Dosi
Structural
- Nov 11, 2011
- 9
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.
