Uplift for Anchored Steel Tanks

[LicensedToPEe]

I am looking at a reference for calculating maximum uplift loads for anchorage (at the outter diameter) for steel tanks. The max uplift for both Wind and Seismic loads are calculated as (4*M/D)-W ,where M is moment due to seismic or wind load, D is dia. of tank and W is dead load of steel shell minus corrosion allowance. Why is there a factor of 4 for the moment?
Any help is appreciated.

 

[whyun]

Define the source of this formula. I have not seen it before.

Also, I have not seen any code (in the US) that relies upon full dead weight as resisting moments.

Not knowing the size of the tank, check two cases:
1) fully loaded tank combined with seismic loads
2) empty tank with wind loads

 

[Tankman650]

The formula is not correct. It should be: 4M/ND-W/N
N=number of bolts

See Structural Engineering Handbook, Gaylord and Gaylord, chapter 23, steel tanks.


The 4 comes from the section modulus of a infinite thin shell s=PI*d^2/4

 

[tincan]

Same 4M/Nd - W/N for AWWA. For a minimum 4-bolt layout with the moment acting thru two of the bolts the resisting tension in the one bolt resisting the moment (neglecting the weight) is 4M/Nd. As the number of bolts increase the formula approaches 3M/Nd. The 4 is retained as an additional safety factor.

Also, if you consider the bolts as an equivalent thin ring, then you develop 4M/Nd.


Best, Tincan

 

[Hookem]

This approach very simply assumes that all uplift forces are applied at the bolts. That's fine, but your efforts do not end there. One must also consider uplift forces beneath the tank because external fluid pressure will be exerted there because the tank edges do not form a seal. Do not forget to look at large sections of the thin plate (see O'Rourks Stress & Strain Book).

 

[JoeTank]

Tincan,
Check your physic and math... the 4 is constant. Three never shows up.

Hookem,
The shell-to-bottom joint of a steel tank is liquid tight. I'm not sure what you are talking about here.

Steve Braune
Tank Industry Consultants

http://www.tankindustry.com

 

[tincan]

SteveBraune
A few years back I wondred about the same question that AdamU had. Made up a little spreadsheet to determine the bolt tension as the number increased:
#Bolts "d"equivalent CG Resulting T
4 d 4M/Nd
5 0.7695d 3.25M/Nd
6 0.8666d 3.46M/Nd
8 0.8047d 3.31M/Nd
10 0.7694d 3.25M/Nd
12 0.7464d 3.22M/Nd
15 0.68d 3.15M/Nd
16 0.641d 3.14M/Nd

As the # of bolts increases T approaches 3M/Nd, I think the correct term is asemtotic ?sp, haven't used that term since calculus 1 too many years ago to remember. You are right, it never becomes 3, but the more bolts you use the more consertative the term 4M/Nd becomes. It's just another good safety factor.

Best, Tincan

 

[JoeTank]

Tincan, I am unrepentent on this one. Consider six anchors oriented such that there are two anchors on the 0 and 180 degree axis. The moment is about the 90 to 270 degree axis. Starting with T = Mc/I. I = Io + Ax^2. In this case Io for each anchor will be negligible.
Anchor 1 at 0 degrees, I = 0 + Ar^2
Anchor 2 at 60 degrees, I = 0 + A(0.5r)^2
Anchor 3 at 120 degrees, I = 0 + A(-0.5r)^2
Anchor 4 at 180 degrees, I = 0 + A(-r)^2
Anchor 5 at 240 degrees, I = 0 + A(-0.5r)^2
Anchor 6 at 300 degrees, I = 0 + A(0.5r)^2
Summing the I's, I = 3Ar^2
Anchor load = Mc/I = Mr/(3r^2) = M/(3r)
Compare to T = 4M/ND... T = 4M/(6*2r) = M/(3r)

The results are the same. The anchor tension is always correct when using T = 4M/ND. This is the tension value for the anchor on the extreme fiber (c = r). The anchor loads at other locations will be a function of the anchors distance from the neutral axis of the group.

Hope this helps.


Steve Braune
Tank Industry Consultants

http://www.tankindustry.com

 

[tincan]

Steve,
Take the six-bolt pattern for the worst case, 2 bolts on X-axis and wind loading moment on Y-axis. Moments about X-axis, only 4 bolts in pattern, 2 in tension & 2 in compression,
# bolts in tension = 2N/6 = N/3
Resisting Arm = dsin60 = 0.866d
Bolt Tension = M/[0.866dN/3] =3.46M/Nd

It's the same as calculating the Mo (moment in pure bending) for the reinforcing pattern for a round conc col
Mo = 0.12AstfyDs
0.12Asfy = total tensil force in reinf, lbs
let As = N/3
Ds =distance cg tensil to comp = 0.866d
Mo = T(N/3)(0.866d)==> T=3Mo/0.866Nd = 3.46Mo/Nd
Tincan

 

[JoeTank]

I think you are getting wrapped around the axle by assuming that all of the anchors on the tension side will carry equal loads. That is not the case. The anchor tensions will follow the Mc/I assumptions. The anchors on the extreme fibers of the tank cross section will carry higher loads than those near the neutral axis.

Steve Braune
Tank Industry Consultants

http://www.tankindustry.com