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Kroon's Method

Kroon's Method

Kroon's Method

I'm designing a tank with the following parameters:

D = 17.2 m
H = 18 m
Hliquid = 17.38 m
SG = 1.48
CA = 2mm

According to API 650 Clause 5.5.3, an elastic analysis must be made to calculate the required thickness of the annular ring. I'm using Kroon's Method to calculate the thickness, and I'm a bit unsure of a couple of things.

1 - Does the calculated thickness used in this method include or exclude corrosion allowance?

2 - What is the minimum width (or internal projection) of the annular ring when performing such an analysis?

If I use L = 493 mm for my annular plate width and Tb = 17.05 mm for my annular plate thickness, then everything works out perfectly as per my attached spreadsheet.

θs = -θc
θb = 0
σ < 75000 psi

This would mean that my annular plate would be a 3/4" plate (17.05 + 2mm CA) at 19-1/2" wide. This works out great since the bottom shell course of my tank is also 3/4" Plate.

HOWEVER, according to API 650, the internal projection of the annular ring needs to be 0.035D as per Clause and E. This means that L' = 602 mm and L = 672 mm.

If I use 672 mm for my plate width, and 17.05 mm for my plate thickness then:

σ < 75000 psi
θb = 0


θs ≠ -θc (in fact they are both the same sign) As a matter of fact once L>576mm then θs > 0 meaning regardless of what thickness you use you can't make θs = -θc and if you do push it to the limit using L = 576 mm then Tb = 103 mm, which is extremely ridiculous.

Does anyone know what governs when using Kroon's Method? Do I need to use the width as per API 650 and not worry about θs ≠ -θc? Or do I need to make sure that θs = -θc and use a more narrow annular ring?



RE: Kroon's Method 2) :
When the bottom plate under the shell is thicker due to wind overturning than the remainder of the tank bottom, the minimum projection of the supplied thicker annular ring inside the tank wall, L, shall be the greater of 450 mm (18 in.) or Lb, however, need not be more than 0.035D.

This says for wind overturning, the minimum is 18" and not more than 0.035D

E. says the same - 0.035D is the max

Note also table 5.1b for minimum annular plate thicknesses and 5.5.2 for 24" minimum projection inwards and length to thickness ratio calculations

RE: Kroon's Method

See attached spreadsheet for a tank I'm quoting.

D = 45.5 ft
H = 98 ft

If I use 18" as the minimum internal projection that means L = 18" + 1/2" wall + 1/4" fillet = 18.75in (476.25)

In order for the math to work out on my attached spreadsheet, the thickness of the annular plate needs to 168.24mm (6.624in) which is outrageous. Notice the stress at the bottom is only 125.7 psi, which tells me it's ridiculously overkill.

I've done a few other iterations with different thicknesses and then solving for L such as:

Tb = 1"
L = 17.977"

Tb = 3/4"
L = 17.473"

Tb = 1/2"
L = 15.928"

As you'll see all of these have an internal projection of less than 18", however the thicknesses tend to make sense.

Perhaps there's something wrong with my spreadsheet?


RE: Kroon's Method

Or is the value of L in the calculation (using Kroon's method) just the MINIMUM length required to reduce the rotation at the inside end to zero?

If that were the case, I can make the annular plate wider (to a minimum of 18" or 0.035D internal projection) and it wouldn't matter?

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