when we calculate failure pressure in anchorage f.7.1 annex

22/10/2014 When we Calculate Failure Pressure in Anchorage - Intergraph CADWorx & Analysis

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When we Calculate Failure Pressure in Anchorage

Jamil Member

Registered:07/18/12 Posts: 21 Loc: Saudi Arabia

as per API650 Table 5-21a/b in faliure pressure case refer note(a)Failure pressureapplies to tanks falling under F.1.3 only. The failure pressure shall be calculatedusing nominal thicknesses.

in (F.1.3)Internal pressures that exceed the weight of the shell, roof, and framingbut do not exceed 18 kPa (21/2 lbf/in.2) gauge when the shell is anchored to acounterbalancing weight, such as a concrete ringwall, are covered in F.7

in my previous experience we can't calculate failure pressure when tank is fallingunder F.7. but now client said you have to calculate failure pressure when todesign your anchor bolt.

Note: tank is non-frengible

could any one give the right direction in above wording that will be thankful for me.

_________________________Jamil AhmadDesign Eng Mech./Struc.Storage TankOlayan Descon industrial Co.

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Re: When we Calculate Failure Pressure in Anchorage [Re: Jamil]

mariog Member

Registered:09/29/07 Posts: 452 Loc: Romania

You quoted correctly F.1.3 and it is clear that F.1.3 means the shell anchored to acounterbalancing weight and design under F.7.

I don't understand why you say "we can't calculate failure pressure when tank isfalling under F.7".

In my understanding:

- F.7 is required by F.1.3.

- F.7.4 says that "The design of the anchorage and its attachment to the tankshall be a matter of agreement between the Manufacturer and the Purchaser andshall meet the requirements of 5.12.", so it is clear that [(1.5Pf 0.08th)D^2785] W3 (or [(1.5Pf8th)D^24.08] W3 in USC) is one case ofthe net uplift to check Bolt stress/ Shell Stress at Anchor Attachment (as5.12/Tables 5-21a/ 5-21b describe).

- F.7.5 must be considered "in addition to the requirements in 5.12" and specificallyrequires to consider the "failure pressure" as criterion for design of thecounterbalancing weight; See F.7.5 point c, where for design of the counterbalancing weight, it is required toconsider "1.5 times the calculated failure pressure applied to the tank filled with thedesign liquid. The effective weight of the liquid shall be limited to the insideprojection of the ringwall (Appendix B type) from the tank shell"

Best regards.

Edited by mariog (02/12/13 02:42 AM)



Jamil Member


Thanks mariog for your quick response and give a answer in detail.now my questionis that in agreement between manufacturer and purchaser there isnothing to mentioned to full fill these requirement of F.7.4. can we ignorerequirement of 5.12


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#52936 - 02/12/13 10:55 AM

#52946 - 02/13/13 12:38 AM

#52955 - 02/13/13 05:01 AM



mariog Member


In my understanding only the design (as details) of the anchorage shall be a matterof agreement between the Manufacturer and the Purchaser. I would consider that we are obliged to consider 5.12 as calculation requirementsand to develop design consulting the Purchaser for constructive details ofanchorage (anchor bolts and attachments to the tank shall).

Best regards.



Jamil Member


Thanks againit is clear now if tank falling under F.1.3 or F.7, failure pressure must be calculate inuplift load case Table 5-21a/b.

one more thing i want to know when we calculate failure pressure as per F.6 "P" isestablished pressure which is calculating from F.4.1 where "A" the area from F.5.which area we will use in F.4.1 equation calculated or provided.




mariog Member


To answer your question, as F.4.1 stipulates:"A = area resisting the compressive force, as illustrated in Figure F-2", or if youprefer, it is the "compression ring area". It is a geometrical calculated area- but theboundary of area came from other considerations not explained there.Anyway you have to consider Figure F-2.

I have to recognize that the Pf calculation procedure described in API 650 is noteasy to be understood.

In fact, failure of the roof-to-shell junction can be expected to occur when thestress in the compression ring area reaches the yield point, as F.6, CalculatedFailure Pressure says. That's why with the notations of Addendum3- 2011, thefailure pressure would be expressed in SI units as:

Pf=8*Fy*A*tan/D^2+4/PI*DLR/D^2 - units in [Pa](where A is the area subject to stress Fy)

or

Pf=Fy*A*tan/(125*D^2)+0.0012732*DLR/D^2 in [kPa]

However, this is not so simple explained in API 650.

As you remarked, API 650/F6 prefers to to express "Pf" in terms of "P".

Just to explain the connection between P and Pf (and hoping I'm not inducing moreconfusion!), F.4.1 considers a safety coefficient of 1.6 applied to the pressureeffects term (i.e. the first term), so the maximum design pressure, P is written as:

P=Fy*A*tan/(1.6*125*D^2)+0.0012732*DLR/D^2==Fy*A*tan/(200*D^2)+0.00127*DLR/D^2

and this is the maximum design pressure, P, for a tank that has been constructedor that has had its design details established; the basis of that derivation was toinclude a coefficient of safety of 1.6 addressed to the pressure term in the "failure"pressure expressions.

Instead to calculate directly, API 650/ F6 prefers to express indirectly Pf in terms of"P", so would be:

Pf=Fy*A*tan/(125*D^2)+0.0012732*DLR/D^2==1.6*P-0.6*0.0012732*DLR/D^2=



Hop to: TANK Go

=1.6*P-0.0007639*DLR/D^2(you can see that they mistyped 0.000746 instead of 0.000764!)

You can conclude that API 650 prefers to calculate first P and later Pf instead tocalculate directly Pf.

You may note also that, in the previous editions of API 650, instead of DLR appearsa term based on the thickness of roof plates and DLR is calculated with a metaldensity of 8000 kg/m^3 (8 times the water density).

Best regards.

Edited by mariog (02/13/13 05:39 AM)


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when we calculate failure pressure in anchorage f.7.1 annex

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