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I've just started learning about axial stresses on solids, primarily soil samples and this is the first time I've encountered proper forces on solids, and we consider this over an area, previously I would be dealing with idealised point-particle based systems where we idealise forces as acting on points, as well as fluid systems, the idea of a force 'acting' at a point and a force acting over an area, of course at a point, the axial stress becomes arbitrarily large.

Do we simply ignore these things in the case of rigid bodies? If we apply a force at a point the pressure is ignored? On a body that is able to be deformed, do forces act on 'points' here as well? Why do some forces act on 'areas' and some on points?

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  • $\begingroup$ A force like that is the average force, and it acts on the center of mass or the point of action, which are also infinitesimally small $\endgroup$
    – RC_23
    Jul 1, 2023 at 19:44

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There are situations, where some bodies can be considered rigid in an analysis, even though they represent deformable objects. Rigid body is a simplification like the force in some cases and also pressure (its really just forces if you go to atomic level and even on macro scale, the pressure can be for example considered constant when its actually not).

If we use rigid bodies and forces, the stresses that would arise in actual object are usually not considered at the moment, because all the forces may not be known. Often we can use simplified analysis with rigid bodies to determine unknown load values and then determine the stresses and deformations to check if assumption of rigid body was ok in the first place. For example if allowable stresses are not exceeded in a steel structure, you could generally consider it rigid (there are exceptions of course like wire ropes etc.).

This reminds me of a George Box quote: "All models are wrong, but some are useful."

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No matter how small it is, the contact Force is always acting over/on a finite area that produces stresses. However, depending on the purpose of computation, and the scale of the matters, we can let it remain a point force, or expressed as the stress resulting from F/A.

In the figure below, each of the 1' x 1' columns supports 10k concentrated force, and in turn, the columns are supported by a 3' x 6' concrete footing. For design, in order to be able to compare to the code-prescribed design values, the engineers are required to get the stress/pressure at each of the contact surfaces, that is P = F/A, or under the column, Pc = 10k/144 in^2, and Pf = 20k/18 sf at the concrete footing and soil interface.

enter image description here

However, now let's say the same columns are supported by a long narrow beam of a warehouse, the structural engineer would carry out the design by treating them as needle-point forces, rather than deriving the pressure/stress at contact faces, because the contact force is of secondary interest for the task.

Wish the examples are useful in advancing your understanding.

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