API 650 11th Edition, Figure F-2
API 650 11th Edition, Figure F-2
(OP)
Hello,
I am designing a tank with no internal pressure, H = 10m, D = 8m.
Due to the Overturning Wind - Moment, it needs to be anchored.
For calculating the Maximum Design Pressure which is needed to get the right Uplift Force, "A = area resisting the compressive force, as illustrated in Figure F-1 (mm2)" is needed.
First of all, I think it must be Figure F-2, like in the 10th Edition. Unfortunetly they haven't changed this mistake from Addendum 1 up to Addendum 3.
First question: Is Area A only the shaded area in Fiure F-2 or do I have to multiply this area with the tanks circumferences?
My question to Figure F-2, detail b:
How do I find "Le"?
If I consider Le as the whole length of the angle (which would make A bigger, so that finally P becomes bigger, too) I calculate like this: The angle is a L 60/60/6mm, shell thickness is also 6mm, roof thickness is 5mm, slope is 10°, Radius is 4000mm
A = 5 * wh + Area of the L-Angle + 6 * wc
with
wh = 0.3*(Rc/sin(theta)*th)^1/2 = 0.3*(4000/sin(10)*5)^1/2 = 102mm
Area of Angle = 2 * 60 * 6 = 720 mm2
wc = 0.6 * (Rc*t)^1/2 = 0.6*(4000*6)^1/2 = 93mm
=> A = 5mm*102mm + 720mm2 + 6mm*93mm = 1788mm2
P = A * Fy * tan (theta) / (200 * D * D) + 0.00127 * DLR / (D*D)
=> = 1788mm² * 138.7 N/mm² * tan(10°) / (200*8006mm^2) + 0.00127 * 29875N / 8006mm^2 = 4*10^-6 ~ 0,00N ??
Am I right?
Thanks for your help.
I am designing a tank with no internal pressure, H = 10m, D = 8m.
Due to the Overturning Wind - Moment, it needs to be anchored.
For calculating the Maximum Design Pressure which is needed to get the right Uplift Force, "A = area resisting the compressive force, as illustrated in Figure F-1 (mm2)" is needed.
First of all, I think it must be Figure F-2, like in the 10th Edition. Unfortunetly they haven't changed this mistake from Addendum 1 up to Addendum 3.
First question: Is Area A only the shaded area in Fiure F-2 or do I have to multiply this area with the tanks circumferences?
My question to Figure F-2, detail b:
How do I find "Le"?
If I consider Le as the whole length of the angle (which would make A bigger, so that finally P becomes bigger, too) I calculate like this: The angle is a L 60/60/6mm, shell thickness is also 6mm, roof thickness is 5mm, slope is 10°, Radius is 4000mm
A = 5 * wh + Area of the L-Angle + 6 * wc
with
wh = 0.3*(Rc/sin(theta)*th)^1/2 = 0.3*(4000/sin(10)*5)^1/2 = 102mm
Area of Angle = 2 * 60 * 6 = 720 mm2
wc = 0.6 * (Rc*t)^1/2 = 0.6*(4000*6)^1/2 = 93mm
=> A = 5mm*102mm + 720mm2 + 6mm*93mm = 1788mm2
P = A * Fy * tan (theta) / (200 * D * D) + 0.00127 * DLR / (D*D)
=> = 1788mm² * 138.7 N/mm² * tan(10°) / (200*8006mm^2) + 0.00127 * 29875N / 8006mm^2 = 4*10^-6 ~ 0,00N ??
Am I right?
Thanks for your help.





RE: API 650 11th Edition, Figure F-2
Le is defined in Note 3 at the bottom of Figure F-2.
The design pressure for a tank is the specified pressure and can be zero if the tank is intended for atmospheric service. You're not obligated to find the maximum allowable pressure and then design anchors accordingly. See the decision tree in Figure F-1.
RE: API 650 11th Edition, Figure F-2
That helps a lot.
So I only design anchors for Wind Load [Pwe * D² * 785 + (4*Mwh/D)-W2], because I start with a clear "no" in Figure F-1? That would make sense to me.
RE: API 650 11th Edition, Figure F-2