Consider a solid torus, which is a circular disk with diameter d with the disk's centre swept around a circle of diameter d, such that there is an open centre also of diameter d. This torus is standing on its side so as to achieve its maximum height. How it is supported so that it does not fall to either side is not an issue for this question.

For a given, arbitrary material, how can I calculate the maximum size that this solid torus could be before collapsing under its own weight in 1g?

For purposes of this question, consider the ground to be of infinite strength.


3 Answers 3


Gravitation will create a stress distribution inside any body. For homogeneous and isotropic materials there is two locations to care about: maximum tensile stress and maximum compressive stress. This is because the behavior of most materials differ when subjected to tensile versus compressive loads. (Ductile materials can be torn apart under tensile load but not under pure compression load; brittle materials usually have a higher compressive strength than tensile strength.)

According to continuum mechanics, the stresses depend on the geometry, the density and the gravitational acceleration only. For a body at rest, conservation of momentum evaluates to $$ \rho f_i + \frac{\partial \sigma_{ij}}{\partial x_j} = 0 $$ where $\rho$ is the material's density, the $f_i$ are the internal forces ($f_i=-g\delta_{i3}$ when the gravitation is $g$ and is acting against negative 3 direction) and the $\sigma_{ij}$ are the stresses. (I'm using Einstein summation convention, so take the sum over $j$ in the $\sigma$-term in the above equation.)

I've done a simple FEM calculations of your problem. Just to make sure, here is the geometry I calculated (dimensions in m):

enter image description here

Note the torus is not perfect, there is a small flat area at the bottom I had to introduce to retain the torus in the simulation (produced by cutting 1 m off the perfect torus). This introduces a small perturbation at the bottom, but you need a finite area to support the torus, otherwise stresses will become infinite at the bottom (see also below).

The maximum compressive stress is at the bottom (where the torus is touching ground) as already stated by the other answers. This is given by the minimum principle stress as shown in the next image:

enter image description here

The picture shows the surface distribution of the minimum principle stress. Note that I have cut the scale at -50 MPa. The maximum compressive stress is at the bottom, and the best way to calculate that is (as given by the other answers) to divide mass by area. (This is the reason you need a finite area to support the torus.)

If the support area is large, the maximum compressive stress may also be achieved at the inner diameter. This maximum compressive stress for the geometry shown is at 32 MPa.

The maximum tensile stress is at the lowest point of the inner cirle. This is given by the maximum principle stress as shown in the following picture:

enter image description here

For this geometry, the maximum tensile stress calculates to approx. 28 MPa. The density I used for the calculations is 2.41 g/cm³ (concrete). The gravitational acceleration is 9.81 m/s² in the simulation. The total mass of the object is 11.9e+9 kg (11.9 million metric tons), its volume 4.93e+6 m³.

For many materials, the compressive stresses at the bottom will most likely be the limiting design factor. However for extremely brittle materials, the tensile stresses may be the limit. E.g. when using concrete you will need to add steel to areas with tensile stresses.

Now this still does not really answer your questions, since you have asked for an arbitrary material. As the others have pointed out, the stresses scale with geometry and density, i.e. doubling the size or doubling the density doubles the maximum stress. Now you have anything you need to calculate the stresses. To evaluate the maximum size compare stresses to strengths.

PS: Before actually building, think about safety factors.

PPS: This stress distribution is also quite intuitive if you think a little about it.

EDIT: I updated the density just now – I had a slightly wrong number in mind.

Also, for ductile materials this is still not the whole answer, since you can allow the material to yield in some areas. The material will then plastically deform in those areas. This is fine, as long as the non-yielding structure can still carry all loads. This approach can be formalized. The associated term in German is 'plastische Stützzahl' which translates to 'plastic support number', but I don't know the correct English translation.

In summary, the yield limit of the material need not be the ultimate limit, depending on your material. But it certainly is a good first estimate.


I think there is no such limit. The only possible factor which comes to mind is the stress at the foot of the torus.

If you momentarily consider the structure to be infinitely rigid, then the region of contact of the torus with the ground will be equal to an infinitesimal point, leading to infinite stresses on the foot of the torus (and the ground beneath it). If you consider deformations, then the torus will no longer be a perfect torus since it will deform at the foot to cover an area such that the stress is lower than the material's yield strength.

Assuming a perfectly elasto-plastic material, this basically allows the torus to deform as much as necessary, so it doesn't offer a limit either. Assuming any other stress-strain relationships, it then becomes pretty much impossible to answer generically.

Internal forces within the torus don't worry me too much. Looking at the top span of the torus, there will be bending there, sure, but the maximum bending moment will be a function of the square of the span, which you have defined as equal to twice the diameter (considering the span equal to the distance between the center points of the different sides of the torus). Bending, however, is resisted by a cross-section's second moment of area, which is a function of the fourth power of the diameter. This means that if you double your diameter, you will quadruple (4x) your bending moment, but your second moment of area will increase sixteen-fold. So, actually, the larger your diameter, the farther from collapse you get. This doesn't even take into consideration the arching behavior of the torus, which will drastically reduce your bending moment and transform it into compressive forces. One could then bring up the possibility of buckling, but given how non-slender this torus would be, that also won't be an issue.

The bottom half of the torus works similarly: the bending moment shouldn't be of great concern for the same reasons as described above. The arching behavior in this case will actually generate tensile forces, but these will be overcome by the compressive forces due to the torus' own weight.

So I don't see where one could find a maximum size for a given material. In fact, the material you need to worry about isn't the torus', but the ground's. You can make a torus out of whatever material you want, but if you put it on weak clay, it will be gobbled up. If you make a Death-Star-sized torus and put it on stone, it will probably crush the stone and be gobbled up, too.

  • $\begingroup$ I'm not considering the torus to be infinitely rigid, I expect it to deform under gravity. So, you're saying that the limit of size would simply be related to the compressive strength of the material - i.e. the point where strain is no longer elastic for that material? Additionally, I am expecting that ground pressure would become an issue with large sizes, however, that is not an issue for this question, so consider the ground to be of infinite strength. $\endgroup$
    – Monty Wild
    Jan 18, 2016 at 1:23
  • $\begingroup$ @MontyWild, the consideration of infinite rigidity only lasted for that one sentence. The next one starts with "If you consider deformations", which implies in a non-infinite rigidity. $\endgroup$
    – Wasabi
    Jan 18, 2016 at 1:30
  • $\begingroup$ So, is the limit of size simply related to the compressive strength of the material - i.e. the point where strain is no longer elastic for that material or where it ruptures under its own weight? $\endgroup$
    – Monty Wild
    Jan 18, 2016 at 1:42
  • $\begingroup$ @MontyWild, not really. When the torus first touches the ground, it will do so at a very small area (an infinitesimal point) with a stress much larger than the yield stress. That area will therefore yield and "sink" into the torus. This will allow a greater area to get in contact with the ground, but still at too great a stress, so this greater area will also yield, increasing the contact surface even more. This will continue until the contact surface is such that the stress is lower than the yield strength. So there isn't really a limit, since the torus will always adapt. $\endgroup$
    – Wasabi
    Jan 18, 2016 at 9:39
  • $\begingroup$ @Wasabi The adaption you describe works for ductile materials only. Brittle materials (e.g. glass) will not yield but rather break. $\endgroup$
    – Robin
    Jan 20, 2016 at 8:10

There is a limit on the size of an object made of a specific material.

The simple reason is that for an arbitrary shape,(assuming that all material properties remain the same) its volume (and hence mass) increases with the cube of the scale factor but it's cross section at any point increases with the square.

So if you consider a cube of side length 1m and mass 1kg (volume 1m^3 and cross sectional area 1) The compressive stress at the base of the cube is its weight divided by area of the base ie 10N/1m^2 = 10Pa

However if you double the length of the side, now we have a volume of 8m^3 and a base area of 4m^2 so the new stress is 80/4 = 20Pa

Double it again (side = 4m) volume = 64m^3 mass = 64kg weight = 640N base area = 16m^2

so stress at base = 640/16 = 40Pa

So...the stress due to the object's own weight is proportional to the scale factor so you reach a scale where this is greater than the yield stress of the material.

Put very simply the weight of an object increases with the cube of its scale and its cross section only increases with the square.

So the maximum size object you can make from any given material is determined by its yield stress. This is the fundamental reason why it is not currently possible to build a 'space elevator'.

In terms of actually calculating the maximum size of a torus, the easiest way would be to use FEA software. Modeling a torus of unit radius should be easy enough then if you find the maximum stress as a fraction of UTS it should be a straightforward extrapolation. In fact many CAD packages will allow you to make a parametric model which you can link to a spreadsheet to get a general solution for any proportions and material properties.

  • $\begingroup$ This relationship is obviously very different for a cube versus a torus of known aspect ratio. $\endgroup$
    – Ethan48
    Jan 18, 2016 at 19:16
  • 1
    $\begingroup$ @Ethan48, well, the square-cube rule works for any structure, so if you double a torus then the stress (assuming there's a non-infinitesimal contact surface) will double, just as with the cube. The issue I see here is that while a deformed cube won't increase its contact area (ignoring transversal deformations) under load, a torus will deform and increase its area and therefore reduce the stress at its foot. $\endgroup$
    – Wasabi
    Jan 19, 2016 at 16:47

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