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Stresses on opposite faces

  1. Sep 12, 2012 #1
    I have a question about the typical stress tensor diagrams (shown here).

    What assumption is made such that [itex] \sigma_{x,x} = \sigma_{x,-x} [/itex]? In other words, why is the normal stress on one face always equal and opposite to the normal stress on the opposite face?

    It seems to me that [itex] \sigma_{x,x} [/itex] and [itex] \sigma_{x,-x} [/itex] do not need to be equal in order for the block of material to be in equilibrium, if the shear forces [itex] \tau_{x,j} [/itex] are allowed to balance it properly.
     
  2. jcsd
  3. Sep 12, 2012 #2
    You have two choices.

    The simple one, which is to say that we are referring to stress at a point and that point is in the geometric centre of the cube.

    So the stress at the centre is actually the average of that of two opposing faces. These become identical in the limit as we shrink the cube to a point.

    Or you can say let the stress on one face be σx and the stress on the opposing face be (σx + δσx) and do about 5 pages of maths to derive the same result.

    I only know of one treatment that takes the second approach.
     
  4. Sep 12, 2012 #3

    AlephZero

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    I think you are confused about the notation. ##\sigma_{x,-x}## doen't mean anything.

    Your diagram is not really showing the faces of a cube, but the stress components (one direct and two shear) on three different planes through the same point in space.

    If you want to find the stress components acting on a plane cutting through a point, you multply the stress tensor by the uinit normal vector of the plane. If the plane has normal (1, 0, 0), one of the stress components is ##\sigma_{xx}##. If you reverse the normal to be (-1, 0, 0) the stress component is ##-\sigma_{xx}## not ##\sigma_{x,-x}##.

    That is just Newton's law that action and reaction are equal and opposite.

    Yes, but remember that the stress tensor is not the same at every point in space, so ##\sigma_{xx}## on one face of the block is not necessarily equal to ##\sigma_{xx}## on the opposite face. To repeat, the diagram is about stresses on different planes all through the same point in space.
     
  5. Sep 12, 2012 #4
    If this is true, then how is the shear force [itex]\tau_{xz}[/itex] any different from [itex]\sigma_{xx}[/itex] if they act at the same point in space with the same direction?
     
  6. Sep 13, 2012 #5
    They are describing different things. σxx is describing the stress normal to that plane, in this case the stress in the x-direction and normal to the zy plane. τxz is describing the resultant stress if a shear force is applied to the xy plane in the x-direction.

    Think of it like this. If I attached a rope to the point that σxx is drawn on the cube and pulled it in the direction normal to the zy plane, then I would get a stress equal to σxx in the cube (tension in this case). If I got another cube and placed it beside this block in the zy plane and rubbed them together I would get a shear stress equal to τxz acting in this direction. The equations will tell you in what direction etc. The deformation shapes will look like a pulled apart rod for the former and a distorted rhombus for the latter. As an example, steel is weakest when in shear.

    Distortion shapes are assuming that the cube is fixed in the zy plane for the former and the xy plane for the latter.
     
    Last edited: Sep 13, 2012
  7. Sep 13, 2012 #6
    AlephZero alluded to how to get an understanding about all this (by dotting the stress tensor with a unit normal, the so-called Cauchy stress relationship) but, to keep things simple, he left out some of the details. This is a problem that everyone who has ever studied stress analysis in solids and liquids has grappled with, and has been confused by. For some reason, academia has felt it is too complicated to introduce the straightforward mathmatics involved in working with the stress tensor. This has made things rough on the students. The stress tensor is what we call a second order tensor. If you take the dot product of a vector with another vector, you get a scalar, but when you take the dot product of a second order tensor with a vector, you get another vector. Dotting the stress tensor with a unit normal vector to a plane within a solid or liquid maps the unit normal vector into the "traction vector" or "stress vector" acting on the portion of the material on one side of the plane (the portion from which the normal is directed) by the portion of the material on the other side of the plane (the portion toward which the normal is directed). This is the Cauchy stress relationship. The stress tensor can be expressed in terms of components and unit vectors, just as in the case of vector. But, in the case of a second order tensor, the unit vectors are written in pairs called dyads:

    σ = σxxixix + σxyixiy + σxyiyix + σyyiyiy + σxzixiz + σxzizix + σyziyiz + σyziziy + σzziziz

    As this equation stands, the summation at the right hand side does not yet have a function. But the stress tensor expression fulfills its primary function in life when it is dotted with a unit normal vector to a plane. For example, if we dot the stress tensor with a unit vector in the x-direction, we obtain:

    σ [itex]\bullet[/itex] ix = σxxixix [itex]\bullet[/itex] ix + σxyixiy [itex]\bullet[/itex] ix + σxyiyix [itex]\bullet[/itex] ix + etc. = σxxix + σxyiy + σxziz

    This is the traction vector acting on the y-z plane. If we dotted the stress tensor with minus the unit vector in the x - direction, the stress vector would be minus this. This is how the plus signs and the minus signs come in. The only thing you need to remember is that, when you dot a dyad on the right with a vector, you just dot the right hand member with the vector, and leave the left hand member alone.
     
  8. Sep 13, 2012 #7
    Well I'm not surprised students are confused.

    They display what is clearly a picture of a cube and ask about opposite faces.

    They are told that their picture is not really a cube because they dared to mention the word tensor.

    Further the answer to the question originally asked is not to do with the normal tractions, but with the need to balance the moments generated by the other tractions about each axis.
     
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