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#52268 - 12/25/12 06:01 AM AWWA D100 - Anchorage
canercaner Offline
Member

Registered: 12/09/10
Posts: 11
Loc: Turkey
Dear Users,

firstly, I kindly ask the forum operator's permission. I know that Intergraph-Tank software can be used for the API related tank design. But,The users,the tank designers might have an experiences regarding AWWA D100 tank design.

My question is about the anchorage requirements in AWWA D100.
You know. In API 650 , Appendix E , we use overturning ratio formula for determining the anchorage is needed or not.
The same criteria is valid for the AWWA D100-05. (Eq 13-36)
But there is a conflict in Section 3.8.9.1, According to the Eq 3-41, you determine the anchorage needs. when we compare the results of Eq 13-36 and Eq 3-41, we come across with a conflict.

If anybody have a comment regarding anchorage in AWWA D100, I would be waiting your answers.

thanks in advance.

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#52281 - 12/27/12 03:03 AM Re: AWWA D100 - Anchorage [Re: canercaner]
mariog Offline
Member

Registered: 09/29/07
Posts: 798
Loc: Romania
Dear canercaner,

Eq 3-42 calculates the seismic uplift force PS; the formula belongs to 3.8.9 Design load, which is under Sec. 3.8 Anchorage.
The anchorage criteria is under 3.8.1 General, 3.8.1.1 Required anchorage that stipulates For ground-supported flat-bottom reservoirs and standpipes, mechanical anchorage shall be provided when the wind or seismic loads exceed the limits for self-anchored tanks. Mechanical anchorage shall always be provided for elevated tanks.

Under 13.5.4.1 Resistance to overturning provisions, the resisting force is adequate for tank stability and the tank may be selfanchored, provided the following requirements are met:
1. The overturning ratio J determined by Eq 13-36 is less than 1.54. The maximum width of annulus for determining the resisting force is 3.5 percent of the tank diameter D
2. The shell compression satisfies Sec. 13.5.4.2
3. The required thickness of the bottom annulus tb does not exceed the thickness of the bottom shell ring per Sec. 13.5.4.1.2
4. Piping flexibility requirements of Sec. 13.6 are satisfied


Paragraph 13.5.4.1 Resistance to overturning details that for J < 0.785, there is no shell uplift because of the overturning moment and the tank is self-anchored. For 0.785 < J < 1.54, there is shell uplift, but the tank is stable, provided the shell compression requirements of Sec. 13.5.4.2 are satisfied.

So it is true that Paragraph 13.5.4.1 considers that in case 0.785< J < 1.54 it is expected an partial uplift under seismic events which may be acceptable if all 4 stability conditions are satisfied. And I guess that for such case using common language we may say that, even the tank is stable, it is not self anchored while 3.8.1 says that anchorage is required when tank is not self- anchored (does not mention stability as anchorage criterion).

Is this the conflict you are considering in your post?

If yes, I would add that it is rather a matter of terminology because the conflict may be consider solved by definitions given in the same standard- just my opinion, of course!
By 13.1.2 Definitions, definition no 7, Self-anchored tanks are Tanks that rely on the inherent stability of the selfweight of the tank and contents to resist overturning forces.. That means the Self-anchored criterion is "stability"/"to resist overturning" rather than "no partial shell uplift".

I would add that the assumptions made for case 0.785 < J < 1.54 (a model with seismic plastic hinges developed in tank bottom, partial shell uplift, etc) and the related overturning resistance calculation are available in an excellent book, "Aboveground Storage Tanks" by Phillip Myers.

Bets regards.


Edited by mariog (12/27/12 04:39 AM)

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#52284 - 12/28/12 06:43 AM Re: AWWA D100 - Anchorage [Re: canercaner]
canercaner Offline
Member

Registered: 12/09/10
Posts: 11
Loc: Turkey
Dear Mariog,

Thank you so much for your help.
Conflicts that I thought were the subjects that you declared. I did not realize the definition in 13.1.2.

By the way.I have still some questions regarding the subject; I would like to understand that;

In AWWA / 3.8.9.1 Anchors;
It says; ....When the calculated net uplift force for both PW and PS results in a negative value,no uplift anchorage is required. For single-pedestal and ground-supported flat bottom tanks, the design uplift forces PW and PS are....
That means.
in negative value --> anchorage is not required.
in positive value --> anchorage is required.
Is it right?
If the J-overturning ratio is above 1,54. We have to check this section and calculate tension forces for anchors.If the j is below 1,54. then do we have to check this section. In API650, this formula is valid only if the anchors are required.


What if we do not provide 5 stability conditions?(In achorage section , E6.2.1) for instance 0,035D limitation. What is the meaning of 0,035D in theory. What is the importance of it? Is it a result of a mathematical model or experimental result.

Thank you so much for your help!

Best Regards

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#52286 - 12/28/12 01:47 PM Re: AWWA D100 - Anchorage [Re: canercaner]
mariog Offline
Member

Registered: 09/29/07
Posts: 798
Loc: Romania
Dear canercaner,

I think the root of misunderstanding is the fact API/AWWA operates with two models.

The first API/AWWA model assumes a simple thin cylinder with constraints which are continuously distributed over the circumference of the base. In some books of strength-of-materials is given that, in case the cylinder is subject to bending of moment M, the "reaction force" over the base- i.e. a thin walled circle- is 4*M/(PI*D^2)*cos(theta) where "theta" is a polar variable.
The extreme values are 4*M/(PI*D^2) and -4*M/(PI*D^2).
This model is the base of various calculations when we are sure that the contact is over that "thin walled circle", i.e. when there is no uplift because tank is stable on bottom or when thank is kept on foundation by anchors. So the calculation is based on the maximum 4*M/(PI*D^2)+wt that gives the compression stress in shell or on minimum -4*M/(PI*D^2)+wt which (in some circumstances) may be an uplift of 4*M/(PI*D^2)-wt. See for example API 650 improved formulas; E.6.2.2.1-1a, E.6.2.2.2-1a for maximum longitudinal shell compression stress and E.6.2.1.2-1, E.6.2.1.2-2 for the calculated uplift load on the anchors.
Please note that in this model we are not considering any liquid resistance to uplift; we are just sure there is no uplift and calculation is conservative made. For this reason, these formulas cannot be used as a criterion for anchorage!
The warning in AWWA is rather a mathematical warning: what to do if "calculated uplift" for anchors is negative?

The second API/AWWA model is more complex and we need it when we have a partial uplift of shell.
Uplift of the tank shell is resisted by the weight of the shell and supported roof plus a band of liquid adjacent to it. The width of this band of liquid is L (and one main parameter in calculation is wa-Fluid Force resisting uplift in annular region, which is correlated with L). "L" is counted between two seismic plastic hinges that develop under seismic event; L depends on the stiffness (or thickness) of the part of the bottom plate inside the shell, which is called the annular ring. The designer can thicken this ring but there are limitations; it cannot be thicker than the shell and L cannot exceed a certain value. If this is not sufficient, additional restraint in the form of anchors must be provided.
So which is the criterion for anchorage? To understand a little bit more, please have a look in Myersĺ book or download this document - and concentrate on "▀" angle; that ▀ is a characteristic of the non-lifted shell but API/AWWA does not consider any ▀ because they are not interested to develop the theory which is behind the results.
When ▀ is zero, there is a total uplift, when ▀=Pi radians we have no partial uplift. The corresponding parameter is " J" that mathematically takes values between PI/4=0.785 (for ▀=PI radians) to PI/2=1.57 (for ▀=0); for calculation practical reasons API/AWWA considers J=1.54 as "total" uplift. You may see also the graph attached which shows J vs. ▀.
For J< 0.785 there is no partial uplift and anchors are not required for this reason, for 0.785< J< 1.54 there is partial uplift and tank may be not-anchored, however the decision is based also considering additional criteria- see API 650/ E.6.2.1.1 Self-Anchored paragraph. In fact, the main concern is the compression in shell because, when shell is partially uplifted, the M seismic bending moment must be compensated by an increased longitudinal stress in shell (due to the reaction force on foundation, only on that part of shell which is not uplifted)- see API650/E.6.2.2.1-2a. Also, the designer may decide to anchor the tank because the attached piping is too sensitive to uplift.

It is true that API 650 says "The maximum width of annulus for determining the resisting force is 3.5% of the tank diameter." However this is not a true criterion, it is just a warning that the decision "tank is not anchored" has been based on the desired algorithm of calculation. In my opinion that sentence can miss there because the same condition is given in various other paragraphs (see E.6.2.1.1) and it is practically impossible to skip it.
I think it is interesting that wa=1.28HDGe (in USC Units) and and L= 0.035D are equivalent by Wozniak & Mitchell model, that;s why API650/ E.6.2.1.1 asks for both.

As limit value, L< 0.035 D appears today as being rather arbitrary; however that limit has been established by Wozniak in 1971 and I guess he had a good reason for it. Today we may understand that the API 650/AWWA algorithm is not an exact one; it is based on some assumptions and has been fine tuned over the years.

My best regards.


Attachments
J vs beta- Wozniak & Mitchell model.pdf (861 downloads)


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#52299 - 12/31/12 06:26 AM Re: AWWA D100 - Anchorage [Re: canercaner]
canercaner Offline
Member

Registered: 12/09/10
Posts: 11
Loc: Turkey
Dear Mariog,
Most of the things are clear. Thank you so much!.

But I am still confused about 0,035D limitation.

Wa(force resisting uplift in annular region) affects on J, it is calculated by using annular plate thickness and minimum specified yield strength of annular plate.
L does not affect on J. ( if we do not use it smaller than Lmin ( E.6.2.1.1-2a))
There is a restriction that L need not be greater than 0,035D. But why it is declared as 0,035D why not 0,05D or something else. Unfortunately I am curious about it. Because I do not believe that, Increasement in width(I mean "L") should not be affect negatively on tank overturning. It should not be required anchorage.

If I am wrong, please advise.

Thank you.

Best Regards

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#52300 - 12/31/12 07:42 AM Re: AWWA D100 - Anchorage [Re: canercaner]
mariog Offline
Member

Registered: 09/29/07
Posts: 798
Loc: Romania
J depends on wa and wa is proportional with L.
Bigger L means bigger wa and smaller J.
You can lower J (changing from "total uplift- anchorage required" to reasonable "partial shell uplift") by increasing wa, which means an increased L.
L can be increased by increasing annular thickness, because the calculated L is between two plastic hinges; a thicker annular means bigger L between plastic hinges. However, this procedure has limits: 0.035D is a limit of L, shell thickness is a limit of bottom thickness.

As limit, L= 0.035D is equivalent to wa=201.1HDGe (in SI) or wa=1.28HDGe (in USC Units). See API 650 (E.6.2.1.1-1a)/ (E.6.2.1.1-1b). In fact it is a limit of the quantity of liquid that has credit to resist overturning.

See also the basic document you can download from
http://mycommittees.api.org/standards/techinterp/refequip/Shared Documents/seismic.pdf; just study it!

The limit L= 0.035D has been established by Robert Wozniak in "Lateral Seismic Loads on Flat Bottomed Tanks" Water Tower, Nov 1971. I haven't that article, I cannot help you more.

In fact, why are you so concentrated on this limit? Are you ready to accept a model with double plastic hinges in bottom (on uplifted shell part) but not a limit of this model?

Happy New Year!


Edited by mariog (12/31/12 08:27 AM)

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#65777 - 03/05/16 04:56 AM Re: AWWA D100 - Anchorage [Re: canercaner]
Robert anto reni Offline
Member

Registered: 06/03/09
Posts: 4
Loc: India
As per API 650 Table 5.21a (net uplift loads)

While calcualting the values in mathcad the units are not matching. Pease kindly clarify and give one example for Desing Pressure case, Design pressure + Wind & design pressure + Sesimic (For valudation puroise)


Regards,

Robert
_________________________
robert anto reni

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