# How large could you theoretically build a freestanding hall?

The Tropical Islands Resort in Germany is the world's largest freestanding hall. But how large could you theoretically build a freestanding hall on Earth?

A freestanding hall is a building with no support pillars to hold the structure up.

• I see a building problem here. Not a worldbuilding one. – L.Dutch - Reinstate Monica Jul 28 at 2:58
• You can't have a building without support pillars (unless you e.g. suspend it from helium balloons). Your example building clearly has support pillars around the edges. One way to have a large clear span would be to pressurize the building, but then your "support" pillars would be holding the roof down... – jamesqf Jul 28 at 17:02

A few kilometers

Planet size objects collapse under their own gravity. That is an absolute limit for any object with structure. Even beyond that you will run into problems with extreme gravity.

Besides that, there is the problem of support columns. These limit the size of any tall structure. Assume your material has a tensile strength of 1000mpa and density of 10 tons/m3, similar to steel. That's one billion newton per square meter of strength.

Gravity has a force of approximately 10 newtons per kg or 10000 newtons for our steel block. That means 100,000 meters of the block could support itself. Naively, this suggests you could make a 100km tall column, but there are axial forces and other considerations that make it less, perhaps similar to the tallest proposed buildings. Anything larger than that will have to be increasingly pyramidal to distribute the load.

The example you gave doesnt have any support columns, but it is still roughly the same situation. You would eventually have material piling on material until it couldn't support it's own weight, and you would have to eliminate interior area until it was no longer a meaningful dome. The fact that it's a dome might actually make it easier to support without columns since it is already pyramidal. The distinction between vertical support columns and very wide curved braces becomes irrelevant after a point because the entire thing is just steel.

The example you gave uses steel. Essentially all large buildings use steel because the forces are too great. If the steel supports take up 1% of the area, that's a lot, considering a wooden shed has the same support area using materials that are ten times weaker. If a 10m high shed has effectively ten times less support than a 100m high dome using the same area for support, you can see how this extrapolates. At 5000m you would probably need half the area devoted to support.

Here is an example of a proposal similar to yours. It requires 3 million tons of steel, or about 0.1% of world production. You see how even without gravity the materials supply would run out eventually.

There is no particular reason why a freestanding hall could not run all the way around the circumference of the Earth, joining up with itself for a total length of 25,000 miles. I don't think you could make it much larger unless you are going to cheat and make it curved.

• Well, then plate tectonics come into play, and after a few hundred miles it would probably have problems. – user77172 Jul 29 at 0:59
• @Mark I - why? The hall need not be taller and wider than any other building, so it will not exert extraordinary pressure. – David Hambling Jul 29 at 13:26
• A single earthquake (which happens all the time) could tear the hall. Of course the majority would be intact but the longer you make it the greater the problems. – user77172 Jul 30 at 20:21

Have Helium balloons pulling up the roof. Give them enough propulsion to fight the wind. Keep the propulsion maintained and fueled. Your hall can cover the Earth.

Note that you have to fight the worst wind your hall will face, not the average one.

## The Tropical Island Resort is an Extended Arch

Built for CargoLifter it is: 220 meters wide, 106 meters high, and 360 meters long.

The shape is single arches. A36 steel has a yield strength 250 MPa, and a density of 7850 $${kg} \over {m^3}$$. The weight of a solid rod arch is $$density \times volume$$. For a constant cross-section, $$volume = area \times length$$.

For a hemisphere, like the tropical island resort, $$length = \pi {{width} \over 2}$$

Ultimately, the weight of the whole structure is borne by the arch. $${{weight \times gravity} \over {area}} << 250 MPa \div 2$$ (because the arch has two legs)

Therefore, then, the biggest arch possible with A36 steel is $${{{7850 \times {area \times {\pi {{width} \over 2}}}} \times gravity} \over {area}}$$ $$\rightarrow$$ $${{{7850 \times { {\pi {{width} \over 2}}}} \times gravity}} << 250 MPa$$

If I've done the math correctly, with no factor of safety, you could get an arch 2,069 meters wide, 1,034 meters high, and an arbitrary length.

Typically buildings have a factor of safety of 2: so 1,034 m width and 517 m high

## Ways to Extend This

A36 is not the strongest steel. If money is no object, you could use 4340 Steel with 1,620 MPa of strength (tougher, but more brittle) and a density of 7870 $${kg} \over {m^3}$$. Since the density is very similar, the width and height would be approximately $$6.4 \times$$ the A36 limits : 6,160 meters wide $$\times$$ 3,308 meters high $$\times$$ arbitrary meters long.

You can also use exotic materials with higher tensile strengths and lower densities to go beyond this, but this is probably a good place to stop.

You can go even further by overpressurizing the interior. By raising the pressure inside the structure a little bit above atmospheric pressure, the skin can be used to partially relieve the weight.

Let's say you overpressurize by 10%. That's $$0.1 \times 101.3 kPa = 10.1 kPa$$ of relief, multiplied by $$\pi {{width} \over 2}$$ and the length of your structure. For the 4340 Steel, this would give you 104 MPa of relief per unit area against 1,620 MPa of peak load -- so about an extra 6% that you could roll back into increasing dimensions.

However, if you use higher internal pressure to keep the structure standing, you will need to constantly be pumping air into the structure. This is a cost in a permanent power requirement.

• It is a good answer, however the pi/2 equation is a simplification because the support bars would thicken until it resembled a solid hemisphere more than a dome. Still, good approximation. – user77172 Jul 29 at 1:02