Is it safe if the diameter of a 50,000-liter storage tank increases during water filling? How to explain it?

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This post was last edited by kareale88 on 2018-11-22 at 16:02. A few days ago, while organizing files on my computer, I remembered an incident from a few years ago: a construction site called to report a problem as mentioned in the title – the diameter of the storage tank increased by approximately 60 mm during filling – and they asked whether there was any issue or if it was safe. At that time, it was analyzed that an increase in diameter was normal, as water filling generated stress, and where there is stress, there is strain. Whether it is safe should depend on whether an increase of 60mm is allowed. The conclusion after doing it is that it is safe. But how to make it convincing? So I wrote something as follows at that time and sent it to the property owner and the construction company; ultimately, there was no response, and the matter was dropped. It is excerpted here for everyone’s discussion to see if there are any issues. Calculation data for the increase in diameter of a 5x104 m3 tank when filled with water: inner diameter = 60 m, wall height = 19.33 m; material used is 16MnR; thickness of the bottom ring plate is δ32 mm. During testing, δt = 32 – 0.3 = 31.7 mm (0.3 represents the negative deviation of the material). The formula for calculating thickness is t = 4.9D(H – 0.3) / [σ] / φ – (GB50341, Equation 6.3.1-2). Stress calculation: σ = 4.9D(H – 0.3) / δt / φ = 4.9*60*(19.33–0.3)/31.7/0.9 = 196.1 MPa. Strain calculation: ε = σ/E, where ε represents strain ； E—Elastic modulus in MPa: 196.1/201/1000. Here, E = 201*1000 (as per GB150) = 0.976/1000. The design perimeter of the tank is L = πD = 3.14*(60000 + 32) (calculated based on the mean diameter) = 3.14*60032 = 188500 mm. The increase in perimeter due to deformation is l = εL = 0.976/1000*188500 = 184 mm. The perimeter after deformation is L1 = L + l = 188500 + 184 = 188684 mm. The mean diameter after deformation is D1 = L1/π = 188684/3.14 = 60090 mm. The increase in inner diameter is 60090 – 60000 – 32 = 58 mm. Note: The values above represent the deformation amount when using the “theoretical diameter”. In practice, it is necessary to take into account the actual diameter deviations during construction and acceptance, as well as the effect of roundness deviations (the minor axis of an ellipse increases in size further due to the ‘rounding effect’ caused by internal pressure). Now I realize there might be a problem: 1. An increase of 58 mm should be the maximum value; could this actually exceed the allowable limits? It’s hard to estimate without specific data provided ； 2. Subtracting a negative deviation of 3 mm gives the minimum thickness; it is reasonable for the actual diameter increase not to exceed 58 mm ； Also: I don’t know how to determine the exact diameter; if anyone knows, please let me know.