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Consider a body that is acted on by a variable external force density $\vec{f}(\vec{r})$.

I want to know what the pressure and shear stress would be within the body as a result of these external forces.

However, the problem as stated seems to be under determined to actually solve for a stress or pressure tensor according to the equation:

$f_i=-\nabla^j\sigma_{ij}$

Furthermore, intuitively I would expect the internal pressures to somehow depend on the variation and therefore derivative of the external force, since if it was uniform the entire body would accelerate in unison (assuming uniform density) without any part having to pull another part.

Are there additional material constraints we must assume (isotropy?) to find the stress tensor, and is my intuition correct and somehow accounted for?

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  • $\begingroup$ You are saying that f is a body force density? $\endgroup$
    – Chet Miller
    Commented Apr 30 at 16:04
  • $\begingroup$ @ChetMiller Yes, force per unit volume, but from an external source. I'm thinking of a charge distribution in an external electric field, or a mass under gravity. Though I now realize the equation I wrote is incorrect because it also accounts for the internal forces within the body. But the question stands for how to solve this in general. $\endgroup$
    – Connor Dolan
    Commented Apr 30 at 16:16

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Assuming that the internal imposed force density components are known as a function of spatial position, the usual way of solving this is to determine the displacements as a function of spatial position. To implement this, the components of the strain tensor are expressed geometrically in terms of the displacements, the components of the stress tensor are expressed in terms of the components of the strain tensor (e.g., Hooke's law in 3D), and the components of the stress tensor (in terms of the displacements) are substituted into your stress-equilibrium equation. To solve this, you also need to specify the imposed diestribution of the displacements and/or stress components on the surface of the body.

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  • $\begingroup$ Is assuming a rigid body consistent with this question? Can you not just say that the object retains it's shape, and each infinitesimal volume provides the necessary compensating force on it's neighbors to retain it's relative position and orientation? $\endgroup$
    – Connor Dolan
    Commented Apr 30 at 17:28
  • $\begingroup$ No, not unless the loading, shape, and boundary conditions are statically determinate (i.e., particularly simple). $\endgroup$
    – Chet Miller
    Commented Apr 30 at 18:48
  • $\begingroup$ Can you elaborate or point me to material for further reading? Perhaps I should resubmit a more concrete question. Let's say I had a small moon orbiting a planet where the internal gravitational forces of the moon on other parts of itself are negligible compared to gravity from the planet. At what parameters for the orbit would the tidal forces break that moon (given a certain specific strength density and size)? That seems a straightforward physical system, what assumptions am I making? $\endgroup$
    – Connor Dolan
    Commented May 1 at 0:48
  • $\begingroup$ Is the moon rotating about its axis? I assume you can quantify the force density variation within the moon, right? And that force variation include inertial forces, right? Then the boundary condition is zero traction vector on the surface of the moon. $\endgroup$
    – Chet Miller
    Commented May 1 at 10:23
  • $\begingroup$ Get yourself a book on Theory of Elasticity. $\endgroup$
    – Chet Miller
    Commented May 1 at 10:26
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In the case of a body with non uniform density, we can say that there is a variable external force density (gravity).

The expression $f_i=-\nabla^j\sigma_{ij}$ is not determined if we don't know how exactly it is supported on the ground (or by a liquid). It is intuitive because in Newtonian model, gravity is a force, and only by the expression, it can even be in free fall, in which case the stresses would be zero (for $g$ uniform).

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