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The way textbooks talk about stress tensor confused me

  1. Jan 10, 2012 #1
    In fluid dynamics, always when some textbook talks about stress tensor, there is a figure like this:
    http://www.fea-optimization.com/ETBX/hooke_help_files/stress.gif [Broken]
    it shows how stress tensor is defined based on a small cubic volume.

    I kind of understand why the shear stress τxz should be equal to τzx: the overall torque of the system be zero, thus no rotation (?).

    But I don't understand why there have to be two σz, one acting on the upper x-y surface and one acting on the lower x-y surface and these two σz have opposite directions, thus they cancel out??? (the same question for σx and σy)

    If the system has these forces acting on it (all shear stress cancel out-->no torque, no rotation; all normal stress cancel out-->no translational acceleration), the velocity of the system will not change at all... but in a moving fluid, there has to be acceleration right? OR what kind of systems is the stress tensor defined for? OR mathematically when you take the limit of volume->0, since fluid is continous, the force on two sides should be balanced???

    Really confusing! Thanks a lot!!!!!:smile:
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 11, 2012 #2


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    This type of diagram is confusing, because it isn't really a "cube". It is a set of 6 planes all through the same point, showing the components of stress acting on each face. The planes are "pulled apart" to draw them all on one diagram.

    You will meet other diagrams which really are a infinitesimally small cube (of size dx by dy by dz) and in that case the stress tensor at x = 0 is not necessarily the same as the stress tensor at x = dx, so the two forces labelled "[itex]\sigma_x[/itex]" are not necessarily equal and opposite. (For example, the resultant of the two forces may be causing the fluid to accelerate).

    From an engineering point of view, you can think of the stress tensor as a 3x3 matrix of numbers. The matrix must be symmetrical because of equilibrium. That can be proved by vector calculus by considering the equilibrium of an arbitrary shaped finite sized body. (for example see http://en.wikipedia.org/wiki/Stress...m_equations_and_symmetry_of_the_stress_tensor).

    To get the stress components acting on a plane surface, you multiply the stress tensor by the vector normal to the plane. For the two faces on the left and right of your "cube", the normals are [1 0 0]^T and [-1 0 0]^T and when you do the multiplications the stress components are
    [itex]\begin{pmatrix} \sigma_x \\ \tau_{xy} \\ \tau_{xz}\end{pmatrix}[/itex]
    [itex]\begin{pmatrix} -\sigma_x \\ -\tau_{xy} \\ -\tau_{xz}\end{pmatrix}[/itex]
    This is exactly the same as cutting a solid body into two pieces, and the forces on the two cut surfaces are equal and opposite.

    So to summarize, forces in your diagram are a "equal and opposite" because the stress tensor is symmetric, and the stress tensor is symmetric because of equilibrium - not the other way round.
  4. Jan 11, 2012 #3
    AlephZero, thanks for your explaination, but I'm still a bit confused..
    1. Why does a point in fliud have to be *at equilibrium*? If every point in a fluid is at equilibrium, then no point will be accelerated, thus no change of motion for the fliud... what about a turbulent flow?
    2. Instead of two planes (e.g. the upper/lower x-y plane) through the same point, are they more like the two surfaces of the same plane...?
  5. Jan 11, 2012 #4
    The picture you've provided is for statics (equilibrium in force and torque) and is meant to depict an infinitesimal fluid element -- though if the stress field is really uniform, it could also apply to a finite cube of fluid. Of course, the fluid doesn't have to be in equilibrium. But you have to understand the stresses in equilibrium before we can even begin to discuss non-equilibrium situations. A more general picture has the normal stress at x written as σx, and the normal stress at x+dx written as σx+dσx. There is a lot to understand even with equilibrium fluid mechanics, e.g. what happens to the stress field when you impose a gravitational field, what happens to the stress field in a bucket rotating with constant angular velocity, etc.

    Once you understand stress in the context of hydrostatics, we can discuss non-equilibrium stresses, e.g. gradients of pressure, gradients of shear, etc. Across an infinitesimal fluid element, the stress will change infinitesimally, and we can write down differential equations that govern the stress field.

  6. Jan 11, 2012 #5
    THANKS! But then
    1. what does stress tensor look like for a non-equilibrium fluid? (The texts usually don't emphasis on on what kind of fluid is the stress tensor defined, why is that so? also, this definition of tensor is used everywhere, on obviously non-equailibirum fluids...)
    2. Is Navier-Stokes equation good for fluid at non-equilibrium? (assuming incompressible flow)
    Last edited: Jan 11, 2012
  7. Jan 11, 2012 #6
    Well, I don't have a picture handy, but imagine that the differential cube has faces at x=0 and x=dx, y=0 and y=dy, and z=0 and z=dz. On the face at x=0 are the normal component σx and the shear components τxy and τxz. On the opposite face at x=dx are the components σx+dσx, τxy+dτxy, and τxz+dτxz.

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