Seismic Overturning Moment - in Tanks
Seismic Overturning Moment - in Tanks
(OP)
I am considering the overturning moment on a tall steel tank (12' dia. X 43' high) filled with a dry product. This is a mild seismic area, so I am using ASCE 7-05, section 12.8 to find the Cs coefficient for the seismic shear equation V=CsW. My question is, even though I have a large overturning moment due to a seismic event, what part of the dead load of the product acts to counter the overturning moment? Would one simply use the total dead load of the product multiplied by the moment arm (centroid of tank to tank edge). This would totally void the seismic overturning moment due to the large amount dead load. I cannot find any direction in the code.
Thanks for any input!
Thanks for any input!






RE: Seismic Overturning Moment - in Tanks
The codes are not a design suicide pact. You don't have to use the full case for lateral compared to the empty for resistance.
RE: Seismic Overturning Moment - in Tanks
for strength design (0.9 - 0.2 SDS)D + ρE;
for ASD (0.6 - 0.14SDS)D - 0.7ρE
RE: Seismic Overturning Moment - in Tanks
In the liquid tank codes, a portion of the tank bottom near the shell is assumed to supply some resisting load due to tank contents, but the amount is limited based on tank diameter and properties of the bottom, and will not be anywhere near the total contents of the tank.
You could presumably use the same approach with a granular product. See AWWA D100 or API-650.
If the tank is anchored to a concrete slab or ringwall, then all of the contents above the slab or ringwall are assumed to resist seismic overturning of the foundation.
If the tank has a suspended/conical bottom, then the entire contents would be assumed to resist overturning as well. Possibly with some adjustment for vertical acceleration.