- the body exhibits a non-spherical shape, like a blob
- its rock has some amount of compressive strength
Irregular Asteroid Stresses
Call the largest dimension the polar axis. This is chosen because it is the axis of symmetry. In that case of planets, the axis of symmetry is typically the smallest one because they are roughly oblate spheroids, whereas 10-km scale irregular asteroids are more like a prolate spheroid. In other words, this is the geometric model I'm using:
This doesn't look very far off for many of the asteroids we've photographed. Eros, in particular, among others. With this in mind, we can consider the stresses that he object's own self-gravity will cause on its interior. For a mental model, this is somewhat similar to a building on Earth. In order to "stick up", it has to have some strength. The only difference between an asteroid sticking out as a prolate spheroid and a building standing up on the surface of Earth, is that the asteroid deviates from a spherical shape, while the building deviates from the flat surface of the ground.
This protrusion will only lead to compressive stress here. After all, it doesn't have to stand up to the wind or any other dynamic forces. If you want a rough formula for exactly how much compressive stress, then consider the gravitational head. For a bad approximation, imagine that the gravitational field only varies with radius. Then it's obvious that the equatorial radius is smaller than the polar radius. That leads to an inconsistency if you imagine the object material is fluid-like. You obtain a different pressure if you measure the elevation drop from the equator surface to the center versus if you measured from the polar surface to the center. It is precisely the difference between these two pressure that is the non-isotropic force, or the compressive force. It has the same units as pressure, because these are both elements within the stess tensor - the bread and butter of civil engineering. After all, the entire proposal basically comes down to civil engineering.
For some math, you can imagine that this stress will be approximately equal to the gravity on the surface of the asteroid (which is an average figure itself) times the elevation difference at the equator and the pole. Without going into details here, you can get a factor of 2/5 to add onto this. This isn't perfect, but it's pretty good.
$$\sigma_z \approx \frac{2}{5} \rho g \left( R_x - R_z \right)$$
To illustrate the occurrence of this compressive force, I've used some arrows here. Imagine that a pressure without any material stress would entail 4 arrows from all directions in this 2D approximation. In reality pressure acts all around you. So instead of that, we have some preferential direction where the pressure squeezes from only two sides. I did my best to illustrate that for the prolate spheroid shape I'm talking about here. This is briefly representative for the natural state of the asteroids I'm talking about.
You can easily extend the idea to a hole in the center. About the same amount of net force will be present over a cross section at the equator. If you drill a hole in the center, that means that there's less area over which to distribute this force. Logically, that means that the stress would be intensified by the presence of this hole.
There are some finer points to this argument - mainly that in the above model the bubble would have to have some internal pressure. This is what we're talking about for the gravity balloon. Specifically, the pressure would have to be exactly enough such that the top and the bottom of the hole in the above image wouldn't experience any compressive stress. You could change that with a different pressure inside the bubble. If you increased the pressure in the bubble enough you would induce tensile stress - stress that tears the material apart, not pushes it together. I have assumed that the pressure set-point would be carefully managed to keep all stresses compressive.
Why would you want to do this? Because of the failure mechanisms associated with breaking. If you compressive force doesn't hold up, then you could see some material rearrangement - just like if you had built your sandcastle too high. That might still not result in loss of atmosphere. Particularly if you had added some membrane to keep the atmosphere from diffusing into the rock to begin with. There still remains a danger that whatever material rearrangement happens would destroy some part of that membrane, but it's still a smaller danger compared to failure of tensile stress holding the atmosphere in. If you relied on the tensile stress of the asteroid, you would risk a catastrophic loss of atmosphere. This is the same sort of event we concern ourselves with the international space station, or any similar design. Breaks are generally fatal.
Impact of a Deformed Central Bubble
Similar thought experiments can be used for imagining that a small bubble is deformed in the shape of a prolate spheroid. Start with the assumption that tensile forces are unacceptable. Then imagine that we deviate from the spherical shape that I've always talked about for the inner bubble of air. If you do this, that is effectively adding material around the equator region and subtracting it from the polar regions.
In doing this, we introduce a quadrupole moment. This behaves as you would expect from a quadrupole field:
But in the case of a gravity balloon, we can only allow compressive stress, by adjusting the pressure of the air bubble lower. Start out drawing the gravitational field lines from the quadrupole gravitational moment. Then, due to the compressive stress argument, draw a compressive stress in all places where two arrows point toward each other. This is what I've done in the image below. You still have to use your imagination to think of this being an prolate spheroid.
There is an obvious utility to this - because the compressive stresses are in the opposite directions to the stress from the asteroid shape itself. That means that an odd shaped bubble may be used to less the compressive forces within the interior of the asteroid. That is exactly what you would do for management of interior forces, building a primitive gravity balloon colony.
In short, the combination of the two above stress diagrams gives a result that is overwhelmingly balanced back to isotropic forces. That is, just pressure, no material stresses. Like illustrated below.
There's no reason to believe this could be done perfectly, but I haven't done the calculations. I imagine it would be quite an involved project to do so. Even as we imagine there will be some residual forces, it needs to be considered what the criteria is. There seems to be every reason to believe that you could start with an asteroid like Eros and establish a colony with breathable air, several kilometers in diameter, all the while keeping the asteroid internal forces less than its natural state.
You would not want to stress the asteroid more than its natural state, because that will give you a virtual guarantee that loss of atmosphere will not happen. That's the type of guarantee needed to have people seriously consider moving there, and that's why a gravity balloon is a competitive concept for space colonies.