. Strain rate is deformation wrt time.

    Where there is a strain rate there is dynamic shearing within the material. I mean, atoms are sliding past their neigbours.

    With the oil film, the atoms at the stationary end (say layer 0), do not move wrt time. Moving outwards towards the shaft end, the atoms in a layer just above the stationary end (say layer 1) are moving at a certain rate wrt to the atoms at the stationary end. The atoms above layer 1, that is, in layer 2, are moving wrt to layer 1 atoms at a certain rate - I think the same rate, as between layer 0 and 1. There could be millions of tiny layers until you get to the layer next to the shaft. Near that point we might say, layer 1,000,000 is moving wrt to layer 999,999 at a certain rate. I think this rate would be the same as between layer 0 and 1.

    If this is all true, there would be a sense of a constant rate of deformation or strain. Deformation is no greater locally between layer 0 and 1, and layer 1,000,00 and 999,999.

    Of course, if you were to measure rate of deformation from the stationary end to a layer some way up, you would be measuring the summation of all these tiny local deformation rates.

    Of course there is also a sense of velocity gradient, if you take as your reference point the stationary end. Velocity of oil atoms increases as you move towards the shaft end.

    The atoms though are not moving from top to bottom of the oil film, they are moving wrt to each other, or a stationary point, parallel to the force or stress.

    I am trying to understand this formula: (Stress) τ = η × (Δy / Δt)

    (Δy / Δt) is the rate of strain. η is the coefficient of dynamic viscosity. How does this fit into what I've said? What does the Δy term corresponds with? ( I think it might be possible to write this as Δl - not entirely sure). Thanks.

    P.S. Δl would only make some sense, if there was an item l, such as the length (or height) of the oil layer, but I don't think that the length/height of the oil layer comes into play with rate of strain. Does it, or not? Of course, there is a velocity gradient across the length/height of the oil layer.
    Last edited: Feb 25, 2015
  5. Feb 25, 2015 #4
    As you're shearing the fluid, the particles attached to the shaft are getting farther from the particles attached to the cylinder that they were originally adjacent to. So, there is stretching occurring in that direction. This is the same kind of thing as occurs in shearing a solid between two plates. In solid mechanics, you also have shear strains. You first need to get a book on solid mechanics and look into how strains are quantified, both in tensile deformations (where the principal axes of strain are aligned with your coordinate system) and in shear (where the principal axes of strain are at an angle to your coordinate system). The same kind of situation occurs in fluid mechanics. In your situation, the principal axes of strain rate are at an angle to the shaft and cylinder.

    By the way, in fluids, the stretching rate is not Δl/Δt. It is Δl/(lΔt) = v/l.

    In summary, studying the kinematics of solid strain deformations, and then also studying the kinematics of fluid deformation rates will clear up all your questions.

  6. Feb 26, 2015 #5
    I'm having difficulty in comprehending the following:

    The shear rate for a fluid flowing between two parallel plates, one moving at a constant speed and the other stationary (Couette flow), is defined as:

    γ = ν / h

    ν is velocity
    h is height

    Rate then is in terms of distance from the bottom of the gap to whatever point you take. I would say, shear rate is (say) 10m/s per 0.05mm. Or, 100m/s per 0.5mm. Whatever, the rate is the same. The unit for this shear rate then would be m/s per m.

    But shear rate is measured in reciprical seconds, 1/s. This is what I don't comprehend. Why, if you have ν / h, do you measure rate of shear in reciprical seconds?

    That would mean the shear rate is simply a time measurement. Not seconds, but the reciprical.

    What am I not understanding. Thanks.
  7. Feb 26, 2015 #6
    You're not misunderstanding anything, but your math is a little rusty.

    m/s per m is the same thing as ##\frac{m}{sm}=\frac{1}{s}##. The m's cancel out. So you're left with reciprocal seconds.

  8. Feb 27, 2015 #7
    Okay thanks. Much appreciated. You have helped a lot.

    So, lets say we had this:

    γ = 20 m/s per metre. Strain rate is 20 meters per second, per metre.

    This is equivalent to saying, the strain rate is 20 reciprical seconds.

    According to my math.
  9. Feb 27, 2015 #8
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