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I'm struggling to get an intuitive understanding of what exactly Stress is, particularly the "direction" associated with it.

In the case of a 1 dimensional bar with just uniaxial loading, the way stress was explained to me was just $\pm\frac{F}{A}$ where $F$ is the force applied to either end, A is the cross sectional area, and the sign refers to tension or compression. This explanation is fine as a formula, but I don't see how it relates to "internal forces".

I've found other sources explaining it with an "imaginary cut" through the material, ignoring one side of the cut, and imposing equilibrium on the other piece. Why can either side be "ignored"? If stress is the internal force per unit of surface, why doesn't the neglected part contribute to the stress? (after all, the neglected part shares the surface).

In the more general case using the Stress Tensor, $$T=\begin{bmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz}\\ \tau_{yx} & \sigma_{yy} & \tau_{yz}\\ \tau_{zx} & \tau_{zy} & \sigma_{zz}\\ \end{bmatrix}$$

Do each of the components describe the stress on the surfaces of some infinitesimal volume? If so, which faces do they describe (there are 2 faces normal to each direction)- is it the sum of the stresses on each face?

Any insight on these questions would be greatly appreciated, thanks for reading

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2 Answers 2

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Do each of the components describe the stress on the surfaces of some infinitesimal volume?

Essentially, yes.

If so, which faces do they describe (there are 2faces normal to each direction)- is it the sum of the stresses on each face?

The assumption is the volume is in equilibrium, both translational and rotational. On that basis, the diagonal terms are the applied external normal stresses on the faces of the cube. There are six faces, but the normal stress on each opposite face is equal and opposite for translational equilibrium, so only three are specified in the tensor. If there was only a normal stress on one pair of opposite faces and no applied shear stress on the faced, you would have your uniaxial stress equation. You can see this in the figure in the Wikipedia link supplied by @nicoguaro, except that $\sigma$ is used for shear stress an $e$ is used for normal stress.

The off diagonal terms are the shear stresses on each face. Six are specified but three pairs are identical, e.g., $\tau_{xy}=\tau_{yx}$. This needs to be so in order that there is not rotation of the cube. For example, in terms of @nicoguaro indices, $\sigma_{21}=\sigma_{12}$

Hope this helps.

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We could start from the Cauchy postulate where we take an arbitrary plane defined by a normal vector $\hat{n}$ and with a finite area $\Delta S$, and compute the traction in that plane $\Delta R$ (here I am neglecting the couples $\Delta M$ since they should vanish in the classical case).

enter image description here

If we take the limit when the size of the are goes to zero we end up with the definition of the traction at a particular point (in index notation)

$$t_i^{(\hat{n})} = \sigma_{ji} n_j\, .$$

What this is telling us is that the effect that we are seeing (traction) is the projection of a tensor $\sigma_{ji}$. That means that the internal "forces" are characterized by this tensor and we probe them via a projection, computing tractions.

Then, we can interpret the components of the stress tensor using the plane over which we are projecting and the direction associated with the given traction vector. As shown in the following figure from Wikipedia's article con Cauchy stress.

enter image description here

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