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I'm not very familiar with continuum mechanics and have a hard time combining my knowledge of forces from simple mechanics with what I read about continuum mechanics.

Let's suppose we have a metal rod of a certain length and a quadratic cross-sectional area $A$ which is put under stress $\sigma$.

1) What exactly is the physical significance of the force $F = \sigma A$? My current understanding of forces is that they need a point of application - where would this point be? Is it a single point? All points of the cross-sectional area at once? The latter seems to conflict with the notion that force gets smaller if I consider only a part of the area:

If I partition the cross-sectional area $A$ into a number of smaller areas $A_i$, I can calculate forces $F_i = \sigma A_i$. Since $A = \sum_i A_i$, we also have $F = \sum_i F_i$. This makes it seem that $F$ is some kind of cumulative quantitiy and raises the question: what is the physical significance of the $F_i$?

Trying to generalize this further, we can also admit different values of stress $\sigma_i$ for the different areas $A_i$ and even make the areas infinitesimal so that we get a stress distribution $\sigma(x,y)$ (in order to simplify things let's suppose that we chose the distribution such that there's no net torque on the rod). What is the physical significance of the "force" $F = \int_A \sigma(x,y) dA$?

2) What's going on mathematically? Does the notion of force as a vector with a point of application need to be replaced by some kind of vector field in continuum mechanics (considering only stress in a single direction and ignoring additional complications related to the tensor nature of stress)? If yes, how can these "area forces" (I've also read the term "surface force") be combined with forces which have a point of application (for example with the weight of a point mass or a rigid body where the force can be described as acting on its center of mass)? In order to be combined, they need to be described by the same mathematical structure.

3) Does the notion of force somehow lose its meaning along the path which I sketched in 1) above? Is the integral quantity $F = \int_A \sigma(x,y) dA$ a force in one of the senses covered above or is it a quantity with units of force but without physical significance as a force?

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  • $\begingroup$ Are you familiar with the mathematical concept of the Dirac delta function? $\endgroup$ – Chet Miller May 31 '18 at 2:57
  • $\begingroup$ Yes. This hints at an answer to part of my question 2) in the direction that the forces of elementary mechanics which refer to a point may be considered to be limiting cases of forces which refer to a surface or a volume. $\endgroup$ – Marc May 31 '18 at 13:01
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    $\begingroup$ Yes. You are talking about the difference between a distributed force on a surface and a concentrated point force. The concentrated point force is the limit of the distributed force as the surface area over which the force is applied becomes smaller and smaller (while the resultant force remains constant). This is very much analogous to the area under the curve f(x) vs x as f(x) approaches a Dirac delta function. $\endgroup$ – Chet Miller May 31 '18 at 13:14
  • $\begingroup$ Ok, so I see now how I can get a point of application if I take the force as constant and shrink the area that it applies to. But I'm still pretty in the dark about my other questions. If I keep tension constant and look at smaller and smaller areas, the force gets smaller and smaller as well. This doesn't square with my intuition. Also I still can't really make sense of the "force integral" which I mentioned. $\endgroup$ – Marc Jun 1 '18 at 15:14
  • $\begingroup$ Tension and force are the same thing. If you keep stress constant over a larger area, this add up to larger tensile force. $\endgroup$ – Chet Miller Jun 1 '18 at 15:22
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The conceptual difficulties largely become resolved by developing the notion of a traction.

Traction ($\vec{T}$) is a vector field that represents a force per unit area acting on a differential oriented surface at some point in a body; I usually think of the traction as the fundamental concept that then becomes a force once integrated, but you can think of it as the limit of the force acting on an arbitrary surface per the surface's area as you shrink the surface to a point.

It is then natural to note that the traction at a point changes when you change the orientation of the tiny surface at that point, but that it has to change in a specific way (tractions balance out at equilibrium, etc.)

This motivates the development of a concept called a stress tensor $\bar{\bar{{\sigma}}}$, which is a tensor field in space, that can give you the traction at a point by selecting the orientation $\hat{n}$ of a differential surface at that given point. The equation through which this is done is:

$$\vec{T} = \hat{n} \cdot\ \bar{\bar{{\sigma}}}$$

It is for this reason that stress is the mathematical object used to describe mechanical states in a continuum system and not the traction/force. A thorough derivation of this can be found in essentially every continuum mechanics book; see A.J.M. Spencer's book or the Mase-Smelser-Mase book for it.

In order to include point forces into this model, you either have to "integrate the model" and use forces at the cost of spatial resolution or resolve the point force into a distributed force over a very small area, which is the true physical scenario. Hope this helps!

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