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I only recently realized this "confusion" despite having cleared all my exams long ago. Consider these 2 diagrams:

  1. http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_II/01_DifferentialEquilibriumAndCompatibility/01_DifferentialEquations_Complete.pdf

and 2. http://users.ox.ac.uk/~kneabz/Stress5_ht08.pdf

Clearly they yield different equations. I have seen both in different sources, but cant reconcile them since in both pictures the element has sides that tend to 0. Anyone sees the problem?

rrkkass
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    You are not clearly explaining you issue here (or the meaning of the symbols - don't assume everyone uses the same conventions). However on the face of it I see different diagrams describing different forces and not surprisingly resulting in different equations. Suggest you use Mathjax to edit in your equations for these systems. What does "the element has sides that tend to 0" mean ? – StephenG - Help Ukraine May 16 '18 at 05:26
  • Hi Stephen, sorry for the missing information. Basically, each square is supposed to be an infinitesimal element on/within a piece of material, and as such its sides (denoted by delta x and delta y in the first picture) should approach zero so that these stresses become point stresses (or tends to such values). – rrkkass May 16 '18 at 09:55

1 Answers1

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In situation 2, for simplicity, we’re considering a region of uniform stress. It doesn’t need to be plane stress; it just needs to not change much locally. This model is useful for establishing the stress tensor and defining the types of stress. (For instance, we can review that normal stresses act perpendicular to faces, whereas shear stresses act parallel to faces; furthermore, $\tau_{xy}$ must equal $\tau_{yx}$ to satisfy static equilibrium.)

In the specific plane stress configuration shown in situation 2, for example, the stress tensor is

$$\boldsymbol{\sigma}=\begin{pmatrix}\sigma_{x} & \tau_{xy} & 0\\&\sigma_{y}&0\\&&0\\\end{pmatrix}=\begin{pmatrix}\sigma_{xx} & \sigma_{xy} & 0\\&\sigma_{yy}&0\\&&0\\\end{pmatrix}.$$

Situation 1 is more complex but essential if we’re going to move beyond just defining stress and start considering how to establish constitutive laws of how the stress changes throughout a finite 3D body in response to applied loads and inertia, for example.

Consider, for instance, the addition of a rightward volumetric body force $B$ and a density $\rho$. Then, I can sum the forces in the $x$-direction and divide by $\Delta x\,\Delta y\,\Delta z$ to obtain $$\frac{\partial\sigma_x}{\partial x}+\frac{\partial\tau_{xy}}{\partial y}+B=\rho\frac{d^2 \mathbf{x}}{dt^2}$$

where $\mathbf{x}$ is the $x$-coordinate of the element. Now I can start solving for how the body deforms and translates.

  • Thank you very much for your answer. As you have mentioned, for the second case, both shear stresses must be equal for moment equilibrium, and normal stresses are identical on opposite faces for translational equilibrium. This clearly is not necessary in the first (more complicated) case from which the theory of elasticity is derived, based on the equilibrium equation you have shown. Does this mean that the second model is simply a dumbed-down version? Or does this have to do with the abstract concept of the "limit", since the element is kind of a point but isn't at the same time. – rrkkass May 16 '18 at 14:55
  • Also, does the second model not allow for variations in the stress field? – rrkkass May 16 '18 at 14:57
  • Correct; the second model is a simplified version that doesn't accommodate variations in the stress field. It's not a matter of limits; you simply can't derive broader continuum constitutive equations using situation 2. – Chemomechanics May 16 '18 at 15:14