mariog123
Mechanical
- Jul 28, 2011
- 48
I kindly ask your interpretation of the "Uplift Load Case/ Failure Pressure" in Table 5.21—Uplift Loads.
I've seen a discussion on this forum (Jann 2008, now "closed") on this subject.
I noted a comment of JStephen "It is not clear to me from the wording of F.6 that the failure pressure requirement of Table 5-21 is only applicable to frangible-roof tanks."
I've checked the last edition of API 650 and I think my questions are still valid.
The note a. in table Table 5.21—Uplift Loads says "Failure pressure applies to tanks falling under F.1.3 only. The failure pressure shall be calculated using nominal thicknesses."
F.1.3 says "F.1.3 Internal pressures that exceed the nominal 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.2 and F.7."
Such tank is anchored and non-frangible versus the criteria of API Tanks. Is there any other interpretation?
1. Which is the intention of API 650 when considers such criteria as
[(1.5 × Pf – 8th) × D2 × 4.08] – W3 as uplift?
Keeping in mind that pf is the pressure when the stress in the compression ring area reaches the yield point, do I need to consider a case of uplift with 1.5*pf as internal pressure?
When the internal pressure is 1.5 × Pf to give such uplift?
2. Is such uplift also a case of designing the foundation? Again, which is the corresponding physical meaning?
3. As a math detail, why API 650 gives such complication of pf calculation?
I tried to review API formulas in SI.
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]
but this is not the API formula, even is sound engineering approach!
Instead this simple formula, API 650/ F.4.1 considers a safety coefficient of 1.6 applied to the pressure effects term (i.e. the first term), to define first the maximum design pressure, "P" 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 later 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 API 650/ F.6 [SI Units] mistyped 0.000746 instead of 0.000764.
So which is the advantage to calculate P and to define pf vs.P, instead the direct calculation of pf?
Thank you very much.
I've seen a discussion on this forum (Jann 2008, now "closed") on this subject.
I noted a comment of JStephen "It is not clear to me from the wording of F.6 that the failure pressure requirement of Table 5-21 is only applicable to frangible-roof tanks."
I've checked the last edition of API 650 and I think my questions are still valid.
The note a. in table Table 5.21—Uplift Loads says "Failure pressure applies to tanks falling under F.1.3 only. The failure pressure shall be calculated using nominal thicknesses."
F.1.3 says "F.1.3 Internal pressures that exceed the nominal 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.2 and F.7."
Such tank is anchored and non-frangible versus the criteria of API Tanks. Is there any other interpretation?
1. Which is the intention of API 650 when considers such criteria as
[(1.5 × Pf – 8th) × D2 × 4.08] – W3 as uplift?
Keeping in mind that pf is the pressure when the stress in the compression ring area reaches the yield point, do I need to consider a case of uplift with 1.5*pf as internal pressure?
When the internal pressure is 1.5 × Pf to give such uplift?
2. Is such uplift also a case of designing the foundation? Again, which is the corresponding physical meaning?
3. As a math detail, why API 650 gives such complication of pf calculation?
I tried to review API formulas in SI.
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]
but this is not the API formula, even is sound engineering approach!
Instead this simple formula, API 650/ F.4.1 considers a safety coefficient of 1.6 applied to the pressure effects term (i.e. the first term), to define first the maximum design pressure, "P" 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 later 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 API 650/ F.6 [SI Units] mistyped 0.000746 instead of 0.000764.
So which is the advantage to calculate P and to define pf vs.P, instead the direct calculation of pf?
Thank you very much.
