1
$\begingroup$

I'm trying to understand the Cauchy-Stress tensor, in which the stress acting on a body at a point is analyzed by considering the cross-sectional area through which a force passes. And my question is - why? Or, for contrast, why don't we consider the force as acting on a differential volume element? Is this a conceptual trick, or does it reflect the way forces actually propagate through the solid? In what way? It just seems rather arbitrary that we look at forces through an element of area when, really, there are an infinitude of area elements passing through a point in different directions.

Furthermore, in the case of pressure, for example, we consider the force acting on an element of area at the surface, and integrating over a section of the surface, for example, we could find the total force acting on that region. But this seems intuitive precisely because it is occurring at the boundary of the object, not at a point within it, and we have a certain area element singled out as being the one we want to integrate over, that is, the element of area along the boundary. Within the volume, it doesn't seem to make sense that we are looking at elements of area along the point.

So... why do we look at it this way?

$\endgroup$
0
$\begingroup$

Ok, so figuring that stress is defined to be the force per unit area around a point, I'd like to know why it's defined that way, rather than say force per unit volume enclosing a point.

Here is the intuitive explanation.

Suppose you have a string S from which a weight W is hanging. We will suppose that the amount of weight is so much that the string is near the regime where it would snap, but is not quite there yet. We're going to think about what happens when we add another length of string S.

If you add it in parallel, so there are two strings holding W, then it seems obvious that you could then increase the mass to about 2W before the combined two-string system would be near-breaking.

By contrast, if you add it in series (W hangs from S hanging from S), then we know that the tension force is the same in both strings, and it seems likely that they'd simply both be near-breaking. The top string might even break, if the weight of the string underneath it is enough to put it past breaking.

This tells us that the material properties which concern us (like when a string breaks) respond to stresses, defined as forces divided by an area perpendicular to that force. The direction that lies alongside the force doesn't matter, qualitatively because it propagates the force rather than responding to it.

You can also examine this two-string thought-experiment with much smaller forces not near breaking, where Hooke's law should hold for lengthening, to find that if the original deforms by $\delta L = \ell$, the two-strings-in-parallel should deform by $\ell/2$ while the two-strings-in-series should deform by $2\ell,$ so that there seems to be a constant "stress / strain" relationship where the "strain" is defined as $\delta L / L.$ In other words you can look at a system with the normal spring-constant $F = k ~ \delta L$ relationship, but it is more helpful from a materials perspective to divide by the length of the "spring" $L$ and also its cross-sectional area $A$ to find $$\sigma = \frac FA = \frac {kL}{A} ~ \frac{\delta L}{L} = \lambda ~ \epsilon.$$ The elastic modulus $\lambda$ then also has units of pressure (force per unit area) and is more fundamental to the material (a material-property) that you're studying than the spring constant (a property of both the material and the setup) is.

Since the elastic modulus is a material-property, this hints that "stress" is the right definition of "force" and "strain" is the right definition of "displacement" at the more-microscopic level when we're peeking inside the substance.

In turn, it becomes very common in materials science to show the "stress/strain" curve of various materials. This starts out of course as a straight line through the origin with slope $\lambda$, but then as a substance deforms it will describe some sort of curve as added stress leads to further strain. So for example the plastic bags from the supermarket will curve upwards; they get stiffer as they stretch more.

Once you know that the stress is the right way to "microscopically" define force, elastic systems start to show off a similar problem: the simplest stretching of a beam consists in that beam not only lengthening but also narrowing. Microscopically, a little box inside the substance is not only feeling a force $+\sigma~dA$ on one side and a force $-\sigma~dA$ on the other side (so it is in force balance and provides tension), but it must also be feeling some forces on its other sides which "pinch" it smaller. So: the force is direction-dependent, and we therefore have not a stress vector (which we already didn't quite have -- the stress is in opposite directions on the top and bottom of the box), but a stress tensor: give me a direction and I'll give you the stress vector on a plane normal to that direction. (This also solves nicely the "stress on the top of the box is negative the stress on the bottom of the box" problem.)

Usually this simplifies a lot because there are eigenvectors of the stress tensor: directions where the stress points exactly normal to the plane it deforms. Those are called the "principal stresses". However there is no reason why they'd have to be orthogonal, especially, for example, in a crystal lattice which is not cubic.

$\endgroup$
1
$\begingroup$

To find out what is happening within any body, it's common to (mathematically) cut it in half (or at any other point), regard each half as a free body, and examine the conditions that must exist across the area exposed by the cut. This is how we get, for example, the stress within a uniaxial bar, or the shear force in a beam.

So, in the same spirit, consider a 3-D body that's in equilibrium under the action of boundary forces. To get the stress field at an interior point, cut the body in half and regard each half as a free body. Each half will no longer be in equilibrium on its own: to maintain equilibrium there must be a force, the so-called traction vector acting across the surface area created by the cut. Regard this as either the force needed to maintain equilibrium of either half on its own, or the net effect of one half on the other. The magnitude of the traction vector is usually rescaled by dividing it by the area of the cut, but it's still a force vector, not stress. The stress state thus has a component vector associated with the direction of the cutting plane, expressible by the normal to the plane.

To fully describe the stress state in the body, we need to cut it in each independent direction in space and find the associated traction vector. A quantity that has a component vector in each direction in space is a (2nd order) tensor. In 3-D space we need to cut the body in three - usually orthogonal - directions. Stress is then a 2nd order tensor that has a component vector in each direction.

Aside: to maintain equilibrium across the cut, there must also be a moment present. In the absence of micropolar moments, this moment tends to zero as we zoom in on the body, which is why the stress tensor is usually (but not always) symmetric.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.