Calculate max depth in water a hollow steel sphere can withstand

In summary, filling a sphere with oxygen at normal atmospheric pressure is not going to have a lot of affect on its crush strength. Calculating the crush strength of a sphere can be complicated, and a rough estimate using a first-order approximation would require doubling the thickness of the wall to meet a 250 mpa yield strength.
  • #1
bgizzle
22
0
Any help is appreciated, I'm really just trying to figure out if it can withstand 300 meters but it would be useful to see how/if a max depth can be calculated.

diameter: 20M
steel density: 7850 kg/m^3
weight: 250,000 kg
shell thickness: 2.54 cm
density of water: 1027 kg/m^3

How much does inside of sphere being filled with oxygen versus being evacuated affect the calculation?

If I put reinforcing bars inside the sphere it would be stronger but I assume figuring out how much stronger is really complicated.

Again, any help is appreciated, thanks in advance.
 
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  • #2
This is a complicated question. The first part is that filling the sphere with O2 (I'm assuming at normal atmospheric pressure) really isn't going to have much affect. The second part involves calculating the crush strength of a sphere. The crush strength can vary all over the place depending on irregularities in its shape. You can, in theory, load the sphere until you reach the yield strength of the material. In practice, it will crush ahead of this due to stress risers and imperfections in the metal itself.

Google something called the "Euler Buckle Strength" to get some insight on this.
 
  • #3
Thanks for your response, I checked out Euler buckle strength along with a bunch of other stuff on the subject.

When you say load the sphere until the yield strength is reached, how would I calculate that? Let's say yield strength is 250 mpa, I know that pressure at that depth is 30 bar, I also know that pressure=density*g*depth or in my example 1027*9.81*300= 3*10^6 n/m^3. If I can get the crush strength depth then I would have the extreme and use that to create a reasonable estimate of a safe thickness.

Right now after all my research the best estimate I can think to use is calculating a rough volume of a submarine I found which was 9032 kg/m^3 and its approximate hull thickness of 2.5 cm's and max dive depth of 700 ft. Extrapolating from that if I know a sphere is better designed for pressure, has a much smaller volume, if I make its hull thicker, say 3.5 cm's I figure that might be a good estimate of a safe thickness.
 
  • #4
This is a first-order approximation which shouldn't be used in any real application, but will give some ballpark figures.

To find the stress in the walls of the vessel, you just need to find the cross sectional surface area of the sphere which will be pi*(20/2)^2 which gives 314 m^2. Multiply this by the external pressure: 314.0 * 3.0E6 = 942.0E6 N. This is the total force on the cross section. Now divide the force by the surface area of the wall itself. A quick approximation since the wall thickness is much smaller that the diameter is to simply multiply the circumference by the wall thickness: circumference = pi * 20.0 = 62.8 m. Now multiply by wall thickness: 62.8 * 0.0254 = 1.59 m^2. Divide this area into the applied force: 942.0E6 / 1.59 = 590.2E6 N / m^2 or 590.2 mpa. This means your wall thickness is too thin to meet your 250mpa goal. You'll need to more than double it.

A word of caution here. The calculation is just a quick and dirty way to estimate the loads. It assumes a lot. The reality is that things like hatches, windows, holes for electrical cables and manipulators, etc are going to introduce stress concentrators in the hull which can multiply local forces dramatically. This is not a trivial mechanical problem. It requires some careful analysis ... more than you're going to find here.
 
  • #5
Thanks for the response, this is exactly what I was looking for. I understand that it is very rough, that is what I am looking for. I am just working on a vague concept, nothing practical or anywhere near any sort of real world application.

A couple of quick thoughts: the object would have no addons, its just a sphere. I'm assuming that because submarines seem to be able to dive deeper with less density part of this could be the materials they are made with and possibly that they have multiple hulls. Also I assume they have support beams throughout, which would also be possibility for the sphere I was thinking of.
 

1. What is the formula for calculating the maximum depth a hollow steel sphere can withstand in water?

The formula for calculating the maximum depth a hollow steel sphere can withstand in water is D = (P*S)/(2*σ), where D is the maximum depth, P is the pressure at the desired depth, S is the tensile strength of the steel, and σ is the stress limit of the steel.

2. How do I determine the pressure at a specific depth in water?

The pressure at a specific depth in water can be determined using the formula P = ρ*g*h, where P is the pressure, ρ is the density of water, g is the gravitational acceleration, and h is the depth.

3. What factors can affect the maximum depth a hollow steel sphere can withstand in water?

The maximum depth a hollow steel sphere can withstand in water can be affected by factors such as the size and thickness of the sphere, the type and quality of the steel used, and the temperature and salinity of the water.

4. Can I use this formula to calculate the maximum depth for other materials besides steel?

No, this formula is specific to calculating the maximum depth for a hollow steel sphere. Other materials may have different formulas or considerations for determining their maximum depth in water.

5. Is there a safety factor that should be taken into account when using this formula?

Yes, it is recommended to use a safety factor of at least 2 when calculating the maximum depth for a hollow steel sphere. This will account for any potential errors in the calculations and provide an extra margin of safety.

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