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

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

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

Note: tank is non-frengible

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

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

I don't understand why you say "we can't calculate failure pressure when tank is falling 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 tank shall be a matter of agreement between the Manufacturer and the Purchaser and shall meet the requirements of 5.12.", so it is clear that [(1.5×Pf – 0.08th)×D^2×785]– W3 (or [(1.5×Pf–8th)×D^2×4.08]– W3 in USC) is one case of the net uplift to check Bolt stress/ Shell Stress at Anchor Attachment (as 5.12/Tables 5-21a/ 5-21b describe).

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

Best regards.

Thanks mariog for your quick response and give a answer in detail.
now my questionis that in agreement between manufacturer and purchaser there is nothing to mentioned to full fill these requirement of F.7.4. can we ignore requirement of 5.12

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

Best regards.

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

one more thing i want to know when we calculate failure pressure as per F.6 "P" is established 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.

To answer your question, as F.4.1 stipulates:
"A = area resisting the compressive force, as illustrated in Figure F-2", or if you prefer, it is the "compression ring area". It is a geometrical calculated area- but the boundary 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 not easy to be understood.

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

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

or

Pf=Fy*A*tanθ/(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 more confusion!), F.4.1 considers a safety coefficient of 1.6 applied to the pressure effects term (i.e. the first term), so the maximum design pressure, P is written as:

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

and this is the maximum design pressure, P, for a tank that has been constructed or that has had its design details established; the basis of that derivation was to include 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*tanθ/(125*D^2)+0.0012732*DLR/D^2=
=1.6*P-0.6*0.0012732*DLR/D^2=
=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 to calculate directly Pf.

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

Best regards.