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Stress, no literally.

  1. Oct 4, 2005 #1
    Ok, I asked this before, but I was not happy with the anwser and I will ask it again. Lets say we have a rigid body. We section the body, and now there is an exposed plane with internal forces present, that become exposed to the section. If we look at a point on the exposed part, we want to know the stress at that point. So WHHYYYYYY do they use an elemental volume to do this???? You can do it much simpler by using a plane. Just find the stress along three mutually orthogonal planes that pass through the point. Then any plane will be some linear combination of these three plains. Why the heck are they using a elemental VOLUME to do the work of a PLANE. If you have an elemental volume, then the point your conrcerned with lies somewhere inside the volume. So that means the value of shear stress you calculate will be some SMALL distance away from the point you actually WANT. (Unless the point lies on one of the faces of the volume). In that case you would get the same anwser as i am stating with the use of planes, but you would get the most error using the other faces to find the shear in an orthogonal direction. This is really starting to bug me. :mad:
     
  2. jcsd
  3. Oct 6, 2005 #2

    Astronuc

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    Staff: Mentor

    What is the objection to a simple cube as a unit volume or dV -> 0, as dx, dy, dz -> 0?

    In Cartesian coordinates (3-D, with mutual orthogonal axes), a cube is the simplest geometry. Stresses and pressure operate on a surface, and a cube has 6 surfaces, 1 facing each of 6 directions (1 + and 1-) in each dimension.

    One could use annular or spherical sections in cylindrical or spherical coordinates.

    Working a square in 2-D is the same as a cube in 3-D. In 2 D, if one assumes a shear stress, then there is an implicit assumption of unit depth and uniform stress in the third (depth) dimension.

    Forces can be thought of as operating on a point - normal and shear stresses operate on a surface.
     
  4. Oct 6, 2005 #3
    yes, allright fine,... a cube is the simplest geometry sure, im not arguing that. It IS a unit volume sure, thats fair too! :-) But im saying, if you want the stress at a POINT, why not just use a SIMPLE PLANE!? It seems more straight forward and logical than using a volume! :-) Do you see what im saying. If I have a plane with normal in the z, and dimensions dx and dy, then i can find the stress on the plane orientend in the z direction, no need for a cube. And this occurs EXACTLY at the point c, whereas a cube, the stress at that face of the cube acts somewhere NEAR the point c, because point c wont lie EXACLTY on that surface, it will be below it since its inside the cube! See the difference.
     
  5. Oct 7, 2005 #4

    PerennialII

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    I'm interested how you'd formulate your continuum theory using only a plane. For a plane you end up with the 'definition' of the stress tensor,

    [tex]
    \overline{t} = \overline{n} \cdot \sigma
    [/tex]

    where [itex]\overline{t}[/itex] is the surface traction vector, [itex]\overline{n}[/itex] the outward unit normal and [tex]\sigma[/itex] the stress tensor.

    To get further from there you need essentially momentum principles or a differential geometric treatment. Or you know something we don't :smile: ?

    (wonder if the latex thing is related to the upgrade?)
     
  6. Oct 7, 2005 #5
    Because you are using a differential cube. A differential cube is a cube whose volume approaches zero so it becomes a point while a plane---by definition---extends to infinity. You need to look at a volume because stresses in one direction will induce stresses in a the other two directions (3D Hooke's law and Poisson's ratio) so one cannot analyze the stresses in the xz-plane only. You have to look at your body---which itself is a volume---as increasingly smaller volumes. Forces act on real volumes not imaginary planes.
     
  7. Oct 7, 2005 #6
    very nice explination faust. very nice....... I realize the plane has that problem, extending to infiinty; however, thats why I placed the stipulation that it has sides dx and dy, and normal in the z direction, (so that it would not extend out to infinity).
     
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