Calculating Stress with Elemental Volumes: Why?

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SUMMARY

The discussion centers on the use of elemental volumes versus planes for calculating stress at a point within a rigid body. Participants argue that while a cube (elemental volume) is the simplest geometry for stress analysis, using a plane could yield more accurate results at a specific point. The conversation highlights the importance of considering the effects of stresses in multiple directions, as dictated by 3D Hooke's law and Poisson's ratio. Ultimately, the consensus suggests that while planes can provide localized stress information, a volume approach is necessary for a comprehensive understanding of stress interactions.

PREREQUISITES
  • Understanding of stress analysis in solid mechanics
  • Familiarity with 3D Hooke's law and Poisson's ratio
  • Knowledge of differential geometry concepts
  • Basic principles of continuum mechanics
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  • Study the application of 3D Hooke's law in stress analysis
  • Explore the concept of differential volumes in continuum mechanics
  • Learn about the formulation of stress tensors and their applications
  • Investigate the limitations of using planes in stress calculations
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Mechanical engineers, structural analysts, and students of solid mechanics seeking to deepen their understanding of stress calculations and the implications of using different geometrical approaches in analysis.

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Ok, I asked this before, but I was not happy with the anwser and I will ask it again. Let's 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:
 
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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.
 
yes, all right fine,... a cube is the simplest geometry sure, I am not arguing that. It IS a unit volume sure, that's fair too! :-) But I am 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 I am 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 won't lie EXACLTY on that surface, it will be below it since its inside the cube! See the difference.
 
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,

<br /> \overline{t} = \overline{n} \cdot \sigma<br />

where \overline{t} is the surface traction vector, \overline{n} the outward unit normal and \sigma[/itex] the stress tensor. <br /> <br /> To get further from there you need essentially momentum principles or a differential geometric treatment. Or you know something we don&#039;t <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /> ?<br /> <br /> (wonder if the latex thing is related to the upgrade?)
 
cyrusabdollahi said:
yes, all right fine,... a cube is the simplest geometry sure, I am not arguing that. It IS a unit volume sure, that's fair too! :-) But I am 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 I am 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 won't lie EXACLTY on that surface, it will be below it since its inside the cube! See the difference.

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.
 
very nice explanation faust. very nice... I realize the plane has that problem, extending to infiinty; however, that's 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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