I have been bugged by this question ever since I studied cauchy's stress theory 10 years ago.
In cauchy's analysis of solids, there two popular geometries.
First one is a unit cube used for deriving equations of equilibrium.
And the second one is a tetraheadon used for deriving Cauchy's stress theorem.
In both the cases limit of force equilibrium as solid geometry shrinks to a point is derived.
My question has been, how come limit of force equilibrium at a point inside a solid depenent on the path taken by limiting process.
If this process were to be rigorous enough, any arbitrary geometry shrunk to a point should yeild the same limit of force equilibrium.
-Somesh
Re: Comments on Cauchy's stress theorem
Because "traction" is not just a function of the point you're looking at; it explicitly depends on the unit normal of the infinitesimal surface it's acting on too. In other words, at a given point you may have infinitely many different traction vectors. What depends on only the point is Cauchy stress (or other equivalent representations of stress tensor).
Regards,
Arash
Yes , THats correct but
Yes , THats correct but ISn't it not true that any geometry although tetrahedron is the best..will give the same result??