Uplift for Anchored Steel Tanks
Uplift for Anchored Steel Tanks
(OP)
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.
Any help is appreciated.





RE: Uplift for Anchored Steel Tanks
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
RE: Uplift for Anchored Steel Tanks
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
RE: Uplift for Anchored Steel Tanks
Also, if you consider the bolts as an equivalent thin ring, then you develop 4M/Nd.
Best, Tincan
RE: Uplift for Anchored Steel Tanks
RE: Uplift for Anchored Steel Tanks
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
www.tankindustry.com
RE: Uplift for Anchored Steel Tanks
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
RE: Uplift for Anchored Steel Tanks
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
www.tankindustry.com
RE: Uplift for Anchored Steel Tanks
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
RE: Uplift for Anchored Steel Tanks
Steve Braune
Tank Industry Consultants
www.tankindustry.com