API-650 Appendix E

[giuliese]

Hi,
I'm an italian civil/environmental engineer (excuse me for any mistakes in writing english), and i have to design a welded storage tank of 10000 cubic meters (353150 cubic feet).
The purchaser wants me to design according to API STANDARD 650 although the construction site is in Italy.
I have the API STANDARD 650 10th Edition (Addendum 4, Dec. 2005), but I've found problems in understanding some formulas.
In E.6.2.1.1: Wa=99*ta*SQRT(Fy*H*Ge) with Ge=G*(1-Av)
I don't understand the physical meaning of this formula:
in particular I don't understand the material wich G refers to (liquid contents or steel bottom plate)and its unit; I also don't understand how to define Av starting from the values of one of So, or Ss or S1.
Furthermore, I've read in a forum dated Dec. 2006, that there are some mistakes in formulas E-18, E-19, E-20, E-21.
So, my question is: does anybody know where I can find an "errata" regarding API650 10th Edition Add.2005?
Thanks to anybody will reply.

 

[TankDude]

Check this link for the published Errata for the 11th Edition:

http://www.api.org/Publications/addenda/upload/650_Ed10wAd4_Er1.pdf

 

[JStephen]

G is the specific gravity of the liquid stored, not the steel.

For the shell to lift up, a part of the bottom plate and the contents over that part of the bottom plate must lift as well, and this formula is calculating the effective weight of the product that resists this uplift.

API did have an errata available for App. 4. Also, the 11th Edition is out now.

 

[giuliese]

API Standard 650, 11th Edition June 2007, Appendix E, formula (E.6.2.1.1.1-a):

wa=[99*ta*SQR(Fy*H*Ge)]<=wa(max) with wa(max)=1.96*H*D*Ge

where:
Ge=G*(1-0.4*Av)

I’ve tried to calculate wa for an existing welded tank that I know to be self-anchored.

My input data are:
ta=8 mm; Fy=250 Mpa; H=13.80 m; G=0.72; Av=0.2g; D=30.48m

The results are:
Ge=0.72*(1-0.4*0.2)=0.662
wa=37840.66 N/m
wa(max)=545.50 N/m

The value of wa(max) is too much smaller than the value of wa and by using it for the calculation of the anchorage ratio, the tank (that is actually stable) would result as not stable.

I’ve read in API Standard 650 - 10th Edition - Addendum3 September 2003 - Appendix E - chapter E.4.1, that the weight of the portion of contents that may resist to overturning is calculated as:

wL=[99*tb*SQR(Fby*G*H)]< wL(max) with wL(max)=196*H*D*G

So my question is:
is the 1.96 coefficient written in (E.6.2.1.1.1-a) wrong? and should it be replaced with 196, on the analogy on the wL formula?

(Obviously, by using the 196 coefficient, I would have more comparable values of wa and wa(max) and my tank would result self-anchored - as it is in reality - having an anchorage ratio J=0.55)

Thanks to anybody will reply.

 

[JStephen]

Working off of 1.28HDGe, with H = 45.28', D= 100', and Ge = 0.662 gives 3,837 lb/ft or 55,995 N/m. So yes, I would say something is off by a factor of 100 or so (answers won't match exactly due to roundoff in the factors used).

 

[giuliese]

Thanks JStephen.
Your reply supports my persuasion in considering 196 as the right coefficient.
Unfortunately, the purchaser doesn't allow me to replace the written wrong coefficient 1.96, with the right (in my opinion) coefficient 196, unless I will provide some official document like an "errata" or something similar.
I've tried to write an inquiry to the API official website, but their responses can take up to 12 months!
Can you (or anybody else who's reading this thread) suggest me other better ways to solve such a problem?
Greetings.

 

[gr2vessels]

giuliese,
I sympatise with you on the error you detected in the API 650;- however, I would imagine that this error was already picked up by the "Tank" software designers, all over the world. Did you contact any of them, to try getting some support from them? If the error was corrected in that software, why don't you compile your calculations on that legitimate software, guaranteed by the software designer? That way you don't have to justify any formulas or method of calculation, the complete software deemed as compliant to the code, rather than your small change of a formula error.
cheers,
gr2vessels

 

[richay]

giuliese,

I think you can justify your alternate/revised/corrected coefficient if you look closely at Addendum 3 and the new Code. In Addendum 3, the coefficient was 196 for SI units. However, if you look at the same equation for U.S. Customary Units, you'll find the coefficient was 1.25.

Now jump to the new Code; the equation for U.S. Customary Units has a coefficient of 1.96. But the new Code also uses the same coefficient for the SI version of the equation. The coefficient can't be the same for both versions of the equation. You should be able to prove this if you work out both versions of the equation.

gr2vessels,

Why do you think that software vendors have the license to alter Code equations? Software vendors suffer from the same confusion as everyone else - and they also wait 12 months or more for interpretations from the Code Committee.

What is "guaranteed software"? Every Engineering Software package I've every seen has a disclaimer right up front that states something like "We've checked this to the best of our ability, but we (the vendor) are not responsible for errors." Software is a tool, not a substitute for engineering or experience. (If you hit your thumb with a hammer, you can't blame the hammer.)


Richard Ay
COADE, Inc.

 

[JStephen]

If necessary, convert the units and work off the non-metric equation- that would be equally as valid if they need something in writing.

You might also try emailing API. Official stuff like this can take forever. But I have gotten responses back quicker, and have gotten erratta back, etc.- definitely worth an email or two there. (It might help to indicate that you don't need an "official" response, but that a prompt response would be very helpful.)

 

[gr2vessels]

Rich,
Thank you for the advice, I read the disclaimer also. I also know that the AI is prepared to stamp and certify the computer calculations as correct, in accordance with the code, for insurance purposes and/or Code stamping, based simply on software designer "assurance" of correcteness.
However, I don't have any proof as you said, of the "correction" I have mentioned. I will now have to take a closer look at this, for the records.
I am surprised, however, to learn that the software designers, who are the first to detect early these errors, will deliberately allow wrong calculations, for the sake of code compliance. Is it that what Coade does?
Best regards,
gr2vessels

 

[richay]

gr2vessels,

You said:

I am surprised, however, to learn that the software designers, who are the first to detect early these errors, will deliberately allow wrong calculations, for the sake of code compliance. Is it that what Coade does?

Maybe - it depends, every instance is different. Take for example the equation in API-650 10th Edition Addendum 4, Paragraph E.6.2.2.3 which reads Fc = 106*ts / D. How is someone who is reading this code supposed to know that this is a typo and the equation should be Fc = 10^6*ts / D? Those familiar with prior editions can readily see this and make the correction.

Contrast this to equation E-1 in Paragraph E.4.8.1, which was corrected by the errata. This is a new equation in Addendum 4 - how was anyone supposed to know that this equation as initially published was wrong?

Or how about equation E-7, where the Errata changed Fa to Fv - how was anyone supposed to know this?

My point is that there can be mistakes in software just as easily as there can be mistakes in Codes and Standards. You're the Engineer in the "drivers seat" - you have to exercise your knowledge and experience on every job performed. Do not assume that because some number came from a computer program that it corrects for any defficiencies in the source or theory.




Richard Ay
COADE, Inc.

 

[giuliese]

Thanks to everybody for your replies.
I think I'm going to proceed as suggested by JStephen: I will convert the units working in the U.S. Customary Units, and at the end, I will re-convert the result into S.I. Units.
So:

being
H=13.8 m = 45.28 ft
D=30.48 m = 100 ft
Ge=0.662

I calculate
wa[lbf/ft] = 1.28*H*D*Ge = 1.28*45.28*100*0.662 = 3837 lbf/ft

now I go back to SI Units:
3837 lbf/ft = 57101 N/m
In this way, I can avoid the difficulty of justifying a non-official interpretation of E.6.2.1.1.1-a formula due to the wrong 1.96 coefficient as written in S.I. Units.

Thank you.
Greetings from Italy