# Storage tank deflection part 2

Posted in real world problems on Sunday, April 12 2015

Continuing on from last time I looked at storage tanks I want to look at how well the one foot method lines up with the equation I derived for a simple cylindrical shell.

As a refresher the one foot method predicts a max hoop stress of: $$\sigma = {{ \rho g D (H-0.3)} \over 2t}$$

Whereas I derived an equation giving: $$\sigma (x) = \frac{\rho g D}{2t} \left[ H - x - H \exp{(-\lambda x)} \left( \cos{\lambda x} + \sin{\lambda x} \right) \right]$$

$$\lambda = \left[ \frac{3 (1 - \nu^2)}{(Rt)^2} \right]^{\frac{1}{4}}$$

I want to compare the minimum shell thickness given by each expression. I'm just arbitrarily going to try all combinations of tank diameters and heights between 1m and 60m, with a 1m resolution. I have run the code using ipython's parallel processing tools to save time since I'm running about 3500 possible combinations of diameters and heights.

But do all these combinations satisfy my criteria for a large tank? If you recall $\lambda H > 6$ was a requirement of the load combinations I used from Roark's. Turns out they do.

I can plot the results and look for any large deviations

Well that's pretty interesting. Whether or not the thickness is calculated by the one-foot method or using the equation from the simple stress analysis the resulting shell minimum shell thicknesses are pretty similar.

But maybe something more interesting can be found by looking at a plot of the difference between the one-foot method and the simple stress analysis.

That's more interesting. We can see that the one foot method often provides for a thicker shell, especially when the wall thickness increases, but overall the difference is less than 3mm at most. Any under-prediction by the one foot method is at most 1mm less than would have otherwise been predicted.